Clarnell Hniaetattg Sitbtratg atlfaca. S^etn Mark BOUGHT WITH THE INCOME OF THE SAGE ENDOWMENT FUND THE GIFT OF HENRY W. SAGE 1891 Cornell University Library TA 350.B29 Analytical mechanics, '='""P"f,'"S.}|jm||!j'" 3 1924 004 085 373 The original of tiiis book is in tine Cornell University Library. There are no known copyright restrictions in the United States on the use of the text. http://www.archive.org/details/cu31924004085373 ANALYTICAL MECHANICS RECENTLY PUBLISHED ELEMENTARY APPLIED MECHANICS (Statics). Introducing the Unitary System. By Alexander Norwell, B.Sc, C.E., Principalof the Greenock (Burgh) Technical School. Crown 8vo, 3^. ELEMENTS OF MECHANICS. By G. W. Parker, M. A. For the use of Schools and Colleges. With ii6 Diagrams and Answers to Examples. 8vo, 4J. td. MECHANICS : Theoretical, Applied, and Experimental. By W. W. F. PULLEN, Wh. Sch. , M. I. M. E. , A. M. I.C. E. With 318 Diagrams and numerous Exapiples. Crown 8vo, 4J. td. ELEMENTARY PRACTICAL MECHANICS. By J. M. Jameson, Head of the Department of Physics, School of Science and Technology, Pratt Institute, Brooklyn, N.Y. Crown 8vo, 6j. net. APPLIED MECHANICS : Embracing Strength and Elasticity of Materials, Theory and Design of Structures, Theory of Machines and Hydraulics. A Text-Book for Engineering Students. By David Allan Low (Whitworth Scholar), M.I.Mech.E. , Professor of Engineering, East London College (University of London). With 850 Illustrations and 780 Exercises. 8vo, 7J. bd. net. MECHANICS FOR ENGINEERS : A Text-Book of Intermediate Standard. By Arthur Morley, M.Sc. , University Scholar (Vict. ) ; Professor of Mechanical Engineering in Univer- sity College, Nottingham. With 200 Diagrams and numerous Examples. Crown 8vo, 4^. net. ELEME!NTARY applied MECHANICS. By Arthur Morley, M.Sc, M.I.Mech.E., Professor of Mechanical Engineering in University College, Nottingham, and William Inchley, B.Sc, Lecturer and Demonstrator in Engineering in University College, Nottingham. With 285 Diagrams and numerous Examples. Crown 8vo, 3^. net. A STUDY OF SPLASHES. By A. M. Worthington, C.B., F.R.S. With 197 Illustrations from Instantaneous Photo- graphs. Medium 8vo, ds. 6d. net. AN ELEMENTARY TREATMENT OF THE THEORY OF SPINNING TOPS AND GYROSCOPIC MO- TION. By Harold Crabtree, of Charterhouse, Godalming. With 3 Plates and over 100 Diagrams. 8vo, 5J. 6d. net. LONGMANS, GREEN, AND CO. LONDON, NEW YORK, BOMBAY, AND CALCUTTA Analytical Mechanics COMPRISING THE KINETICS AND STATICS OF SOLIDS AND FLUIDS BY EDWIN H. BARTON D.Sc. (LOND.), F.R.S.E., A.M.I.E.E., F.Ph.S.L. PROFESSOR OF EXPERIMENTAL PHYSICS UNIVERSITY COLLEGE NOTTINGHAM WITH DIAGRAMS LONGMANS, GREEN, AND CO. 39 PATERNOSTER ROW, LONDON NEW YORK, BOMBAY, AND CALCUTTA igi I H PREFACE This text-book on theoretical mechanics is intended for those possessing, or concurrently acquiring, an elementary knowledge of the calculus. Within these bounds it is fairly complete, dealing with the kinetics and statics of solids and of fluids. Further, in the kinematical portion, mechanisms and strains are included, and the work closes with a short chapter on elas- ticity. In the transition from kinematics to kinetics, Newton's principles and the views subsequently held respecting them are passed in review. This critical treatment culminates in a set of proposed enunciations. To minds thus prepared these enunciations, though brief enough to be easily remembered, may serve to recall a sort of central position of modern thought on dynamical axioms. But no finality has yet been reached on these philosophical topics. Their discussion is accordingly con- fined to a single chapter. This leaves the formal mathematical developments equally readable to those holding the most diverse views as to the foundations underlying this super- structure. Probably most users of the book will bring to its study some previous knowledge of the subject, the amount and fresh- ness of which differ widely in individual cases. To provide for such variety of preliminary attainment, the elementary parts are briefly outlined to serve as a revision or reference and for logical completeness. Similarly, parts lying beyond the central scope of the work are often indicated and sources of fuller information quoted. The work is not written narrowly to any one examination syllabus. But its general scope and treatment will be found to meet the needs of degree candidates of London and other vi ANALYTICAL MECHANICS universities at home and abroad, also of those offering the third and honours stages of the Board of Education. The arrange- ment is such that any candidate not requiring the whole would usually find his course provided for by a certain selection of chapters taken entire, the others being wholly omitted. This fact conduces to a simplicity and coherence in these special courses which could not be attained if the chapters themselves needed minute subdivision into parts to be read or omitted. As to order of "treatment : after a short introductory part there follow Kinematics, Kinetics, Statics, Hydromechanics, and Elasticity. But the detachment of Chapters II., III., and XI., giving respectively formulae, geometrical basis and physical basis, will enable a student or teacher to take the other sec- tions of the book in a different order if preferred. Through the body of the work, at frequent intervals, are given sets of examples mostly of a simple character and strictly on the text. At the end occur additional examples of a harder or more varied character, some classified, some miscellaneous, also subjects for essay-writing. These bring the total number of examples almost to eight hundred. It was intended to give hints for the solution of the problems set, but considerations of space forbad the inclusion of such a section in the present volume. In addition to the great classic authors, too well known to need mention here, many other authorities have been con- sulted and quoted, as may be seen by the index, in which proper names are italicised. Acknowledgments are hereby tendered to the holders of copyright for their courteous permissions, as follows : — (i) To the Controller of H.M. Stationery Office for permis- sion to print {a) questions set at the examinations of the Board of Education, and {b) the mathematical tables as issued to candidates thereat ; (ii) To Messrs. Macmillan & Company, Limited, for similar permission respecting part of those tables, viz. the logarithms of numbers from looo to 2000 ; PREFACE vii (ill) To the University of London ; and (iv) To the University of Calcutta for permission to print their examination questions. To Mr. H. T. H. Piaggio, M.A.(Cantab.), cordial thanks are offered for his careful reading of the proofs, and for his valuable comments thereon. If any readers noticing errors, omissions, or obscurities would communicate such, their kindness would be highly appre- ciated, and subsequent editions in consequence improved. Nottingham, /2^()' 1911. CONTENTS PART I.— INTRODUCTORY CHAPTER I PRELIMINARY SURVEY ARTS, 1. Scope of Mechanics, 2. Illustration of Projectile, 3. Order of Treatment, Examples — i., . PAGE 1-3 3 4 4 CHAPTER II FORMULAE 4. Object and Use of Collection : Algebra, 5. Plane Trigonometry : Spherical Triangle, 6. Plane Co-ordinate Geometry : Solid Triangle, 7-8. Differential and Integral Calculus, 9. Differential Equations, . . . . S-6 6-7 7-9 9-11 11-12 PART II.— KINEMATICS CHAPTER III GEOMETRICAL BASIS 10. Space and Position, II-I2 Units and their Dimensions, 13- Displacement, 14. Scalars and Vectors, IS- Composition of Vectors, 16. Localisation of Vectors, Examples — n., . 17- Time and Motion, 18. Velocity, . 19- Acceleration, 20. Displacement Graphs, 13 I4-IS 15-16 16-17 17 17-18 18 18 19-20 20-21 21-23 ANALYTICAL MECHANICS 21-22. Velocity or Speed Graphs, etc., . Examples — in., . . . ■ 23. Composition of Velocities and Accelerations, 24-25. Vectorial Polygons, 250. Moments,. .... lib. Composition of Angular Velocities, 25c. Curvature, i^d. Centroids, .... 26. Constraints and Degrees of Freedom, . Examples — iv., .... PAGE 23-24 24-25 25 25-27 28 28-29 30 30-31 31-32 33 CHAPTER IV RECTILINEAR MOTIONS 27. Uniform Acceleration : Low Falls, Examples — v., .... 28. Uniform Acceleration by the Calculus, . 29. Simple Harmonic Motion, Examples — vl, .... 31-33. Composition of Collinear S. H. Motions, Examples — vii., 34-36. Acceleration inversely as Distance squared. Examples — viil, 37-38. Falling Mist, 39-40. Falling Hailstone, 41-44. Rise and Fall of Shot, . Examples — ix., . 45. Damped Vibration, 46-48. Forced Vibration, Examples — x., . 34-35 35 35 35-37 37-38 38-40 40 40-43 43 43-45 45-46 46-51 51-52 52 52-55 55 CHAPTER V plane motions of a point 49-51. Projectiles, ... Examples — xi., ..... 52-53. Constrained Motions ; down Incline and Pendulum, 54-57. Simple Pendulum in Finite Arcs, Examples— xiL, 58-60. Motion in Vertical Cycloid, Examples — xin., 61-65. The Brachistochrone, Examples — xiv., 66-68. Rectangular Vibrations, . Exampi,es— XV., . 69. Tangential and Normal Accelerations, 56-59 58-59 59-60 60-62 62-64 64-66 67 67-70 70 70-71 72 72-73 CONTENTS ARTS. 70-71. The Hodograph, 72. Conical Pendulum, . Examples — xvi., . 73- Angular and Areal Velocities, 74-75- Radial and Transversal Velocities and Accelerations, Examples— XVII., ..... 76-78. Areal Velocity Constant with Central Acceleration, 79- Differential Equation of Orbit, 80. Curvature of Orbit, ... 81. Formula for Central Acceleration, . 82. Velocity in Orbit, . Examples — xviii., ... 83. Orbits under Natural Law, , 84. Hodograph a Circle, 85-86. Orbit a Conic, 87. Hyperbolic Orbit and Hodograph, . Examples — xix., ..... 88. Initial Conditions for Orbits, 89. Velocity and Period in Elliptic Orbit, 90. Acceleration, Period, and Velocity in Elliptic Orbit, 9o«. Simultaneous Orbits : Kepler's Laws, 91. Focal Acceleration for any Conic, Examples — xx., . . . . 73-74 75 75 76 76-78 78 78-80 80-81 8i 81-82 82 83 83 83-84 84-86 86-87 87 87-88 88-89 88-90 90-91 91 91-92 CHAPTER VI plane rotational motions 92. Uniform Angular Acceleration, .... 93-94 93. Variable Angular Acceleration, ... 94 Examples — xxi., ...... 94 94. Composition of Coplanar Rotation and Translation, . 94-95 95. Instantaneous Centre of Rotation ; Rolling Motion, . 95-96 96. Analysis of Plane Motion of Rigid Body, . . . 96-97 97. Composition of Linear and Angular Velocities, . . 97-98 98-99. Composition of Angular Velocities, etc., ... 98 100. Analytical Treatment of Coplanar Motions, . . 98-100 loi. Instantaneous Centre : Body and Space Centrodes, 100 102. Summary of Coplanar Motions, .... loo-ioi Examples — xxn., ...... 101-102 CHAPTER VII solid motions of a point 103-104. Solid Co-ordinates, 105. The Straight Line, Examples— XXIII., 103-104 104 105 ANALYTICAL MECHANICS ARTS, 1 06. 107. 108. Cylindrical Motion, . . . • Conical Motion, .... Spherical Motion, . Examples — xxiv., . . . ■ 109. Spherical Motions under Uniform Acceleration, 110-11 1. Spherical Pendulum, H2. Accelerations in any Curve, Examples — xxv., . PAGE loS . los -106 . 106-108 108 . 108 -no no -III III -112 112 CHAPTER VIII SOLID ROTATIONAL MOTIONS 113-114. One Point fixed : Euler's Theorem, 1 1 5-1 17. Rodrigues' Co-ordinates and Theorem, Examples — xxvl, .... 118. Composition of Angular Velocities, 119. Angular Velocity and Acceleration compounded, 120-122. Precessional Motions, 123-124. Angular Accelerations about Moving Axes, 125. Most General Motion with Point fixed. Examples — xxvil, 126-129. General Displacement of Rigid Body, 130. Central Axis : Twists and Screws, . 131. General Motion of Rigid Body, Examples — xxviii., • 132. Velocity of any Point of Rigid Body, 133. Resultant Twisting Velocity, Examples — xxix., . 113-114 114-117 117 117-118 118-119 119-122 122-125 125 125-126 126-128 128-130 130 130-131 131-132 133-134 134 CHAPTER IX MECHANISMS 134. Subdivisions and Treatment, 135-137. Inextensible Cords and Belts, 138. Incompressible Fluids, Examples — xxx., . 139. Links and their Motions, 140-141. Lower and Higher Pairing of Links, 142. Non-Rigid Links, . 143. Plane Linkages : Inversions, 144. Criterion of Deformability, etc., 145. Use of Instantaneous Centres of Rotation, Examples— xxxL, . ' . 146. Quadric Linkages, . 147. Velocity Ratios ; Polar Diagrams, . 148. Conditions for Crank and Lever, . 13S-136 136-137 137-138 338 138-139 139 139-140 140 140-141 141-142 142 142-143 143-144 144 CONTENTS ARTS. 149- Double-Crank Linkwork, . 150. Change and Dead Points, . 151. Watfs Parallel Motion, Examples— XXXII., 152. Criterion for Rotation in Linkworks, 153. The Pantograph, . . . . 154. Peaucellier's Cell, . . . . 155-156. Hart's Cell, . . . . . 157. Parallelogram Linkages : Lazy-Tongs, Examples — xxxin., 158-161. Slider Crank Chain : Velocity Ratios, etc., 162. Screw Pairs, . . . . . 163. Higher Pairing, .... Examples — xxxiv.. xm PAGE 144-145 145-146 146-148 148 148-149 149 149-150 150-152 152 '52-153 153-156 156 156-157 157 CHAPTER X STRAINS 164. Simple Strains, 165. Typical Pure Strains, 166-169. Composition of Small Coaxial Pure Strains. 170-172. Shear viewed as Slidings : Amount of. Examples — xxxv.,. 173. Homogeneous Strains, 174. Strain Ellipsoid, 175. Nine Constants of Homogeneous Strain, 176. Rotation in Homogeneous Strain, . 177. Conditions for Pure Strain, . 178-179. Pure Strain derived and simplified, Examples — xxxvi., 180. Equation of Strain Ellipsoid, 181. Cone of Constant Elongation, 182-183. Shear EUipsoid by Slidings, 184. Composition of Pure Strains and Rotations. 185. Restricted Strains, . Examples— XXXVII., 158-159 159-160 160-162 162-165 165-166 166 166-167 167-169 169-170 170-171 171-173 173 173-174 174-175 175-177 178 178-179 179 PART III.— KINETICS CHAPTER XI physical basis 186. Physical Conceptions, 187. Mechanics before Newton, . 188. Newton's Principles, 180-181 181 181-182 ANALYTICAL MECHANICS ARTS. PACK I89-I9I Newton's Enunciations, . 182-184 192. Newton's Disciples and Critics, . 184-185 193- Criticisms by Mach, .... 185-186 194. Enunciations by Mach, . . . . 187 I9S- Tribute of Mach to Newton, 187 196. Retrospect by Mach, . 187-188 Examples— XXXVIII., 188 197. Kari Pearson's View, . 188-189 198. Love's Treatment, . . 189-190 199. Lodge on Axioms, . 190 200. Universal Gravitation, .... 190 201. Friction : Coulomb, Morin, Tower, I9O-I9I 202. Laws of Hooke and of Boyle, 192 203. Relative Character of Motion and of Mechanics, . ■ 192-193 Examples— XXXIX., .... 193 204. Measurement of Time, .... 193 205. Attitude, towards Physical Axioms, . • 193-194 206. Mass at High Speeds, .... 194 207. Quantities usually proportional to Mass, . ■ 194-195 208. Retrospect, ...... 195 209. Brief Enunciation of Chief Mechanical Bases, 196 210. Concluding Remarks, . 196-198 211. Bibliography, 198 Examples— XL., . 198-199 CHAPTER XII KINETICS OF PARTICLES 212. Mass brought into Equations, 213. Choice of Units of Mass, Force, etc., 214. Established Systems of Units, Examples— XLL, . 215. Potential Energy and Transformations, 216. Work in Oblique Displacements, . 217-220. Impact : Direct, Oblique, etc.. Examples — xlil, . 221. Angle and Cone of Friction, 222. Motion on Rough Incline, . 223-224. Atwood's Machine, and with Friction, 225. Connected Particles on Rough Inclines, Examples — xliii., . 226-228. Pendulum in Accelerated Chamber, 229. Elevation of Exterior Rail, . Examples— xLiv., . 230. Three connected Masses, 231-232. Falling Chains, 200-202 202-203 203-205 205 205-206 206-207 207-210 2 10-2 1 1 211 2II-2I2 212-213 214 215 215-217 217 217-218 2l8 219-220 CONTENTS 233. Fall of Growing Raindrop, . 234. Slip of Snow on a Slope, 235. Note on Vibrations, EXAMPI,ES — XLV., XV PAGE 220 220-221 221 221 CHAPTER XIII PLANE KINETICS OF RIGID BODIES 236. Acceleration of Rigid Body about Fixed Axis, 237. Moment of Inertia and Torque, 237a. Rotation about Fixed Axis treated generally, 238-240. Moment of Inertia Theorems, Examples — xlvi., . 241. Oblique Axis Theorem, 242-247. Evaluation of Moments of Inertia, . Examples — xlvii., ... 248-252. Triangular Lamina : Moments of Inertia of, 253. Routh's Rule : Typical Moments of Inertia, 254. Graphical Method for Moments of Inertia, Examples— xlviil, 255-256. Well Roller and Bucket, 257. Atwood's Machine with Inertia of Pulley, . Examples— XLix., . 258-261. Compound Pendulum, .... 262-263. Torsional Pendulum, etc., . Examples— L., . ... 264. General Plane Motion of Rigid Body, 265-269. Independence of Translation of Centre of Mass and Rota tion about it, . 270. Centre of Percussion, 271-272. Fall of Trap Door, . 272a. Sudden Fixation of Point in Rotating Body, 273. Ballistic Pendulum, . Examples— LL, . . . . ■ 274-277. Pure Rolling down IncHne, . 278-284. Combined RolHng and Sliding, 285. Rolling Oscillations, Examples — lil, . . . • ■ 222-223 223-224 224-225 225-227 227-228 228-229 229-233 233 233-238 238-239 240-241 241 242 243 243 243-248 248 249 250 250-255 255-256 256-257 257 258-259 259-260 260-263 263-269 269-270 270-271 CHAPTER XIV solid rigid kinetics 286. Motions of Rigid Body with one Point fixed, 287-289. Precession of Top maintained, Examples— LiiL, . • • • 290. General Expression for Angular Momenta, 272 272-275 275 275-277 xvi ANALYTICAL MECHANICS ARTS. 291. Angular Momentum is a Vector along Axis, 292. Axes of Resultant Velocity and Momentum, 293. Analogous Relations of Mechanical Quantities, Examples— Liv., . 294. Torques about Moving Axes, 295. Moving Axes fixed in Body, 296. Euler's Dynamical Equations, Examples— Lv., 297. Steady Precession of Top, . 298. Conical Precession without Torque, 299. Kinetic Energy of Rotations, Examples— LVi., . 300. Starting Precession, 301. Nutations in the Azimuthal Plane, 302. Precessional Velocities at Limiting Inclinations, 303. • Tilting Velocity of Top, 304. Minimal Velocities for Top to Spin and ' Sleep,' 305. Examples of Gyroscopic Motion, Examples — lvil, . 306. Centrifugal Reactions and Torques, 307-309. Independence of Translation of Centre of Mass Rotation about it, . Examples — lviil, ..... 310. Equations of Motion for Axes of Fixed Directions, 311. Hayward's Equations, .... 312. Motions of a Quoit, . Examples— Lix., . ... and PAGE 277-278 278-279 279-280 280 280-281 281-282 282 282 282-284 284-285 285-286 286 286-288 288-289 289-290 290 290-291 291-293 293-295 295-296 296-297 297-298 298-299 299-301 301 PART IV.— STATICS CHAPTER XV statics of particles 313. Forces on Body devoid of Acceleration, 314. Composition and Resolution of Forces, Examples— Lx., 315. Conditions of Equilibrium, . 316. Inclined Plane, 317. The Wedge, 318. The Multiplied Cord or Tackle, . Examples — Lxi., . 319-320. Work for given Force and Displacement, 321. Virtual Work in Zero for Equilibrium, Examples— Lxii., . 322-323. Equilibrium of Cord, 324-325. Uniform Cord under Gravity, Examples— LXiii., . 302-303 303-304 304 304-305 305 306 306-307 308 308-309 309-310 310 311-313 313-31S 31S ARTS. 326-327. 328. 329- 330- 331- 332-333- CONTENTS Equations of Common Catenary, . Elementary Properties of Common Catenary, Examples— Lxiv., . . . _ Approximations to Catenary, Sag and .Excess Length in Wires, . Parameter of Catenary, . , Loads for Parabola and Circle, Examples — lxv., ... PAGE 315-316 316-317 317-318 318 318-319 319-320 320-322 322 • CITAPTER XVI ATXRACTKJNS AND POTENTIAL 334- Attractions of Two Particles 335 Gl-avitational Field, . 336-339. Attractions of Filaments, Exampl:es — lxvi., . 340-343- Attractions of Discs, • 344-345- Thin Spherical Shell, 346. Sudden Change' of Field through Shell Examples — lxvil, 347- Solid Sphere arid Thick Shells, 348. Graphical Representation of Field, 349. Field in Eccentric Cavity, . ' 350. Newtonian Constant of Gravitation, Examples — Lxrviir., 351. Solid Anigles, Lines, and Tubes of Force, 352. Gauss' Theorem, .... 353. Potential Introduced, * 354. Potential and Field, 355. Potential and Work, 356. Equipotential Surfaces, Examples — lxix., .... 357. Laplace's and Poisson's Theorems, 358-360. Potentials of Disc, Shell, and Sphere, 361. Graphic "Representation of Potentials, etc.. Examples — lxx., .... 323 323-325 ■ 325-328 328 328-330 330-332 332-333 333 334 334-335 335 335-336 336-337 337-338 339-340 340-341 341-342 342-344 344 344-345 345-346 346-349 349 349-350 CHAPTER XVII PLANE STATICS OF RIGID BODIES 362. Resultant of Parallel Forces, 351-352 363- Couples, .... 352-353 364- Resultant of Coplanar Forces, 353-354 36s- Change of Axes, ... 354-355 366. Poinsot's Analogy between Statics and Kinematics, 355-356 367-368. Conditions of Equilibrium, . 356-357 Examples — lxxi., .... b 357-358 xviii ANALYTICAL MECHANICS ARTS. 369. Determination of Centroids, T. 370-373. Elementary Examples of Centroids, Examples — lxxii., 374-375. Centroids by Integration : Lines, . 376-381. Centroids by Integration : Plane Surfaces, Examples — lxxiii., . 382-383. Centroids of Surfaces of Revolution, 384. Centroid of Solids of Revolution, |. - 385. Centres of Mass where Density varies, 386. Pappus' Theorems, . . Examples— lxxiv., 387-388. Virtual Work for Rigid Bodies, . 389. Lever : Wheel and Axle, . . 390. The Efficiency of a Simple Machine, 391. The Screw, . Examples— Lxxv., 392. Stability of Equilibrium, . 393. The Balance, Examples— LxxvL, 394. Graphical Statics, . 395. Reciprocal Figures, 396-397- Frame and its Force Polygon, Examples— Lxxvn., 398. Funicular Polygon for Resultant of Coplanar Forces, 398a. Link Polygons with Different Poles, 399. Graphical Conditions of Equilibrium, 400. Reactions found by Link Polygon, 401. Roof with Asymmetrical Load, 402. Evaluation of Stresses apparently Indeterminate, Examples— LxxviiL, 403. Reactions at Joints, 404. Separation of Bars, 405-407. Reaction inside Bodies, 408-409. Bending Moments and Shearing Forces in Beams, 410. Bending Moment Diagram is a Link Polygon, , Examples— Lxxix., .... 359 359-363 363 363-365 365-370 370 370-372 372-374 374 374-375 375-376 376-378 378 379 379-381 381-383 383-385 385-387 387 387-388 388 388-391 391 392-393 393-395 395-396 396-397 397-398 399-401 401 401-403 403-404 404-405 405-409 409-411 411 CHAPTER XVIII SOLID STATICS OF RIGID BODIES 4ii-4ii«? Poinsot's Reduction of Forces, . 412-414 412. Conditions of Equilibrium, 414 413- Six Components of a Force, ■ 414-415 414. Moment of a Force about a Line, . . 415-416 Examples— Lxxx., 416-417 415. Conditions for a Single Resultant, 417 416. Its Line of Action, .... . 417-418 ARTS. 417- 418-419. 420. 421-422. C 01^ TENTS Reduction to Two Forces, ..... Equilibrium with Points fixed or on Plane, Examples— Lxxxi., . . . . ! Reduction of Forces to a Wrench, Tripod, Pile of Spheres, and Bifilar Suspension by Virtual Work, ... Examples — lxxxii.. xix PAGE 418 420 420-421 422-424 42s PART v.— HYDROMECHANICS CHAPTER XIX HYDROSTATICS 423. Natures of Fluids, Liquids, and Gases, . 424. - Hydrostatic Pressure Independent of Direction, . 425. Fundamental Equations, .... 426. Pressure and Depth in a Liquid, . 427. Force on Submerged Plane, Examples — lxxxiii., 428. Centre of Pressure, 429. Force on a Closed Surface, 430. Floating Bodies, 431-432. Metacentre, Examples — lxxxiv., 433. Heights by Barometer, 434. Pressure on Tense Curved Membrane, 435. Soap Bubbles and Films, . 436-437. Capillary Ascent, . Examples — lxxxv., 438. Equilibrium Form of Large Drop, 439-440. Liquid in Accelerated A'essel, Examples — lxxxvl. 426-427 427 427-428 428-429 429-430 430 430-431 431-433 433 433-437 437-438 438-439 439-440 440-441 441-442 442-443 443-444 445-446 446 CHAPTER XX HYDROKINETICS 441. Equation of Continuity, .... 447-448 442. Equations of Motion, ..... 448-449 443. Relations between Pressure and Density, . . 449-450 444. Steady Flow under Gravity : Bernoulli's Theorem, . 450-452 445. Torricelli's Theorem of Outflow Velocitj-, . . 452-453 446. "\'ena Contracta, . . 453 447. Liquid Rotating as Rigid Solid, . . 454 Examples— LxxxviL, . . . 454 448. AngTilar Velocities of Fluid Elements, 454-455 449. \'elocity Potential, .... 455-456 450. Angular Accelerations of Fluid Elements, 456-457 XX AIVA1.YJ1LA1. MJl.i^MAjyH,:i ARTS. PAGE 451. Coaxial'Circular Currents, . • . • 457-459 452. Steady Flow past Cylinder, • 459-460 453- Water Waves, • . . ■ . . 460-463 454- Viscous Flow through Cylinder, . ■ 463-465 Examples— Lxxxviii., . . 465-466 PART VI.— ELASTICITY CHAPTER XXI STATICS OF ELASTIC SOLIDS 455- Nature of Elastic Bodies, . 467-468 456-457. Stress and its Relation to Strain, . 468-470 458. Homogeneous Stress and its Six Components, 470-471 Examples— Lxxxix., 472 459- Stress across any Plane, 472-474 460. Composition of Stresses, . 474 461. Two Aspects of Shearing Stress, . 474-475 462. Elasticities and their Relations, . 475-477 Examples— xc, . 477 463- The Work of Stress and Strain, . 477-478 464. Bent Bar, . 478-480 465. Twisted Cylinder, . 480-481 466. Elongation of Helical Spring, 482 Examples— xci., . 483 ADDITIONAL EXAMPLES Examples— XCII. : Chiefly Kinematics, .... 484-487 XClll. : Chiefly Particle Kinetics, 487-490 , xciv. : Chiefly Rigid Dynamics, 490-492 , XCV. : Chiefly Statics, . 493-495 , xcvi. . Chiefly Attractions, 495-496 , xcvil. . Chiefly Hydrostatics, . 496-499 , xcvill. : Chiefly Hydrokinetics, 499 , XCIX. : Chiefly, Elasticity, 499-500 , c. : Miscellaneous, . 501-504 , CL : Do. . . 504-507 , CIL : Do. . . • 507-510 , CIIL : Do. . 510-512 , CIV. : Do. . . 513-51S , CV. : Do. . . 515-516 , cvi. : Essay Subjects, 517 , cvil. : Miscellaneous, . . - • • 517-519 , cvin. : Do. . 519-520 Mat HEMATICAL TABLES, , . 521-526 INDE X,. . . . • 527-535 ANALYTICAL MECHANICS PART I.— INTRODUCTORY CHAPTER I PRELIMINARY SURVEY 1. Scope of Mechanics. — The theory of Mechanics deals with space, time, and mass. But if left with so vague a description it would be found to include both Physics and Chemistry, instead of being but the simpler part of the former as it really is. In order to see more precisely how Mechanics is limited, it will be a convenience to note what the above broad statement might include and how it must then be reduced. This may be done as follows : — Take, in imagination, a number of columns and head each with the name of some material system, such as a single particle, two particles, rigid body, etc. Next, divide the columns into a number of lines or rows, reserving to each row some one definite type of attracting or other influence or con- straint, under which the systems might be placed, such as gravita- tional attraction, one point fixed, and so forth. We should then obtain in this table a number of spaces or squares each referring to a specified system placed under given conditions. And, with respect to each such square, Mechanics would be concerned with two types of problems, viz. : — (i) What motion ensues for each possible initial con- figuration or motion? (2) What forces or initial configurations are necessary in order that the subsequent motion may be of some given type, including the special case of rest? Thus, under the first type of problem, we may be asked what happens if a pendulum bob be pulled aside and let go. While, under the second, we may be asked (a) what the length of the pendulum must be so that it shall beat seconds, or (b) what force is required to keep it pulled aside a given amount. Now, in our supposed table, the columns and the lines each extend without any definite limit; hence the possible number of squares is doubly infinite, the total number of cases being further complicated by the double or treble nature of the problems attachmg to each square. But all the above applies to the initial vague statement as to the scope of Mechanics, and which really includes also Physics and Chemistry. We exclude these two branches of science by restricting to their simpler forms the material systems contemplated. And speci- ally we restrict our attention to systems whose constituent parts are of known shape and number. The line between the systems retained and those excluded is somewhat arbitrary, and when drawn so as to relegate ANALYTICAL MECHANICS [art. to Physics and Chemistry all the possible cases, it still leaves to Mechanics a large number of columns. Neither is there any definite hmit to the number of lines referring to the separate sets of forces and constraints under which the systems may be placed. Further, since the motions depend also on the initial circumstances, it is a convenience to let the separate lines stand for the various conceivable motions actual or possible. We thus obtain a system of subdivision of the subject which, on grouping together as far as possible the columns and lines, yields the traditional scheme of subdivision shown in Table i. Table I. Subdivisions of Mechanics. STATES SYSTEMS particles ' RIGID BODIES^ FLUIDS ELASTIC SOLIDS MOTION* kinetics 2 OF PARTICI,ES KINETICS 2 OF RIGID BODIES HYDROKINETICS KINETICS OF ELASTIC SOLIDS REST STATICS OF PARTICLES STATICS OF RIGID BODIES HYDROSTATICS STATICS OF ELASTIC SOLIDS In this condensed form each column replaces or includes an in- definite number of those in our imaginary schemes and calls for a little explanation. Thus, in Table i., the first column is headed particles. If we pass from two particles to a countless number, we pass from Mechanics to Thermodynamics as treated in the kinetic theory of gases. The problem of three attracting particles is certainly one of Mechanics, but seems to have defied general solution hitherto. The wave motion of a stretched string may be treated under Mechanics or relegated to Acoustics. The wave- motion of the ether is studied under optics or electromagnetism. The attractions of the sun and earth are studied under Mechanics, those of hydrogen and chlorine and of the other so- called elements are dealt with under Chemistry. But when we restrict ourselves to those simpler systems in which 1 Motion purely without regard to its cause is studied under the title of Kinematics which IS a necessary preliminary to kinetics. ' 2 Kinetics is often studied under the title Dynamics, which is often, however u«ed to embrace statics also. ' s Under the heading Particles we may include simple systems of connected particles and under the headmg Rigid Bodies we may include jointed frames, etc. , some or all of whose parts are rigid. ART. 2] PRELIMINARY SURVEY 3 the main occurrences are mechanical, subsidiary ones are often of a physical nature. And even when the physical and chemical phenomena are ignored, the truly mechanical problems that remain are often too complicated for solution. So that at the outset of many problems we must further neglect the less important mechanical phenomena, and deal only with the simpler and sahent features of the case abstracted from the almost bewildering tangle of total occurrences in which they are involved. 2. Illustration of Projectile. — These principles may be easily seen in many familiar phenomena. Take, as an example, the firing of a rifle and the course of its projectile to the target. The explosion of the powder is a chemical process. The production of sound and heat at the target are physical processes ; the whistling sound of the shot on its way also falls under the latter category. But the description of the trajectory is the subject of Mechanics, and at first sight seems a very simple affair. In strictness it is highly complicated, and perhaps, in all its generality, still awaits solution. It must be approached step by step. Gravity deflects the shot downwards from the line of original projection, and this consideration affords the first approximation to a solution or description of what happens. A second approximation might be obtained by taking into account the resistance of the air. A third by considering the tendency of the shot to set its length at right angles to its direction of motion. A fourth by considering the resistance to this tendency, due to the spinning motion of the projectile produced by the rifling of the gun. Yet further steps remaining are the allowances for the facts that (1) the friction between the shot and the air drags the latter, disturbs the distribution of its pressure, and may deflect the shot; and (2) when great heights are reached, changes occur in the values of gravity and of air pressure, density, and resistance. And these are all legitimate subjects of Mechanics. Further, the friction between the shot and the air produces heat, expands the shot and the air, and again the phenomena are affected in consequence, but these disturbances pass over into the domain of Physics. The principles that have been noticed for the shot in flight apply to every occurrence, however simple it may appear at first sight. Thus, apart from the limitation of our consideration to the purely mechanical parts of any occurrence, there is also the initial limitation to the simpler and more important features of the case. And afterwards, at some stage in the process of successive approximations to a complete solution, there is usually a limitation imposed by the student's lack of mathematical weapons competent to deal with the subject in hand. Hence any course of study or treatise on Mechanics must be planned with regard to the mathematical proficiency assumed. Thus the present text-book supposes that a knowledge of the elements of the differential and integral calculus is possessed or is being acquired by the reader. When differential equations are introduced, they are so far 4 ANALYTICAL MECHANICS [art. 3 explained as to be intelligible to those without any previous conversance with them. 3. Order of Treatment. — Since we are so much concerned with motions, we shall consider them first apart from their causes. This necessary preliminary branch, called Kinematics, occupies Chapters iii.-x., forming the second part of the work. Kinematics rests on the conceptions of space and time only, and leads naturally to the third part, called Kinetics. In this we study motions, having regard to the circumstances under which they may be expected to occur. But, to form a basis for this, something beyond conceptions of space and time are needed. The conception of mass must be introduced. And for the part it plays we must fall back upon universal experience as generalised, co-ordinated, and formulated by thinkers ancient and modern. ArSsumS of this in Chapterxi. accordingly introduces the third part ; the kinetics of particles and rigid solids occupying Chapters XII. -XIV. Statics is next dealt with in Chapters xv.-xviii., the treatment concluding with brief chapters on Hydromecfianics and Elasticity. This order has been adopted as appearing on the whole most Con- venient. But no possible sequence is free from objection. Readers wishing to take the several parts in a different order will find their task simplified (a) by the collection of preUminary notions and theorems which occupies Chapter iii., and \b) by the mathematical formulae given for reference in Chapter ii. Examples — I. 1. State what you understand by Mechanics, showing clearly how it is dis- tinguished from Physics. 2. Make a scheme of the subdivisions of Mechanics, dealing with the various possible systems and their states of motion or rest. 3. Analyse some familiar mechanical phenomena, indicating the various approximations which may be made in the endeavour to treat it mathe- matically. ART. 4] FORMULAE CHAPTER II FORMULAE 4. Object and Use of this Collection.— When solving mechanical problems a number of mathematical formulae are needed. Many of the.se are usually remembered readily as required. But to each student at times there may occur a lapse of memory, or his knowledge in a certain essential may be found incomplete. To obviate the necessity of reference to other books in the first case, and to direct attention to the defect in the second case, the collection of formulae composing this chapter is placed here. No one student is likely to require every one of these formulae. But while only the formulae of an advanced character may be referred to by the stronger readers, the more elemen- tary examples may be useful to those less highly equipped. Further, the collection as a whole serves to indicate the scope of the mathematical knowledge which should be possessed or soon acquired by the student. Thus the ground indicated by these formulae in algebra, plane trigonometry, and plane co-ordinate geometry is supposed already familiar to the student of this book. The study of the elements of the differential and integral calculus must be undertaken concurrently with the reading of this book, if not already possessed. While the systematic study of differential equations may be deferred for a time, though, of course, its possession is a great advantage. Algebra. Binomical Expansions. / , \« 1 1 n(n—i) „ , nin—i)(n — 2) , UJf.xY=\-\-nx+^ -'a- + -^ f^ '-x^+ . . . |_2 1 3 For n, a. positive integer, the right side is clearly a finite series, and then correctly expresses the left side for any value of x. For n, a negative integer or fractional, the right side is an infinite series; \,vX, provided x<\ numerically, it is a convergent series which truly expresses the value of the left side. When X is very small, h say, whose square is negligible in com- parison with unity, we may write {\-\-hY=\-\-nh nearly. Exponential Series. e=i-\-}-+^+^+ ■ ■ ■ =271828183 nearly. \l If l3 6 ANALYTICAL MECHANICS [art. 5 |2 l3 [4 [? Logarithmic Series, etc. x^ x^ ^* loge(i+*)=^ 1 V ■ ■ ■ when.«L 3 3. e+27r . . b-V2-. or r''' I COS —1 l-«sin 3 3 / or.'=(cos^+,-sin^±l"). ^Sm^ fli«rf Cosine Series. cose=,-^ + ^-^+ . . . sine=e-fl + fl_fl+ ... l3 LS 17 Exponential Values of Sine and Cosine. 2 cos 0= «'*+«- '"9. 2»sin 6 =<;'9— «-'■«. cos^+«sin^=«±'*. Hyperbolic Sine and Cosine. 2Cosh6=e^-\-e-^. 2 sinh 6=e^—e~^. Spherical Triangle of spherical angles A, B, and C and opposite sides subtending at the centre of the sphere the angles a, b, and c. cos a=cos b cos f+sin b sin c cos A. 6. Plane Co-ordinate Geometry. a b 00 V The Straight Line may be represented \:-j y=mx-]-b, 1-— =i, or X cos a-\-y sin a=/. The general form is ax-\-by-\-c=o. The perpendicular on this from (h, k) has the length /= — , „ i its equation being 3.«—a)'+a/4—^/4=o. Two lines through the origin are given by Px-+a-y' = o. Polar Equation, r cos {d—a)=p. The Circle of radius a with centre at {h, k) is {x-hf+(j-ky=:a\ The tangent to it at {x', V) is {x-h){x'-h) + {y-k)(y'-k)=a\ Polar Equation is r^ — 2Rr cos (fl— a)+i?'-fl^ = o when the radius is a and the centre at {R, a). The Parabola is represented by j-' = ± Js^ax, or x' = + ^ay, with origin at the vertex. 8 ANALYTICAL MECHANICS [art. 6 If the vertex is at (h, k), and the axis along the negative direction oiy, the equation is {x-hy+Aa{y-k) = o, the focus being at {h, k—a), and the directrix being j'=^+fl. The tangent at {x', y') is {x-h){x'-h)+2a{y-k-lry'-k)=o. If h'=^ak, the parabola passes through the origin P, and the tangent there is hx—2ay=o. Calling the focus S, Ji denoting the point {x', y') on the curve, and Q the point of abscissa x' on the tangent through the origin JP, we have FS=a+k, PQ^=x'\a-\-k)la, and QR=x"l^a, whence FQ^ = ^PS.QR — a useful relation. x^ y' The Ellipse —s-\-jt-=t- has at the point {x',y') xx , yy' the tangent — j-+-^=i. a b The semi-axes a and b satisfy b'=a^(\ —e^)~al, where e is the eccen- tricity and / the semi-latus rectum. If the foci are denoted by 5 and S, the corresponding extremities of the major axis by A and A', r and r being the focal radii to a point P on the curve, and p and /' the perpendiculars from the foci on the tangent at P, we have the following properties : — AS.SA'=b^=^pp\ plp' = rjr', and r-\-r'=2a. Also, if p be the radius of curvature of the ellipse at P, and c the semi-axis conjugate to that through P, then ^ = rr', and c'^abp. The Hyperbola and its conjugate are represented by " yl a" 'b^ i-7-»=±i, the asymptotes of both being -^—■j^=o. General Equation of a Conic. ax'' + 2hxy-\-by'' + 2gx ■\- 2jy-\-c=o. Geometrical Relation fulfilled by any Conic. SPIPM=a. constant ratio, e say, where S is the focus, P a point on the conic, and -Af the foot of the perpendicular from ^on the directrix. The conic is an ellipse, parabola, or hyperbola according as e< i, e=i, or e>\. Polar Equation of a Conic, its Focus being the Pole. llr=i—e cos 6. Solid Co-ordinate Geometry, the axes being rectangular. Co-ordinates of a Point {x, y, z). e ^'^'^- 7] FORMULAE 9 Equations of a Plane. General form, Ax-\-By-^ Cz+B=o. Perpendicular form, /;c+w;/+«z=«> ntn^ ^ !f /^^ length of the perpendicular from the origin on to th plane, and /, m, n are the direction cosines of that perpendicular V. S^l cosines of its angles with the axes). pwpenaicuiar (t.e. the T.\^Jj!'J"''''^'t- ^''", T"^ ^^ represented as the intersection of two planes of equations, lx+»,y= i and ny+p,= r ; or we may write for a I m ~ n ' rilnti^'r/'/ ''^ ^"^ ^^' ^'' '^ ^'^ P°'"'' °" 'he line whose direction cosines are /, m, and n. I J^« fr^li ''T'Z ^"'" '-^ S'"''^" ^y ^""^ 6=li'+„,m+„n, where th?other '^^"■^'=*'°" ^°^'"es of one line, and /', m', n those of Sphere, x' ->ry"--\.z''=.a''. Ellipsoid, J+-^+j=i Cone, vertex as origin, Ax--\-By^->rCz"=o. 7 Differential and Integral Calculus.— The following list may be regarded as giving on the right the differential coefficients of the functions on the left, or as giving on the left the result of inteerating the functions on the right. Simple Algebraic Functions : — Integrals Differential Coefficients y^^'ltc^ ■^^. dy_ dx Jm logc* be^ a^logea I log« X X I x\oge.a ax ax u V du dv dx dx .Jx^-\-d^ X Jx- + a^ ^ From this we have, for integration by ^axls, /vdu=uv ~/udv. 10 ANALYTICAL MECHANICS [a Trigonometrical Fumtions : — Integrals Differential Coefficients dy dx sin a; cos X cos^ — sm X tanx sec '^x cot « — cosec X sec a; sec X tan x cosec x — cosec X cot JK Inverse Functions : — Integrals Differential Coefficients di dx sin X cos ~^x tan '^x cot " ^x 8. Hyperbolic Functions : Integrals y or \^^x J dx + + + + Ji—x' I Jl—X' I i+x' I I -j;^^2_j + ^ xjx'-i Differential Coefficients dy dx sinh a: coshx tanh X coth X sech a; cosech X coshjc sinh X sech 'x — cosech 'x — sechx tanh* — cosech :» coth* Harder Miscellaneous Functions : — Integrals y or 1 -^dx J dx Differential Coefficients dy^ dx I ^ X — tan — a a d'+x'' ^'^T. 9] FORMULAE — log ^^"^^ I It -';<; cosh -orlog,(x+V*'-a) sinh -^orlog,(^+ ^x^-\-a^) ^ ^ _ — sec _ + , Ax>a) J_l tr ■a^ I Conception of Definite Integral.—The. following paragraph may be of service to some who have to use definite integrals in mechanics before they have reached them in their systematic study of pure mathe- matics. Consider the area 0PM between some curve OP, a portion OM of the axis of jv and the ordinate MP, and denote this area by u—x^ (i), the curve OP being such as to fulfil this relation. Then, if MP shifts to M'P' by the very small increment MM.'=h or dx, we may write du Area MPP'M' ,,„ , , ■:r— innnr, = MP=_j' (2), dx MM ^ ' but by (i) —=«*""', so j = «;e""^ (3). Take now any value a of x, and erect the ordinate aA, cutting the curve OP in A. Then the area of 0«A is a", but it may be regarded as made up of vertical strips each of area>'^ or ydx. Thus we have the summational formula, 2>a;»-'/% = a» (4); or, in the notation of the integral calculus, /: «.v''-Vx=«" (5). 9. Differential Equations.- — Many mechanical problems lead to differential equations in which the variables are separable. Hence, after 12 ANALYTICAL MECHANICS [art. 9 the separation, such an equation may be integrated and the solution readily obtained. Thus the differential equation dy may be written ■ — ^ + adx = o. ay-\-b And this integrates to \ogeiay-\-b)-\-ax=C, which is equivalent to ay+6=Be~'^, the B and C being constants. Other differential equations often occur in mechanics whose solu- tions are very easy. Thus, some may be dealt with by multiplying by an integrating factor or assuming a trial solution of the form e"^, or of some other form suggested by mechanical considerations. It must suffice here to notice the following important types, the solutions of which may be verified by differentiation. When any differential equa- tions occur in the subsequent text, they are dealt with simply so as to be understood by students not having any previous knowledge of them, though, of course, such knowledge is a great advantage. Thus, the differential equation is easily found, by trial of j)'=e™*, to be satisfied by y = Ae'V^-\-BeP^, the quantities A and B being arbitrary constants to be fixed by the initial conditions. Again, the differential equation is satisfied hy y=-A %vc\px-\-B cos/^, A and B being arbitrary constants depending on the initial state. If x refers to time, this is obviously a to-and-fro motion or vibration, the former case being a subsidence if B vanishes. For the differential equation d'^y . j,dy , .i the solution may be written y — g-lcx(^j^ sin qx-\-B cos gx), where q^=p^ — k^. This gives a diminishing vibration if x is time. The differential equation --%+/ y=fsmnx is satisfied by _fsmnx y- p^-„^ ■ ART. lo] GEOMETRICAL BASIS 13 PART II.— KINEMATICS CHAPTER III GEOMETRICAL BASIS 10. Space and Position. — Geometry treats of space; and the space of ordinary human experience has three dimensions only, often referred to as length, breadth, and thickness. By a process of abstraction we can conceive of space of two dimensions only (or even of one only). We thus have the geometry of two dimensions as well as that of three. Or we have plane geometry as well as solid. And it is convenient to deal first with the position of points in a plane. To fix the position of one point relative to another, both being in a given plane, we need either (i) two distances along given directions, or (2) one distance and an angle with a given direction in the plane. These methods form respectively the cartesian and polar systems of co-ordinates. They are illustrated in Fig I. Thus, on the cartesian system, the position of P with respect to O is fixed by the two distances or co-ordinates, OM called x and ON called y, MP and NP being re- spectively parallel to the two direc- tional axes OY and OX. On the polar system, P's posi- tion is fixed by the distance OP called r, and the angle called Q which OP makes with the fixed direction OX. It is accordingly obvious that the following relations hold when as usual OX and OY are at right angles :— .T=rcos 5 and^ = r sin 5 ■ (i)- ^2 = a:'-f/- andtan 0=j'/a; (2)- The pair of equations (i) give the cartesian co-ordinates x and j in terms of the polars, while equations (2) give the polar co-ordmates m terms of the cartesians ; hence either transformation can be readily ^ ^Looking at the two systems, we see that the specification of the position of a point in a given plane requires two quantities, of which one at least must be a length, the other being a length or an angle. Y N p ^^^^ y ^>y^\ X f A Fig. I. Plane Co-ordinates. 14 ANALYTICAL MECHANICS [ARTS, n-12 11. Units.— The facts just noted lead us to inquire how lengths and angles can be measured and specified. Obviously any given length can be specified only by stating the number of times it contains some other given length taken as a standard. Thus the measure of a length consists of two factors — (i) a pure number, and (2) a given length called a unit. The unit may be the British yard, foot, inch, mile, etc., or the French metre, centimetre, millimetre, kilometre, etc., as may be convenient. These units are defined by the respective governments by reference to standards which are preserved, as far as may be, in their official departments. The above remarks as to stating the magnitude of a length apply equally to the measure of an angle. It must consist of two factors {expressed or understood), of which one is a pure number and the other an angle taken as the unit angle. The unit angle may be the degree, of which 360 correspond to one complete revolution, or the radian, which is the angle whose arc equals its radius. In analytical geometry the units, whether of length or angle, are often omitted, the numbers (or letters representing them) being used alone. It should be borne in mind, however, that such numbers or letters, apart from the unit understood, fail to completely express the length or angle in question, but are simply the working factors with which we are concerned in the analysis. The complete expression for the measurement of a physical quantity may be represented symbolically. Thus, suppose some length / contains the standard length or unit [Z] m times, then we may write ^=4^1 (i). Similarly for an angle i* which contains B radians, the unit angle or radian being denoted by \R\ we have «=^m (2). Suppose we change our unit of length to one of a third the size, equation (i) may then be written ^=^-[f] (3). thus showing that the new number expressing the given length is in- Creased threefold. Or, generally, the number measuring a given quantity varies inversely as the size of the unit in terms of which it is specified. Now let it be required to express the angle a in degrees [Z?], of which 180 correspond to tt radians. Then 7r[i?J=i8o[Z)], which put in (2) gives "=^°^[^] (4). Or in other words, the new units (degrees) being tt/iSo times the old units (radians), the new number measuring the given angle a is iSo/tt times the old number. 12. Dimensions of Units.— Consider now the measurement of ^'^T- '3] GEOMETRICAL BASIS ,5 an area, say of the rectangle ONPM in Fis i Tf Fai k^ •.. r the unit of length, the are! in questionTs etlde^ntl/L^sseVb;"'" '" a=x\L-\y.y{L'\=xy{L'\=xy{A-\ ... (5) oHhe Kl.'' """""" ^"^ '^' ""'' °^ ^'■'' """^ ^ '■°'" "^^ '=°'"P'^'« '"^-^"'e one^^hird thfSe!" Sn S^b^o^m^^ ""' °^^^"^'^ ^^ '^^^"^^'^ ^ -3.[f]x3;^[f] = 3=4f] = 9^[f] . . (6). An inspection of (5) and (6) shows that the unit of area \A\ equals the square of that of length [ZJ, and that consequently, when [Z] IS reduced to [Lli\, then {A] is reduced to [^/s^. Hence the new number expressing a is j'' times the old number. If we wish to dis- tinguish between the directions of the lengths in the above two cases we may replace (5) by a=x\X\y{y\ = xy\XY\ The unit of length is called z. fundamental unit, and that of area a derived one, since it depends on the former. Thus we see that if a derived unit equals the «th power of a fundamental unit, and the latter is changed in the ratio r, the former is changed in the ratio r". Or, in symbols, if Q—L", and L' — rL, that the corresponding derived unit ^' is expressed by (2' = (rZ)" = r" (2 (7). In this case the unit Q is said to be of n dimensions in Z, or to involve Z to the «th degree. It is obvious that we may extend this principle to the case where a derived unit is founded upon several fundamental units — each, it may be, raised to a certain power. Thus, if P=A':B^.Ci (8), where /"is a unit derived from the units A, B, C, we have a'^b^cyP^{aAY{bBY{cC)i (9), or, P'=a'-b?(nP (10), in which a'^b^ct is the factor affecting the derived unit when the fundamental units are respectively affected by the factors a, b, and c. It is specially noteworthy that, if any of the indices are negative, an increase of the corresponding fundamental unit will involve a decrease of the derived unit. In equations (8) and (9) the derived unit on the left side is said to be of the dimensions a, /J, and 7 respectively of the three fundamental units on the right side. Thus we may write for the unit of volume V={L}JI}W\ = \_V\orV=\XYZ\ . . . (ii), either of which shows that volume is of three dimensions in length. 13. Displacement. — Suppose that the position of a point undergoes a definite change. How may this change of position, step, or dis- placement be specified ? Obviously one method is to specify by the usual co-ordinates the initial and final positions, for from these data the i6 ANALYTICAL MECHANICS [art. 14 change of position is ascertainable. A more usual way is to specify the step directly and leave the final position to be ascertained if required. But how may it be directly specified ? Suppose its magnitude to be 2 inches. We have here the two factors which specify a length, but these fail to specify a displacement. For if the point in question is free to move in any direction, its final position is simply any point on the surface of a sphere of radius 2 inches, and whose centre was the initial point. Or, if it was free to move in a plane only, then its final position is simply any point on a certain circle, namely, where that plane inter- sects the sphere previously mentioned. Hence^ to specify a displace- ment, we must have something in addition to its magnitude. It is evident that the magnitude and direction suffice to specify any displace- ment of a given point. On looking back at the specifications of position, it is evident that this method of specifying a displacement is simply the polar-co-ordinate method of specifying a position. For we have simply to take the origin of co-ordinates at the initial position of the point, and then the r and B, which specify the final position, also specify the displacement, if it is understood to be in the plane of the diagram. Of course, the dis- placement could be expressed by the equivalent cartesian co-ordinates X and 7. But, though each system is available for each class of speci- fication, the cartesian system is usually preferable for specifying posi- tions and the polar system for specifying displacements. 14. Scalars and Vectors. — We are thus led to observe that though a length may be thought of apart from any definite direction, as for example when we say the earth's diameter is 8000 miles, there are other cases in which the direction of a length is just as vital as its magnitude, as in the case of specifying the displacement of a point or other figure. These are examples respectively of the classes of quantities called scalars and vectors, of which the latter have direction, while the former have not. Many other examples of these two classes of quantities will occur later. It is evident that vectors may be represented by straight lines. For a vector is specified by magnitude and direction, and the length of the line may represent to a certain scale the magnitude of the vector, while one of the two possible directions along the line, viz. that of the order of naming its terminal letters, represents the direction of the vector. Thus any straight line OP could adequately represent the displace- ment of a point in the direction OP, and of a magnitude represented by OP on a certain scale. The displacement called PO would be equal in magnitude but opposite in direction to the displacement called OP. Another device is to put an arrow head on the line representing a vector so as to indicate the direction of the vector. It should be noted that some writers use ' direction ' with a wider meaning than that employed above, namely, to denote both ways along a given line. They then say that the line's ' direction ' represents the ' direction ' of the vector, which needs also the ' sense ' in which the line is supposed drawn to be indicated for the complete specification of the vector. Which- ARTS. 1S-16] GEOMETRICAL BASIS 17 ever phraseology is employed, the fact to be borne in mind is that a line fails to represent a vector precisely until an order of naming its termmal letters (or some equivalent device) is supplied. Thus, if the centre of a sphere is taken as the point through which to draw lines representing various vectors, a vertical diameter is ambiguous, but its upper and lower halves, if supposed drawn from the centre, specify vectors equal in magnitude but opposite in direction (or, as some would prefer to say, equal in magnitude and direction but opposite in sense). 15. Composition of Displacements. — Suppose a point to suffer the displacement represented in Fig. 2 by OP, and then the displacement represented by OQ. To find its position after both displacements, or to compound them, it is obviously necessary and sufficient to allow the second displace- ment to operate upon the position due to the first. In other words, we must draw on the figure from P the line PR equal and parallel to OQ. Or, we may complete the parallelogram on OP and OQ by draw- ing PR and QR parallel to OQ and OP respectively pic. z. Compo- and intersecting at R, thus making PR equal to OQ sition of Dis- as well as parallel to it. If we are now asked to state placements. to what the two displacements are equivalent, i.e. what is the result of their composition, the reply may be, the point originally at O is, after the two displacements, at R. Or, more formally thus, the composition of the two displacements repre- sented to scale by OP and OQ yields the resultant displacement repre- sented to the same scale by OR. This is a very simple example of the important operation called the Addition of Vectors. It may be repre- sented symbolically as follows : — OP + OQ = OR (i), where the sign over the plus indicates that the addition is vectorial, i.e. having regard to direction and not simply algebraic. If the dis- placement OQ occurred first, and then the point at Q received the dis- placement represented by OP, it would as before be found finally at R, as is obvious. Thus we might write OQ+OP = OR (2). Other examples of the addition of vectors are the theorems of the parallelogram and triangle of forces, familiar to the student of elementary statics. 16. Localisation of Vectors. — A comparison of equations (i) and (2) and the operations they respectively represent will bring out important points as to the component vectors and their results in the two cases. Thus the equations seem to express that the order of quantities added is indifferent, and the result therefore the same, in the two cases. But, on going into details, we see that in (i) OP is applied at O and then OQ is applied at P. Whereas in (2) OP is apphed at Q, OQ having been previously applied at O. In other words, the points of application or localisations of the vectors were different in B i8 ANALYTICAL MECHANICS [art. 17 each case. Again, though the resultant displacements were the same in each case tlie paths of the points were different, being OPR in (i) and OQR in (2). Hence we see that although the equations (i) and (2) may express all that is needed for certain purposes, they do not give the entire details of what we suppose to have happened. Vectors may be spoken of as unlocalised or localised to various degrees, as for example in a point or a line. In concrete cases they are probably always localised to some extent. Thus, if a ship in harbour is said to be raised 10 feet by the tide, the vector has a magni- tude 10 feet, a direction vertically upwards, and is localised to the volume occupied by the ship in question. If one of the masts be similarly referred to, the same vector is localised to the volume of that mast, or if the thickness of the mast is regarded as negligible, the vector is localised in a line. Or, again, if the top of the mast be in like manner spoken of, the vector is localised ra what may be regarded as ^. point. Hence, in compounding displacements and adding vectors in general, care must be taken that the appropriate degree of localisation is in each case present in the component vectors thus dealt with. Examples — II. 1. Explain the distinction between fundamental and derived units, and show how the size of a unit of one class is related to the sizes of those of the other class. 2. Define the terms scalar and vector. Give examples of each, and show how to add vectors. 3. Show by illustrations what you understand by the localisation of vectors. What bearing has the localisation of vectors upon their composition .? 17. Time and Motion. — The physicist, as such, regards time as that familiar though inscrutable one-dimensional something which separates changes in bodies and individual sensations and extends without known limits from the past to the future. Just as space is the abstract of all relations of co-existence, so time is the abstract of all relations of sequence. Each is a primary conception that cannot be rendered in terms of anything simpler. When we combine, in a certain manner, the conceptions of space and time, we have the conception of motion, a point being said to move if, at different instants of time, it occupies different positions in space. To measure a given quantity of time we need a unit of time in which to measure it and a number to express how many of these units the given quantity contains. As units, the familiar second, minute, hour, etc., are used. To fix an instant in time with respect to another instant we need to state only (i) the duration, or quantity of time, separatmg them, and (2) which instant is the later. Or, we may accomplish both algebraically by prefixing to the statement of the duration the sign of plus if the instant to be fixed is later than that takeri as origin, or the sign minus if it is earlier. This is like fixing the position of a point on a line, no divergence into solid or even plane space being permitted. ART. 1 8] GEOMETRICAL BASIS 19 18. Velocity. — The velocity of a point is its rate of change of position. This is a brief definition serving to introduce the subject, which we shall now examine in detail. Thus let a point move fjom O along the axis OY, and let the times when it has the displacements °i yii y'.i Ji's, • • • > be respectively o, ^,, 4, t^, . . . t. Consider the quotients j^'i/A; y 1.1^1 y^lhi ■ • ■ yjt- And suppose these quotients all have the same value, v say. Then this common value v measures the velocity of the point or its rate of change of position with time while passing from O to P. In what units is this quantity measured and expressed? To answer this we fall back upon the dimensional equations previously used. Thus, writing [ y] for the unit of length, [J'] for the unit of time, and [ F] for the unit of velocity, we have v\V]^y[Y]-ri{T] (i), whence [V] = [Y'T~'] _. . . . (2). Or, the unit of velocity is of plus one dimension in length (in some definite direction) and of minus one dimension in time. Of course, the actual size of the unit of velocity depends upon the units adopted for length and time. Let us now consider the significance of the supposed equality of the quotients jc,//i, y^ft^, y^lt,, ... and yjt. It is clearly only when this equality holds that we can apply to the common value v the phrase velocity of the point from O to P, and leave it unquahfied by any further restriction. If, on the other hand, knowing nothing of jVi, y^, )'s, we simply knew the values J)' and / having the quotient v, we should then say that v was the mean velocity of the point from O to P. Again, if we knew a very large number of such intermediate values of the distances j>'i,ji'„jC3, ...:►'«..., and their corresponding times t^, t^, t^, . . . tn. . ; and found all the quotients jCi/A • ■ -ynltn . . . =v, we should then say that the velocity of the point between O and P was uniform and of the value v. The rigour with which the term uniform would apply would depend upon the number of the intermediate positions and times known, and would become absolute only in the ideal case of all such intermediate information being available. Finally, if the quotients had different values, thus y^jt^^v^, y^lt^=Vi, etc., then the velocity would be variable, v^, v^, etc., each expressing the mean velocity over the range in question. But even though the velocity is varying, the idea forces itself upon us that at each instant of time (or position in space) the velocity must have some definite value ; just as when a point is moving it has at each instant of time some definite position. How shall we in thought attam and measure this instantaneous value of a varying velocity ? Take shorter and shorter durations, t„ t„ t^ . . ., each includmg the instant in question, suppose the corresponding displacements jy,, iq^, '?a • • • known, take the respective quotients i?i/ti, ^uIt^, -q^JT, . . ., and suppose these quotients to continually approach a limiting value v, then V is the value sought, and expresses the instantaneous velocity at the instant in question. 20 ANALYTICAL MECHANICS [art. 19 It is obvious to students of the calculus that an instantaneous velocity is the first differential coefficient of a distance (in some given direction) with respect to the time. Or, in symbols v—dyldt=y, vfhere the dot denotes a single differentiation with respect to time. It should be noted that in scientific usage velocity involves the idea of direction just as displacement does. If we wish to speak of the magnitude of a velocity apart from all idea of its direction we use the word speed. Thus if a point described a circle so as to pass over an arc of 10 cm. length in each second, i cm. in each tenth of a second and so forth, we should say that the speed was uniform, but that the velocity varied because one of its elements, viz, direction, was con- tinually changing. It should be noted, however, that in speaking of the speed of a point moving along a given path we may use the positive or negative sign to indicate the opposite directions in that path. 19. Acceleration. — Since velocities may vary it is incumbent upon us to consider the changes of velocity, and also the time-rate of change of velocity, which is called acceleration. Just as we passed from the conception of displacement to that of velocity by using time once as a divisor, so we may pass from velocity to acceleration by using time once more as a divisor. Thus, let a moving point be considered, having at times o, Aj h, l>, ■ • ■ I, the velocities o, Vi, v^, v^, . . . v, all along the same straight line, OY say. Then, if the quotients v^jt^, v^lt^, Vsjti, . . . v/t all have the same value a, this common value a measures the acceleration of the point, which in this simple case is also directed along the line OY in question. If it is only known that after time / the velocity has increa'Sed by the amount w, then the quotient, vji=a say, measures the mean acceleration during the time t. If, on the other hand, a large number of corresponding intermediate values of v and t are known, each pair yielding the same quotient a, then a measures the uniform acceleration, the uniformity being ascertained with more and more rigour as the data increase in number. If, however, the intermediate values of v and t yield varying quotients the acceleration is variable. Its instantaneous value may be ascertained in thought and expressed in symbols, as was done for a velocity. Thus, let shorter and shorter times, t,, t^, etc., be taken, each including the instant in question, the corresponding changes in velocity being respectively &,-», v^—v, etc. Then, if the quotients (»i-z')/T,=rt„ (v^-v)lr^ = a^, etc., approach a limiting value a, that value a measures the instantaneous acceleration at the instant in question. Or, in the notation of the calculus a=dv\dt=d^y\df=y, where the two dots denote two differentiations with respect to time. From what has gone before it is easily seen that an acceleration is of plus one dimension in length and minus two dimensions in time. ART. 20] GEOMETRICAL BASIS Or, if its unit is denoted by \A\ then a\A-\=v\V\^t\T-\ = l,{YT-'\ Accelerations may accordingly be measured in centimetres (or other units of length) per second per second. It is often convenient to abbreviate the units thus : cm. per sec.^ It should be noted that acceleration is a vector, since it has magnitude and direction. If we wish to speak of the magnitude only of an acceleration we may say 'rate of change of speed' or quickening. This is, of course, a scalar having magnitude and sign only. The relations of the various quantities hitherto considered may be exhibited compactly as shown in Table 11., their order of development being followed by reading down the columns. Table II. Relations of Kinematical Quantities. QUANTITY (POSITION) VELOCITY CHANGE OF QUANTITY DISPLACEMENT CHANGE OF VELOCITY CHANGE DIVIDED BY TIME VELOCITY ACCELERATION 20. Displacement Graphs. — The motion of a point along a straight line may be usefully represented graphically on a displacement - time diagram or by a displacement graph as follows: — Take distances along the axis of y to represent the re- spective displacements, and dis- tances along the axis of x to represent the corresponding times. Then each such pair of co-ordinates will define a point on the diagram, and a continuous line drawn through those points will give the graph required. It should be noted here that some amount of discretion must be exercised in drawing this line be- tween the points furnished by the data of the case, and that only more or less probability, but not certainty, attaches to any such intermediate portions of the line so drawn. Ex- amples of such graphs are shown in Table iii. 6 seconds Time Fig. 3. Displacement Graphs. Fig. 3, plotted from the data of ANAL YTICAL MECHANICS [ART. 20 Table III. Data for Displacement Graphs. GBAV^ A. Graph B. Gkaph c. Displacement Time in Displacement Time in Displacement Time in in Feet. Seconds. in Feet. . Seconds. in Feet. Seconds. 2 I I I 0-54 I 4 2 4 2 2 4 2 3 10 s 9 3 6 7-46 4 ! 5 12 6 8 6 i Graph A illustrates a uniform velocity or speed of 2 feet per second, and even if only the final point (12, 6) were given we should have a mean velocity of the same amount, though no knowledge of the uniformity. Obviously the equation of the graph is y=2x, in which the 2, being the tangent of the angle between the graph and the axis of time, expresses the rate of increase of displacement with time, i.e. the speed. Graph B, for the first second, shows only half the mean velocity represented by A, but by the end of two seconds, where the graphs intersect, their mean velocities are the same. Beyond that point the graph B is higher than A, i.e. indicates a higher mean speed from the start than A does. It therefore gives an example of a variable speed. Let us now inquire what is the instantaneous speed at some instant, say two seconds from the start. A glance at the corresponding point on the diagram shows that the speed there is of the order 4 feet per second, or double that of graph A. A closer examination would con- firm this result. We may also arrive at the same conclusion by another method. Thus we see from Table 111. that graph B has the equation y=x^, hence the tangent to it at the point {x'y') is represented by the equation ^(^y+y')=xx'. Thus for the point in question (2, 4) the geo- metrical tangent to the graph is y=4x—4. Hence 4, the co-efificient of*-, represents the trigonometrical tangent of the angle between the axis of X and the geometrical tangent to the graph at the point in question. In other words, the slope of the curve at this place, estimated so as to measure the speed, is represented by the number 4. Similarly after three seconds from the start, when the displacement is 9 feet, we find the speed to be 6 feet per second. If we apply the method of the calculus we must differentiate y with respect to x in the equation of the graph to find the ratio of the very small increase of displacement y to the corresponding increase of time represented by x. Thus, the graph being y = x', we find dyjdx=2x^v, which agrees with the two results already dealt with. ART. 2l] GEOMETRICAL BASIS 23 Again, if we ask at what instant is the speed denoted by graph B equal to a certain assigned quantity, say that of graph A, then by the graphical method we must find the point on B which is touched by a line parallel to A. It is evident that point is in the neighbourhood (i, i), i.e. the instant one second from the start, and when the displace- ment is I foot. And this is confirmed by the analysis. Consider now the graph C ; it indicates by the horizontal portions, at times o and 6, zero velocities or instantaneous states of rest. It also indicates continuous increase or decrease of speed between these instants, and a maximum speed at three seconds where the graph is steepest. It may accordingly be recognised as representing a motion which at any rate resembles that of a pendulum bob. The equation of the graph may be written ^' = 4— 4cos(7r.T/6) ox y—%s\n'{iTxji2). For its slope we have dyjdx=:^w%\n (jra:/6). Although the ordinates in the displacement graph only refer to steps taken along a direction otherwise specified, it is clearly legitimate and convenient to let negative ordinates mean displacements just opposite in direction to those represented by positive ordinates. These examples sufficiently illustrate that in a displacement graph slope represents speed, and therefore change of slope per unit distance along the axis of time represents acceleration. But acceleration can be more clearly and simply exhibited on another type of graph now to be dealt with. 21. Velocity or Speed Graphs. a point moving along a straight line as the quantity to be plotted along the y axis, the time being represented by distances along the X axis as before. The curve thus obtained may be called the velocity graph or speed graph of the moving point. In order to facilitate comparison of the two types of graph we shall plot the speed graphs corresponding to the data in Table in. and the infer- ences as to speeds deduced from them. These are shown in Fig. 4. It is thus seen that graph A, representing a uniform velocity, is now a horizontal line, whereas graph B, representing a velocity proportional to the time, is now a straight line inclined to the -Let us now take the velocity of 7 secoTids Time Fig. 4. Speed Graphs. a siruigni line iin-iinwvi i-w .."V, , horizontal, i.e. it represents a uniform acceleration. Fmally, the graph C rises from the origin and afterwards falls to zero. The equa- tions of these three speed graphs are respectively jc= 2, jj^^ 2a:, and ^=§5rsin(7r^/6), as shown in the preceding article. 24 ANAL YTICAL MECHANICS [art. 22 It may be noted here that in a speed graph the area below the graph between two given ordinates represents the length, distance, or space described by the moving point in the time defined by the ordinates. For each narrow vertical strip of this area has an area which is the product of height into width. But the height of the strip or ordinate represents the instantaneous speed, and the width of the strip or increase of abscissae denotes a short time. Also the product of these is the distance described in that short interval. Hence the whole area, or sum of these strips, represents the sum of such distances, and is therefore the whole distance described in the finite time under consideration. Thus, the abscissae being times and the ordinates speeds, the slope of the graph represents acceleration and the area the distance described. The speed graph is accordingly often very useful, embracing as it does so conveniently the four quantities with which we are concerned in the case of a moving point. Though the ordinates in a speed graph denote the speeds of a point along a line whose direction is otherwise specified, it is convenient to use negative ordinates to denote speeds in the opposite directions along that same line. 22. Other Graphs for a Moving Point. — Since out of four quantities we may choose six different pairs, it is evident that for a moving point we may construct graphs in six different ways, viz. with co-ordinates denoting s and /, v and t, a and t, v and s, a and s or a and v, where s, t, V, and a denote space, time, velocity, and acceleration respectively. But most of these other graphs have only a very limited usefulness, or may be described as curious rather than useful. We may perhaps refer to some of them in special cases later. It may be just noted here that plotting accelerations and distances as co-ordinates brings out the fact that in the graph C of Figs. 3 and 4 the acceleration is proportional to the displacement from a certain point, the graph being a straight inclined line cutting the axis of distances at the point 4. Examples — III. 1. Define velocity, uniform velocity, 7nean velocity, and instantaneous velocity. 2. Exhibit in tabular form the relation of displacement, velocity and acceleration, giving also the dimensions of each. 3. Plot a displacement graph from the following data :— Displacements in yards, o, o, 8, 19, 29, 39, 49, 58, 67, 76, 88, 100, 103. Tzmesrn seconds, o, i, 2, 3, 4, 5, 6, 7, 8, 9, 10 i, 12. If the graph represents a man running a race, indicate what you believe are the start and finish. 4. A point moves so that its displacement after t seconds is 16/2 piot its displacement and speed graphs, showing the corresponding features of each graph, and find its acceleration. 5. Explain the special advantages of a speed graph. Plot one to represent some variable motion, find the space described, and indicate anv important features. ^ 6. ' Define the mean velocity of a point in any interval of time. Prove by a ARTS. 23-24] GEOMETRICAL BASIS 25 graphical method (or otherwise) that if the acceleration be constant the mean velocity is equal to the arithmetic mean of the initial and final velocities, and also to the velocity at the middle instant of the interval. Show by examples that in any other case these three quantities are usually different.' (LoND. B.Sc, Pass, Mixed Math., 1904, i. i.) 7. ' The times /j, ^2, t^ at which a particle moving with constant retardation passes three points P^, P^, P^ of its path are recorded ; find the acceleration, and the velocity at P^, having given P^P^ = a^ PJP^ = b? (LoND. B.Sc, Pass, Applied Math., 1905, 11. i.) 23. Composition ofVelocities and Accelerations.^ — A little reflec- tion will show that the operation of the addition of vectors which is valid for the composition of displacements applies also to the composi- tion of velocities, provided that the point in question is so circum- stanced that, no matter how far it goes due to one velocity, it is still equally affected by the other. Thus, after a short time t, the two velocities v^ and v^ will have imparted to the point displacements in their own directions and of magnitudes v-^r and v^r respectively. The position of the point after time t is accordingly determined by the vectorial addition of these two displacements, and may be denoted by VT, Now let the position after a finite time / be considered. It will evidently be determined by the vectorial addition of the displacements vj and v^t along the same directions as at first, and is therefore denoted by vt. Hence we have the same diagram as before, but magnified in the ratio of / to r. Thus the point has uniform velocity V, and in the direction first determined. Or, in other words, the point moves with a resultant speed and direction determined by the vectorial addition of its two component velocities. Similar reasoning would apply to show that two accelerations simultaneously possessed by any point may be compounded by vectorial addition to give the resultant acceleration. Suppose now that a velocity and an acceleration have to be dealt with, even then one aspect of the problem can be treated by vectorial addition. For the acceleration, though possibly varying in magnitude and direction, will impart to the point in time t some definite velocity, z/i say. Thus, if the original velocity of the body were »»> the final velocity v would be determined in magnitude and direction by the vectorial addition of z'„ and z'„ or w=z'o+»i- The aspect of the problem not here dealt with is the path of the point during time /. Unless the acceleration is for the whole time m the direction of the original velocity, it is evident that the path miist be curved. And even if the velocity and acceleration are always colhnear, we have still left undetermined the position of the point at each instant of time. Such aspects of the cases will be dealt with in their proper places later. 24 Vectorial Polygons. — For the composition of more than two vectors it is evident that we might take any two first and find their resultant, then compound this resultant with a third of the components 26 ANALYTICAL MECHANICS [art. 24 for a new resultant, and so forth, until all had been dealt with. A simpler method, however, of compounding n vectors is to construct a polygon n of whose sides taken in order represent in magnitude and direction the vectors to be compounded. Then the remaining side required to close the polygon will, if taken in the reverse order, repre- sent both in magnitude and direction the resultant vector sought : and this holds true whether the components are all in one plane or are in various directions in solid space, i.e. the vectorial addition is valid whether the polygon by which it is effected is of necessity plane or gauche. This is easily seen by considering the simplest example of a vector, namely, a displacement. Thus, to illustrate the gauche polygon, if three displacements at right angles and of magnitudes equal to 9 inches, 4I inches, and 3 inches respectively are given to a point at one corner of a brick of that size, the point would be carried to the opposite corner of such a brick if rightly situated with respect to those dis- placements. It is sometimes a great convenience to be able to write down the result of the addition of vectors without actually drawing the poly- gon to scale. The formulae for a plane polygon may be easily de- rived by reference to Fig. 5. In this figure the component vectors, which may be thought of as displacements, are represented to scale by 0P„ P,P„ P.P,, P,P. The resultant is obviously repre- sented to the same scale by OP. Further, let the magnitudes and directions of the components be respectively denoted by r^Q.^, r^d^. A P y^' /% aa. N3 r, / ? i N2 j( k/ X ^-P Xi\ N. ^ 'L*-— -"^ ^■^^^^ Ai M, M2M3M Fig. 5. Polygon of Vectors. rji^, and r^Qi, their resultant being r&. Then, by the figure, we have OP' = OM'-t-MP''andtanMOP = MP/OM (i) ButOM = OMi + M,M2-|-M,M3-MVr3M, and MP = MN.-f NiN.-hN^Na-FNsP. Or, using the symbols — 0M=/-, cos6l,-|-raCos6»2-f>-, cos^s-l-^, cos6i^=2/-cos6' . (2), and MP=riSinej-f/-2sine2-f^3sin6*3-|-r,sin6li = Sysin6' . . (3)'. Hence, by (2) and (3) substituted in (i), we obtain /-' = (2^cos5)= + (2rsin6l)^ (4) and tan(9=(2rsin6l)/(2;rcos(9) \ i'\ In equation (5), giving the value of tan B, it should be noted that an ambiguity arises unless the algebraic signs of numerator and de- nominator of the fraction on the right side be each maintained in their original positions. Thus if the fraction in question is +9/(+io) the angle is uniquely determined as of the order 42°. But if the fraction is ART. 25] GEOMETRICAL BASIS 27 written +(9/10) we are uncertain whether the angle is 42° or 222° whose tangent should have been preserved in the form ( — 9)/(— 10). Thus the angles 45°, 135°, 225°, and 315° are uniquely determined by their tangents if written respectively ( + 1 )/( + 1 ), ( + 1)/( — i ), ( — i )/( — i ), and (-!)/(+ 1). 25. Gauche Polygon of Vectors. — If the vectors to be compounded form sides of a gauche polygon, then the closing side which represents the resultant will be the diagonal of a parallelepiped whose adjacent edges are built up as were the sides OM and MP of the rectangle in Fig. 5, whose diagonal OP represented the resultant of the plane polygon. This is illustrated in Fig. 6, in which the axis of z shown in perspective is to be understood as at right angles to the plane of xy. The separate vectors are not shown in the figure, but are supposed to have magnitudes ri, r^, r,, etc., and to make with the axes of x, y, z the angles u,„ ^„ -y,, a^, /3j, y^, aj, ^3, 73, etc., the resultant OP having magnitude r, and making with the axes the angles a, ^, and 7 as shown. Z Fig. 6. Addition of Vectors in Solid Space. It is thus seen that OP is the resultant of OA, OB, and OC, i.e. its components along the co-ordinate axes are r cos a, r cos ^, and r cos 7 respectively. Similarly each of the component vectors, r, say, contri- buted along the axes of x, y, and z the components r^ cos u.^, r, cos /?, and ri cos y,. Hence we have OP= = OA^+OB= + OC and cos AOP = OA/OP, cos BOP = OB/OP, cos COP=OC/OP But 0A=riC0sa,+r2C0S o„ + . . . = 2rcosa .... OB = r,cos/3.-fr,cosj8o + . . . = 2rcos/3 .... OC = ?-,cosy,+r2COS72+- • . = 2rcos7 . . . • Thus, by substituting in (i) and (2) the values from (3), (4)= and (5), we obtain r- = (Zrcosay + {^rcosPy- + {^rcosy) . . ■ ■ .6), also cos a= IVcos a)/r, cos I3=(2r cos ^)//-, and cos 7 = (2;-cos y)/r (7). It may also be seen by the geometry of Fig. 6 that we have cos'a + cos°-i8+cos'7=i K^)- (2). (3), (4), (5). 28 ANAL YTICAL MECHANICS [arts. 25a:-25^ 25a. Moments. Definition. — The moment of a vector with respect to a point is the product of the vector into the perpendicular from that point upon the line in which the vector is localised, and is reckoned positive when the direction of the vector about the point is counter- clockwise. Theorem. — If two vectors are localised at a point O, then the algebraic sum of their moments with respect to any point P in their plane is equal to the moment of their resultant about P. In Fig. 6a, let OA, OB represent the component vectors, OC their resultant, found by the parallelogram law, and all localised in O ; also let P, in the plane of OABC, be the point about which the moments are to be taken. Proof. — Then, by definition, the moment of any vector is re- presented to scale on the figure by twice the area of the triangle O A Fig. 6a. Theorem of Moments. whose base in the vector and whose vertex is the point P about which its moment is taken. Thus, half the moment of OC =area of AOCP= AOBP-|- AOCB-|- ABCP = AOBP-|-AOAP =half-sum of moments of OB and OA. This accordingly establishes that case of the proposition represented in the figure. When P is differently placed the needed demonstration follows in like manner. 25b. Composition of Angular Velocities.— Measuring angles in radians, we naturally measure angular velocities in radians per second Thus, if a point P moves with an angular velocity ai = 61/^ about a given axis from which it is distant by the radius r, we have the relations iii = 6lt=sjrt or s=mrt, where s is the arc described by the point in time t. Hence in a given time the displacement s is proportional to the product co/-. This ART. 25^] GEOMETRICAL BASIS 29 suggests an analogy with the moments of a vector with respect to a pomt. Thus, referring to Fig. 6a, consider the point P in the plane of the diagram as having an angular velocity about O A and proportional to the length of OA=cui say. Then in time dt, P would suffer a dis- placement perpendicular to the plane of the diagram of amount ^j, = (o,^/X perpendicular from P upon OA=(//x double area of OAP Similarly if the point P had a coexistent angular velocity o^ proportional to OB about OB, its displacement in virtue of that in time dt would be given by ^ff2 = w2(f^x perpendicular from P upon OB. Hence the sum of these displacements would be represented to a certain scale on the diagram by the sum of the areas OAP and OBP. But, as we have just seen, this sum equals the area of OCP, which consequently represents to the same scale the displacement ds which P would experience in time dt under the influence of an angular velocity about OC of amount to proportional to the length OC. Thus, so far as a point P in the plane of the diagram is concerned, the resultant of the coexistent angular velocities whose axes are OA and OB and magnitudes OA and OB respectively is an angular velocity whose axis is OC and whose magnitude is OC. Suppose now a point P' is taken out of the plane of the diagram such that PP'=/ is perpendicular to that plane. Then the displace- ments we have just considered for P would also be the displacements perpendicular to that plane for P'- The displacements parallel to the plane for P' would obviously be vtipdt, which would appear in the diagram as perpendicular to OA, w^pdt appearing perpendicular to OB, and (updt appearing perpendicular to OC. Thus these displacements would form in the diagram a parallelogram of the same shape as OACB, but of different size, and with sides perpendicular to the original parallelogram. Hence these component displacements due to r . . . Xn). But some of the points might have abscissae of the same value. Thus, let w, points have the abscissa x^, and Wj points the abscissa Xj, and so on ; the same notation with y's holding for their ordinates. Then the co-ordinates of their centre would be x='2{mx)/'2,m and y—'S{my)/'2fn (i). We may extend the application by supposing the m's to refer to the magnitudes of elements of length, area, volume, or other scalar quantity situated at the points defined by the corresponding x and y. The ART. 26] GEOMETRICAL BASIS 31 point {x,y) so determined is called the centroid of the system, figure, or body in question. If the co-ordinates of points with respect to the centroid are .r', >■', we have ■»=-v-f-*' and_)'=j'-|-y . (2). Hence Imx = 5c2.m + '^mx and '^my =ylm + "Lmy . But, by (i), these reduce to Q=-1mx' = ^my' . . (-,)_ Thus, if equations (i) are taken as defining the centroid, we may regard (3) as giving some of its properties. It is, however, often con- venient to take (3) as giving the definition of the centroid and equa- tions (i) as forming the working rules for the determination of its position. For quantities distributed throughout space of three dimensions we have simply to add the corresponding equations, involving z by analogy with (i), (2), and (3), thus giving the following additional relations : — z = 'l.vizfZm z=z+z' '- (4). and o = ~f/iz' 26. Constraints and Degrees of Freedom.— If a point is not con- strained in any way, it is said to have tAree degrees of freedom, since it is obviously free to move parallel to the three co-ordinate axes of solid space. If the point is constrained to remain on a plane surface, say that of xy, it has then lost one degree of freedom, namely, the motion parallel to the axis of z, and retains two degrees of freedom only, namely, those along the axes ol x and y. Similarly, if the point is con- strained to remain in a line, say the axis of x, it is evident that it has lost two degrees of freedom and retains only one. With any actual small particle these two cases may be represented by floating on still water and confinement in a straight tube respectively. Let us now contemplate an extended body whose parts are debarred from any relative motion j this is called a rigid body. Then obviously, if it has no constraints whatever, it has what may be called six degrees of freedom, viz. translation parallel to each of the three co-ordinate axes and rotation about each of them. It should be noted that what is styled a single degree of freedom of a rigid body may (namely, if it is a rotation) involve the two-dimensional motion of its points. And if with this rotation we combine a translation along the axis of rotation, we have a three - dimensional motion of its points. Whereas two translations and a rotation about the axis per- pendicular to both of them involves only a two-dimensionsd motion, for all the motions occurring are parallel to the plane containing the two translations. Thus the classification of motions according to their constraints and degrees of freedom, though very useful and needing to be borne in mind, dififers materially from that according to dimensions in 32 ANALYTICAL MECHANICS [art. 26 space, which is usually simpler, and will be followed in developing the subject later. To take away any one of the original six degrees of freedom of a rigid body, one constraint is needed, so that six such constraints are needed to fix its position. A few illustrations of simple cases may be given here. Let possible translations in three rectangular directions be denoted by x, y, and z, the corresponding possible rotations being u, v, and w. J Top-spinning on level ground. — One point of the body touches the xy plane, and so loses motion in the z direction up or down. Degrees of freedom left are x, y, u, v, and w. Top with spike in groove. — Two points of the body touch the sides of, the V-shaped groove parallel to axis of x, and so lose motion in directions oiy or z. Degrees of freedom left are x, u, v, and w. Top with spike in hole. — Three points of the body touch the sides of the quasi-conical hole, and so it loses motion in directions x, y, and z. Degrees of freedom left are u, v, and w. Beam of Balance. — Instead of knife edges, let the beam have two blunt screw points, one of which touches at three points in a quasi- conical hole and the other at two points in a V-shaped groove pointing to that hole, i.e. along the axis ol x ; it thus loses all three translations and two rotations. The sole degree of freedom left is u, or the rotation about the axis of x. Instrument on 'hole, slot, and planed — Three points of the apparatus touch the sides of a quasi-conical hole, two points touch the sides of a V-shaped groove pointing to that hole, one point rests on the plane containing the hole and groove. In the case of physical and other apparatus requiring levelling and leaving set up without shake, or re- placing in exactly the same position after temporary removal ; this method is adopted, the blunt points of the levelling screws resting on the ' hole, slot, and plane ' respectively. It may easily be seen that a rigid straight line, if unconstrained, has five degrees of freedom. Thus, denoting the numbers of constraints and degrees of freedom by C and irrespectively, we have the following scheme : — For a point, C-f F= 3. For a rigid straight line, C-\-F= 5. For other rigid bodies, C-\-F= 6. Where no great pressures or speeds are to be used, the above arrangements of constraints, called geometrical clamps, attain ideal results in spite of the inevitable imperfections of human workmanship or machining. The ordinary arrangements for sliding and rotating motions in machinery involve surfaces which are approximately plane and cylindrical But though such surfaces fail of ideal perfection, they attain an approximation sufificient for the purpose and in combination with a facility for maintaining that degree of accuracy in spite of wear. Thus the physicist and the engineer have different ends in view, and rightly take different methods of reaching them. ART. 26] GEOMETRICAL BASIS 33 Examples — IV. 1. A boat steams due north at 15 miles per hour, while a man walks her deck m various directions at 2 miles per hour. Find graphically the man's velocities when his directions of walking the deck are [a) south-east, (*) north-north west, {c) west. From what body as base are your velocities reckoned ? 2. Show how to compound graphically three or more velocities. Will this construction serve for any other quantities ? if so, name some such. 3. Obtain analytical expressions for the lesultant of any number of coplanar vectors. 4. Determine graphically or analytically the resultant velocity of a point which has simultaneously a vertically upward velocity of 5 cm. per second, a horizontal westerly one of 45 cm. per second, and a north-easterly one of 1000 cm. per second. Ans. 968 cm. per sec. nearly, making with the northerly, easterly, and upward directions the angles whose cosines are 707/968, 662/968, and 5/968. 5. Taking the point P in a different position from that shown in Fig. 6a {e.g. inside the parallelogram OACB), show that the theorem of moments still holds. 6. Establish the construction for the composition of simultaneous angular uelocilies. 7. Consider the turning of a parallelepiped through a right angle first about one axis OA, and then about another axis OB, then reverse the order of turning about these axes. Hence show that the construction of question 6 does not apply to the composition oi successive finite angular displacements. 8. Determine the angular velocities of the earth about two rectangular diameters, one of which meets the surface at Greenwich (lat. X) and the other in the meridian of Greenwich, so as to compound to the earth's angular velocity cos X about the axis through Greenwich and the perpendicular one. Or, taking X = 5i° 29' and o) = 27r per day, the velocities are 4'9i6i5 and 3-912815 radians per day, i.e. one com- plete turn in 1-278075 and 1-6058 days respectively. 9. Discuss the degrees of freedom of particles and bodies with and without constraints and obtain expressions for the numbers of degrees of free- dom of various bodies. 10. If a railway track turns uniformly at the rate of 9°-55 in a length of one furlong, what is its radius of curvature .-' Ans. 3/4 mile. 11. Define the term ceniroici, and obtain general expressions to locate it in a plane over which any number of points are distributed. I-' State what arrangements can be made to permit only the following geometrical motions of a rigid body :— (a) translation along a given horizontal line, {b) rotation about a given horizontal axis, {c) rotation about a vertical axis. . . . . , • v 13. Name five geometrical and mechanical quantities, giving their dima,- sions and stating for each whether it is a scalar or a vector. 34 ANALYTICAL MECHANICS [art. 27 CHAPTER IV RECTILINEAR MOTIONS 27. TJnifomi Acceleration : Low Falls. — In this chapter we re- strict ourselves to rectilinear motion ; hence, when the acceleration is given as uniform of value a, any initial velocity of magnitude u possessed by the point under consideration must be along the same line as the acceleration, though it may have either algebraic sign. Let the velocity have magnitude v at time t, and the space described be .f, then it is required to establish relations between v, u, a, s, and /, and to use them for the solution of any problem relating to motions of the type in question. By definition it is clear that in time i the change of velocity occur- ring is ai, which must be added to the initial to obtain the final velocity. Thus, we have ....... . (I). This is illustrated by the speed graph of Fig. 7, in which OK is the initial speed u, MP the final speed equal to MN+NP or u+a(. It is clear that the equation of KP isy=u+ax, so that NP=fl/. Consider next the space s, which is represented in the dia- gram by the area of KPMO. It is evidently given by the rect- angle OKNM plus the triangle KNP, i.e. by uf plus INPx/. But NP is ai, thus we have s=uf+^ai' . . . (2). On eliminating / between (i) and (2) we have • ■ ■ (3). V = = u-\-at Y 1 In ^ p fr _^_^ K N M T/ME . i X \ Fig. 7. Uniform Acceleration. 11 =u- + 2as (For by squaring (i) we find v-=u'' + 2i(ai-\-a-i\ and from (2) we see that 2as=2uat-\-a'^f.) These equations, (r)-(3), are the required relations between the five quantities concerned, and serve for the solution of any cases of recti- linear motion with uniform acceleration. For example, for ' low ' falls or rises, i.e. motions in a vertical line near the surface of the earth, we may take that surface as the origin and measure j or w upwards as positive, then the acceleration due to the earth being downwards is to ARTS. 28-29] RECTILINEAR MOTIONS 35 be accounted negative. It has the approximate numerical value 32-2 feet per second per second or 981 centimetres per second per second. Thus denoting either of these positive numbers by ' g,' as is usual, we must insert — ^in the equations instead of +a. If, on the other hand, we find it convenient in other problems to take positive quantities to denote downward displacement, velocities, etc., then the acceleration due to the earth becomes positive, and is accordingly represented by -|-^ in the equations. Examples — V. 1. Find the uniform acceleration which in 5 seconds changes an upward velocity of 64 feet per second into a downward velocity of 96 feet per second. Ans. 32 ft. per sec.^ 2. A tram passes in 18 minutes between two stations 6 miles apart, stopping at each. If the train at first increased its speed uniformly under steam, and then immediately, with steam off and brakes on, decreased its speed uniformly at twice the former rate of increase, find the accelera- tion and retardation, and make displacement and speed graphs of the journey. Ans. + 1/18 and — 1/9 mile per min.^ 3. What is the rate of increase per foot of the square of the speed of a point under uniform acceleration of 20 ft. per sec.^ ? and what velocity is attained in 35 feet from rest ? Ans. 40 ft. per sec.^ ; 37^4 ft. per sec. 28. Uniform Acceleration by the Calculus. — Using the notation of the calculus for the motion of a point whose distance from the origin is s and acceleration a, we have d'slde=a ... . (i). Thus, on integrating, we find MS)=«M * or dsldt=(it-\-b . . . . . (2), where b, the constant of integration, is evidently the initial velocity previously denoted by u. By a second integration we have /ds=^J{at+b)dt, or s^i^^^+bt (3). Obviously equations (2) and (3) of this article correspond respec- tively with (i) and (2) of article 27. 29. Acceleration proportional to Displacement. Simple Har- monic Motion, i— We pass now to cases of varying acceleration, taking first that in which it is proportional to the displacement but oppositely directed. , , , , , Let the displacement OM, Fig. 8, be denoted by y, and suppose 1 If desired at this early stage, this motion may be treated without the calculus, as is often done by students of physics. (See. e.g., the writer's Sound, Arts. 13-16. ) 36 ANALYTICAL MECHANICS [art. 29 Then we have as the equation of motion (i)- Fig. 8. Simple Harmonic Motion. the acceleration to be — m y. of the point M ^ + 0,^=0 Now it is obvious that the acceleration, being always opposite to the displacement, will retard every motion of M from the origin and reverse it, thus causing the point M to pass through O in the opposite direction. But on pas- sing through O the displacement reverses, and therefore also the acceleration. Thus this motion from O must be annulled and re- versed as before. Hence we see that the motion of M is a to-and- fro movement or vibration along YOY' about the origin O as centre. The simplest way to ex- press such a motion analytically is by a sine or cosine function of the time. To make such a func- tion as general as possible we need three constants, denoting (i) the amplitude (a) or maximum value of OM, (ii) an angular velocity {n say), and (iii) the epoch (e) or phase angle at the beginning of the time. We thus write as a trial solution jl'= a sin («/+«) .... (2). Substituting this in (i) we have { — tii -\- (i>^)a sin {nt-\- c) = o. Hence (2) is a solution of (i) provided that 2 I 2 — « +0) =0, i.e. when n=+(a .... (3). Now reversing the sign of n is only equivalent to changing « into r— e. Thus we may write the solution of (i) as follows : — j'=a sin((o^4-£) (4), in which a and € are as yet undetermined, whereas w shows that the motion passes through its complete cycle of changes in the time 27r/(«. The meanings of the various symbols and the solution itself are illus- trated in Fig. 8. In this figure the angle XOH = e, the angle HOP= tat, and the auxiliary circle passing through H and P has centre O and radius a. PM is drawn at right angles to YOY', cutting off the dis- placement OM—y. Hence by construction OM corresponds to the value of y expressed by (4). It should be noted that by differentiation of (4) twice with respect to time we have y= — w°a]sin {iat-\-f). ART. 30] RECTILINEAR MOTIONS 37 that is, the acceleration is a sine function of the time. Hence if we had begun with that problem instead of acceleration directly propor- tional to displacement, the solution would have been the same. 30. Initial Conditions.— We have seen that of the three constants in the solution only one is determined by the given law of acceleration, the other two are dependent upon the initial values of displacement and speed. Let these be respectively y, and z)„. Then from (4) we have ^ .1'0 = «Sin€ . (r\ Also, by differentiating (4) with respect to time and then putting / = o and equating, we obtain for the speed at time t j = (oacos (w^+e) and Wi,/(i)=acos£ (6). Hence by squaring and adding (5) and (6) we find «'=^?+w?K. . . . . (7). Also, bydivisionof(5)by(6), we havetanc=:fc)j'„/w„ (8). Thus (7) and (8) when with (4) complete the solution of (i) for any specified initial conditions. Two cases of special importance and simplicity may be noticed. First. The initial displacement is zero, the initial speed being w„. Then, putting these values in (7) and (8), (4) becomes 1) r=— sinw/ (o). Second. The initial speed is zero, the initial displacement being j„. Then, from (7), (8), and (4), we find as the solution j'=j„ sin f 0)/+ — J = v„coso)/ (ro). If we illustrate the motion under consideration by a bullet suspended by a cocoon fibre about a metre long, then these two cases of initial conditions correspond (i) to striking the bullet a horizontal blow while it hangs at rest, and (ii) to pulling it aside and letting go while at rest in the displaced position. In the simple harmonic motions noticed the time of complete execution of the cycle once is called the period. Denoting it by t, we obviously have from (4), etc. T^lTrjm. ... . . ... (11). Examples — VI. 1. Define simple harmonic motion and obtain expressions for the velocity and acceleration of a point executing it. 2. Given that a particle P in a straight tube has acceleration directed to a fixed point O in the tube and of magnitude proportional to OP, deter- mine the motion in general terms. 3. If the acceleration of a point M in rectilinear motion is always - 16 x OM, where O is a fixed point, find the equation locating M, its initial displacement and velocity being respectively 5 cm. and 10 cm. per sec. Ans. OM = a sin (4/ + ^), where (z=5 \^2 cm. and tan e = 2. 38 ANALYTICAL MECHANICS [art. 31 4. A point executes a simple harmonic motion of amplitude 3 cm. in a period i,n seconds. Find {a) the maximum velocity, {b) the velocity at half-displacement, (c) the acceleration at full displacement, {d) the acceleration per cm. displacement. _ Ans. [a) ±3/2 cm. per sec. (6) 3 s^s/4 cm. per sec. (c) ± 3/4 cm./sec.2 (rf) ± 1/4 cm./sec.2 31. Composition of Collinear Simple Harmonic Motions. — A kinematical problem of considerable interest and importance is pre- sented in the composition of two or more simple vibrations. The case in which they are collinear naturally falls in the present chapter, and we shall first treat two vibrations only, their periods being equal. Thus, let the component vibrations have displacements n and v along the axis of j>', and be expressed by u^ asm {(at -\- a) . . .... (r) and v=bsm{ii>t+li) (2). Then their resultant, of displacement y, is given by or ^ = r sin (oiZ-f^) say/ " ' ^'" We are here assuming that the resultant vibration is of the same type as its components, an assumption which remains to be justified or condemned. To test the matter, expand the right sides of (i), (2), and (3) and equate. We thus find that the assumption leads to (a sin a -f (J sin (i) cos tai-{- {a cos a-\-l> cos /8) sin lat = /-sin ^cos w^-|-rcos 6sint and those of sin t. We thus obtain two equations, namely r&\nd = asvaa.-\-bsvi\li (5) and r cos 0=a COS a^b cos (i . (6). Whence, squaring and adding, r'' = a^-l-/,= + 2«^cos(a~/3) . ... (7). Also, dividing (5) by (6) ^^^^^asina+/^sin^ a cos a -1-/5 cos p ^ ' Equations (7) and (8) show that for any real values of a, b, a and fi, corresponding real values are possible for ;- and 6. Accordingly the assumption in the lower line of equation (3) is justified and, together with (7) and (8), affords the solution sought. It is seen from (7) that the resultant amplitude r usually lies be- tween the limits a+b, reaching them for a~£i=o and jr respectively. Further, it may be seen from (8) that 6=(a+fi)J2 when a=b. For more than two collinear vibrations of same period we see from (5) and (6) that it is easy to generalise and write rsin = 2asina (g) and rcos^=2acosa (10^. the component amplitudes and phases being denoted respectively by «i) <^s! (r» • ■ ■ and a„ Oj, a, . . . ARTS. 32-33] RECTILINEAR MOTIONS 39 Tlios, squaring and adding, we have Again, by division, we have tan e= (Ea sin a)/(Sr cos a) . (II). (12). Fig. 9. Graphical Composi- tion OF 'iWO COLLINEAR A'lBRATIONS. 32, Graphical Composition.— The composition of collinear vibra- tions may be illustrated or performed graphically, and this view of the matter well deserves notice. Fig. 9 illustrates the composition of two simple harmonic motions supposed to occur along YOY', their com- ponents being as already specified in equations (i) and (2) of article 31. The ordinate OF represents the displacement u at the instant /=o due to one vibra- tion, and is the projection of OP of length a and inclination a to OX. Similarly, OG gives the value v of the other dis- placement at the same instant, being the projection of OQ of length i and inclina- tion p. The ordinate OH of length y is the sum of OF and OG, and is also the projection upon YOY' of OR, the dia- gonal of the parallelogram upon OP and OQ. ■ Hence OH represents at ^=0 the sum of the component displacements. But since the periods of the two com- ponent vibrations are equal, the radii OP and OQ must move with equal angular velocities in describing the auxiliary circles through P and Q corresponding to the two vibrations in question. Hence the angle POQ=a~^ must remain constant as well as the lengths OP and OQ themselves. Thus the parallelogram OPQR remains of fixed size and shape. Therefore H, the projection of R, executes simple harmonic motion upon YOY' of amplitude OR = r say, the phase angle at /=o being XOR = ^ say. Further, it is easy from the figure to confirm or obtain the relations analytically deduced in article 31 and expressed in equations (7) and (8). 33. For the construction of the above figure it is evident that OR could have been obtained with fewer lines by putting PR of length l> and inclination /8 with OX instead of first drawing OQ and then completing the parallelogram. We should in that case draw one half only of the parallelogram to obtain R instead of both halves. This, which is a small matter when only two components are con- cerned, is a distinct advantage when three or more component vibra- tions are to .be dealt with. This me'thod is illustrated for four collinear vibrations in Fig. 10. We suppose the vibrations to occur along YOY', the component dis- placements being represented by the projections upon that line of OP„ P1P2, P2P3, and PjP,. The resultant vibration is accordingly represented by the projection upon YOY' of OP4, the line which closes 4° ANALYTICAL MECHANICS [ART. 34 the polygon. If the components are as specified by the right side of equations (9) and (10) in article 31, it is seen that (11) and (12) give the amplitude and angle for the resultant. Thus the resultant of col- linear simple harmonic motions of equal periods may be ob- tained by the addition of vectors according to the parallelogram or polygon method, the vectors for components and resultant being in each case the radii of the corresponding auxiliary circles at some one instant, say for t=o. In other words, if the am- plitudes and phase angles of component collinear simple har- monic motions be represented respectively by the lengths and inclinations of the sides of a polygon, then shall the closing line of the polygon represent by its length and inclination the amplitude and phase angle of the resultant simple harmonic motion, which is of the same period as its components. For the composition of rectangular vibrations see Chapter v. Fig. 10. Graphical Composition of FOUR Collinear Vibrations. Examples — VII. 1. Establish the general expression for the resultant of two collinear simple harmonic motions of the same period. 2. Compound two collinear simple harmonic motions of equal periods, their amplitudes being 2 and 3 cm. and their phase angles 7r/4 and 7r/3 respectively. Ans. y = 4'g6 sin {a>f+ 54°). 3. Confirm graphically the results obtained for the preceding example. 4. Compound analytically or graphically collinear vibrations of equal periods whose amplitudes are 8, 6, 4, and 2, their phase angles being 0°, 30°, 45°, and 60° respectively., Ans. Resultant amplitude 187 and phase angle 23° 51' nearly. 34. Acceleration inversely as Distance Squared. — We now consider a second example of acceleration varying with position, but this time it is inversely as the second power instead of directly as the first. As this case is of great importance and occurs early in the course we shall treat it first by elementary methods. Thus, students only just starting the calculus may defer the analytical method till a second reading. Let the acceleration be towards a fixed point O in the line along ART. 34] RECTILINEAR MOTIONS 41 which the point P under notice is to move. And let P and P', distant J and / from O, be two very near positions of the point at times t and /', the corresponding velocities being v and v'. Then, by definition of velocity, we have '-^-{t'-t) = s'-, (l). Now since the acceleration is towards O, and we are taking distances, velocities, and accelerations positive when from it, we may write — /^A" for the acceleration at P, where /* is some constant. Similarly the acceleration at P' is —it-js'^. Thus, for the mean acceleration while PP' is described, we may write and equate the two expressions ^^=-A (3). t-t ss' ^ ' Hence, multiplying these two equations, we obtain z;"-z;' = 2/x^^-y) (3). This is the general expression for the very small step PP'. Let us now take a finite step PQ, where 0Q=.5 and the velocity at Q is V. Divide this step PQ into a large number of very small ones, the inter- mediate distances and velocities being s^, s^, s,, . . . Sn and Vi, v„ . . . v„. Then from (3) we may write I'i — V =2/1. Vl—V\=2fi. (f.-7.) ■■-"•■-""'(s-i;) r-rt=,,(i-i;). And, on adding these, we see that on each side terms cancel out diagonally, giving F»-.-2/.(i-j) (4). On comparing with (3) we see that the relation for a small step is valid for a finite one also. Suppose now that V^o for S=r, then (4) gives the general formula for a fall from rest — »-=K;-J) <* We may now obtain the same result by the calculus. Thus the equation of motion is fx. _dv _ds ^ dv_^dv ~7~di~Jt' ds~ ds' 42 ANAL YTICAL MECHANICS [arts. 35-36 Separating the variables and integrating between the appropriate limits we have .whence Jo Jr ^ z;' = 2/i(---) . • • (6), \5 r/ in agreement with (5). If r=oD, the corresponding w is + J21J./S. 35. High Falls. — We shall see later, in the section on Attractions (Chapter xvi.), that this law of acceleration is that which applies outside a spherical gravitating body of uniform density, or concentric shells, each of which is of uniform' density, its centre being the point O. We may accordingly apply it as an approximation to what would happen in the case of a body falling to the earth from a great or infinite distance, all resistances being supposed negligible. In this case it is convenient to express the general constant fi in terms of the particular one g giving the acceleration on the earth's surface, and J? the radius of the earth. The relation is evidently — yu,/.^^= —g or ii.=gR^. Hence, if Foo is the velocity acquired in an unresisted fall to the earth's surface from an infinite height, we have VSo=2gR (7). Suppose now the velocity V is acquired at the earth's surface by a fall from a height h above the surface. Then by (6) introducing g we find ^'=^^^i)<-jk-^ («)• Further, we may put this in the forms the latter expression being an approximation obtained by neglecting ^7-^" in comparison with unity. This accordingly applies where the height is not too great. Obviously, if hjR is negligible compared to unity, the square of the velocity reduces to the familiar 2gh as for uniform acceleration of magnitude g. 36. Time of Fall.— We have hitherto dealt with the relations be- tween velocity and distance. Let us now change to space and time. Thus, from equation (6) of article 34, remembering »=^.y/(//', taking the square root, rearranging and integrating, we have /■' ds ,— /■' /■' J7sds U.ls-^lr-'^H''' = lU7^s (^°)- To evaluate the right-hand integral, put s^rcos'e, then J^s= Jr sin 9, ds— — 2r cos 6 sin 6 dd, and the lower limit becomes zero in the new integral. Hence , -,-- ds ^'^'^- 37] RECTILTNEAR MOTIONS 43 /V2^=r"^jf (i+cos 2Q)de=r'i\re+r%xT,eco^ 6). giving the time of a fall from rest at distance r to the distance s under the acceleration —fi/s'. If we have r=co, all times of fall to any finite distance j- become inhnite also But we may obtain the times (/-/„) over the distance {s„-s) as follows :— Beginnmg with equation (6), and putting r=ao, we obtain Hence, separating the variables and integrating / ^•'Vi-=- ,/2/X di,OTi-f, = ^{s,"'--s"') . . (12). K Jto 3 J/j.^ ' ^ ' Examples— VIII. 1. Establish the general relation between velocity and space for a point moving along a straight line under acceleration inversely as the square of its distance from a point in that line. 2. From the result obtained for question i, pass to the relation between space and time. 3. Find the velocity acquired by a fall to the earth's surface from rest at a height of 400 miles, the earth's radius being reckoned 4000 miles, and g as 32-2 ft./sec.2 Ans. 2To6 miles per sec. (or 2"096 by the approx.). 4. Calculate the time for the fall of question 3. Ans. 5 min. 5"8 sec. • 5. Determine the vertical velocity which, in the absence of resistances, would suffice to carry a particle away from the earth. Ans. 6'98 miles per sec. 37. Acceleration diminished proportionally to Speed : Mist. — We now treat cases in which the acceleration over the given region is uniform except as it is changed by the speed of the moving point or body. And in the first place this change of acceleration shall be a diminution proportional to the speed. This applies to very small bodies and to very slow-moving bodies falhng through the air. In these cases the uniform acceleration in the region in question is due to gravity and the diminution of the acceleration to the resistance of the air. Thus tiny spherules of water, as in the case of mist or very fine rain, are always falling, with respect to the air surrounding them, but are also resisted so that their speed relative to the air is never great. Speed and Time. — Let the space co-ordinate s, the speed ?', and the acceleration a be all reckoned positively in the same direction. Also let the diminution of acceleration be such that at speed k it equals a. ■"•"^■(l^) <->• 44 ANALYTICAL MECHANICS [art. 38 and therefore suffices to annul the acceleration which affects bodies at rest. Then at speed v the diminution of the acceleration will be avlk. Thus we may write the equation of motion in the form f=l<*-' «■ Let the particle start from rest at the origin of co-ordinates when /=o. Then, separating the variables in (i) and integrating, we have and, on evaluating, at V which gives the time t in terms of the speed v. If we wish to have v given explicitly in terms of t, we may raise e to the powers given by each side of equation (2). Thus k — V Whence z;=/J(i -«-«'*) (3). We see from either (2) or (3) that the speed v only rigorously reaches its limiting value k in an infinite time. But for 0=^=981 cm./sec.'' and ,5= 0-981 cm./sec, we have «-««/*=e-'°''°'. Hence when ^=one hundredth of a second v will differ from k by only «"'° out of i, i.e. by less than one part in twenty thousand. 38. Speed and Space. — We have obtained relations between v and t, and now pass to obtain them between v and .f, the speed and space passed over. Thus, referring to equation (i), we see that the first term may be transformed as follows : — , dv ds dv dv dt~ dt ds ds The whole equation may accordingly be rewritten ^Is-%'~^) (4)- Thus, separating the variables and integrating, we have \\'ds^{fL^-[''-=^dv. kk k k — V Jo V — k Hence, on evaluating, we find |.= -.+^log.^^ ... . (5), which gives the space .f in terms of the speed v acquired in it, the start being from rest. Here again it is seen that the rigorous limiting value of the speed is only attained after an infinite space is passed over. But with the ARTS. 39-40] RECTILINEAR MOTIONS 45 numerical values ^='981 and 0=981, as previously used, a very small distance suflfices for an approach to the limiting speed. Thus, put- ting the second term on the right side of (5) equal to k, i.e. making v=k(\ — j, which is distinctly more than half its limiting value, we have s=k^lae=-c)62/g8i X 2-718 . . . =1/2,772 of a cm. Thus less than four ten-thousandths of a centimetre are passed over, while the speed attains nearly two-thirds of its limiting value. 39. Diminution proportional to Square of Speed : Hailstone. — In the case of quicker-moving bodies, such as large raindrops, hail- stones, or shot, the diminution of the acceleration is approximately as the square of the speed. It thus furnishes us with another slightly different problem for attack. Taking as before the space j- and speed V positive when reckoned in the same direction as the acceleration a, and again indicating by k the limiting value of the speed, we see that the acceleration is diminished by the amount av'jk' when the speed is V. Speed and Time. — The equation of motion may accordingly be written S-J**'-') <■)■ Separating the variables and integrating we have a\'dt=v[J^. = ^{^^^\dv, k k k'-V 2X V-v k+v/ which, on evaluation, yields . k . k-\-v / V t— — log^T — v2;. thus giving the time t in terms of the speed v. If the speed is required explicitly in terms of the time we can transform the equation exponentially as before (see equation (2) of article 37). We thus obtain pdHk g — at'k ''+e- as the relation required. ^=>^^SX^-^tanh(aV^) (3) 40. S^eed and Space.— U, however, the speed is required as a function of the distance, or vice versa, go back to equation of article 39, and note, as before, that dv/dt=vdv/ds. Hence (i) may be transformed into /J!. = l^(k'-v') (4). ds K 46 ANALYTICAL MECHANICS [art. 41 Whence, on separating the variables and integrating, we find a ('_ r vdv _ . rd {k''—v^) And therefore on evaluating 5 = — l0g.rs Tfl (S). which gives the distance s passed over in acquiring from rest the speed V. If now we wish to express the speed explicitly in terms of the dis- tance, take exponential values of each side as before. We thus have v^=k\Y-e-'^'^>''^') (6). 41. Bise and Fall of Shot. — Let us now consider the rise and fall of a body in a region where the uniform acceleration is changed by an amount proportional to the square of the speed, that change being always opposed to the speed so as to diminish its numerical value. Rise. — Take the origin of co-ordinates where the particle starts upwards at ^=0 with a speed U, and let it reach a height s and have speed u at time /, its utmost height being S with zero speed at time T. Thus though the space and speed are reckoned positively upwards, the acceleration g due to gravity, and the diminution of the speed gu'fk"^ due to air resistance, are both downwards. Speed and Time. — The equation of motion for the ascent may accordingly be written l=-|(*-+"') <■'• Thus, separating the variables and integrating, we have Whence 15=— i tan^'y , .=|{tan-f-tan-|}. . . . (.). Thus, at the summit of the motion, we have ^=|tan-'|^ (3). This ends the ascent in terms of speed and time to which (i) applies. But before taking the descent we may with advantage take the ascent again in terms of Speed and Space. — Thus, transforming the first term of (i), we may write du •S=-|.(*-+--) (4). ART. 42] RECTILINEAR MOTIONS Separating the variables and integrating gives g [\ _ 1 f 2udu Wo ^L^^+^''■^ Whence ^^^-^^\og(^/i'+u') 47 ^-^l°g-^'=+Tr (5)- Thus, at the summit of the motion, we have ■^=-108. ->- (6), which completes the consideration of the ascent. 42, J^a//. — Leaving the zero and co-ordinates of space and time as before, we will now write v for the numerica/ vahie of the speed in the descent, so that v is positive throughout the fall, just as u was in the rise. Thus the falling speed v is increased by gravity g and dimin- ished by the air resistance gv'ji'. Speed and Time. — Thus, we may write as the equation of motion for the descent S-f(^^-^^) (^)- On separating the variables as before and integrating we find |(/-r;=^iogf±^. Thus /=r+- loge -:!^, ^g ^-^ I (8). or by use of (3) ^-_^tan-'-^+ Alog.^ I Speed and Space.— 'Lei us now treat the fall in terms of speed and space passed over. Then on simply reversing the sign of ds m equa- tion (4) of article 40, or deducing it from (7) of the present article, with similar regard to the fact that v^-dsjdt, we have as the equation of motion .^^^=Uk^-v-) (9). as M Thus, by the usual steps, we have in succession p- /•« /•" vdv and -^-|l°S^I^ • • • • ^^°)- 48 ANALYTICAL MECHANICS [art. 43 When the shot again reaches the origin, i.e. for j=o, let the speed be V ; then we have Equations (6) and (11) enable us to find a relation between U, the speed of ascent through a given point, and V, the speed of repassage downwards through the same point. For obviously by the two equa- tions for 5 we have Whence ^=^j+^ (12), ""-^ jtAw • ■ ■ : ■^''^- These equations show that V cannot exceed k, and will only reach it, for U==ulk, we have gtlk=Q.-u>, tan (0 = tan (J2 —gtjk), and -- ^M-tan^//^ , , k i-\-{Ulk)tSiX^gtlk *' 5;. giving the ascending speed in terms of the time. Again, from (5) we have by the exponential transformation u'+k'={k'-\-uy-^^i''' (16), giving the ascending speed in terms of the space risen. For the descent we have from (8) of article 42 by the exponential transformation Whence _=____= tanh M/-r)/>J} . . . (17), giving the descending speed in terms of the time. ART. 44] RECTILINEAR MOTIONS or 49 Again, from equation (lo) of article 42, by the same method, we find z;^=/J^(i_e-Ws-s;*2)| .... (18), giving the descending speed in terms of the space {S-s) fallen throueh cannot fnthI^;™^'•"' '''' V' reckoned posilively if upwards and cannot m that direction exceed ^. Hence when the origin is passed through agam m the descent, . changes to a negative value Thus!-. lecToTLH^^h '""V '^V'"' °^ '^" "^^^^^"^ '° ^ ^' '^' Voim of pro- throu h '^^""fo-^^ard increases towards infinity, being positive all 44. Graphical Treatment.— We have thus expressed speeds of \ •ff \ \ L \ ■s 3 •2 ■/ \ \ \ \ \ \ T r M £ / 2 \ 3 * S 6 7 Sec 'onds \ ■3 \ \ s \ ^ \ ■« \ \ V t ■~~- ^ Fig. II. Speed-Time Graph FOR Shot. rise and fall in terms of the time occupied and distance traversed and vice versa. But no general direct relation between space and D 5° ANALYTICAL MECHANICS [art. 44 time has been given. Neither is it easy to obtain such relations analytically. But since time and space have been given each in terms of the speed, we can assign to the speed a number of possible values, place them in a column, and then in neighbouring columns insert the corre- sponding values of time and space calculated from the formulae already developed. We should thus derive a number of corresponding values / ■9 •s ■7 ■6 •5 ■■* 1^ N s S N N N s. ^ S \ . « - \ \ ■2 •/ ■1 ■2 \ \ feet 5 o / 2 3 4 s 6 u 7 P a a O 10 1 1 / / y / ••i ^ y ■ft ^ ^ ■7 _ ^ ■fi •9 1 k Fig. 12. Speed-Space Graph for Shot. of space and time, and could then plot a space-time graph. Or, we could begm by plotting a speed-time graph and a speed-space graph from the formulae. Then, selecting any one value of the speed on the ordmates of each, their abscissae would give an ordinate and an abcissa for a third curve forming a space-time graph of the motion. This graphical method of expressing the meaning of the equations obtained, ART. 44] RECTILINEAR MOTIONS 51 and deriving a new relation from them, is exhibited in Figs 11 12 and 13, which will repay careful examination.! & ■ > . 1 too to / N \ fto / \ 70 ~tt — / \ dp 5- / \ SP / \ 40 / \ 1|7 / \ \ ?(< / \ w / \ 10 SO 30 <0 SO 60 70 / Time / 2 3 'i- s \ 6 Seco ndsl \ ■? \ \ ?i \ flO Fig. 13. Space-Time Graph for Shot derived from the former two. Examples — IX. 1. ' Investigate the motion of a heavy particle which falls vertically from rest in a medium whose resistance is proportional to the velocity.' (LoND. B.Sc, Pass, Applied Math., 1909, 11. 5.) 2. ' The position of a point which describes a straight line is defined by its distance .*• from a fixed point of the line. Show that its acceleration is d^xjdt^. The motion of a particle projected with velocity Kis retarded at a rate which varies as the velocity. Find the time which elapses before its velocity is halved, given that the retardation is X, when the velocity is V.' (Lond. B.Sc, Pass, Applied Math., 1906, 11. i.) 3. For a point with initial velocity U vertically upwards under constant ' In these the limiting speed k is 96-6 ft./sec., which is about the value for a golf ball (Tait). The speed of projection is also equal to k. 52 ANALYTICAL MECHANICS [arts. 45-46 acceleration downwards and retardation proportional to velocity squared, find the relations between speed and time and speed and space. 4. For the point of the previous question trace its motion after the summit of its path is reached, and find the speed Fwhen at its original level again, 45. Acceleration varying witli Displacement and Speed : Damped Vibration. — Let us now consider the motion of a point whose accelera- tion is the sum of two parts which are directly proportional to its dis- placement and its speed respectively, but each oppositely directed. Then that part which is opposite and proportional to the displacement would, if operating alone, yield a simple harmonic motion. But this motion will evidently be diminished by the continuous operation of the other part, which is opposed to the speed and proportional to it. These reflections give us a clue to the motion, which we shall presently see is that diminishing vibration called a damped %\K&^\e. harmonic motion. Let the displacement of the point be called y, and let its accelera- tion he — {p''y+2Kdvldf). We can then write as the equation of motion ^+"l+^--" (■)•; The simplest way of representing analytically a damped vibration is to insert in its coefficient a factor of the form e'™*. We accordingly try as a solution .y = ae'™* sin (^f+e) . . . ... (2), in which the unknown ^ is purposely written instead of/ to provide for a possible difference of period between the ensuing motion and that which would occur if k were absent ; a and e are also inserted to make the expression as general as possible. By substituting (2) in (i) and differentiating we find it is satisfied, provided that ;» = k and ?'=r-'<' (3)> We may accordingly write as the solution sought y—ae-"' sin{^i+e) . . ... (4), in which g is defined by (3), a and e being dependent only on the initial conditions. These equations indicate that if « is small the factor £'"<■ may be appreciable, while ^ is practically/. It can be shown that the motion analytically expressed by (4) may also be regarded as the projection upon the axis of jc of a point F which describes with angular velocity q the logarithmic spiral r=ab-» (S), in which \og,b=K\q and the angle Q of the polar co-ordinates is reckoned from the inclination € to the axis ol x. \ie^=b'', A is some- times called the logarithmic decrement; in other cases the phrase is applied to /x where ei^ = bi''. Thus the term denotes the logarithm to the base e of the ratio of one amplitude to the next on the other side or the same side respectively. For the further development of this aspect of the subject of damped vibrations the reader is referred to the writer's Text-Book on Sound, Arts. 56-58. 46. Accelsratiou varying with Time, Place, and Speed : Forced ART. 47] RECTILINEAR MOTIONS 53 Vibration. — Having considered cases where the acceleration depends upon position only, upon speed only, and upon both, we now finally treat a case where it depends upon time also in addition to the other two variables. We shall thus suppose the acceleration to be made up of three terms, two of them being as in the previous article, the new or third term being a sine function of the time. We may accordingly write for the equation of motion d^v dv ^ + 2K^+/_y=/sin;2^ (0- Or in words, the acceleration is made up of three parts, of which one is a sine function of the time of amplitude /and period 27r/«, another is —p"' times the displacement, and the other is — 2k times the speed of the moving point or particle in qaestion. Now the remark at the end of article 29 shows that the part of the acceleration which is a sine function of the time would of itself produce a motion in which the dis- placement is a sine function of the time of the same period, opposite phase, and generally of different amplitude. We may well doubt if like results will follow now where the acceleration has in addition two other terms, but it is clearly easy to test the matter by trying as a solution a sine function of the period of the fluctuating part of the acceleration and of undetermined amplitude and phase. Thus let us try as a solution of (i) the expression 7i=(Z sin («^— 8) .... . . (2). Then, inserting it in the left side of (i), performing the differentiations, regarding «/ on the right side of (i) as («/-8) + S, and expanding, we obtain {p°—ti')a sin {nt—S) + 2Kna cos («/— 8) =/cos S sin («^— 8)+/sin S cos (nt— S). But since for a valid solution this equation must hold for every in- stant of time, it breaks up into two on equating the coefficients of sin (ni—S) and cos (nt-S). We accordingly derive from it 2Kna=/s'ni 8 (3) and {p''-n'')a=f cos 8 (4)- , s /sin 8 /.\ Thus by (3) "--J^ .... • (5). and by taking the quotient of (3) and (4) we see that .. S 2K« . (6). tan6 = --5 -J- y"/- p —n Hence using (5), (6), and (2) we see that /sin 8 . , . ,. /sin(^;'-8) , . V.—- sm(«/— 0)= — ■„ ^^ (,7; is a solution of (i) when 8 has the value given in (6). 47. Complete Solution.— But this, though a solution, is not neces- sarily the complete solution. Thus, if we write _y2=«e-'='sin(y/-fe) (°) 54 ANAL YTICAL MECHANICS [art. 48 where g'^p^ — i^^ we see, from equations (i), (3), and (4) of article 45, thatjVa put in the left side of equation (i) of the present article would reduce to zero. Thus if jj were added to^i it would not disturb the solution, for jCi in the left side equals the fluctuating term on the right side of the equation, while ^2 on the left gives zero. Indeed, while not disturbing the solution, the presence of y^ serves to complete it, for it contains two arbitrary constants a and «, and the theory of differential equations shows that this is the number required in the »?»//«/« solution of equation (i). A little reflection will show that these two constants are the unknowns to be determined by the displacement and speed respectively of the vibrating point at the instant /=o. We may accordingly write as our complete general solution of (i) the expression which is the sum ofji andjCa from (7) and (8), viz. r^- sin(«/— 8) +««-'<' sin (^/+e) .... (9), ;' in which tan 8= --. ~^, q^=p'' — K^, a and e are arbitrary. 48, Discussion of Solution. — Of these two parts of the motion shown on the right side of (9),ji'i ax\dy^ respectively, the first, ji, depends, as we have seen, solely on the fluctuating or periodic part of the accelera- tion, it is the response of the moving point to that acceleration, and is maintained by it of the same period and ceases if it is withdrawn. It is called 9. forced vibration, being of the period of this imposed accelera- tion, and not that due to the accelerations dependent on displacement and speed. The second term, or y^, is the vibration of damped harmonic type studied in article 45, and is the free^ or natural vibration peculiar to the (conditions which impose the accelerations given as dependent on displacement and speed. These natural vibrations might be present prior to the application of the periodic part of the acceleration, and the resultant of the forced and natural vibrations is competent to represent any initial conditions since the amplitude a and phase € are arbitrary. It should be noticed, however, that if the periodic part of the accelera- tion lasts long enough the natural vihiaXion represented hy y^, being of the damped or diminishing type, will practically disappear, leaving the forced vibration alone in the field. If then, after a time, the periodic acceleration were withdrawn, the displacement and speed obtaining at that instant would have to be regarded as a new initial state giving corresponding values for amplitude and phase of the natural vibrations, which would thenceforward ensue and continue till they died away by the effect of the damping factor «"«'. It is seen by equations (6) and (7) that the nearer p and n are to each other the greater is the amplitude of the forced vibration. Indeed, but for the presence of «, the amplitude would become infinite for n=p. In acoustics, wireless telegraphy, and wireless telephony this is a fact of far-reaching importance, referred to under the terms resonance, tuning, etc. For a fuller treatment of the subject of forced vibrations, sym- A RT. 48] REC TILINEA R MO TIONS 5 5 metrical and asymmetrical, and with single and double forcing, the interested reader may consult articles 91-116 of the author's work on Sound. It may be noted, in conclusion, that all the examples dealt with in this chapter are special cases of the one general diffierential equation d^s/di^ = C+S+ V+ T, where s is the space co-ordinate, C is a constant, and S, V, and J" are functions of the space, velocity, and time respec- tively. Examples — X. 1. A point moves in a straight line with an acceleration whose components are respectively proportional and opposite to its displacement and speed. Write down the corresponding equation of motion and solve it. 2. If a point executing simple harmonic motion becomes subject to a further acceleration opposing its motion and directly proportional to its speed, what changes follow in the amplitude and the period ? Can one of these quantities suffer an appreciable change while the other is prac- tically unaffected ? 3. A rectilinear motion is executed under an acceleration whose components are numerically- 169 times the displacement and- 10 times the velocity respectively. Find the equation, period, and logarithmic decrement of the motion. Ans. y = ae~'''^sm{i2i + f). r = 77/6, X = Stt/i 2 or /i = 57r/6. 4. Plot a displacement - time graph of the motion of question 3, putting a= 10 and c = o. 5. Discuss the case of rectilinear motion in which the acceleration has com- ponents depending on the displacement, the speed, and the time, and show what will follow after the cessation of the third component of the acceleration. 6. If a point initially without displacement or speed becomes subject to acceleration represented by y+ i69j/ = 25 sin I2t, show that its motion may be thenceforward expressed by y = s\n I2t- \'l sin 13^. 7 Show that all the cases of motion dealt with in the present chapter may be represented by one differential equation, giving illustrative examples, and indicating the character of the solutions. S6 ANALYTICAL MECHANICS [art. 49 CHAPTER V PLANE MOTIONS OF A POINT 49. Uniform Acceleration : Projectile. — In dealing with the motions of a point in two dimensions we begin with the case in which the acceleration is uniform over the space under consideration. It thus applies approximately to the motion of an ideal projectile whose path is so small that no variation in gravity need be considered, the projec- tile itself being regarded as a mere point or particle devoid of rotation and suffering no resistance from the air. Directrix.. Fig. 14. Parabolic Trajectory. Referring to Fig. 14, let the origin of co-ordinates Obe the point of projection, the axes of x and j* being respectively horizontal and vertical. Then the general conditions of the problem are expressed by stating that the horizontal acceleration is zero and that the vertical accelera- tion is —g. Or, if we denote the co-ordinates of any point P in the trajectory by x and^, the component velocities and accelerations by x, y, and x, y, we have as the acceleration components x=oJ=—g (i). Thus, if the angle of projection is a and the initial speed u, the velocity components after time t are given by i;=?^cosaandj>=«sin a— ^^ . . .... (2). ART. 50J PLANE MOTIONS OF A POINT 57 »'=-*'+jP'andtan^=j>/x /-\ For the cartesian co-ordinates of P at time t we have from {2) a:=«/cosaandj)/ = ?,:^sina — J^i'= /.\ The corresponding polar co-ordinates are obviously given by ^==a:'+y andtane=j/* _ i^\_ Summit.— To find the co-ordinates of the summit of the trajectory we must write j>=o in (2) and obtain from (4) the corresponding values of X and y. r o Thus we find from (2) that for the summit t=(u sin a)/?-, which put in (4) gives " ^ z ^' sin 2a tt^sWa ^, ^ Jiange and Time of Flight.— lo obtain the range OH on the horizontal plane we write .y^o in (4) and obtain the corresponding value of X. We see that the time of flight is _ 2« sin a ^^-^- (7)' and that the range is given by X=^^'^^0^ g ' Thus, for a given value of u, the maximum range is obviously obtained for a—trl^, under the ideal conditions supposed to exist. 50. Equation of Trajectory.— From equations (4), eliminating i, we have >'=.«tana .^5 — 2U' cos a or / 2<^2 sin acosaV_ /«"cos^a\/ z^" sin"a\ , > V"^ 2g ) ~ ^\ 2g JV 2g J ' ^ '' equations which show the path of the particle to be a parabola of latus rectum (27/° cos"'a)/g, and with directrix and focus respectively above and below the summit A by the distance («" cos"a)/2^. The equation to the directrix may accordingly be written y'=ir/2g. . . . . (10), the co-ordinates of the focus being ?^^ sin 2a J a" / • 2 - \ 7^^( — C0S2a) / > and — (sin a— COS a) = -'^ ' . . (11), 2g 2g' 2g and those of the vertex as already shown in equations (6). Velocity due to Fall from Directrix. — Referring now to equations (2), (3), and (4), we see that v°' = u' — 2gy . ■ (12). But by (10) this could be put in the form v- — 2g{y'—y) (13). 58 ANALYTICAL MECHANICS [ART. 51 i.e. the velocity at any point P in the trajectory is that which would be acquired in a free fall from rest in the directrix to the level of the point in question, for y'-y is obviously the length of MP on Fig. 14. 51. Range on an Incline.— The range J? of a projectile on an incline Q may be found from the polar co-ordinates given by equa- tions (4) and (5) by putting the given incline in the expression for tan Q, and thence deducing R and the time of flight T. Thus R= s= a- (^4)- cos d cos V Also -= ^^- = tan^, X U cos a ^^^«sin(a-^ (IS). g cos (t Whence by (15) in (14) we obtain i?=^^{sin(2a-0)-sine} (16). ^cos^ Thus for a given speed u of projection the range on the incline 6 is a maximum for 2a — 6=w/2, that is, for an angle of elevation which bisects the angle between the incline and the vertical. An alternative method is to take new axes of co-ordinates parallel and perpendicular to the incHne. The accelerations are then x= — ^ sin ^ and J'=— ^ cos^ (17). Still retaining a as the angle with the horizontal made by the direction of projection we have for the co-ordinates at time / x' = uicos{a — 9) — ^gfsin9\ , o\ and y = «/sin(a-6l)-|^^''cos6l/ ''"*^- Thus, writing y=o, we obtain again J" as in equation (15), and this substituted in the expression for x' gives for the range sought x=-R as in equation (16). Examples — XI. 1. Obtain, both in cartesian and in polar co-ordinates, general expressions for the velocity and position of a point moving in a vertical plane under uniform vertical acceleration. 2. Show that the trajectory of an unresisted shot is a parabola, find its equation, and write down its focus, directrix, and latus rectum. 3. With a given initial velocity of projection determine the angle of elevation for maximum range on an incline. 4. ' A gun is firing from the sea-level out to sea. It is then mounted in a battery h feet higher up and fired at the same elevation a. ' Show that its range is increased by the fraction H(-K&y-} of itself, K being the velocity of projection.' (LoND. B.Sc, Pass, Applied Math., 1908, 11. 2.) ARTS. 52-53] PLANE MOTIONS OF A POINT 59 5. ' Prove that the least energy of projection of a particle, in order that it may have a given horizontal range, is such as would carry it vertically to a height equal to half the range.' (LoND. B.Sc, Pass, Mixed Math., 1904, i. 4.) 6. A bullet describing a nearly horizontal path with a constant retardation moves over a space of 500 feet whilst its velocity falls from 1200 fs. to 1000 f.s. ; where was it when its velocity was 1100 fs. ?' (LoND. B.A., PA.SS, Mixed I\Iath., 1906, i. 5.) 7. ' Give a simple geometrical construction for the velocity at any time of a point whose acceleration is constant in magnitude and direction. ' Find by kinematical principles an expression for the radius of cur- vature at any point of a parabola.' (LoND. B.Sc, Pass, Applied Math., 1906, 11. 3.) 8. ' A projectile has a horizontal range of 1 50 yards, and the time of flight is 5 seconds ; find the velocity of projection, assuming that the resistance of the air may be neglected.' (LoND. B.A., Pass, Applied Math., 1906, i. 7.) 52. Constrained Motion in a Region of Uniform Acceleration. — We now consider the plane motion of a point or particle in a region of uniform acceleration, but with conditions imposed which constrain it to a specified path along which we have accordingly only a component of the acceleration which obtains in free space. Motion down Incline. — Thus, if a particle be constrained to motions at an angle 6 with the direction of the acceleration a which obtains in the region, we have along the direction of motion (by the resolution of vectors) an acceleration of value a cos 0. There is accordingly a motion with uniform acceleration of this amount, and the case falls under the methods of articles 27 and 28. In particular, if the motion in question is down a slope inclined at a with the horizontal, and the acceleration is that due to gravity, then the acceleration down the slope is obviously g sin a. It is, of course, supposed that the constraint in question imposes no check in any way upon the motion except that of keeping it confined to a given path. Suppose a particle to slide a distance s from rest down an iricline of angle a with the horizontal, the vertical height descended being k and the speed acquired v. Then we have v' = 2as . (i) = 2^ sin 0..S = 2g{/l/s)s or v'^2gh . • ■ ■ • (2) independent entirely of the inclination. 53. Simple Pendulum in Small Arcs.— Take now the case of a particle confined to motions near the lowest point of a vertical circle under the uniform acceleration of gravity. This is most easily realised by attaching a shot or bob by a fine thread to a fixed point, the arrange- ment being known as the simple pendulum. The hmitation to a vertical circle instead of to a spherical surface is then obtained by the method of starting the pendulum. _ • r> ^u u„k Referring to Fig. 15, let S denote the point of suspension, P the bob. 6o ANALYTICAL MECHANICS [art. 54 and let the thread SP have length /. Consider it when SP makes as shown an angle with the vertical SO. Then the component of gravity which is effective, being along the arc at P, is — ^sin d. But since we are limiting the motion to very small arcs, we may write this —gO nearly. Let the displacement OP measured along the arc be s, then 6=s//. We accordingly have as our approximate equation of motion »+f=° <■)■ But this is equivalent to equation (i) of article 29, hence the solution may be written s—SoSin(JgJ/i+i) . . (2). This shows that along these small arcs the motion is simple harmonic of period r^ 2Tr J Ijg (3). Thus the period varies as the square root of the length, and is independent of the amplitude to this approximation. In equation (2) J„ and «, the amplitude and phase, are arbitrary constants to be determined by the initial conditions as in article 30. 54. Simple Pendulum in Finite Arcs. — Let us now consider the case in which the arcs or angular displacements of the pendulum are not so small as to admit of writing sin 6= 6. Then, again referring to Fig. 15 and the notation of article 53, we have as the equation of motion Fig. 15. Simple Pendulum. iP0 (!)■ And let the initial conditions be for i= o. -^=oand6'=a (2), that is, the pendulum is let go from rest with an angular displacement a radians. It is required to find the period t corresponding to this amplitude a. Multiply (i) by 2M and integrate; then we obtain. or •2^ cos d+C=o (3). Applying (2) to (3) we find that the integration constant is C= 2g cos a. Equation (3) may therefore be written \^) ~2^(<=os^— cosa) = o. ARTS. 55-56] PLANE MOTIONS OF A POINT 61 = ->/y(cOS 6^-008 a) ... . (4), or dt in which the negative sign of the square root is chosen because the angle d decreases as time increases. Separating the variables and integrating over a quarter of the period t, starting from the equiUbrium position, we obtain h ' ^h Vcosf— cosa ^^' The algebraic sign before the square root is now changed to positive because the limits of integration have been written o to a instead of a to o as they were supposed to actually occur. Using the identity i —cos 6= 2 sin^^/2, and the same for a, we may rewrite (5) in the form a/ sin' — sin"- V 2 2 (6). 55. Transformation of the Integral.— To deal with the integral on the right of (6) it is desirable to introduce a new variable ^, defined by sin-=:sin -sin (^ (7). 2 2 This relation is obviously legitimate, since t> is never greater than a. The transformation shown by (7) obviously affects the function to be integrated, the differential, and the limits, ^^'e accordingly note that y sin - — 2 sin — = 2 = sin "cos 2 2 sin - cos 2 4>d4> y. . ..a — sin"- 2 sm°4> (8). The limits o and .* for 6 become o and 7r/2 for ^.i Hence, substituting from (8) in (6), we transform the integral to ^liere ^ = sin 0/2' • .... (10). Obviously, for k=o, (9) becomes T=z7r J/jg, as found before for infinitely small arcs. 56 Evaluation of the Integral.— To evaluate (9) note first that by the binomial theorem we have the expansion Jo 62 ANALYTICAL MECHANICS [art. 57 Also, by the integration of even powers of sines, we have I sin*<^^<^-^^g _ ^^, •- • • ■ • ^12;- Hence (9) becomes giving the period in terms of the ampUtude. It may be noted that, if h is the height fallen through by the pendulum bob from its highest to its lowest point, we have ,, . ,a I — cos a h /t.\ /4' = sm'-= = — , (14)- 2 2 2/ Hence, by help of (14), (13) gives t in terms of / and h. In many cases, though the oscillations occurring are not infinitely small, they are fairly so, and it is then usual to stop at the term involv- ing k^ in (13). Thus to this approximation, and writing a/2 for sin a/2, we have T=2.y^(i+A)^2.y^(i+y nearly . (15). Various relations may be obtained geometrically for a pendulum swinging in finite arcs. Some examples are given later to afford the student exercise in such problems. 57. Motion in a Vertical Circle. — Reverting now to equation (9) of article 55, we see that the integral may be immediately evaluated if k'' = i, i.e. if a=n-. This corresponds with a motion of a particle from the highest point of a vertical circle under gravity. It is convenient, however, to go back to the form of equation (6) of article 54 and rewrite it without the limits of integration. We thus have ^\dt=JJlg\^-^ (16). This gives /=( V^) log/tan —W const. (17). It should be observed that, since log tan jr/2 = Qrj, it would take an infinite time for the particle to just reach, or to fall from rest at, the highest point. If, after starting from rest at the top, a fall is noted from ^i to d^ then, by (17), the corresponding time ^,2 is given by ''a.= V^(log=tan^'-logetan'^±^^) . .(,8). The speed at any point, whether the fall started at the top or else- where, may be found by equation (4) of article 54. Examples — XII. I. Show that a particle descending a smooth curve in a vertical plane under gravity will experience the same change of velocity as in a vertical descent between the same levels. ART. 57] PLANE MOTIONS OF A POINT 63 2. Find the approximate period of a simple pendulum from the differential equation of its small motions. 3. Fit the equation which expresses the displace .nent of a simple pendulum (a) to the case where it is pulled aside a and let go, and (b) to the case where the bob receives a velocity v at the central position. Ans. s = al cos '^gllt. .= ^^^sinViy7/. 4. Discuss the equation of motion of a simple pendulum when executing finite arcs and find the period for an amplitude of 5°. 5. ' Prove that the period of a complete .oscillation of a simple pendulum of length / is 27r^/( //§■). If the bob of a pendulum 100 feet long be drawn aside from the position of rest through a space of 3 feet and then re- leased, find the velocity with which it passes through the equilibrium position.' (LOND. B.Sc, Pass, Mixed Math., 1904, i. 6.) 6. ' Prove that the length of the pendulum to beat seconds in London is 39'i4 inches. 'To gain or lose one second in one hour, or 24 hours in a clock, the length must be altered o'02i75, or o'ooo9o6 inch.' (LoND. B.Sc, Pass, Mixed Math., 1903, i. 9.) 7. ' Prove that the time of a single swing of a plummet at the end of a thread / feet long is - seconds when the oscillations are small, and that for the plummet to beat seconds the length of the thread must be 39'i4 inches. ' Prove that as the plummet swings through the arc BAB of an angle 2a from B to B on the horizontal chord BDB' of the circle of which ADE\s the vertical diameter, the point Q on the circle on the diameter A D will follow P at the same level with velocity 'V; V' I "^^ AE and thence show that the time of a swing lies between tta/— and TT'W-sec^a.' (LoND. B.Sc, Pass, Mixed Math., 1901, 11. 6.) ' Prove that the time of swing of a pendulum / feet long is undistinguish- able from ■rrij{llg) seconds when the oscillation is small. ' If a light is placed at E, the upper end of the vertical diameter ^^ of the circle" on which the plummet oscillates, to throw the shadow T of the plummet on the floor, moving from Fio F', and if TR is the ordinate of the circle on the diameter FF, prove that the velocity of R varies as ET, and fluctuates beween iFF''\jj a.nd ^FF' cos -^ ^|j, la denoting the angle of oscillation, not restricted to be small ; and thence show that the period of oscillation lies between 27r a/- and 2jr sec -^ '\^ -J (U3ND. B.Sc, Pass, Mixed Math., 1902, 11. 8.) 64 ANALYTICAL MECHANICS [art. 58 9. ' Prove, by analogy with Harmonic Vibration, that the bob /" of a circular pendulum of length /, oscillating through a finite arc BAB', moves with velocity nsJ{BP.PB'),n'=gll. ' If AE is the vertical diameter of the circle on which P moves, and if EP drawn from the highest point E cuts the horizontal chord BB' in R, prove that the velocity of R is n^^^{BR.RB'\ and that the period of oscillation lies between the limits (■""■a)" VI' (LoND. B.Sc, Pass, Mixed Math., 1903, 11. 9.) 58. Motion in a Vertical Cycloid. — A cycloid is a curve described by a point in the circumference of a circle which rolls without sliding upon a fixed straight line. The point is called the tracing point, the circle the generating Y «V<:/if,andthestraight r line the base. Thus, referring to Fig. 16, P is the tracing point, DPEF is the generating circle with centre at C, and AEB is the base of the cycloid, part of whicli is shown by OPB. It is at once obvious that a cycloid con- sists of a number of precisely similar por- tions, and certain points called vertices are most remotefrom the base, O being a vertex in the figure ; while other points called cusps lie on the base, B being a cusp of the cycloid OPB in the figure. The lines through the vertices at right angles to the base are called axes, one of which, AO, is shown in the figure as an axis of OPB. Intrinsic Equation of the Cycloid.— T&Ymg the origin of co-ordinates at the vertex O in Fig. 16, and the axis of x parallel to the base as The Cycloid. ARTS. 59-60] PLANE MOTIONS OF A POINT 65 shown let the co-ordinates of the tracing point P be {x,y\ it being understood that P was at O when the opposite end of the diameter F was at A 1 hen we can readily obtain the intrinsic equation of the cycloid For taking angle DCP=^ and CP=a, the co-ordinates of V are obviously A=a(^-|-sin ^) and^ = a(i— cos6i) . . . (i). Hence, on differentiating, we have (2). dx^ = a{i-\-zo%e)de=i,azo%--£- 2 2 and dy^^a^wedd = ji,a%xvi-co%-/- 2 2. 2J Thus on division, and calling the inclination to the horizontal at P = <^, we have ta.r\=dy/dx=::.ta.nd/2 or y-, or ds—+ 4a cos 4>d . . . (4). Thus integrating, and remembering that the value of the arc x and that of the angle vanish together, we have I ds = 4a cosff'df, Jo Jo or s=4as\n (5), the intrinsic equation required. 59. Period of Cycloidal Oscillation. — Suppose now that a particle is constrained to describe a cycloidal curve in a vertical plane with the axis vertical and vertex downward. Then, calling the displace- ment from the vertex along the arc .f, we have the component acceleration along the curve due to gravity denoted by —g sin 4>. Thus, by (5), the equation of motion of the particle may be written g+-^^ = ° (6). di 4a But the solution of this, as already seen, is of the form s=s,sm(^g/4af+e) ...... (7), thus giving a simple harmonic motion of period T—2i!-J4alg ... (8) entirely independent of amplitude. 60. Cycloidal Pendulum. — The constraint on the particle to make it describe a cycloid might be in the form of a tube, along the smooth interior of which the particle slides. A special property of the cycloid enables us, however, to conveniently regard it from another point of view. By unwinding a thread from one curve, a second curve is 1 Some writers studiously avoid the differenlials dx and dy and use always the difterential coefficients dxIdB. dyidd, etc. Others use freely the separate symbols dy and dx presumably, on the understanding that they represent small increments whose ratio is the limit of hylSx as Sx aff roaches zero. For the formal justification of the separate use of dy and dx, modern works on the calculus should be consulted. E 66 ANALYTICAL MECHANICS [art. 60 obtained called the involute of the first, which is itself termed the evolute of the second. Thus, to describe a cycloid by the unwinding of a thread from some curve, we have to determine what that curve is ; in other words, we have to find what is the evolute of the cycloid. It may easily be shown that it is two halves of an equal cycloid. Thus, referring again to Fig. 16, the evolute of the half-cycloid OPB is that half-cycloid SQB shown above it. And a thread fixed at S and wrapped round Q to B, if unwrapped while kept stretched and carrying a pencil starting at B, would describe the half-cycloid BPO. To establish this it is necessary and sufficient to show (i) that PQ lies along the normal to the cycloid OPB at P ; (ii) that the length PQ is the radius of curvature of the cycloid OPB at P ; (iii) that PQ is tangential at Q to the cycloid SQB ; and (iv) that the length PQ equals the length from Q to B along the cycloid SQB. (i) We have shown, in equation (3), that ^ = 61 2, but it is evident also from the geometry of the figure that PED is ^/2 also. Thus PQ is perpendicular to the tangent PT or is along the normal to the curve. (ii) Again, from equation (4), and calling the radius of curvature of the cycloid p, we have /D = ii!f/(/<^=4acos^=2PE = PQsay .... (9). Thus Q is defined as lying along PE produced so that EQ=EP. It is thus on the circumference of the circle GQE, equal to DPEF, and standing vertically over it. Now if this upper circle rolls along GS it is evident that Q will reach S, because it is exactly like the point F in the lower circle, which by rolling reaches A. Accordingly the vertex of the upper cycloid is at B, the cusp of the lower one. (iii) PQ is readily seen to be tangential to the cycloid SQB at Q since, by construction, it is at right angles to the normal QG, the angle GQE being in a semicircle. ^ (iv) Finally, to establish the fourth point, let the angle with the horizontal made by the upper cycloid at Q be xp, and the length from B to Q along the curve be s'. Then we have by (5) / = 4asin ^=4acos ^ (lo)- Or, on comparing with (9), •f' = P (11). as needed to be shown. Thus the length of the thread which wraps along the cycloid from cusp S to vertex B is seen from its central position SAO on the figure to be double the diameter of the generating circles. Or, referring to equations (10) or (5), and writing tt/z for the angle and / for the length of the thread, we have /=4« (12). Thus, putting this value in the expression for the period, equation (8) becomes T=2i-^^ (13). ARTS. 61-62] PLANE MOTIONS OF A POINT 67 Examples — XIII. 1. Draw a cycloid by carefully rolling on a straight edge a disc of card carrying a pencil on its circumference. Indicate on this diagram the base, a vertex, a cusp, and an axis. Also state the lengths of the base and the curve from cusp to cusp in terms of the radius of the generat- ing circle. 2. Obtain the intrinsic equation of the cycloid from its definition. 3. Calculate the period of oscillations in a cycloid under uniform accelera- tion parallel to an axis and directed from the base towards the vertex. 4. Show that a cycloid is the involute of a precisely equal cycloid, and indicate on a carefully drawn figure the relation of the two curves. 5. Assuming that a cycloid is the involute of another, find the period of a cycloidal pendulum, and explain without mathematical symbols why the period is independent of the amplitude in this case, though it is not for a simple pendulum. 6. ' Establish the isochronous property of cycloidal motion. ' Show that if the particle oscillates from cusp to cusp, tue direction of motion rotates with constant angular velocity.' (LoND. B.Sc, Pass, Applied Math., 1906, 11. 5.) 7. ' A circle rolls on the inside of a fixed circle of twice its size ; prove that every point on the circumference of the rolling circle describes a diameter of the fixed circle.' (LoND. B.Sc, Pass, Applied Math., 1905, 11. 5.) 61. The Brachistoclirone. — A notable problem in the history of mechanics is that of the curve of swiftest descent, or the brachistochrone. To deal with it in its general form requires the calculus of variations, which is beyond the scope of the present work. We shall accordingly restrict the treatment to the problem in its simplest form, which may be stated as follows : — Enunciation of Problem. — Two points being given which are neither in a vertical nor in a horizontal line, to find the curve joining them down which a particle sliding under gravity, and starting from rest at the higher, will reach the lower in the least possible time. In theyfrf/ place, we can see that the required curve must lie in the vertical plane containing the two points. For if it deviated therefrom it would thereby both make the acceleration along the path less and the path longer ; hence for both reasons the time would be greater. Secondly, if the time through the entire curve is a minimum, each portion of the curve must be such that a change in it would increase the time in that portion. 62. Problem Attacked. — We have now to find by a simple method some clue to the type of curve required. For this purpose we use the almost self-evident fact, that if a curve exists of minimum time of descent it must be possible to draw near it on each side curves for which the times are slightly greater but equal to each other. And what applies to the whole curve applies to elementary portions of it. Refer- ring, then, to Fig. 17, let H and R be near points in the brachisto- chrone, HKR and HLR being very near alternative paths down which the times are equal, the distance KL being very small in comparison 68 ANALYTICAL MECHANICS [art. 62 with HK and KR. Then the element of the brachistochrone re- quired must lie between HKR and HLR. Let KL be on the same level ; then in the paths H K and HL, since the speeds are everywhere the same at the same levels (see equation (2), article 52), the average speeds down such path must be equal. Hence the time in either path equals the length of that path divided by the same value, v say, which is the average speed. Thus the extra time, t say, in HK over that in HL equals the extra path length divided by V. Let fall from L the perpendicular LM on HK ; then since KL is very small compared with KH, HM = HL nearly, and the extra distance in question is de- noted by KM, which equals KL cos <^, where <^ is the angle HKL. We accord- Derivation of Brachistochrone. ingly have KL cos COS <^' V ~ v' (3)- The reader should note that it is by no means implied that the times of descending the paths HM and HL are equal, neither is the speed over MK equal to v (but possibly double this). It is simply the extra time in the one path over that in the alternative one, that is, extra path length divided by that average speed between the levels H and KL (or KL and R) which is the same for both parts above and below KL. Again, the speeds over the elements MK and LN are not distinctly different, but differ by an infinitesimal quantity only ; but neither of these speeds are represented by v nor by v', which are only average speeds above and below KL as defined. ARTS. 63-6S] PLANE MOTIONS OF A POINT 69 63. Equation of Curve. — Now let the points K and L of Fig. 17 approach and coalesce so that the one path substituted for the two is now an element of the curve sought. Then the property possessed by the paths in common is possessed by this also. But the and now apply to the single inclinations of the upper and lower parts, the speeds v and v as before being average speeds in those upper and lower parts. Thus we may represent by UVVV in Fig. 18 two consecutive elements of the curve. In the upper element UV the average speed is v and the path inclined to the hori- zontal, in the lower element VW the average speed is v and the inclination <^'. And to these elements equation (3) still applies. Hence we may write as the equation to the curve which is obtained by infinitely reducing the lengths of UV Fig. 18. Elements and VW of Brachisto- „„„ J, /.\ CHRONE. W0CCOS9 (4). 64. Cycloid is a Brachistochrone. — We can now easily show that this equation is satisfied by the cycloid with base horizontal, vertex downwards, the start being from a cusp. Thus in Fig. 16 a particle descending under gravity from rest at B would reach P quicker along the cycloid than by any other path. To establish this note first that v^ = 7.gh if h is the vertical height descended through in acquiring the speed V (see equation (2), article 52). Hence (4) may be written Ao (5)- We have accordingly to find from the equations to the cycloid values for A and (f>. Obviously k the depth of P below AB equals OA minus the ordinate y of P, or h—2a—y .... . . (6). And from equation (5) of article 58 we have s = 4asin'f>=4adylds .... (7). Thus, by integration, we obtain / 2sds:=^Sal dy, k k or s- = Zay (8). Hence (6) and (8) yield h=2a—y=2a{\—s-l\(ia-) . . (9). Andby(7) cos^—i—s°-/i6a°- (lo)- Thus (9) and (10) show that the cycloid satisfies the condition (S) for the brachistochrone, the fall being from a eusj>. 65. Construction for the Cycloid as Brachistochrone.— Suppose with a given initial point B a brachistochrone is required to pass through a lower point R, join BR by a straight line, and describe on a 70 ANALYTICAL MECHANICS [ART. 66 horizontal base through B and in the vertical plane cycloid with any generating circle of radius a. Let BR in P. Then, since all cycloids, like all circles, other only in size, to make a cycloid pass through R change a in the ratio BP to BR. Thus let the radius circle for the required cycloid to pass through R be r. ^=aBR/BP- . . . . containing R a this cycloid cut differ from each we need only to of the generating Then we have . . . (II). Examples — XIV. Define the brachisiochrone, and show that every. point of the curve must satisfy the condition z/occos^, v being the speed of the point describing it where it is inclined <^ to the horizontal. Prove that a certain part of a cycloid in a specified position forms a brachistochrone, and construct the curve properly for two points lo feet apart, the line joining them being inclined at an angle of 45°. 66. Central Acceleration Proportional to Radius. — We now con- sider the plane motions which may be executed by a point P subject to an ac- celeration directed to a fixed point O in the plane and proportional to the distance OP. Thus, referring to Fig. 19, let the point P have co-ordi- nates X and J', speed compo- nents u and V parallel to OX and OY respectively, and let the central acceleration be — /V. Then, denoting by X and y the accelerations parallel to the co-ordinate Y i V P (x.y) X-"^^^^ ,^'-^; *z* X Fig. 19. Central Acceleration pro- portional TO Radius. as (1) of • • (0 . . (2). article 29 ; axes, we have as our equations of motion i:=— /Vcos^=— /^a; . . and j'=— /Vsin0= — /^j/ . . But these equations are the same in form hence by (4) of the same article the solutions may be written «=a sin (/)^+o) /,■) and y=bw\{pt-\-p,) (4). In these equations the constants a, a, b, and ^ are to be determined from the initial displacements and velocities as shown in article 30. The motion is therefore to be regarded as represented by equations (3) and (4), in which all the quantities are known. We thus see that it consists of two simple harmonic motions at right angles to each other, of same period, but differing in amplitude and phase. The problem accordingly reduces to the composition of rectangular vibrations. ARTS. 67-68] PLANE MOTIONS OF A POINT 7, 67. Composition of Eectangular Vibrations l.^ ST^^r'^'f T^y changing the origin of time equations (3) and (4) of the last article transform to \i>) "^ ^ a: = asin(/)/+8) and>'=^sin// .... (5), where S=a-/3. Thus expanding the expression for x and using that for J', we eliminate t between the two equations, and obtain ---^cosS=^i-^sinS. Or, on rationalising, .V" 2.vy „ V' ^^-^cosS+-^=s,n=S . ... (6), which is the equation of the path of the point P. Case I. — For S=7r/2, this becomes X' V' ^^=^ (7), the equation of an ellipse with axes along OX and OY. Case II. — For S=o or t, (6) becomes (=Tf)'-° (S), which represents two coincident straight lines through the origin, sloping one way or the other according to the sign in the brackets, which again depends upon the value of 8. In the general case (6) represents an ellipse with inclined axes, which evidently, however, lies inside and is tangential to the rectangle of sides 2a and 2b, their equations being x— +a, y= +^. 68. Different Periods. — So far we have supposed the acceleration to be directed towards the centre. It then follows that the acceleration parallel to ic or to ^ is the same for the same displacements in each direction. If, however, the acceleration per unit displacement parallel to the x axis is «° times its value parallel to the y axis, it is seen from equations (i)-(4) of article 66 that the period of the vibration parallel to the X axis will be i/«th of its value parallel to the y axis. Thus, in this case, to find the resultant motion we should have to eliminate t between equations of the form x—a%m(npt-\-^\ / \ and j= ds d4> ,, .= thehm,tof^ ^_^=,_|^,_.^^,=/, . . (3), since — = ijp and dsjdi=v. The quantity d4>lds, or rate of change of direction per unit length along the arc, is called the curvaiufe, and is, as we have just seen, the reciprocal of the radius of curvature. The centre of curvature denoted by C in the figure is the intersection of consecutive normals. Thus, if the total acceleration affecting a moving point is at any time normal to the direction of the motion, and of value c, we have no change in its speed, but only a curvature of the path produced, whose radius is P=irlc (3). Denoting by fi the angular velocity d4>ldt, and remembering that vlp=dslpdt=d4>ldt, (2) may be written c=v^^pQ,v/p=pil' (4), forms which are often very useful. 70. The Hodograph. — Instead of using the analytical method of the preceding article for the component accelerations, it is often convenient to represent graphically the total or resultant acceleration derived from the consecutive velocities. This is done in a very elegant manner by the use of a curve called the hodograph, introduced into kinematics by Sir W. R. Hamilton. Referring again to Fig. 20, we see that the effect in time 5/ of the total acceleration is to change the velocity OS into OT. Hence, by the addition of vectors, the effect in question must be represented by ST. The same would apply to any other lines all drawn from O and representing the velocities of the point in the path PQS. Thus, if such lines were drawn from a point O called the pole, and their ends S, T, etc. 74 ANALYTICAL MECHANICS [art. 71 connected by a continuous curve, the speed of describing this curve called the hodograph would represent the total acceleration of the point describing the path PQS. For, taking the first element ST, we have ST represents the total acceleration X S/. .•. speed of describing ST =ST/8/,which represents total acceleration (i). And this is the fundamental property of the hodograph which makes it so useful. If a point move so that when its inclination to a fixed line is 6 its speed is v=f{6), it is obvious that the polar equation of the hodograph may be written r^kf{e) (2), where k is some constant chosen for convenience when drawing to scale. 71. Uniform Circular Motion. — As the simplest example of the use of the hodograph, let us apply it to find the acceleration of a point describing a circle of radius p at uniform angular speed co. Then the linear speed is always p(o=»say. The polar equation of the hodo- graph may in conse- quence be written r=v. Or, the hodo- graph is another circle described in the same direction and in the same time, but the corresponding points in each are distant a quarter of the cir- cumference. These r „. . , . , , points are clearly seen trom J:'ig. 21, m which the circle of radius p at the left shows the actual path and the circle of radius v at the right shows the hodograph. The time T of describing each is obviously given by T=2ir/ia = 2'irplv=:27rv/c (5) in which c, the to(a/ acceleration of the point in the actual path, is repre- sented by the linear speed of describing the hodograph. Thus we have p\v=vlc, °' ^=vyp=pu>- (4). And, since the total acceleration is here entirely normal, we see that this result agrees with (2) and (4) of article 69. The points P' and Q' are another pair of corresponding points in path and hodograph. Sometimes from the conditions of the motion the hodograph is virtually given, and its use forms the readiest means of investigating the speed and direction of the moving point after a given time. Thus for an unresisted projectile, the hodograph is evidently a vertical straight line described downwards with uniform speed numerically equal to r the acceleration due to gravity. Actual Paxh Hodograph Fig. 2:. Uniform Circular Motion. ART. 72] PLANE MOTIONS OF A POINT Fig. 22. Conical Pendulum. 75 r.2h °°f "^^ Pendulum.-We may approximately realise the above case of uniform circular motion in a horizontal plane by the arrange- ment known as the conical pendulum. In this a small bob P is attached by a strong fine thread to a fixed point S, just as in the case of the simple pendulum, but the motion imparted to P at the start is such as to obtain in a horizontal plane a circular motion, of radius p say, described with angular velocity u. But this, as we have just seen, is only possible under the influence of a central acceleration of value poi', which in the present case must be derived as a component from the acceleration g due to gravity. To find the re- lations which ensure stability of this motion, let the length SP of the thread be / and the angle it makes with vertical 0. Then, resolving the vertical acceleration g into two components, one along the thread and the other of value c hori- zontally to the centre (see Fig. 22), we have £/g^i&ne=CF/SC=pl/cosd . . . (5). But since c—pw% we obtain from (5) <«7^=i//cos 9, or (0= n/^//cos 6 . . (6). Hence the period of rotation t is given by T= 27r/a) = 27r \/(/ cos S)/^ . The component of g along the thread is, of course, ineffective, as the thread is supposed inextensible. Examples — XVI. 1. 'Define the hodograph of a moving point. Find the hodograph in the case of a projectile moving under gravity, and the law of its description. ' Sketch the hodograph, as accurately as you can, in the case of a simple pendulum swinging through s. finite angle.' (LoND. B.Sc, Pass, Applied Math., 1905, 11. 2.) 2. ' Define the hodograph of a moving point, and prove that the velocity in the hodograph varies as the acceleration in the original orbit. ' Hence (or otherwise) prove that the accelerations of a point along and at right angles to its direction of motion are i) and v^ respectively, v being the velocity at any instant and ^ the angle which this velocity makes with a fixed line in the plane of the motion.' (LoND. B.Sc, Pass, Applied Math., 1908, 11. i.) 3. ' Obtain expressions for the tangential and normal components of the acceleration of a particle which is describing a plane curve. ' Prove that, if these two components are constant throughout the motion, the angle -f through which the direction of motion turns in a time t is given by ■^ = A log(i +Bt)? (LoND. B.Sc, Pass, Applied Math., 1900, 11. 2.) 4. ' If a particle describes mi ellipse under the action of a force directed towards its centre, find its velocity at any point of its path, and show that the ellipse is its own hodograph.' (LOND. B.Sc, Pass, Applied Math., 1900, 11. 3.) (7). 1(> ANALYTICAL MECHANICS [arts. 73-74 73. Angular and Areal Velocities and their Relations. — It is now desirable to consider more generally the relation between angular and linear volocities and to introduce the conception of an areal (or sectorial) velocity. Thus let a point move in the plane curve PQR, Fig. 23, and have at P the linear speed v and, with respect to the fixed point H, the angular velocity w. Also, let the rate V, at which the radius vector HP describes area as P moves in the curve, be called the areal velocity (or sectorial velocity) of the moving point P with respect to the fixed point H. Let the infinitesimal path length PQ be called ds, the angle PHQ be M, the radius vector HP be denoted by r, the angle UPT between it produced and the direction of P's motion be <^, and let / denote the length of the perpendicular HN upon the tangent to the curve at P. Then the length of the perpendicular PM upon HQ is ds sin <^, and the area HPQ=^5 say, described or swept out by the radius vector in time dt as P moves to Q, is given by dS^ \rds sin <^ = \pds. Similar expressions are easily obtained for the angular displacements and velocity and the areal velocity. These are collected in Table 11. Table II. Linear, Angular, and Areal Velocities. Fig. 23. Angular and Areal Velocitibs. LINEAR ANGULAR AREAL displacement elements ds ^Qjs%in<^ r =pdslr'^ dS = ^rdssm^ = \pds velocities v = dsldt=s (0 = 6 = ~ r =pvlr^ V=S=\rv%va.^ = \h say 74. Badial and Transversal Velocities and Accelerations. — Let r and Q be the polar co-ordinates of a moving point P referred to the fixed axes or ' frame ' OX, OY. And suppose u and v to be the radial ART. 75] PLANE MOTIONS OF A POINT 77 u*6u and transversal velocities of the point, that is, the component velocities respectively parallel and perpendicular to OP. We shall also denote by / and j respectively the component radial and transversal accelerations, which are effective on the moving point and change its velocity in magnitude or direction with respect to the fixed axes or 'frame' XOY. Then let it be re- quired to express «, w, /, and j in terms of r and Q and their differential co- efficients. Referring to Fig. 24, let P {r, 6) be the position of the moving point at time f and P' {r+Sr, d+Sd) be that at time t-\-Si. by the figure « = tne limit of i — ! — ^ = — =/■ . . It dt Radial and Transversal Velocities AND Accelerations. Then we have I lu 1- •.. r (''+S/')sin80 rdO a and V— the limit of '^ — ! — 4^ = — — rd bt dt (i) (2)- Again, if u+Su and v+Sv be the radial and transversal velocities of the point when at P' at time t+St, we have from the diagram the radial acceleration given by 7= the limit of >^ — ■ '- 5 — ^^ — ■ — '- ot du '"dt' dd ■■"Tt (3)- Also, from the diagram, we have for the transversal acceleration , ,- - f (z)+8z))cosS^— z'-J-(«+S«)sin6^ / = the limit of ^^ — ■ '- J- — =^ dv . dd dt +"^ (4). By use of (i) and (2) in (3) and (4) these latter transform into the compact expressions f=r-re"' ... (S) and j=re+2rd (6). Thus, though r truly represents the velocity of P along OP, r does not represent the acceleration of P's motion with respect to OX and OY in this direction, but it represents only the rate of change of rate of change ofOV. This will be made clearer by the following illustrations. 75. Circle uniformly described. — As an example of radial and 78 ANALYTICAL MECHANICS [art. 76 transversal velocities and accelerations, consider first uniform circular motion. In this case r=u=o, v is constant, so 0=o. Tiien we see that these agree with (i) and (2), and that from (5) and (6) we obtain f=—rd^=—ria'^a.\\dj=o (7). Thus/ and /here agree with our b and c of articles 69 and 71, as they should do for the circle with constant speed. Straight Line unifonnly described. — Consider as a second example the point P moving with constant speed V'm the straight linfe x=a; or, in polar co-ordinates /■cos ^=(2 . . . . (8) and ata.n6=Vi .... (9). We then obtain from these by elimination and differentiation r^Vsind, e=aV/r\ \ ,. r=a' V'/r\ and 6= -{2a V sin (?)/r' f ' " '^^°>- Thus, substituting these values in (5) and (6), we get f=a'Vyr'-ra'yyr* = o (11) and _;= ^ + -5 =0 .... (12). Or in words, the accelerations are zero in each direction; as should be the case, since the velocity varies neither in magnitude nor direction. Examples — XVII. 1. Exhibit in tabular form the relations between the linear, angular, and sectorial displacements and velocities of one point about another. 2. Obtain expressions for the radial and transversal velocities and accelera- tions of a point in terms of its polar co-ordinates r and 6. 3. A point moves from rest with acceleration g vertically downwards. If, at any instant t, its polar co-ordinates in a vertical plane are (r, 6\ their initial values being (a, o), state in terms of these the radial and trans- versal accelerations of the point. Ans. d!y=a^sin5-^a:^'^cos^fl-rz/2cos^fl c?j= {a(gr+ 2v^) - 2rv^ cos 6} sin 6 cos^fl, where/andyare the radial and transversal accelerations and v^ = 2gata.n 6. Thus/andy each vanish with g. 4. 'Two points are moving with uniform velocities u, v in perpendicular lines OX, OY, the motions being towards O ; when t=o they are at distances a, b respectively from O. Calculate the angular velocity of the line joining them at time /, and show that this angular velocity is greatest when au + bv , u^ + v^ (LOND. B.Sc, Pass, Applied Math., 1907, 11. i.) 5. If the radial and transverse velocities of a point in plane motion are both constant, show that in time / the angle described by the radius vector and by the direction of motion may be expressed by ^ = A\og{i+Bt). 76. Areal Velocity is Constant under any Central Acceleration. — When a moving point describes a plane curve with an acceleration ART. 77] PLANE MOTIONS OF A POINT 79 towards a fixed point in the plane, then its areal or sectorial velocity about that fixed point is constant. This theorem, though so simple, is yet of such fundamental importance in connection with planetary motion that we shall offer three proofs of it, one geometrical and two analytical. Geometrical Proof.— ^^iQiurig to Fig. 25, let the moving point have at P a velocity u represented by OS, T and at the very near ^'-^\'0--'^ point Q reached in ^^^ Oc'^ /"' time It let it have a velocity v repre- ^^^^^ 1 \ f. sented by OT, O ^-^i-""'^ / \^ 1 being the intersec- ^ ^^ tion of the tangents H r'^ fdt ^^^ to the path at P and Q. Then ST will /P represent the vector which, added to OS, g|X?^ .the resultant p.^. ,5. Areal Velocity Constant in OT, I.e. ST repre- Central Orbit. sents in magnitude and direction the change of velocity produced in time &t by the acceleration, / say, directed towards the fixed point. Thus a line from O parallel to ST will pass through the fixed point, H say. Take on OH, OR = ST=/S/', and join TR, TH, and SH. Then, if the initial and final areal velocities at P and Q are respectively U and V, we have by definition and the figure 27=AHOS and r=AHOT. But the triangles HOS and HOT are equal, being on the same base HO and between the same parallels HO and TS. Therefore V= U, as was to be proved. 77. Proof by Moments. — Turning now to analytical methods for a proof of the constancy of areal velocity in a central orbit, we first recall the theorem of article 25a, viz. that the moment of a resultant localised vector about a point equals the algebraic sum of its component vectors about that point, all being in one plane. For the resultant localised vector in the present case take the acceleration of the moving point P directed to a fixed point, O say. Then its moment about O being zero, the sum of the moments for its components, however chosen, will be zero also. As components take the cartesian co-ordinates of the moving point P {x, y), the origin being at O. Then using dots for differentiations with respect to time, we have xy—yx—o . (i), or ^(-^1'-^^)=° (2)- Whence, on integrating, a:;)—^.r= constant (3). 8o ANAL YTICAL MECHANICS [arts. 78-79 But the left side of (3) represents the algebraic sum of the moments of the component velocities of P about O. Hence by the theorem of moments it represents the moment about O of P's velocity, which is thus seen to be a constant. But this moment is double the areal velocity. So, if the linear velocity of P is v, and the perpendicular upon its direction from O is /, the areal velocity about O being F, we have 2 F=/w=/^ say, a constant. . . . (4). 78. Proof by Radial and Trassversal Accelerations. — Recalling now the expressions obtained in article 74 for the radial and transversal accelerations, we can easily give another proof of the constancy of areal velocity in the description of a central orbit. Thus from equation (6) of article 74 we have for the transversal acceleration j=rB^^re=.i^-j^ye) (S). But this quantity is here zero, because the acceleration is wholly radial. Also the right side has in the brackets r'Q, which is twice the areal velocity. Hence with the former notation (5) becomes i(/.»)-f=|(=n=o (6). In other words, the areal velocity is constant, as shown by equation (4). 79. Differential Equation of Orbit. — Consider the case where a point P moves in a plane under an acceleration of numerical value f directed towards a fixed point S in the plane, and let it be required to find the differential equation of its path or orbit. We again use the equations (5) and (6) at the end of article 74, which for our present case become, S being the origin of r, r-re\=-f (i), rd+zr6 = o (2). The second of these, as we have previously seen, gives at once '■'^=^ (3), ^ being the constant value equal to double the areal velocity of P about S. But these equations contain the variables r, d, and /, from which we must eliminate i. Thus dr _dr dQ _f^dr _h dr , , Jt~dd'di'~^~dd~7'd~e • • ^'^'■ Now introduce the new variable u, defined by «='/'■ ■ • . (5)> so that du= —drjr^ and dr= —du\u-. Then (4) becomes r=~hdu\d6 ... . ... (6). So, on differentiating again, we obtain r=-h{d'ulde')e=-h-u''{d^uldd^) . (7). Hence (3), (6), and (7) substituted in (i) give d'u , / ^ + "=F^^ (8). ARTS. 80-81] PLANE MOTIONS OF A POINT gi the dififerential equation required, in which i/«( = r) and 6 are the polar co-ordinates of the orbit and / is the acceleration towards the origin. Thus when /is given in terms of u the orbit can be deter- mined. Or, on the contrary, if the orbit is given, the form of / as a function of r can be found. We shall return to these cases a little later. 80. Curvature of Orbit. — Suppose we now wish to obtain an expres- sion for the central acceleration in terms of the areal velocity, the radius vector, and the perpendicular / upon the tangent. For this pur- pose we need to use the relation rdr=pdp, where p is the radius of curvature of the orbit at the point defined by r and /. This relation is proved in text-books on the cal- culus, but because of its import- ance, and also to save references, it will be briefly given here also. Thus, referring to Fig. 26, P and P', Si- apart, are consecutive positions of the point describing the orbit under acceleration to S, SN is the perpendicular / on the tangent, C the centre of curvature of the orbit at P, <^ the angle be- tween the radius vector SPR, and the direction NPT of the point's motion, Si/- the angle subtended at C by PP'. Then from the figure we have the following five relations : — p—rsm(t> (i)' S=SP=SsIp, FM = rSe^{sm)Ss, Sr=Sscos (2). Thus, by differentiating the first and using it and the others, we obtain dp =^r cos d + sin (j) dr=^r cos {iiiy- d6) -f {filr)dr Fig. 26. Curvature of Orbit. :/-cos<; pdp^rdr was to be shown, thus giving the curvature in terms of r and/ (3). 1 81 Formula for Central Acceleration.— Referring again to Fig. 26, we see that the normal component along PC of the acceleration / to S is given by <:=/sin , or, using (i), ':=fPlr (4)- 1 Or more briefly thus ■.-Sf=(y cos ^)5^ and Sr=Ss cos ^ ; hence dfldr=rdi>lds=rlp. F 82 ANALYTICAL MECHANICS [art. 82 But in article 69 equation (2) showed that c=v'Ip (s)- Thus from (4) and (5), and writing as before /» = ,4, we have fplr=v'lp=h'jp'p,sothaXf=±h'rlp'p. . . (6), the plus and minus signs being used to suit cases of either curvature. Further, using in addition equation (3) of article 80 we obtain /=^.± {6a), equations (6) and (6a) being the desired expressions giving the central acceleration in terms of the areal velocity, radius vector, and perpendi- cular on tangent. 82. Velocity in Orbit. — In Fig. 26, let the osculating circle to the orbit at P cut PS (produced if necessary) at K, and let PK be denoted by ^. It is called the ckord of curvature in the direction of the accelera- tion. We see from the figure that its length is given by PK=2CPcosCPK, or ^=2/)sin=2/)///- (7). Now by (4) and (5) of article 81 v-=fpplr (8). Hence w'=/f/2 = 2/(^/4) (9). But this is the velocity acquired from rest under an acceleration / while passing over a space q\/^. Hence the velocity of the moving point when at P in the orbit is that which it would acquire by moving from rest under a constant acceleration / equal to that at P, through a space equal to one quarter of the chord of curvature in the direction PS. Examples— XVIII. 1. Show, both graphically and analytically, that under central acceleration of any law the sectorial velocity of the moving point about the centre of acceleration is constant. 2. Obtain the differential equation of the orbit described by a point under central acceleration of any law. 3. Derive an expression for the curvature of an orbit in terms of the radius vector r and the perpendicular from the centre of acceleration on to the direction of motion of the moving point. 4. Establish a formula giving the central acceleration in terms of the con- stant sectorial velocity, the radius vector, and the perpendicular upon the tangent. Also show that the dimensions of the expression so obtained are those of a linear acceleration. 5. Find a general relation between the velocity of a point describing any orbit and the chord of curvature in the direction of the acceleration. 6. ' A point moves so that the radius vector describes equal areas in equal times. What may be inferred as to the acceleration under which the path is described ? ' If the path is an ellipse, and the centre of force is the centre of the curve, show that thft eccentricity is fJ\i-{V'lV)^'\ where V, F' are the maxi- mum and minimum velocities in the orbit.' (LoND. B.Sc, Pass, Applied Math., 1906, 11. 4.) ARTS. 83-84] PLANE MOTIONS OF A POINT 83 83. Orbits under Natural Law. — Let us now consider the orbits which may be described under the natural law of central acceleration, viz. /oci/r', ox f=ixu- (i), where ^ is a constant, u being as before equal to ijr. Then the differential equation of the orbit, given by (8) of article 79, may be written ^>^) + (.-,M=) = o . . (.). Then, by comparison with article 29, we see that the solution may be written u-txlk'= A co%(e-y) (3), in which A and y are arbitrary and depend on the initial conditions. Thus, on writing l=h^ll^ (4), and using e for ~Al, we may transform (3) into llr-i-ecos{e-y) (5), which is the well-known polar equation of a conic, the focus being the pole, / being the semi-latus rectum, and e the eccentricity. Thus, for the natural law of central acceleration, that of the inverse square of the distance, the orbit is a conic; but whether an ellipse, a parabola, or a hyperbola depends upon whether the value of e is less than, equal to, or greater than unity. And the value of e depends in turn upon the initial circumstances of the motion. Before, however, discussing this further by the purely analytical method, let us take another and more geometrical point of view, applying the hodograph to the problem of planetary orbits. 84. Natural Orbits by Hodograph, which is a Circle. — To show that the hodograph is a circle we have to prove that its curvature dxpjds is constant. We accordingly need the values of 8s and 5\jj, the element of the curve and the angle between the tangents at its v; c' P\ S -r ' /P '0 Fig. 26a. Corresponding Elements of Orbit and Hodograph. extremities. By the principle of the hodograph (article 70) the element &s of its arc described in time 8t, shown by QQ' in Fig. 26A, repre- sents in magnitude and direction the change of velocity of the point m 84 ANALYTICAL MECHANICS [art. 85 that time 6/, and while describing the corresponding element PP' of the path. Thus, for our present case, we have the relations hs=J^t^it.Ulr' (6), where/ is the acceleration along PS. Further, since each element of the hodograph is parallel to the acceleration obtaining for the corresponding portion of P, the tangent to the hodograph at Q in our case is parallel to the radius vector PS = r. Thus QF and PS both change direction with the same angular velocity Q. But, by article 76, the areal velocity \r''Q of P about S is constant=/%/2 say; hence the angular velocity of the radius vector and of the tangent to the hodograph are each given by e=hlr\ Thus, in the time element S/, the angle ^ through which the tangent QF to the hodograph turns is Sxl'=eS(=/iSi/r' (?)• Hence, by (6) and (7), the curvature and radius of curvature of the hodograph are respectively 8i/'/8.f=,4//x=i//D and/)=/i//% (8). But these values are constant. Hence, for orbits described under a central acceleration proportional to the inverse square of the distance, i/ie hodograph is a circle. The pole of the hodograph may, however, be inside the circle, on the circumference, or outside. Suppose first the pole to be inside, as re- presented by O in Fig. 27, and consider the point Q on the hodograph. Then AQ, the radius of the circle, is p—fi-lh; OQ=w, the velocity of the point in the orbit; Fig. 27. Circular Q-^j ^^^ velocity in the hodograph, is the ac- HoDOGRAPH FOR celcration / of the point in the orbit, and so is Planetary Orbit. parallel to the radius vector. Thus we see that the velocity v in the orbit, represented by OQ, may be regarded as the resultant of two parts OA and AQ, each constant in magnitude, OA being fixed in direction also ; while AQ, always perpendicular to the radius vector, moves round, Q describing the circle, while the actual point P describes the orbit. This consideration is very useful in the graphical treatment of certain problems. 85, Orbit a Conic. — We may also resolve the velocity » in a different manner. Thus, in Fig. 27, let fall OM perpendicular to QA, produced if necessary ; also make QN perpendicular to OAN. Then OM represents that component of the velocity of the point P which is parallel to FQ. But FQ is parallel to the radius vector to P. Hence OM represents the speed with which the radius vector changes in length. Or, in symbols OM^drjdt (9). ART. 86] PLANE MOTIONS OF A POINT 85 Again, NQ represents the component, perpendicular to the fixed line OAN, of the velocity of the point P in the orbit. So, if the co-ordinate of P perpendicular to OAN is denoted by y, we have NQ=dy/d^ (10). But we see by the figure that OM/NQ=OA/AQ=a constant, e say . . .(11). Thus by (9), (10), and (11) drjdy^e . . (12). So, on integrating, we have --ey+b=e(v-\--'\ = 'y' (i3)> showing by the focus-directrix property that the orbit is a conic, b being the constant of integration. It is clear from (13) that this conic is an ellipse, a parabola, or a hyperbola according as e is less than, equal to, or greater than unity; or, by (11), according as O is within the circle, on the circumference, or outside it ; that is to say, according to the velocity OQ at a given instant. 86. Alternative Proof that Natural Orbit is a Conic. — We can also show that the orbit is a conic by use of the properties referred to at the end of article 84, viz. that the velocity of P is resolvable into two parts each fixed in amount, one in direction also, and the y other always perpendicular to the radius vector. Thus, in Fig. 28, let P be the point describing the orbit under acceleration, /=/t/r°, directed to S, the origin of co-ordinates. Let PV be the velocity v of P. Then it may be regarded as consisting of the two parts PA' and A'V, equal and parallel respectively to OA and AQ in Fig. 27. Hence, if VN' is let fall perpendicular to PA'N', which is parallel to SX, we see that /. VA'N'= ^PSY' = /3 say, A'V= p, YA!—ep. Thus, if x and y are the co-ordinates of P, x and y the components of its velocity, we have by the figure « == PA' + A'N' = P A' + A' V cos ^, ^ ^ V K ^ y /, ■ ^XQ / \- -s>^'f A-- — ' ^ PUyl A' r i' Fig. 28. Planetary Orbit is a Conic. x—p{e—y/r) (14)- (is)- (16), or Also y=KY = px/r Thus, since r^'^x'+y' we have, by differentiation and use of (14) and (15), rr= XX +yy = xpe — ery, or r^ey .... So, by integrating, we have as before r=ey+b (17). (18), 86 ANALYTICAL MECHANICS [art. 8; giving the orbit as a conic by its focus-directrix property, b being the constant of integration. A comparison of Figs. 27 and 28 shows that the line OAN in the hodograph is parallel to the directrix of the conic, and therefore at right angles to its axis. 87. Hyperbolic Orbit and its Hodograph. — From what we have seen as to the relations of the hodograph and planetary orbits, it is easy to draw the two diagrams and mark off on them a series of positions for the points Q and P that correspond to each other. So for the simpler cases of the ellipse and parabola this exercise will be left to the student. But, because of its special interest in one particular, the pair of associated diagrams will be given for the case in which the orbit is a hyperbola. fFig. 29 represents this case, for e = 2, so the equation of the hy- perbola x" y' _ becomes ~.-—.=^ . . (19). Also, the semi-latus rectum /=a(e^— i) be- comes 3a, and the asymptotes are .)'=+-*=+ V3* (20), and therefore make angles of 60° with the axis of X. The P's with sub- scripts on the hyperbolic orbit correspond with the Q's with the same subscripts on the circu- lar hodograph in the lower part of the figure, whose pole is O, where OA equals twice the radius of the circle. It will be seen that the right branch of the hyperbola is described downwards by P, while the lower part of the hodograph from L, to Lj is described counter-clockwise by Q. This is effected under acceleration towards S. Then, while Q describes the upper part of the hodograph (again between the tangents from O) still counter-clockwise, Hyperbolic Orbit and Hodograph. ART. 88] PLANE MOTIONS OF A POINT 87 P describes the left branch of the hyperbola still downwards. But it will be noticed that the smallest value of the speed of P is reached at P5, corresponding to OQ5. Hence this left branch is described under acceleration j^^^w S. Indeed, this may be seen directly from the figure. For, not only is the tangent at any P parallel to the corresponding OQ, since both represent the velocity of P ; but also the tangent at any Q is parallel to the corresponding PS (or SP), since both represent the direction of the acceleration. For example, the forward tangent at Qs represents the acceleration in the same direction //-— (12). 89. Velocity and Period in Elliptic Orbit. — When the orbit is an ellipse of semi-axes a and b we have the semi-latus rectum given by l=a{i-^) = h'll^ (13). Thus, by (3) and (13) we find , J. vY sm'/3 a(i—e)=-^ -. But by (10) , ,. /2vlt:sin^B vysm'B\ a(i—e) = a{ = — ^)- ART. go] PLANE MOTIONS OF A POINT 89 Hence, equating the right sides of these we obtain ^«=Kj-^) (H). As we ma-y consider any point in the orbit the point of projection, we have for the linear velocity v at any instant when the radius vector has the value r, the expression ■-c--») (is). Again, the period r of describing the ellipse is area divided by areal velocity. Thus using (13), and remembering that b'':=a-l,\.-^)=aL we have Trab 2-Kab zirab 2ir irab 2irab 2irab 277 ] or T-oc(j», for jj. constant. By putting (15) in (16) to eliminate the semi-axis-major a, we find 4tVV'' '''-{2ti.-rvy ('7), showing that the periodic time is independent of the angle of projection (^ of last article) provided the conditions are such as to give an elliptic orbit. 90. Focal Acceleration, Period, and Velocity for an Elliptic Orbit. — We have considered the orbits possible for a central acceleration varying as the inverse square of the distance. Let us now regard the matter from the other standpoint, and supposing that a point is known to describe an ellipse with constant areal velocity about a focus, let it be re- quired to find the direction and law of the acceleration. This gives the essential features of the astronomical problem of the elliptic orbit of the earth round the sun. Thus, let the ellipse in Fig. 31 represent fig. 31. Focal Accelera- the orbit of semi-axes a and b, semi-latus tion for Elliptic rectum /, and let c denote the semi-diameter Orbit. CD conjugate to CP, and p the radius of curvature at P. Draw the tangent at P, and let fall upon it from the foci S and S' the perpendiculars SY and S'Y' of lengths / and /' respectively. Then, by the known properties of the ellipse, we have p^^lab,plr=p'lr\pp'=b'',rr'=c\b^=al,!Lndr+r' = 2a . (18). Thus^^<=y^ = ^ (19). r r y rr c Also if hfz is the constant areal velocity of P about S, and / the acceleration directed to S, we have by equation (6) of article 81, and using (18) and (19), go ANALYTICAL MECHANICS [art. 90a where /t is written for ^'//. Thus the acceleration is directed towards the focus S, about which the areal velocity is constant, and is seen to vary inversely as the square of the radius vector from that focus. Period. — For the period t of description of the orbit we have iral> 27raja/ 2ff Velocity. — Also, remembering that pv=h, and using (18), we have for the velocity at any point in the orbit J y^" / b'^ pp' r 2a — r (H) <">. or V =(!■ as previously found in equation (15) of article 89. This result may also be easily obtained by using pv=h and the tangential-normal equation for an ellipse, the focus being a pole, which is p'' r a It may be noticed that articles 69, 76, 81, 84, 85, 87, and 90 give the more elementary parts of central accelerations in a form practically free from the calculus. It may, therefore, be an advantage to some students to take these articles consecutively, omitting the other parts till these are understood. 90a. Simultaneous Elliptic Orbits. — Although an intrusion upon the pure kinematics with which the present part of the book is con- cerned, it is perhaps as well to note here that, strictly speaking, we have not in nature a point S occupied by one body void of acceleration while the other body at P describes an orbit round it. On the con- trary, the acceleration of the body at P should be reckoned with respect to a point G on SP, and then the body at S has also an acceleration towards G which may be regarded as at rest. Further, these two simultaneous accelerations are proportional to the segments into which SP is divided. Thus at any instant the acceleration of P towards G is to that of S towards G as GP is to GS. And each acceleration is in- versely proportional to the square of SP as it changes from instant to instant. Hence we have in any actual case an ellipse described by S about G as a focus, while P describes the similar ellipse also about G as a focus. In the case of the sun and the earth or other planets G almost coincides with S. Assuming this to be so, we have the acceleration ju.//-" for any of them, and consequently from equation (21) T^oca" (22a), which is one of Kepler's laws symbolically expressed. In the case of ART. 91] PLANE MOTIONS OF A POINT 91 equal double stars G would be midway in SP. These points may receive further attention when we deal with attractions in Chapters xi. and XVI. 91. Focal Acceleration for any Conic. — Let the polar equation of the conic be lu—i—e cQsd (23), / being the semi-latus rectum, u the reciprocal of the radius vector r, and e being the eccentricity, and let it be described with areal velocity A/2 about the pole. Then we have, on differentiating, d-u . o d'u ,1 , , ^° ^+"=7 (^4). But by equation (8) of article 79 the left side of (24) equals //h''u^ whereyis the acceleration directed to the pole or origin. Thus f=h'llr'^iilr (25), that is, the acceleration varies inversely as the square of the radius vector independently of the value of e, and therefore this is the law whether the conic is an ellipse, parabola, or hyperbola. Examples — XX. 1. Discuss the initial conditions of a point describing an orbit under central acceleration of the natural law, and show how to determine the type of conic described and its orientation. 2. Prove that when an ellipse is described with constant areal velocity about a focus the acceleration is directed to that focus and varies inversely as the square of that focal distance. 3. Obtain expressions for the velocity at a point in an elliptic orbit under focal acceleration and for the period. 4. Derive expressions for the velocity at a point in a parabolic orbit and in a hyperbolic orbit. Ans. ■i'2_2^/;-and7'' = /i(= + i). \r aJ 5. ' If the velocity of a planet describing- a circular orbit about the sun be suddenly diminished by a slight amount, show by a figure the relation of the new orbit to the old one. ' Will the periodic time be increased or diminished ? ' (LoND. B.A., Pass, Applied Math., 1906, i. 10.) Ans. Point where the velocity is checked becomes the end of the major axis of new elliptic orbit, the sun being at the distant focus. The periodic time is diminished according to the expression t/to = (2 - z/77/;)-^'^, where the subscripts o refer to original period and velocity. 6. 'Prove the formula v^ = \i.{^-^^ for the velocity at any point of an elliptic orbit described under a central acceleration /i (distance) "2. Ob- tain corresponding formulae for parabolic and hyperbolic orbits described under the same attraction. ' Show that, if a parabohc orbit and rectangular hyperbolic orbit have 92 ANAL YTICAL MECHANICS [art. 91 the same latus rectum, the velocity at the end of the latus rectum is V3/2 times as great in the latter case as in the former.' (LoND. B.Sc, Pass, Applied Math., 1908, 11. 4.) 7. ' Investigate the law of force tending to the focus under which an ellipse may be described, and show that the periodic times in the orbits de- scribed by particles projected from the same point with the same velocity are equal.' (LoND. B.Sc, Pass, Applied Math., 1909, n. 4.) 8. ' A particle P describes a parabola under the action of a force tendirlg to the focus ; determine the law of force, and prove that the kinetic energy is inversely proportional to the distance from the focus. ' Prove also that, if M is the foot of the perpendicular from P on the directrix, the motion of M is the same as that of a particle constrained to slide without friction along the directrix and attracted to the focus by a force proportional to the inverse fifth power of the distance from the focus.' (LoND. B.Sc, Pass, Applied Math., 1910, 11. 4.) ART. 92] PLANE ROTATIONAL MOTIONS 93 CHAPTER VI PLANE ROTATIONAL MOTIONS 92. Uniform Angular Acceleration. — Consider an assemblage of points in a plane, each of which moves in the plane with the same uniform angular velocity w about the same fixed axis perpendicular to that plane, no other motion but this being possessed by such point. Then obviously the relative distances of any of the pairs of points would remain unchanged. Further, if this common angular velocity were changed in any way to some other value u', so that at each instant its value was the same for every particle, still the relative distances of the particles would be unchanged by the motion about the axis. Such an assemblage of particles under the conditions named may accord-' ingly be referred to as a rigid body or system, since it changes neither shape nor size. Strictly speaking, no absolutely rigid bodies are ever met with, but the conception and the term are useful, and the state of things in question is often closely approximated to. If the assemblage of points or particles, instead of being confined to a plane, are distributed in solid space and all move parallel to a fixed plane with the same uniform angular velocity about a fixed axis per- pendicular to that plane, we have, as before, a rigid system in plane rotational motion. Let us now suppose that a rigid body has an initial angular velocity (Oo about a certain axis fixed with respect to the body and also with respect to our co-ordinate axes, and let the body have also a uniform angular acceleration a about this axis of motion. It is required to determine the subsequent motion. A little reflection will suffice to show that this case is closely analo- gous to that of the rectilinear motion of a point under uniform linear acceleration. Hence by the meanings of angular velocity and angular acceleration (rate of increase of angular velocity) we obtain at once a set of equations for rotations like those obtained in article 27 for trans- lations. Thus, using B for angle described in time / and repeating the former equations, we have the two analogous sets as follows :— Translations. Rotations. v=v,+at w=u>,->r„t+^at' . . (2), v- = vl+2as w" = wl+2ae . . (3). Further if the perpendicular distance of any particle from the axis of rotation is r, it is obvious that the relations between its Imear and angular displacements, velocities, and accelerations may be written r=s/6=v/ia=afa (4)- 94 ANALYTICAL MECHANICS [arts. 93-94 These equations accordingly serve to solve any problems concerned vfith the uniform acceleration of rigid bodies about a single axis fixed both in space and in the body. 93. Angular Acceleration proportional to Displacement. — Con- sider a rigid body capable of motion about afixed axis under conditions which impose an angular acceleration proportional but opposite to its angular displacement d. Then, using dots to denote differentiation with respect to time, we have for the equation of motion 0+j,'e=o (.). The solution of this (see article 29) may be written e=6»„ sin (//+.) (2), in which 9„ and «, the amplitude and phase angle, are to be determined by the initial conditions as shown in article 30. We thus see that the motion performed under the specified con- ditions is a simple harmonic rotation of period given by ■r=2Trlp (3). Examples — XXI. 1. With angular acceleration 2 radians per sec.,^ find (i) the angular velocity 5 sec. after it was 3 radians per sec. ; (2) the angle described in that 5 sec. ; and (3) the increase in the square of the angular velocity while one revolution is described. Ans. 13 radians/sec, 40 radians, Sir. 2. If the angular velocity of a rigid system about a fixed axis increases from 12 to 13 radians per sec. while it turns through 2| radians, what was the angular acceleration if uniform ? Ans. 5 radians per sec.^ 3. Find the time required to describe 28 radians with an initial angular velocity of 3 radians per sec. under an angular acceleration of 2 radians per sec.^ Ans. 4 seconds (or — 7). 4. A rigid figure turning about a fixed axis is subject to angular acceleration - 9 times its angular displacement. If displaced through 30° and let go, what will be the subsequent motion ? Ans. fl = ^ cos 3/. 6 94. Composition of Coplanar Rotation and Translation. — The composition of rotations about a given fixed axis is clearly a matter of algebraic addition, so may be dismissed without further remark. But the case of a rotation or angular displacement, and a translation or linear displacement, requires consideration. Thus, in Fig 32, let the linear displacement or translation of a point A in the body be AA'=5, and let the angular displacement or rotation of the body be 6 also in the plane of the diagram. Construction. — Bisect AA' in M, and draw MR and AC at right angles to AA'. Also draw AB, making the angle —OJ2 with AC, and produce BA to meet MR in R. Draw RA'B'. Then A'B' is the new ART. 95] PLANE ROTATIONAL MOTIONS 95 position of the line in the body originally at AB. That is A'B' represents the effect on AB of the rotation and translation which were to be compounded. Proof. — By construction RA'=RA, and is inclined at an angle 6' with it. Hence by rotation about R, the line AB moves to the position A'B', in which the point A' has the prescribed translation AA'=.r, and the line AB has the pre- scribed rotation BRB' = ^. We see by the figure that j/2 = MR tan (9/2 = ARsin 6>/2, and that the angle MAR is the complement of 6/2. Fig. 32. Rotation and Tkanslation Compounded. 95. Instantaneous Centre of Rotation: Rolling Motion. — The fore- going example of composition shows that a coplanar rotation and trans- lation are equivalent to an equal pure rotation about a particular fixed point. Thus, as regards initial and final positions, any prescribed dis- placement in a plane may be regarded as a rotation about some point. (In the special case where the motion is translation only this point is at infinity.) Hence, by taking the steps small enough, we may approxi- mate to any motion in a plane by a series of motions, each of pure rotation, about certain points. Each such point, while in use, is called the instantaneous centre of rotation. Thus any finite motion of a rigid system in a plane could be regarded as a series of infinitesimal rotations about an infinite number of instantaneous centres of rotation. The locus of the instantaneous centre forms a curve in the plane called the space centrode. Its locus in the rigid body or system forms another curve called the body centrode. Thus, the instantaneous centre of rotation may be regarded as the point of contact of the space centrode and the body centrode. Hence the whole motion of the body may be represented as the rolling of the body centrode on the space centrode, and this con- ception is often of great value. Thus, as the rolling proceeds, that mathe- matical point called the instantaneous centre has in general a velocity both in space and in the body. But the point in the body which for the instant coincides with the instantaneous centre, has for that instant no velocity either in space or in the body. And this distinction must be clearly grasped. As an example, in Fig. 16 of article 58, consider the circle EPDF as rolling along the straight line AEB, and let its centre advance at uniform speed u. Then AEB is the space centrode and EPDF the body centrode. Also E is the instantaneous centre of rotation in the position of the figure. And clearly the point E will advance along the straight line and along the circle with uniform speed u. But consider now the point P, fixed with respect to the circle and carried with it so as to reach B, a cusp of the cycloid which it is describing as the circle 96 ANALYTICAL MECHANICS [art. 96 rolls. Then we see that B is always at rest in space, that P is always at rest with respect to the circle, and, when it reaches B, the point P is at rest in space also. For it is then at the cusp of the cycloid, and is at the pause before describing the next branch of the curve. But, although E moves along the straight line at the uniform speed «, without any check or pause, it passed through B j ust when P coincided with it. And as soon as E has got beyond B, P has moved perpendicularly away from AEB, and so has ceased to coincide with the instantaneous centre. Thus P's first motion, after the pause at B, destroyed its claim to be the instan- taneous centre, because it took it out of contact with the line AEB. If the rigid system extends into the three dimensions of space, but all the motions are still parallel to one plane, it is more fitting to speak of the instantaneous axis of rotation, and of the space axode and body axode, which are the loci in space and in the body of the instantaneous axis. 96. Analysis of Plane Motion of a Rigid Body. — In article 94 we dealt with the composition of a rotation and a coplanar translation. Let us now take a different aspect of the matter and, the initial and final positions of a line in the system being specified, find the centre of rotation, also the equivalent rotation and transla- tions. Thus, in Fig. 33, let AP and A'P', be respectively the initial and final positions of a line in the rigid body. Join A A', bisect it in M, and draw MS at right angles to AA'. Then obviously the centre of rotation must lie in MS. Similarly by joining PP', bisecting it in N', and mak- "j^ ing NT at right angles to PP', Fig. 33. ANALYsrs of Plane Motion ^^ ^^^^ another line on which OF A Rigid Body. the centre must lie. It is accordingly at R, the intersection of MS and NT. We may therefore describe the motion from AP to A'P' as a rotation about R through the angle ARA'= PRP'=PAQ, where AQ is parallel to A'P'. But we may also describe the motion from AP to A'P' as either (i) a rotation through the angle PAQ about A together with a translation of A from A to A'; or (2) a rotation through the angle PAQ about /"together with a translation of P from P to P'. When AA' and PP' are parallel, the above construction for R obviously fails. In this event two cases arise according as AP and A'P' are parallel or not. In the latter case, illustrated by Fig. 34, R is found as the intersection of AP and A'P' Whereas when the initial and final positions of the lines are parallel as well as those which represent the displacements of their ends as ART. 97] PLANE ROTATIONAL MOTIONS 97 AP and the dotted line A'P" in Fig. 34, the centre of rotation is thr^S ^;J"" V^^ intersection of MU and SV, the perpendiculars through the middle points of AA' and PP". In other words, as is obvious, the motion in ques- tion is one of pure translation only, the displacement of each point being equal to AA' or PP". If the character of a rigid body's motion is variable, it is obvious that the instantane- ous centre of rotation lies in the line which bisects at right angles any known small dis- placement of a point in the body. Thus, the intersection of two such lines gives the instantaneous centre of rota- tion. A consideration of the circumstances of the case, or repeated constructions, then determine the space centrode as the locus of this centre. The body centrode is often found most easily by 'fixing' move. Examples of the application of these principles will be found in connection with mechanisms treated in Chapter ix. 97. Composition of Linear and Angular Velocities. — Velocities, whether linear or angular, being the rates of increase of the corre- sponding displacements per unit time, the results obtained for the composition of translations and rotations still hold, but admit of some simplification when the velocities in question are instantaneous ones and not mean velocities merely. Thus, referring to the end of article 94 and supposing that in Fig. 32 the line A A' shrinks to the infinitesimal element ds described in time dt, we have ds/2 = AR.d6l2 and MAR=n-/2 nearly. So, if the linear and angular velocities are respectively v and w, we obtain ds/di=ARde/df, or AR=Z'/o). That is, the resultant of an angular velocity o about a given axis and a linear velocity v in the perpendicular plane is an equal angular velocity co about a parallel axis distant v/(o, such distance being per- pendicular to that of the linear velocity v and in the direction shown in Fig. 32. We see that if & is in the positive direction along the axis of j' and G Fig. 34. Special Construction for Centre of Rotation. the body and supposing the ' space ' to 98 ANALYTICAL MECHANICS [arts. 98-100 01 denotes the positive rotation in the plane of xy, that the distance AR is in the negative direction along the axis of -r. Conversely, any given angular velocity w (say in the xy plane) may be resolved into an equal angular velocity about a parallel axis distant p (say positively along the x axis), together with a linear velocity »= +/*>, perpendicular to the axes of rotation and to the direction of/ (i.e. positively along thejy axis). The student should carefully note that although he is provided with the power of compounding a translation and a rotation into an equal rotation about some other point, it is not advisable in every case to use it. On the contrary, it will sometimes be found preferable to solve a pro- blem by directing attention to the specified components of translation and rotation especially if accelerations be required. (See Examples — xxii.) 98. Composition of Angular Velocities about Fixed Parallel Axes. — Let us now determine the resultant of two component angular velocities Ui and Wj about parallel axes which intersect the perpendicular plane at A and B a distance p apart. Take a point C on the straight line joining AB, and consider the motions there. Then by the previous article we have 0), about A= w, about C and linear velocity (+0),. AC) perpendicular to AB, also (i)j about B = Uj about C and linear velocity ( — Wj.CB) perpendicular to AB. Now let C be so chosen that the linear speeds there are equal as well as opposite. Then we have as the resultant motion an angular velocity about C of value oj = (o, + a)5 (i), C being defined by AC.coj^CB.oij, or AC CB AB , , — = — = — j — (2). And, if other points are taken in the plane, it will be found that the motions due to the resultant angular velocity about C is the resultant of the motions due to the component angular velocities about A and B. It should be carefully noted that these results do not apply to successive finite angular displacements but to simultaneous velocities. 99. Linear and Angular Accelerations. — As, in article 97, we passed from linear and angular displacements to the corresponding velocities, so we may pass from velocities to accelerations, the same laws of composition and analysis still holding true. 100. Analytical Treatment of Coplanar Motions. — Let us now treat the coplanar motions of a rigid system or body analytically. Referring to Fig. 35, let the motions occur in the xy plane, let x', y' be the co-ordinates of a definite point O' in the body and x, y those of any other point P. Take also axes O'X' and O'Y' meeting in O', fixed in the body and moving with it. Also let P have co-ordinates g and i\ referred to these moving axes O'X' and O'Y', and at time / let these axes ART. lOo] PLANE ROTATIONAL MOTIONS 99 be inclined at the angle >/■ with the fixed axes OX and OY. Then we have from the figure x=;i;' + ^cos i/'— rjsin i/',jK=y+^sin i/'+tjcos ^. Thus, by differentiation, remembering ^ and 17 are constant, we obtain jc=x' — (^sin ^+17005 i/')w'\ and .i'=/ + (|cosi/' — ijsin\t)w'j' (")' OT ^' X X Fig 35. CoPLANAR Motions of Rigid System. where to' =(/^/(//= angular velocity of body about a perpendicular axis through the point x'y'. These may be put in the form .T = .r' — (jy— y)cD' and J>=J>' + (.« — x')(o' .... (2). These show that the velocity of the point xy is made up of two parts, viz. (i) one of translation expressed by x and y\ and (ii) one of rotation in the xy plane about the point x'y' with velocity w', expressed by — C^*— y)w' along the x axis and -\-(x—x")iii' along the v axis. For any other definite point O" in the body, having co-ordinates x", _)'", we have similarly .-v=a"— (j— y')(u"andj)>=/'+(;c— x")fo" . . . (3),* " —{x— «')( say Or, in words, the angular velocity is the same whatever the point through which the axis is supposed to pass. (5)- (6). loo ANALYTICAL MECHANICS [arts. 101-102 The linear velocity of the axis, on the other hand, depends upon the position of that axis. Thus from (5) and (6) we have ^'-*"+(y-y>=°\ /,) and y —y" — {x' —x")(i>=oj ^"' 101. Instantaaeous Centre; Body and Space Centrodes. — By putting x=y—o in equation (i) we obtain the co-ordinates of the instantaneous centre of rotation R referred to the axes X'OY' fixed in the body. Thus, calling these co-ordinates fo and 170, and remembering o)':=a), we have f„ sin ^+»jo cos ^=x'//o- This would be the equation of the curve described by the instantaneous centre of rotation with respect to the body. In other words, it would specify the body centrode already referred to in article 95. Again, on substituting the known functions of / in the two equations (9) and eliminating / between them, we should have the equation of the curve described by the instantaneous centre with respect to the fixed axes XOY. In other words, we should obtain the space centrode. 102. Summary of Coplanar Motions. — The chief results we have obtained for the coplanar motions of a rigid body or system, together with others which easily follow from them, may now be summarised as follows : — 1. The motion of a rigid body parallel to a given fixed plane at any instant consists in the general case of — (i) An angular velocity about an axis perpendicular to the plane and passing through any arbitrary point in the body, the magnitude w of this velocity being independent of the position of the axis ; and (ii) A linear velocity v parallel to the plane, the magnitude and direction of which are dependent on the position of the axis. (See articles 94, 96, and 97, also 100.) 2. At each instant there is, in general, an axis called the instan- taneous axis of rotation, such that the whole motion of the body is one of rotation about it with angular velocity u. (The intersection of this axis with the plane in which the motion occurs or to which it is referred is called the instantaneous centre of rotation^ In other words, a motion of rotation only with angular velocity *> about one point is equivalent to one of rotation with same angular ART. 102] PLANE ROTATIONAL MOTIONS lot speed 0) about another point x distant, together with a linear velocity xbi parallel to the axis of y, i.e. perpendicular to that of x. (See articles 95 and loi.) 3. Two coexisting angular velocities round parallel axes fixed in space are equivalent to a single angular velocity equal to their algebraic sum about an axis parallel to and in the plane of the two former, and dividing the distance between them inversely as the component velocities. (See article 98.) 4. Two equal and opposite angular velocities whose common magnitude is w, round parallel axes / apart, are equivalent to a linear velocity /w, perpendicular to the plane of the axes. (See equation (2) of article 98.) Examples — XXII. 1. ' Prove that any displacement of a plane figure in its own plane is equivalent to a rotation about some finite or infinitely distant point. ' If the figure be rotated through 90° about a fixed point A, and then through 90° (in the same sense) about a fixed point B, the result is equivalent to a rotation of 180° through a certain fixed point C \ find the position of C' (LoND. B.Sc, Pass, Applied Math., 1905, n. 4.) 2. ' Prove that when a lamina is moving in its own plane there is in general one point of it which is instantaneously at rest. ' A rod moving in a vertical plane with one end on the ground remains constantly in contact with a small peg. Construct geometrically the tangent to the path of any point of the rod ; and show that, when the inclination of the rod to the vertical is a, the velocity of the point which is moving vertically is to the ^'elocity of the point which is moving horizontally in the ratio of tan a to unity.' (LoND. B.Sc, Pass, Applied Math., 1907, 11. 3.) 3. 'Es giebt fiir jede Bewegung eines Systems in einer Ebene immer zwei bestimmte Curven oder Polbahnen, eine feste in der Bewegungsebene und eine in der beweglichen Ebene, welche beide den Pol enthalten. ' Translate, prove, and give illustrations of the proposition.' (LOND. B.Sc, Pass, Applied Math., 1906, 11. 7.) 4. ' Explain what is meant by relative velocity and relative acceleration. 'A circle of radius a rolls on a fixed horizontal straight line with velocity v and acceleration / Find the accelerations of the highest and lowest points of the circle.' (LoND. B.Sc, Pass, Applied Math., 1907, 11. 2.) 5. ' Translate and prove the following statements • — 'Jede Bewegung einer ebenen Figur auf einer festen Ebene von der Stellung 5 auf Zeit t zu der Stellung S auf Zeit t' ist einer einzigen Drehung gleichwertig. Die Bewegung einer solchen Figur auf einer Ebene kann sich als RoUen einer zur beweglichen Figur gehorigen Kurve auf einer festen Kurve darstellen.' (Lond. B.Sc, Pass, Applied Math., 1908, ii. 3.) 6. ' Explain the terms (i) angular velocity, (ii) instantaneous centre, m the case of a lamina moving in its own plane. 'The centre of a disc falls vertically with constant acceleration, while the disc rotates in its own plane (which is vertical) with constant angular velocity. Prove that the locus in space of the instantaneous centre is a parabola.' (Lond. B.Sc, Pass, Applied Math., 1909, 11. i.) 102 ANALYTICAL MECHANICS [art. 1 02 7. The centre of a lamina moves in its own plane with constant linear speed «, while it rotates in its own plane with uniform acceleration a. Show that the space centrode is the rectangular hyperbola xy = u fa. 8. A ladder of length L rests on horizontal ground and leans against a vertical wall. Show that as it moves parallel to a plane perpendicular to the intersection of ground and wall, its space centrode is a quadrant of radius L and its body centrode a semicircle of diameter L. g. A joiner's square is moved in a plane with its blade in contact with one fixed circle and its head or stock in contact with another fixed circle. Show that the space and body centrodes are similar curves, the former having double the curvature of the latter. 10. Show how to compound velocities about fixed axes parallel or inclined. 1 1. Discuss analytically the subject of coplanar motions. ARTS. 103-104] SOLID MOTIONS OF A POINT I03. CHAPTER VII SOLID MOTIONS OF A POINT 103. Solid Co-ordinates. — To represent the position of a point in space of three dimensions, in addition to the rectangular co-ordinates X and y, we may use another, z, parallel to OZ and therefore perpen- dicular to XOY, and thus expressing the perpendicular distance of the point from the xy plane. We are then adopting the solid cartesian system of co-ordinates, which is the one most frequently employed. Taking another view of the matter, we see that this system defines a point as the intersection of three planes respectively perpendicular to the three axes of co-ordinates. For to say that the co-ordinates of a point are a, b, and c is equivalent to saying that it is on each of the three planes x=-a, y=b, and z^c, i.e. on the planes parallel to YOZ, ZOX, and XOY, and distant from them by the respective amounts a, b, and c. Obviously any three intersecting surfaces would define a point, and other systems of co-ordinates are founded upon this principle. Thus, on the polar system of solid co-ordinates a point is specified as the intersection of a sphere, cone, and axial plane ; on the cylindrical system a point is given as the intersection of a cylinder with planes through and perpendicular to its axis. But these systems will be but little used in this work ; we leave the further consideration of them to the instances of their occurrence. 104. Eight-handed System. — Reverting now to the cartesian system of co-ordinates, we have still to decide which way to draw the axis oiz with respect to those of « andj'. In this work we shall adopt the right-handed system, as uniformly adopted by Lord Kelvin for sixty years (see Baltimore Lectures, pp. 445-6), and as advocated and used by Maxwell {Electricity and Magnetism, 1873, § 23). In this system 'the motions of translation along any axis and of rotation about that axis ' are 'assumed to be of the same sign when their directions correspond to those of the translation and rotation of an ordinary or right-handed screw. ' For instance, if the actual rotation of the earth from west to east is taken positive, the direction of the earth's axis from south to north will be taken positive, and if a man walks forward in the positive direction, the positive rotation is in the order, head, right-hand, feet, left-hand. ' If we place ourselves on the positive side of a surface, the positive direction along its bounding curve will be opposite to the motion of the hands of a watch with its face towards us.' I04 ANALYTICAL MECHANICS [art. 105 Hence if (as usually seems to the writer most natural and con- venient) the axis of ;*; is drawn horizontally to the right, and that of y vertically upward on the blackboard, just as in plane geometry, then the axis of z must be supposed to project from the board towards the spectator, and may be represented in perspective accordingly. Also the positive direction of rotation, as always used in trigonometry, from X to y, or counter-clockwise, corresponds on this right-handed system with the positive direction of advance along the axis of z. Of course, the axes may be turned about into any convenient position provided their mutual relation be preserved. ' Thus, if x is drawn eastward and y northward, z must be drawn upward.' These two positions are illustrated in Fig. 36. Of course, they may be shown in other ways to suit the case in hand. It should be noted that the positive direction of rotation in any co-ordinate plane is that which corresponds with the cychcal order of the letters indicating the axes in Z 'X Fig. 36. Right-handed Co-ordinate Axes. that plane, as YZ, ZX, XY. Also translation along the positive direction of the remaining axis OX, OY or OZ in each case bears the right-handed relation to that rotation. 105. The Straight Line. — If a straight line pass through two points {a, b, c) and {x, y, z) distant r apart and make with the co-ordinate axes angles whose cosines are /, m, and n respectively, it may easily be seen that we have {x-ay + {y-by + {z-cr = r^ (,), x—a_y—b_z—c I ~ m ~ n ~^ ^^'' and r-f-;«''-f«==i (3). Equation (i) is useful as giving the distance between two points whose co-ordinates are known, (2) gives the equations of a straight line in space of three dimensions, {x, y, z) being* the current co-ordinates, while (3) gives an important relation between what are called the direction cosines of the line. This brief notice will suffice to introduce the system of solid geometry and render what follows intelligible. For further information rec6urse must be had to the text-books on the subject. .ARTS, 106-107] SOLID MOTIONS OF A POINT 105 Ans. "=^ = ^U = 3 Examples — XXIII. State the relations between the three rectangular cartesian axes which form a right-handed system, and sketch such a set in various positions. Show in perspective the points P and Q, having the co-ordinates (i, i, i) and (4, 3, 2) respectively. Write the equations of the hnes OP, OQ, and PQ, O being the origin and P and Q as defined in question 2. _y _z _ X- I _y- \ _z-\ "2 ' 3 ~ 2 ~ I 4. Find an expression for the angle 6=POQ as defined by questions 2 and 3 (see Chapter 11.). Ans. cos fl = 9/^87. 106. Cylindrical Motion. — Consider the motion of a point P on the curved surface of a right circular cylinder of radius c with axis along the axis of 2, and let 6 represent the angle between the fixed plane ZOX through the axis of z and another plane con- taining that axis and the point P. This is shown in Fig. 37. Let us note expres- sions for the velocities and accelerations as the point moves in any way on the cylindrical surface. The velocities par- allel to the axis, to the tangent to the base, and to its radius, i.e. parallel respectively to PW, PT, and PC, are easily seen to be Fig. 37. Cylindrical Motion. s, cd, and zero (i)- The accelerations in the same directions are respectively 2, 2r t^ J B Fig. 39. Spherical Motion. Consider now the motion of the point P on the surface of the sphere, its position being defined by 6 and <^. Then adopting what we have just established for conical motion, we easily obtain from the figure the accelerations in three directions at right angles, viz. ON, NP, and PT. Thus, the acceleration parallel to ON = i3(«cos>^) =-—{ — a sm.^}=— a cos4>-' — a sin (i). The acceleration parallel to NP = 3-2(asin<;f)) — asin<^.^''z=— (flC0S<^.^)— asin <^.^" = — flsin^.^^+flcos (^.<^— asin "^.^^ (2). The acceleration parallel to PT, i.e. perpendicular to OCPR (see equa- tion (6) of article 74) = a sin cos 4>.0^ (4). Lastly, the sum of the products, equation (i) multiplied by cos <^ and (2) by sin gives the acceleration parallel to OP = -a^''-asin=^.^' (5). On replacing the constant radius a by a variable, and differentiating io8 ANALYTICAL MECHANICS [art. 109 it when necessary, these results could be extended to any motion in space of three dimensions. Examples— XXIV. 1. Obtain general expressions for the accelerations in the principal directions for a point moving on the surface of — A cylinder ; 2. A cone ; 3. A sphere. 4. A point moves with constant speed v on the surface of a right circular cylinder of radius a and traces out a regular helix whose axial advance per radian is^. Find the velocities and accelerations of the point radially, tangentially, and axially. Ans. Velocities, zero, va{d?+f)-''i\ vpia^+p^)-'"'' . Accelerations, - av\a^+p^)~^, zero, zero. 5. A circular wire 10 cms. in radius is in a vertical plane and rotates with constant velocity 3 radians per second about its vertical diameter ; at a certain instant a bead on the wire is 30° from the top of the circle, and is moving downwards along the wire with speed 4 cm./sec. and accelera- tion I cm./sec.^ Find the vertical and radial accelerations of the bead. Ans. Vertical acceleration is downwards i "93 cm./sec.'' Radial acceleration is inwards i"82S cm./sec.^ 109. Spherical Motions under Uniform Acceleration. — ^The full treatment of this problem is beyond the scope of the present work ; it must accordingly suffice to indicate here the chief features of the rnotion executed under these conditions. It is easily seen to be the general case in three dimensions of which the simple pendulum was the special case in a vertical plane. The present problem is, therefore, often known as the spherical pendulum. Referring to Fig. 40, we take the uniform acceleration g to be vertically downwards parallel to OCG. The notation and lettering are as in Fig. 39, but the figure is now in- verted. Since the particle P is supposed constrained to remain on the spherical surface, it has no motion normal to the surface, and we are therefore concerned chiefly with its component accelerations in two directions tangential to that surface at P. It is • convenient to take these directions in the horizontal and vertical planes through P, thus giving as the tangents PT and PG respectively. The vertical acceleration at P being along PM has clearly no component in the perpendicular direction PT. We can accordingly equate to zero the corresponding expression. Thus from equation (3)'of article 108 we have a sin <^.^+ 2a cos i \ "^ i^ p / C G ^ M Fig. 40. Spherical Pendulum. ART. 109] SOLID MOTIONS OF A POINT 109 where a is the radius of the sphere. If now we write NP = >-=a sin , the above equation becomes re+2re='LA(r"-d) = o. r at Hence r-e=a^ s\n''.e=h (i), where h is double the areal velocity of the horizontal radius vector NP about the vertical axis of the sphere OC. We have now to consider the component acceleration along PG, the other tangent. It is clearly g cos MPG=^ sin . We may accord- ingly change its algebraic sign and equate to the expression for the acceleration along GP given in equation (4) of article 108. We thus have a4>—a sin cos <^.^'= — ^sin (j>. Multiplying by 2aii and integrating, this becomes 2(7 'j^d(^^^ + 2a'e'-jsmd(sm)=2gajd(cos4>), or a'ij>^+a-sin'' + «'(^- i) = 2ia{cos -cos a) (7), \sm / expressing ^ in terms of <^ and the initial state of things. To find the highest and lowest places, put il>=o in (7), when we see that either <^=a as in (3), or else we may reduce the equation to the forms , sin'a — sin^<6 ■ 2 , a' ?^ ^ — ^= 2ga sm

— cos a and «'(cos(^+cosa) = 2^(a(i— cos'-^) . (8), from the latter of which the factor (cos <^— cos a) is removed. ANALYTICAL MECHANICS [arts. It accordingly gives other values for cos <^. It is easy to see that equation (8) gives one value for cos ^ lying between — i and + 1. Thus, for example, if we make cos ^ change continuously from — i to +i, the left side of (8) changes continuously from the negative value —u''{i—coz a) to the positive value +u^{i -|-cos a). The right side of (8) for the same range of values of cos 4> changes from zero to the maximum positive value zga and falls again to zero. Hence for some value of cos <^ between + 1 the equation is satisfied. That is to say, there is some value, /3 say, of other than a, for which, in equation (7), the angular velocity vanishes. Hence the whole motion of P lies between the two horizontal circles on the sphere defined by o and p. Fig. 41. Spherical Pendulum ey Rectangular Vibrations. 110. Alternative View of Spherical Pendulum. — A useful insight into the character of the motion by a spherical pendulum is obtained by the following alternative view of the matter: — Take two vertical planes at right angles to each other through the centre of the sphere as shown by AOA'C and BOB'C in Fig. 41. Now sup- pose a pendulum suspended at O and of length / (the radius of the sphere) to perform simultane- ously oscillations of amplitudes CP and CQ in these two planes respectively. Also suppose that both these amplitudes, though in the figure shown large for clearness' sake, are really so small that the period of vibration is in each case represented with sufficient approximation by the simple expression given in article 53, viz. T=2-irJJfg (9)- On these assumptions we may regard the actual motion of P on the spherical surface as represented with close approximation by the result- ant of two rectangular simple harmonic vibrations of equal periods. But this is an ellipse, in the general case which we now require (as was shown at the end of article 67). Further, since this ellipse, or quasi- ellipse, is to be traced on a spherical surface with its centre C at the lowest point on the sphere, it is evident that the ends PP' of the major axis will be higher than QQ', the ends of the minor axis. Hence the special extreme values u and j8 of ^ found in article 109 correspond to the angles COP and COQ in Fig. 41. 111. Rotation of Quasi-Ellipse. — Let us now proceed to a further approximation, and suppose that while the amplitude CQ is still small and the period t represented by (9), that the amplitude CP is so large that the period of the vibration in this plane has a period T (10). ART. u 2] SOLID MO TIONS OF A POINT , , , articles 54-56. To consider the effect of this on the motion, take a plan of the quasi-elhpse described. This is shown by the broken ■A- "! A^^ '^^ ^^ P°'"' """'"S '■o""^ in the clockwise direction as indicated by the arrows. Suppose the moving point to start from P then when it next approaches that position it will have completed its vibration in the vertical plane ^ through QCQ' in the time t, a little earlier than the instant at - which it will have completed its ^,''' vibration in the vertical plane p' (- through PCP', for this takes the slightly greater period o-. The "~~--- moving point accordingly crosses Q i=r^ the line PCP' at R a little inside Fig. 42. Rotation of Quasi-Ellipse. P, and reaches its maximum elongation at S a little on the Q-side of P, as shown by the full line in Fig. 42. In other words, the quasi-ellipse w/a/M about the vertical axis of the sphere in the direction of its description by the moving point. Further, the angle PCS, whose magnitude depends partly on (o-— t), is itself described in the time a- nearly. Thus the curve is not immediately re-entrant, but forms a series of loops in radial symmetry about the centre C. It can be shown by higher analysis that the rate of this rotation of the quasi-ellipse is proportional to its area. By an elementary view of the matter it is at once seen that the speed of rotation is increased by an increase of the larger amplitude CP, since this augments the excess of the periods (t—t, thus shifting R and S each farther from P. The increased speed of rotation when the amplitude CQ is increased, and therefore R shifted nearer P, is accounted for by the fact that the ellipse is now broader at the ends, and S is still shifted farther from P in spite of the approach of R. Of course, when the quasi-ellipse becomes a horizontal circle the rotation of the figure ceases to be of consequence, so the two views are harmonised. The rotation vanishes only when the area of the ellipse vanishes, i.e. when the motion is initially confined to a vertical plane. And this accords with our elementary treatment of the problem. The phenomenon of the rotation of this quasi-ellipse in the direc- tion of its description may be easily noticed in the case of a plummet suspended from a fixed point and properly started. With a pendulum started truly in a vertical plane, Foucault showed by the apparent shift of that plane that the earth was rotating. 112. Accelerations of a Point moving in any Curve. — A plane curve has at each point a tangent, and a normal in that plane and a radius of curvature along that normal. But what is called a tortuous curve has also a change of its plane by rotation about the tangent; it therefore extends to three dimensions in space, and is the general 112 ANALYTICAL MECHANICS [art. 112 example of a curve. The tortuosity of a curve is measured by the rate of rotation of the plane about the tangent per unit length along the curve. Thus, if this angle is denoted by <^, the length along the curve by s, and the tortuosity by o-, we have >)=b, and '2{nia)=c . . . (3). Hence, putting the values from (3) in (i) and (2), we obtain the resultant angular velocity of magnitude '^ about the axis, whose direc- tion cosines are A, [i, and v. In dealing with a numerical case it is well to arrange the quantities in seven columns headed, w, /, m, n, /w, m and a. are easily dealt with by the formulae of article 92. We have therefore to consider here the composition of the angular velocity to about OX with the angular acceleration ^ about OY, which leads us to a somewhat new con- Precession derived from Acceleration. ART. i2o] SOLID ROTATIONAL MOTIONS jig ception. Let us begin to consider the motion at the instant t=o. Then at this instant the angular velocity possessed by the body IS ft) about OX, and is represented by OM in Fig. 46, there being at that instant no angular velocity about OY. After the infinitesimal time S^, the angular acceleration fi about OY will have produced the angular velocity ^ht about OY, this being denoted by ON on the figure. Hence to find the state of things at the instant t=U we have to compound the angular velocities represented by OM and ON. Their resultant of magnitude "^ about OR inclined 8^ with OX is represented by OL and, according to article 118, is expressed in magnitude and direction by ■5'^ = (oH(/3S/)= (i) and /t=cosYOL=:sinXOL=/3S//->F . (2). Since there is no acceleration to produce angular velocity about the axis of z, the third direction cosine, v, is zero ; that is, the resultant axis OR is in the xy plane. In the limit where the square of /SS/ vanishes in comparison with <^)^ have from (i) *=- (3). And from (2) and (3), when the sine of XOL may be assimilated to the angle, W=^htjui or, in the limit, lil^=deidt=Q. say (4), where fi represents in radians per second the angular velocity of OR in the plane of XOY, i.e. about the axis of z. 120. Precessional Motion. — Hence, the initial effect on the angular velocity la of the perpendicular acceleration jS may be represented by leaving the magnitude of w unchanged, while changing the direction of its axis by a rotation 12 radians per second in the plane of the axes of ft) and /8 and from o) towards fi. This rotation of the axis is called precession, and fi is called the rate of precession. It is useful and suggestive to rewrite equation (4) in the form ^.^'-fi (5), and to compare it with equation (4) of article 69. We then see that there is an analogy between the familiar conception of uniform circular motion and the present new phenomenon of precession. Thus the two equations under consideration may be put in words as follows : — Uniform Circular Motion. Precession. Without changing its magni- Without changing its magni- tude, to rotate a linear velocity v tude, to rotate an angular y&XocAy at angular velocity fi requires a ft) at angular velocity fi requires a perpendicular linear acceleration perpendicular a«^/«r acceleration vQ.. <"^' We may further notice as to the relation of the directions of angular velocity, angular acceleration, and precession that (for the right-handed system of axes used in this work) if the first two are positive about the I20 ANALYTICAL MECHANICS [art. 121 axes X and y respectively, the last is positive about the axis of z. These directions are indicated by arrows in the figure. It should be noted that, in the ideal case just considered, the axis of rotation precesses not only in space, i.e. with respect to the co-ordinate axes, but with respect to the body also ; for we postulated neither velocity nor acceleration round the axis OZ, so there can be no rotation of the body itself about OZ. Thus, if at the instant ^= o a certain point K is on the axis of rotation and therefore on the axis OX, it does not remain on the axis of rotation and describe a circle in the plane XOY. This would involve an angular velocity of the body about OZ, which does not exist, as it was not there at first and there has been nothing to generate it. On the contrary, when the instantaneous axis of rotation has passed from the axis of x, the point K would be no longer on the axis of rotation but would be circling round that axis. It may naturally be asked how could the motion under discussion be realised. The following scheme would accomplish it:^Let the body be a sphere with centre at the origin of co-ordinates and the diameter KOK' lying initially along the axis of x ; let two cylinders with axes parallel to OZ touch the sphere at K and K' and be capable of a motion round OZ. Give to the sphere the angular velocity w about the diameter KOK' coincident with the axis of x, then let the cylinders roll upon the sphere without slipping as though the sphere were at rest, and so that their axes move round OZ with the angular velocity fi=/3/(«. Then, though the sphere itself has no rotation about OZ, its instantaneous axis of rotation which replaces KOK' would be moving round OZ at angular velocity 12 with respect both to space and to the sphere itself. 121. Rotation about a Moving Axis. — Let us now consider the subject of articles iig and I20 from another point of view, as this throws an additional light on the matter.. We now start by supposing the body to have an angular velocity u about an axis OA, which axis has itself a precession or angular velocity J2 about a fixed perpendicular axis OZ. Let it be re- quired tofind what angu- lar acceleration, if any, this velocity and preces- AccELERATioN ENTAILED BY PRECESSION, sion entail. Let OA move in the plane XOY, and consider the moving axis OA when it makes an angle \lt with the fixed axis OX as shown in Fig. 47. ART. 122] SOLID ROTATIONAL MOTIONS 121 Let OL on the axis OA represent the angular velocity w. Then, on the same scale, OM = a)cosfi^ and ON = m sin 12/ represent at the instant in question the component angular velocities about the axes OX and OY respectively. Hence, by differentiation, we may obtain the angular accelerations ^ and »; about these axes. We thus have g= — 12(osin fl/=OPonFig. 47 . . . . (i) and 7/= fia)cosfi/=OQ do. (2). Whence the resultant acceleration, found by compounding the vectors OP and OQ, is given by OR on the axis OB, defined in magnitude y8 and direction YOB by /3^=f + ,f = a)^J2^or;8 = (ofl (3) and tanYOB=tanfi/=tanXOA, or ^YOA=: ^XOA . (4), i.e. OB remains perpendicular to OA, and therefore rotates with angular velocity fi about the axis OZ. Thus the result may be summed up in words as follows : — If a body has an angular velocity o) about an axis OA, which axis rotates at angular velocity fl about OZ perpendicular to OA, then there is an angular acceleration /i? = coJ2 about the axis OB, which rotates at angular velocity i2 about OZ so as to be always at right angles both to it and to OA. This proposition holds whether (i) the axis of rotation OA moves both in space and in the body, or (ii) moves in space only, being fixed in the body. But, whichever is the state of things at starting, so it will remain if there is no angular acceleration about OZ to change the zero or uniform rotation which is initially postulated. The cases of motion considered in this and the two previous articles are somewhat ideal, and the student is warned against hastily concluding that they apply rigorously to any special case he may have in mind in which the constraints and masses involved may need somewhat detailed consideration. The subject will be referred to again in Chapter xiv. 122. General Precessional Rotation. — The example already con- sidered illustrates precession in its simplest form, the axis of rotation moving in a plane because it is at right angles to the axis OZ about which the precession occurs. In the general case of precessional rotation the axis of rotation OC may make any angle ZOC with the axis of precession ; it accordingly follows that OC describes a conical surface of semi- vertical angle ZOC. It is also easily seen that any desired relation between the angular velo- city (0 about OC and the rate of precession fi, or angular velocity of OC about OZ, can be represented by the rolling of a moving circular ^^^ ^g Precessional cone of axis OC and semi-vertical angle p on a rotation represented fixed circular cone of axis OZ and semi-vertical by Rolling Cones. ancle , the sum of d and ^ being ZOC and their common vertex being at O. This is shown in Fig. 48, representing the plane ZOC, which accordingly contains also the line 01 along which ANALYTICAL MECHANICS [art. 123 the two cones are instantaneously in contact. Suppose OC is advanc- ing from the plane of the diagram towards the reader. Then, since the moving cone is rotating about the instantaneous axis 01, the point P will in time St move perpendicularly to the diagram a distance MPa)8/=a>rsin^S/ (i), where r denotes OP. Similarly, as OC is moving round OZ at angular velocity J2, in time S/ the point P will move perpendicularly to the diagram through the distance NPfi8/=I2/-sin(e+^)8/ (2). Hence, equating these two expressions for the element of path described by P, we have ) (3). For the simpler case of the previous articles, in which the conical surface described by the moving axis is a plane, we have d+=Tr/2. Hence, equation (3) reduces to wsin^=n (4). Comparing this with equation (4) of article 119 we find sine=^K (5), which with el>=ir/2 — d (6), gives the angles 6 and <^ for the rolling cones to represent this simple precessional motion when osi- tions instantaneously occupied by OA, OB, and OC being re- spectively ^1, 6^, and ^s- Let the body or figure under consi- deration have angular velocities o),, Mj, w, about these moving axes, all as repre- sented in Fig. 49. It is required to obtain expressions for the angular accelerations a, ^8, 7 about the axes OA, OB, and OC respectively. Fig. 49. Angular Accelerations about Moving Axes. ART. 124] SOLID ROTATIONAL MOTIONS 123 It must be noted that though the six angular velocities referred to are a.11 about the instantaneous positions of the moving axes, the velocities are all of magnitude reckoned with respect to axes fixed in space. Thus, if ^3 had the magnitude o-i, this would mean that at the instant in question the axes OA and OB were moving in their own plane away from axes OX and OY fixed in space at the rate of o-i radian per second. Again, if (o, had the value 0-5, that would mean that with respect to \.)\& fixed axes OX and OY the body rotated at the speed 0-5 radian per second about OC. Thus the body would have the angular velocity (0)3 — ^3) =0-4 with respect to OA and O3. Consider first the angular acceleration y8 about OB. By reference to equation (3) of article 121, it is clear that it will contain the term + (i)i^a, since we have an angular velocity tu, precessing towards OB at the rate ^3. But by a second application of the same principle to the axis OC, it is clear that the acceleration about OB must have another term — (03^j, due to the angular velocity (03 precessing at rate Q-^ from OB. There is also the term (bj contributed by the rate of increase, if any, of the velocity about the axis OB itself. We accordingly obtain for j8 the algebraic sum of the three terms just mentioned. But it is evident from the symmetry of the notation and the figure that the other accelerations may be written down by suitably changing the subscripts to ft> and 0. We accordingly obtain for the angular accelerations about the moving axes the following expressions, each consisting of the rate of increase of one angular velocity and a pair of products of the other angular velocities concerned : — a = (Uj — Mj^j + 0)3^2! /3 = a)2 — ft)3^i + (Ul^3 J- (l). These results are valid for any actual case whether the moving axes move both in space and in the body or always coincide with certain lines fixed in the body. In either case the relations between the w's and the 6's will follow from the conditions prescribed. Thus, if the axis OA is fixed in the body, then 0^=iU)^, ^3 = ^3 j hence a=r(Ui. 124. Angular Acceleration for Steady Precession. — We are now prepared to deal with the determination of the angular acceleration occurring in the general precessional rotation of article 122. We suppose the body to have an angular velocity of constant magnitude w about an axis OC fixed in the body, while this axis describes a conical surface by moving with an angular velocity of constant magnitude J2 about a fixed axis OZ, to which it is inclined at the constant angle 6. We take OZ vertically up and other fixed axes OX and OY horizontally as shown in Fig. 50; also two moving axes at right angles to OC, namely OB horizontal and OA in the same meridian as C. It will be convenient to regard OX, OY, OZ, OA, OB, and OC as each of unit length, so that O is the centre of a unit sphere on whose surface the other six points always lie. 124 ANALYTICAL MECHANICS [art. 124 Then, using the notation of the preceding article, we easily see that <0i = 6i, 6j = oi2 = o, 0)3=0), (i3=d)=o. We have thus to find 6^ and 0^. Suppose that in time S^ C moves to the near position C Then the displacement CC may be regarded (i) as the result of the angular velocity SI about OZ, the radius being sin 6, or (2) as the result of the angular velocity 0^ (or o)i) about OA, the radius being OC, which equals unity. We accordingly find that —fl sin 0=^1 = o)i, the minus sign occurring because the displacement due to a positive value of J2 is negative, while that to ^1 or 0), is of the Fig. 50. Angular Acceleration for Steady Precession. BB' has the value Q.ht, Now consider OC the same sign as the velocity. Again, in time 8/, let B move to B', then since it is described by unit radius about OZ. polar axis of our unit sphere, and draw the quadrant from C to B ; also another meridian from C through B', and an arc of an equator through A and B cutting the meridian CB' in B". Then the angle between the two equatorial arcs BB' and BB" is that between their respectively polar axes OZ and OC, namely 6. Hence BB"=BB' cos 0. And, since the arc BB" is described in time &t with the angular velocity 6, and unit radius about OC, we have 6138/= BB"=BB' cos e={mt)cos 6, or 03 = i2cos 0. We may accordingly write the following scheme of values : — About the axes : — the angular velocities of \ the body are / the angular velocities of \ the axes are / We have also OA, OB, and OC ; i)j:= — flsin 6, 0)2=ro, 0)3 = 0) . ?,= -S2sin6l, ^, = 0, e^^nco%6 (2); (3). (4). 0) J = 0)2 =: 0)3 ^ O Hence, to determine the angular accelerations, we substitute from equations (2), (3), and (4) in the right sides of equations (i) of article 123. We thus find a^o /8 = (o)-J2 cos 0)n sine . and ■y = o . . It is seen that if 9=7r/2, (6) reduces to P=->^ (8), (5), (6), (7). ART. I2sl SOLID ROTATIONAL MOTIONS 125 in agreement with the results of articles 119 and 121 for the case first considered. Further, (6) shows that if co=J2cos6', then jS=o . . . (9). 125. Most General Motion of Rigid Body with One Point Fixed.— The example of precession considered in article 122, in which the rolling cones are both circular, though representing any purely precessional rotation, falls short of complete generality. Now, if the fixed point of the rigid body is taken as the centre of a fixed spherical surface, it is obvious that the most general motion of that body can be represented by the most general motion of a spherical figure on this fixed spherical surface. But as in the case of a plane moving in a plane (dealt with in articles 94-96 and loi), so here this most general motion of the spherical cap on the fixed sphere may be represented by the rolling of some line, fixed relatively to the moving cap, on some other line, fixed on the stationary spherical surface ; these lines being respectively the body and space centrodes, both, however, being spherical now instead of plane. Extending our thoughts to the interior of the sphere, it is evident that the body and space centrodes are simply the base lines of the moving and fixed cones (with common vertex at the centre of the sphere) which, by the rolling of one on the other, represent the motion of the rigid body in question. These guiding cones are not necessarily circular, nor need their bases be bounded by re-entrant curves. On the contrary, the term cone is here to be understood in its widest sense as a conical surface generated by a straight line passing through a fixed point and a fixed line circular, re-entrant, intersecting and re-entrant or even non-re-entrant. With these provisos then, the most general motion of a rigid body with one point fixed may be represented by the rolling of a cone fixed in the body on a cone fixed in space, the fixed point of the body being the common vertex of the two cones. Examples— XXVII. 1. Discuss the composition of angular velocities about intersecting axes. 2. If a simple pendulum were set vibrating at one of the geographical poles of the earth, how would you expect its plane of vibration to move with respect to the earth ? How would the plane of vibration of a pendulum in latitude X appear to move with respect to the earth's surface ? Ans. The plane of vibration would appear to rotate at the rate of (2ir sin X) radians per day, or one complete revolution in (cosec X) days. 3. Trace the analogy between the effect of an angular acceleration on a perpendicular angular velocity and the corresponding case of linear acceleration and velocity. 4. What acceleration, if any, is needed to maintain an angular velocity of constant magnitude about an axis which is itself rotating so as to describe a plane ? . ■ ■.■ ^ 5. Show that the motion of a body consisting of rotation about an axis which is describing a conical surface may be constructed as the rolling of one cone on another, and obtain the relations between the various quantities involved. 126 ANALYTICAL MECHANICS [arts. 126-127 6. Obtain the complete scheme of expressions for the angular accelerations about a set of rectangular moving axes with fixed origin in terms of the various velocities involved. 7. Assuming the general expressions for the angular accelerations about moving axes, find the acceleration required to maintain a steady conical precessional motion. Under what special conditions may this accelera- tion vanish, and when may it reduce to a simpler form ? 8. Explain carefully how to construct the most general motion of a rigid body with one point fixed. 126. General Displacement of a Bigid Body. — We now pass to the consideration of the most general displacement possible to a rigid body having no point fixed, and shall closely follow the treatment adopted in Routh's Rigid Dynamics (i. pp. 186-8, sixth edition, 1897). It is first necessary to establish the following proposition : — Enunciation. — Every displacement of a rigid body may be repre- sented by a combination of (i) a motion of translation, or linear displace- ment, whereby every point in it has a displacement equal in magnitude and direction to those of any assumed point P rigidly connected with the body; and (2) a motion of rotation or angular displacement oi the whole body about some axis through this assumed point P, which may be referred to as the base point. Proof. — It is evident that the change from initial to final position may be effected by shifting P from its old to its new position, P' say, by a motion of translation of the body as. a whole, and then retaining P' as a fixed point while moving any two other points of the body not in a straight line with P' into their final positions. But any displacement of a rigid body with one point fixed has been shown in article 113 to be equivalent to a determinate angular dis- placement about some axis through that fixed point. This accordingly establishes the proposition. Since the above displacements, linear and angular, are quite in- dependent, their order may be reversed, i.e. we may rotate the body first and then translate it. Further, the motions might occur simul- taneously. It may easily be seen that any point P could be chosen as the base point of the double operation, the translation and rotation being defined accordingly. Hence the given dis- placement may be constructed in an infinite variety of ways. 127. Change of Base Point : Axes Parallel.— Let us now find the relations between the axes and angular displacements when different Fig. 51. General Displacement pomts P, Q are taken as base points. OF A Rigid Body: Axes Parallel. Suppose that the displacement in question may be represented (a) by the angular displacement about an axis PR together with the linear displacement PP', or {b) by the angular displacement <^ about an ART. 128] SOLID ROTATIONAL MOTIONS 127 axis QS together with the linear displacement QQ' as indicated in Fig. 51, in which PP' and QQ' are not necessarily in the plane of the diagram. Consider the method (a) of constructing the total displacement. Then it is clear that any point, B say, has two displacements — (i) a translation equal and parallel to PP', and (2) a motion through an arc in a plane perpendicular to the axis of rotation PR, the length of this arc being rQ where r is the length of the perpendicular BM let fall from B on to the axis PR. This arc is zero only when r is zero, that is, when the point B is on the axis PR. We accordingly conclude that the only points whose displacements are the same as that of the base point lie on the axis of rotation corresponding to that base point. Thus, any line all points of which have the same displacement must be an axis of rotation for any point in it taken as base point ( i ). Through the second base point Q draw QABC parallel to PR. Then, for each of these points Q, A, B, C, etc., on this parallel, the displacements by the method {a) are a translation PP' ; and an arc (in planes perpendicular to PR) of magnitude r6 where r is the length of any perpendicular QK, AL, BM, CN, etc., between the parallels PR and QABC. Hence, by the statement (i) above, we see that QABC parallel to PR must be the axis of rotation QS corresponding to the base point Q. Thus, as P and Q are quite general, we conclude that the axes of rotation corresponding to all base points are parallel (2). 128. Rotations Equal. — We have still to find the relation be- tween the angular displacements Q and <^ which occur about the parallel axes PR and QS respectively in the methods (a) and (b) of represent- ing the total displacements. For this purpose let us take a new dia- gram in a plane perpendicular to these axes PR and QS, as shown in Fig. 52- Then, by method (a) of construct- ing the displacement, we see that the rotation through the angle 6 about PR carries Q to q, where the chord Qf=2^- sin =^ (S). Therefore, since P and Q are quite general, we have the result that the angular displacements corresponding to all base points are equal. 129. Azial Projections Equal. — From equation (3) we may find the translation and rotation for any given base point, Q say, when those for any other base point P are known. For, since the displacement Q^ is produced by rotation about the axis PR, it must occur in a plane perpendicular to PR, and consequently its projection upon PR is zero. Thus, taking projections upon PR, from equation (3) we find Projection of QQ' upon PR— projection of PP' upon PR . (6). Hence, the projections upon the axis of rotation of the displacements of all points of the body are equal. Thus QQ' is fully determined. And we already know that QS is parallel to PR, also that ^=.6. Hence, referring to Fig. 52, if P' is at any distance, 3 cm. say, from the plane of the diagram towards the reader, so is Q' at that same distance and in the same direction. If the projections of PP' and QQ' upon PR are each zero, then all axial projections of displacements are zero, and the whole displacement reduces to an example of coplanar rotation and translation, as already treated in articles 94-102. 130. Central Axis : Twists and Screws.^It is often important to choose a base point such that the direction of translation may lie along the axis of rotation. We shall now examine how this may be done. Let the specified displacement of the body be a rotation 6 about PR and a translation PP'. And, if possible, let this displacement be represented by a rotation i> about QS and a translation QQ' along QS. These lines are shown in Fig. 53, in which also P'L and Yp are drawn perpendicular to PR, and therefore parallel to each other ; M and N are the middle points of P/ and LP' respectively. Then, from articles 127-129 we have the following relations: — QS is parallel to PR (7), '^=^ (8), and QQ'==PL=MN (9). Also, with the former notation, P/ will represent the displacement of P due to the rotation ^ = d about QS. Thus, Q/>=QP \ .. and P/=2QPsine/2 = LP'/' " ' ^^°^- Hence, QS lies in the plane QQ'NM which bisects LP' at right angles; it is also parallel to PR and at a perpendicular distance PQ from it, such that 2PQsine/2 = LP'. And this forms the solution of the problem. The distance of the central axis QS may be stated by giving MQ or NQ', which are the perpendiculars from the plane PLP' ; thus ART. 130] SOLID ROTA TIONAL MOTIONS 129 2MQtan (9/2 = LP'. As to the direction of this distance, it should be noted that NQ' must be measured from the middle point of LP' in the direction in which that middle point is moved by its rotation round PK.. Fig. 53. Determination of Central Axis. Having found the only possible position of QS, it is still desirable to show that the linear displacement is really QQ' along QS. The rotation d about PR will carry Q through an arc Q^, whose chord is parallel to P'L and equal to 2PQ sin ^/2. That is, this chord is equal to P'L. Then, taking the displacement PP' in the two steps P/ and /P', the first step brings q back to Q, and the second step carries it from Q to Q' along QS, as was to be shown. It thus follows that the most general displacement of a rigid body can be represented by a rotation about some straight line and a translation parallel to that same line. Such a displacement is like that of a nut on a screw, and is called a tivist. Hence we may say that a rigid body may pass from one given position to any other by means of a twist. In order to specify a screw we must state (i) the direction and position of the line round which the rotation is effected, i.e. the central axis or axis of the screw ; and (ii) the ratio of the tratislation to the rotation, which occur along and about that axis. This ratio is a linear magnitude called the pitch of the screw. In the present theoretical work the pitch of the screw is measured in units of length per radian of rotation ; in workshop practice it is measured in units of length per revolution of the thread. To specify a twist we must further state the amplitude, i.e. the I I30 ANALYTICAL MECHANICS [art. 131 magnitude of the rotation or angular displacement occurring on the screw in question about which the twist is effected. It should be noted that the twist by which a rigid body may pass from one position to another is, in general, unique. For, if possible, let there be two central axes PR and QS; see Fig. 51 of article 127. Then by that article these axes are parallel. Also, taking PR as the central axis, the displacement of any point A on QS is found by turning the body round PR and moving it parallel to PR. Thus A has one displacement perpendicular to the plane PRA, i.e. to QS, and another parallel to QS, and accordingly cannot move solely along QS. Hence QS cannot be a central axis. When the rotations are indefinitely small, the construction to find the central axis is simply stated by Routh somewhat as follows : — Let the displacement be represented by a rotation lodt about an axis PR and a translation vdt in the direction PP'. Measure a distance _j'=»(sinP'PR)/(u from P perpendicular to the plane P'PR on that side of the plane towards which P' is moving. A straight line parallel to PR through the extremity oi y is the central axis. This construction may be illustrated by reference to Fig. 53 of the present article. From this figure it may be seen that when PP' shrinks indefinitely, MQ and PQ each coalesce with P<7, which corresponds with Routh's construction for the central axis QS. 131. General Motion of a Rigid Body. — On consideration of article 126, it may be seen that the most general motion of a rigid body, being a succession of an infinite number of elementary displacements of the most general type, may be represented by the rolling of a cone, fixed in the body, on a cone unattached to the body, the latter cone having a motion of pure translation, the two cones having their vertices in coincidence. For this rolling of cones gives at each instant the combination of translation of a point and rotation about an axis through it, together with a possible variation of every element of that motion. The motion of the body would be completely determined by (i) the dimensions of the two cones and their initial positions, (ii) the path and velocity at each instant of their common vertex, and (iii) the rate of rolling at each instant of the body cone on the moving space cone. Hence the whole motion falls into two parts, that of a point and that of rolling cones. Examples— XXVIII. 1. Discuss the general displacement of a rigid body with no point fixed, showing that if the base point is moved the axes are parallel and the rotational displacements equal. 2. Reduce the most general displacement of a rigid body to a rotation about a determinate axis and a translation parallel to it. 3. A point P in a rigid body has a displacement 5^/2 cm. at an angle of 45° with the horizontal axis, about which it rotates through 60°. Draw a diagram indicating (i) the central axis, (2) the displacement along it, and (3) the rotation about it. ART. 132] SOLID ROTATIONAL MOTIONS 4. Establish the proposition that any displacement of a rigid body may be represented as a twist about a screw. 5. Show that the most general motion of a rigid body may be constructed as the rolling of a body cone on a space cone which has a motion of translation only, the vertices of the cones being always in contact. 132. Velocity of any Point of a Rigid Body in most general Motion. — We have already seen (in article 126) that the most general displacement of a rigid body may be represented by a linear displacement or translation of a base point, O say, and an angular displacement or rotation about some axis through O. Further, we have seen (in article 131) that the most general motion of a rigid body is that of translation of some point O combined with a rotation about some axis through O. But the magnitude and direction of the linear velocity of O may change from instant to instant, also the magnitude of the angular velocity and the direction of its axis through O may change in like manner. We may compactly provide for these changes by supposing the magnitude and direction of each of these vectors to be given by their rectangular components. It is then a problem, in terms of those components, to specify the velocity of any point in the body. This we now treat, following the method of Routh. Choose any three rectangular axes OX, OY, OZ, meeting at the base point O and moving with O, but keeping their directions fixed in space. Let n, v, w be the components of the linear velocity of O and I, 7;, f be those of the angular velocity of the body. The usual con- ventions as to the relation of the positive directions of these six com- ponents apply as shown in Fig. 54. Fig. 54. Velocity of any Point in a Rigid Body. Let U, V, W be the velocity components of P, its co-ordinates being x, y\ z. ' Consider first the expression for U, the x component of P's velocity It obviously consists of three terms, viz. those due to translation of O in the OX direction and to the rotations about OY and OZ. These are seen to be u, +riz, and -Cv respectively, and so 132 ANALYTICAL MECHANICS [art. 133 give the first equation in the following scheme, of which the others follow also from the figure, or may be written down by observing the cyclical order of the letters : — U =u +r]Z— M V=v+Cx-$z\ (i). Each of the products in the above equations expresses that part of the linear velocity in the given direction, which is due to the angular velocity denoted by the Greek letter and the perpendicular distance denoted by the co-ordinate. Suppose now it is desirable to change the base point from O to O', whose co-ordinates are a, d, and — b) — rf{x — a) . . (3). But this equation holds for any position of P, thus we may change any one co-ordinate in any way we please with or without alteration of the others. If, therefore, we put y=o, in the above equation (3), it still holds. But we have thereby reduced the left side by ^y and the right side by ^'y. Hence these two angular velocities are equal. And this can in like manner be shown to hold for the others. Hence f=^, V=r;, andf=f (4). Or, in words, whatever the base point chosen, the component angular velocities for a given resultant motion remain the same. By substitution of (4) in (2) and equating the result to (i) we obtain the linear velocity components of the new base point O' in the form «'= u-\-r]c— y>\ v'-=v+^a~\c\ (S). w = 7V + ^b — rja] It is seen that this just agrees with what we should obtain directly from (i). 133, Resultant Twisting Velocity. — Let the motion of a rigid body be specified by the linear velocities {u, v, w) of some base point O, and the angular velocities (f, ?j, {) about axes (OX, OY, OZ) meeting in O and moving with it but keeping their directions fixed in space. It is required to find the central axis, the linear velocity along it, and the angular velocity round it. In other words, the three rectangular ART. 133] SOLID ROTATIONAL MOTIONS 133 screws and the twisting velocities about them being given, it is required to determine the resultant or equivalent screw and the resultant twisting velocity which occurs about it. In the treatment of this problem we shall again follow Routh's method. Let the direction cosines of the central axis be A, /i, v, the linear velocity along it be V, and the angular velocity round it be fl. Then we have to determine these five quantities and make certain deductions from them. Let P be any point on the central axis, then if P were chosen as base point, the components of the angular velocity would be the same as for the base point O (equation (4) of article 132). Also we have seen (articles 25^ and 118) that angular velocities are vectors, and therefore compounded by vectorial addition ; hence we obtain fi^ = f+V + r (1), and \=$/n,fi=y,/[l,v=(/il . (2), or a=^/X = r,/f.=Clv . . . (3). Again, we have seen (in articles 129-130) that the velocity of every point resolved in a direction parallel to the central axis is the same and equal to that along the central axis. We accordingly obtain by projection V=uk-\-viJ,-ir'ui\> . . . . (4). Taking the product of (3) and (4) we have VQ.=u^^-vr^+w^ . . ... (5). Also, dividing (5) by (i), we obtain as the piU/t of the equivalent screw round which the resultant twist occurs !^~^' + ri'+C . Let X, y, z be the co-ordinates of any point P on the central axis. Then the linear velocity of P is along the axis of rotation. Hence its components, given by equation (i) of article 132, are proportional to the direction cosines A, /i, v of the central axis, and accordingly proportional to the components f r], C of the angular velocity about that axis. We may accordingly write u+i}z-(v _ v+Cx-^s _ w+^V—>-lx _V ^^y ^ ~ V C ^ ' And these form the equations of the central axis. Hence equations (i), (2), (4), (6), and (7) present the solution of the problem under consideration. If we shift the base point, or change the direction of the axes, it may be shown from (5) that the value of Tfi remains constant. The product Kfl may therefore be called the invariant of the components. The resultant angular velocity has already been seen to be constant, and may be called the invariant of the rotation. If the motion is such that rfl=o • • (8). then it follows that either V=o or i2=o. Accordingly equadon (8) 134 ANALYTICAL MECHANICS [ART. 133 expresses the condition that the motion is equivalent either to one of translation only or to one of rotation only. For either motion to be not zero we must, of course, have its components not all zero. The foregoing very brief introduction to the subject of screws and twists must suffice here. For fuller treatment the reader is referred to the works on Dynamics by Routh and by Williamson and Tarleton, also to the original memoirs of Sir Robert Ball, to whom the theory of screws is principally due. Examples — XXIX. 1. Obtain expressions for the velocity components of any point of a rigid body in the most general motion possible to it. 2. Having given that a certain point O in a rigid body has linear velocity components u, v, w parallel to the axes OX, OY, OZ, and that the body also has angular velocity components ^, r), f about these axes, it is required to determine the resultant motion as a twist about a screw. "^^^^ '34] MECHANISMS ^ CHAPTER IX MECHANISMS 134. Subdivisions and Treatment.— We now pass to the consideration ot detormable figures, a simple treatment of which will occupy this Chapter and the next. A fairly comprehensive scheme indicating a set of possible subdivisions of this subject is given in Table iii. Thus by supposing any one of the six types of systems i-6 to be subject Table III. Kinematics of Deformable Figures. Deformable Systems :— a. Mechanisms, or Partially Deformable Figures. 1. Inextensible Cords and Membranes. 2. Incompressible Fluids. 3. Linkages, etc., with Rigid Links. B. Elastic Bodies, or Generally Deformable Sicbstances. 4. Extensible Cords and Membranes. 5. Compressible Fluids. 6. Elastic Solids. Possible Deformations and Motions :— a. Displacements and Strains. b. Steady Flow or Currents. c. Reciprocating Motions and Vibrations. d. Wave Motions. e. Vortical Motions. in turn to each of the five types of motion a-e, where such motions are possible to them, we obtain the various subdivisions. Some points in this full scheme have, however, been already suffi- ciently touched upon under other headings. Others again lie beyond the scope of this work. Of the remainder, the various displacements and motions of mechanisms occupy this chapter, the strains of elastic bodies being dealt with in the next. The consideration of a few points of an advanced character will be deferred to those later chapters, where they are needed in connection with the corresponding kinetical or statical problems. It may be well to explain here that the 'partially deformable 136 ANALYTICAL MECHANICS [art. 135 figures ' in the table denote those bodies or systems which by their nature or arrangement absolutely preclude or render negligibly small certain conceivable deformations, other deformations being possibly very large. The 'generally deformable figures,' on the other hand, are those of a nature such that no conceivable deformation is thereby precluded or kept negligibly small, though some deformations may still be much smaller than others. Thus, the extensible cords could suffer much greater deformation by bending them than by stretching, though the latter is not prohibited as for the inextensible cords. 135. Inextensible Cords. Multiplied Cord. — The most striking mechanical use of a cord which is not appreciably extensible but very readily flexible occurs when it is passed round and round two parallel cylinders or similar bodies, one end of the cord being fastened to one cylinder, while the other is free. We then have an example of the reduplication of a cord or of the multiplied cord. Regarding the cylinder to which one end of the cord is fastened or fixed, we may inquire what is the ratio of the displacement s of the free end of the cord to the displace- ment r of the centre of the moving cylinder, round which say n plies of the cord pass, all practically parallel to each other. It is easily seen that we have the relation s=nr ... . . (i), since for every element of the cylinder's displacement an equal element of each of the n plies of cord is set free, which total to a displacement nr of the free end. If the displacement r affects the free end of another multiplied cord, we have only to repeat the equation (i) to find the final dis- placement ratio. Or, we may combine two or more such equations in the form ^ = «l''l = «l(«2'-2)=«I«2(«3'-3) = etC. . . (2), where «i is the number of plies of the cord whose end has the dis- placement s corresponding to the displacement ;-, of the centre of the first moving cylinder ; n^ is the number of plies of the cord attached to the centre of the first cylinder, whose displacement r, corresponds to the displacement r^ of the centre of the second cylinder round which the «j plies pass ; and so on for the others. The above considerations form the first step in the theory of that simple machine or mechanism often miscalled 'the pulley.' The rnachine in question is seen to derive its essential character from the kinematics of the multiplied cord, which settles its displacement ratio as shown above. The other details of the machine, such as the provision of a pulley to lessen friction, are really only modifications of the ideal case now under notice, though they may be of great practical import- ance. If, in the case of equation (i), we inverted matters and regarded the cylinder to which the cord was fastened as moving, the other being ARTS. 136-138] MECHANISMS 137 our basis of reference, then the n would be replaced by («+i). Thus we might write, using accents for distinction, 5' = («+i)r . . . . (3). 136. Connecting Belts. — In the case where an endless cord or belt passes over two cylinders or other shaped wheels, we are concerned with the velocity ratio of the two wheels thus connected. The motion is supposed to occur without slipping between the wheels and the belt. Thus, if a and b are those radii of the wheels at which no slipping occurs, and .f is the length of the belt passing for the angles of rotation and <^ respectively, we have or eiij>=ei4,=bla (4). That is, the angular velocities of the belt-connected wheels are inversely as their effective radii. The above is on the supposition that the belt is uncrossed or open, so that the angular velocities have the same sign. If the belt is crossed the displacements and velocities are of opposite sign, so that we should have e/ii>=e/4>=-b/a . . . (4a). 137. Analytical Conditions of Inextensibility. — Consider an element Ss of a cord, and let the tangential velocity of the cord be i at one end of the element and s+5s at the other. Then, in time Si, the tangential displacements of the two ends will be sSi and {s+Ss)Sf. Thus the total increase of length of the element is the difference of these quantities, viz. SiS/. Hence the rate of increase of length per unit length per second will be Ss/Ss, or ds/ds . • (s)j and, for an inextensible cord, this must always be zero. Taking cartesian components of velocity, we see that the tangential velocity of the cord at a point is .dx , .dy , .ds /,> since the terms on the right are the projections on the cord of the com- ponents of velocity. Hence the general analytical condition, for the inextensibility of a cord may be written d's _dx dx dy dy dj^dz _ / n • ds~ ds ds ds ds ds ds 138. Incompressible Fluids.— Imagine a chamber prismatic or cylindrical throughout, having two portions, of which one is small and the other large in cross-section, the respective areas being a and b. Let pistons fit tightly in these two portions and enclose between them a volume v of incompressible fluid. Then we have, as the expression of the condition of incompressibility, w= constant. Hence, if a normal 138 ANALYTICAL MECHANICS [art. 139 A A f/JJMli ^ BB displacement s of the smaller piston occurs, and a simultaneous dis- placement r in the same sense of the larger piston, we must have as=br, or i/r=j/r=^/a (i)) which gives the displacement and velocity ratio of the pistons as the inverse ratio of their areas. The arrangement is illustrated in Fig. 55, in which AA! = s and BB'=^-, the lines at A, A', B, B' representing the inner faces of the pistons. It is easily seen that this rela- tion is the essential principle of the hydraulic press and the so-called w/y/y/^A/y/^j//- ft) Fig. 55. Displacements of Incompressible Fluid. hydrostatic paradox. Examples — XXX. 1. Discuss the subdivisions of mechanisms, and draw up a scheme showing the chief topics needing treatment. 2. Sketch three different arrangements, each showing an application of the multiplied cord, and calculate the velocity ratio for each. 3. Obtain the analytical condition of inextensibility of a cord in motion and passing round curves in solid space. 4. Explain the so-called hydrostatic paradox, and obtain the velocity ratio of pump plunger and ram in a hydraulic press. 139. Links and their Eelative Motion. — As its name implies, a linkage is an aggregate of separate parts called links, whose nature, con- nection, and motion must now be studied. The term linkage is, how- ever, usually restricted to a certain class of connections which is but one example of a more general type, called a kinematic chain by Reuleaux, whose classic treatment of this subject will, in the main, be followed here. In a kinematic chain the connections of the parts may be of the most general type, and when one part of the chain is fixed the arrangement is called a mechanism. Hence the title of the present chapter. Let us first suppose the links to be rigid, and then inquire how they may be connected, and what relative motions are thereby permitted. Consider the following simple examples ol pairs of elements, which repre- sent the types of contact of the touching parts of the adjacent links : — 1. Square prism in square hole. 2. Cube on plane. 3. Circular cylinder in circular hole. 4. Circular cylinder in circular hole, but with collars to stop end play. 5. Cylinder with generator touching plane. 6. Cylinder with generator and base touching perpendi- cular planes. 7. Sphere touching plane. 8. Sphere touching two perpen- dicular planes. 9. Screw in nut. 10. Spherical cap on sphere. * 11. Spherical capon sphere, but working against a circular rib. 12. Spherical cap on sphere, but working on a pivot. ARTS. 140-142] MECHANISMS 139 It will easily be seen that the relative motions permitted are recti- linear m example i, plane in examples 2,4, 6, 11, and 12, but are soltdm. examples 3, 5, 7, 8, 9, and lo. 140. Lower Pairing of Links.— On further examination of the twelve examples just given it may be noticed that some pairs are in contact throughout the surfaces of identical geometrical form, both being plane, or the one convex and the other equally concave (often termed solid and hollow). Such pairing of links is called by some writers lower pairing, and is illustrated by examples i, 2, 3, 4, 6, 9, 10, II, and 12. When, on the other hand, the contact is along lines or points only, the arrangement of links is termed higher pairing. This form of connection is illustrated by examples 5, 6, 7, and 8, and will be dealt with in article 141. Reverting now to the lower pairing, we see that in examples i, 4, 9, iij and 12 only one degree of freedom is left. Such cases of lower pairing are said to be closed. In examples 2, 3, and 10 the pairs were not closed, for in each case two or rnore degrees of freedom were left. We have yet to assign names to the closed examples of lower pair- ing. Following the nomenclature of Reuleaux and others, we call these respectively (fl) A sliding ^&ir. (Example i.) {b) A turning pair. (Example 4.) (c) A screw pair. (Example 9.) 141. Higher Pairing of Links. — Passing now to the higher pairing of examples 5, 6, 7, and 8 of article 139, we find that a greater variety of relative motions is possible. But at each instant the moving link, being supposed a rigid body, is generally rotating about some axis. It may also have a motion of translation. The consideration of these two possible motions affords a clue in the examination of the matter in hand. Thus, simple rolling occurs if the instantaneous axis lies in the common tangent plane at the point of instantaneous contact. But, when the instantaneous axis is the common normal at the point of con- tact, simple spinning occurs. If, however, the relative motion is such that the instantaneous axis passes through the point of contact, but is neither in nor perpendicular to the tangent plane, then that motion is combined rolling and spinning. Lastly, if the instantaneous axis does not pass through the point of contact, there is sliding combined, it may be, with rolling or spinning according to the inclination of the axis. Thus if rolling, spinning, and sliding be denoted by the letters R, N, and D respectively, then the possibilities of motion for a sphere on a plane would be denoted by {R), {N), {D), (R and JV), {N and D), {D and R), and (R, N, and D). Similarly, the possible motions of a cylinder with a generator in contact with a plane would be denoted by {R), {D), (iV and D), {D and R), and {R, JV, and D). 142. Non-Eigid Links.— In the classification shown in article 134, 140 ANALYTICAL MECHANICS [arts. I43-I44 Table in., A. 3, the term linkage was restricted to deformable aggre- gates of rigid parts, but in the classification adopted by Reuleaux the term linkage includes all the partially deformable figures of Table in. Hence Reuleaux calls ropes, belts, chains, and the cylinders on which they wrap, examples of tension pairing, since no motion can be communicated except by pulling. On the same principle the water in a hydraulic press is called by Reuleaux an example of pressure pairings since the desired motion of the piston can only be produced by pres- sure. Thus, these non-rigid or deformable links give example^ of motions which can be communicated in one way only, and not in the reverse way also through a given single link, whereas in the case of rigid links both motions may be transmitted through any single link. Simple cases of these non-rigid links have already been dealt with in articles 135-138. 143. Plane Linkages. — We now commence the treatment of that most important class of kinematic chains consisting solely of rigid links whose whole motion is plane and the relative motion of adjacent links always an angular one. In other words, we have now a number of rigid links in a plane, all their contacts being turning pairs. This is the form of kinematic chain called a linkage ; or, if one link is fixed, a linkwork. The discussion of various types of this class extends to article 157. Inversions. — To represent the relative motion of the links we must reckon the displacements from some frame of reference. Now in any use of a linkage some one link is usually fixed. Hence, the problem of its motions is in that case obviously simplified, if lines in the fixed link are chosen as the frame of reference or axes of co-ordinates. If now, instead of the previous one, a second link is fixed and the motion of the others redrawn with reference to it, we may have a motion apparently quite different from the first, and perhaps, in some respects, really so. The linkage is now said to be inverted. Thus we have for a given linkage first, second, third, etc., inversions, the motions being perhaps apparently very different in each case. For, it should be noticed, that although the inversion of a pair of adjacent links may cause no alteration in the relative motion of that pair, yet it may and generally does alter the motion of the moving link with respect to other bodies. 144. Criterion of Deformability and Rigidity.— If we cpnsider linkages or jointed frames typified by the capital letters V, A, N, and W, it is evident that the first is deformable, the second is rigid, while the third and fourth have motions which are indeterminate, if any one link is fixed and only one displacement is specified. Again, if we have a quadrilateral linkage, first without diagonals, second with a diagonal, and third with both diagonals, it is obvious that we have three examples which are respectively deformable, just rigid, and over-rigid. In the last case we see that there is a redundant link, or one beyond the number necessary to make the arrangement just rigid. We are here almost solely concerned with those arrangements of ^^T. I4S] MECHANISMS 141 fl"kf T'^'^u ^""^ deformable, and that in a determinate manner speci- tiable by the relative position of one pair of adjacent links. It IS of interest, however, to seek an analytical criterion for the deforrnabihty or rigidity of a plane linkage. Thus, following some- what the method of Henrici and TMxviQx {Vectors and Rotors, pp. 183-4, i9o3)> let us consider n points confined to a plane but without other constramt, then they possess two degrees of freedom each or 2« ^" .}■.. '^^^^ ^s' 'liese points coincide with the centres of the eyes of rigid hnks, each link having two such eyes and two only. Then each link will, in general, introduce a constraint and remove one degree of freedom. Thus, if there are m bars fulfilling this general condition, the number of degrees of freedom of the « points is expressed by 2« — m. But a rigid body in plane motion has thi;ee degrees of freedom, viz. two translations and one rotation. Hence if the aggregate of links is just rigid, we have 2«-W = 3 . . . . ... (l). We may now find a relation between n and m depending upon the number of links meeting at an eye. Thus, of the n points or eye centres, let a number h^ each have only one link there, let h^ points each have two links meeting there, let h^ be joints of three links each, and so forth. Then n = h^-lh^-^h:,-\-h^-^h^-\-h^ (2). Also, since each bar has two eyes, we have for the number of bars M = i{//, + 2/i,-f3//3-|-4y4,-f 5^5 + 6/^-1-. . .} (3). Hence, by (2) and (3) in (i) we have |/5,+/,, + ^ + 0-^l-//e-...=3 (4), 2 2 expressing the condition that the frame is just rigid. If the left side of (4) exceeds 3, then the linkage is deformable; if the left side of (4) falls short of 3, then the frame is over-rigid. It must not be supposed, however, that its degrees of freedom have fallen below 3, because the member or link added beyond those necessary for rigidity only forbade a motion which was already forbidden, and consequently introduced no new constraint. Thus equation (4), or cases where the left side has a value differing from 3, must be interpreted with caution, as they only apply to certain assumed possibilities which, though usual, are not universal. For another way of obtaining a criterion of determinate deforma- bility the interested student is referred to S. Dunkerley's Mechanistn, §9- 145. Use of Instantaneous Centres of Rotation. — Linkages are used to derive from one motion another motion of a different kind or magnitude. Hence, we must be able to deal with this derivation or conversion and determine the ratios of the displacements and velocities of the various parts. For this purpose we may begin at the fixed link and pass to those immediately connected to it ; we can thus find the motions of certain points in other links not directly connected to the 142 ANALYTICAL MECHANICS [art. 146 fixed one. Then, knowing the instantaneous coplanar displacements of two such points, we can find their instantaneous centre of rotation, and thus determine the relative displacements of all points in the link in question. The conceptions here involved have already been dealt with in articles 95-97 and loi. The method of applying these principles will become suflficiently clear as the various typical cases are dealt with in order. Where a linkage is used to produce some particular motion of one point, say, e.g., a straight-line motion, a special treatment may be necessary to demonstrate this property Examples — XXXI. 1. Give examples of the contact of rigid pieces or links which illustrate relative motions that are rectilinear, plane, and solid, and also lower and higher pairing. 2. What is meant by tension pairing and what by pressure pairing? Draw some example of a mechanism illustrating one of these connections, and calculate the velocity ratio involved. 3. Define plane linkage and linkwork, also state what is meant by the inversion of a linkage, giving drawings in illustration. 4. Discuss the various possible states of an aggregate of rigid links as to deformability, rigidity, etc., and obtain an analytical criterion for them. 5. What method is available for the determination of the motions of the various points in any moving part of a linkwork ? Illustrate your answer by a diagram. 146. Quadric Linkages.— Let us now consider the relative motions of a plane quadric linkage having joints which admit of rotation only. A typical case is obtained if the links have lengths as 2 and 4 for one pair of opposite sides, 5 and 6 for the other pair, as illustrated in Fig. 56 by KLMN. The instantaneous centres of relative rotation of adjacent links obviously coincides with that of the pin which connects them. Hence we have immediately four of the re- quired centres at the angular points K, L, M, N. It can easily be shown that the instantaneous centre for either pair of opposite sides is the point of intersection- of the remain- ing sides. Take for example the side LM as fixed, then K must move at right angles to KL, whence it follows that the instantaneous centre of KN lies on LK, produced if necessary. Similarly since N must move at right angles to MN, the instantaneous centre of KN lies along MN, produced if necessary. Thus the centre sought is the point P where LK and MN intersect. In precisely the same way it is seen that Q, the intersection Fig. 56. Quadric Linkage and Instantaneous Centres. ART. 147] MECHANISMS 143 of ML and NK, is the instantaneous centre of relative rotation of KL and MN. It is noteworthy that if any three adjacent links be taken the three instantaneous centres of relative rotation of the three pairs, which can be taken from that set of three, all lie on a straight line through the two joints of the three links. Thus, choosing the three links KL, LM, and MN, or a, b, and c, the three pairs are ab, be, and ca, and the three corresponding instantaneous centres are L, M, and Q, all lying on the straight line LM, produced where necessary. The same applies to the other three sets of three adjacent links. Suppose now that LM is the fixed link or frame, then we may call LK the crank since complete rotation is possible to it, MN the lever since it can only move to and fro in a limited arc, while NK may be called the coupler since it couples the crank and lever (see Fig. 57). If, on the understanding of LM fixed, we examine the six instantaneous centres, it is evident that only two, L and M, remain stationary, K moves in a circle with the crank round L, N swings with the lever in a limited circular arc round M, while P and Q describe curves which may be found by the following construction :• — Draw N displaced to N' in the circle round M as centre, K displaced to K' in the circle round L, and making K'N'=KN. Then producing MN' and LK' to their intersec- tion we have P', the new position of P. In other words, we have P', the instantaneous centre for LM and K'N'. It will be found that this locus of P may be a curve of two branches which intersect at M and have points at infinity. It is obviously the space centrode for the motion of KN. In like manner, by taking the .-^^ locus of P as though KN were ^^^^^^^^ fixed, we should obtain the body \ / ^-^^^ centrode for KN moving with LM fixed. We might also find the locus of Q, but it would not represent either a body or space centrode while LM is fixed. 147. Velocity Ratios : Polar Diagrams. — Continuing our sup- position that, in the quadric link- work, LM is fixed as shown in Fig. 57, let us determine the linear and angular velocity ratios for the lever MN = (r and the crank LK = a. Let d, <^, and u represent the instantaneous angular velocities of the coupler KN about P, of the lever MN about M, and of the crank LK about L respec- lively. Also let u and v be the instantaneous hnear velocities of N and K. Then we have, from the figure. C-.-.-- K' Fig. 57. Velocity Ratios of Quadric Linkwork. 144 ANALYTICAL MECHANICS [arts. 148-149 « = 6i.PN=^.MN (1), and z/=6».PK={c-\-d')\ ,,-. (2) for passing the upper point K" in the circle, {a-\-d)1^{b+c)] '^' provided that, as here supposed, c + cj (12), where also a and dm are small Q is far away, and therefore QK = QL, hence the last ratio at the right in (14). Again, by considering the small angles at Q, we have ^_KF^LG._KF^KE ,. d NK, of which KLMN form a parallelogram of sides a and b re- spectively. In the position shown, KRS is a straight line, and it al- ways remains so. For first, the lengths KL, LS, RM, MS are all constant, and second, the two angles ^^^^ ^ ^;:^S marked, viz. KLM and RMS, al- ways remain equal, since KLMN is a parallelogram of fixed sides. Hence, the ratios KL : LS and RM : MS being once equal (when KRS is straight) always remain so, i.e. KRS remains straight. Further, the ratio KR : KS, or say rjs, remains constant, for it is seen to be equal to LM : LS. Thus while the point R traces about K as pole the curve whose polar equation is r=/(^), the point S simul- taneously traces the same curve, and similarly placed about K, but enlarged in the ratio sir. . ^ , It may be noted here that in actual practice for beam engmes Watt's parallel motion was combined with the kind of pantograph just dealt with, so that one point, like E in Fig. 60, being constrained to an approximate straight-line motion, a second point copied it as K would copy R, in Fig. 62, if S were fixed. 154. Peaucellier's Cell.-This eight-part linkwork due to M. Peaucellier was devised in 1864, and solves the problem of drawing rccurately a straight line by geometrical --nst is represented m Fig. 63, and, while the point B describes he circle GB A, the point L describes the straight line HC at right anges to AFGH The link AF is fixed, and the equal link iB rotates about R The two links AD and AE are equal, and finally the four hnks BD, DC, ISO ANALYTICAL MECHANICS [art. 155 CE, and EB are all equal. Hence, by symmetry, ABC: is always a straight line, also B and C always lie on a circle whose centre is at D, thus by the properties of the circle AB.AC=AD'-BD'=aconstant .... (i). Produce AF to H, cutting at G the circle ABG whose centre is F, and Fig. 63. Peahcellter's Cell. let fall from C on AH the perpendicular CH. Then we have by the construction AB/AG=cos BAH = AH/AC. Or AG.AH = AB.AC. Thus by (i) we see that AH is a constant (2). Hence the conditions imposed on the points A, B, C by the linkwork are precisely those which correspond to the accurate description of the straight line CH by C, while B describes the circle of radius FB = FA about the centre F. Thus C and B' show another pair of correspond- ing positions of the tracing points C and B. It is obvious that if A passes inside the rhombus BCDE, the point C still traces a straight line. 155. Hart's Cell. — Let us now consider some of the properties and uses of a linkage in the form of a contra-parallelogram as shown by KLMN in Fig. 64, in which KL=MN and KN = LM. The contra-parallelogram may be regarded as derived from the ordinary parallelogram by folding one half through two right angles about a diagonal. Or, it may be considered as consisting of the inclined sides and diagonals of a symmetrical trapezoid. Thus LN is parallel to KM. To examine the motion and properties of the contra- parallelogram we may draw certain lines which, though not corre- sponding to any material in the links themselves, behave exactly hke the rhombus in Peaucellier's cell (shown in Fig. 63). Thus in Fig. 64 bisect KN at B, LM at C, LN at D, and KM at E. Then, by con- struction and the properties of similar triangles, we see that BDCE is ART. 156] MECHANISMS 151 and always remains a rhombus, its sides being also of constant leneth name y each equal to the half of either of the Hnks K^ or MN luS a^ AB^'the " '' """" '"^"^'^^ ^^^ '' ^ ^'^S^* "- equal links AF and BF, AF being fixed, also draw, CH per- pendicular to AF. Then the six - part linkwork, consisting of AF, FB, and the contra -parallelogram KLMNK, consti- tutes the Hart Cell. It has the property that as B moves in the circle BA of cen- tre F and radius FB, the point C traces ac- curately the straight line CH at right angles to the fixed link AF. That it Fig. 64. Hart's Cell with Central Points. possesses this property is easily seen by comparing in Figs. 63 and 64 the points AFBDCE. In the former figure these points are connected by the eight actual links of the Peaucellier cell, while in the latter figure, showing the Hart cell, beyond the links AF and FB we have the contra-parallelo- L N gram KLMNK which, as already shown, serves the purpose of maintaining the distances re- quired, though there are no links at AD, AE, BD, DC, CE, and EB, as in the PeauceUier ar- rangement. 156. Alternative Propor- tions for Hart's Cell. — The pro- portions and use of the Hart cell just discussed are not the only ones possible. The points A, B, and C might have been taken so as to divide KL, KN, and ML in some other constant ratio instead of being points of bisection as chosen. Thus, let KA/KL=KB/KN = MC/ML=«. Then, it is seen from the pro- perties of similar triangles that ABC are in a straight line parallel to KM and LN as shown in Fig. 65. We can also easily show that AB.AC is constant. L' N' Fig. 65. Hart's Cell with General Ratio. 152 ANALYTICAL MECHANICS [art. 157 Thus, using again the properties of similar triangles, we have AB = ^LN, AC = ( I - «)KM, or AB.AC = «(i-«)KM.LN (3)- But from the figure we have KM = KN'+N'M=^cos<^+acos6l'| ,. and LN = KN'-KL'=3cos^-rtcose/ ^^'' where a = KL=MN,/5 = KN = LM, 6I=LKM, .^=NKM. Also (!!sin^=L'L=N'N = ^sin^ (5). Thus, substituting (4) and (5) in (3), we obtain AB.AC=«(i-«)(<5''-a')=const (6). Thus, as shown for the PeauceUier cell in equation (2) of article 154, if B moves in a circle passing through A, its centre being at F, C moves in a straight line CH at right angles to AF. (See also Figs. 63 and 64.) 157. Parallelogram Linkages. — If we have a parallelogram of sides a and b including an angle 6, the diagonals being c and d, the ordinary expressions for each of the triangles into which the diagonals divide it give c^=ia'-\-b' — 2abco5 0, d'' = a'+b''-2abcos{Tr-6). Thus ^'+(jr' = 2(a"+^')=constant (i). Hence if the sides are equal and the diagonals lie along the co- ordinate axes and are denoted by x and y, we have x^ +y^ = cons tan t, and xdx-\-ydy=o . . (2), or dy/dx=-.x/y (3), which gives a useful relation between the corresponding small changes in the respective diagonals. If we have a succession of rhombuses, the sides of one figure passing for an equal length beyond the crossing to form half of the next figure, we obtain the linkage called i/ie lazy-tongs. If the succession of figures lies along the axis of x, which accordingly coincides with one diagonal of each, then obviously a given change dy in the y diagonal results in changes of magnitude dx in each of the x diagonals, which lie end to end. Hence at a distance of m rhombuses from the origin we shall have a displacement -\-mdx consequent upon the change —dy in any one of the J diagonals. This we might express by writing dX=mdx . .... (4), or X=mx. A familiar example of a parallelogram linkwork is afforded by the parallel cranks of coupled driving wheels on locomotives. Examples— XXXIII I. Apply the graphical criterion for a double-crank linkwork to a linkage of sides 2, 4, 5, and 6. Will it still be a double-crank linkwork (i) if each side is increased by 2, and (ii) if each side is doubled ? ARTS. 158-159] MECHANISMS " ^'^vfo^lTffel^Tar^'' ^"' ^^^^^''^'^ ''^ P™P"'>' °f -Py-^ any 3- Explain carefully the linkwork called Peaucellier'.; r^ll ,„^ .t,„ .v . . ^r^\^^'^'t^-'- ^ straight line while anothe describes a arclV °"' '• fclr'nnks'' ^'"^"'^ °' Peaucelher's cell is attat^d'^y Harttcell with ^' ^shown."" P"'' °^ ^"^y-'°"g=' =^"d find its velocity ratio for the position 158. Slider Crank Chain : Instantaneous Centres.-This important kinematic Cham IS derived from the quadric linkage by substituting a shdmgfair for one of the four turning pairs. As each of the four links IS fixed in succession it produces four distinct mechanisms, which we shall deal with presently. We may, however, first note with advantage the positions of the six instantaneous centres of the relative motions of the inks in this slider chain. Thus, using the same notation as in article 146 for the quadric linkage, and the same methods as there employed, we obtain the results shown in Fig. 66. In this figure the links d, a, b, and c are united by turning pairs at K, L, and M ^ respectively, the corresponding in- / 1 stantaneous centres being obvi- /' \ ously their centres of junction / | denoted by these letters. The / ' link or block c shown by M slides / ! on the link or bar d denoted by '^T ~;;^^\ a ' KM ; hence the instantaneous cen- i y^ ^'^^^ 'M tre for the relative motion of c and y ^^ ) >J,-- 1 d is on the line drawn through M '^ '^ i at right angles to KM, but is situ- I ated at infinity in either direction \ along that line. It is accordingly ' indicated on the diagram by N at » + 00. Take next the centre for \ b and d. It is evidently at P, the N|czi^°° intersection of KL with the line Fig. 66. Slider Crank Chain and through M at right angles to KM. Instantaneous Centres. For L must move at right angles to KL and the centre lie along KL, while M moves along KM, and the centre accordingly lies along the perpendicular to KM at M. Finally, consider the centre of instantaneous rotation of a with respect to c. If we imagine the block c fixed, K must move along KM, so the centre in question lies along the perpendicular to KM at K, while L must move at right angles to ML, so the centre lies along ML, produced if necessary. The instantaneous centre sought is therefore as shown by Q. Thus of the six centres five, K, L, M, *P, and Q may usually be shown in the diagram, but the sixth, N, corresponding to the shding pair of the block c on the bar d, is always inaccessible. 1 59. Velocity Eatios obtained Analytically. — We shall now con- 154 ANALYTICAL MECHANICS [art. i6o sider what may be called the first inversion of the slider crank chain, in which the link or bar d is fixed, and so becomes the frame. The resulting mechanism then illustrates the case of the direct-acting engine, the block c ■'~--X ,Y becoming the cross- " ' head, b the coupler or connecting rod, r and a the crank. The important relatidns as to displacement and velocity between the crank and coupler will be first treated analytically and afterwards graphically. Referring to Fig. 67, let the crank KL, of length a, make at a given instant the angle d with the frame KOMX, the inclination of the coupler LM of length b=na being at the same instant <^. Let fall the perpendicular LL' from L to the frame KOX, where O denotes the central point in the traverse or stroke of M, and denote OM by x. Then we have by construction L'L/a = s\nd=n sin 4>, whence n cos <^= \/V^— sm^ . (i). Also ic=KM-KO = KM-LM Fig. 67. Analytical Diagram for Direct-Acting Engine. ncos4>= ijn^ — sm^6 . ■ ic=KM-KO = KM- C'^a'cosd ^na^QS.^^na j . . . (2). Thus (i) in (2) gives x = a{cos6 — n+ Jn' — sin'O) . . . (3). Now differentiate (3) with respect to time, writing v for the linear speed of M and w for the angular speed of L. We thus obtain «=-«a.sine(i+-^=^) . . . (4). And if n' is large compared with unity, as is usually the case, this reduces to the approximate formula v—usind(i+ — j. ... . . (s), in which ti is the linear speed of the crank pin L without regard to sign. Referring to (3), we see that a cos 6 would be the displacement for simple harmonic motion, so that the other terms a{ J»' — s'm'9—n) are corrections for the obliquity of the connecting rod or coupler. Of course, if n is -jj these corrections reduce to zero. And quite low values of n make the corrections small ; thus if n is only 5, even then Jn' — sin'O — n never exceeds— cii. 160. Velocity Ratios Graphically Treated. — Referring now to Fig. 68, we will treat the same problem graphically. Thus, the link d being the frame as before, and b the coupler, we produce the line KL ART. l6l] MECHANISMS 155 of the crank and draw from M a perpendicular MP to KM, thus finding at their intersection P the instantaneous centre for b'^ motion with respect to d. Thus, the ratio of M's velocity to that of L or v\u is given by the ratio PM/PL. Produce ML to meet at V the line KV drawn perpendicular to KM. Then the triangles PLM and KLV are similar. Hence (6). vK^ Fig. 68. Graphic Treatment for Direct- Acting Engine. y_PM_KV «~PL~KL Accordingly if u is constant and represented to scale by KL, v, at the instant to which th e diagram applies, is represented to the same scale by KV. As it is inconvenient to have Ay any variable angle like that shown between the lengths KL and KV, we may describe about K as centrewith radiusKV the arc VR, thus trans- ferring V to R. It is thus seen that R is one point of the polar dia- gram of velocities of M. Another method of graphically exhibiting the velocity of M is to drawVS parallel to KM, cutting MP in S. Then c ^ :■ ■ i,vi, fi,» S is one point of a cartesian diagram of velocities m which the abscissae are the displacements of M and the ordinates the velocities of M. 161 Other Inversions of the Slider Crank.— In articles 159 and 160 we have considered what may be called the first inversion of the s ider Trank chain, namely, that in which the bar d is fixed, the resulting mechanism being that utilised in the direct-acting engine. Three new Sfchanisms res'ult if each of the other links is msuccessio^fixed^ Thus if the link b is fixed (see Figs. 66, 67, or 68), we obtain tne mechanism used in the oscillating engine, in which a is the ciank as Se^ S now the piston rod, and c is the osciUatmg cylinder IJain if th^ link « is fixed, we obtain the Wh.tworth quick-retuin motttin wSch the link b may rotate -^-j^' -'j'^^^-H^m ^n variable motion in complete rotation xi b>a, but ^^.th a swinging '"iTstW^Tfixed we have the mechanism used in the pendulum ThPlinkThere takes the form of steam and water cylinders 156 ANALYTICAL MECHANICS [arts. 162-163 former point and swings about the junction of b and c. It is this swinging ox pendulum motion of the link b which gives its name to the mechanism. The displacement and velocity ratios in any of these mechanisms yielded by the various inversions of the slider crank may be dealt with as already explained in articles 159 and 160 for the first inversion. The fuller discussion of the subject belongs rather to special treatises such as the classic by Reuleaux (translated by A. B. W. Kennedy), or the excellent and more recent Kinematics of Machines by R. J. Durley, which should be consulted by those wishing for further information. 162. Screw Pairs. — The last example of lower pairing that we shall consider is afforded by screw pairs, in which the relative motion of the parts so paired is obviously a twist &s defined in article 130. Kinematic chains involving screw pairs frequently occur in machines. Cases of special interest are presented when two screw threads of different pitches,/ and q say, are cut upon the same cylinder, each such thread engaging an appropriate nut, the nuts sliding without turning in a frame which also carries the screw and allows it to turn without sliding. We have thus a chain of four links involving two screw pairs, two sliding pairs, and one turning pair. Hence on turning the screw, while one nut advances endwise the distance p the other advances the distance q, so that the relative motion of the nuts is {p~g). An example of this character is often seen in the screw couplings of railway carriages, in which q— —p, or the screw threads are of the same pitch numerically but of opposite hands. If we extend our survey so as to include fluid links, we may note as further examples of screw pairs — (i) ships' screw propellers, (ii) turbine water wheels, (iii) windmills, (iv) the rifle barrel and its pro- jectile, and (v) steam turbine engines. For in each of these cases we have screw surfaces in use whose relative motions are accordingly of the type called a twist. In case (iv) the screw pair consists of the projectile and the rifled bore of the laarrel, the fluid link being the gas which drives the projectile endwise. In each of the other cases the fluid link assumes the form of one surface of the screw pair so as to fit the other surface, which is of solid material. 163. Higher Pairing.— We now conclude the treatment of kinematic chains by a brief reference to examples of higher pairing, the contact between the elements here allowing them greater freedom of relative motion because they touch only along lines or points as already mentioned in articles T39-141. Taking first examples of plane motion, the case of toothed wheels formed from right circular cylinders needs consideration. These are called by engineers spur wheels. It is shown in technical treatises that if the teeth of both wheels in gear are involutes of the circle of constant obliquity, or have their faces formed of appropriate epicycloidal curves, and their flanks of corresponding hypocycloidal curves, then the velocity ratio of the pair is constant and inversely as the radii of certain circles known as the pitch circles. These circles are concentric with ART. 163] MECHANISMS 157 the wheels, and lie rather nearer the tips of the teeth than their root. They are purely geometrical lines, but are in contact at the pitch point when the wheels are running in gear, and ih& pitch of each wheel (that is, sura of tooth and space) is measured on this circle, and is, of course, of the same value in each of the wheels gearing together. Hence the radii of pitch circles of any two wheels which are to gear together are proportional to their respective numbers of teeth. We accordingly have the working formula for relative speeds : — Angular velocity ratio of gearing wheels is the inverse ratio of their numbers of teeth. It is evident that this still applies approximately if the connection is not by direct meshing and contact of the teeth but by the inter- vention of a chain, as is usual in pedal bicycles. But the conditions for accurate constancy of velocity ratio may not be the same as before, and are perhaps rarely fulfilled in either case in actual practice. Other examples among mechanisms of higher pairing and plane motion are afforded by cams, ratchets, locks, and escapements. The case of the possible motion of an inclined ladder is an example of plane motion and higher pairing, and is easily dealt with by reference to its instantaneous centre. Perhaps the commonest examples of higher pairing in solid motion are given by (i) bevel wheels^ in which the teeth are formed on cones, their axes being inclined and intersecting ; and (ii) the worm and wheel, in which the axes are at right angles and not intersecting. The worm is simply a short screw, and the worm wheel resembles a spur wheel, but has the teeth set obliquely (and often hollowed out also) to suit the worm with which it gears. It is evident that the velocity ratio for bevel wheels is simply the inverse ratio of their teeth numbers, while that for the worm and wheel is the inverse ratio of teeth number and number of threads. Thus if the wheel has fifty teeth and the worm but a single thread, the worm turns round fifty times to the wheel's once. Many other examples of both higher and lower pairing are dealt with in treatises on mechanisms, but are beyond the scope of the present work. Students requiring further information on this subject may with advantage refer to Dunkerley's Mechanism. Examples — XXXIV. I. Explain by a diagram what is a slider crank chain, and find its instan- taneous centres. , . , , , r ..t. -^ Sketch a shder crank chain with a coupler four times the length of the crank, and obtain an expression for its velocity ratio. What does this approximate to when the coupler is much longer ? 3. Obtain a graph for the velocity ratio of a slider crank chain whose coupler is six cranks long. , . , . . , 4. What forms are assumed by the slider crank in other inversions ? 5. Give several familiar examples of screw pairs, and state the velocity ratios which hold. <• , . , 6. Enumerate and discuss several examples of higher pairing. 158 ANALYTICAL MECHANICS [art. 164 CHAPTER X STRAINS 164. Simple Strains. — In dealing with elastic bodies the terms stress and strain were introduced in 1854 by the late Professor Rankine and have been found very useful. In the following year Kelvin modified the original usage as regards stress, and gave ^ definitions of both terms, that for strain being as follows : — Definition. — 'A strain is any definite alteration of form or dimen- sions experienced by a solid.' It is easily seen that a slightly modified form of this definition will allow us to apply the term to the compression or dilation of a fluid. The term has been so used by Kelvin and Tait in their Natural Philosophy (vol. i. p. 116, 1890), and we shall follow that precedent here. It is obvious that in the case of a fluid we cannot so easily identify the individual particles in their primitive and strained positions, but we can note the volume change, and that is all we require. Indeed for our present purpose we may say that a strain is a change in the dimensions of any given figure. The mathematical theory of elasticity and even that of strains, which forms the kineraatical preliminary to it, are both beyond the scope of this work. The subject of strains will accordingly be con- sidered here at first in a very restricted form, the lines of a fuller treatment being just indicated later. Imagine a unit cube with its edges parallel to the co-ordinate axes and centre at the origin. Now let all lines in the cube parallel to the axis of X be elongated by the very small amount a, so that the faces parallel to the yz plane each move normally from it by the distance fl/2, remaining parallel to their former positions. Also let it be under- stood that this elongation of lines parallel to the x axis occurs pror portionally throughout the length of each line as well as uniformly over the yz faces. Let a similar uniform and proportional small elongation of amount e occur parallel to the y edges, and finally one of amount / parallel to the z edges. Then, if the primitive position of a point P in the unstrained cube has co-ordinates {x, y, z), and shifts to P' with co-ordinates {x', y', z) in consequence of the strain, the opera- tions we have described may be represented by the equations x' — x-^ax^ y'-y=eyi (l). z'—z= iz J 1 Encyclopaedia BritannUa, ninth edition, vii. p. 819. ART. 165] STRAINS 159 Thus the first three vowels of the alphabet here denote t\\Q fractional elongations (or briefly elongations) occurring parallel to the axes x, y, and z respectively. The form of these equations shows that the elongation occurs propor- tionally along each line, and also that each face moves normally to its own plane and remains parallel to itself, for the change of x depends on the X co-ordinate alone and not on y or z, and so for the other co-ordinates. As will be seen more fully later, the above strain is of the type called homogeneous, because it is all over alike. It is also called a. pure strain, that is, devoid of rotation of the body as a whole, because the three diameters of the cube elongate simply without rotation. Lines inclined to the diameters may change their inclination even in this case. Further, since the directions or principal axes of elongations are parallel to the co-ordinate axes, the strain as expressed by the equations ( I ) assumes a very simple analytical form, and can be very easily dealt with. Fractional Change of Volume. — If we now consider a parallelepiped of edges .v,j»', and 5, in the unstrained state, we see that by (i) it has edges (i+«)"'*^i (i+^)>'j and {i-\-i)z in the strained state; hence its strained volume is {i-{-a){i -\-e){i-\-i)xyz = {i -^a+e-ir i)xyz nearly, if the elonga- tions are so small that their products are negligible, which we shall always suppose to be the case unless otherwise stated. Hence, to this approximation, 8, the fractional change of volume, is given by the sum of the elongations, or &=a+e+i . . . (2). Thus, the condition for no change of volume is obviously a+e+i=o (zrt). 165. Typical Pure Strains. — A few strains that it is important to notice are collected in Table iv. as given in article 68 of the writer's Sound. Table IV. Typical Pure Strains. Elongations parallel TO AXES OF Cases. ' y - I a e i 2 a i c 4 d d d 5 d d 6 e - e 7 a -i -i Strains. General Pure Strain about co-ordi- nate axes. Plane Strain. Simple Elongation. Uniform Dilatation, S = 3(/. Shape unaltered. Uniform Plane Strain. Simple Shear. \ 8 = Volume unchanged) X — e-{-e) = 2e Elongation a with lateral contrac- tions /, or axial strain. i6o ANALYTICAL MECHANICS [arts. 166-167 The first strain in Table iv. is the general pure strain with elonga- tions parallel to the axes of co-ordinates, the second the same reduced to two dimensions, the third an elongation merely. All these have involved change of both size and shape ; the fourth is a case in which, though the size is increased, the shape remains unaltered. The fifth is but a simpler case of the second. The sixth, called a simple shear, is of special importance, since it presents the case of a change of shape without change of volume, the e being supposed small. Of course, the negative elongation denoted by —e in the 2 column represents a con- traction parallel to the z axis and equal in amount to the elongation occurring parallel to the x axis. The seventh and last case in the table is also specially important, since it is what occurs when certain solids are pulled parallel to one axis, the other two axes being unacted upon. Thus, as shown by the symbols, an endwise elongation a is accompanied by— /, — ?', i.e. lateral contractions, in the two perpendicular directions. The ratio of i to a in these cases is called Poissoris ratio, and will here be denoted by e>i, the semi-axes of the ellipsoid in order of decreasing magnitude are i+a, i-fs, and i-f «, along the axes of ^, y, and z say. Then, along the axis of x the elongation is a maximum, along that of s a minimum, while along the axis ol y the elongation is a minimum for all directions in the xy plane, but a maxi- murn for all directions in the j'.' plane. A contraction is to be counted as a negative elongation, and the above statement taken m the algebraic '^"Tf two of the elongations are equal, the eUipsoid becomes an ellipsoid of revolution, i.e. a spheroid, oblate, or prolate. Jf/ " tfiree elongations are equal, it becomes a sphere, and the strain is a uniform dilation or contraction. 175. Analytical Representation of Homogeneous Strams.-Let us now express a homogeneous strain by a set of equations. Take O, the 1 68 ANALYTICAL MECHANICS [art. 175 origin of cartesian axes, at a point of the body that remains unmoved by the strain. Let x, y, z be the co-ordinates of any point P before the strain, and x, y' , z' those of P', its new position in consequence of the strain, which we will suppose to be as general as possible subject to its being homogeneous. Now we have seen that any three perpendicular diameters of a sphere in the primitive state become three conjugate diameters of the strain ellipsoid, and are consequently in general changed in length and in mutual inclination. Hence to fully express a homo- geneous strain we need to indicate what becomes of three lines not originally coplanar. For simplicity's sake we will take these along the co-ordinate axes and of unit length. Obviously each line can suffer a change both of length and of inclination, and the latter needs two angles to specify it. We accordingly need three constants to state what happens to each of our three unit lines, i.e. nine constants in all. We may conveniently take these as constants expressing the displacements of the ends of the three unit lines parallel to each of the co-ordinate axes respectively. Thus, let the end of the unit line from the origin along the axis of x shift by a, d, and g parallel to the axes x, y, and z respectively. Let the end of the unit line along the y axis have like shifts 6, e, and A. Finally, let the end of the unit line along the 2 axis have shifts <-, /, and i. Then, if we multiply the shifts for a unit line by the value of a co- ordinate in the same direction, X say, we should obtain the shifts for the end of a line of original length x. This there- fore applies to the x co-ordinate of P, similar remarks holding for the y and z co-ordinates. But the strained values and positions of these three co- ordinates meeting in O, and originally x, y, and z, are . the three adjacent edges of the oblique parallelepiped whose opposite corner is P', the strained position of P. The shifts from P to P' parallel to the fixed co-ordinate axes are accordingly given by the expressions x' — x=ax+ly-\-c::^ y'-y=dx+ey+/zi (14). z'—z = gx+Ay+iz] It is seen that the coefificients on the right side are the first nine letters of the alphabet taken in order. We also see from (14) that the shifts of any point in a body experiencing a homogeneous strain are linear functions of its co-ordinates. The equations can easily be verified by reference to Fig. 73, which shows the meaning of the nine constants. Y y 1 R v1 6 p''>r > P Q z y ja yiK-A z Fig. 73. Nine Constants of ilOMOGENEOUS STRAIN. ART. 176] STRAINS 169 and represents by PQ, QR, and RP' the values of x' — x, y'—j, and z' — a respectively. Of the nine constants in (14), it should be noted that the vowels a, e, i denote the elongations parallel to the axes of x, y, z, as in Table iv. and elsewhere. The consonants b, c, d, f, g, h, on the other hand, show the amounts of the shears reckoned as the relative slidings of the three pairs of planes to which the co-ordinate axes are normal, one of each of these pairs of planes having slidings in the two directions parallel to its edges. Thus, referring again to Fig. 73, d shows the amount of the shear suffered by the body by the sliding of the plane Px parallel to the axis of y, i.e. d is the relative slide parallel toji', of planes parallel to yz, per unit distance apart along the axis of x. Similarly g is the amount of the shear reckoned as the sliding of the same planes parallel to r, and so on for the other four consonants, as may be seen from the figure. 176. Rotation in Homogeneous Strains.— Let us now inquire if the homogeneous strain expressed by (14) involves any rotation ; if so, what modification in it would correspond to the elimination of that rotation, and so reduce it to a pure strain. To deal generally with the problem requires the use of solid analytical geometry and the discussion of the result- ing cubic equation. But the following simple geometrical treatment gives a useful prelim- inary insight into the matter. Consider a cube of the un- strained substance, and take co- ordinate axes parallel to its edges with origin at its centre, also let its sides be of length two units. Take a section of this cube in the xy plane as represented in Fig. 74. Hence, since z—o in the plane of the diagram, the equations (14) reduce to Fig. 74. Unequal Slides involve Rotation. x —x^ax-^-hy'X y'-y=:idx-\ey\ (iS)- z —o=gx-\-hy] But as the constants a and e express elongations without rotation along the rectangular axes of x and y respectively we may omit them^ from our present consideration. Further, smce the components of will only involve a rotation of our sectional plane about the axes ol y and X respectively, and be invisible in the diagrani, we ignore them also, and confine our attention to the constants which may express rotation I70 ANALYTICAL MECHANICS [art. 177 in the plane of the diagram that is about O (or about the axis OZ). Hence for this plane case (15) finally reduces to x —x—by a.ndy—y=dx (16). Thus, following the equations (16), the square PQRS, as shown in Fig. 74, is changed by the strain to the parallelogram P'Q'R'S'. And it is seen that not only are the diagonals of the square PQ and RS each rotated, but that they are 6oiA rotated in the same direction, viz. counter- clockwise, if d>b. Now, if the original square had been subject to unequal elongations parallel to the axes of .* andj', the diagonals 'would have been each rotated, but in opposite directions ; P and S approaching and P and R separating, or vice versa. Thus we see that opposite rotations of certain lines originally perpendicular are consistent with a pure strain. But rotations of such lines in the same way are not con- sistent with a pure strain, for they clearly involve a rotation on the whole in the direction in question. 177. Conditions for Pure Strain. — If now (/and b interchange the values assigned to them in the diagram (Fig. 74) ; or, keeping the same values, the positions of these consonants in equations (16) are inter- changed ; then, in either of these equivalent cases, the pure strain in- volved is the same in character and magnitude as before, and the rotation is the same in numerical value, but reversed in sign. Hence, if d and b have the same values, the rotation, if any, is still reversed by their interchange. But since the interchange of equals has no effect, there cannot be in that case any rotation to reverse. This could also be easily seen by drawing d equal to b on Fig. 74, when obviously the diagonal PQ would be simply lengthened without rota- tion, and the diagonal RS shortened only, also without rotation. Referring again to equations (14) of article 175, we have now shown that the equality of d and b means no rotation about OZ of the sections parallel to the plane of XY. Similarly, therefore, the equality of g and c would mean no rotation about OY of the section parallel to the plane of ZX. Finally, the equality of h and/ would correspond to no rotation about OX of the planes parallel to YZ. Thus, with all the three equalities fulfilled, we have a strain devoid of rotation. We may accordingly write as the cotiditions for a pure strain d=b, g=c, axiA h=f (17), the pure strain being itself analytically expressed by x' — X — ax -\-by+cz^ y-~y = 6x+ey+fi\- (18). :,' — z=cx+/y+iz] Thus, the ni/ie constants for the homogeneous strain (equation (14) of article 175) being reduced by three on introduction of the conditions for a pure strain of equation (17), leave us the six constants of equation (18). And it may easily be seen that six consonants are necessary and sufficient to specify any quite general pure strain, since three constants are needed for the three principal elongations, and three more to define the principal axes about which they occur. For, with respect to the Al^T. 178] STRAINS ,71 co-ordinate axes, two angles give one principal axis of elongation, and a third angle then suffices to fix the other two principal axes of elongation since all three are mutually perpendicular. 178. Pure Strain analytically derived from Homogeneous Strain —Let us now derive the conditions for a pure strain in a more formal analytical manner. In the primitive state, let x, y, z be the co-ordinates ot a point P distant r from the origin, and, by the strain, let this become P of co-ordinates x, /, z distant r' from the origin, but without angular displacement of the line OP, which accordingly suffers elongation only in the ratio r : r'= i : A say. Then we have x'lx=y'ly — z'lz=k (15). Also, both before and after the strain, the line OP, or OP', has direction cosines xjr.yjr, zjr (20); or, the same letters with accents all through. If we were to write x'=x\, etc., from (19) in equations (14) of article 175, we should obtain three linear equations, and on eliminating from them the two ratios of x, y, and z we should leave a single equation, namely, a cubic in A, which must accordingly have three roots. Of these roots one must be real, and the other two may be both real or both imaginary. But, for our present purpose, it is unnecessary to write these equations and derive the cubic. We need only to take the case in which the three roots are real, Aj, A,, A, say. These correspond to three directions along which elongation free from rotation occurs, and we shall further suppose them to be mutually perpendicular, points on each line being denoted by x,y, z, and r, with subscripts i, 2, and 3 like the A's. We have accordingly to determine the condition that the roots should be real and correspond to mutually perpendicular lines. Then, making in equation (14) the substitutions a' = A.v, etc., for two of the Unes, and writing a for (t-|-«), € for (i-|-f), and i for (i-|-/), we find Aj.T , = axj -\-l>yi + czA X.^, = dx,+^y^+fz.A . . . .(21), X„z^=gx.,+/iy., + iz.,] and A3A-S = a.v3 + /n'3 -f cz^ 1 A,^j'3=^.r,-|-£)'3-f-/:3 - ■ • • -(22). -^sSs =gX:, + ^yi + '^='.j Now multiply the three equations of (21) by .\\,y,, and ^3 respectively and add the results. From this sum subtract that obtained by multi- plying the three equations of (22) by -v,, y.,, and z, and adding. We thus find as the difference of these two sums (A-/)(y,z,-y,z,) + {c-g){z,.v,-z,x.;) + {cl-l'){x.j,-x,j',) ^(!^^.-K){--^--2x^+y'.y^+"2Zz)- ■ (23)- _ But, remembering that the cosine of the angle between two lines is the sum of the products of their corresponding direction cosines, and 172 ANALYTICAL MECHANICS [art. 179 noting (20), we see that the condition for perpendicularity of the lines r^ and r^ is x.2X^+y,yi+ZiZ^ = o (24). Hence the right side of (23) vanishes, and consequently the left side also. The conditions that r^ is perpendicular to each of the Unes r^ and r, must also be introduced, and may be written XiX3+y,ys+ZiZi = oj From these two equations (25), by the ordinary algebraic elimination, we obtain *• _ ->'' — ^-1 , . (26). y^z^—yiZ^ z^x^—z^x^ x^y^—Xsy^ Substituting (24) and (26) in (23) we have {A-/)x,+{c-g)y, + {d-6)z, = o . . . . (27). But, since the order of subscripts is indifferent, this relation must hold for the co-ordinates of points on each of the three mutually rectangular lines along which elongation occurs free from angular dis- placement. This can be true only when each of the coefficients of x,.y, and z vanish. We accordingly find as the condition for the reduction of a homogeneous to a pure strain h=f, c=g, and d—b '. (28). And this agrees with (17) of article 177, and reduces the expression of the strain to the form shown in (18), requiring six constants only. 179. Pure Strain along Co-ordinate Axes. — Of the six constants for a pure strain, we have seen that three were required to define the principal axes of elongation. Hence, if these are chosen as the co-ordinate axes, the corresponding defining constants disappear, and therefore, in order to fully specify the strain, we then need only the three constants which express the principal elongations. We have thus returned to the simple case with which we began in article 164 as expressed in equation (i). This result may be deduced analytically also from the equations of article 178. Thus, if we put r^, ;-,, r^ along the axes of x, y, and z respectively, we have yi=Zi = o, x, = z^ = o, A-3=j3 = o. Then by regarding equations (21), (22), and (28), it may be seen that d=g=6=h=c=/=o and We may accordingly compactly summarise, as in Table v., the characteristics of the homogeneous and pure strains already noticed, and may refer to any one of these strains by simply quoting the corresponding set of co-efficients. d=g=b=h=c=f=o \ . . ART. 1 80] STRAINS Table V. Coefficients for Homogeneous Strains. 173 1 The Body of the Table shows the coefficients of the co-ordinates General Homogeneous Strain General Pure Strain Pure Strain along the co- ORDINATE Axes X, y, AND z TO EXPRESS— -v, y, z. -r, y, z. X, y, z. and z - z. a b c d e f g h i a b b e c f c f a £ ;■ Examples- ■XXXVI. •I. Define a homogeneous strain, and enumerate some of its properties. A bar of india-rubber about three inches long and half an inch square is in turn pulled, bent, and twisted in the fingers ; describe the strains in each case, showing by sketches what becomes of straight lines, circles, etc., drawn in the substance in its unstrained state. 2. Show that a homogeneous strain can be specified by stating the figure produced by it from a sphere in the primitive body. 3. Define strain ellipsoid, and explain how it is derived, illustrating your answer by some simple examples. 4. Represent a homogeneous strain by a set of equations involving nine constants, and illustrate your answer by a sketch showing the meanings of each constant. 5. Show that a homogeneous strain of the most general type involves rotations in addition to a change in dimensions, and simplify the expressions for the strain so as to remove the rotations. 6. Represent by equations homogeneous strains requiring for their specifica- tion 9, 6, and 3 constants respectively, and show by diagrams and descriptions what each class of equations really denotes. 7 Show analytically that the nine constants of a homogeneous strain reduce to six if rotations are absent, and to three if the co-ordinate axes are tak^n along the principal elongations. 180. Equation of the Strain Ellipsoid.— The principal axes of elongation being now taken as the co-ordinate axes, the equation of the strain ellipsoid assumes in consequence a simple form. Thus if, in the primitive figure, we take a sphere of unit radius, we may denote it by the equation , ' .v=+r+2 =1 (3°)- Then, on subjecting it to the strains of elongations a, e, t, this sphere becomes the strain ellipsoid, whose equation is y'' (31). or a^'^?'^!.' of semi-axes a, ., ., which equal {i+a), (i+e), and (1+/) JT^Pfctively^ By giving to a, e, and i any of the values shown in Table iv. of 174 ANALYTICAL -MECHANICS [ART. i8i article 165, we obtain the strain ellipsoid corresponding to the particular type of strain in question. Thus for case i we have the general type, the ellipsoid having three unequal axes as shown in (31). In case 2 only two of the axes are changed from their original unit values ; in case 3 only one of them is changed. For case 4 the ellipsoid is a sphere. For case 5 the sections parallel to the zx plane are circles, the semi-axes along y remaining of its original unit value. For the simple shear shown in case 6 the axis of y remains unchanged, that of X has the elongation e, and that of z the equal contraction, or negative elongation —e. Finally, in case 7 we have an elongation a along the x axis, and contractions i along the, other two axes. Thus for cases 3, 5, and 7 we have ellipsoids of revolution. 181. Cone of Given Constant Elongation. — Let us now consider the possible directions in which the elongation has a given constant value, say i : A. These will evidently be given by the directions from the origin to the intersection of a sphere of radius A., with the strain ellipsoid of semi-axes a, e, and t. Hence we may write the equation of this sphere in the form x^ y^ z^ F+x^+r=^ (33)- Subtracting (31) from this we obtain, for the required locus, the equation '*(p-J)+/(?-?)+-(f-?)=° ■ • <^* or Ax" + By"" -^ Cz- —o (34). These last two equations represent 2i general comc'A surface with vertex at the origin. Various special cases need notice. Case I. — Let a>A>e = t. The ellipsoid is then one of revolution about the axis of x, and from (33) we see that the locus becomes Ax^-B{f+z-^)=o (35), which is a right circular cone about the axis of x, as might have been anticipated on geometrical grounds. If we now reduce A so as to equal e and 1, it is evident that the cone shrinks to two planes coincident with the plane of yz and represented by the equation ■^' = ° (3S4 Case II. — Let a>A=£>i. The ellipsoid is now general, and the elongation A equals the medium principal elongation. Then we see from (33) that the locus reduces to Ax'-Cz'' = o (36). That is, the cone reduces to two planes intersecting on the axis of y. But, since these planes are also sections of the sphere of radius A, we see that they are the tivo central circular sections of the ellipsoid. Case III. — Let a.={\\e\ A = £=i,and 1 = 1— e, ebeing very small. Then this gives us the simple shear of elongations e in the plane of zx. Referring again to (33), we see that the locus now reduces to '^■'-2'=° (37). ART. 182] STRAINS 175 which represents two planes intersecting each other perpendicularly on the axis of y, and inclined at angles of 45° to the axes of z and x. Further, since these planes are the intersections of a sphere of radius A=i, they are the two central circular sections of the ellipsoid and of Xheh primitive size. And this fact, of no distortion caused by the strain, holds for all planes parallel lo those of (37). 182. Shear Ellipsoid derived by Slidings.— But we have seen in articles 170 and 171 that a simple shear may be viewed as a pro- portionate sliding of undistorted planes parallel to each other. Hence the strain ellipsoid representing a simple, shear should be capable of derivation from the primitive sphere by this method of sliding, and we may now easily see that this is the case. As a preUminary, let us note that all the sections of the ellipsoid parallel to the central circular sections are also circular, since all parallel sections of an ellipsoid are similar. It remains then to show that the circular sections of the sphere and ellipsoid by a given plane are equal, and that the same amount of shear suffices to slide any such section of the sphere to the position it must occupy as the corresponding section of the ellipsoid. To illustrate these points clearly in a diagram it is desirable to deal with a finite shear of elon- gational ratios t, i, and i/e along the axes A, J, r l • •.• u and z respectively. A section m the zx plane of the primitive sphere and the strain ellipsoid is shown in Fig. 75. In this figure the central circular sections are shown by EOF and GOH Draw OT perpendicular to EOF and meeting the circle in T, then draw through J the line JK parallel to EOF and tangential to the circle at J and to the ellipse at K, also jom KO and produce to K. Then KOK' and EOF are conjugate diameters Hence OJ and OK bisect in the sphere and ellipse respectively all chords parallel to EOF one of the central circular sections. Thus the amount of the shear when viewed as a sliding parallel to EOF as ^^own by the arrows is measured by JK/OJ. Draw Aa, and B^. parallel to EOF and hrSigh the ext^reinities of the major and minor semi-axes of the ellipse^ ThenV see that the primitive lines .,0^. must -J^i^^/l°-kwise o their final positions AOB through the angle A0«: or EO^i, Thus the Fig. 75- Strained Ellipsoid derived BY Slidings. 176 ANALYTICAL MECHANICS [art. 183 sliding of the parallel undistorted planes involves a rotation and derives A from ai and B from b-, ; whereas/if the same strain ellipsoid is derived from the sphere by the elongations i : « parallel to the axis of x and the contraction e : i parallel to the axis of z, the A is derived from a and B from b, the strain being pure, that is, devoid of rotation. It should be noted that in either case lines parallel to the axis of y perpendicular to the plane of the figure suffer no elongation or contraction. Thus, for the pure strain, unit lengths along the axes of x, y, and z change to lengths e, I, and i/e respectively, these three rectangular axes suffering no angular displacements. Whereas for the shear when occurring as a system of slidings parallel to EOF, only the axis of y remains without angular displacement, and is the axis about which the lines Ooj and Ob^ rotate to the final positions OA and OB. Of course, the slidings might occur parallel to the other circular section GOH, as shown by the dotted arrows, in which case the points A and B would be derived from a^ and b^ respectively, and the rotation involved is equal and opposite to that in the former case. Thus, if a diametral section, like EOF, of the primitive sphere is maintained at rest, the shear produced by a set of slidings involves a rotation in the sense of those slidings in addition to the pure strain as expressed by the elongation ratios. . If, on the other hand, a point on the surface of the sphere, say J, is kept at rest while a shear occurs by these slidings, then we have a shift of the centre besides the above rotation and the pure strain. 183. Analytical Treatment of Shear Ellipsoid. — In the preceding article certain relations between the circle and the ellipse were referred to in general terms and without any formal proof. Some of them rest on well-known properties and need no further proof, others requiring proof may be treated in various ways. The ordinary cartesian treat- ment will be outlined here and the necessary quantitative relations established. The primitive sphere of unit radius and the strain elUpsoid of semi- axes e, I, and i/e have for their sections in the zx plane the respective equations ^"'+^' = 1 (38), and 6V-H.r7£^=:i (39). The equations of EF and GH derived from these are respectively ez-|-a: = o and e.z=.x (40). The radius OJ, perpendicular to EF, is f^* (4i)> J has co-ordinates (1/ \/e''-|-i, «/ ve^ + i) . .... (42), and JK, parallel to EF, and tangential to the circle at J, is ez+^=V6'-fi (43). But this fulfils the condition for tangency to the ellipse, which it touches; at the point K, whose co-ordinates are (c7V7+7, i/-ev/r + i) (44). AJ^T. 183] STRAINS 177 tTveT' '^^ '^"g^"'^ °f the angles XOK, XOH, and XOJ are respec- lA^ i/e, and e (45). Also JK IS f-i/e, and since OJ is unity, we see that the amount of the sheQris givtn hy T,. .,,^ , . X = JK/OJ = .-i/c (46). inus, It the elongational ratio e=i+«, where e is vanishingly small, we find ^ •' X=i+'?— 1/(1 +«) = 2« nearly . . .(47), as shown in article 172 of equation (13). Also, we see by (45) that where e is practically unity, as for the small shear just considered, the lines OK, OH, OJ all coalesce and make with the axes of z and x the angle 77/4. Hence we see, as before, that the two sets of planes of no distortion are perpendicular to one another and bisect the angles between the axes of elongation and con- traction. That the shear is of the same amount everywhere really follows from the fact that we could make the sphere and ellipse of any size we like within the volume subject to the strain in question, but we can also prove it analytically thus. Take the line PQRS parallel to EF, then its equation may be written ea+A-+/=o. ... ... (48), where /=:L0, and the difference of the x's for either (i) P and Q or (ii) R and S is, by (38), (39), and (48), /(.= -i)/(.= + i) (49). Thus, denoting by 9 the angle between EF and OC, we have tan 6—e, cos 0=ij ve^ + i, sin ^=e/ Ve'+i, and so find PQ=NP/cos d=l{^- 1)/ J?^i, and OM=/sin e^/e/ J^' + i. Hence, the amount of the shear which carries P to Q and R to S is given by X = PQ/OM-(.= -i)/e=.-i/. .... (so), and is thus seen to be independent of /and to agree with (46). Consider now the rotation involved when the shear is produced by slidings parallel to EF, which remains at rest. It is easily seen that A is derived from a^ and B from i-^, the equations of Aa^i and oBi^ being respectively ez+.x=£ and €2+*= I (Si). The co-ordinates of a^ and 6^ are also seen to be {2e/{^+i), (.»_i)/(.^ + i)) and (-(.= - x)/(.^+i), 2e/(.= +i)); thus 0^1 and O^ are at right angles, and the angle >/■, through which each rotates about the axis of jc from its initial to its final position, is given by tanv!'=(t=-i)/2€=(£-i/e)/2 = x/2. ■ • • (52). Therefore, in the case of a vanishingly small shear of elongation e, and with slidings parallel to EF, we should have the rotation expressed by tani/'=(r (SS)- M y = 6x+€y+fzj- (54), 178 ANALYTICAL MECHANICS [arts. 184-185 184. Composition of Pure Strains and Rotations.— Although the composition of strains and rotations in general {i.e. when they are about axes inclined in any way) lies beyond the scope of this work,, it seems desirable to point out that the resultant of two finite pure strains occurring in succession may involve a rotation in addition to the dis- tortion. Hence on afterwards applying another pure strain to undo that already produced, we reach the striking result that three successive pure finite strains may yield a rotation only, without relative change of the parts of the figure. This is shown by Tait in his Dynamics as follows : — Let the first pure strain be represented by Z =CX+/)>+I.Z ) and the second pure strain by x" = ax' +b'y' -{■ c'z' 1 y" = b'x' + i'y'+fy\ (5S), z = c'x +f'y' + i-'z'] in which the Greek letters a, e, and 1 represent as before the elongation ratios, and are respectively equal to i+a, T.+e, and i+z. Hence, by substituting in (55) the values of x',y', and z' expressed in (54), we can express x", y", and z" in terms of x, y, and z. In other words, we can express the resultant of the two pure strains when applied in the order given. To show that this resultant involves a rotation, we need only fill in two of the coefficients. Thus x" = { )x + {a'6 + 6'c+c/)y + {. . .)z] y'==(/,'a + ^'d+fc:)x + i )y + (. . .)z\ . . . (56). 2" = ( )^+( > + (• • ■>] Now the criterion of a pure strain is that three equalities exist between the nine constants as proved in articles 177 and 178, shown in Table v. article 179, and as illustrated by the b, c, and / in (54) and (55). But it is clear that in the general case these equalities will not be satisfied in (56), consequently a rotation is, in general, involved. For example, an elongation along one axis followed by an equal con- traction along another axis not at right angles to the first involves a rotation. The reader should also note that the resultant of a pure strain and a rotation usually depends on the order in which they occur, for these operations are not in general commutative. 185. Bestricted Strains. — We have hitherto dealt with strains chiefly in three dimensions, and supposed that the individual points of the primitive and strained figures can be identified and followed during the occurrence of the strain. Of course, these assumptions apply only to solid bodies extending throughout this tri-dimensional space. But the modifications required for other cases need only brief mention. Thus, {or fluids, the strain reduces to a change of volume simply, and it may become impossible to identify any points and follow them in their ART. 185] STRAINS 179 motion during the strain. Again, in the case of thin extensible membranes, we are almost reduced to a surface, and often a plane surface, in which case we have only plane strains to consider. Lastly, for elastic cords of negligible thickness, if straight, we have simple elongation. If, however, the cord passes round constraints so as to occupy solid space, it is sometimes of importance to find an expression for its rate of stretching per unit length per unit time. Equation (7) of article 137 was simply this rate of stretching equated to zero. Hence we have for our present case where it may be finite d£_dx dx dy dy di dz , . ds~ls'Ts^Js'Js^Js'Js ' ' ' ■ \51)- We have now dealt sufficiently for our purpose with pure and homo- geneous strains. Whatever theory of heterogeneous strains may be needed later will be then developed as required, and with immediate reference to the special problem under discussion. Similar remarks apply to other possible motions of deformable bodies as vibrations and waves, for since we are only subordinately concerned with them their kinematics may be taken along with the kinetics of each such problem. Examples— XXXVII. 1. Define strain ellipsoid, and write down its equations for (a) a uniform dilation of 0'003 fractional volume change, {b) a shear of amount o'oo4, (f) an axial strain of elongation o'ooS and lateral contractions o'oo2. 2. In the case of a unit sphere in the primitive body becoming an ellipsoid in the strained body, find the locus of the constant elongations, and discuss some of the chief cases which arise. 3. What special significance have the two circular sections of the strain ellipsoid in a certain case ? 4. Draw carefully the strain ellipsoid for a shear of elongations one-tenth and amount one-fifth nearly, showing how it may be regarded as derived from the unit sphere by progressive slidings of undistorted planes. 5. Show analytically that the shear ellipsoid may be derived from the sphere by an elongation and equal perpendicular contraction, or by slidings at angles of 45° with the above directions and of an amount double the elongation. 6. Show that the successive application of two pure strains may result in a rotation. Is this possible if each of the pure strains involved only elongations about the same co-ordinate axes ? i8o ANALYTICAL MECHANICS [art. 1 86 PART III.— KINETICS CHAPTER XI PHYSICAL BASIS 186. Physical Conceptions. — In Chapter in. the fundamental con- ceptions or intuitions of space and time laid the geometrical basis for the kinematical development which followed in Chapters iv.-x. In these the combination of displacements, velocities and accelerations was sometimes seen to tally more or less closely with various natural occur- rences familiar or rare. But the accelerations were in all cases simply postulated without any conditions being prescribed under which alone they might be expected to obtain. And that which was supposed to move was often only a mathematical point, linear, plane or solid figure or a combination of such ideal parts. In order, therefore, to make our description of phenomena at once more extensive and more precise, we must now introduce new and physical conceptions. We must conceive bodies as moving and know something as to the conditions under which their accelerations may be expected to occur. Then, having laid this physical basis in the present chapter, we can build upon it in combination with the kinematical theorems already developed, any subsidiary experimental data being introduced where required. The chief new physical conceptions that need introduction here are the following four : — (i) Mass, or Inertia ; (ii) Gravitation, or the tendency of all particles to approach each other ; (iii) Friction, or resistance to relative sliding of bodies in contact ; and (iv) Elasticity, or resistances to change of volume or shape. Two limiting cases of elasticity are especially noteworthy. If the resistances to change of shape and volume are both infinite, the body is said to be rigid. If the resistance to change of shape entirely vanishes, the substance is a frictionless^w/i/. Following the fundamental physical conception of mass and arising from it, others naturally occur, being products and quotients of the new and old quantities, such as force, momentum, work, etc., and need- ing definition merely. The consideration of some of these can be deferred till they are needed in the development of the subject. But one of them, force, had better be taken now along with mass. And as ARTS. 187-188] PHYSICAL BASIS 181 this conception of mass was only slowly developed, and very various views have been held concerning it and force, a brief historical account may be useful, in which of necessity other topics must be referred to. 187. Mechanics bafore Newton. — The earliest mechanical theories related wholly to statics of solids and fluids, dynamics being founded by Gahleo (1564-1642), and continued by Huyghens (1629-1695). Galileo experimentally investigated the motion of falling bodies, timing by a water-clock the rolling of a ball down an inclined groove. He thus discovered that the distances descended on a given incline were proportional to the squares of the corresponding times. This he had previously shown theoretically would be the case if the velocity was simply proportional to the time. He thus established that bodies fall with constant acceleration, acceleration being an entirely new con- ception to which he was led by this investigation. He further showed that bodies by falling acquired a velocity such that they were able to ascend to the level from which they fell. If this ascent were inclined and made less and less steep, the time of ascent was more and more increased. So that if the slope were continually diminished, this time might be indefinitely prolonged. He thus groped his way to the fundamental conception of inertia. For motion from rest with constant acceleration, Galileo thus used the two relations which, in our notation, are written v=at, and s=afl2. The importance of the third relation, derived from these, namely : — v' = 2as, was perceived by Huyghens, who thus laid the foundation for important advances. For, soon after Galileo, it was noticed that a body having velocity had a ' something ' in virtue of which a resistance could be overcome. Was this something, this efficacy, proportional to the velocity simply or to its square? Huyghens seemed to have seen quite clearly that a doubled velocity enabled a body to ascend against its weight for a douik time, but through a fourfold distance. So that the efficacy as regards time is proportional to the velocity simply ; but, as regards distance, is proportional to the square. Huyghens solved problems on the dynamics of several connected bodies, whereas Galileo restricted himself to a single body. Thus the compound or bar pendulum was treated, the centre of oscillation determined, and the acceleration of gravity found by pendulum observations. But it should be noted that throughout this period before Newton, the conception of mass had not been clearly formed. ' It did not occur to Galileo that mass and weight were different things.^ Huyghens, too, in all his considerations, puts weights for masses. . . .' 188. Newton's Principles.— To Sir Isaac Newton (1642-1 726) we owe (i) the enumeration of those principles of mechanics which still form the basis of its formal development, and (ii) the discovery of universal gravitation. Perhaps it was in connection with the latter that the distinction between weight and mass was first felt ; weight being i82 ANALYTICAL MECHANICS [art. i8g something different for a given body on the earth and on the moon, but its mass, the difficulty of starting it and of stopping it, being the same everywhere. The enumeration of the mechanical principles Newton arranged in a number of definitions, axioms, or laws and corollaries, all interpersed with remarks. These were contained in his Principia^ written in Latin; hence the very different versions of them which appear in various text- books. The definitions and laws are given here with an indication of the scope of the corollaries and some of the more important explana- tory remarks, the English version of Evans and Main being chiefly followed. 189. Newton's Definitions. ' Definition i. — Quantity of matter is the measure of it arising from its density and bulk conjointly. ' This quantity of matter is, in what follows, sometimes called the body, or mass. It is known for each body by means of its weight ; for it has been found, by very accurate experiments with pendulums, to be proportional to the weight. ' Def. 2. — The quantity of motion of a body is the measure of it, arising from its velocity and the quantity of matter conjointly. ' Def. 3. — The innate force of matter is its power of resisting, where- by every body, so far as depends on itself, perseveres in its state, either of rest, or of uniform motion in a straight line. ' This is always proportional to the body, and differs in no respect from the inertia of the mass, except in the manner of viewing it. To the inertia of matter is due the difficulty of disturbing bodies from their state of rest or motion ; on which account the innate force may be called by the very suggestive name, force of inertia. ' Def. 4. — An impressed force is an action exerted on a body, tending to change its state either of rest or of uniform motion in a straight line. ' Def. 5. — A centripetal force is one by which bodies are drawn, impelled, or in any other ivay tend from all parts towards some point as centre. ' Of this kind is gravity, by which bodies tend to the centre of the earth ; magnetic force, by which iron approaches a magnet ; and that force, whatever it may be, by which the planets are perpetually drawn away from rectilinear motions and forced to revolve in curves. The quantity of this centripetal force is of three kinds, absolute, accelerative, and motive. 'Def. 6. — The absolute quantity of a centripetal force is a measure of it which is greater or less according to the efficacy of the cause which propagates it from the centre through the regions of space all round it. 'Just as magnetic force is greater in one magnet and less in another, according to the mass of the magnet, or the intensity of its magnetism. 'Def. 7. — The accelerative quantity of a centripetal force is a ' Philosophiae Naturalis Princifia Mathematica (London, 1686). ART. 190] • PHYSICAL BASIS 183 measure bj it proportional to the velocity which it generates in a given time. ' Just as the power of the same magnet is greater at a less distance, less at a greater. Or, as gravitating force is greater in valleys, less on the peaks of high mountains, and so less the greater the distance from the earth ; but at equal distances the same on all sides, because it accelerates equally all falling bodies. 'Def. 8. — The motive quantity of a centripetal force is a measure of it proportional to the motion which it generates in a given time. ' Just as weight is greater in a greater mass, less in a less mass ; and, in the same, is greater near the earth, less in remote space. ' These quantities of forces may for brevity be called motive, accelera- tive, and absolute forces ; and, for the sake of distinctness, may be ascribed severally to the bodies which tend to the centre, to the positions of the bodies, and to the centre of the forces ; so that, in fact the motive force is ascribed to the body, as if it were the effort of the whole composed of the efforts of all its parts ; the accelerative force to the position of the body, as if there were diffused from the centre to all places around it, some power efficacious towards moving bodies which are in those places ; and the absolute force of the centre, as if at this point there were situated something which was the cause of motive forces being propagated through space in all directions ; whether that cause be some central body (just as a magnet is at the centre of magnetic force, or the earth at the centre of gravitating force) or any other cause which is not ascertained. This is simply a mathematical conception ; the physical causes and seats of the forces are not here considered.' 1 90. Newton's Axioms, or Laws of Motion. ' Law I. — Every body perseveres in its state of rest, or of uniform motion in a straight line, except in so far as it is cotnpelled to change that state by forces impressed on it. ' Projectiles persevere in their motions, except in so far as they are retarded by the resistance of the air, and driven downwards by the force of gravity. A hoop, whose parts continually draw each other from their rectilinear motions by cohesion, ceases to roll only in conse- quence of its motion being retarded by the air. But the larger bodies of planets and comets, whose motions, both progressive and circular, take place in less resisting spaces, retain these motions longer. ' Law IL — Change of motion is proportional to the moving force im- pressed, and takes place in the straight line in which that force is im- pressed. ' If a force produce any motion, twice the force will produce twice the motion, thrice the force three times the motion, whether it has been impressed all at once, or by successive gradations. And this motion (since it must always take place in the same direction as the force which produces it) is— if the body was originally in motion— added to i84 ANALYTICAL MECHANICS [arts. 191-192 its original motion if that motion was in the same direction, subtracted from it if in the opposite; or if in an inclined direction, is added to it in an inclined direction, and compounded with it, the position of the body being determined by the motion in such direction. 'Law III. — An action is always opposed by an equal reaction; or, the mutual actions of two bodies are always equal and act in opposite directions. ' Whatever presses or pulls something else, is pressed or pulled by it in the same degree. If a man presses a stone with his finger, his finger is also pressed by the stone. If a horse draws a stone tied to a rope, the horse will be (so to speak) drawn back equally towards the stone: for the rope being stretched at both ends will by the same attempt to relax itself urge the horse towards the stone and the stone towards the horse ; and will impede the progress of one as much as it promotes the progress of the other. If a body impinge on another and by its force change the motion of the other in any way, the latter will in its turn (on account of the equality of the mutual pressure) undergo the same change of motion in a contrary direction. To these actions are equal the changes, not of velocities, but of motions ; that is, in bodies not hindered in their motions by other forces. For the changes of velocities, which also take place in the same direction, are — since the motions are changed equally — reciprocally proportional to the bodies. This law holds also in attractions.' 191. Newton's Corollaries. — To his three laws Newton appended six corollaries. Of these the first and second relate to the principle of the parallelogram of forces (or impulses), the third to the conservation of momentum in spite of mutual actions, the fourth to the inability of mutual actions to disturb the motion of the centre of gravity of the system, while the fifth and sixth refer to relative motions. 192. Newton's Disciples and Critics. — From the foregoing transla- tion of the definitions and laws laid down by Newton, in this one branch of his varied activities, some notion may be formed of his tran- scendant greatness. The doctrines thus formulated have been accepted as a sufficient and satisfactory basis of dynamics and statics by numerous writers, including the authors (Kelvin and Tait) of the classic work on Natural Philosophy. On the other hand, some, while agreeing in the main with the in- formation Newton gave, have seriously criticised the verbal forms in which he gave it. Others have gone further than this, and taken a distinctly different view of most of the points in question. But possibly Newton would have been unintelligible to his contem- poraries had his thought and language been abreast of the most advanced thought of the present day. Of matters so fundamental, probably no statements logically perfect can be humanly invented. Certainly no such statement can be at once brief and full, conveying to friends and foes alike the self-same ART. 193] PHYSICAL BASIS 185 message. Yet the brevity seems highly desirable, to facilitate memoris- ing and quotation. Hence, for any audience in any age, the best attainable enunciations of such principles are probably those that briefly convey the essential truths to the audience in view, even though they may be slightly redundant in parts needing emphasis, or logically incomplete in others of minor importance. Thus, much of the modern endeavour to modify Newton's enunciations must not be taken either as any disparagement of his greatness or as any claim to present-day superiority. It is rather an attempt to restate very similar contents in forms more suitable to the audiences now addressed. Before going into details, four general lines of criticism of Newton's enunciations may be noticed. Firstly, it has been the task of modern criticism to disentangle the mere definitions from the statements of natural fact. Secondly, it has been urged that the first law is logically unnecessary because only a special case of the second. Some defend the first law as being necessary to remove the pre-conceived notions from men of Newton's time, while others regard it as permanently necessary. Thirdly, there is a strong body of opinion that Newton's definition of mass is incomplete and illogical, and must be replaced by one in which the ratio of masses is the negative inverse ratio of mutual accelerations. On this view Newton's third law becomes unnecessary. Fourthly, it has been pointed out that since all motion is relative, force is relative also, and that accordingly to complete the laws some statement is necessary as to the base, axes, or frame of reference to which they must be referred, and for which alone they are valid. We may now fitly pass into details, noticing at some length the criticism and constructive scheme of Mach, and more briefly the views of some other writers. 193. Criticisms by Mach. — In his Science of Mechanics (Prague, 1883, American Edition, Chicago, 1902), Dr. Ernst Mach, after devoting fifty pages to the achievements of Newton, passes on to a synoptical critique of the Newtonian Enunciations. After quoting the Definitions, Mach writes (p. 241 of American Edition) :— ' Definition i is, as has already been set forth, a pseudo-definition. The concept of mass is not made clearer by describing mass as the product of the volume into the density, as density itself denotes simply the mass of unit of volume. The true definition of mass can be deduced only from the dynamical relations of bodies. 'To Definition 2, which simply enunciates a mode of computation, no objection is to be made. Definition 3 (inertia), however, is rendered superfluous by Definitions 4-8 of force, inertia being included and given in the fact that forces are accelerative. ' Definition 4 defines force as the cause of the acceleration, or tendency to acceleration, of a body. The latter part of this is justified by the fact that in the cases also in which accelerations cannot take place, other attractions that answer. thereto, as the compression and distension, etc. of bodies occur. The cause of an acceleration towards i86 ANALYTICAL MECHANICS [art. 194 a definite centre is defined in Definition 5 as centripetal force, and is distinguished in 6, 7, and 8 as absolute, accelerative, and motive. It is, we may say, a matter of taste and of form whether we shall embody the explication of the idea of force in one or in several definitions. In point of principle, the Newtonian definitions are open to no objection.' Mach then quotes the laws and deals with them and the appended . corollaries as follows : — ' We readily perceive that Laws I. and II. are contained in the definitions of force that precede. According to the latter, without force there is no acceleration, consequently only rest or uniform motion in a straight line. Furthermore, it is wholly unneces- sary tautology, after havmg established acceleration as the measure of force, to say again that change of motion is proportional to the force. It would have been enough to say that the definitions premised were not arbitrary mathematical ones, but correspond to properties of bodies experimentally given. The third law apparently contains something new. But we have seen that it is unintelligible without the correct idea of mass, which idea, being itself obtained only from dynamical experience, renders the law unnecessary. 'The first corollary really does contain something new. But it regards the accelerations determined in a body K by different bodies M, N, P, as self-evidently independent of each other, whereas this is precisely what should have been explicitly recognised as a fact of experience. Corollary Second is a simple application of the law enun- ciated in Corollary First. The remaining corollaries, likewise, are simple deductions, that is, mathematical consequences, from the con- ceptions and laws that precede. ' Even if we adhere absolutely to the Newtonian points of view, and disregard the complications and indefinite features mentioned, which are not removed but merely concealed by the abbreviated designations "Time" and "Space," it is possible to replace Newton's enunciations by much more simple, methodically better arranged, and more satis- factory propositions. Such, in our estimation, would be the following.' 194. Enunciations by Mach. 'a. Experimental Proposition. — Bodies set opposite each other induce in each other, under certain circumstances to be specified by experimental physics, contrary accelerations in the direction of their line of junctions. (The principle of inertia is included in this.) ' b. Definition. — The mass-ratio of any two bodies is the negative inverse ratio of the mutually-induced accelerations of those bodies. ' c. Experimental Proposition. — The mass-ratios of bodies are inde- pendent of the character of the physical states (of the bodies) that con- dition the mutual accelerations produced, be those states electrical, magnetic, or what not ; and they remain, moreover, the same, whether they are mediately or immediately arrived at. ' d. Experimental Proposition. — The accelerations which any number of bodies A, B, C, . . . induce in a body K, are independent of each other. (The principle of the parallelogram of forces follows immedi- ately from this.) ARTS. 195-196] PHYSICAL BASIS 187 ' e. Definition.— Wosm^ force is the product of the mass-value of a body into the acceleration induced in that body. _ ' Then the remaining arbitrary definitions of the algebraic expres- sions " momentum," " vis viva," and the like, might follow. But these are by no means indispensable. The propositions above set forth satisfy the requirements of simplicity and parsimony which, on econ- omico-scientific grounds, must be exacted of them. They are, more- over, obvious and clear ; for no doubt can exist with respect to any one of them either concerning its meaning or its source; and we always know whether it asserts an experience or an arbitrary con- vention.' 195. The Tribute of Mach to Newton. — 'Upon the whole, we may say, that Newton discerned in an admirable manner the concepts and principles that were sufficiently assured to allow of being further built upon. It is possible that to some extent he was forced by the difficulty and novelty of his subject, in the minds of the contemporary world, to great amplitude, and, there- fore, to a certain disconnectedness of presentation, in consequence of which one and the same property of mechanical processes appears several times formulated. To some extent, however, he was, as it is possible to prove, not perfectly clear himself concerning the import, and especially concerning the source of his principles. This cannot, however, obscure in the slightest his intellectual greatness. He that has to acquire a new point of view naturally cannot possess it so securely from the beginning as they that receive it unlaboriously from him. He has done enough if he has discovered truths on which future generations can further build. For every new inference there- from affords at once a new insight, a new control, an extension of our prospect, and a clarification of our field of view. Like the commander of an army, a great discoverer cannot stop to institute petty inquiries regarding the right by which he holds each post of vantage he has won. The magnitude of the problem to be solved leaves no time for this. But, at a later period, the case is different. Newton might well have expected of the two centuries to follow that they should further examine and confirm the foundations of his work, and that, when times of greater scientific tranquillity should come, the principles of the subject might acquire an even higher philosophical interest than all that is deducible from them. Then problems arise like those just treated of, to the solution of which, perhaps, a small contribution has here been made. We join with the eminent physicists, Thomson and Tait, in our rever- ence and admiration of Newton. But we can only comprehend with difficulty their opinion that the Newtonian doctrines still remain the best and most philosophical foundation of the science that can be given.' 196. Retrospect by Mach.— 'If we pass in review the period in which the development of dynamics fell,— a period inaugurated by Galileo, continued by Huyghens, and brought to a close by Newton,— its main result will be found to be the perception, that bodies mutually i88 ANALYTICAL MECHANICS [art. 197 determine in each other accelerations dependent on definite spatial and material circumstances, and that there are masses. The reason the perception of these facts was embodied in so great a number of principles is wholly an historical one ; the perception was not reached at once, but slowly and by degrees. In reality only one great fact was established. Different pairs of bodies determine, independently of each other, and mutually, in themselves, pairs of accelerations, whose terms exhibit a constant ratio, the criterion and characteristic of each pair. 'Not even men of the calibre of Galileo, Huyghens, and Newton, were able to perceive this fact at once. Even they could only discover it piece by piece, as it is expressed in the law of falling bodies, in the special law of inertia, in the principle of the parallelogram of forces, in the concept of mass, and so forth. To-day, no difficulty any longer exists in apprehending the unity of the whole fact. The practical demands of communication alone can justify its piecemeal presentation in several distinct principles, the number of which is really only deter- mined by scientific taste. What is more, a reference to the reflections above set forth respecting the ideas of time, inertia, and the like, will surely convince us that, accurately viewed, the entire fact has, in all its aspects, not yet been perfectly comprehended. ' The point of view reached has, as Newton expressly states, nothing to do with the "unknown causes" of natural phenomena. That which, in the mechanics of the present day, is caWed/orce is not a something that lies latent in the natural processes, but a measurable, actual circumstance of motion, the product of the mass into the acceleration. Also, when we speak of the attractions or repulsions of bodies, it is not necessary to think of any hidden causes of the motions produced. We signalise by the term attraction merely an actually existing resemblance between events determined by conditions of motion and the results of our volitional impulses. In both cases either actual motion occurs, or when the motion is counteracted by some other circumstance of motion, distortion, compression of bodies, and so forth, are produced' {Science of Mechanics, Chicago, 1902, p. 246.) Examples— XXXVIII. 1. On what points are physical conceptions needed to enable us to pass from kinematics to kinetics ? 2. What do you know of mechanics before the time of Newton? 3. Give an outline of Newton's definitions and critically examine them. 4. State Newton's laws of motion and carefully comment upon them. 5. Enumerate some respects in which Mach criticises Newton's principles. 6. What enunciations does Mach propose in place of Newton's laws of motion ? 7. Write a critical essay on Mach's position with respect to Newton's principles of mechanics. 197. Karl Pearson's View.— The attitude of Professor Pearson to the laws of motion as shown in his Grammar of Science (London, 1892) is, in some respects, distinctly radical, and may have to wait ART, ig8] PHYSICAL BASIS 189 long for wide-spread adoption. It must accordingly suffice to quote here his own summary of the chapter of about fifty pages in which his five laws of motion are developed and those of Newton are criticised. ' Summary. ' The physicist forms a conceptional model of the universe by aid of corpuscles. Those corpuscles are only symbols for the component parts of perceptual bodies, and are not to be considered as resembling definite perceptual equivalents. The corpuscles with which we have to deal are ether-element, prime-atom, atom, molecule, and particle. We conceive them to move in the manner which enables us most accurately to describe the sequences of our sense-impressions. This manner of motion is summed up in the so-called laws of motion. These laws hold in the first place for particles, but they have been frequently assumed to be true for all corpuscles. It is more reasonable, however, to con- ceive that a great part of mechanism flows from the structure of gross " matter." ' The proper measure of mass is found to be a ratio of mutual accelerations, and force is seen to be a certain convenient measure of motion, and not its cause. The customary definitions of mass and force, as well as the Newtonian statement of the laws of motion, are shown to abound in metaphysical obscurities. It is also questionable whether the principles involved in the current statements as to the superposition and combination of forces are scientifically correct when applied to atoms and molecules. The hope for future progress lies in clearer conceptions of the nature of ether and of the structure of gross "matter."' 198. Love's Treatment. — In his Theoretical Mechanics (Cambridge, 1897), Prof. A. E. H. Love expresses indebtedness to Kirchhoff, Pearson, and Mach, and adopts for the basis of the subject a set of rules, of which he then speaks as follows : — ' The system of definitions and rules which we have laid down lead to a system of differential equations for determining the motions, relative to a frame, of a system of particles, or of a body or a system of bodies, conceived to be made up of particles. It may be regarded as a purely ideal system, and its vahdity is unaffected by the question whether it has or has not any relation to the observed motions of natural bodies. The subject, so treated, is known as Rational Mechanics. The objects of which it treats are pure objects of thought. Its development consists in the logical deduction of particular results from the general principles laid down. The application of Rational Mechanics to the formulation of the Laws that govern the motions of natural bodies consists in the state- ment that it is possible to assign masses to the bodies and to choose a frame of reference determined by parts of natural bodies, such that the observed motions of natural bodies, relative to the frame, obey the Laws of Rational Mechanics with certain limits of exactness ; that m fact the observed motions coincide with the motions described in the igo ANALYTICAL MECHANICS [arts. 199-201 phraseology of Rational Mechanics so closely that no discrepancy can be observed.' 199. Lodge on Axioms. — Sir Oliver Lodge, in a paper to the Physical Society of London (Proc, vol. xii. pp. 291-292, 1893) on The Foundations of Dynamics, spoke as follows on fundamental laws or axioms : — ' The setting forth of an axiom I regard as a kind of challenge, equivalent to the statement — "Here is what seems to me to be a short summary of a universal truth ; disprove it if you can. I cannot prove it; it is too simple and fundamental for proof; I can only adduce hundreds of instances where it holds. I have indeed critically examined a few special cases and never found it fail, but a single contrary instance will suflSce to overthrow it ; hence, though it be hard to prove, yet if not true its disproof should be easy : find that contrary instance if you can? If no disproof is forthcoming for a few generations, the axiom is likely to get accepted. Meanwhile its un- deniable simplicity is a practical advantage, even though in the course of centuries a flaw or needful modification in its statement may be dis- covered.'" Of this nature are some of the principles already given, and the following also by Newton on gravitation which now needs notice. 200. Universal Gravitation. — Kepler (1571-1630) analysed the observations of Tycho Brahe to find the true motion of Mars. After years of labour emerged his first two laws, and subsequently (in 16 18), his third, which relates to other planets. Kepler's laws may be stated as follows (see Pioneers of Science by Sir Oliver Lodge, 1893, p. 56) — Law I. Planets move in ellipses, with the sun in one focus. Law 1L The radius vector sweeps out equal areas in equal times. Law III. The square of the time of revolution of each planet is proportional to the cube of its mean distance from the sun. From these laws and other considerations of his own, Newton passed to his grand conception of universal gravitation. This idea is to be gathered from various parts of the Piincipia and by Tait {Properties of Matter, p. 113, 1890) is expressed thus : — Law of Gravi- tation. ' Every particle of matter in the universe attracts every other particle with a force whose direction is that of the line joining the two, and whose magnitude is directly as the product of their masses, and in- versely as the square of their distance from each other.' It is now very questionable whether the law of inverse squares holds for small distances of the order of those between adjacent molecules. But the law still serves our purpose, as we are here concerned only with much greater distances. 201. Friction: Coulomb, Morin, Beauchamp Tower. Leaving for a little the most fundamental bases of mechanics, let us now notice in order the subsidiary matters enumerated in article 186. The essential laws of friction for dry surfaces, with which we are chiefly concerned, seem to have been first enunciated by Coulomb in 1781, and ART. 20 1 ] PHYSICAL BASIS 191 were confirmed by the experiments of Morin, 1 830-1 834. For know- ledge respecting lubricated surfaces we are greatly indebted to the ex- periments of Beauchamp Tower (see Froc. Inst, of Mechl. Engineers, 1883-1888). The subject of friction is fully treated by Prof. J. Goodman {Mechanics applied to Engineering, 1908, pp. 240-260), from whose tabulated comparison of dry and lubricated surfaces the state- ments in Table VI. are abridged. Table VI. — on Friction. Dry Surfaces. ; i.Thefrictional resistancebetween i surfaces in relative motion is nearly j proportional to the normal force (or total pressure) between the two sur- faces. The upper limit of the ratio, resistance -H normal force, is called the coefficient of friction (/i). 2. The frictional resistance is nearly independent of the speed for low pressures. For high pressures it tends to decrease as the speed in- creases. 3. The frictional resistance is not greatly affected by the temperature. 4. The frictional resistance de- pends largely on the nature of the material of the rubbing surfaces. 5. The friction of rest is slightly greater than that of motion, but may be reduced by vibration, to that lower \alue. 6. When the pressure between the surfaces becomes excessive, seizing occurs. 7. The frictional resistance is greatest at first, and rapidly de- creases with the time after the two surfaces are brought together, per- haps due to polishing. Lubricated Surfaces. I. The frictional resistance is al- most independent of the pressure with bath lubrication, and approaches the behaviour of dry surfaces as the lubrication becomes meagre. 2. The frictional resistance varies directly as the speed for low pres- sures. But for high pressures the friction is very great at low velocities, becoming a minimum at about 5/3 ft. per sec, and then increases nearly as the square I'oot of the speed. 3. The frictional resistance de- pends more upon the temperature than upon any other condition. 4. The frictional resistance with a flooded bearing depends bitt slightly upon the nature of the material of the rubbing surfaces. 5. The friction of rest is enor- mously greater than that of motion, especially with thin lubricants, which are then probably squeezed out. 6. When the pressure becomes excessive (requiring much higher pressure than for dry surfaces) the lubricant is squeezed out and seizing occurs. 7. The frictional resistance is least at first, and rapidly increases with the time after the two surfaces are brought together, perhaps due to parti al sque ezing out of the lubrican t. Since we are here concerned chiefly with dry surfaces, paragraphs i, 2, 4 and 5 in the first column are the most important. And, owmg to the remark about vibration under 5, we may always use the smaller value for the frictional resistance. 192 ANALYTICAL MECHANICS [arts. 202-203 202. Laws of Hooke and Boyle. — We must now pass to the necessary physical conceptions respecting the simple elastic bodies with which we have to do. For our purpose these are sufficiently expressed (i) for gases, by Boyle's Law, and (ii) for solids and liquids, by a generalisation of Hooke's Law. What we now know as Hooke's law was expressed by him in Latin in the form ' ut tensio sic vis.' This may be translated as the extension so is the force. Let us pass from this particular case of strain to strain in general. And let the single word stress be used to denote any set of equilibrating forces applied to a body, i.e. any set of forces which would have no apparent effect on a rigid body. Then we may restate the law in a generalised form thus, strain is proportional to the corresponding stress. This law only holds within very small, limits, which we shall suppose are not passed over in the strains and stresses under treatment. Hence, for a given substance and a given type of stress and strain, the quotient, stress divided by strain, is a constant which measures the particular elasticity of the substance in question. Obviously, therefore, the elastic behaviour of bodies, under the limitations mentioned, depends upon and is sufficiently specified by its various elastic constants, the details of which we shall examine later. These constants are called moduli of elasticity, as bulk modulus, or bear special names, e.g. rigidity. For gases, Boyle's law states that the volume of a given mass of gas, kept at a constant temperature, is inversely as its pressure. This is only a very approximate statement, but is near the truth for moderate pressures and for temperatures far above the point of liquefaction of the gas in question. It accordingly serves our purpose here. 203. Relative Character of Motion and Mechanics. — Since motion is relative, force and mechanics usually have a like relative character. Hence, in each class of problems, it is important to notice that the laws of motion, gravitation, etc., should be construed with respect to axes appropriate to the phenomena under discussion. It may be difficult to give an instruction, at once general and precise, as to the choice of co-ordinate axes. Yet no difficulty in this respect usually occurs in the development of the subject. Especially is this the case if, at the outset, the fact has been recognised that discretion in this matter must be exercised. We then obtain, for each problem, the system of mechanics possessing just that kind of relative character which it needs. Thus, for terrestrial motions, comprised within a few miles and a few minutes, the co-ordinate axes may be fixed in the earth. This gives us at once the ordinary mechanics of the factory, the field, the road and the railway, which may be called terrestrial mechanics. When concerned with planetary rotations or their orbital motions, or in dealing with Foucault's pendulum, the tides, the seasons, etc., the axes may be directed by the so-called fixed stars, yielding a system that might be called planetary mechanics. Hence, proceeding in this manner, as larger spaces and longer ARTS. 204-205] PHYSICAL BASIS 193 times were surveyed, a number of systems of mechanics might be in turn developed, each suitable and sufificient for certain problems, each succeedmg system embracing new problems and conferring a deeper insight into the old ones. We might thus repeatedly approach, though perhaps never attain, a system of mechanics deservedly regarded as absolute and universal. In the meantime, the relative systems serve for all practical purposes, although perhaps the formulation of an absolute one may be a legitimate problem for philosophy. Examples— XXXIX. 1. What is Karl Pearson's attitude towards Newton's principles of mechanics ? 2. What do you understand by rational ?nechamcs ? 3. Explain precisely what you mean by an axiom, and state on what under- standing it is accepted. 4. Enunciate the law of universal gravitation. Do you believe it applies under all conditions ? 5. Give a brief outline of what is known about friction. 6. State and explain the laws of Hooke and Boyle. 7. Explain what you mean by the relative character of mechanics and gi\ e illustrations. Can we reach or approach an absolute mechanics ? 204. Measurement of Time. — Similar remarks to those in Art. 203 apply to the measurement of time. For all ordinary purposes we may be content to adopt the second of mean solar time as the unit of time ; just as, for terrestrial mechanics, we may fix our axes of co-ordinates in the earth. But for astronomical purposes sidereal time is used. And, if we wish to express the retardation of rotation of the earth (due to the tidal friction acting on it like a band brake), it is evident we must go a step further and choose some measurer of time which is believed or imagined to suffer no change throughout centuries, as e.g. a perfect clock truly rated by the earth centuries ago. Kelvin and Tait suggest {Natural Philosophy., Art. 406) as such a timekeeper ' a carefully arranged metallic spring, hermetically sealed in an exhausted glass vessel.' The period of vibration corresponding to a given spectral line (say the D, line for sodium) might also be used, and has been suggested by Maxwell (Electricity and Magnetism., vol. i. Art. 4, page 3, Oxford, 1873 ; see also Kelvin and Tait's Natural Philosophy, part i.. Art. 223, p. 227, Cambridge, 1890). 205. Attitude towards Physical Axioms. — We may now fitly revert to the subject of axioms and the place they fill, which was just referred to in Art. 199. Though such physical laws or axioms cannot be formally proved, we may legitimately trust them provisionally, at any rate as first approximations. But since their acceptance is based on their inherent probability and the lack of any disproof, it must be noted that our trust in them should be coextensive with the experience N 194 ANALYTICAL MECHANICS [arts. 206-207 upon which their acceptance is based. Any pushing of beUef in them beyond such limits should be of the nature of an experiment. Thus, the Newtonian principles are accepted as consistent with an experience having certain bounds of space and time and relating to gross or ponderable matter with speeds within certain limits. May we push them and legitimately build upon them outside these limits ? May we apply them to that medium called the ether, conceived as co-extensive with the physical universe and supposed endowed with inertia and elasticity but not with gravitation ? May we apply them to the very small corpuscles or electrons of modern science, moving at speeds comparable with that of light ? This may be done, but only tentatively and in an exploring manner, the results being continually tested by experimental checks. In fact we are now in a position to appreciate the following quotation from E. Mach {Science of Mechanics, pp. 237- 238, Chicago, 1902): — ' The most important result of our reflection is, however, that precisely the apparently simplest mechanical principles are of a very complicated character, that these principles are founded on uncom- pleted experiences, nay on experiences that never can be fully completed, that practically, indeed, tluy are sufficiejitly secured, in view of the tolerable stability of our environment, to serve as the foundation of mathematical deduction, but that they can by no means themselves be regarded as mathematically established truths but only as principles that not only admit of constant control by experiment but actually require it.' 206. Mass at High Speeds. — In illustration of the preceding article we may note that the electron (or elementary negative electric charge) now so prominent in physical research is believed to behave as having different inertias at different very high speeds, this inertia being more- over different along and perpendicular to the direction of motion. According to the theories of Prof. H. A. Lorentz of Leiden, if the so- called longitudinal and transverse masses at speed v are denoted by m-^ and /«2 respectively, and that at infinitesimal speeds by ;«„ we have mi = m„ly' a.nd m2 = mo/y, where -/= ^f i—v"'jc', c being the speed of light. The motions are here reckoned with respect to the ether which, on this view, is considered not to share any of the translatory motions of gross matter. It is possible that the ether, if in the above sense ' stagnant,' may prove to be the best base in which to fix our co-ordinate axes for an absolute system of mechanics. 207. Quantities usually proportional to Mass. — ^It is perhaps desirable to note here that though Newton defined mass as quantity of matter proportional to product of density and volume, most other writers have regarded mass as the inertia or measure of sluggishness of a body. And it thus appears in the definitions of Mach and of Pearson. Really, in the general use of the terms ounce, pound, or ton, for mass or weight, we have more or less clearly in our minds several different ART. 208] PHYSICAL BASIS 195 quantities which we often vaguely assume or believe to be proporLional. Thus, in buying food of a standard quality we are concerned with its nutritive or heat-generating properties and believe these quantities to be proportional to its weight. We might say we were here, if any- where, concerned with quantity of matter. Again, if material is dis- posed in different places on a cricket-bat or a golf-club, we may be chiefly concerned with the inertia of that material, whether wood or metal. Thirdly, if we hang a piece of iron or lead to balance a sash window or pull a door to, we are concerned with the weight of that iron or lead. There are other cases in which we are concerned with the elastic resistance of a given piece of a certain substance, or with its use- fulness as a conductor of heat or electricity, etc. Now the first four quantities are the most important to us mechanically, viz. : quantity of matter, inertia, iveigkt, and elasticity. The first three we often take to be strictly proportional. The last we take to be proportional to any one of the others for a given definite shape in the simplest cases, like the stretching of a wire of given length. But, since it may prove that none of these quantities are strictly proportional in any perfectly general view of matters, it must be clearly grasped in which sense the word mass is used in mechanics. It is used fundamentally in this work to denote inertia or sluggishness of a body to change its velocity in magnitude or direction. It is so used in the theory of Lorentz just noticed in Art. 206. From which it appears that at those high speeds, mass (or inertia) is not proportional to quantity of matter, if electrons are counted as matter and quantity is gauged by number of electrons, as Nature's identical elemental units. 208. Betrospect. — It is now time to glance in thought over the various views advanced as to the basis of mechanics, and to endeavour to extract some simple rules for guidance, some sufficient foundation to build upon. Nothing can be said or written on this topic which is not open to attack from several quarters. Nothing can be stated which shall be at once full, precise and brief. Yet it seems desirable that some short enunciations should be made which, to the sympathetic student, will convey a view of the foundations of mechanics that, while yielding much to modern criticism, both as to form and substance, avoids going to an extreme in any direction. With much diffidence, an attempt in this direction is submitted in the next article. Each statement is accompanied by a brief symbolic expression of the law or definition under its title. The notation in each case will be readily understood from the context. Of course no permanent or widespread value can attach to any such brief statements. At best, they can but suit a Hmited number for a limited time. To such, for the present, it is hoped they may be of service. Possessing no possible permanency, they, or any of hke nature, should be under constant criticism and revision. And if their presence here serves to stimulate an interest in the subject which shortly leads to their supercession, their formulation will not have been in vain. 196 ANALYTICAL MECHANICS [arts. 209-210 209. Brief Enunciation of Chief Mechanical Bases. 1. Law of Motion. Accelerations occur only in opposite pairs, (— rtoca') whose ratios are constant for given particles. 2. Definition of Mass. The masses of particles are positive con- {mjm'=~a'ja) slants, inversely as their mutual accelera- tions. 3. Definition of Force. Force is the product, mass into acceleration, (F= ma) and has the direction of the acceleration. 4. Law of Gravitation. Every particle attracts every other with a {Focmm' jr'') force, along their joining line, directly as the product of their masses, and inversely as their distance squared. 5. Law of Friction. With dry surfaces in contact, relative sliding (7'4>/*iV) is resisted by tangential forces, whose limit- ing values are proportional to the normal forces. 6. Law of Elasticity. Within narrow limits, the strain of a body is {Strainoc Stress) proportional to the equilibrating set of forces, or stress, applied to it. 7. Gaseous Law. The volume of a given mass of gas is in- (^» = const, for ^ const.) versely as its pressure at constant tempera- ture. 8. Choice of Axes. Since motion is relative, force and mechanics are relative also. Hence, the foregoing and any problems based upon them, should be referred to axes which, in each case, yield a mechanics most appropriate to the phenomena under discussion. 210. Concluding Kemarks. — We may now with advantage glance over the enunciations of the previous article and note certain points concerning them. In the first single statement the endeavour is made to embody all we know, of an experimental or axiomatic nature, as to motion generally. It contains within itself Newton's first law and the qualitative aspect of his third law. It also asserts that proportional aspect of all the simultaneous accelerations possible to any given pair of mutually interacting particles, which forms the basis of the modern definition of mass, as inertia. This definition accordingly follows ; mass being stated to be a positive constant characteristic of any given particle, and such that the products, mass into acceleration, for any pair of interacting particles have opposite signs but equal magnitudes. This supplies the quantitative aspect of Newton's third law, and so completes the statement of its substance. It is a convenience to have a name for these opposite but equal products, and this is next supplied by the modern definition of force, which replaces Newton's eighth definition and second law, those two being practically identical. But, by this definition, force is seen to be a vector, hence the law of ART. 2io] PHYSICAL BASIS 197 addition of vectors naturally applies. We thus have, at once, the triangle, parallelogram and polygon of forces without further proof. It has not been considered necessary to include among these brief enunciations any specific statement either (i) as to the mass of any particle or body being the sum of the masses of its parts, or (ii) as to the mass of any particle being the same whether derived directly or indirectly by comparison with some given particle. For, it is supposed all through that the system being enunciated is a self-consistent one, and also that it satisfies any experimental checks that can be applied to it. These suppositions imply that the second point holds. As to the first point, it may be naturally inferred that the quantity called mass, since primarily it expresses the sluggishness of a particle and is char- acteristic of it, will increase to some larger quantity of the same kind characteristic of the sluggishness of a body when conceived as built up of particles. Also, since mass is a positive constant without refer- ence to direction ; and force, the product of mass and acceleration, has the direction of the acceleration ; we see that mass is a scalar quantity. Hence masses, being scalars, are susceptible only of arithmetical addi- tion. Thus, consistently with the foregoing enunciations, we obtain the mass of a body by the simple arithmetical addition of the masses of its parts. And, unless this agreed with experimental checks, the enun- ciations would be invalid as a description applicable to the physical universe as perceived by means of our senses. Hence, though it is conceivable that the masses of particles might not add to that of the body, the fact that they do so add may be naturally inferred from the enunciations without formal statement. It should also be noticed here that, in the laws of motion and gravitation and in the definition of mass, only particles are mentioned. Then, the behaviour of particles being axiomatically formulated, that of extended bodies and systems of various constitutions is to be analyti- cally derived. Further, it is specifically mentioned in the fourth statement that the gravitational accelerations between particles occur in the line joining them. But, in accordance with the general principles adopted in these brief enunciations, the fact that in the so-called contact of particles the consequent opposite accelerations occur along the same line, is not mentioned in the Law of Motion, but is left to the natural inference of each reader. Of the remaining enunciations, Nos. 5-7 call for little remark. Obviously those readers omitting parts of the present course may omit the corresponding enunciations. But the last, on the choice of axes, may need noting by those not concerned with some of the inter- mediate ones. This eighth statement is purposely placed last for two reasons : First, because it may apply to all the rest ; Secondly, in accordance with the principle that although such a proviso may be needed somewhere, it should not be obtruded too early, and thus mtro- duce ditificulty in the preliminary axioms, where all should be kept as broad and simple as possible. . . We may perhaps with advantage note here an application of this 198 ANALYTICAL MECHANICS [art. 211 principle of choice of axes, expressed in 8, to the law of gravitation given in 4. Thus, to jreduce the proportionality to an equality, intro- duce a constant 7 and write F=ymm'lr''. Then by the definition of force in 3, we see that the accelerations of m and m' axe respectively ym'jr^ and ymj?-'. In other words, the accelerations have magnitudes inversely as the masses, which agrees with the definition of mass in 2. We thus see that these accelerations are noi reckoned for each mass relatively to the other, but are each reckoned with respect to an origin which lies between them and divides their distance apart inverselyi as the masses. We shall see afterwards that this point is called the centre of mass or centre of gravity of the two bodies. Obviously the re- lative acceleration of the two masses is the sum of their separate accelerations relative to the same point between them. Thus adding the above expressions, we obtain for this total or mutual acceleration the value y{m-\-m')lr^. The units used for measuring any of these mechanical quantities will be dealt with as occasion arises throughout the rest of the work. Some of the topics of the present chapter are of a highly contro- versial nature. Thus, a partial discussion of them here and there throughout the book might well prove irritating to some readers and make the corresponding parts of the detailed treatment less acceptable. Accordingly, to obviate this drawback, the discussion of the physical bases of mechanics has been confined to this single chapter, from which mathematical deductions are excluded. Hence, the remainder of the work, presenting the formal developments of kinetics and statics, is left equally open to all classes of teachers and students whatever their views or convictions on the debatable matters underlying them. 211. Bibliography. — Considerations of space preclude any further treatment of these and kindred topics, for which the reader is referred to the following works, placed in alphabetical order of their authors : — H. Hertz. Principien der Mechanik (Leipzig, 1894.) Sir Oliver Lodge. The Foundations of Dynamics (Proc. Phys. Soc. London, 12 pp., 289-236, 1894.) A. E. H. Love. Tiieoretical Mechanics (CaxnhnAge, 1897.) Ernst Mach. The Science of Mechanics (Chicago, 1902.) Sir Isaac Newton. Philosophiae Naturalis Principia Mathematica (London, 1686.) Karl Pearson. The Grammar of Science (London, 1900.) H. PoiNCARE. Scietice and Hypothesis (J^onion, 1905.) Bertram Russell. The Principles of Mathematics (Cambridge, 1903-) Herbert Spencer. First Principles. (London, 1898.) Alexander Ziwet. The Relation of Mechanics to Physics (' Science,' 23 pp., 49-56. New York, Jan. 12, 1906.) Examples— XL. I. State how time is usually measured and by what inaccuracy the method is affected. What other methods have been proposed with the view of diminishing the present inaccuracy ? ART. 21 1] PHYSICAL BASIS 199 2. What do you consider should be our attitude towards mechanical and physical axioms ? 3. Explain how the inertias of an electron vary with its speed. 4. What three mechanical quantities are often tacitly assumed to be propor- tional to one another ? Give examples showing the distinctions between them. 5. Do you consider the mechanical creed as enunciated by Newton needs revision ? If so, state in your own words by what you would replace it. If not, defend Newton from his various critics. 6. Translate and comment upon : — L'acceleration d'un corps est t^gale k la force quit agit sur lui divisee par sa masse. Cette loi peut-elle etre verifiee par I'experience ? Pour cela, il • faudrait mesurer les trois grandeurs qui figurent dans I'enonc^ ; acceleration, force et masse. J'admets qu'on puisse mesurer l'acceleration, parceque je passe sur la dififi- culte provenant de la mesure de temps. Mais comment mesurer la force, ou la masse ? Nous ne savons meme pas ce que c'est. Qu'est-ce que la masse ? C'est, repond Newton, le produit du volume par la densite. II vaudrait mieux dire, repondent Thomson et Tait, que la densite est le quotient de la masse par le volume . Qu'est-ce que la force ? C'est, repond Lagrange, une cause qui produit le mouvement d'un corps ou qui tend k le produire. C'est, dira Kirchhoff, produit de la masse par l'acceleration. Mais alors, pourquoi ne pas dire que la masse est le quotient de la force par Taccdldration ? ' (London B.Sc, Pass, Applied Math., 1908, in. 7.) ANAL YTICAL MECHANICS [art. CHAPTER XII KINETICS OF PARTICLES 212. Mass brought into Equations. — To pass from the kinematics of a point to the kinetics of a particle we suppose our point to be endowed with mass, m say, and multiply the appropriate kinematical equations throughout by that mass {m). Thus, for the rectilinear motion of a particle with uniform accelera- tion, we take (from article 27) the kinematical equations representing the increase of velocity from « to w in time / under acceleration a ; also the increase of the square of this velocity while describing space s. These may be written v—u = at . . (i), and v'' — u^ = 2as .... . (2). Now, multiplying by m and writing F for the force concerned in place of the product ma, we obtain jnv—mii = Fl. . ■ ■ . (3), and ^inv' — \mu''=Fs ... . (4). We see that these equations involve certain products, and since these often occur names have been adopted for them which are now given and defined, each being accompanied by a convenient symbol shown in brackets. Momentum. — The product of a mass into its linear velocity is called its linear momentum {7nv=^F). Impulse. — The product of a force into its duration is called its impulse {Ft= Q; or, if F is the variable value of the force during the time T, the impulse is (2= I Fdf.) Jo Kinetic Energy. — Half the product of a mass into its velocity squared is called its kinetic energy (|OT»^=y). Work. — The product force into distance described by the accelerated particle in the direction of the force is called work {^Fs^ W ; or, if F\s, the value of the variable force over the space s, IV= I Fds). Jo symbols, we may put (3) and (4) : ^-^«=G (s), Using these ideas and symbols, we may put (3) and (4) in the forms ART. 212] KINETICS OF PARTICLES 201 ^'^^ T-T,= W (6) qiesTion!'^ '^''° subscripts denote initial values of the quantities' in We may state the above important relations verbally thus •— .,,nlT ^,^7^f ^f equals impulse and Change of kinetic energy equals work The first of these is equivalent to Newton's second law but may be here regarded as derived from the single law of motion and the defimtions of mass and force of article 209. The second statement is important in connection with the conservation of energy, that funda- mental doctrine so well known to physicists. From (s) we see that Q is of the same nature as P, i.e. mass into ve ocity. And we have hitherto supposed P to change by change of velocity only, the mass bemg constant. But it is evident that P might, under some circumstances, change by a change of 7nass, all these increments of mass having the same velocity change, from o to w say. Or, both changes may occur together. We might thus write as a more general relation dQ=Fdt= mdv-{-vdm—d{mv)=dP, whence F= dPjdt . (7). That is, force is the time-rate of increase of momentum. Similarly, from (6), we find .^'=rf77^.r . . (8); or, force is the space-rate of increase of kinetic energy. Some other current forms of speech used in connection with the above quantities and relations may be noted here. The acceleration a of a particle of mass m is said to be due to the action of the force F=zma. The particle, assimilated to a point, is called the. point of application of the force. In the case of more extended bodies, the point of application becomes a surface or volume of application, according as the force is applied by contact of gross matter over a surface or by other means throughout a volume, as in the case of an attraction like gravitation. The change of momentum from P^ to P is said to be due to or produced by the impulse Q, which equals P~P^. The change of kinetic energy from T", to T is said to be due to or produced by the expenditure of work W, which equals T— T^. Some writers object to these expressions as implying ctz^jfl/ relations of which we have no proof. But it is difficult to avoid these current and convenient expressions, even though we may regard all mechanics 2LS purely descriptive of obset^ed phenomena. The relations involving impulse and work were derived from those for uniform acceleration, and so correspond to a unifortn force, but we may now with advantage remove that restriction. Thus, if F is the variable force for time t and space s, the speed meanwhile changing from u to V, we may write mdvldt=F=!nvdvlds. From these, multiplying up and integrating, we obtain m{v-u)=i' Fdt^Q . . . (9), 202 ANALYTICAL MECHANICS [art. 213 and \m{v^-u^)=^ Fds^W (10). Thus, on eliminating m between them, we have the general relation w=(p:^ (lO; ^ 2 or, the work of a variable force is its impulse multiplied by the arithmetic mean of the initial and final velocities. 213, Choice of Units of Mass, Force, etc.— The above new quantities need appropriate units for their measurement. These units are not entirely settled by the relations between the quantities expressed above in words or syfnbols; but, certain units being chosen, the others then naturally follow from the relations in question. Thus, corre- sponding to the three indefinable quantities with which mechanics is concerned, viz. Space, Time, and Matter, we may choose units of length, time, and mass ; the corresponding units of force, momentum, and impulse, energy and work, then follow. The units of length, time,, and mass are then called fundamental units and the others derived units. Or, a different selection of fundamental units may be made, say length, time, and force, and then the others derived from them. We shall use both plans, and in the above order, but it will be well first to consider the relation between the mass and weight of a body near the earth's surface. Let a particle of mass m fall freely at some place on the earth and be found to have an acceleration g. Then by our definition the force concerned, called its weight w, is given by tv^mg, %o thsX m = wlg (12). The acceleration g and the weight w of a given particle or body are both known to vary in different parts of the earth, as we shall see later, but the quotient wlg=m, called the mass of the particle or lump, is believed to be practically constant, as stated in the definition of mass. (See article 209.) Now, if our units of length and time were such that g were unity at some place, then, for that place, the mass m and the weight IV for any body would be represented by the same number. But this is not the case for any system of units of length and time at present in vogue. Hence in all such systems the weight of a body is not represented by the same number as its mass, while we retain the definition F=ma. Thus, while it would be a great simplicity to retain this definition, have a standard lump of stuff as the unit of mass, and its weight at some standard place as the unit of force, this is in- compatible with our present units of length and time. There are thus three simple courses open to us in choosing units of mass and force. I. Choose a standard body as the unit of mass, and let the unit of force follow from the definition F=ma. We thus have the weight in these units of force given by ■w=mg. That is, the weight of a body is expressed by a number g times the n^imber which expresses its mass. ART. 214] KINETICS OF PARTICLES 203 Hence the unit of force is i/^th of the weight of the unit mass. The centimetre-gram-second system and the so-called British absolute system are both of this type. 2. Choose a standard body and specify a place at which its weight shall be the unit of force, and let the unit of mass follow from F—7na. Thus as before w=mgor m = w/g. Hence in these units the number expressing the mass of a body is i/^h of the number which expresses its weight at the standard place, g being the acceleration of gravity there. That is, the unit of mass is g times that of the standard whose weight is unit force. The British engineers' system is of this type. 3. Choose a standard body as unit mass, its weight at a specified place as unit force, and abandon the definition F—tna, replacing it by FoQma or F=kwa. Then k would have to be ifg, and the force would be given by F^wajg. The use of any system of units requires care in passing from the theoretical expressions to the concrete ones, namely, the care that all the units introduced are of the same system. But the adoption of this third choice for a system of units calls for greater care, for it involves a change in the definition of force (from an equality to a proportionality), and thus changes all the theoretical expressions which explicitly or implicitly involve force. It will accordingly be no further considered here, but attention confined mainly to the first and second methods of choice of units, though occasionally special units may be introduced to suit special problems. 214. Established Systems of Units. — In Table vir. are given the chief units on the three systems already referred to, while Table viii. gives the ratios for conversion from each system to the others. Of these systems, the C. G. S. was fixed by an International Confer- ence at Paris in 1875, t-^e standard Cii length being the metre, which is the length at the temperature of melting ice between the ends of a platinum rod, made by Borda and preserved in the Bureau des Archives in Paris ; the standard of mass being the kilogramme des archives, made in platinum by Borda and preserved in the Conservatoire des Arts et Metiers, Paris. The British standards of length (bronze) and mass (platinum) are the yard and pound respectively, and are in the charge of the Warden of the Standards. [Tables. 204 ANALYTICAL MECHANICS Table VII. Mechanical Units. [art. 214 Systems OF Units. Units in each System. Length. Time. Mass. ; Force. Momentum and Impulse. A gram at a cm. per sec. or a dyne for a second. Kinetic Energy and Work. International or C. G. S. ^=981 nearly. Centimetre. Second of Mean Solar Time. Gram. Dyne = igm. cm./sec.2 =~ of a gm. wt. Eirg =Jgm.cm.2/sec.2 = icm. dyne. British or P. P. S. ^f=32'2 nearly. Foot. Second of Mean Solar Time. Pound. ' Poundal ' = ilb. ft./sec.^ ■ =-of a lb. wt. g A pound at a foot per sec. or a poundal for a second. Foot Poundal =Jlb.ft.2/sec.2 Engineers' or P. S. S. ^0=32 'igi^- Foot. Second of Mean Solar Time. ■Slug' =gi) lbs. Weight of a Pound at Sea-level in London. I lb. wt. —gfi poundals. A slug at a foot per sec. or alb. wt. for a second. Foot Pound Weight = lslugft.2/sec.2 Table VIII. Conversion of Units. Metric to British. British to Metric. I cm. = 0-03280899 ft. I gm. = 0-0022046 lb. I dyne = 0-0000723432 poundal. = 0-0000022473 lb. wt. I ft. = 30-48 cm. I lb. = 453-59265 gm. I poundal = 0-0310644 lb. wt. = 13,823 dynes. I lb. wt. = 32-1912 poundals. = 444,979 dynes. In the two British systems of units, the first in Table vii., often referred to as the British Absolute, takes the pound as the unit of mass, the unit of force being called ih^ poundal, a term suggested by the late Professor James Thomson (see Kelvin and Tait's Natural Philosophy, Part I., p. 229, 1890). In the British engineers' system the pound weight at sea-level in London is taken as the unit force, the unit mass being 32-1912 pounds, for which unit the term slug was suggested by Professor A. M. Worthington (see his Dynamics of Rotation, p. 9, footnote, ART. 215] KINETICS OF PARTICLES 205 1904). It seems desirable to note here that some writers who freely use one or other of these systems object to the terms poundal and slug and prefer to leave the units in question without a name. In the table F. S. S. after the title engineers' signifies/t?^/, slug, second. Corresponding to these two British systems there are two Con- tinental ones {M. K. S.) using the metre, kilogram, and second as the units from which all else are derived. But the one system (called absolute) takes the kilogram as the unit of mass, derives 10° dynes as the unit force and 10' ergs, called a. joule, as the unit of work. The other system takes the weight of a kilogram as the unit of force, and then derives 9' 81 kilograms as the unit of mass. Much controversy has arisen as to the relative advantages of some of these systems of units. The view here taken is that, in spite of any personal preference, it is incumbent on serious students to become conversant with the chief systems that have attained any considerable vogue. Examples — XLI. 1. Accepting the definitions of mass and force, transform the kinematic equations of rectilinear motion under uniform acceleration to the corre- sponding case of kinetics of a particle. Define carefully any new quantities that now enter into the equations. 2. From the fundamental equations of kinetics of a particle derive two new expressions for a force. 3. Describe carefully what you mean by an impulse, and show to what other quantity it may be equated. Obtain the value of an impulse as the area of a curve, and show approximately the proportions of such curves (i) for the blow of the racquet on a tennis ball, and (2) for the blow of a hammer on a nail in hard wood. 4. Assuming that force is equal (or proportional) to mass into acceleration, derive and critically discuss the systems of mechanical units in use among the Enghsh-speaking nations. 5. 'A shot having given size and shape, how will its penetrative power depend (i) on its weight, and (2) on its velocity, the resistance to penetration being supposed uniform ? Give reasons for your answer. 'A bullet fired with a velocity of 2000 fs. penetrates to a depth of 18 inches in wood ; what would be its velocity of emergence if fired through a board i inch thick?' (LOND. B.A. AND B.Sc, Pass, Mixed Math., 1904, i. 3.) 6. ' Determine the tension in any position of the thread by which a body is whirled round in a vertical circle. 'Prove that in a bicycle track looped round in a complete vertical circle of 30 feet diameter, the velocity at the highest point should be due to a fall of 7 feet 6 inches, or about 15 miles an hour; and the reaction of the ground on entering the circular track at the lowest point will be increased to about si.x-fold.' (LOND. B.A. AND B.Sc, P.ASS, MiXED MaTH., I902, I. 8.) 215. Potential Energy and Transformations to and from Kinetic. — Consi'der a particle of mass m moving in a region where it will have uniform acceleration a. Let it have at P, Fig. 76, a velocity u m the direction of the acceleration and w at right angles thereto, its kinetic 2o6 ANALYTICAL MECHANICS [art. 216 energy being then Ta = \m(u^ -^-w"). When it has moved through a space s parallel to the direction of the acceleration, let it be at Q, with velocity components v and w as shown in the figure, its kinetic energy being now T. Then we have or T—T^ = mas=Fs=W (i), where i^is the force ma and W\s, called the work done on the particle by the field while it moves from ^ PI/ P to Q. Also, by suitably chang- r^^ ing our equations, we should * ^\^ obtain the result that the reversed N^^^ velocity at Q would restore the I \\ particle to P with the reverse of "■ i \ its original velocity. That is, the \ kinetic energy would here change Ajc from Tx.0 T^ while the work W f'-T'*W V=v'-w"\'" was done by the particle against " \ the field, for the reversals of the ^ velocities have no effect on their Fig. 76. TRANSFOR.VIATION OF Energy, squares, which alone enter into the expressions for kinetic en- ergies. And, since the cross velocity w disappears from equation (i), it is clear that what we have obtained applies not only to P and Q but to any other corresponding pairs of points on their levels. But though the particle in passing from Q to P loses kinetic energy to the amount T— T^, it acquires an advantage of position in the field, which enables it to gain the like amount of kinetic energy on passing back to the level of Q. This advantage of position is called potential energy, and increases to the extent of the work done by the body against the field. Thus, calling the potential energies of the body Fo and V at the levels of P and Q respectively, we have V,= V+W . . . (2). But by (i) this becomes F„= F+ T~ T^. Hence r-1- F= r„+ r„=constant . . (3). This is an example of the eftnservation of energy, the energy of one kind which is lost by the body being exactly balanced by the energy of the other kind gained by it. Also during the motion either from P to Q or in the reverse direction, we have the transformation of energy occurring from potential to kinetic or the reverse. These transforma- tions and reverses occur automatically in many familiar instances, e.g. a stone thrown into the air, a pendulum in motion. Here, in the rise, kinetic energy is lost and potential energy is gained ; at the summit of the motion the transformation ceases for an instant; in the fall the reverse transformation from potential to kinetic energy occurs. In the case of a pendulum there is also a reversal of the transformation to the opposite kind at the lowest point where the velocity is greatest. 216. Work in Oblique Displacements. — Let us now consider the ART. 217] KINETICS OF PARTICLES 207 work involved when a particle is constrained to move at an angle 6 with the force F. Then, if the element of work dW corresponds to the actual displacement ds, whose component is dy, in the direction of the force F, we have dW=Fdy=Fco&eds^F'ds . . . . (4), where F' = FcosO. This is illustrated by Fig. 77, in which the force ^acts parallel to PN, along which y is measured, but the particle is constrained to move along PQR, along which s is measured. Thus if PQ is ds and PM is dy, then dlV re- presents the work from P to Q. It is seen that the F' in (4) is the component of the force along the path ; hence we may state the equation in words thus : — When a particle is constrained to describe a speci- fied path under a given force, the element of work is expressible by any of the three following products :— (i) the total force into y,^^ 77". ^oj^i, jj, qblique the component displacement in direction Displacement. of force ; (ii) the total force into the total displacement into the cosine of the angle between them ; (iii) the total displacement into the component force in direction of displacement. The total work along any finite portion of the path, say P to R, is given by the corresponding integrals, viz. W=/Fdy=/Fcoseds=:/F'ds . . . (5). If /^varies from point in magnitude or direction or both, /■'and 6 may be expressed in terms of v or 5 and the evaluation effected. If^is constant in magnitude and direction and 6 varies simply owing to the curvature of the constrained path, we may take the first expression to the right of (5), and find JF ■-\'Fdy=Fy (6), where y is the length of PN. It is seen that this use of the constrained path corresponds in this respect with the motion (supposed free) considered in article 215. 217. Direct Impact. — Let a perfectly smooth particle of mass m with velocity u strike a second similar particle of mass m with velocity u, both velocities being along the same line which is also the line of centres of the particles when in contact. Then by Newton's laws or the definition of mass in article 209, the accelerations of the particles are in the negative inverse ratio of their masses. Thus from the equation of article 209 (2) we have mlm'=—a'a ox ma + m'a'=o (i), the fl's denoting the accelerations of the particles. But since this relation holds for any and every instant, the total changes of velocity must be proportional to these accelerations. Hence we have m{v—u) + m'{v'—u')=^o (2) 2o8 ANAL YTICA L MECHA NICS [art. 2 1 8 That is, the algebraic sum of the changes of momenta is zero. Or we may write this in the form mv-^-niv' =-mu-\-m'u' (3), which is equivalent to the statement: — The algebraic sum of the final momenta equals that of the initial. This is often referred to as the principle of the conservation of momentum. The passage from equation (i) to (2) could be expressed fully in symbols thus : — ' m a' f , , f ,, v' — u -,= — = \adt-=r \adt= ma] J v—u Equation (3) is, however, insufficient to determine the two final velocities v and tl . We accordingly require another condition, which is supplied by stating the ratio of the velocity of separation to that of approach. For this ratio, called the coefficient of restitution, is found to be approximately a constant for given bodies. Thus, denoting it by e, we have — v-\-v' = e{u — u') (4). In deriving the above equations it should be noted that all velocities are reckoned positive when in one given direction and negative if in the opposite direction. Care must be exercised if the conditions of any problem are stated in terms u u' which violate this convention. * Fig. 78 illustrates the case of direct impact in question, the V V velocities before impact being in- FiG. 78. Direct Impact. serted above the particles and those after impact below them, any example being directly solvable from equations (3) and (4). 218. Oblique Impact. — To pass from the case of direct impact of smooth particles to that of oblique, it is obvious we have simply to compound with the velocities along the line of centres that which occurs at right angles thereto and which is not changed by the ^ — yr, impact. Denoting these cross U/ j/v'i components by zc/'s, and retaining A/l < ! » the previous notation, we have the y-~// ^ !** complete scheme as shown in — r , — -r-r^-GrC*^ — '•7 — -S — Fig- 79- , Tv\ Tfl ^Xyy-m. -m' In this case, therefore, the v s j J^^Xi are found as for the direct impact, ii^r^T^.i/ and the unchanged zf's being compounded with them give F'«- 79- Oblique Impact. the final velocities in magnitudes and directions, though often it is just as convenient to retain the expressions for the components simply. Where the actual velocities are required, denoting them by capital ART. 219] KINETICS OF PARTICLES 209 letters, and calling the angles with the line of centres before and after impact a and Q respectively, we obviously have ? tan ;8, then 7'>7' and there will be acceleration of the body no matter how small F may be. On the other hand, iftan6l:J>tan/?,then 7>r' (4), and there is no acceleration of the body no matter how great the force F may be. The angle /3 defined by tan j8=ya . ... • • • (s) is called the a7igle of friction, and the cone whose semi-vertical angle is j8 and axis the normal to the surfaces at the point of contact is called the cone of friction or friction cone. Since the vertical makes the same angle with the normal to a surface that the surface makes with the horizontal, it is evident that a surface can be inclined from the horizontal up to the angle of friction before a body will start to slide down it under gravity. So the angle of friction is sometimes called the angle of repose. This relation also furnishes a ready means of finding the values of y8 and i>. for a given pair of surfaces. 222. Motion on Rough Incline. — Consider a body placed on a rough surface inclined at an angle a with the horizontal, the coefficient of (3)> Fig. 81. Motion on Rough Incline. 212 ANALYTICAL MECHANICS [art. 223 friction being /u.= tan/8, where ;8 214 ANALYTICAL MECHANICS [art. 225 Fig. 83. Connected Particles on Inclines. 225. Motion of Connected Particles on Rough Inclines.— Consider now the case of two particles or bodies connected by a thread and each resting on a rough incline as shown in Fig. 83. We shall suppose the masses of thread and pulley and the effect of friction of pulley axle negligible. Also let us at first suppose that the masses, inclinations, and roughnesses are such that motion with acceleration a occurs to the right, M^ descending the incline a, and dragging M^ up the incline a^, the coefficients of friction being ^^ and /tj, and the tension in the thread T. Then considering each mass in turn, we have yl/'i^(sinai— /iicostti)— 7'=J/ia (12), and 2"— j?/2^(sina2+/i2Cos 0^) = J/jfl (13)1 whence, by addition and transformation, we have ^_J^i(sinai— /Xj COSai) — 7l/2(sinaj + /i2 COSOj) And, by substitution of this in (12) or (13), we find „ M^M^ , . , . , , /. One Mass descending vertically drags the other on a Horizontal Plane. — The case just dealt with is very general, and by insertion of suit- able values for the angles applies to various special cases. Thus for the present case we have only to write 01^:90° and 03 = in equations (14) and (15), and we find a __Mi — M^ , ,. -g-M.+M, ^'^^■ (14). (XS)- and M,M,g (1+/^.) (17), 'M,+M, results which are easily obtained by direct consideration of this problem. We can, of course, pass to the case of a smooth horizontal plane by writing 1^-^ = in (16) and (17). //. Jioth Masses hang vertically. — We now write aj = 02=:9o°, and find on substitution -^^-^^~ (18), and T=. M^ + M,^" 2MiM„_g (19). ' M^ + M., which, by their agreement with (3) and (4), serve as an additional check. ART. 226] KINETICS OF PARTICLES 215 Examples — XLIII. 1. Explain the phrases angle of friction and cone of friction, and illustrate by a sketch. 2. Discuss the sliding of a body down a rough incline when by a thread and pulley it pulls another body up a second incline also rough. Check your result by reducing the second plane to a horizontal one. 3. ' Describe the use of Atwood's machine for the experimental verification of the Laws of Motion.' (LOND. B.Sc, Pass, Mixed Math., 1902, 11., istpart of 6.) 4. Show how to allow for friction in Atwood's machine. 226. Pendulum in Accelerated Chamber. — A simple pendulum of length / hangs from the roof of a chamber which has accelerations a vertically upwards and b horizontally, let it be required to find the circumstances of the motion in the plane of a and b, and the tension T of the thread. A brief consideration will serve to show that the zero position will be displaced and the value of the effective g, and therefore also of the period t, altered. But perhaps a detailed ex- amination is desirable. The case is repre- sented in Fig. 84, in which S denotes the point of suspension, P the bob of mass m, SL and SM vertical and horizontal lines which move parallel to themselves. The pendulum is shown dis- placed from the vertical by the angle Q, and is supposed to have angular velocity and accelera- tion d and Q respectively. It will be convenient to write expressions for the component radial and transverse ac- celerations and equate their values to the quotients (resultant force/mass) for the corre- sponding directions. Thus from equations (5) and (6) of article 74 we have for radial and transverse accelerations about a fixed point the general expressions . . . f^r—rd"- a.ndj = r9 + 2rd. In the present case of r=/= constant, these would become f=-Id' iind/=ie if 5 were at rest. Thus, taking into account the accelerations a and b, we have for the total radial acceleration in the direction S to P and the corresponding force divided by mass the expressions /, —T+mg cos 6 Fig. 84. Pendulum in acceleeated Chamber. ■ 19' = - or Tlm — {g+a.)cos,6-b%\nd+l6"' . . . (i). We have similarly for the-transverse acceleration, reckoned in the direction from SL to P, and the corresponding force divided by mass asme+bcos9+W=- ^ Qf (^+a) sin ^-|--5cos ^+^^=0 m (2). 2i6 ANALYTICAL MECHANICS [arts. 227-228 227. The similarity of form of these equations suggests the possibility of a simplification, and obviously in (2) on putting 0=^o the remainder of the equation defines the displaced zero line which takes the place in the moving chamber of the vertical in the chamber when at rest. And on examination of equations (i) and (2) we find they are susceptible of the form Tlm=gfco%<}>+lil>' (3), and ^'sin^+/^=o (4), where ^= J(g+af+b' (s), 4>=e+a. ... . . (6), and tana=^/(^+a) (7). Thus, as the new variable angle 4> exceeds the former one by a, we see from (4) that an angle must be measured from the vertical in the negative direction and equal numerically to a to represent the zero line. This is shown by SN in Fig. 84, in which SL and SM are pro- portional to g+a and b respectively. If we now confine ourselves to small oscillations about this zero line, for which sin ^=<^ nearly, equation (4) is replaced by ^'<^+/^=o,or^=^^^ (8). Hence the motion is simple harmonic of period T'=27rV/// (9), and is fully represented by (t>=6,cos{jYl/i+S) (10). In this the <^o and S are arbitrary constants to be determined by the initial conditions. Thus, if the pendulum start from rest with amplitude yS, we have for t=o, 4'=P ^nd (f> = o. Hence therefore 4>(,=P and 8=0. Hence (10) becomes 4>=lizo^U7Tit) (11). the velocity being given by s('=-^x/i7^sin(V//J/) .... (12). 228. Results and Applications. — Thus, the initial conditions being known, equations such as (11) and (12) substituted in (3) give the tension at any instant, e.g. for ^=0, '4>=o, <^=ft and T—mg'cosP. Hence, to sum up, when a pendulum makes small oscillations in an accelerated chamber, they ar« performed (i) about a displaced zero defined by (6) and (7) ; (ii) in a disturbed period t' given by (9) ; (iii) as though gravity had the disturbed value g', in (5) ; and (iv) the tension of the thread ioWo-^s from this disturbed^, as shown in (11), (12),' and (13). We have accordingly confirmed by strict analysis the view which might have been conjectured at the outset, and may be stated thus : — To find the behaviour of a pendulum in an accelerated chamber compound the acceleration due to gravity with the reversed acceleration of the point of suspension ; this gives in direction and magnitude the ART. 229] KINETICS OF PARTICLES 217 disturbed or effective gravity. Then the oscillations occur about the direction of this effective gravity as zero position, the period and tension being expressed in terms of this new gravity just as they are for an ordinary pendulum with the actual gravity. Obviously this examination applies to a pendulum inside the carriage of a mountain railway when starting or stopping, also to a carriage or cage slipping under gravity down a steep incline and whether or not it is pulling another up. Of course, uniform velocity of the carriage along a straight line has no influence on the pendulum. In the case of a train, with acceleration b, on a horizontal straight track, we have only to put a = o and let Fig. 84 be the longitudinal section of the carriage. For the case of a lift, with vertical acceleration a, we have simply to put b=Q. We may note also that the zero line is not now displaced. Thus, the presence of the vertical acceleration a alone has no power to displace the zero line, although it can alter the displacement produced by the horizontal acceleration. See equation (7). 229. Pendulum in Carriage round a Curve : Elevation of Exterior Bail. — For a train moving at uniform speed v along horizontal rails curved to a radius R, we must take Fig. 84 to be a cross section of the carriage or an end view inside. Then, putting a=^o and b-^v'^fR, we find by (7) is.na=v''IRg . (13). And, since this is the angle of the effective gravity to the vertical, it should be the angle of elevation of the road crosswise, with the hori- zontal. Or, in other words, the exterior rail should be elevated u as seen from the interior one, in order that the train may -proceed as if going straight along a level track. This result could have been seen from the first, but serves here as a useful check on the methods of the pendulum problem. Examples — XLIV. 1. 'A railway carriage is moving with a constant acceleration of a feet per second per second along a straight road, and a particle is suspended by a fine thread from the roof of the carriage. The acceleration ceases at a certain point, while the rectilinear motion continues ; then the carriage moves with a velocity v i.js. on a curved part of the line where the radius of curvature is p feet. 'Trace the motion of the pendulum through these changes.' (LoND. B.Sc, Pass, Mixed Math., 1904, 11. 8.) 2. ' Prove that a man, weighing w lb., moving about in the cage of a lift equilibrated by a counterpoise of IV lb., will experience an apparent relative field of gravity 2 Wglii W+ iv).' (LoND. B.Sc, Pass, Mixed Math., 1902, 11., 2nd part of 6.) 3. ' The gauge of a railway being 4 feet 8 inches, calculate the elevation (m inches) of the outer above the inner rail, so that there shall be no flange pressure when trains travel at a speed of 30 miles per hour, at a curved part of the line where the radius of curvature is J mile. ' Show that if trains travel at this part of the line with a speed of 45 m./h.. 2i8 ANALYTICAL MECHANICS [art. 23° there will be a flange pressure against the outer rail equal to 11 W\liyv]=g{a' + 5a-di+5ad'-t°'+i''f)di. Whence, on integrating, we obtain (a + bifv = ^(4fl'iJ/-l- (iO'bH- + Aab'f -f b't% ,^^[,+,,_^^] . . . . (3), or _ the constant of integration being zero for v=o when the radius was a. 234. Slip of Snow on a Slope. — Consider the case of a uniform layer of snow on a slope inclined a to the horizontal, the adhesion being just sufificient to hold it while at rest. Next, suppose the upper line of snow next the ridge (of roof or mountain-side) to be moved downwards, the friction between it and the slope being negligible. Then the snow will move down from the top and start other portions till the whole mass is sliding down. Let us . find the circumstances of this motion. Consider a portion of the snow of length b parallel to the ridge, and suppose a breadth x of it down the slope to have started and to have velocity v. Then, neglecting friction, the equation of motion may be written -j(fibxv)=-ijbxg^ma. (3), where a- is the mass of snow per unit area. If a be written for ^sin a, this becomes ART. 235] KINETICS OF PARTICLES 221 x-+v=ax (4). But dv_dv dx __vdv _ d{v^) dt dx'dt~ dx~^^dx' '5)' So, introducing (5) in (4) and multiplying by 2x, we obtain x'^dv'' , —f-r + 2*» = 2ax , or d{fev')^2ax''dx ... . . (6). Whence, by integration, we obtain ■" =2~x=~{g^\no)x . -(7). The constant of integration is zero, since v and x vanish together. We thus see that, in this imaginary ideal case, the acceleration is one-third that of a compact body sliding freely down the same slope. The substance of articles 230-234 is derived from the elegant treat- ment of these topics by Dr. Besant {Dynamics, London, 1885). 235. Note on Vibrations. — The chief vibrations of a particle with which we are concerned have been dealt with already in Chapter iv. (see articles 29-33, 45-48) and Chapter vii. (see articles 109-111), The conditions under which some of them are possible may be inferred from the later treatment of elasticity in Chapter xxi. In the case of forced vibrations (articles 46-48) we may approxi- mately realise the phenomena in question by attaching a small bob of mass m to a very large one of mass M, both the pendulums being of about the same length and period. Then, if the large bob be set in oscillation, the consequent obliquity of the suspension of the smaller one below will supply the impressed force which gives to that small bob its acceleration varying as a sine function of the time, in addition to its own acceleration proportional and opposite to its own displace- ment. But, in this actual case, we must note that with masses of a finite ratio the small mass m will, by its motion, react upon the larger mass M. It is therefore only as the ratio MIm approaches infinity that we approach the ideal case of article 46, in which the impressed acceleration is unaffected by the forced vibrations of the particle. Examples — XLV. 1. A ring weight is let fall so as to strike and lodge on the small ascending weight of an Atwood's machine. Discuss the motion which ensues in the case where the small weight plus the ring weight are together heavier than the large weight of the machine. 2. Show that the acceleration of an ideally simple avalanche is of the order one-third that of an ordinary compact solid on the same slope. 3. Find the pressure exerted at any instant by a chain falling on a table. 4. Determine the velocity of a raindrop which picks up stationary moisture so that its radius grows proportionally to the time. ANALYTICAL MECHANICS [art. 236 CHAPTER XIII PLANE KINETICS OF RIGID BODIES Fig. 85. Rotation of Rigid Body. 236. Accelerated Rotation of a Rigid Body about a Fixed Axis.— Let us consider the rotation of a rigid body about a fixed axis under the action of forces. Then because the body is supposed rigid all the motions are parallel to a given plane, which is taken as that of the diagram in Fig. 85, the fixed axis being perpendicular to it at O. Consider first a particle of mass m-^ at a point whose projection is Pi, distant perpendicularly r^ from the axis and having a linear acceleration CTi, which is of course in, or parallel to, the plane of the diagram and perpendicular to OPj. Then the force on this particle is given by Fi = m^a^ (i), and is the direction of the accelera- tion fli. Similarly, for another particle at Pj, we may write F^=m^a^ (2), and so forth for all the particles in the body under examination. But it is obvious that equations of this kind cannot be added arithmetically, for the corresponding pairs of a's and F'% are in various directions, though all parallel to the plane of the diagram. A little alteration, however, enables us to effect this simple addition. We tlQ)\& first that, though the linear accelerations denoted by the a's may all have different magnitudes and directions, the angular acceleration about O of all the particles is the same in magnitude and direction, and may be denoted by a—a^lr-^—a^jr^, etc. Thus a may be made a common factor on the right sides of the equations. Secondly, let us pass from the forces F^, F^, etc., to their moments about O, which are represented by i'l/'i, F^r^, etc. Then, since each of these moments is a product of two vectors, each may be represented by a vector perpendicular to the plane of its components. But the components are in the plane of the diagram, so the resultant is perpendicular to tliat plane. Further, the direction of this resultant or moment vector is related to the direction of the force about O, as the advance of a right-handed screw is related to its rotation. Thus the products F^r^, F^r^, etc., may be added algebraically, for they are all representable by vectors along the same line, viz. the perpendicular ART. 237] PLANE KINETICS OF RIGID BODIES 223 to the plane of Fig. 85, and out towards the reader if the moment is counter-clockwise. Lastly, when the equations are thus transformed, we have on the right side, as coefficients of the angular acceleration a, terms of the form Wi^?, m^ri, etc. But it is known by the theory of vectors that the product of collinear vectors is a scalar quantity, hence all these terms may be simply added. We accordingly find, from (i) and (2), F,r^-=miria.] ' ' ' ' ■ ■ ■ (3;. Whence, on addition l(Fr)=aImr'' (4). Now the forces denoted by the F's on each of the particles of the body have been hitherto treated as the resultant forces on these particles. So each /"may be made up of one force F due to external bodies, and another force F due to the interaction of the particles of the body itself. We should accordingly have F=Fe+Fi {4a). But, since these internal forces of mutual interaction occur in pairs whose members are applied along the same line, opposite in direction but equal in magnitude, it is obvious that in the summation their effect will disappear, i.e. '2{Fir) = o (4^). Hence the summation in the left side of (4) may be taken as applying to the external forces just as though the others did not exist. 237. Moment of Inertia and Torque. — Equation (4) contains, under the signs of summation, two new quantities of great importance which need definition and symbols to represent them. On the left side we have 2(/>-), which is the sum of the moments of all the forces about the axis through O. We call this briefly the total or resultant torque about O and denote it by G. On the right side we have ^(inr^), which is called the moment of inertia of the body about the axis through O, and may be denoted by I. Thus (4) may be rewritten in the form G=Ia . . (s). We may further write from (5) and (4) as analytical definitions of moment of inertia I=Gla = l,{mr') (6). The first of these forms (6) may be regarded as the dynamical definition, like (s). Whereas the second expresses / in terms of the masses m of the particles and their perpendicular distances r from the axis, and may be called the geometrical definition. It, of course, forms the basis for evaluations of the moment of inertia of any given body, while the first indicates the dynamical significance of the quantity itself. The meaning of moment of inertia will become clearer when we introduce it in the equations of uniformly accelerated rotation. Note first how the dynamical equations (3) to (6) of article 212 were derived 224 ANALYTICAL MECHANICS [art. 237a from (i) and (3) of article 27, by the introduction of mass. Next, refer to equations (i) and (3) of article 92 and multiply throughout by /, writing G for la.. We then obtain Iw—Iw„^Iat=Gt . (7), and ^/w''-^/i4=/a0=Ge . ... (8). Equation (7) presents two new products which require names. The first, /\ = Ia6=G6,\ ' oxT-T^=W. ] Radius of Gyration. — It is often convenient to express the moment of inertia as the product of the total mass M of the body and the square of a length k called the radius of gyration. We accordingly have the relations Y.mr'^i^m)k\\ I=Mk'' / or (9)- 237a. Rotation about fixed Axis: General Treatment. — Let us now consider the rotation of a rigid body about a fixed axis, the point ART. 238] PLANE KINETICS OF RIGID BODIES 225 OP in the .«>; plane have length r in- clined e to OX, as shown. Then, as the body rotates, r will remain con- stant for the given particle, while d, X, and y will change. We shall de- note the angular velocity ^ by w and the angular acceleration 6 by a. Then, since the angular mo- mentum ^ about any axis is the sum of the moments about it of all the linear momenta, and the latter are expressed by the algebraic sums of their rectangular components, we have for the angular momentum about OZ H='2m(yx—xy) (i). But, on reference to the figure, we see that .»=rcos0, y = rsmd \ / \ x=^ — rsin 9.6= —(i)y,yz=rcos6.d=(Dxj ^^'' Hence by (2), (i) becomes //= 'Zm(o)x^ -\-<^y^) = <.i1m(x''+y'), H=I<. ... . . ... (3). Fig. 85A. Gkneral Rotation ABOUT Fixed Axis. or Thus, on differentiating (3), we have li=Ii>z=Ia . . But, on differentiating (i), we find Jjr= -r ^m{yx — xy) - --^m{vx — xy) (4). (S)- Hence, on writing for the products, mass into acceleration, the symbols for the corresponding force components, X and Y due to external bodies and X' and Y' due to mutual interactions, we find from (s) Ji=.mY+Y')x-{X+X')y} = l,{Yx- Xy)+l.{Y'x- X'y), or B'=^Yx-Xy) = G .... since the summation for the internal forces vanishes. (6) give G^Ia ■ ■ ■ (6), Thus, (4) and • • • (7) in agreement with (5) of article 237. We shall see later that, if rotations are occurring about various axes, (s) and (6) still hold, but not (7), for If is then no longer reducible to Iio. 238. Parallel Axes Theorem.— As a preliminary to the evaluation p 226 ANALYTICAL MECHANICS [art. 239 Fig. 86. Parallel Axes Theorem. of the moments of inertia of typical figures, several theorems are needed, and will now be given. Consider the moments of inertia K^ and AT of a body about two parallel axes, the first being that of z and the second passing through a point A on the axis of x. Let Fig. 86 represent the xy plane, C being the origin of co-ordinates. Suppose the body of total mass Mto have a particle of mass m at B{xy), the sides of the triangle A, B, C being denoted by the corresponding small letters a, 6, c as usual. Then, beginning with the geometri- cal definition of moment of inertia, and using the well-known trigonometrical relation, we have K=lmic' = ^m{d'-\-b'' - 2ab cos C) = '^ma^ -^-^mb^ — 2bllma cos C, or, K=K,-\-Mb''-2b1mx (i). Hence, if the axes are so chosen that '2.mx=o (2), equation (i) becomes K=K,+Mb' (3). If, in addition, we have also '2my=o, the axis of z passes through the point defined by '2,mx='2,my=^mz=o (4\ which point is termed the Centre of Mass of the body, and possesses various important properties, as we shall see later. In this case equation (3) embodies the theorem to be established, which may be worded as follows : — Theorem. — The moment of inertia of any body about any axis is the sum of that about a parallel axis through its centre of mass plus the product of the mass of the body and the square of the perpendicular distance between the axes. Hence, in evaluating the moments of inertia of typical bodies, it often suf- fices to take the axis through the centre of mass and with the required orienta- tion. The value for any parallel axis then follows from this theorem. 239. Lamina Theorem. — Let the body be in the form of a thin plane lamina and take the axes of xy in this plane as shown in Fig. 87. Let the moments of inertia about the perpendicular axes x and y in the plane of the lamina be I and / respectively and K denote that about the axis of z through the same t^\j) Fig. 87. Lamina Theorem. ART. 240] PLANE KINETICS OF RIGID BODIES 227 Then it is re- origin but perpendicular to the plane of the lamina quired to show that I-\-J=K. Take a point Y{xy) distant r from O, and let a particle of mass m be situated there. Then we have ""^ . ^ K^I'rJ (5), as required. "' . We see at once from this that if different axes OX' and OY' are taken in the plane and through O, though the moments of inertia about them may change, their sum remains constant. For r+J'=K=I-\-J (6). 240. Rectangular Axes Theorem have now a body with any dis- Y for any Body. — Suppose we tribution of matter in solid space, the moments of inertia about the axes oix,y,nx\d z being f, J, and K. Let any radius vector of length r be drawn from the origin to the point P, where there is a particle of mass m. Then it is required to show that /+/+7r=22w^r^ Let the co-ordinates of P be X, y, and z as shown in Fig. 88. Then we have by definition and the geometry of the figure I=^m{v'+z-), /=Ym{z'-+x-), Jir=i:?n(x'-+y'). Hence, by addition, we obtain the result Rectangular Axes Theorem. as sought. /+/-f 7r= 2^m{x'+y' + z'') = 2-Emr' (7), Examples — XLVL 1. For the rotation of a rigid body about a fi.\:ed axis obtain an expression analogous to E= ma for the kinetics of a particle. Explain carefully what quantities now replace F and. ni. 2. Define moment of inertia, torque, angular momentum, and impulsive torque ; and write equations exhibiting relations between these quan- tities and others involving kinetic energy of rotation and the corre- sponding work. 3. Show that of all parallel axes in a body that about which the moment of inertia is a minimum passes through the centre of mass. Also find a general relation between all the moments of inertia about these axes. 4. Any number of particles of equal mass are arranged equidistantly on the circumference of a circle and rigidly connected to the centre by bars of negligible mass. Show that about a central axis perpendicular 10 the plane of the circle the moment of inertia is only half that for a parallel axis through a particle. 228 ANALYTICAL MECHANICS [art. 241 Moment of Inertia about AN Oblique Axis. 5. Show that for a square lamina of uniform thickness and density the moment of inertia about any central axis in the plane of the lamina is the same. Prove the corresponding relation for a regular octagonal or duodecagonal lamina. 6. Consider the rotation of a uniform filament about perpendicular axes through the centre and through an end, and thence show that the moment of inertia for the latter case is one-third mass into square of length. 241. Theorem for Oblique Axis.— Let us now determine the rela- tion between the moment of in- ertia of a body about any axis through the origin, the moments of inertia about the co-ordinate axes, and other necessary constants being known. Let I denote the moment of inertia about the axis 00', whose direction cosines are A, /i, and v. Let a particle of mass m be situated at the point P, whose co-ordinates are x, y, and z, and join OP, as shown in Fig. 89. Let fall the perpendicular PN upon OO', also dr^w PM parallel to the axis of z meeting the xy plane in M, and ML parallel to the axis of y meeting the axis of X in L. We may note some relations that will be required as we proceed. Thus, we have at once by the geometry of the figure OP==x^+/+«' (8). Also, by regarding ON as the projection upon 00' of either OP or its components OL, LM, and MP, we see that ON=Aar-f /y-f-v2: . .... (9). Finally, X'+jliH..' = i (10). Then, by definition, we have /=2»?NP' =2w(0P' -ON") or I=^m(y''-^z')X^^-2.m{z'^-x'Y-{-^m{x^-^y''y — 2^myzii.v — 22mzxv\— 2S,mxykfi (n)- In the three summations of the first line of (11) we recognise the moments of inertia of the body about the co-ordinate axes, which we may denote by A, B, and C respectively. The second line of (11) contains three other summations, Imyz, etc., in which the products of co-ordinates occur instead of their squares. These are called products of inertia, and will be denoted by D, E, and F respectively. Thus (11) may be written briefly I=Ak''-\-Bli.''^Cv'-2Dll.V-2EvX-2F\lX . . (12). The foregoing follows the treatment of Routh, who calls this the theorem of the six constants. ART. 242] PLANE KINETICS OF RIGID BODIES 229 When three rectangular lines meeting in a given point of a body are such that, if taken as co-ordinate axes, we have '2/myz = ^mzx=^mxy = o (13), then these are said to be the principal axes at the given point. Also, the moments of inertia about the principal axes at any point are called the. principal moments of inertia at that point. The constants in (12) are reduced from six to three if the co- ordinate axes are taken so as to be the principal axes at the origin. In this case (12) becomes I=A\'' + Bix^ + Cv-' (14). In some simple cases we can determine the position of these axes by considerations of symmetry. Thus, if a body is symmetrical about the plane oi yz, then for every particle of mass m at {x, y, z) there is an equal mass at { — x, y, z). Hence the summations 'Emzx and 'Zmxy would vanish. If, in addition, the body were symmetrical about the zx plane, we should have a particle at (x, y, z) balanced by one at (x, —y, z), so that "Zmyz would vanish also. Hence the conditions of (13) are fulfilled, and (14) becomes valid for the case in point. 242. Typical Moments of Inertia Evaluated.— The moment of inertia of a continuous body, often hitherto represented by 2;»r^ is really an integral, and must usually be treated as such and evaluated accordingly. For when the distance from the axis varies continuously the small particle m at any point must then be replaced by the product of the density of the material, and the infinitesimal space occupied by the portion of it under consideration. Then the space integral must be taken between such limits as will include the body under treatment, In different cases different kinds of space and density occur. Thus, for a wire or filament, we are concerned with length and linear density. For a thin sheet or shell we have surface and surface density. And for any ordinary solid extended in the three dimensions we have volume and volume density. Some cases of moment of inertia require rather complicated working, while some are so simple that they can be written down at once. We shall begin with such simple ones, and using the theorems as required proceed to the more difficult examples. Students, without knowledge of the integral calculus, may evaluate most moments of inertia by taking on faith the formula n where a Sind 6 are the Hmits of r, and .f is a small increase of r, which is the distance from the axis. Circular Wire and Cylindrical Shell— L&t the mass be iJ/ and the radius a, the moment of inertia about the geometrical axis bemg /. Then obviously since the r in. 2^?^' is everywhere the same, we have , , I=^Ma^ (O- 530 ANALYTICAL MECHANICS [arts. 243-244 243. Filament about Perpendicular Axis. — Take now the case of a straight filament OA of mass M, and length a about a perpendicular axis through one end O. Then the linear density is Mia, and we may take as our element PQ of length dx situated at x from the origin where the filament meets the axis OY about which the moment of I I •■ inertia is to be taken, as shown in Fig. 90. —Br ^ dx >* ^ Then the mass of our element is Mdx/a, Fig. 90. Moment of In- and this replaces the m in 'Zmf^, the ^ ERTiA OF Filament. being replaced by x^. Further, the limits of integration are clearly o and a. Thus we have for the moment of inertia I=^r^^dx=^.^^ = ^Ma^ (2). «7o « 3 3 It is obvious that this result applies equally to a rectangular lamina of mass M about one edge as axis, like a door of width a and negligible thickness turning on its hinges. For, the lamina may be regarded as made up of a large number of filaments whose masses add to give that of the lamina, the moments of inertia similarly adding to the expression in (2). Again, equation (2) expresses the moment of inertia of a filament of length 2a about a perpendicular axis through its centre, provided the total mass is M. It should be noted that in this case the linear density is Mjia, just half the. previous value. Suppose now we take a filament of length a-\-b about a perpendicular axis through the point leaving a and b on either side, the total mass being still M. We then have for the moment of inertia /=— — x'dx-^X — ! — )-—{a' — ah-\-b^). (3). a^rb]-a a^-b\ 3/3^ Obviously the cases covered by (2) and (3), where the axis passes through the filament instead of at one end, apply equally to a rectangu- lar lamina about an axis in its plane parallel to one edge instead of coincident with an edge. 244. Lamina and Parallelepiped. — Consider now a rectangular lamina of mass J/ with edges 2a and 2b parallel to the axes of x and j/ respectively, the origin being the centre of the figure. Then, if the moments of inertia about the axes of x, y, and z are /, J, and K respectively, we have by (2) of last article and the lamina theorem (see equation (5), article 239) I=\Mb\J=\Ma\K=\M(a-^b') (4). But it is obvious that the last of these results, giving K, would apply equally if we passed from the single lamina to a parallelepiped of any thickness, 2c say, and of total mass M. Thus, by symmetry, we have for the parallelepiped ARTS. 24S-245«] PLANE KINETICS OF RIGID BODIES 231 and (5). Fig. 91. Moment of Inertia OF Disc. Hence we have 245. Circular Disc about Axis and Diameter r^^ncVi 1 required to find its moments of inertia /about a diamet;r and^about the axis perpendicular to its plane. If "'"'ecer ana a aDout we take a second diameter perpendicu- lar to the first, it is obvious from sym- metry that the moment of inertia/about this is equal to /. Hence the lamina theorem applied to the disc yields /^=^ (6). It IS simpler now to determine K directly by integration and deduce I, thus reversing the procedure followed for the rectangular lamina. For our element we take a ring of radius r and radial width dr, as shown in Fig. 91. The area of this element is 2'Krdr, its mass 0- times this, and its moment of inertia r" times the previous product. dK=- {2Trrdr)(Tr'^ = 2Trcrr'dr. The limits of integration are o and a, thus we obtain K—2it(t\ r^dr= = JJ/a= (7). Whence by (6) we see that I=lMa- (8). It is obvious that (7) will apply also to a cylinder of any length rotating about its geometrical axis. For a cylindrical tube of radii a and b outside and inside and density p, we have from (7) K'=l■Ko■{a*-b') = l{■K<^{a■'-by|{a^+b^') = \M{a'+b'){^a). 245a. Elliptic Lamina — Consider now a very thin lamina of mass Mm the form of an ellipse with semi-axes a and b along the axes of X and y respectively, and let the corresponding moments of inertia be I and/, that about the axis of z being K. Then, since this elliptical lamina differs from a circular one of radius b only in the extension of material parallel to the axis of x, which makes no difference to the moment of inertia about this axis, we have from (8) I=\Mb'- (9). Similarly J—\Ma- (10). Hence, by the lamina theorem, K=\M{a'+b'') (11), which is seen to agree with (7) if b=-a. 232 ANALYTICAL MECHANICS [ARTS. 246-247 246. Spherical Shells and Solid Sphere. — Consider a very thin spherical shell of radius a and total mass M, and let /, J, and K denote its moments of inertia about three rectangular axes meeting at its centre. Then by the three axes theorem, equation (7) in article 240, we have I+/+X=2'2mr' = 2Md' (12), since every r=a. But, by symmetry, I,/, and .AT are all equal, so that each is one-third their sum. Thus we have /=/=X=jMa' (13), giving the moment of inertia of the shell about any diameter. We may now easily pass to the moment of inertia about its diameter of a solid homogeneous sphere of radius a and density p by using a shell of radius r and radial thickness dr as our element. The area of this shell is 47^/-^ its volume dr times this, and its mass p times the product. Hence by (13) we have o dl= — (4Trr'drp)r' = — irpr^dr . .... (14). Thus, taking the integral between the limits o and a, we have _ 8 f 4 , 8 a' 2/4 3 \ 2 3 Jo 3 5 5 ^ 3 ^ f=~Ma' (15), where Mis the mass of the solid sphere. For a shell of finite thickness we have obviously only to integrate the same expression from the internal radius, d say, to the external radius a. Thus we have I=-^P r*dr=-^n~^- (16); 3 Jb 3 5 or, introducing the mass M of the shell, which is —Trp(a'—d'),lhis becomes /= — J/-5 — jg .... (17). 247. Eight Prism about Perpendicular Axis. — Consider the moment of inertia of any right prism of homogeneous material about any axis perpendicular to its geometrical axis. Take the axis of 2 along the axis of rotation and the axis of x parallel to the geometrical axis of the prism, the plane of xy being that of the diagram Fig. 92, in which ABCD shows the prism, GG' its geometrical axis. Suppose a particle of mass m of the body to be at the point P of co-ordinates {x, y, z), then for the moment of inertia about the axis of z we have = 'Smx^+'S,my', or i:=F+S (18), R Y S A ^^^.z) C ^.---^ Ci ^--^ 1 P F „ C S' D ^ ART. 248] PLANE KINETICS OF RIGID BODIES 233 where /^ and i" represent respectively the moments of inertia about the ^is of z of the bodies produced by condensing the prism first to the filament FF' along the axis of x, and second to the slice SS' in the yz plane. Thus, if for the prism itself were substituted these two ideal figures, the moment of inertia would be unaltered, although the mass would be doubled. Usually the K is required for the case of O at the centre of mass ; the corresponding ex- pressions for F and S are then simpler than for an asymmetri- F'g- 92- Moment of Inertia of Prism. cal case. Thus, for the moment of inertia of a right circular cylinder of mass M, length 2a, and radius c about a central axis perpendicular to its geometrical axis, we find from (2), (8), and (18) K=^Ma'+~M^ .... (19). 3 4 Also, for a rectangular prism of length 2a and width 2b about a central axis perpendicular to these directions, we find, from the prin- ciple of this article, the result already given in equation (5), viz. K=^-M{a' + b"-) (20). Examples — XLVII. 1. Find the moments of inertia about its three edges of a brick of size 9 inches by i^^ inches by 3 inches, the mass being 9 lbs. Ans. 871, 270 and 303! lbs. inches'' about the 9 inch, 45 inch, and 3 inch edges respectively. 2. Show by direct integration that the moment of inertia of a uniform cyhnder about its axis is half mass into radius squared. Thence show that for a disc about a diameter the moment of inertia is one quarter mass into radius squared. 3. Find about a diameter and about a tangent the moments of inertia of a spherical shell and of a solid sphere, the densities being uniform in each case. 4. Obtain a general expression for the moments of inertia of any prism about a central axis perpendicular to the geometrical axis. 5. Find the moment of inertia of a right circular cone of uniform density about an axis through the vertex parallel to the base. Ans. ^M{na^ + b^), where a is axis and b base radius. 20 248. Triangular Lamina about Axis parallel to Base. — Let the triangular lamina be represented by ABC in Fig. 93, its sides being a, b, and c, its mass M, and its surface density o-. As seen in the figure, the corner C is taken as the origin of co- 234 ANAL YTICAL MECHANICS [art. 248 ordinates, the axis of x being along the side a. Let us first find the moment of inertia / about the axis of x. Take as the element the strip PQ of ordinate J and width dy. Then, if the perpendicular EA of the Fig. 93. Moments of Inertia of Triangle. triangle is denoted by p, the length of the strip PQ is a{p—y)lp. Thus we have 'jp-^y^'-jYi-'Z 12 or o (21). Take now a parallel axis through the centre of mass G, which is easily shown to be one-third up the median DA. Then, calling the corresponding moment of inertia /„, we have from the parallel axes theorem =^^'(5-")- So i.=-^Mr (22). Further, take a parallel axis through A, and call this moment of inertia J'. Then we have again by the theorem r=i^+^(^/)-Mf{±+^\ or 2 (^3)- ARTS. 249-250] PLANE KINETICS OE RIGID BOT)lES 235 Bai^^tn??.^^*" J*^ p-""^ ^^?* °°P^*''^'^ ^^^ perpendicular to Base—Still referring to Fig. 93, let us denote CE by y and CF by n JavT'frim (""' '""■''" ^''°"' '"' '"'' °^ "^ ^"^"g called /,Ve __r. y(y + a)o- _y'' + a° 6 2 y+a ' or J=^^M(a'-ay^f) (3^) Transferring now to a parallel axis through G, whose abscissa is we find for the moment of inertia \ 6 9 /' whence /„ = ^(«+«7+/) (25). Passing now to the parallel axis through A, and calling the moment of inertia y, we find M "* whence 7'=-^ (a' + 3«r + 37') /'=f(^^+/37+7=) (26), where y8 is written for fl + y, /.e. for EB on Fig. 93. 250. Triangular Lamina about Axes perpendicular to its Plane. — With the usual notation, let K denote the moment of inertia of the lamina about the axis of z perpendicular to its plane, K^ and K' standing for the moments about parallel axes through the centre of mass G and through the vertex A respectively. Then, by the lamina theorem, we have M ^.=/o+/o=-o(/*H«=+«r+r-) 18^ M, 36 or JiTo^^ia'+i'+c') (27)- To obtain -K' from this result and the parallel axes theorem, let us denote by m the median AD. Then we may easily see that 4m^ = 2^' + 2c'-a' (28)- 236 Thus Whence by (28) ANALYTICAL MECHANICS [art. 251 36 + 9 /■ =m{^- ^' = ^(3^= + 3/-a^) (29). We may obtain the same result by using the relation K'=J'+J'. Also by either of these methods, or by symmetry of notation merely, we find K=f{?,a^-V?,b'-^) (30)- method 251. Direct Methods for Triangle about an Axis perpendicular to it will be convenient to take the axis through the vertex A of the triangle and take this point for the origin of co-ordinates, though the moment of inertia will still be called K' to agree with the previous notation. Thus, referring to Fig. 94, we take the strip PQ parallel to the axis of x as our element. Its ordinate being y, its length is obviously ayfp. Its moment of inertia about an axis perpendicular to the plane of the diagram and pass- ing through -its middle point R is accordingly given by Fig. 94. Direct Method FOR Triangle. i/aydya\/ay\^ 3\ / )\2p) ■ But, for its moment of inertia about the parallel axis through A, we must add the product mass of the element into the square of AR. Now AR is evidently given by myjp, where m as before denotes the median AD. Hence «'=("7")(i(g)"+(?)T Then, integrating between the limits o and /, K'=^,{a^ + i2m')^/dy 12 4 or 24^ ' • (31), which agrees with (29). ART. 252] PLANE KINETICS OF RIGID BODIES ,37 -fC^'^H-t-V).. 3f ^0 2 6 = J/ .2j8^ + 2/3y+2yH6/^ 12 = -|3(/ + r=)+3(/+/3^)-(^^-2/3y + /)|. Hence, as before in (31), we have (32). K'=~J.5l>' + Zr-a^) 252. Triangle about Central Perpendicular Axis by Dimensional Method. — We may now illustrate the use of the B method of dimensions or dynamical similarity in calculating moments of in- ertia. Referring to Fig. 95, let ABC be the tri- angular lamina of mass M, its moment of inertia about an axis through its centre of mass G and per- pendicular to its plane being K^. Take the middle points DEF of the sides and join them to each other and to A, B, and C. Then we have four triangles all similar, each of half the linear dimensions of the original triangle, and therefore of one quarter the area and mass (■M/4). Hence, if we write k for the radius of gyration of the triangle ABC, we have x^=MA'- (33)- A Fig. 95. Moments of Inertia of Triangle BY Dimensional Method: 238 ANALYTICAL MECHANICS [ART. 253 Whereas the moments of inertia of the small triangles about axes perpendicular to their planes and passing through their centres of mass G, P, Q, and R, will be expressed by 7^gy=^o/i6 (34). Hence, using the theorem of parallel axes, we can build up the moment of inertia K^ of the whole triangle from those of its com- ponent triangles each of one-fourth the mass. We thus have K,= SH-(S+f0P.) + (§+^.GQ.) + (^-^f.G4 or3vr.=7l/(GPHGQ^ + GR^) (35). We can now with advantage transform this expression by noting some of the geometrical properties of the figure. ThusAL=JAD = LD, pl=|al=^ad, lg=|ld=jad, GP = PL+LG=JAD = GD = - say. Hence GP' = GD'' = 4^;z' = ^^' + ^f "''' (36). And obviously similar expressions may be written for GQ, GR, etc., by interchanging the letters. Whence GP' -1- GQ^ + GR' =-''' ■'■^' "'"'^^ 12 1" (37)- rr^GD' + GEHCF' Substituting (37) in (35) we have ^„ = -(GD''+GE=-f-GF') (38), which shows that the moment of inertia in question is equivalent to that of particles each of one-third the mass of the lamina and placed at the middle points of the sides. Again, by (37) in (35) we find K.=M'2±i±l ,3,) in agreement with (27) of article 250. 253. Eouth's Rule. — The following very useful rule is given by the late E. J. Rouih in his Rigid Dynamics, and is a valuable help in recall- ing many of the results established. Moment of inertia about an axis of symmetry == (mass X sum of squares of perpendicular semi-axes)-;- (3, 4, or 5). The divisor is to be 3, 4, or 5, according as the body is a rectangular or elliptical lamina, or an ellipsoidal solid. The chief typical cases of moments of inertia already dealt with are summarised in Table x. and the included Figs. 96-100. All the axes, about which the moments of inertia are taken, pass through the centres of mass of the bodies dealt with. ART. 253] PLANE KINETICS OF RIGID BODIES 239 Table X. Typical Moments of Inertia. Moments oi' Inertia. bOniES AND AXES. about OX. /o about OY. ^0 aboui OZ. Rectangular Lamina in xy plane. Y si 1" ^2 > + 3^) X ^ ' 2a. -i Fig. 96. Parallelepiped. i fi'^f) M ^V+'^^) Y / 2b ^-b- / X -^ -' 2a, Fig. 97. Elliptical Lamina in xy plane. i r^ ^\ f- ^2 ^(«'+*=) X- T—^ X ^ Fig. 98. Ellipsoid, Y ^ ^-- ^i^' + c') ^{c^ + a^) 'yV+^^) ^Z Fig. 99. Triangular Lamina in xy plane. K Y P I \ If ^(«2 + «y + 7') f(«'^+*''+^') \ " \ y ^ X 7 "■ '< Fig. 100. 240 ANALYTICAL MECHANICS [art. 254 Fig. ioi. Related Figures. 254. Graphical Method for Moments of Inertia of Laminae. — In problems as to the bending of beams of various cross sections we require what is usually called the moment of inertia of those sections. That is, we need to know the moment of inertia of a lamina of uniform surface density and in the shape of the section in question. In some cases the graphical method which follows is very useful for this purpose, as it reduces the calculation to mechanical drawing and arithmetic, though the proof of the method requires more than this. Let us first consider a certain possible relation between two plane figures and the method of deriving one from the other. In Fig. loi the original figure is an oval, of which we take an element PQ parallel to the axis of x and of width dy. Let us then mark off on PQ a smaller element P'Q', whose length is ylb of PQ, where b is here the extreme width of the figures from OX. It is evident that P'Q' can be found graphically by drawing the parallel lines P/, Q^ and then the converg- ing lines /P'O, ^Q'O. Drawing a line through all such points as P'O', we obtain the first derived figure. By dealing with P'Q' as we have just dealt with PQ, it is obvious that we may obtain points P"Q", such that P"Q" is ylb of P'Q', and therefore is j'^/iJ" of PQ. Drawing a line through all such points as P"Q", we have the second derived figure. Both these figures have valuable properties with respect to the axis OX, in relation to which they were drawn. Thus, let the areas of the original and first and second derived figures be respectively A, A', and A', and the corresponding masses of some lamina occupying those positions be M, M', and M", then we have MIA=M'IA'=M"IA"=fsa.y, e and / would need substituting for o and b in the limits of integrations in (42), all else in (42) to (44) remaining the same. The shapes of the first and second derived figures are shown by single and double shading in Fig. 102, the original figure being a rectangle of height b and length a. The line limiting the first figure is given by x' = aylb, while the equation of the parabolic curve for the second is given by x" — x'yjb = ay^/b', or y^ = b''x" ja. It is seen that the point O is here taken at one side, which is quite legitimate. We may find occasion to apply this method later. For fuller details, with method of moving the pole O and many illustrations, the reader is referred to more technical works, such as Professor Arthur Morley's Strength of Materials. In connection with the torsional pendulum, dealt with a little further on in this chapter, we shall see how the moment of inertia of any body whatever may be experimentally determined. Examples— XLVIII. 1. Obtain expressions for the moments of inertia of a*triangular lamina about axes (i) parallel to a side and (ii) perpendicular to a side. 2. Find, by any method, the moments of inertia of a uniform triangular lamina about axes perpendicular to its plane. 3. Give Routh's rule for the moments of inertia of symmetrical figures, and show in tabular form the typical bodies and their moments of inertia about their chief axes. 4. Explain how the moment of inertia of an irregular lamina or surface about an axis in its plane may be obtained by a graphical method. Q 242 ANALYTICAL MECHANICS [arts. 255-256 255. Well Roller and Bucket.— The descent of a bucket down a well under gravity affords a simple example of the uniformly accelerated rotation of a body of revolution, namely, the roller. Let the bucket have mass m, the roller mass M, radius r (to the centre of the rope), and radius of gyration k. Suppose at first that the mass of the rope is negligible, also the resistances due to its stiffness or the friction of the axle in its bearings. Then the equations of linear and angular motion may be written mg—T=ma .... ... (i), Tr=Mk^a, (2), where T\s the tension of the rope and a and a the linear and angular accelerations occurring in bucket and roller respectively. Dividing (2) by r, and noting that a=alr, we find T=Mk'alt^ (3). Then, adding (i) and (3), we obtain -mg={m-\-Mk^lr')a, or a — — , ,°,„, „ (4). Substituting this value of a in (3), we find mr'+Mk^ ^^'' 256. Motion modified by Friction of Axle. — Let us now introduce the coefificient of friction /* of the axle in the bearings, the radius being p. Then the reaction of the bearings on the axle, if taken to be vertical, is Mg-^ T, the frictional resistance fi times this, and the torque due to friction p times the product. In reality the axle on turning would roll up in the bearings a little before slipping, and the above expressions would need modification for strictness. This aspect may be dealt with rigorously later ; the approximation is sufficient for the present case, where the friction only introduces a slight modification in the motion. Hence, to our present approximation, we may write the equations of motion thus : — mg— T=ma, or T=m(g~a) (6), and Tr—{Mg+ 7)p/*= Mk''a=Mk''alr, T^m]a^gr?v) ( ). r(r-pp) Then, equating the right sides of (6) and (7), we find _ fiir^ — {M-\- m)rpp. And by (8) in (6) we have ^mr'+Mk^—mrpp. " ' " W- It is easily seen that, if the friction is negligible, we recover the previous values of a and T in (4) and (5) by writing pp.=o in (8) and (9). ARTS. 257-258] PLANE KINETICS OF RIGID BODIES 243 257. Atwood's Machine allowing for Inertia of Pulley. — In articles 223-224 Atwood's machine was considered without and with friction. But, in each case, the mass of the pulley was supposed negligible. We now proceed to allow for this, friction being supposed absent. Let us write for the larger and smaller suspended masses M-^ and J/j, respectively, for that of the pulley M, its radius of gyration being k and the radius to the centre of the thread r. Further, let the tensions of the thread supporting the two masses be Ji and T^. Then, for the equations of motion, we may write as follows : — M,g-T,=M,a (i), T,-M,g=M,a (2), ( r, - T^)r=Mk'a=Mk'ajr, or T,-T,=^Mk''alr- (3). Hence, by addition of (i), (2), and (3), we eliminate the T's and find {M,-M,)g={M,+M, + Mk'lr,)a, _ {M^-M,)g . Then, by (4) in (i) and (2) successively, we obtain T-Mg-~'^^^^^- (5) ^"<^ ^^=^^^M;+M,+Mk^^ ■ ■ ^^> Here again it is seen that these expressions reduce to the simpler ones of article 223, if the inertia of the pulley is ignored by writing Mk-'lr'^o in (4), (5), and (6). Examples — XLIX. 1. Determine the accelerations and tension when a mass of 10 lbs. hangs by a cord from a solid cylindrical roller of i foot diameter and 30 lbs. mass. (Take ^=32-2 ft./sec.^) Ans. 12-88 ft./sec.^, 2576 radians /sec.^, 193-2 poundals or 6 lbs. wt. 2. In the previous problem, show how to allow for the friction of the axle, assume values for the radius of the axle and its coefficient of friction, and find the new accelerations and tension. 3. Suppose the roller of the previous questions to be replaced by a roller and fly-wheel, and the apparatus to be used for finding g; investigate the required relations, and indicate the procedure. 4. In an Atwood's machine the pulley is a single disc of mass 100 grams and runs on ball bearings, the moving masses being 510 and 490 grams. Find the linear acceleration and the error in the determination of g if the pulley's mass is ignored. Ans. 2g/ios. Spurious determmation would be iooj^/105. 258. Compound Pendulum.— Having considered, sufficiently for our purpose, uniformly accelerated motion about a fixed axis, we now pass to examples of variable accelerations, namely, those cases which occur under gravity or elastic conditions. 244 ANALYTICAL MECHANICS [art. 258 We take first the compound or physical pendulum, which is a bar or other rigid body suspended on a fixed horizontal axis at the end, or elsewhere, and capable of oscillating freely about this axis. For dynamical considerations it is characterised by three important constants : its mass M, the distance h from its axis S to its centre, of mass G, and the radius of gyration k about a parallel axis through G (see Fig. 103, which shows the vertical plane of oscillation). To find the total torque about the axis S when SG is inclined at 6 with the vertical, consider a particle m which is then situated at Q distant horizontally QN=j; from the vertical SY. Then the torque due to this particle is — mgx. Thus the total torque is given by G= —gS,mx= —g5cZm=- —Mgh sin 6. Or, if only small oscillations are considered, and for them we write sin Q=0, then we have the approximation G=-MgM (i). But, as seen before, torque= moment of inertia into angular acceleration. Hence, using the parallel axes theorem for the moment of inertia about S, we find G=M{h^-^k'')d . . . (2). Thus, equating the right sides of (i) and (2), we have (h-^k')e^ghe=o (3). Further, on writing h' = k^lh,ox k''=hh' . . . (4), equation (3) may be put in the form e Fig. 103. Compound Pendulum. h+h' (5)- Hence, by article 29, we see that the motion is simple harmonic of period given by T=27rl^=2^^ ^l-J- = 2wJ -^ (6). By comparison of (3), (4), and (6) with article 53 on the simple pendulum we see that h+k''lh=h + h:=l (7), where / is the length of the equivalent simple pendulum, i.e. the one of equal period with the compound one under discussion. Hence, if we lay off" h' from G to O along SG produced, then S0 = /. In other words, O is such a point that, if all the material of the pendulum were collected there and yet connected by massless rigid bonds to the axis S, we should have an ideal simple pendulum that ART. 259] PLANE KINETICS OF RIGID BODIES 245 would oscillate in the sarae time as our actual compound pendulum This point O defined by SGXG0 = ,4' (8), (SGO being straight) is called the centre of oscillation. Thus, though the simple pendulum is an ideal one which can never be realised, we have found how to specify its equivalent length in terms of the constants of any rigid body which may be experimented with. 259. The Centres of Oscillation and Suspension are Convertible. — Let us now suppose the pendulum suspended about a parallel axis through O, its period being denoted by t'. Then bv (6) and (4) we have or r ^^r ] We thus see that the periods are equal about parallel axes through a given point of suspension S and through the corresponding point of oscillation O. It is, however, easily seen (as pointed out by Professor Gray in his Dynamics and Properties of Matter, pp. 147-149) that the parallel axes for the given period t are not confined to ihe points S and O. On the contrary, they may move parallel to themselves to any positions distant h and h' respectively from G. For, from (4), (6), and (9), we see that precisely the same expressions hold for t in whatever directions h and h' are measured in the plane of Fig. 103. But, of course, it is only when h and K are taken in opposite directions from G that SO represents their sum and is the length of the equivalent simple pendulum. Thus it is correct to state that if the period of a compound pendulum about an axis through S is t, and SO, being SG produced, is the length of the equivalent simple pendulum, then the period of the compound pendulum about a parallel axis through O is t also. But it is not correct to state that any two points on a line through G, about parallel axes through which the periods are equal, are distant apart by the length of the equivalent simple pendulum. For this to hold, the points S and O must be on the straight line through G and on opposite sides of G. For evidently a point O' may be found between S and G and distant h from G, about a parallel axis through which the period will be t, but SO' will be h — K instead of h-\-H ; and it is the latter and not the former which is equal to /, the length of the equivalent simple pendulum (see Fig. 104 in article 260). We may also state this in an analytical form ; thus from (7), putting X for h, we have x''—xl-\-k- = o (10), whence x = — (ii)i 2 showing that h has in general two values for any given / or t. Since for oscillations x must be real, we note that the limiting value of / is the minimum l=2k (12)- 246 ANAL YTICAL MECHANICS [art. 260 And for this value of / we have X = ll2=k (13). This, therefore, corresponds to the minimum period, which by (13) or (12) is T—2ir^2klg (14). 260. Variation of Period with Axis. — To study the variation of the period t with ordinate h of the axis, it is convenient to write _)» and X for these quantities and plot the curve defined by them as co-ordinates. Thus, from equation (6) of article 258, we derive gxf^6,'rr\x'^k') (is), in which J is the period and x is the distance GS from the centre of mass to the axis of suspension. Differentiating (15) with respect to x, we find dx xjgx{x^+k^) ^ -'' This shows that as x increases from o to k, dyjdx is negative, or y is decreasing; for x:=k, dyjdx=o, or y is stationary; as X increases beyond k, dyjdx is positive, and therefore y in- creases indefinitely with x. Thus the stationary value of y {orx=k{&ee. GK in Fig. 104) is a minimum as shown before in (12) and (13), the corresponding value of y being that given in (14). Equation (15) shows that for x=o, y—fa, as we should expect, for there is no torque to produce motion when the pendulum is disturbed. In fact, the meaning of disturbed position disappears when S coincides with G. We see also from equation (16) that for a;=o the value ol dyjdx is j;oo. Thus the curve asymptotes to the axis OY. If we write x negative in (16), y becomes imaginary, which may be interpreted that oscillations are no longer performed with the pendulum the same way up as at first. The pendulum topples over, a new x, again positive, may be taken in the inverted position and the periods found as before. It is convenient, however, in the diagram to show an inverted curve for this portion, as indicated by the broken line in Fig. 104. In this figure SO = /= S'O', the length of the equivalent simple pendulum, while GK=GK'=,4 is the distance of the axis from the centre of mass to give minimum period of oscillation. The figure is drawn on the assumption that the pendulum is a uniform bar of length SS' and of negligible thickness. Thus GO = GO' = /%'=/4/3=J of GS, and GK = GK'=^=/J/s/^. S X Fig. 104. Variation of Period of Compound Pendulum. ART. 261] PLANE KINETICS OF RIGID BODIES 247 And the ratio of periods with axis at K and at S is x/a 73/2 = 0-93 nearly. 261. Rigid Pendulum by Energy.— We may now take the motion of a compound or rigid pendulum as an example of potential and kinetic energy and the transformation from one kind to another. Thus, consider the pendulum to have zero potential energy when the centre of mass G is vertically below the axis of suspension S, and take from G as origin the axis of x along SG produced, the axis of y being perpendicularly to the right. (See Fig. 105.) Let there be a particle of mass m at P, whose co-ordinates are (x, y). Then in the standard position this particle is at a distance h-\-x below S. But when the pendulum is displaced through an angle 6 as shown in the figure, the par- ticle is below S by the smaller distance {h-\-x) cos d—y sin B. Thus, the potential energy of the pendulum in the displaced position is given by V='^mg{(}i-\-x){\-co% 6l)H-j|/sin Q) ^=gh{\ —cos ^)2»2-|-^(i — cos 6'^mx-\- ^sin Q^my. But 'Zm—M, the mass of the pendulum, and 'S.mx- G is the centre of mass. Hence the above becomes V=Mgh{i-cos6) (17). Again, since the pendulum is rigid, the angular velocity of every point in it is the same, 6 say. Hence, writing r for SP, the linear velocity of the mass m at P is r6 and its kinetic energy ^m{rOy. Thus, we have for the kinetic energy of the pendulum Or, writing as before k for the radius of gyration of the pendulum about a parallel axis through G distant h from S, this becomes T=\M{h' + k')e'' (18). But the sum of Tand f^is constant, as the pendulum is supposed to move under gravity without any other supply or withdrawal of energy. Hence, from (17) and (18) we obtain {k^ ■\-k^)d-' + 2gh{i-cos e)=con5t (19). Thus, on differentiating with respect to the time, and dividing out by 0, we have {h''+k')e+ghsme=o (20). And, on writing sin^=^ nearly for small oscillations, this reduces to the approximate equation (3) of article 258. The inquiry as to initial conditions and oscillations in finite arcs given in articles 30 and 54-56 for rectilinear oscillations and for the Fig. 105. Pendulum by Energy. -o=l^my, since 248 ANAL VTICAL MECHANICS [arts. 262-263 simple pendulum, etc., apply equally to the compound one of equiva- lent length l=zh-\-h', and so need not be repeated here. 262. Torsional Pendulum. — Imagine a body suspended from a fixed point by a wire or other elastic arrangement which introduces a torque proportional and opposite to any angular displacement 6 of the body about a vertical axis. Then, if the body is symmetrical about its vertical axis of suspension, and is displaced and let go, it will obviously oscillate. For the body has the same relation to its vertical axis as the compound pendulum with small amplitudes had to its horizontal axis. Thus, if the body has moment of inertia / about the vertical axis, and the restoring couple is E6 for a displacement 6, the equation of motion is ie+E6=o (i). Hence the solution may be written (9=eoCOs(25r//T-|-e) (2), the period being given by r=2ir'JljE (3). Or, for a given suspension of elasticity independent of the mass hung on it, as is the case for a single wire, we may write I=ET'Uir'=cr'- (4), where f is a constant. 263. Moment of Inertia Table. - Fig. T, 106. 75 Ti Moment of Inertia Table. By using as a torsional pendulum a cylindrical disc and stalk and mounting upon it other bodies of known and unknown moments of inertia, the latter can be found in terms of the former by timing the periods in each state. Thus, by refer- ence to Fig. 106, we see the pendulum in three states : the bare table, the table and a body K of known (or easily calculable) moment of inertia, and finally with a body X of unknown moment of inertia. Let the moments of inertia of the whole suspended masses in the three cases be /,, /„ and T3 as shown. and Z, and the respective periods t^, t. Then, from (4), we have for the three cases Ii=ct\, J„=c7l, a.r\A I,=ctI ... . . Whence A — A _t|^^i„ moment of inertia of X I,-I~t\-t\ that of K • • • thus giving the required ratio in terms of the periods observe^. (s). (6), ART. 263] PLANK KINETICS OF RIGID BODIES 249 Examples — L. 1. A circular hoop of radius h of uniform wire hangs on a knife edge. Find the periods of its oscillations (i) in its own plane and (ii) perpendicular to that plane. Thence show that the addition of weights at the bottom point of the hoop is without effect on one period but increases the other. Ans. Periods are iir \l2klg and 2v s'3^/20; 2. Explain what you mean by the centre of oscillation of a pendulum, and find its position (i) for a thin uniform rod suspended at one end, and (ii) for a uniform disc oscillating about a tangent. Ans. (i) Two-thirds total length from point of suspension, (ii) A quarter of the radius below the centre. 3. Establish the theorem of the convertibility of the centres of suspension and oscillation. 4. Investigate the variation of the period of a pendulum with the position of its axis, and illustrate your answer by curves for some actual pendulum. S- Obtain by any method the period of small oscillations of a rigid pendulum. 6. Find the period of oscillation of a unifilar torsional pendulum, the torque per radian being assumed. Thence show how the moment of inertia of any irregular body can be determined. 7. 'Find the smallest time of oscillation of a uniform bar of length 2a about an axis fixed perpendicularly to the bar at any point in its length. ' Find the same for a uniform plate in the form of an equilateral triangle, the axis to be fixed perpendicularly to the plane of the plate. 'For what position of the axis is the time greatest.!" (LoND. B.Sc, Pass, Applied Math., 1906, in. 5.) 8. ' If at equal intervals, a, there are fixed n equal particles on a straight rigid wire AB, whose mass is negligible (there being no particle at A), and a pendulum is made by suspending the system from A, prove that it will oscillate in the same time as a simple pendulum of length {2n+i)a/3. Is this true if the wire is replaced by a flexible thread ? (Give reasons.)' (Lond. B.Sc, Pass, Applied Math., 1907, iii. 4.) 9. 'Show that the time of oscillation of a compound pendulum is 2ir JK'-Igh where h is the distance of the centre of gravity from the axis of suspen- sion and k is the radius of gj'ration of the pendulum about that axis. ' A heavy particle is attached to a straight compound pendulum at a distance x-^ from the axis of suspension, and the length of the simple equivalent pendulum is then /, ; it is then attached to a point distant x<^ from the axis, and the length of the simple equivalent pendulum is now l.,. Show that if it be entirely removed, the length of the simple equivalent pendulum will be ;ri/i - xjlj - 1-J.ix,_^ - jTj) , (Lond. B.Sc, Pass, Applied Math., 1909, iii. 3.) 10. ' Explain the phrase simple equivalent pendulum as applied to the motion of a heavy rigid body about a fixed horizontal axis. Find the length of the simple equivalent pendulum when a uniform circular disc moves about a fixed horizontal axis in its plane, in terms of the radius of the disc and the distance of the axis from the centre of the disc' (Lond. B.Sc, Pass, Applied Math., 1910, in. 2.) Y /^ w p 1 y / * r/ /" / / >^ r y y c \ / / a X V • « y a: X Fig. 107. Plane Motions of Rigid Body. 250 ANALYTICAL MECHANICS [arts. 264-265 264. General Plane Motion of a Rigid Body.— Let us now consider any rigid body, its general motions of translation, rotation, or both combined all parallel to a given plane, the momentum and energy possessed by it, the impulses, forces, and torques acting upon it, and the work done on it by those forces. Let the motions and forces be all parallel to the xy plane. At time i let the centre of mass G of the body be at x, y, and let there be a particle of mass m of the body at P, whose coordinates are x,y, z with respect to the fixed axes XOY, and a, b, z with respect to the axes X'GY', which remain always parallel to the fixed ones but move with G. Further, let r be the length of the projection of GP upon the xy plane, and let this projection make the angle 6 with GX'. Then, on reference to Fig. 107, and remem- bering that for a given particle r is invariable, but that a and b vary with time, we can. find the follow- ing preliminary relations, in which « and V are written for the component velocities of G and a> for 6 : — x=x-\-a,y—y-\-b (i), '2ma=o=^mb ... . (2), a=rcosd, b=rs,md, a'-\-b'=r'' .... (3), a=—rsir\d.d=—b,'b=rcos6.d=(i>a . (4), x = u — ii>b,y^v-\-b>a (5), x-^it — iib — u!i^a,y^v-\-uia — u>^b .... (6), Not only in these differentiations but throughout the subsequent work it must be remembered that, since the body is supposed rigid, r does not vary with tivie for a given m at P, but only varies from point to point, i.e. with m in the summation over the body. Further, while Q, 0, all the x's, y's, and their derivatives may vary with time, x, y, u, v, to, and d)(x-\-a)—(u—ii)b){y-^i>)) =:'Em{vx—uy)+') +(v-{- (i>x)2ma —{u— (oyySmd = (vx—uy)'2im+io2mr'', or '2,m(jx—xy)=M{vx—uy)+Ji!'a<^ (12), where K„ is the moment of inertia of the body about an axis GZ' through the centre of mass and parallel to OZ. Hence, we see that the angular momentum of a rigid body about a given axis equals that of the whole mass at the centre of mass plus that of the body about a parallel axis through the centre of mass. (See also article 238.) If we now turn to the impulses under which these momenta may be supposed to have been produced from rest in time ^„ we have, in the former notation, for theii; torque with respect to OZ ^Yf,x-Xi^y)='E{yx-Xy)t,+'E{ya-Xi>)t^ .... (13). Thus, the moments of the impulses or the impulsive torques split up in the same way as the momentum. Further, on comparing (12) and (13) and recalling the fundamental relation, impulse equals change of momentum, we see that these equations agree term to term. We may accordingly write '^(Yx—Xyyi='Sm(yx—xy) (14), 'zlYx-Xy)f^=M{vx—uy) (15), and ^Va-Xl>y^=Ko<^y (16). It should be further noted that (15) splits into two equivalent terms on each side, giving ^Xf, = Musind'2,Yi=Mv (17). Thus, equating the left sides of (12) and (13), we may say that a given angular momentum about OZ is acquired under a certain impulsive torque about OZ. Or, using (16) and (17), we may say that the velocity of the centre of mass is due to the linear impulse and the angular momentum about it due to the corresponding impulsive torque. 267. Grrowth of Angular Momentum. — We may now notice that the rate of increase of the angular momentum of a rigid body also falls into two terms, as is the case with the momentum itself Thus, we may take the differential with respect to time of the angular momentum just found, or we may take the moment about OZ of the product (mass into acceleration) of any particle, and then sum for the whole body. Let us adopt this full method first and use the other as a check on the result. As before with momentum, we may replace the moment of an acceleration by those of its components. Hence we may write as follows : — Rate of Increase of Angular Momentum of Body about 0Z= '2m(yx—xy) = 2w{(w+ wa — a>'d)(x-\-a) — (u — £)i—'a)(y-\-6)} = 'Em{vx — uy) + (ai,m(a^ + ^^) + (?) + ^^ + tii^y)S,ma — {u — ^y-\- ti>'x)^mfi, or 'Sm{j''x—xy) = M{v.\—uy) + Ji!'oO) (i8). And this answers the check of differentiating the equation (12) with respect to time. ART. 268] PLANE KINETICS OF RIGID BODIES 253 Turning now to the forces under whose action these accelerations occur, we see that their moment about OZ may be expressed and split up as follow^s : — 1{Yx-Xy) = 1{Yx-Xy)-ir^{ya-Xb) . . . (19). And here again, not only are the left sides of (18) and (19) equal, but also the terms on the right are equal, each to each. Thus, on splitting the equations involving x and y, we have ^X=Mu, ^y=Mv . . (20), 'E{Va-Xd) = Ji:,6> . ... (21). We have usually hitherto treated the total mass M as invariable. But in certain problems particles become adherent on a rigid body or leave it, so it is desirable also to provide for a variable AI. Thbs, differ- entiating (17) and (16) with respect to / on this understanding, we have 2Zrf/= d{Mu) = Mdu + udM, 2 Ydt=d{Mv) = Mdv + vdM (22), and 1(^Ya-Xb)dt=d{K,is>) = K,d^-^i>>dK^ (23). Hence, when forces are absent, we have J/«=constant, Af»=constant . . . (24). And when torques are absent .^,(0= constant • (25). Equations (24) and (25) are the symbolical expressions of the very important principles called the Conservations of Linear and Angular Momenta. 268. Kinetic Energy of Plane Motion of a Rigid Body.— Con- sider now the kinetic energy 7" of any rigid body whose centre of mass G has the velocity components u and v, the body also rotating about GZ', parallel to OZ, with angular velocity w (see Fig. 107). Then, with the previous notation, and the relations of article 264, we have 2T=l^m{x■'^-y-)^^m{{u-i^b)■'^r(v^<^aY] = («=■ + w')2w -f a)^2w(a' -f -^') -4- 2V, and the velocity of S along SX is , Xt^( hp\ Xb> Fig. 108. Centrb of Percussion. 256 ANALYTICAL MECHANICS [art. 271 But this is to be zero, hence k'' = hp • (2). as before in (i). ' 271. Fall of Trap Door from Vertical Position. — It is instructive to consider also cases where the impulsive pressures at the axis do not vanish. A simple example of this is presented by the fall of a trap door from the vertical to the horizontal position, consideriiig the door to be a uniform lamina turning freely about a horizontal hinge at one edge and stopped suddenly by an impulse along the edge parallel to the line of the Jiinges. Thus, referring to Fig. 1 09, let the motion be in the plane of x, y, the hinges being at S, the centre of mass falling from rest at G' in the quadrant to G, the edge B' striking the frame at B. Let the trap door acquire in its fall an angular vel- ocity (1) and be arrested by two i impulses D and E acting vertically at B and S. Then, calling the Fig. 109. Fall of Trap Door. jj,ass of the door M and its half width /4=SG, the square of its radius of gyration about a parallel axis through G is k''=k''l^. We thus see that its potential energy when upright is Mgh (see equation (7), article 264) reckoned from S; and that its kinetic energy when down, just before striking the blow, is ^M{h^+k^), or D-E=M—io . . ... (s). Thus, by addition and subtraction of (4) and (5) and using (3), we find D=^M^lghl2 (6), and E=-^M^2,ghl2 (7). These two blows are evidently equivalent to a single one D+E applied at P, where Q,Y=p=k'lh=hli (8), as might be expected. ARTS. 272-272fl] PLANE KINETICS OF RIGID BODIES it^j 272. Fall of Trap Door from Horizontal Position. — Referring again to Fig. 109, let us now suppose the support at B to be suddenly removed when the door is at rest in the horizontal plane. We may then inquire what is the force exerted by the hinges, and what is the angular acceleration, each immediately after removal of the support. Let this initial force at the hinges S have components X and y, and let the initial angular acceleration be y. Then, by (28) and (29) of article 269, we have X=o, Y-Mg=M{hi) (9), and Y{-h) = {Mh^li)y (10), since .^0 = J/Zi^S- From these we find Y=Afg^i,^.y^A'i=-lgl^,h (11). The initial linear acceleration of the centre of mass is accordingly -Zgl^ (12)- We might also treat this problem by taking moments about S, yielding -Mgh = M(h'^h'\l)y, which with (10) determines F and 7 as before. Before the removal of the support at B, it is obvious from symmetry that the support at each edge was Mgl^. It is thus noteworthy that the removal of one support has the immediate 'effect of changing the force exerted by the other from a half to a quarter of the weight of the door. 272a. Sudden Fixation of a Point in a Rotating Body.— The fore- going example of the trap door illustrate the type of problem in which some point or line is suddenly fixed or suddenly freed, the motion being under gravity. If the motions occur in a horizontal plane we may have a point given which is to be suddenly fixed, the subsequent velocity and the impulse to be found. Or, we may have a relation between velocities before and after given, and from that be required to deter- mine what point is fixed. . , . , , It is clear that the impulse acts through the pomt which becomes fixed, so has no moment about any axis through that point. This principle is very useful in attacking such cases. Thus let a uniform thin rod of mass M and length ih be rotating about its centre in a horizontal plane with angular velocity co„, when by the sudden fixing of a point P in it the angular velocity is reduced to half. Find the position of P and the impulse Q exerted there when P is fixed. , ,. , ,, • c At the instant in question let the rod lie along the axis of y with centre at the origin and P at (o, y) and be rotating counter- clockwise. Then, equating angular momenta about P before and after the fixation, we have (J//i73)<«o=Vl/(/+/i73)<«o/2, whence j. =J_/3 or jy = (o-577 ...)/5. Accordingly the centre of the rod starts off with velocity and 6, we note that the pendulum starts with a certain kinetic energy, and as it swings this is gradually changed into potential energy. Hence we may equate the gain of potential energy to the loss of kinetic, thus obtaining Mgh{i-co5 e) = \Mk'W, or 2gh{2 sm-6/2)=^''(i>^, or 2 Jgh%m.0J2 =k'ia (2). Then (i)-i-(2) gives _ ( 2Mk' Jgh \sm 6/2 . . '~\ b ) m ^^>' the quantity in brackets being a constant for the pendulum, the second factor expressing the variables for the projectile under examination. Sometimes the angle Q is estimated by allowing the pendulum to pull out a cord or tape. Thus, if a length c is pulled out at radius r on swinging through the angle 6, we have sin^/2=f/2/-. ... (4). Then (3) becomes ART. 273] PLANE KINETICS OF RIGID BODIES 259 -e showing that u varies as c. Hence if the tape were graduated uniformly the values of u could be found on multiplying by a constant and dividing by the mass of the bullet. Or, for a given m, the tape could be graduated to read values of u directly for the pendulum in question. By allowing the pendulum to swing we could observe the period, given by r=2-n-k'l Jjh (6). Thus, introducing this value to eliminate k', (5) becomes \2'Kbr)m ^''' in which Mgh is the torque exerted about S by the pendulum in a horizontal position, and so is easily ascertainable. For fuller details as to construction and use of ballistic pendulums the student may consult Routh's Rigid Dynamics, i. pp. 98-101, 1897; and Sir G. Greenhill's Notes on Dynamics, pp. 190-19 1, 1908. Examples — LI. 1. Establish the independence of rotation of a rigid body and the transla- tion of its centre of mass as regards linear momenta and kinetic energy. 2. Show that the angular momentum of a rigid body, about an axis perpen- dicular to the plane of its motions of rotation and translation, may be divided into two parts, one being that of the body as though condensed to a particle at its former centre of mass G, and the other being that of rotation of the actual body about a parallel axis through G as though it were stationary. 3. 'A uniform bar AB, 6 feet long, mass 20 lbs., hangs vertically from a smooth horizontal axis fixed at ^ ; it is struck normally at a point 5 feet below ^ by a blow which would give a mass of 2 lbs. a velocity of 30 feet per second ; find the impulse received by the axis, and the angle through which the bar will rise.' (LoND. B.Sc, Pass, Applied Math., 1906, in. 6.) 4. 'A system of particles is moving in one plane. Show how to compound their momenta, and reduce the result to its simplest form. ' Three equal particles are attached to the corners of an equilateral triangular area ABC, whose mass is negligible, and the system is rotating about A. A is released, and the middle point of AB is suddenly fixed. Prove that the angular velocity is unaltered.' (LoND. B.Sc, Pass, Applied Math., 1905, in. 6.) 5. 'If a rigid body is revolving with angular velocity m about a fixed axis, find— («) its kinetic energy ; {b) its moment of momentum about the axis. 'A uniform rigid bar, AB, is rotating in a smooth horizontal plane about its centre with angular velocity a ; if suddenly a point P in the bar is fixed, find the position of P for which the new angular velocity will be ^m.' (LoND. B.Sc, Pass, Applied Math., 1907, m. 2.) 6. ' Obtain a formula for the kinetic energy of a rigid body in terms of the motion of its centre of mass and its motion relative to the centre of mass. . . , . ' At a point P of a uniform circular hoop there is attached a particle of mass equal to that of the hoop. The hoop rolls, in a vertical plane, on a perfectly rough horizontal table. Prove that if the system starts from 26o ANAL YTICAL "MECHANICS [art. 274 rest when P is at the highest point, the angular velocity a when the radius to P makes an angle 6 with the downward vertical will be given by iJ*2-C0S^' (LoND. B.Sc, Pass, Applied Math., 1908, iii. 4.) 7. ' A disc of mass M and radius a is moving in its plane, the velocity of its centre being u and the spin about its centre <». One of the points of the disc distant a/2 from the centre in a direction at right angles to the direction of motion of the centre is suddenly fixed. Show that the' loss of energy is —M\p.u±aaj: (LoND. B.Sc, Pass, Applied Math., 1908, iii. 6.) 8. ' Obtain an expression for the kinetic energy of a lamina moving in its own plane about a fixed point. ' A uniform rod of weight W, free to turn about a fixed smooth pivot at one end, is held horizontally and released. Prove that when, in the subsequent motion, the rod makes an angle 6 with the vertical, the pressure on the pivot is (LoND. B.Sc, Pass, Applied Math., 1909, in. 5.) 9. ' In connection with the uniplanar motion of a rigid body, explain what is meant by the principle of the independence of the motions of transla- tion and rotation. ' A lamina moves in its own plane, being subject to a single force constant in magnitude and direction which always acts at the same point of the lamina. Find the motion.' (LoND. B.Sc, Pass, Applied Math., 1910, iii. 5.) 10. Explain the construction and action of a ballistic pendulum, and obtain a formula involving the quantities concerned. 1 1. Obtain the following expressions for the reactions on the axle of a rigid body swinging from rest at the inclination of its centre line a above the horizontal, to 6 above the horizontal : — X=— v-( — 2 sin a cos 5 + 3 sin 5cosfl), Y= — p( T - 2 sin a sin ^ + 3 sin^d - I j, where W is the weight of the body, h the distance of centre of mass from axis, and / the length of the equivalent simple pendulum. 12. If, in question 11, the body starts from rest in the vertically upright position, show that, for any shaped body whatever, the maximum horizontal force exerted towards the side to which the body swings occurs ai 6= +63°'25, and that the maximum horizontal force exerted in the opposite direction is for 6= - 34°. 13. For the case of question 12, plot curves for X and V, the angles 6 being the other co-ordinate. 274. Pure Rolling down an Incline by Energy. — We now pass to the consideration of motions which are special examples of combined translation and rotation. Let us take first the rolling, without slipping, of a solid of revolution down an incline. Then, the solid and incline being specified, the accelerations are required. It is instructive to treat this problem several ways, each possessing its own advantages. We ART. 275] PLANE KINETICS OF RIGID BODIES 261 commence by applying the conservation of energy, equating the potential energy lost by the body in a descent to the kinetic energy gamed. Let the body on the incline start from rest in the position snown m i< ig. 1 1 1, where it makes contact at C. Let the body have mass M, and moment of inertia about its geometrical axis K^^.bMr'-jc, where b and c are mere numbers depending on the form of the body and r is the radius of the rolling part. Then its moment of inertia about a parallel axis through C is K={b+c)Mr'lc. Let the plane make an angle a with the horizontal, and let x represent distances upward on the slope, u linear speed, and a acceleration. Also let 0) and y be the angular velocity and acceleration in the plane of xy. Then, if the body rolls till the point of contact C has moved a distance x down the plane, it will have lost potential energy of the amount Mgx sin a. But, if its angular velocity in the meantime changes from zero to -'^=^1:7=^;' ^7>- In cases where the distance between the instantaneous centre and the centre of mass is changing another term is needed in the above analysis. It is then simpler to u.se the method of article 276, which is invariably safe. 276. Rolling by Properties of Centre of Mass. — We again consider the problem of the pure rolling of a body down an incline, applying this time the properties of the centre of mass summarised in article 269. Still referring to Fig. 1 11, we talce successively the forces parallel toy and X and the moments about the perpendicular axis through G, equating each to the appropriate product, of inertia and acceleration factors. Thus '2.Y=N-Mgco%a.=o (8), 'lX=T-Mg€\'a.a.=Ma (9), ^{Yx -Xy')=Tr=U^Mr^y . . . . (10), where x' and y' are here used for the co-ordinates of the points of application of the forces. We have also the condition expressing no slipping y=.-ajr . ... . (11). Then (11) in (10) gives -T=—Ma (12), and this added to (9) yields the accelerations eg 5m a. Again (13) in (12) determines the frictional force T=j-^Mgsma (14). Thus, though this method may be longer, it affords a closer insight into the phenomena, and gives the values of TV and J^-the normal and tangential reactions at the incline on the rolling body. 277. Condition for Pure Rolling on Incline. — When the rolling body is also just on the point of sliding on the incline the full frictional resistance is called into play, so that we have J'=/iiV" where ju is the coefficient of friction. Hence, by (8), we have T=flMgCOSa (15). Thus, equating the right sides of (14) and (15), we have the condition ART. 278] PLANE KINETICS OF RIGID BODIES 263 for the limit between pure rolling and rolling combined with sliding, namely, Hence, for pure rolling, MCOSa= , , sin a, h-\-c l>. _ b tana b-^c tana ^b-\-c {16). (17)- The values of the linear acceleration and this limiting relation are given for a few typical figures in Table xi., also the constants in the expression bMr^jc for their moments of inertia. Table XI. Pure Rolling down Inclines. Figure Rolling. Constants in Moment of Inertia. Limit for Pure Rolling, Acceleration Ratio. i>. - tan a b+c a/(jf sin a) _ c _ Cylindrical Shell . . . Solid Cylinder . . Spherical Shell . . . Solid Sphere .... I I 2 I 2 3 5 1/2 1/3 2/5 2/7 1/2 23 35 5/7 278. Combined Rolling and Sliding down Incline. — When the condition expressed in the inequality (17) is violated sliding occurs. Hence the geometrical relation expressed in equation (11) of article 276 no longer holds. But since the full frictional resistance is called into play we may replace it by T=ix.N (18), where ft is the coefificient of friction. We have thus for the solution of the present problem this equation and (8), (9), and (10) of article 276. By substituting in (10) the value of J" from (18) and (8), we find ,_Cl^ gC05a ^j^^^ y-- br Then the value of T put in (9) gives a= — ^(sina— /tcosa) . ... . (20). If at time / from rest the linear and angular velocities are u and • we have 2^ = a/z= — rf (sin tt — ju- cos a) (21), CIJ.s:t cos a and '^~y^~ '° br ^^^^' Also the speed of sliding of the body over the plane at the part of contact is (23)- b-\-c . u-\-Mr=-—gi{svaa. T-/*cosa) 264 ANALYTICAL MECHANICS [art. 279 It may be noted that a body in pure rolling motion has a smaller linear acceleration than if sliding without friction. Here, where there is sliding combined with rolling, the linear acceleration is precisely that of a body sliding simply down the rough incline. We may naturally inquire whence comes the energy of rotation. The answer is easily found in the saving of work against friction which that rota- tion effects. For the kinetic energy of rotation after time t is lK,.^^l{LMr^^J^l^P^=\Mif'l^ . . (24). And the work against friction saved by the rotation is resistance into space, or y.N{iLt^^^MgcoUJ^^r- = lMie'^^^ . . . (25). Or, to test the matter a second way, the potential energy Mgi^—^af) sin a lost in descending for a time t may be equated to the sum of the kinetic energy gained, \M{atf-\-\Ka{yff, and the work done against friction, which is \^N{ — u — iar)t. The result is an identity, the value of each side being |j^g^/°(sin^a— /a sinacosa). 279. Frictional Couple for Axle in Bearings. — We may now fitly consider the value of the torque or couple required to overcome the friction of a cylindrical axle in somewhat loose cylindrical bearings. A little reflection will show that the contact between axle and bearing will not be at the lowest points of each when the axle is turning, for the axle will at first roll up on the inside of the bearings without slip- ping, and only when a certain steep- ness of slope is reached by the point of contact will slipping begin. Let this occur when the angle with the horizontal is Q, and to maintain the axle in uniform rotation let the torque or couple G be required, consisting of two equal unlike para- llel forces of value F and each ap- plied at a perpendicular distance p from the axis. Then it is required to determine B and G in terms of the mass .^of the axle and its loads, r its radius, and /n(=tan ;8) the coefficient of friction between the axle and its bearings. Let P'ig. 112 represent the position and distribution of forces when slipping has just commenced and the rotation is uniform. Then, resolving parallel to the axes of y and x, and taking torques about the axis through the centre G perpendicular to the plane of the diagram, we have ht mt a 1 \ N=MgzQ%» (i), T=ix.N=Mg%\nO (2), G=2Fj>=Tr— iJ.Mgr cos 6 . ... . (3). Fig. 112. Frictional Couple FOR Axle. 265 ART. 280] PLANE KINETICS OF RIGID BODIES Thus (2)-^(i) gives tan6'=/i=tan/3, or 0=/:} .... (4) as might have been anticipated. Putting this value in (3) we find for the torque G^Mgrsmfi ..... (c) which is slightly less than if the contact had been at' the lowest points of axle and bearing. ^ 280. Sphere with Initial Rotation on Rough Level Plane — Considering now a solid sphere, we will imagine it to have an initial rotation about a horizontal axis but no translation, and to be then gently placed on a rough level plane and immediately let go. It is required to determine the subsequent motion. Let the motion be in the xv plane as represented in Fig. 113, the' sphere having mass M, radius r, and initial angular velocity (o„, the coefficient of friction between it and the plane being /i and the reactions between them being N and T. Then, since the only forces and torques available are finite, finite changes of linear and angular velocities can occur only in finite times. But at the start there is at the point of con- tact a slip of the sphere over the plane at the velocity —oi^r. And this can only be removed in a finite time t say, during which the full friction must be called into play. We. accordingly have the following equations of motion during this stage of the phenomena. (See article 269.) ^=A*^ (0, N-Mg=o (2), T=Mb . . . (3), T{-r)=^^A£r-y . ... . (4), where b is the horizontal hnear acceleration of the centre of mass G and 7 is the angular acceleration of the sphere about an axis through G and perpendicular to the plane of the diagram. These equations give ^>=H-g (S). and y=~c^ng'2r . . . (6). But if at time t the linear and angular velocities are v and w, and slip ceases then, we have v—tar=o . (7). Hence, substituting from (5) and (6), and remembering that the initial values of v and w were o and Woj we obtain Fig. 113. Sphere with Initial Rotation. 266 ANALYTICAL MECHANICS [art. 280 Whence l>.gt-{i^,-Sl^gtl2r)r=o. t=- TH-g And by (5), (6), and (8) we find v= —(o„r 0>= — o 7 49f^S ■ ■ (8). • ■ (9). The space passed over in time\ /is given by j^' It is noteworthy that v and w are independent of ft but that / and j^ diminish with increasing jj.. We have now to inquire what occurs after the instant when the sh'p ceases. While the slip speed was negative the frictional force T was posi- tive, the linear acceleration 6 was positive, and the angular acceleration y was negative. Suppose first that when the slip ceases T still retains any positive value. Then obviously the accelerations retain their foriner signs and the speed of slip of the form given in (7) becomes positive. But considerations of friction show that this involves a negative T contrary to the hypothesis, which is therefore untenable. Imagine secondly that when the slip ceases the frictional force T suddenly assumes some negative value. That would reverse the accelerations, making the linear acceleration nega- tive and the angular acceleration posi- tive. In other words, it would put us back to the state of things obtaining a little be- fore the slip ceased. But we have already seen that Ihe state of things in question involved a negative slip and a positive T, again contrary to hypothesis. Hence T must be zero, both accelerations henceforth zero (in the absence of rolling friction and air resistance), and the velocities expressed by (9) and (10) are thus retained. The phenomena of this problem can be illustrated graphically by a speed-time diagram as in Fig. i r4. The fact that the force T must vanish after the time t may also be seen from this diagram. For the difference of the ordinates of the two speed graphs expresses the slip speed. And beyond the point where this slip vanishes at time t, if the accelerations remain as before the graphs cross, the slip speed changes 114, Speeds of Sphere with Rotation. TIME Initial ARTS. 281-282] PLANE KINETICS OF RIGID BODIES 267 sign ; hence the frictional force would have to be reversed and the accelerations reversed instead of remaining as before. 281. Cylinder with Initial Kotation. — It is easily seen that if the body were a solid cylinder instead of a sphere the moment of inertia is Mr^/2, and we should have ^=I^S^ (5«). (6a), (9«). (loa), (iia). y=-2iJ.g/r . t^b,,rlsfig .... v=u>„r/2 . ■ and s=„=l>/{b+c) (12), which confirms (10) and (loa), for the right side of (12) becomes 2/7 for a sphere and 1/3 for a soUd cylinder. We may also check this result by consideration of energy. Thus the kinetic energy at the start should equal the sum of that left when pure roUing has commenced and the work expended on friction. But the kinetic energy for our generalised body is, at the start, bMr'i^/^c, and when rolling it is (b+<:)Mr-b>y2c. Also for the work done in slipping we have the product, frictional force into distance slipped. But the distance slipped is ,^a^ initial speed of slip (r(o„) into the time of slip given in (8) and (8a). We accordingly find the following expressions which furnish the desired check : — Final kinetic energy+work done 268 ANAL YTICAL MECHANICS [arts. 283-284 -Mr''i^\-\- -Mr'^ '2c{b+c)'" '" ■ 2{b+c) -.M—Mr'^\i»l=lmtiA\ kinetic energy (13)- 283. Motion on a Steep Plane with Initial Rotation. — Let us now suppose a body with initial rotation about a horizontal axis to be gently placed on a plane whose steepness and roughness are such that the body if without initial motion would descend by combined rolUng and slipping as in article 278. Or, in symbols, we have the condition (17) of article 277 violated. The problem presents two cases. Case I. Initial Slip is down the Incline. — On referring to article 278 it will be seen that this initial rotation leaves the forces just as they were. Hence the accelerations remain as they were, and the motion presents simply this difference, that all through, the angular velocity now exceeds that in article 278 by its tnitial value, a)„ say. Thus the body fails to ascend the plane, although the slip is down and frictional force up, the linear acceleration being from the outset downwards. Case II. Initial Slip is up the Incline. — Here the frictional force is initially downwards, and the linear acceleration during the first stage is down and numerically greater than in article 278. But obviously the moment of this force about the centre of the body is such as to retard the angular velocity. Hence a point must be reached when the speed of the slip vanishes ; it then changes sign, the frictional force reverses, and the accelerations become as in article 278. The whole motion might be represented by a graph as was done for a different case in Fig. 114. This is left as an exercise to the student. 284. Motion up and down a Rough Plane slightly inclined. — We now suppose the condition (17) of article 277 for pure rolling in free descent to be fulfilled and place our body with initial angular velocity cii„ so as to endeavour to roll up the plane at first. Just as on the level plane it will here, after an initial shp, reach the pure rolling stage ; it may then ascend with diminishing speed, pause, and return, the accelerations for pure rolling being obviously those of articles 274-276. It is accordingly the initial stage that requires consideration. This dif- fers from the pure rolling, not in a reversal of the frictional force but in its being increased if neces- sary to its full limiting value instead of having the perhaps smaller Fig. 115. Motion up or down Rough Plane. ART. 285] PLANE KINETICS OF RIGID BODIES 269 value of the same sign, which satisfies the geometrical condition of pure rolling. But, during the initial stage of slip, the body might fail to ascend the plane but simply descend with a smaller acceleration; or, even remain without motion of its centre of mass, till the slip had ceased. This will appear more clearly from the analysis. Let us take axes and forces as shown in Fig. 115. Then calling the linear and angular accelerations bi and yj, and distinguishing by subscripts i the values of accelerations or forces which apply to this first stage, we have the following equations for a general body : — N—MgCOia. \ TT, — Mgs\na=Mb^ \ (14). Whence b ^= g{ii. cos a— sm a), \ Thus, by the condition for pure rolling, (17) of article 277, b^ may be either positive, or negative of value just reaching that found for a in articles 274-276. We also have from (14) c u-gcosa. , ,, y^=- -n;^ ('^)- Hence, whether b^ is positive or negative, y^ is negative, showing the initial upward rotation will be destroyed. If at time t^ the linear and angular velocities are v and 10, the speed of sHp is then v—iar. So on substitution of their values from (15) and (16) we have • s / c ii.gt cos o.\ Speed of slip =gi{l'- cos a - sm a) - ^^ a)„ -^ — J r = ('^^/iC0sa — sinaW— (o„r . . . (17). Hence the slip vanishes at the time /= /b + c \ g{ —j~ fJL cos a — sin a 1 b(or fj. b tana b-\-c. (18). And by the condition for pure rolHng, the term in square brackets ^it be negative. Thus, the slip must vanish, and after this the cannot be negative. . _ accelerations must be as in articles 274-276. 285. Boiling OscilIations.-Let us now consider the pure rolling 270 ANALYTICAL MECHANICS [art. 285 of a cylinder in a cylinder, a sphere in a sphere, or other body of revolution inside a circular surface so that the motion is parallel to a given vertical plane. Then it is obvi- ous that oecillations are possible about the lowest position as centre. Referring to Fig. 116, let the body have angular displacement 0, its mo- ment of inertia about the central axis perpendicular to the plane of the dia- gram being bMr^jc, and the radius SC of the surface on which it rolls being R. Then since 6 is the inclination to the horizontal of the slope on which the body is now situated, it follows from articles 274-276 that the linear ac- celeration of its centre of mass G is approximately given by —cg6j(b-{-c) if we restrict ourselves to small arcs for which sin d=d nearly. Thus, since the linear acceleration may be expressed by {R~r)d, we have for the equation of motion Fig. 116. Rolling Oscillations. {R-r)e=- b+c' ^6 nearly (i). Whence, the motion is simply harmonic of period, given by T=27rJ{i+c){R-r)/cg (2). Thus, for a cylinder in a cylinder this becomes ■^i = 2Tjs{R-r)/2g . . . . (3). And for a sphere in a sphere we have T,=z27r jT(R-r)/ss . . . (4). Assuming g and observing t, (R—r) may be determined by these relations. Of course, the condition for pure rolling must be fulfilled, but for small oscillations this is practically always the case. Examples — LII. 1. Obtain an expression for the linear acceleration of a body of revolution rolling down an incline, and show what condition must be fulfilled in order that sliding shall be absent. 2. Discuss the conditions and motion occurring when both sliding and rolling of a body on an incline are possible. 3. Investigate the behaviour of a solid wheel and axles rolling down a pair of parallel inclined bars on which the axles rest. What advantage in the determination of ^ would this form of experiment possess over the rolling of a sphere on an inclined plane ? 4. A sphere rolls down a V groove of uniform width and shape in an inclined plane. Determine the linear and angular accelerations. 5. ' Obtain the expression for the kinetic energy of a rigid body whose motion is parallel to one plane, in terms of its angular velocity and the velocity of its centre of mass. ' A thin spherical shell of negligible mass, quite smooth internally, is filled ART. 285] PLANE KINETICS OF RIGID BODIES 271 with water, and allowed to roll down a length / of a rough plane inclined to the horizon at the angle i ; find the time taken. ' If the water freezes and becomes rigidly attached to the shell, what will the time be ? ' (LoND. B.Sc, Pass, Applied Math., 1906, in. 4.) 6. ' Find the moment of inertia of a uniform solid sphere about a diameter. ' A sphere of radius r, rotating with angular A-elocity lT.' (LoND. B.Sc, Pass, Applied Math., 1907, in. 10.) 7. ' Explain what is meant by the principle of energy, and show how it can be used to obtain the motion of any system having only one degree of freedom. ' Two rough solid cylindrical rollers of radius a and mass M are placed with their axes parallel and horizontal upon a fixed plane inclined at an angle i to the horizontal. A log of mass 2M is laid across the rollers. If there be no slipping between either the log or plane and the rollers, find the acceleration with which the log mo\es.' (LoND. B.Sc, Pass, Applied Math., 1909, in. i.) 2 72 ANALYTICAL MECHANICS [arts. 286-287 CHAPTER XIV SOLID RIGID KINETICS 286. Motions of a Rigid Body with One Point fixed. — In dealing with the kinetics of a rigid body we shall follow the order used in Chapter viii. on the kinematics of the subject, taking first the motions when one point is fixed and afterwards those when no point is fixed. And, for the kinetics of the case when one point is fixed, there are two methods of attack open to us which are characterised by the use of fixed axes and by moving axes respectively. 'Y\\.\^'s,, first, we may find the angular momenta (of the form ^mrv) about each of the three fixed axes at right angles with origin at the fixed point of the body, and then equate the rates of change of these momenta to the corresponding torques about these fixed axes. Or, second, we may refer the motions to moving rectangular axes with origin at the fixed point, use expressions like those for angular accelerations (see equation (i), article 123), but, for the angular velocities on the right sides thereof, substitute angular momenta ; the left sides then become the corresponding torques about these moving axes. But in this second plan we have to note that the angular momentum about any axis is not necessarily the product of moment of inertia and angular velocity about that axis. It is given fundamentally and always by an expression of the form ^mrv, where ot is a particle of the body at a perpendicular distance r from the axis about which the momentum is taken and v its component of velocity perpendicular both to r and to the axis. And this expression only reduces in specially simple cases to the product of moment of inertia and angular velocity, as we shall see presently. As to choice between these methods, the first is often unnecessarily long, though for a simple case it is very instructive, and by those wanting only one or two simple examples may be preferable. The second method is more generally useful, and is often very expeditious when the necessary preliminaries have been discussed and the required expressions deduced. We shall accordingly give here a single concrete example of the first method and then pass on to the details of the second. 287. Maintenance of Rectangular Precession of Top. — As an example, then, of the fixed axes method of treatment let us consider the case of a body of revolution rotating with uniform angular speed u about its geometrical axis OA, which is horizontal, while this axis is ART. 287] SOLID RIGID KINETICS 273 itself turning with uniform angular velocity 12 about OZ, which is vertical. AVe here make no use of the theory of moving axes nor do we assume that angular momentum is a vector, though incidentally that fact is illustrated. Take, as shown in Fig. 117, OXYZ as fixed axes, OABC as moving axes, OC being coincident with OZ and vertical, the other four axes being horizontal ; and consider the top when as illustrated the axis OA makes the angle <^ with OX. Then we may write <^ = <^„+J2/ (i). Prece.ision il C lar Momemtum =Iw Fig. 117. Gyroscope. Let us now take in the top a particle of mass m situated at D of co-ordinates x,y\ z with respect to the fixed axes OXYZ, and co-ordinates a, b, c with respect to the moving axes OABC. Further, let the radius FD have length r and be inclined to the horizontal radius FE. Then we may write (9=6l„ + a)/ (2), and using it we obtain b=rco%B, c=rsvc\.0\ / s ^=— fcwrand (r=<«i5 / ^^'' A\'e may further find relations between the two sets of co-ordinates. Thus, making EJ parallel to YO, we have OJ = OFcos<^ — EFsinc^ and JE = OFsin <^-f EFcos <^, or .v=flcos<^— ^sin<^, _i' = asin<^+i5cos <^, 2=(r=ED (4). Then, differentiating (4), remembering a is constant, and using (3), we obtain i- = ( — Q.a + wf) sin ^—Q.bzo%i>\ v={Qa — (oc) cos — 9.b sin j- ... (5). 'i = cos^m{d°+c') = /(ocos(t> (7). Similarly Q='2m(xz—zx) = '2m{ — ilea sin+— ^bc cos <^ — tuab cos <^ + uiJ' sin <^), or Q=(usin<^2»z(iJ=4-c^)=7(usin^ (8), and R=-'^m(yx—xy) = 2ot/ (^^cos0— cu^cos<^— I23sin<^)(rt:cos<^— /5sin(^) \ \ — ( — na sin (^ + ft)f sin <^ — fi^ cos 0)(a sin i + iJ cos ,M=Q=/, J\r=P=o . .(10). If we now choose the instant when the axes AOB coincide with XOY, then <^=o and equation (10) reduces to Z=o, M=/il about OY will maintain an already established positive precession Q. about OZ. But it is only for the instant when the axis of the top OA coincides with OX that the torque G is about an axis which coincides with OY. This can be seen more clearly if we go back to equation (10) and compound the values of L and .^ there expressed. For, as shown in Fig. 117, laying off OL and OM to represent L and M to scale as vectors, it is clear that their resultant is OG of magnitude G^I^Q. (12), and that its axis is coincident with OB at angle 4>_ with OY, i.e. at right angles to OA. Consequently the torque axis, being OB, rotates about OC with angular velocity Q. just as OA does. 289. And, in the case shown in the figure, if the top's spindle is supported at the origin O only, and the torque G is due to gravity, we easily see that G=gSma=M^gd (j^^^ if M^ is the mass of the top and a is the co-ordinate of its centre of mass; and that the torque axis automatically remains at right angles to the axis of the top. Hence the rate of precession i2 which, if started ART. 290] SOLID RIGID KINETICS 275 by other means, can be maintained by this torque due to gravity, is from (11), (12), and (13) given by I2=(?//a,=Ar„^a//a, (14). Thus, if the top consists mainly of a disc of mass 72 lbs., i foot diameter, with 5=8 inches, and spins at 6000 revolutions per minute, we have /=9 lbs. ft.^ and to = 2oo t; whence 12 = o'273 radian per second, or one revolution in about 22^ seconds. Referring again to (12), we see that corresponding to a quicker precession we have a larger G ; hence if the precession be hurried beyond the value in (14), while only the gravity torque is available, the axis OA will rise from the horizontal. Conversely, if the precession be slowed or prevented, the torque remaining constant, the axis OA will fall. The foregoing is perhaps the simplest way to account for these possible rises or falls. Another instructive way of regarding the matter is to treat the hurrying or retarding of the precession as a (positive or negative) torque about OC which corresponds to a precession about OB. Both views are useful, and the phenomena in question have important practical applications, as we shall see later. Examples — LIII. 1. What chief methods are open to us in the discussion and treatment of the motions of a rigid body with one point fixed ? Indicate what you consider to be the relative advantages of the various methods. 2. Find the torques necessary to maintain a steady precession about one axis of a constant angular momentum about a perpendicular axis. 3. Give a numerical instance of question 2, in which the torque is supplied by gravity, using either c.g.s. units or British. 4. Enumerate several familiar illustrations of the above-mentioned preces- sional phenomena, accounting in each case for the sign of the precession which occurs. 290. General Expressions for Angular Momenta. — The length of the foregoing example of the first method of treatment, simple as the problem was, well illustrates the necessity for generally using the second method involving moving axes. We now take a step towards the second method by obtaining expressions for the angular momenta ^1, }u, and K about each of three rectangular axes OXYZ which we will at first suppose fixed in space, the simultaneous components of angular velocity about them being denoted by n^x, <^y, and uiz- Then we shall find that the angular momentum about any one axis depends, in the general case, upon all the three components of angular velocity, and is not necessarily the simple product of moment of inertia into angular velocity about the axis in question. The fundamental expression for the angular momentum about the axis of z say is ^m(}'x—xy). And, although when there is rotation about OZ only this reduces to Cwz where C is the moment of inertia about the e axis, this is by no means the case when there are simultaneous rotations about the other axes. For in the geiieral case these other rotations modify the motions of many of the particles in the body, and some of these changes of velocity have moments 275 ANALYTICAL MECHANICS [art. 290 about the axis of z, and therefore modify the angular momentum about it. Fig. 1 18. Angular Momenta. Thus, consider the point P of co-ordinates x, y, z in Fig. 118. Then from equation (i) of article 132, the origin being now at rest,' we find y — iazX — u>.xZ \ (l). z^in^y — fiiyx] Putting these values in the expressions for angular momenta, we have h 1 = 2»i(zy — jz) = (u^m(y' -\-z )— (Uy^mxy — ui^'Zmzx. = 2ot( u>yZ^ — oi^yz — (D^yx -\- toyx' ) = —lo^mxy+iDT^miz" -\-x'') — ti)^myz. hi = 'S,m{yx — xv) = 1,m(b>zX^ — lo^zx — uiyzy -\- ("zy") = — la^mzx — (a^myz + (o^m{x^ +J>'^)- If we now write for the moments and products of inertia of the body about these axes at the instant in question A, B, C and D, E, F, we then have A^l,m(y''+z-),£=y.m{z''+x''),C='S.m{x''+y-) ) ,s D=2.myz, £—'Zmzx, F=1,mxy j ' ^ '' Hence, on substituting these abbreviations, in the above expressions for the angular momenta, they become ■.+Au ■Juo,, — £(ji (3)- h„^—Fwx,-\-BMy — Du>z \ /^3= — Eilix — Diay-\- Claiz J These expressions are easily remembered by noting that the A, B, and C appear in order as positive coefficients in a diagonal, the D, E, and i^ occurring twice each as negative coefficients in order returning round from C to ^. The above formulae (3) are quite general, and give the instantaneous ART. 291] SOLID RIGID KINETICS 277 angular momenta whether the axes are fixed or not. For, if the axes of reference are moving, the motion of the body in an element of time IS constructed by using the components of motion as if the axes were instantaneously fixed. And the above axes may be any whatever. Hence, if they are chosen to coincide for the instant with any given set of movmg axes, the above formulae give the instantaneous angular momenta about them. We may note here a few typical cases and the values to which the general expressions for angular momenta then reduce. _ Thus, if the products of inertia vanish, the axes in question are called principal axes for the given origin, and we ha\e h^ = Aia^,h.=£i,j,, /l^=C(a^] ' • ■ • ■ • (4), the subscripts i, 2, and 3 being now used for the cu's, since we have seen the results apply to moving as well as fixed axes. If the body is a plane lamina perpendimlar to the third axis, then two of the products of inertia vanish and one moment is the sum of the other two, we thus have D^E^o, C=A+£ -1 h, = A^-Fia^, h.,^-Fi»^+Bw^, h, = {A+B)ui,] ' (S)- For a body of revolution about the third axis, all the products of inertia vanish and two moments are equal. Thus we have n = E=F:=o,A = B \ hj,—A(o^,hi=A(o^,h3=Cfa^j ■ ■ ■ .... {b). For a homogeneous sphere or a spherical shell or a sphere built up of concentric shells each of which is homogeneous with origin at centre, we have n=E^F=o,A=B=C \ ,. hi — A(»i,h.2=Aw,,h3=Au)sj ' '^■'" 291. Angtilar Momentum is a Vector directed along its Axis. — From the fundamental expressions for angular momenta, it may be seen that they are vectors directed along the axes about which the momenta are taken. Thus, for the value of hi the fundamental expression is 1,m{zy—y3). So that, apart from mass and time, we have the directed quantities y and s which define the j^z plane characterised by its normal, the axis of x, which is therefore the direction of the vector h^, the angular momentum about the axis of .r. The same treatment of the like expressions for the other angular momenta, or for the general expression ^mrv, shows that the angular momentum about any axis is a vector directed along that axis. If, however, we now turn from these fundamental expressions in terms of velocities v and arms r to the general expressions in equation (3) of article 290, a doubt may be felt by some as to whether each term on the right side of any equation is a vector along the axis of the corresponding angular momentum. Indeed, at first sight, it might be imagined that each angular momentum was expressed as made up of three parts directed along the three rectangular axes. But this is 278 ANALYTICAL MECHANICS [art. 292 not the case. Each term on the right in any one equation has the same direction, namely, that of the axis indicated by the subscript to the h on the left side. To establish this we may conveniently apply the method of examining the dimensions of the factors that constitute the terms on the right, but we must retain the idea of the direction of each one and not represent each length by L simply. Thus making this dimen- sional or directional inspection of the terms for /^i, and indicating masses, direction, and time by M, XYZ, and T, we find / Z Y\ The dimensions oi Au>x are M{ F'+Z-)(-r™ or -^\ Z The dimensions of Fwy are (JOrF)rr;r^ Y The dimensions of Eidz are (MZX)^n-^ Z_ Y\_ MYZ \ YT°^ Zt)~ T MYZ '' T MYZ T ] (8). We thus see that each term in this general expression for h^ is a vector directed along the axis of x, the normal to the plane defined by YZ. Similarly h, is expressed by three terms each of which is directed along the axis of jc, and h^ consists of three terms each directed along the axis of a. It therefore appears by this method that angular momentum about any axis is a vector directed along that axis. 292. Axes of Resultant Velocity and Momentum. — Thus, since angular momentum is a vector, if we have given the angular velocities of a body about each of the three axes, also the moments and products of inertia for these axes, we may find the three angular momenta, and can then determine both the resultant angular velocity and the resultant angular momentum, using vector addition in each case. At first sight it may seem startling to find that the directions or axes of these two resultants are not necessarily coincident. A numerical example of this is illustrated in Fig. 119, which shows angular velocities and mo- menta about the three axes OXYZ of the ellipsoid of semi-axes 3, 2, and r, with centre at the origin and geometrical axes along the axes of co-ordinates. Thus its equation is 3 2 + p=i. Its products of inertia are all zero on account of symmetry, and its moments of inertia are A=iJ, B = 2j, and € = 3^ if the mass is \. Thus, if the component angular velocities are 60, 48, and 40, the corresponding component Fig, 119. Resultant Angular Velocity and Momentum. ART. 293] SOLID RIGID KINETICS 279 angular momenta are 75, 120, and 130. Then, the resultant angular velocity J2, represented by Ofi on the figure, has magnitude fi= V6o=+48= + 4o' = 86-6 . . . and direction cosines 6o/86-6, 48/86-6, and 40/86-6. Whereas the resultant angular momentum, represented by OH in the figure, has magnitude H— 775'+i2o'-+-i3o'=i92'i • • • and direction cosines 75/192-1, 120/192-1, and 130/1921. Thus the angle between the directions of the two resultants is given by cos H0n = ^°^-71+48X 120+40X130^^ 86-6 X 192 ^ ^ Whence the angle H0fi=2i° 35' 30" nearly, as may be seen also by taking the logarithmic cosine. 293. Analogous Relations between Mechanical Quantities. — Looking again at Fig. 119, we easily see that, if any component angular velocity receives an increase, the point fi, which defines the resultant angular velocity, moves a corresponding distance and parallel to the axis of acceleration. Also, if any component angular momentum receives an increase, the point H, defining the resultant angular momentum, moves a corresponding distance and parallel to the axis about which the increase occurred. Also the rate of increase of angular momentum about any axis may be equated to the torque about that axis, which is therefore equal to the velocity of the point H in the diagram. (See Notes on Dynamics, by Sir G. Greenhill, p. 195, 1908.) Thus we may construct for angular velocities and angular motnenta loci analogous to the hodograph of a point moving in any way. We may also collect in a compact form the analogous relations as to linear velocities and momenta. These are shown together in a form easily remembered in Table xn., and are capable of numerous applications. [Table. 28o ANALYTICAL MECHANICS [art. 294 Table XII. Analogous Changes in Directed Quantities. Quantity. Continuous change in magnitude only at rate dldt requires a collinear :- Continuous change in direction only at rate fl rad./sec. requires a perpendicular : — Velocity. ■ >v Acceleration. AcceleratioDi Momentum. • >p Force. Force. pa. Velocity. Angular Acceleration. -St . CO Angular Acceleration. iQ Angular Momentum. Torque. Torque. hQ A continuous change of any quantity in both magnitude and direction requires a suitable combination of the above collinear and perpendicular components. Examples — LIV. 1. Obtain a set of general expressions for the angular momenta of a rigid body. 2. Prove that an angular momentum is a vector directed along its axis. 3. Show that the axes of resultant angular velocity and momentum are not necessarily in coincidence. Take a numerical illustration, and find the angle between the two axes. 4. Discuss linear and angular velocities and momenta and their changes, and show that a useful analogy may be traced. Exhibit your chief conclusions in tabular form. 294. Torques about Moving Axes.— If, in equations (i) of article 123, we now substitute the values of the angular momenta (from (2). ART. 295] SOLID RIGID KINETICS 281 equation (3) of article 290) or denote them by h^, h^, and h^, we obtain expressions for the torques about the set of moving rectangular axes OA, OB, and OC. These are M=h,-h,e^-\-h^e\ . ... . . (i). and N=h-h^e^-\-h,B^] Substituting the general values for the /z's, we have for the torque about the moving axis OA L = -jiAm^ — Fi = wa;, '»>2 = 'i'!/, and (b3 = fc)z . . . . • (5) ioT moving axes fixed in the body at the instant of their coincidence with the fixed axes. The great advantage of using moving axes fixed in the body lies in the obvious fact that the moments and products of inertia which appear in the equations are then constant quantities. 296. Euler's Dynamical Eguations. — We now introduce a third simplification, namely, that the moving axes OABC fixed in the body are also the principal axes for the fixed point O. Then we have the products of inertia all zero, or D=E=F=o (6). Accordingly, the angular momenta reduce to hi=A(ii^,hi = Bu hi=Cto^ .... (7). Thus (3), (4), (5), and (7) in (i) yield for the torques about these moving principal axes fixed in the body I,:=A(J)i — (£ — C)t02(i)A M=£a^-(C-A)i»,(aA . . . . (8). J\/'= Ca — {A — £)iii^ia^ J These are the well-known Dynamical Equations of Euler. But in- stead of using them it is often found preferable to take a set of moving axes of which only one is fixed in the body, one fixed in space, the other moving with respect both to space and the body. Examples— LV. 1. Obtain an expression for the torque about one of a set of moving axes for any specified angular velocities of a given body. From what draw- back does this treatment suffer ? 2. If the set of rectangular moving axes are chosen so as to \)& fixed xa the body, establish the relations and state the simplifications which then follow. 3. Assuming the results of questions i and 2, establish Eulei-'s equations. 4. Apply Euler's equations to prove that the maintenance of the steady precession a about a vertical axis of the angular momentum Aa about a horizontal axis requires the torque AaO. about a perpendicular horizontal axis. 297. Steady Precession of Top. — Let us now apply the method of moving axes and the torques about them to the case of the steady pre- cession of a body of revolution, with one point fixed. Let the moving rectangular axes OABC have OC inclined at an angle 6 with the vertical, OA inclined Q with the horizontal, and OB horizontal. Let the fixed point of the top be at the origin of co-ordinates, the axis of the top coinciding with OC, about which its angular velocity is w and ART. 297] SOLID RIGID KINETICS 283 Its moment of mertia C. Then the other two moments of inertia are alike, and the products of inertia all vanish since the body is one of revolution about OC. We may thus write, in the usual notation for moments and products of inertia, A^B,D^E=F=o (i). Further, let the axis OC move at constant angular velocity fi round the vertical fixed axis OZ say. Then, by the kinematics of the case, worked in article 124, equations (2) to (4), we have for the angular veloci- ties w„ M„, (03 of the top'i'J about OA, OB, and OCj^| and for the angular velo- St cities ^i, ^2, ^3 of the axes about their instantaneous positions 3 = Thus, by substitution of these values in the general expressions for the angular momenta h^, h„, and h^, equation (3) of article 290, we obtain Steady Conical Precession OF Top. (3)> (3), (4). h.i = o (5)- (6). hi = C(o J And then, using the expressions given in equation (i) of article 294 for the torques about the axes OA, OB, and OC, we find Z = o \ 7l/=Ca)Osine-^fi'sin6lcos(9[ . These results should be compared with equations (5), (6), and (7) of article 124, which gave the corresponding angular accelerations. We easily see that, if the torque M\% due to the weight ^of the top, its centre of mass being at a distance h from O, M=WhwyB (?)• Hence by combining (6) and (7) fVA=:C(oa- Ail' cose (8). Accordingly this is the condition for the maintenance of the existing motion by a torque due to gravity. This may be compared to equation (14) of article 288, to which it is equivalent on putting 6»=7r/2 and C=/. We may further note from (6) and (8) that since these are quadratics in fi, there are in general iwo rates of precession, either of which, if 284 ANALYTICAL MECHANICS [art, 298 established, could be maintained by a given torque. We see from (8) that these are given by Cm+ J Coi^-4A Wh cos e ,. "- ^A^^^ .... ^9;. The two roots are, however, coincident when the quantity under the radical sign vanishes. And, for a smaller value of la, the rate of spinning, it is obvious that the rate of precession is imaginary. Hence we have as the minimum speed of spinning for the maintenance by gravity of the steady precession '^ = ~r 'J-AWhco% d . (10). 298. Conical Precession without Torque. — On reference to equa- tions (6) or (8) of article 297, we see that if Ctu— ^ficos ^=0 (11), the torque usually required to maintain the precession vanishes. Fig. 120. Automatic Precession. Hence, the motion being once established, it continues without any impressed torque. Sir G. Greenhill has shown that a loaded bicycle wheel is very useful for demonstrating various gyroscopic phenomena (Notes on Dynamics, p. 197, 1908). ART. 299] SOLID RIGID KINETICS 285 One mounted as shown in Fig. 120 is in use at Nottingham. It shows very well this conical precession without torque, just referred to. For, when balanced carefully by sliding the balance weight along the tube at the side remote from the wheel, the wheel spun smartly by hand, and then the far end of the tube moved with the right direction and speed in a circle as found by a few trials, the motion is maintained when the tube is let go. Many other gyroscopic properties can also be demonstrated with this apparatus more effectively than with the smaller gyroscopes sold as toys. For details of the various experiments the reader is referred to such works as Worthington's Dynamics of Rotation, or Crabtree's Spinning Tops and Gyroscopic Motion. 299. General Expressions for Kinetic Energy of Rotations. — Consider any body which has simultaneous angular velocities Wx, <^y, and z about the axes of x, y, and z, the origin of co-ordinates being at rest. Then, from equation (i) of article 132, we have for the velocities of a point whose co-ordinates are .r, y, and z the following expres- sions : — .V = layZ — 0)^ \ y = ta^x — io^z \ (l). Z=lilxV — Wyx] Thus, if we imagine there is a particle of mass m at the point in question, the kinetic energy is given by = }^m{{wyZ — u>;^Y + {<»zX — <>)xZ)- + {iaj^—U)yXy] = W^m{v'+z') + W^m(z"- +x') + ia>^2»2(.r= +/) — (OyUiz-myz — (o^oix^mzx — oj^'^ii^ffixy. Or, with the usual notation for the moments and products of inertia, this becomes T=iAa>l + ^Bu)l + hC(i>yOiz—-Ei^z'^x—-Pz (4)- When the axes in question are principal axes, the products of inertia are all zero, and we have the simplified expressions These rectangular axes of rotation may be either hxed or the instantaneous positions of moving axes. As a check upon the above relations we may reduce to a simple case Thus, let the kinetic energy T of rotation about a single ajas be produced by the expenditure of work ^in the form of a steady torque G exerted through an angle 6, accordingly giving a final 286 ANALYTICAL MECHANICS [art. 300 angular velocity w (or at) and angular momentum H about this axis. Then, the moment of inertia in question being /, we have T=W=Ge=lGti^^iiH<^=lI'»'' (6), since 6=\is>t and H=Iu>, the initial values T^ and H^ being each zero. Examples — LVI 1. Show that the maintenance of a steady precession a about OZ of the plane 20 C while an angular momentum Ca> exists about OC requires about an axis OB, perpendicular to the plane ZOC, the torque CaO. sin 6 - AQ?' sin 6 cos 6, where A is the moment of inertia of the body about the axes OA or OB perpendicular to OC. 2. From the result of question i prove that, if the torque is due to gravity, OZ being vertical, different values of 6 will require different values of Q. Also find the minimum speed of spin for the maintenance of the preces- sion by gravity. 3. Account for the fact that, if a body be thrown into the air spinning, its axis of spin is often seen to describe a cone about a line of fixed direction moving with the body. Do you know of any apparatus which illustrates the same phenomena ? 4. Obtain general expressions for the kinetic energy of rotation of a body, and check this by considering some simple case. 300. Starting Precession. — We are now in a position to attack a simple example of the un- steady motion of a gyroscope in which the precession and inclination are each variable. We take the case of a body of revolution initially rotat- ing at the speed Wo about its geometrical axis OC, which is then inclined at the angle ^„ with the vertical OZ, the body being sup- ported at the origin only and having no other motion than that specified, but free to acquire any other mo- tions. This initial state of things is indicated in Fig. 121, in which OXYZ are the fixed rect- angular axes, OABC the moving rectangular axes, ^ being the angle between the planes ZOX and ZOC. Let the weight of the top be W, its centre of mass be distant h from the fixed point of support O. Then, neglecting all frictional resistances, the sole torque acting upon the top is that due to gravity, and is about the horizontal axis OB, which is perpendicular to the plane ZOC. This gravity torque has accordingly no component torque Starting Precession. ART. 300] SOLID RIGID KINETICS 287 about either of the axes OC or OZ. Hence, the angular momentum about OC is constant, or in symbols, w^o and (i) = a)|, .... .... (i). Further, the angular momentum h about OZ is constant, or yi=o and /« = constant (2). But, to solve the problem, we need the values at any time t of the three angular velocities (o about OC, B about OB, and ■^ about OZ. We accordingly need a third equation. This could be obtained by equating the torque about OB to the corresponding rate of change of angular momentum. It is, however, somewhat simpler to use the principle of conservation of energy. Thus, finding general expressions for the kinetic and potential energies, we may equate their sum to that of their initial values. This gives the third equation required, which we may now write as r+r=r„+r„ (3). We have next to substitute the actual values in (2) and (3). Let C be the moment of inertia of the top about OC, B=A that about either OB or OA. And, since the body is a solid of revolution about OC, its three products of inertia, D, E, and F, all vanish. It is seen from the figure that the angular velocity ^ about OZ has a component — ■^sin ^ about OA. We accordingly obtain for the angular momenta about OA, OB, and OC the values h,=A{-i5in.0), h^=Bd=Ae, a.n6. Ih = Ci->. Further, the angular momentum h about OZ receives no component from OB, which is perpendicular to it, but only from those about OA and OC. Thus, for the angular momentum about OZ, we have h=hi{— sin 6)-\-h^cos0 — A^sWd+Cwcos 6. And, since by (2) this is constant, we may write A-ir 5\n- 6 +€>>> cos 6 =Ci» cos 6 „ . . (4) as the equation of conservation of angular momentum about OZ. We have still to form the expressions for the energy and substitute in (3). Thus, from article 299, we have 7^^i.j(_-f sin ey- +^Be- +^c<^\ Also the potential energy, reckoned from the level of O, is obviously V= Wh cos 9. Thus, substituting in (3), and cancelling Cw" from each side, we have A{f'' sin- e+&-) + 2 Wh cos, d= 2 Wh cos d„ .... (5). Equations (4) and 5 determine the whole subsequent motion of the top. From (4) we obtain the rate of precession in terms of 6, viz. . _ C(0(COS dj, — COS^) /g\ ^~ A sin'<^ Thus, at the start when 6=6,, we have for the initial value of the precession ^^^^ 288 ANAL YTICAL MECHANICS [art. 301 But (6) shows that for any larger value of Q than 5„, the precession ■^ has a finite value. Thus, though the top does not immediately precess, it cannot fall without doing so. If now we differentiate (6), we find for the precessional acceleration .. Cw^(l — 2COS^„COS(9+ cos°6) , > ^= \ "A^^xTQ _ ■ ■ (*)• Thus since at the start 5=o, the initial value of ^ is zero also. Hence the first motion must be an increase of Q simply, that is, a falling of the axis OC from the. vertical in the initial position of the plane ZOC. But, immediately has a finite value, '^ has a finite value also, and if accordingly grows; that is, the plane ZOC acquires a velocity about the vertical OZ. For, if & is finite while Q is still practically equal to ^0, (8) reduces to ^=CiaeiAsme, (9). 301. Nutations or Oscillations in the Azimuthal Plane. — We have just seen that the starting of the precession is inseparable from a change in the value of 6, that is, a fall or rise of the inclined axis OC of the top. We may next naturally ask whether this motion, which is expressed by d, and which lies in the azimuthal plane ZOC, is initially a rise or a fall, whether it has any limits, and what is its general character. To answer these questions we begin by eliminating ■^ between equations (5) and (6) of article 300. Thus substituting from (6) in (5), we find Cw^cos e,- cos e)-+4'e'' s[n'e= 2 a wk(cos e,- cos e) ^m-e. The stationary values of 6, if any occur, will be obtained from this by putting in it d=o. We thus find (cos e„-cos d){2A Wk sin's -C'b^^cos, 6l„-cos 6)\^o. Hence, the stationary values of 6 are given by ^=^0 ' (lo), and the roots of sin"^— 2A(cos ^„ — cos ^) = o ... (ii), where 2X.= C(a^/2AIV/i. Equation (ii) may be written cos'^— 2Acos ^+2A.cos ^„— i=o . . (r2), giving for cos 6 the values cos^=AHt Vi — 2Acos ^„ + A^ . (13). Since cos 6, is less than unity, it follows that the quantity under the radical sign is greater than ( r — A)^ Thus, if the radical has the value + (i — X+k), taking the upper sign would give for cos 6 the value 1+^; and this value, though real, gives no real cosine. We are therefore limited to the negative sign of the radical, and calling the corresponding value of the angle 61, we have cos6i = A— ^i — 2ACOS S„ + A^=:2A— I— /4 say . . . (14). Thus, as the azimuthal plane ZOC rotates about OZ with the variable velocity ijf, the axis OC of the top also oscillates in the azimuthal plane between the positions defined by the values 6„ and S, ART. 302] SOLID RIGID KINETICS 289 Fig. 122. Nutation. Of the angle ZOC, Q, being the initial value when OC was at rest, Q, the other limiting value reached at some later instant. This oscillation ot the axis m the azimuthal plane is called nutation _ It IS obvious that the inclination Q, is below Q,, for the additional kinetic energy of the preces- sional motion -f , which we saw was associated with the change in 9, could only be obtained at the expense of potential energy. Thus Q^ and 6^ are the mini- mum and maximum values of assumed by the axis OC. These limits are indicated by Co and Ci in Fig. 122. The path described on the surface of a sphere by a point C on the axle is also shown by Co, D, E, F. It is thus seen that the curve has a cusp for each time the upper limit is reached, as at C„ and E. This motion may be easily observed if a gyroscope be spun slowly, placed in an inchned position with its point in the cup, and then suddenly let go. If, instead of letting go simply, the top be given a forward or backward impulse, the curve described will be a wavy curve or a looped one respectively, instead of the cusped one just dealt with. This experiment for showing nutation is much more striking if shown with the bicycle wheel apparatus represented in Fig. rzo, article 298. 302. Precessional Velocities at Limiting Inclinations. — Let us now obtain the values of -^ for the highest and lowest positions of the axis OC, that is, for the limiting values of Q, viz. Q„ and Q^. Re- write here equations (4) and (5) of article 300 in the form ^■^sin^^=Cu)(cos ^0 — cos ^) (15), and ^f=sin'^-f^^' = 2/f-7z(cos^„-cos(9) . . . (16). Then (15) multiplied by -^ gives Af°'sin'0=Co)i/-{cos6„ — cosd). . . . (17). Hence, for the limiting inclinations, when ^ = 0, the left sides of (16) and (17) are equal. Thus, on equating their right sides, we have (cosO.-cos e)(Cwf -2 UA)=o . . .(18). Accordingly, eit/ier (i) 0=0„, in which case we have from (15) ^0 = (19), or (2) Cb>f—2lV/l — 0, that is, f^ = 2 1171/ Ceo (20). T ago ANALYTICAL MECHANICS [arts. 303-304 The subscripts to the ^'s here correspond to those of the ^'s, to which positions they refer. Equation (19) agrees with what was previously found in (7) of article 300, the result in (20) being new. It may be noted here that this maximum rate of precession ^, at the limit 6-^ of inclination exceeds the steady rate of precession which, if established, the given torque is able to maintain. Thus if 6t is r/a, we reduce to the case of rectangular precession, in which, as shown in equations (14) of article 288 and (8) of 297, we have a=WhlCi.> (21), so that the steady fi is in this case only half the maximum value of '^. On the other hand, the minimum value of ■^ is zero. 303. Tilting Velocity of. Top. — From equations (15) and (16) we may now find the value of 6, the angular velocity about OB, that is, the velocity of tilting in the azimuthal plane ZOC. Thus, substituting in (16) the value of ■^ from (15), we have CV(cOS^„ — C0SS)V .M jj,,, a m ^ . .\a +A6'' = 2lVk(cos9a — cosd). A sm ^ ' Whenc 6= '^ ^^ Wh{co% gp-cos 6) sin'g- C'a)'(cos6l„-cos^)"^ A sin Q ^^^'' And dividing this by the equation obtained from (15) expressing the precessional velocity ■f, we obtain the ratio of the two velocities , ■ .. / ^'0 , , yjr ^ '\ 2A(C0S6„ — cos S) ^ ^" where 2A. is used as before to denote C^iu^jzA Wh. This equation gives, in terms of 6, the direction of the path traced on a spherical surface by any point on the axis OC. Its projection on the horizontal plane xy lies between two circles which correspond to the limiting inclinations 0^ and 6^ of the axis. Further, the path meets the inner circle with cusps tangential to the radii, because there 0/'^=oo, as shown by (23), and touches the outer circle, because there 5/y=o, as shown by (6) of article 300 and the fact that B is stationary at ^1. Thus, before any precession is established, the top yields slightly to the tilting torque, but soon the precession exceeds the steady value which the torque could maintain, and the top rises to the next cusp. 304. Minimal Velocities for Top to Spin and to ' Sleep.' — The foregoing considerations lead us to note that if the value of Q^ is such that the body of the top catches the horizontal plane, it will cease to spin. Suppose the inclination when the top thus catches the plane is Qi, then for spinning we must have 6^ > 9^. And this may be shown to necessitate a minimum value of the angular velocity o) of spin. Thus (following the method of Crabtree) we may write from (15) of article 302 !_Cm(l-~C0s6—K)_Cb)f I '^ \ / \ '^~ A{i -cos'WT ~l4\i~+coIe~ I -cos'ef ' ^'"^'' ART. 305] SOLID RIGID KINETICS 291 where k is the positive constant i— cos d^. Hence 1^ increases with 6. Thus if ^3 corresponds to 6„, we have fi>fi. But, reverting again to (15), we have : _C(u(c0S ^„ — cos ^a) , , '^'~ A sin^(9, ......... (25). And by (20) we have f, = 2JVA/C(o (26). So, on substituting the values of these two precessions in the inequality connecting them, we have C4AlV/i/C' .... (29). Of course, in the case of any actual top whose spinning is not maintained, friction will check the speed until it falls below the value in (29), and the top then begins to wobble, and finally falls. 305. Examples of Gyroscopic Motion. — Perhaps the most important example of gyroscopic motion is that of the earth itself, which we may regard as a top spinning about its polar axis and subject to fluctuating torques or couples due to the attractions between its equatorial pro- turberances and the moon or sun. Fig. 123 illustrates this effect of the sun or moon when in its most favourable position for producing it. For the attracting body B is shown in the plane containing the polar axis of the earth. Now since the earth is not spherical we may regard the i,B'_; attractions on it as separable into three components, viz. (i) the main one correspond- ing to the sphere on the polar Fig. 123. Cause of Precession of the diameter and acting at the Equinoxes. centre; (2) a small force of •, / \ r like sign on the eastern equatorial protuberance; and (3) one of opposite sign on the western protuberance. These smaller forces are obviously due to the one protuberance being nearer to and the other farther from the attracting body than the earth's centre is. We have thus a counter-clockwise torque or couple acting on _f. 292 ANALYTICAL MECHANICS [ART. 305 the earth as shown in the diagram. Then by the phenomena of pre- cession, as already dealt with, we see that since the earth is rotating from west to east, instead of simply yielding to this torque, it will turn about the third rectangular axis WE, and in such sense that N comes towards the spectator and S recedes as indicated by the conventional signs © and ©, representing respectively the point and the feathers of an arrow. If now we transfer the attracting body to the same distance on the other side as indicated by the dotted circle B', it is evident that though the main attraction is reversed, as shown by the dotted arrow, the couple represented by the small arrows remains unchanged. But, if B is brought to a position on the normal to the diagram through the centre of the earth, then the couple vanishes whether the attracting body is before or behind the plane represented. For the W and E protuberances shown are in those cases equidistant from the attracting body and the near and far protuberances (not shown in the figure) are exactly in the line of centres. Thus we see that as the moon goes round the earth the torque alternately waxes and wanes, vanishing twice in the period of the moon's orbit, but has always the same sign. This torque produces the lunar precession. The sun also produces the solar precession ; and similar remarks apply to this as to waxing and waning and constancy of sign, the period being now that of the earth's orbit, or a year. The solar and lunar precessions are roughly in the ratio of 3 : 7. The combined effect is sometimes called the luni-solar precession. In consequence of this the pole P of the earth on the celestial sphere slowly describes a circle of radius 23° 27' round K, the pole of the ecliptic on the celestial sphere. The average angle described in a year is about so"-2, and the time of a complete revolution is of the order 25,800 years. But, superposed upon this precession, there is also nutation ; and this is due partly to the sun and partly to the moon. The lunar nutation is due to the change of the moon's nodes, and has a period of about 18 years 220 days, being that of a sidereal revolution of the moon's nodes. The solar nutation is due to the variation of the sun's declination, and has a period of half a tropical year (which year is the time between successive vernal equinoxes). The amplitudes of these lunar and solar nutations are of the order 9" and i"-2. Hence the pole P of the earth's equator traces a wavy line on the celestial sphere round the pole K of the ecliptic. There are many other examples of gyroscopic motions, some of which are quite familiar. Thus, the turning of a hoop when rolling along and made to lean, or of a coin, are cases in point. The rearing or plunging of a very fast single-screw turbine steamboat, instead of answering its helm, is another. Two cases in which gyroscopic actions have been applied may be mentioned. Thus Otto Schlick has arranged to steady a vessel at sea by means of gyroscopes, a rolling amplitude of 15° each side the vertical being thus reduced to an arc of rolling (out to out) of 1°. ART. 306] SOLID RIGID KINETICS 293 Louis Brennan has applied gyroscopes to balance a special monorail car, both when going on the straight and round curves, the principle underlying his automatic devices being that of hurrying the precession, and thus causing the car to rise from a tilted position. It is beyond the scope of this book, however, to enter into details of these ingenious inventions or the many other problems of gyroscopic motion. The interested reader may consult H. Crabtree's Spinning Tops and Gyroscopic Motion, A. M. Worthington's Dynamics of Rotation, and Sir G. Greenhill's Notes on Dynamics; also Engineering, vol. Ixxxiii., 1907, pp. 442, 448, 623, and 794, and p. 797 of June 24, 1910, and Nature, Professor Perry's article. The Use of Gyrostats, vol. Ixxvii. pp. 447-450, March 12, 1908. Examples — LVII. 1. A top is spinning about an inclined axis with its highest and lowest points supported, discuss what happens when the upper point is released and the axis is accordingly set free to move. 2. Explain what are meant by the terms precession and nutation, and give examples from an experiment and from the solar system. 3. In a typical case of the motion of a top under gravity, find an expression for the velocity of nutation or movement in the azimuthal plane. 4. Explain how the rate of precession changes when nutation is occurring, and obtain the limiting expressions for tliis rate. 5. Show that there are minimum velocities of spin for a top to clear the ground and for it to ' sleep.' 6. Give several examples of gyroscopic actions ; some of them presenting difficulties to be overcome, others being useful applications of this action. 306. Centrifugal Keactions and Torques. — Referring now to Euler's equations, obtained in article 296, we note that the whole changes occurring in the values of the component angular velocities Wi, Wj, and (03 are not entirely due to the direct action of the external torques Z, M, and N, but are due in part to the centrifugal reactions which occur in virtue of the motions of the. particles and the rigidity of the body. Thus, taking the. last equation of (8) in the article referred to, it may be written ,dtlC is due to centrifugal xe- actions. Following Routh, this may be enunciated and proved thus : — If a rigid body be rotating about an axis 01 with an angular velocity ,, OZ being the instantaneous position of the moving axis OC, the component velocities being w^, w^, and 0)3. In Fig. 124, let P be the position of any particle m of the body, its co-ordinates being x, y, and z, which are represented re- spectively by OR, RQ, and QP. Let PS=r be the perpendicular 294 ANALYTICAL MECHANICS fART. 306 upon 01 and let OS = «. Then, as the particle m rotates round 01 at angular velocity w, the centripetal force needed on it is »wV directed from P to S. Accordingly the reaction exerted by the particle is oppositely directed and of equal magnitude. Hence, it is muP'r acting at P and directed from S. Since the vector SP=ris equivalent to the vector sum of SO, OR, RQ, and QP, this force /wmV is equivalent to the four forces — mm^u, muy'x, mw^y, and muy'z, all acting at P and parallel to u, X, y, and z respectively. The moment of m,, —muwo)„, and —viuamig re- spectively, since the direction cosines of 01 are loi/o), Wj/cu, and Ws/o). The moment of these forces round OZ is mu(i>{^x-\-(i>iy-{->i>gz)((j>jy — ut^x) = m{(OiO)^{y'—x') + o)sti>^z—ta^w^zx+{^ = N+N' . . (8). 307. Independence of Translation of Centre of Mass and Rotation about it.— We have hitherto in this chapter supposed one point of our rigid body to be fixed. Let this restriction be now removed. Then the subsequent work is often much simplified by the fact that the trans- lation of the centre of mass and the rotation about any axis through it are quite independent of each other. That this is so might have been inferred from what was previously shown for coplanar motions. But it is more satisfactory to have an independent proof. We shall deal in turn with linear momenta, kinetic energy, an- gular momenta, and their rates of change. Linear Momenta. — Refer- ring to Fig. 125, let OXYZ denote fixed axes and GABC axes whose origin G is at the centre of mass of the moving body and whose directions re- main parallel to those of .r, y, and z. Let a particle of mass m of the body have co-ordinates X, y, z with respect to the fixed axes and a, b, c with respect to GABC. Further, let the moving point G have co-ordinates x, y, and z and velocity components u, v, and w, the body having angular velocity components tu^, Wy, and w^. Then we have the following relations among the co-ordinates and velocities : — x=x + a, y=y-\-l>, 125. Independence of Trans- lation AND Rotation. -ia,l>. h=^w2a — coj;(r. --z-\-c (0- x = ic-\-iOyC—ii)J),y=:V-\-uiza — ^7fi, i = w-\-Wx,b — Wya Let the linear momenta h& px,fy, and/z, then we have p^ = '2mx = u'2m + a,i'2mc— hi^mh. But since G, the origin of a, b, and c, is the centre of mass, ^mc=o ='2tnb='2,ma. Thus, writing M lor the total mass, we find px=MH,py=Mv,ax\Apz—Mtv (2). In other words, the linear momenta of the body are not affected by any rotations and equal those of the whole mass at G. 296 ANALYTICAL MECHANICS [arts. 308-309 308. Kinetic Energy of Rigid Body in General Motion. — Let us now form the expression for the kinetic energy of a rigid body whose centre of mass has velocities u, v, and w, the body at the same time having rotations (u^, Wy, and Wz about axes parallel to the fixed axes OXYZ. Then, using (i) above, we have for the kinetic energy = ^'2!n{{u + ii>yC—iOzby + {v-\-'^za — l+iC>^l-\ . . . / ) the letters A, B, C denoting the instantaneous moments of inertia and .D, E, F the corresponding products of inertia. It is easily seen that the first term on the right expresses the kinetic energy of the whole mass as if concentrated at G, while the other six terms give the kinetic energy of rotation about the axes through the centre of mass. (See equation (2) of article 299.) But it should be noted here that, since the axes GABC are not fixed in the body, but only the point G, the values of the moments and products of inertia in (3) may be continually changing. 309. Angular Momenta of Rigid Body in Greneral Motion. — Using again Fig. 125 and equations (1) in article 307, we may form the expressions for the angular momenta hx, hy, and hz about the axes of X, y, and z. Thus, for the first we have hx=^m{zy—yz) = '2m{(w+(axi — Wya)(v+l') — (v+Oza — id.j:c){z+c)}. And, on performing the multiplications, omitting as before the vanish- ing terms, we find /ix=(7vy — vzyS,m-\-uix'Sm(d'+c') — ay'Zmai — (D^'2mca . (4). Hence, on substiruting the usual symbols for the mass, moments, and products of inertia, and writing the other momenta by symmetry, we obtain Ax=M(wy—vz)-\-Ax-\-Bi!)y—Du)z\ (s). hz = M(vx — uy) — Ev>x— Dwy +C'i>z ] These again show that each of the angular momenta splits into two parts, one representing about the given axis the angular momentum of the whole mass concentrated at its centre G, and the other expres- sing the angular momentum of the actual body about a parallel axis through G. (See equations (3) of article 290.) Examples — LVIII. 1. In Euler's equations show that the increases of angular velocities are partly due to centrifugal reactions, and determine the resultant axis of these reactions. 2. In the case of a rigid body rotating under torques as expressed by Euler's equations, prove that the moment of the centrifugal reactions ART. 310] SOLID RIGID KINETICS 297 about any principal axis is the continued product of the difference of moments of inertia about the other two axes and the two corresponding angular velocities. 3. Show that the linear momentum of a rigid body moving generally in solid space is that of the whole mass as if at its centre. 4. Prove that the kinetic energy of a rigid body in general motion in three dimensions splits into two terms, one expressing that of the whole mass at its centre of mass and the other that of the rotation of the actual body about an axis through the centre of mass as though it were at rest. 5. Show that the angular momentum of a rigid body in any motion is the sum of that of the whole mass at its centre and that of the actual body about a parallel axis through that centre. 310. Equations of Motion for Axes of Fixed Directions.— We have just shown that, as regards linear and angular momenta and ' kinetic energy, a rigid body moving in any way is replaceable by the total mass at its centre of mass together with the actual body in its actual motions relative to the centre of mass. It is clear without a formal proof that the same independence will hold for rates of change of angular momenta since it holds for the momenta themselves. Hence, adhering to our co-ordinate axes GABC moving parallel to themselves with their origin G always at the centre of mass of the body, we can now write general equations of motion of the rigid body. For each product, mass of a particle into its acceleration, corresponds to a force, and each change of linear momentum to an impulse. Thus, if the force components are represented by X, V, and Z acting at the point X, y, z, we may write, by the results of articles 307-3°9i the foUowmg equations : — 1Xdt=Mdu,^Ydt=Mdv,^Zdt=Mdw . . ■ (6), y:{Xdx-\-Ydy+Zdz)^dT . ... (7). y.{Zy- Yz)dt=dh^,1{Xz--Zx)dt=dhy,'2.{Yx-Xy)dt=dh (8), ^X—Mji,^Y=Mv,'^Z==Mw (9)> ^Zy- Yz) = h,, ^{Xz-Zx)=ky, 3( Yx-Xy)=h, . . • (10), 1{Zb - Yc) = jfi^'^x - Fi^v - E<^z) '2.{Xc-Za)=^j(^-Fw^+B<,>y-n<^z) (11). ^{Ya-Xh) = jj, - ^«x- - Dmy + Ceo,) where a, b, and c are the co-ordinates with respect to the axes GABC of the point of application of the force components X, J, Z If we perform the differentiations mdicated by h^ m (10), using tne values from (5) of article 3°9, a^d remembering that a, b, and c are variables as shown in (i) of article 307, we obtain ^(Zy-Yz)^M{wy-vz)-^Ai>x-Fi>y-E^z \ (12). The other two equations for hy and 4 could then be written by syin- metry The cumbrousness of these expressions (due to the fact that 298 ANALYTICAL MECHANICS [art. 311 A, B, C, D, E, F, are varying) shows that it is usually desirable to adopt moving axes so that the above coefficients become constants. Thus, since we have already seen that the translation of the centre of mass and the rotation about any axis through it are independent, we may use for the latter Euler's equations and for the former the equations for a particle. Another system is given in the next article. 311, Hayward's EcLuations. — We have just been discussing the equations of motion with respect to axes with origin fixed in the moving body and at its centre of mass but with directions always parallel to those of the axes fixed in space. Let us now refer the general motion of a rigid body to rectangular axes whose origin is fixed in space but whose directions vary by rotat- . ing at angular velocities B^, 0„_, and 0^ about their own instantaneous positions, being, in fact, the system usually referred to as ' moving axes.' We have already found (articles 290 and 294) the form assumed in this case by the expressions for the angular mo- menta and their rates of change. And it might be inferred that precisely similar expressions would hold for linear momenta. It is, however, desirable to give here an independent proof of this. At the instant in question let the moving axes OABC coincide with the fixed axes OXYZ as shown in Fig. 126, and consider the point P of co-ordinates .r, y, z with respect to OABC and velocity components u, v, w in the direc- tions of the moving axes, but reckoned with respect to the fixed origin O, Then the velocity u is the algebraic sum of (i) that of P relative to Q; (2) that of Q relative to R; and (3) that of R relative to O. Thus, writing these values in from the figure and by symmetry the similar expressions for v and w, we find Fig. 126. Linear Momenta Moving Axes. (!)■ --i-x9,+ye,j If we now wish to pass from the velocities to the corresponding linear accelerations a, b, and c, parallel to OABC but with respect to the fixed origin O, let OR, RQ, and QP in Fig. 126 now represent to some scale the velocities w, v, and u. Then, to the same scale, the component velocities of P will represent the accelerations required. Hence we have, by substitution in (i), ART. 312] SOLID RIGID KINETICS 299 b=v—ivQ-^-\-uG^\ . . (2). c=w—uO^-\-vB^] If now we suppose a particle of mass m to have these accelerations, we may sum the products of mass into acceleration over the whole body and equate to the corresponding sum of force components X, V, and Z. Thus Or, if the sums of the force components are denoted by U, V, and Wand the linear momenta by/,, /j, and /a, we have Let us now quote here the equations (1) from the beginning of article 294,_expressing the relations between the torques Z, M, and N about moving axes and the corresponding angular momenta h^, h^, and h^. M=L-h,e,+h,eA (4). It is now seen that these two sets of relations between forces and linear momenta in (3) and torques and angular momenta in (4) are precisely alike in form. Further, by the independence we have already seen to exist between the translations of the centre of mass of a rigid body and its rotations about any axis through the centre of mass, equations (4) still hold when the body has a motion of transla- tion provided the origin of these axes now coincides ivith the moving centre of mass of the body. The equations of motion in the forms shown by (3) and (4) were first given by R. B. Hayward, F.R.S. (see Camb. Phil. Trans., Part I. vol. X., 1856). Thus, in the preceding and present article, general equations have been obtained for the translations and rotations possible and referred in each article to different systems of moving co-ordinate axes. It is not, however, by any means necessary or desirable to use either system in its entirety for any given problem. On the contrary, we may often with advantage treat the translation of the centre of mass as though the whole body were concentrated there and the rotations by equa- tions involving moving axes (sa> Euler's) as though the centre of mass were at rest. This is illustrated in the next article. 312. Motions of a Quoit. — Following the treatment of Tait, let us now consider the motions possible to a quoit when thrown. Let the moment of inertia about OA, the axis of figure, be A, and B and C those about any two rectangular axes OB and OC in the plane of the ring. Then obviously A>B==C, and by Euler's equations we find Ai = o, or Wj is constant . . . (i), Bb)2—{-B—A)o)sx'a.^o (3)- 300 ANALYTICAL MECHANICS [art. 312 For brevity write A-B = n, a. constant (4). Then (2) and (3) become respectively bi^+n(o, = o. . . (S), and 0)3 — «u2=o. ... (6). From (5) we obtain (o^=—o}^ln (7), and this differentiated and substituted in (6) gives a)2 + «^(i)2 = o (8). But the solution of this may be written v>^ = (a^cos{nt+„ and <^ depend upon the initial conditions of the throw. This value of i)^ put in (7) gives for the other angular velocity (1)3 = (o„ sin («/+ <^) (ii)- Thus, the resultant of m^ and (Os about the perpendicular axes in the plane of the quoit is an angular velocity (o„ about an axis OD in the plane of BOC and making at time t the angle {nt+ (f) with OB, as shown in Fig. 127. Hence, compounding this angu- lar velocity (o, with Wj, we find the magnitude m and axis 01 of the in- stantaneous angular velocity to be given by ui^=a>go° ; for minimum i? we have Q + Pcos a=o, the minimum /? being x'/"^ - Q^. 5. Give a graphical construction for the composition of three or more forces, and write its analytical equivalent. 315. Conditions of Equilibritun. — If we have several forces acting on a particle, it is obvious that the condition of equilibrium is that their resultant vanishes. But the resultant is represented by the line which closes the polygon, whose sides represent the forces as if placed end to end. Hence for equilibrium this closing side must vanish, or the polygon of forces must be itself closed. We may also give the same ideas in analytical form. Thus, let forces Ft, F^ . . . he applied to the particles in the plane of xy, their components being Zj, Y^, X,, Fj, etc. Then, for equilibrium we obviously have the conditions SZ=o, 2F=o (i). ART. 316] STATICS OF PARTICLES 30^ since the resultant R is given by and accordingly only vanishes when (i) is satisfied. n,ri-?>i^' ^°°"°^*^ Plane.-Let us now consider the equilibrium of a particle on a rough inclined plane H"'"ur'um ot a under the action of a force at an angle with the plane. Let the weight of the particle be W, the coeiScient of friction between it and the plane At=tan/3, the inclination of the plane to the horizontal u,, the angle be- tween the force P and tlie plane 6. Take the axis of x down the plane, the origin at the particle, call the normal reaction of the plane N, and suppo.se the equilibrium to be such that the particle is just on the point of moving down the plane. Then the frictional reaction on the plane is upwards and equals its limiting value ^i-N. The forces mentioned are shown in Fig. 129. Then on reference to the figure and using the conditions of equili- brium as expressed in (i), we have ^X—W%\x\a.-iLN—P<:.o%6=Q> (2), 2y=- ff'cosa+iV-/'sin6l=o (3), two equations from which to find P and N. Eliminating N between the two equations we find ,sin (a— /S) Fig. 129. Equilibrium on Rough Incline. P— W"- cos(e-/3) Then, substituting for P in (2) and (3), we have N: '^wL COSa + sin ^sin(a— /8) (4). (5). cos (6-/3) Equation (4) shows that for a minimum supporting value of P we must have 5— /?=o, when /'min.= fFsin(a-/3) (6). If the particle is to be on the point of moving up the plane instead of down, we must reverse the sign of /x in (2). We then obtain the other Hmiting case in which the force, now distinguished by a dash, is sin(a+^ '^cos(e+/3) ^^''■ And now for the minimum value of P', which is on the point of dragging the particle up, we find 0-\-/i=o, i.e. P' is pushing at angle (i below OX or pulling at the angle /8 above ft-N in the figure, the corre- sponding value of the force being /"min.= rFsin(a-f/3) (8). 3o6 ANALYTICAL MECHANICS [arts. 317-318 317. The Wedge.— For P' horizontal put ^=a in equation (7), and we obtain F= fFtan (a+/S) (9)- Further, on putting two such planes of inclination a base to base, we pass to a wedge of angle 20=7 say, the total horizontal force being W -P -P' w Fig. 130. The Wedge. 2P' 130. W=7P Q say. We have then the state of things represented in Fig. The relation between Q and W is obviously represented by F=Wta.n{a-\-P), ^ (2=2fFtan0+/3U ('°)- Of course, if the friction is negligible, this reduces to Q=2W\.a.nyl2 (n), or if the angle is small Q— Wy nearly . . (12). If the angle is not small, and the resistances are taken R, R normally to the inclined forces of the wedge, we have instead of (i i) (2=2.ffsin7/2 (13), as may be seen at once from the triangle of forces, which is a figure like the wedge shown but turned through a right angle, the base representing Q and the sides the equal resistances R and R. 318. The Multiplied Cord or Tackle. — The com- bination of several plies of a cord, rope, or chain with pulley blocks is called a tackle or purchase. The pulleys are of great practical importance in certain cases as they lessen friction at the places where the directions of the cord change considerably, but the esseiitial efficacy of the contrivance lies in the re- diiplication of the cord or its disposition in repeated plies or parts side by side, and the pulleys are in some cases omitted with advantage. Many arrange- ments and combinations may be made and are in use. It wiH suffice here to consider two illustrative types. Take first the system represented in Fig. 131. In Fig. 131. Tackle of Single Multi- plied Cord. ART. 318] STATICS OF PARTICLES 307 Lenrff 51^''"'' °"'^,°u' ""'^ '^ "^^'^ ^"d P"l'eys are provided at the bends, the tension will be practically the same throughout if we reeard friction and stiffness of the cord as negligible. Hence the re si tf nee W, which may be balanced by the applied force P at the free e^d of the hauling part of the rope, is given by the expression W=nP , , where « is the number of plies supporting the /.ze-./or running block Thus in the figure fF=6^. If we write W for the load puF by the upper or fixed block on the beam supporting it, obviously W'= W+F=n'P / X ^iseltel '^Th^S'' °\ P'^'' ^^"u^'"S from the fixed block, in this case seven. The TV is, of course, the total re- sistance which the plies of the cord balance, and includes that of the running block itself. Thus if this block has the weight w, only an additional load W—w could be supported by the force P. Consider now a system with several cords and several running or movable pulleys. This is shown in Fig. 132, from which it is seen that the first movable pulley is supported by two plies of the cord to which the force P is applied. Hence the cord from it has a tension 2F. But this cord has two plies which support the next movable pulley, which' can accordingly exert on its cord a tension 4P. Finally, as shown, two plies of this rope support the weight W, which is accordingly 8/". Thus, as we take for the number of separate cords and movable pulleys I, 2, 3, . . . «, we see that the ratios, weights which can be supported to the force F applied, are given by 2^ 2', 2', . . . 2". Hence the general relation for this system of plies and pulleys W=P.^- (3), where n is the number of separate cords and of movable pulleys. If here, again, W is the weight put on the beam by the upper or fixed block, we have W'^W+F=F{2- + i) (4), which is the relation required if the system is inverted. Of course, if the weights of the pulleys themselves are comparable with the tensions in the plies they must be taken into consideration. Thus if w-^ is the weight of the pulley supported by the first cord of tension F, the tension of the second cord is 2F—W1. If the pulley supported by this second cord has weight w^, the tension of the third cord is 2{2F—Wi) — Wi, and double this value is available to support the weight w^ of the third pulley and the load hung on it. QP Fig. 132. Tackle of Several Moltu'lied Cords. 3o8 ANALYTICAL MECHANICS [arts. 319-320 Examples — LXI. 1. Express analytically and graphically the conditions for the equilibrium of a particle under a number of coplanar forces. 2. Find the force required to support a body on a rough inchne and plot force against inclination, thus confirming the analytic result that the minimum result is W sm (a - /3), where JV is the weight of the body, a the inclination of the plane, and tan j3 the coefficient of friction. 3. Obtain the ratio of resistances to force in the case of a wedge (i) when the resistances are in opposite directions along the same line, and (ii) when they are each normal to the respective faces of the wedge. ' 4. Explain why a wedge when driven into wood does not slip out again. Give a numerical instance, and work it out to support your explana- tion. 5. Sketch two arrangements of ' tackle ' or ' purchase,' and find in each case the relation of load to applied force, allowances being made for the weights of the pulleys. 319. Work for given Force and Displacement. — Consider the work done on a particle when it has a displacement OD under the action of a force P inclined at an angle 4" with OD as shown in Fig. 133. From J' let fall PQ perpendicular to OD produced, and from D the per- pendicular DE on OP. Then, if OP represents to scale the force I', OQ will represent to scale the component force Q along OD, the actual displacement. Also OE represents the component dis- placement in the direction of the force P. Now, by a definition in article 212, work is the product force into dis- placement in the direction of the force. Hence we have three forms for the work ^in question, viz. fr=/'.OE=(2.0D=/'.ODcos<^ (i). These are easily seen to be identical, for OE = OD cos 4> and Q=Pcos 4>- If the component displacement in the direction of either force be denoted by the corresponding small letter, we may state the above results still more compactly thus : — W=£p=Qq=Pq cos 4> (2). 320. Eesultant Work is Algebraic Sum of Components. — Let now a particle have a displacement OS while under the action of forces, say A, B, and C, whose resultant is Ji. Estimate the works of each of these forces as in the last article. S L M N Then it may be readily seen fr,„ „ t, that thp wf,ri r.( ^u^ ^ u\ ^'°- '34- Resultant Work is Sum that the work of the resultant of Components. lorce Ii is the algebraic sum of the works of the component forces. Thus, let the forces A, B, and C be Fig. 133. Work for given Force and Displacement. B A__- — ■ — ^ ^-^.^ T| -V-^ c Af. — 'n <2( ^^=ra^'' ■ Mr -T ART. 321] STATICS OF PARTICLES 309 represented to scale, end to end, by OA, AB, and BC, their resultant by OC = J?, and the displacement by OS along OT. Let fall from A B, and C the perpendiculars AL, BM, and CN on OT, and denote by a, /3, y, and <^ the angles which the forces A, B, C, and Ji make with OT. Then, by the geometry of the figure, we have OL+LM + MN = ON, or ^cosa+^cos^+Ccos7=^cos^. Multiply throughout by 0S = ^ say, giving ^(j cos a)+^(j cos /S)+ C(^ cos 7) = .ff(j cos (^). But the quantities in brackets are the component displacements in the directions of the various forces. Hence, denoting these by the corre- sponding small letters, we may write Aa+M+Cc=J?r . . . (3). Or, in a general case, where the forces are F^, P„, etc., and the com- ponent displacements /i,/2, etc., we have F,p,+F,p, + ... = ^Pp=.Rr .... (4), for what holds for these forces is obviously valid for any number. It may be noted that the forces ^, ^, C or /'ijT'j, etc., are not restricted to a plane but may be distributed in any way in solid space. Further, for these results to hold each displacement must be small enough for the forces to be practically constant in magnitude and direction throughout it. 321. Virtual Work is Zero for Equilibrium. — Suppose a single free particle at the point O is in equilibrium under the action of given forces Pi, P^, etc., and let it be imagined to have an infinitesimal displacement ^j, whose components or resolved portions in the directions of the forces are respectively dpi, dp^, etc. Then ds, dpi, dp^, etc., are called virtual displacements of the particle in the given directions ; also, the products Pidpi, P^dp^, etc., are called the virtual works of the corresponding forces. (Sometimes the phrases virtual velocities and virtual moments are used instead of the above.) If, for an instant, we imagined the forces to have a resultant R, along whose direction the virtual displacement was dr, we should have from (4) Pidpi+PJp,^-... = ^Pdp=Pdr . . . . is). But, since for equilibrium R must be zero, it follows that the total virtual tvork is then zero for any virtual displacement. Or, in symbols, 2/V/=o (6). Conversely, if the total virtual work vanishes for any virtual displace- ment, the particle is in equilibrium. This dual statement expresses what is called the Principle^ of Virtual Work as applied to a single free particle. The principle also applies to a particle on a curve or surface (as we shall presently see), and even to rigid bodies, as we shall note in a later chapter. Thus, for a particle in equilibrium on a smooth curve or a smooth 3IO ANALYTICAL MECHANICS [art. 321 surface, there would be the reaction Q of the curve or surface in addition to the other impressed forces Pj, i^, etc. Hence by (6) we should have for any small displacement l.Pdp-\-Qdq^o (7). If, however, we take ds along the curve or in the surface, dq vanishes, because Q and ds are at right angles, and (7) accordingly reduces to (6). Conversely, consider the particle constrained to move alonga curve, the virtual work for a tangential displacement being zero. Then, the resolved part in that direction of the resultant of the impressed forces is zero also. Hence the particle is in equilibrium. Also, for a particle constrained to move on a smooth surface, if the virtual works for any two displacements, not in the same straight line, are each zero, the particle is in equilibrium. Because in that case obviously the resultant force had no component in either of those directions, and accordingly vanished. If the curve or surface were rough, the frictional forces would need taking into account. And usually the principle of virtual work, now under discussion, is not then of any service, it being as hard to find those frictional forces as to solve the whole problem by some other method. A more general statement of the principle of virtual work is the following, quoted from Todhunter's Statics : — ' If any system of particles is in equilibrium, and we conceive a dis- placement of all the particles which is consistent with the conditions to which they are subject, the sum of the virtual moments {i.e. works) of all the forces is zero, whatever be the displacement. And conversely, if this relation hold for all the virtual displacements, the system is in equilibrium.' It is easily seen that this form applies to all systems of multiplied cords or tackles, and gives the results already obtained. Examples — LXII. 1. When the displacement of a point is inclined to a force acting upon it, as in the case of a canal boat towed by a horse, find several equivalent expressions for the work done. 2. Prove that the resultant work is the algebraic sum of the component works in the case of a particle displaced under the action of various forces. 3. State clearly the principle of Virtual Work, and establish it for a particle. 4. ' Enunciate the principle of Virtual Work. If a material system is in equilibrium under the action of gravity and smooth constraints, under what condition will it rest in all positions into which it can be placed without violating the constraints ? Apply this to the following case : — A is a fixed smooth pulley ; a light flexible cord passing over the pulley has a freely hanging mass of weight P attached to one end and a ring of weight W to the other ; if this ring is constrained to move along a smooth fixed wire in the form of a certain conic having A for focus, the system will rest in all positions.' (LoND. B.Sc, Pass, Applied Math., 1905, i. 6.) ARTS. 322-323] STATICS OF PARTICLES 311 322. Eqiiilibrium of Bent Cord.-If between two points P and P on a stretched cord there are no forces applied (except the tensions), It IS almost obvious that this portion of the cord would remain straight and of constant tension throughout. This has already been assumed in dealing with the multiplied cords and tackles. If, on the other hand, isolated forces occur at various points P, F, etc., of a stretched cord, it is evident that the cord will change its direction or the magnitude of its tension at each of these points, re- maining straight and of constant tension between them. The equili- brium at each of these points may obviously be dealt with by the poly- gon of forces. If, however, a fine inextensible flexible cord is subject to forces Q per unit length between the points P and P', it is obvious that the tensions J" and 7" at the points must differ in magnitude or direction or both. And the precise mode of these differences requires examina- tion. Thus, referring to Fig. 135, and resolving along and perpendicular to OK, we have r'cosi;^— r-fPP'.2(2cos^ = o\ , . r'sini/'-fPP'.S(2sin<^ = oJ ^^'' the sign of summation being introduced before Q to cover the cases where various forces occur, of which Q is the type and all are stated per unit length. The above equations are written in the form which applies when the angle ^ between the tangents at P and P' is finite. Now let P' and P approach and the angle become the infinitesimal one (f^^, then cos di^ is unity and sin d'^ becomes d'^. Also T —T is dT and PP' is the infinitesimal ds. Thus equations (i) transform into Fig. 135. Equilibrium of Bent Cord. -v- +2(2 cos <^=o dtj/ '2,Q sin <^_ (2). These are the general differential equations for a cord in equilibrium under coplanar forces. They may be expressed in words thus : — Space rate of change of tension equals tangential component of forces. Curvature equals quotient of normal component of forces divided by tension. It should be remembered that the forces are estimated /^r unit length. We easily see from (2) that when Q vanishes dTjds—o, and d^lds=o, or p—, that is, the cord is of constant tension and straight. 323. Cords wrapped on Curves. — Let us now suppose the forces Q 3,3 ANALYTICAL MECHANICS [ART. 323 to be due to the reaction of a curved surface on which the stretched cord is wrapped. At first suppose the surface to be perfectly smooth so that the angle <^ becomes 37r/2, and the force Q may now be called N since it is normal to the surface. Thus equations (2) reduce to the following :— dT , ^•/'_ N _ I / X -;-=oana -5-=-^; = — \3A ds ds T p These show that round a smooth curve the tension is constant and that the normal reaction N is Tjp, where p is the radius of curva- ture. Consider now a rough surface, and denote the normal reaction by iVand the tangential frictional force by F as shown in Fig. 136. Suppose the cord to be on the point of slipping in the direction of J", the tension applied at P'. Then F=ij,N numerically and its Fig. 136. Cord on Rough Curve. <^=^^ thg ^ for JV being still 317/2 as for the smooth curve. Hence for the rough surface equations (2) give -=/.iV^and- = - = - (4). Eliminating ds between the two equations of (4) we find dTld^=,.T (5). Thus, if the tension is T„ at A and T at P, the angle' between these directions being a, we have by integration of (5) between these limits /:?-r* Whence \ogeT-\ogeT,=p.a.=\oge{TjT,). Thus, on raising e to the powers indicated by the two sides of this equation, we have or T=r,ei^'^) ^' It is noteworthy that the ratio of the tensions is independent of the form of the curved surface, but depends only on the coefficient of friction and the Ma/ angle involved. Also, owing to the exponential form, the ratio of the tensions increases very quickly with the angle a. Thus a turn and a half of a cord round a cylinder will give a threefold tension if the coefficient of friction is slightly over one-tenth (say o'ii66). And three turns would give a ninefold tension. This explains the possibility of hauling against a great resistance T with a rope which has a couple of turns round a capstan, the slack of the rope being taken off with a small but ART. 324] STATICS OF PARTICLES 313 indispensable tension TJ,. Thus, with a rope round a wooden post, a single complete turn may yield a twenty-fold tension. 324. Uniform Cord under Gravity. — Let us now consider the equilibrium of a uniform, inextensible, and flexible cord under gravity. If attached at one point only, it is clear that its position of equilibrium is that of the vertical line below that point, the tension at each section being the total weight of the portion of cord below that level. If, however, the cord is attached at two point's not in the same vertical line, the cord will hang in a curve which needs investigating. There are various problems to be solved and different ways of attacking 7"+d T them. We will first apply the methods A., already used and then pass to others //'%(' ^ which are desirable for the sake of ^'//' \ the further light they throw upon the a'-X'' '^ subject. p,' yg \ Let the cord have weight w per '^TJ *" 14 unit length, and consider an element ^ X^ tj of length 'P'P'=ds inclined at an angle V^ i/- with the horizontal, the tensions at pm, j^;. Uniform Cord under its ends being J" and T+dT, as shown Gravity. in Fig. 137. Then we see that, in the notation of article 322, the angle <^ between POK and the weight Ow is 37r/2-i/'; thus cos <^= -sin !/■ and sin = -cos ^i-. Hence equa- tions (2) of that article reduce to dT . , \ ^l/'_WCOSl/' and Ts^^^~ ] These may be put into the rather more useful form dT—wds%vo.-^ = wdy I _ . . (g), and T=pw cos i' f ' ' ' where y and p are the vertical ordinate and radius of curvature respec- tively at the point P to which T applies. From the second of these it is clear that for ^-=0, i.e. the bottom point of the cord, we have for the tension there at>, the letter c denoting the radius of curvature at this point. But from the first of (8) we see that the tension increases proportionally to the vertical height; hence if we chose the origin of y so as to make the constant of integration zero, we have l=^y\ (9), and ^.-«'^/ , u Ki showing that the horizontal axis of x must be taken a depth c below heTowest point of the cord at which ^=0. The tension at any point P is then tL weight of cord whose length equals the verUcal ordinate of P reckoned from this axis of x. The curve assumed by a uniform cord under gravity is called the (ommon catenary. 314 ANALYTICAL MECHANICS [art. 325 f^ Y G t To C / / > X r W Elementary Relations FOR Catenary. 325. Elementary Relations for Catenary. — We began treating the problem of a cord by taking a finite length, which was then indefinitely reduced. Another method is to pass from the finite portion to the infinitesimal ele- ment by differentiation, which will be convenient now in de- riving further elementary rela- tions from the uniform cord at rest under gravity. Thus, referring to Fig. 138, let us begin by considering the equilibrium of the finite portion h.V-=s of the' cord from its lowest point A, under the forces To at A horizontally to the left, y at P upwards to the right at inclination ^t, and its weight JF' vertically downwards through its centre of mass G. The origin is taken at the dis- tance c vertically below A, so that J'o=<= iocosh(.*-/io) ; find, by using the tables, the tension and the vertical component of tension at the point for which x=2c, feet.' (Lond. B.Sc, Pass, Applied Math., 1907, i. 9.) 2. ' If a string hangs in equilibrium in the form of a catenary, show that the tension at any point is equal to the weight of a length of the string equal to the height of the point above the directrix of the catenary. A kite is flown with 600 feet of string from the hand to the kite, and a spring balance held in the hand shows a pull equal to the weight of 100 feet of the string inclined at an angle of 30° to the horizon ; find the vertical height of the kite above the hand.' (LoND. B.Sc, Pass, Applied Math., 1908, i. 9.) 3i8 ANALYTICAL MECHANICS [arts. 329-330 3. 'Show that the length of an endless chain which will hang over a circular pullfey of radius a so as to be in contact with two-thirds of the circumfer- ence of the pulley is J 3_ +4^1.' lloge(2H-V3) 3 J' (LOND. B.Sc, Pass, Applied Math, 1905, i. 9.) 4. ' Prove that the difference between the tensions at two points of a uniform chain hanging under gravity is equal to the weight of a portion of the chain whose length is the vertical distances between the points. 'A uniform chain is stretched between a point on the ground and a point 100 feet above the ground. The tension at the ground is the weight of 3100 feet of chain, and the inclination there of the tangent to the horizon is cos "'(16/31). ' Find the length of the chain to the nearest inch.' (LoND. B.Sc, Pass, Applied Math., 1910, \. 9.) 5. Draw carefully a catenary, indicating its vertex, axis, and directrix. Obtain graphically the length of the curve from the vertex to any point P, also the radius of curvature of the catenary at P. 6. ' A uniform chain 100 feet long is to be suspended from two points in the same horizontal line with such a span that the tension at the ends is to be three times that at the middle. ' Find (with the help of tables) the required span to the nearest inch.' (LoND. B.Sc, Pass, Applied Math., 1905, \. 9.) 329. Approximations to the Catenary. — When it; is so sma/lthat its fourth power is negligible, we may see from equations (16) or (21), on expanding the exponentials, that the approximate value of the ordinate is 2 { 2C ) Thus y—c+x^l2c, \ , , or x^ — 2c(y — c) nea.r\y j ' ■ ■ ■ ■ ■ (.23;, which is the parabola' of latus rectum 2 f with vertex at {o,c) and axis vertically upwards, which approximates to the catenary at its lower portion. On the other hand, when x is large the negative exponential e'"'" vanishes in comparison with e^'". Thus, for the higher parts of the catenary the curve approximates to the positive exponential only. Or, y= — e'^'" nearly (24). There is an intermediate portion of the catenary for which neither (23) nor {24) furnishes a close approximation. 330. Sag and Excess Length in Telegraph Wires.— We may now find the sag and extra length due to it in the case of wires tightly ART. 33i] STATICS OF PARTICLES 319 stretched between points on the same horizontal level and at a distance / apart, as in the case of telegraph or telephone wires. Clearly we are here concerned with a relatively small portion at the bottom of a very large catenary. Hence we may commence with the parabola approximation of (23). Thus putting x=pl2 and denoting by q the sag or dip {y — c) in the middle, we have ?=/78^ (25)- For the unknown c on the right side we may use the approximate value /= Tj^iii, the tension T at the ends in terms of the weight w per unit length being supposed known. We accordingly write as the working approximation ?=/78^ (26)- For the excess length we may conveniently calculate ^ from its true expression as in (17) and (20). Thus, expanding the exponentials, c I X X- x' 2 V^c^T^^ec^ X x" a," \ -'+-c—2?+6?----r Hence, omitting powers of x/c beyond the cube, we find s=x+x'/6c\ Accordingly, for X-PI2, we have for the excess length in a single span , , 2S—p=p'/24C^ nearly ... {-V- For the £ on the right we may here use c^y=t nearly, or the closer approximation , , c=y-q=i-q (28)> a being approximately known from (26). , ■„ . . j k The case in question may be more conveniently illustrated by a numerical example than by a diagram. Thus, let the poles be 88 yards apart and the tension of the order, weight of a mile of wire. Then, by equation (26), the sag is given by ^=8878x1760 = 11/20 yard=i9-8 inches. Again, by (27) and c=t nearly, the excess length due to this sag may be written 25-/ = 88724/= =11/1 200 yard = 33 inch! By (28) we note that c is 17 59-45 yards, hence the directrix of the catenary is practically a mile below the wires. 331 Parameter of Oatenary.-Where the foregoing approximations are not sufficiently close for the purpose - -ew^J ^^y be necessa^ to determine the parameter c of the catenary from the data and then 320 ANALYTICAL MECHANICS [art. 332 use the true curve. This may be done graphically as follows :— Suppose the X and j- for a given point are known ; then, using (17) or (20), j-/i;=sinhic/r. Transform this by writing ,F=y'K\i.\o%A — - \ . . . (13), But if the filaments be brought together so that f=o, the attraction becomes infinite in any of the cases (iia) to (13). On the other hand, it is seen that the attraction vanishes with a or b, as should be the case. The problem may be solved directly by a double integral. Thus taking the origin at A, let the elements be dr at a distance r in AL and ds at a distance s from A in BM. Then we have for the attraction ■ between the filaments whence ^= ^A/. loge^^^^^ (14), as before. ExAMPLiiS — LXVI. 1. Explain carefully the expressions : — attraction between two particles and field of one particle. What word may be substituted for field in tlie second phrase? How must the axes be chosen for the motions of particles of comparable masses under their mutual attraction ? 2. Show that the attraction of a uniform straight filament for an axial particle is proportional to the geometric mean of the distances of the particle from the ends of the filament. 3. Find the attraction of a straight filament of density X per unit length for a particle not on the direction of the filament. 4. ' Find the attraction of a thin uniform circular arc whose mass per unit length is p, upon a particle of unit mass situated at the centre of the circle. 'Write down conditions necessary for the equilibrium of a particle situated at the centre of a circular wire whose mass per unit length at any point is a given function of the angular distance of the point from a chosen diameter.' (LoND. B.Sc, Pass, Applied Math., 1906, 11. 8.) 5. ' Find the resultant intensity of attraction at the point whose co-ordinates referred to rectangular axes are {a, a) due to attracting matter of line density W2 placed along the co-ordinate axes from x=a \ox=^a, and from^ = a to^ = 2a.' (LoND. B.Sc, Pass, Applied Math., 1909, n. 6.) 6. ' Prove that the attraction of a uniform rod AB on any particle P bisects the angle APB! (LoND. B.Sc, Pass, Applied Math., 1910, 11. 9.) 340. Disc and Particle on Axis. — Consider now the attraction of a thin circular disc of radius a and surface density o- on a particle of unit mass on the axis of the disc and distant z from its centre. Take in the disc a ring element of radii rand r-\-dr2& shown in Fig. 147, ARTS. 341-342] ATTRACTIONS AND POTENTIAL 329 hfcenTrf 'VSn fh" '^'™'".' ?•'" 1^' ""§ subtending the angle M at his and the n^fl. ^'"^ °^-*^'' ^'"'"^"^ '^ '^^(^'^^)' •'« ™a^^ - times rnis, and the attraction on unit mass at P equals Ja^-\-b^-\-a)\ C21') or ^=-25ry(rA(BO-AB + AO)/ ^ '' If we introduce the masses M and m for the disc and filament respectively, we have for the general case 7^=-^5^(BC-AB+CA) (22). a V 343. Cylinder and Coaxial Particle. — Referring again to equation (15) of article 340, we see that on writing pdz instead of tr, we may interpret it as referring to the attraction between a slice of thickness dz in a cylinder of radius a and density p and a coaxial particle of unit mass. And, if the cylinder has length b and the particle is distant c from the centre of the near base, we have then to integrate between the limits c and b-\-c to sum the effects of all the slices composing the cylinder. Hence and F=-27ryp r'idz ^^LJ\ = -2-Kyp{b- ^{t+cy+a'+ Jr+ar) . . . .(23). Thus, if c vanishes, the particle being at the centre of an end of the cylinder, we have F=:-2Tryp{b- Jb^+^' + a) (24). Again, if b, the length of the cylinder, is infinite, (23) reduces to F—-2Tryp{J7+a'-c) (25). Of course, if we wish, the mass M of the cylinder may be intro- duced. The expression in the general case is then ^=--Jf{b-J'(b+?f+^'+J?+7-) . . . .{25a). 344. Thin Spherical Shell and Particle.— Let the spherical shell have radius a and surface density o-, the particle P being of unit mas s ART. 345] ATTRACTIONS AND POTENTIAL 331 Fig. u SrHERicAi. Shell and Particle. and distant c from O, the centre of the shell. Take any point Q on the shell, the plane OPQ being that shown in Fig. 148. And at Q let there be a small mass dm of the shell. Its attraction on unit mass at P is then ydmjr" where r denotes the distance QP. But it is obvious from symmetry that the resultant at- traction between the shell and the particle will be along PO. We may accordingly consider the component in this direction simply. It is derived from the former expression by the factor cos QPO, and this is MP/QP or {c—a cos d)lr, where 6 is the angle POQ. Take now as our element of the shell a ring with OP as axis and passing through Q. Then its radius is a sin 6 and its width adO. Hence, combining the above considerations, we have as the axial attraction of this ring on the unit mass at P ,„_ {2TTa5\n6){ad9)(r c— a cos 6 , ■. 7 ^2 ■ r But we have here, so far, the two variables 6 and r. We accord- ingly eliminate 6 by aid of the relation which holds for the triangle OPQ, viz. r'^ = c'^-\-a'' — 2ca cos 6. From this we have, by transposition, c—acoso= .... 2C and may derive, by differentiation, 2rdr= 2ca sin Q dO, so that a sin 6 dO=rdr/c Now, substituting (2) and (3) in (i), we obtain (2)> (3)- (4). 345 Case I. Particle Outside.— BeioxQ integrating this equation we must decide where the particle is to be. We take first the case where it is outside the shell, as shown in Fig. 148. We then find for the attraction required Thus where M is the mass of the shell, which accordingly has for an ex- (5), 332 ANALYTICAL MECHANICS [art. 346 ternal particle the same attraction as though its whole mass were concentrated at its centre O. Case II. Particle Inside. — For the particle inside, the inferior limit of integration changes sign. We thus have --^n-'^"')- (6). That is, Jio attraction is experienced by a particle within a uniform spherical shell of material attracting according to the law of the inverse square of the distance. Case III. Particle on Shell. — We consider finally the case of a particle on the outer surface of the shell. Here c=a, and the limits of integration are o and 20 ; but if we attempt to integrate (4) with these inserted, a difficulty is experienced with respect to the term (^ — a^)lr^. It appears at first sight to be zero. But this is not the case where r is very small near the lower limit of integration. Perhaps the simplest method is to write first for c the sum a-\-h, and then finally make h vanish. The limits of integration are accordingly h and 2a-\-h, and we have from (4) ^^_7yJ-(,+(^+^r-^')^, 2a/4T-"+ r J;. -y\Ka^c>b. We must accordingly divide the thin shells into two sets : — (i) those whose radii exceed c, which clearly contribute nothing to the field, and (ii) those whose radii do not exceed c, and which may be replaced by their masses at the centre (equation (5), article 345). Thus we find, for the. field in question, P=-^^y{c'-b')pl^. . . (11). Case IV. Field in Solid Sphere. — On putting i5=o and f=rin (11), we pass to the field at any point in the substance of a solid sphere, which is accordingly ^ , , > F=-^-Kypr (12), or Fccr. To make this applicable to the earth, supposed solid and homogeneous, on the surface of which at radius P the field is —g, we have 4 D / \ g^^-^yp^ (13)- Hence (12) may be written F=-grlR . . . . (14). Of course, g, r, and P must be expressed in terms of the same system of units to give F correctly in any system. 348. Graphical Representation of Fields. — It is often very desir- able to represent fields graphically, their intensities being plotted as ordinates and the distances of the points as abscissae. In other words, the distance of the point P from the centre of shell or sphere is plotted as x and the corresponding value of the field as y. Figs. 150 and 151 show the fields in this way for the thin spherical shell and solid sphere. Since the field is to the left when the distance is to the right, the ordinates are always negative when the abscissae are positive. The student may with advantage draw other fields for himself, preferably ARTS. 349-350] ATTRACTIONS AND POTENTIAL 335 using squared paper on which to plot to scale. Fig. 150 gives an interesting academic example of a discontinuity in the field. This discontinuity would not, however, occur in nature, for the shell to have Fig. 150. Field of Thin Shell. Fig. 151. Field of Solid Sphere. a sensible value of the surface density o- would really need an appreci- able thickness also. And in this thickness the change of field 4Trycr would occur without actual discontinuity. 349. Field in Eccentric Spherical Cavity.— Consider now a homo- geneous sphere of radius a and density p with an eccentric spherical cavity of radius b, and let it be required to find the field at any point in the cavity. Let the centre of the sphere be S, that of the cavity C, and consider the field at P (Fig. 152). Regard the eccentric shell as built up of a complete sphere of radius a and density p, and a second sphere of radius b and density minus p, their centres being at S and C respectively. Then the field at P has com- ponents in the directions PS and CP due to these component spheres and expressed by -i^rypSP and -|ry( Hence these components may be represented to scale by PS and CP and their resultant F by PF parallel and equal to CS. Thus we have for the field at P F= -i^ypSC . . ...... (is). And since this value is independent of the co-ordinates of P, it applies to any point in the cavity whose field is consequently uniform throughout. 350. Newtonian Constant of Gravitation. — For detailed accounts of the determination of the Newtonian constant of gravitation the reader is referred to physical text-books. It may, however, be remarked here that many of the methods consist essentially in finding the attrac- tion between given bodies at a specified distance apart in terms of the weight of a standard body or unit. For this determination Professor C. V. Boys used a special torsion balance. Professor J. H. Poynting Fig. 152. Field in Eccen- tric Spherical Cavity. -P)CP. 336 ANALYTICAL MECHANICS [art. 350 used a special form of the ordinary beam balance, as did also Drs. F. Richarz and Krigar-Menzel. Suppose by any method it is found that the force of attraction between masses M and m a distance r apart equals the weight of a mass n. Then we have F=yMmlr'—ng (i). But, if the earth be taken as a nearly homogeneous sphere of radius E and mean density A, we have (by equation (13) of article 347) g^^^yMi .' (2). Thus, by division, we find the density in terms of measurable quantities : — A=^^ (3). The value of the Newtonian constant may be found from (i) or (2) provided i is known from experiments with the pendulum or by other means. Thus from (i) or (2) y=Mm ^'^- The values found for y and A by the investigators referred to are given in Table xiii. Table XIII. Gravitation Constant and Earth's Density. Investigator. Newtonian Constant. 7 in c.g.s. units. Earth's Mean Density. A in gm./cc. C. V. Boys . . . J. H. Poynting . Richarz and Krigar-Menzel 6-6576 X io"8 6-6984 X IQ-' 6-685 X 10-8 5-527 5'4934 5'5o5 Examples— LXVIII. 1. Find the fields inside and outside a solid sphere of homogeneous material, and plot them as graphs. 2. Show that the field is uniform in an eccentric spherical cavity of a homogeneous sphere. 3. ' Prove that the attraction exerted at any external point by a homogeneous solid sphere of gravitating matter is the same as that which would be exerted by a particle of equal mass situated at the centre. Find the law of attraction in the interior of the sphere. ' Show that, if a smooth straight tunnel could be cut between any two places on the earth's surface, it would be traversed by a particle, starting from rest and influenced only by gravity, in about 42^ minutes.' (LoND. B.Sc, Pass, Applied Math., 1907, 11. 8.) 4. ' Prove that the gravitational force within a uniform solid sphere varies directly as the distance from the centre. ART. 35i] ATTRACTIONS AND POTENTIAL 337 ' A uniform solid sphere of mass M is cut in two by a diametral plane ; show that the resultant attraction between the halves is 16 where a is the radius of the sphere and y the constant of gravitation.' (LoND. B.Sc, Pass, Applied Math., 1908, 11. 9.) 5. ' If the volume density of a sphere varies uniformly from zero at the centre to 2p at the surface, show that the intensity of its attraction at a point of its surface is half as great again as that of a sphere of the same radius whose volume density is uniformly equal to p.' (LoND. B.Sc, Pass, Applied Math., 1909, 11. 7.) 351. Solid Angles, Lines and Tubes of Force.— Let ABCDE (Fig. 153) be any closed figure, curved or polygonal, plane or not, and let lines be drawn through each point of the closed figure to any other point O. Tiien these lines form a cone, and the closed figure is said to subtend at O a solid or conical angle, which we will denote by u. To define it quantitatively, describe about O as centre a sphere of radius r intersecting the conical surface at abcde, and let the area of the spherical surface thus enclosed be .S. Then '-«• ., ^ .^ If we take an infinite number of such lines side by side, say those which practically constitute the conical surface between ABCDE and abcde (Fig 153), we then have what is usually called a tube offeree. Here again tube of the field would be a better phrase Now each line of the field has at each point the direction of the attraction experienced by a small particle placed there, and the wall of the tube is composed of these lines. Hence no attraction crosses the wall of such a tube from inside to outside or vice versa. Y 338 ANALYTICAL MECHANICS [art. 351 Consider now the field ^ at a distance r from a particle of mass m. It is given by F^ymlr' ........ (2), which accordingly expresses the field at the spherical surface in Fig. 153. Let us now take the product, field into area, for a portion of the spherical surface, say that contained by the cone. Then this product is FS; or, substituting from (i) and (2) the values of ^andi^, we have FS=ymu) . . . . . . (3), which is a constant for the given cone of solid angle w with the mass m at its vertex O, independent of the distance r from O. Thus, if we describe another sphere about O with radius r, the field there being F' and surface enclosed by the cone S', we should have the product of the same numerical value as before. But if we were taking the pro- duct, field into area, positive when entering this frustum of a cone, we should have to give the two products for its bases opposite signs. And we have seen before that no field crosses the sides of a tube of force. Accordingly for the whole product entering such a tube of force we have FS-]-F'S=o (4). If we drew the bases of the frustum oblique, making an angle Q say with the spherical surface, then an equation like (4) would still hold if we took the component of the field normal to the new surface. For obviously this component would be the true field multiplied by cos 6, and the new surface would be the corresponding portion of spherical surface divided by cos 6. Hence (4) holds for any surfaces provided the .^'s always mean components normal to those surfaces. It is easy to give to these relations a graphical aspect. For, since the product FS remains constant for the cross section of a tube of force, it is evident that we may represent the intensity of a field at any section by the number per unit area of continuous lines drawn through it. Thus the product FS will then be the total 7iumber of such lines in the tube, which by their continuity remains throughout a constant quantity, as required by equation (3). Hence, these lines may pass continuously from a given mass without annihilation or creation in space devoid of ponderable matter. Another and perhaps better plan of graphic representation is to imagine the tube of force divided into a number of smaller tubes lying side by side, such that the cross section of any such small tube shall have its value of FS equal to unity. Such tubes may be called unit tubes. If, for our conical tube of force round a mass m, we take the whole external space, we must write for where/ refers to the field due to m only. Thus, when we sum m to obtain the total internal mass tJ/,/ becomes J^, and we have / MS=4wyM (8), as was required to be shown. From this result it follows that for any tube of force, straight or curved, containing no masses, the sum or normal Qux/FdS is zero for the whole surface of the tube. But since there is no J^ through the sides of the tube, we have for the ends J^S+F'S'=o, as was obtained simply in (4). We may further notice that, if a mass m" is situated at an ordinary point on the surface itself (say at K, Fig. 153), then the whole surface 5 subtends at this point the solid angle 2 7r. Hence for such a particle the total value of the integral would be 2-n-ym". Thus for a mass M" consisting of particles on the surface at ordinary points, we should have generally / FdS=2TryM" ... (9). This applies only to particles at points like K which may be touched by single tangent planes. At internal or external conoidal cusps resembling dimples or spikes, as L and N, the above value would obviously be modified, and might range anywhere between the extreme values of ^i^ym and zero. Calling the left sides of (6), (7), (8), and (9) the Gauss integral, we may summarise the results of this article thus : — ■ The Gauss integral for a closed surface .S receives no contribution from masses outside it, and has the value 47ry times the total mass in- side it. Particles on the surface itself contribute to the integral the value 27ry times their mass if at points of the surface each of which may be touched by a single tangent plane, but a larger or smaller value if situated at the vertex of an internal or external conoidal cusp. We sometimes for brevity use the \.&xmflux instead of Gauss' integral. 353. Potential Introduced. — As we have already seen, the gravi- tational field in any region may be specified by stating its magnitude and direction at a sufficient number of points. It may also be re- presented graphically by unit lines or unit tubes, the closeness of these lines or tubes per unit area being directly as the intensity of the field. Further, if at a certain point P the field has a magnitude F and a given direction, then the component of the field in any other direction inclined Q to the lines of the field at P is clearly given by F cos Q, since the field is a vector quantity. Hence, the field being specified by its ART. 354] ATTRACTIONS AND POTENTIAL 341 magnitude and direction, we can find the components of the field in any other directions whatever. All the foregoing may be illustrated by a possible method of indi- cating the slopes of a hilly district on a map. Thus we might state, for each of a number of points on the map, the direction of the steepest slope there and its gradient (analogous to the direction of the field and its magnitude). And from these supposed details on the map we could then deduce the gradient in any other direction at any one of these points. But, instead of carrying out on maps this idea (like our specification of a field), an entirely different method is generally taken to present quantitatively the features of the hills and valleys. Thus, we usually have the height above sea-level stated for a number of points, and often series of lines are shown, each point of any one line denoting the same height above sea-level. These are called contour lines, and are given on many ordnance and tourist maps, sometimes with the intervening bands in special colours. Then, the heights being known, the gradient in any direction is inferred as the rate of change of height per unit horizontal distance in that direction. Here, then, there is substituted a scalar quantity, height above sea-level, in place of the vector quantity, steepest gradient. A distinct advantage in simplicity results from this method of describing or specifying the region, and very little dis- advantage, if any, is entailed in consequence. For each point requires only the magnitude of the scalar quantity to be given, whereas both magnitude and direction are needed for a vector quantity. And, as to the readiness of using the map, it is practically as easy to read the gradients in any direction from the heights and their distances apart as it would be from the steepest gradients if shown. A like advantage is often obtained in the theory of gravitational fields, if instead of specifying them as heretofore we state for each point a scalar quantity called the potential, whose rate of change in any direction gives the corresponding field component. The maximum rate of change of the potential at any given point gives by its direction and magnitude the lines and intensity of the field at that point. This potential, from which the gravitational field is derived, is often called the Newtonian potential to distinguish it from the electric or magnetic potentials from which the corresponding fields are in like manner derived. 354. Potential and Field.— Let V denote the potential at any point P, and let X, Y, and Z be the field components parallel to the axes of co-ordinates. Then, by the relations between them, we have xJJ^, y=^-Z, zJ-^ . . (:). dx dy dz In any direction defined by the co-ordinate s (see Fig. 155) the field F\& given by fJ-^ (^). ds . The resultant jR of X, V, and Z gives the field, or the direction and 342 ANALYTICAL MECHANICS [ART. 355 magnitude of maximum space rate of change of V. Thus, we have the field given by i?'=X^+F''+Z' (3) stating its magnitude, and the following set of direction cosines XIR, YIR,ZIR (4) defining its direction. If we denote by r the co-ordinate along the direction of ^, we may write -f «■ If we suppose the direction cosines of the co-ordinate s to be (l,m,n) Fig. 155. Potential and Fields. /, m, n referred to the axes of x, y, z, and the angle it makes with that of r to be Q, we have by projecting R upon s F=R cos e . (6). But also by projecting upon s the X, Y, Z components of R, we obtain F=Xl+Ym+Z?i (7). Hence the right sides of (6) and (7) should agree. This is easily seen to be the case if we write the usual expression for the cosine of ROF. Thus, from (4), . X.^Y , Z cose=^/-f-;«-F~« (8), so that the two expressions for F in (6) and (7) are seen to be identical. 355. Potential and Work. — We have hitherto written for the field F, as derived from the potential dVjds simply, without regard as to whether the positive or negative sign was appropriate. This question must now be examined. We see from (i) that change of potential equals field multiplied by distance. If we place a particle in the field it experiences a foice. So the passage of a particle from one potential to another involves the product force multiplied by distance. Accord- ingly such a passage invohes work. Thus, if we have the signs rightly arranged, a body must have work done on it to pass to a higher potential, and then possesses the exact equivaktit of that work in its potential energy. To secure this convention of signs, the potential and potential energy must increase when the motion is against the field. ART. 355] ATTRACTIONS AND POTENTIAL 343 That is, when the force exerted on the body is -Rm, where R is the field and m the mass of the body moved in it. Thus, we must have for the increment of work done on the body d\V=mdV={- Rm)dr\ or R^-dVjdr f (9)- And, if the field R is due to a mass M at distance r, we have R=-yMlr' (10). Thus, by (lo) in (9), ,^ Mdr dW , ^ dV=:y 2-= ■ (11), and we may well note here that the equality of the first and last quanti- ties in (11) forms the basis of the definition of potential in elementary electrical theory. Now let the zero value of V be arbitrarily chosen to correspond to the infinite distance from the mass JSI to which it is due. Then we may integrate (11) between the limits as follows: — Whence V=—yM\r (12). Thus we see by (9) and (12) that, with these conventions for mutually attracting material, the field is the rate of decrease of the potential, and that the potential is itself negative for all finite values of the radius r. But as mutual repulsions also have to be dealt with in electro- statics and magnetism, it is customary in mechanics to calculate the numerical values of field and potential without inserting the negative signs in either of the relations shown in (9) and (12). Hence, in the following working of the potentials and fields for various cases, we shall usually regard them as expressed by V=yMlr . . (13), and F=dVl'ds (14), the appropriate signs being afterwards inserted if required. But, in plotting curves for V and F, the correct relations are observed in this work (see Figs. 160 and 161 of article 361). It may be noted here that, if the velocities of two attracting bodies are derived from their loss of potential energy in approaching each other, then the velocities of each body with respect to the centre of mass of the two bodies will be so determined. The kinetic energy of either body as found from its velocity relative to the other will not necessarily equal the loss of potential energy of the system by the mutual approach of its parts. The forms of (13) and (14) suggest some other expressions for potential and change of potential. Thus, from (13), we see that if the potential at a point P is due to various masses nh, nu_, etc., distant r„ ^2, etc., from P, then we may write V^y^- (^S). ' r 344 ANALYTICAL MECHANICS [art. 356 For obviously, each element of the potential contributed by any quotient mlr, being a scalar, they all add arithmetically. If, however, the mass is continuously distributed with a density p at the distance r from P, then V=-^{{{-^dxdydz . . . (16) gives the resulting potential, the integration extending over the whole volume occupied by the matter in question. One of the forms, (15) and (16), is the basis of any calculation of the potential produced by specified masses. Another view of potential difference, viz. the /tne integral of the field, is suggested by (14). Thus, transforming it, integrating between s' and J- along the path taken in the field, and calling the two potentials V and V, we find V- V'= CdV= i'pds (17). In this equation F is the field component along the curve s. 356. Equipotential Surfaces. — We have already seen that the re- sultant field Ji has the direction and magnitude of the maximum rate of change (or gradient) of the potential. And further, that the value Fo[ the field inclined at an angle 6 to ^ is i? cos 6. Hence at right angles to H the field vanishes. Thus, since the potential difference is the line integral of the field, there can be no potential difference if we move in any direction at right angles to the resultant field at the place. By moving in such a manner we should describe an equipotential surface. And it is evident that a region in which gravitational or other attractions are experienced may be graphically represented by plotting the field lines or tubes and the sections of the equipotential surfaces. Thus, for a single particle, the field lines or lines of force are radial and equally distributed, the tubes of force are of equal conical form with a common vertex at the particle, and the equi- potential surfaces are concentric spheres. They may be drawn to represent a common difference of potential by use of equation (13). Examples — LX IX. 1. Explain what you mean by lines and tubes of force, and how a field may be specified in terms of them. 2. Defining potential a.s that function of the co-ordinates whose space rate of change is the field, prove that the difference of potential at two points is the quotient W/tn, where IV is the work done on a particle of mass m as it passes from one of those points to the other. 3. ' Define the terms tu6e of force and equipotential surface. Prove that in a field 6f force due to gravitating matter systems of tubes of force and of equipotential surfaces can be drawn, such that at any point of free space the force varies inversely as the cross section of the tubes, and also inversely as the distance between consecutive surfaces. ' Will these properties be affected if the attraction does not follow the Newtonian law ? ' (LoND. B.Sc, Pass, Applied Math., 1905, 11. 7.) ART. 357] ATTRACTIONS AND POTENTIAL 345 ^' ' ^f^l'"-";'^'' '[ *f ^^"^ of attraction be that of the inverse square, the sur- face integral of normal attraction over any closed surface = 4^7 (mass inside), where y is the constant of gravitation.' , , T-, c • • , (LoND. B.Sc, Pass, Applied Math., 1907, 11. 10.) 5. Define equipotential surfaces, lines of force, tube of force. Show that if ^ be the force intensity along any tube of force, *- ■ : ■ • <■"• But, by Gauss' theorem, we already know that this flux may be expressed in terms of the matter in the volume enclosed by the surface ^S (see equation (8) of article 352). 346 ANALYTICAL MECHANICS [art. 358 First, suppose that this infinitesimal parallelepiped contains none of the gross matter to which this potential is due. The above sum is accordingly zero. This leads to Laplace^s Theorem, which is then expressed by ^+^^-+^=° (^°^- Second, suppose that there is gross matter of density p within the parallelepiped and to which the potential is wholly or partly due. Its mass is therefore pdxdydz, and, by Gauss' theorem, equation (19), then gives d^V . tPV d'V , , - + ^ = 4'r7P (21), dx'^'dy'' which expresses Foisson's Theorem. In accordance with the convention of signs here adopted, the right side of (21) is positive. In using the theorem for an electric field, we have i^= —dVjds, and the right side 2 is negative. 358. Axial Potential of Disc. — Consider now the potential V at a point P distant z along the axis from the centre of a disc of radius a and surface density o-. Take, as the infinitesimal element, a ring of radius r and width dr passing through Q as shown in Fig. 157. Now, for each part of the ring, as that near Q and subtending the angle dO at O, the con- tribution to the potential at P is simply (mass-f-PQ), and all such parts add arithmetically. Hence for the whole ring element we may write the mathematical expression for this and integrate between the appropriate limits. rv fa j-dr Fig. 157. Axial Potential of Disc. Thus I d V:^ 2TVy„), and (fla^a) shall lie on one straight line. So, unless this is fulfilled, (6) must be satisfied, in which case, by (3), G also vanishes ; or U^V=G^o (7). Thus, if the algebraic sums of the moments of a system of coplanar forces on a rigid body vanish with respect to three points in the plane not in a straight line, that system is in equilibrium. Examples — LXXI. 1. Sketch a set of three parallel forces in equilibrium, indicating the relation between their magnitudes and positions and showing also what con- struction gives the resultant of any pair of the forces. 2. Explain fully what you understand by a couple, and show how it may be represented by a line, and thus facilitate the composition of coplanar couples. 3. Reduce any set of coplanar forces to a single force and a single couple. What may the system reduce to in certain cases ? 4. How are the resultant force and couple of question 3 affected by a change of axes .'' 5. Enunciate one of Poinsot's analogies between statics and kinematics. 6. Give two forms of the conditions for equilibrium of a rigid body in circumstances allowing motions in a given plane, and establish one of these forms. 7. Show that any system of coplanar forces may be reduced to two forces if not to one, and indicate the conditions of each case. 358 ANALYTICAL MECHANICS [art. 369 8. ' Show that a system offerees acting in one plane on a rigid body can be reduced to a single force or to a couple. ' ABC is an equilateral triangle and P is the foot of the perpendicular from C on AB. Find in magnitude and line of action the resultant of the following forces : — ' 10 acting from ^4 to ^, 8 from B to C, 12 from A to C, and 6 from Cto P.' (LOND. B.Sc, Pass, Applied Math., 1905, i. i.) 9. ' Show that any number of couples acting simultaneously on a rigid body can be replaced by a single couple. ' (The proof may be confined to the case of couples all acting in the same plane.) "■ABC is a triangular plate ; A', B', C are respectively the middle points of BC, CA, AB ; forces represented in magnitude and sense by k.AB, k.BC, k.CA, \.B'A', XA'C, and X.C^' keep the plate in equilibrium ; what is the relation between i and X ? ' (LoND. B.Sc, Pass, Applied Math., 1907, i. i.) 10. ' Show that a system of coplanar forces may be reduced to two com- ponents along chosen axes at right angles to one another together with a couple in the plane of the axes. ' The algebraical sums of the moments of a system of coplanar forces about points whose co-ordinates are (i, o), (o, 2), and (2, 3) referred to rectangular axes are G,, G2, Gg respectively ; find the tangent of the angle the direction of the resultant forces makes with the axis of jr.' (LoND. B.Sc, Pass, Applied Math., 1908, i. i.) 11. 'Show that a system of forces in one plane is in equilibrium if its moment, about three points of the plane not lying in a straight line, is zero. 'The moments of a system of forces about two points A, B axe P and Q respectively. Construct a point in the line of action of their resultant.' (LoND. B.Sc, Pass, Applied Math., 1909, i. i.) 12. 'ABCD is a square lamina which is acted upon by forces of 5 units along BA, 3 units along BD, 7 units along DC, and by a couple of moment 8a in the sense ABCD, where 2a is the length of a side of the square. Find the equation of the line of action of the resultant of the system referred to AB, AD as axes of co-ordinates.' (LoND. B.Sc, Pass, Applied Math., 1910, l i.) 369. Determination of Centroids. — The conception of a centroid was introduced in article 25^/, was used a little in Chapter xiii., but must now be more fully dealt with and its positions calculated for some typical cases. As before stated, the term centroid may be applied to the centre of a number of points, to the centre of given lines, surfaces, or volumes, or to the centre of any scalar quantity distributed discretely or con- tinuously in space. We are here particularly concerned with the centroid in its use or application as the centre of mass or centre of inertia ; often referred to as the centre of gravity. But it should be noted that every distribution of mass has a single definite centre of mass fixed relatively to that distribution ; whereas, strictly speaking, it is the exception for bodies to have a true, single, fixed centre of gravity, when they are large enough with respect to the distance of the attracting body to make the attractions of their separate particles sensibly inclined. Bodies that have this exceptional property are called centrobaric. ART. 370] PLANE STATICS OF RIGID BODIES 359 Homogeneous spheres afford an example of this class, as we saw in the sixteenth chapter on attractions. Of course, any small body, commonly used in experiments on the earth, has its centre of gravity then practically fixed at its centre of mass. But the two conceptions are entirely distinct. See, e.g., equation (i) of article 336 and the remark following it. It is the centroid that we usually find in what follows, whether of particles supposed condensed at points, or material condensed into lines or surfaces, or material distributed in solid space. If we abstra.ct the idea of the associated matter, we have the centroid of the given points, lines, surfaces, or volumes. If we keep the notion of the matter in the foreground, we may then fitly speak of the centre of mass. If the body is small, the forces of gravity on its particles, though vectors instead of scalars, may be regarded as parallel, and hence the single invariable point found as the centroid may be taken as representing with sufficient accuracy the really movable point which is the true centre of gravity. The working rule for finding the centroid may be quoted here from article 251^. From this it may easily be seen that if the centroids of certain portions of the system are known, then, for the centroid of the whole, each such portion may be replaced by its magnitude at its centroid. Hence, we may sometimes easily find the centroid of the whole from those of its parts. To formally prove this, let the ^'s have the subscripts i and 2 to distinguish two parts of the system. Then, finding the centroid for the whole from its two parts, we have (2). or x—^ — , as found from the direct treatment of the whole system and given in equation (i). Before passing to the use of integration for centroids we notice a few simple cases which may be solved more quickly by other methods. 370. Elementary Examples of Centroids. TAree Equal Masses at the Corners of a Triangle.— Take the origin of co-ordinates at one corner and the axis ofy parallel to the opposite side. Then the co-ordinates of the masses may be written (o, o), (a, b), and {a, c), and each mass denoted by m. Hence by (i) we have - „ - b+c .v=|a,.r=— . Thus the centroid is tivo-thirds along any median from the vertex, which agrees with the result found by taking first the centroid of two masses and supposing the doubled mass to be concentrated there. Hence we see that the three medians intersect at one point. 36o ANALYTICAL MECHANICS [art. 370 Four Equal Masses at the Corners of a Tetrahedron. — Following the analogy of the previous case, we see that the centroid of the four masses must be at three-fourths from any vertex along the line to the centroid of the corresponding base. Another way to regard the matter is to take the centroid of one pair of masses at the middle of the edge joining them, then similarly the centroid of the other pair at the middle of the opposite edge. Thus the centroid required will be found by bisecting the line joining these two middle points of edges. But as the tetrahedron has six edges, three such lines may be drawn. Obviously therefore they bisect one another, which is another geometri- cal property derived from the conception of the centroid. Sides of Triangle. — Consider the sides of a triangle as though they were very thin uniform wires or filaments, and let it be required to find the centroid of this ideal triangular framework. Clearly we may replace each side by a mass proportional to its length and placed at the centre of the side. The co-ordinates of the centroid may then be readily found by the usual relations or by simple geometrical con- siderations. Take the latter method first. Hence, in the triangle ABC shown in Fig. 166, we first replace each side by a particle at its middle point and of equivalent mass, i.e. proportional to the length of that side. We thus obtain particles at D, E, and F of masses proportional to a, b, c, the sides of ABC, but these are proportional to the sides EF, FD, and DE. Hence, to find on DE say the centroid ^3 of the particles at D and E, we must divide it inversely as the masses of those particles, which is directly as the sides FD and , „ , FE. Hence, according to the well-known theorem, we must bisect the angle DFE in FG^,. Similarly to find g-, on EF, we must bisect the angle FDE by DQfi. Thus, the intersection G of these two bisectors is the centroid re- quired. In other words, it is the centre of the circle inscribed in DEF, the triangle whose corners are the middle points of the sides of the original triangle. Taking now the analytical method, let us denote the points A B and C by the co-ordinates (o, o), {h, k), and {h, k+a) in accordance with Fig. 166. Then we easily find for the co-ordinates x, y of the centroid G A /i X Fig. 166. Centroid of Triangular Frame. where ^sx—h{2a-\-b+c) "I '^sy=k{2a+b+c)-ira{a+b)] ' " 2S=a-\-b-\-c, the sum of the sides. (3). ARTS. 371-372] PLANE STATICS OF RIGID BODIES 361 371. Surface of a Triangle.— Take any triangle, as ABC in ^ig. 167, and draw two medians AD and BE intersecting at G. Then U IS the required centroid, for each median bisects all elements of the triangle parallel to the corresponding base, and therefore contains the centroid of the whole surface. By construction and similar triangles, we have CD^CE_ED_Gp_ CB CA~AB~AG~^' Whence AG=|ofAD (4). Volume of a Tetrahedron.— 'L&t ABCD in Fig. 168 represent the tetrahedron, and take in it the plane CDE through the edge CD and the middle point E of the opposite edge AB. Then by symmetry this plane must contain the centroid sought. Take in CE and DE the centroids F and G of the faces ABC and DBA; join CG and DF intersecting at H. Then since by symmetry CG passes through the centroids of all slices parallel to ABD, it contains the cen- troid of the tetrahedron. Similarly, so does DF. Accordingly their intersection H is the centroid re- quired. Join FG, then we have, by construction and similar triangles, the following equal ratios : — EF_EG^FG^FH^i EC~ED""CD~HD *' Centroid of a Triangle. Whence Fig, 168. Centroid of a Tetrahedron. DH=fDF (5)- Cone or Pyramid. — We thus see that for a tetrahedron, a cone, or any pyramid, if we draw the hne DF from the apex D to the centroid F of the base, then the centroid of the whole volume is H where DH = |DF. 372. Frustum of a Pyramid. — As shown in Fig. 169, let the vertex of the completed pyramid be O, A and B being the centroid of larger and smaller ends respectively. Let F and G be respectively the centroids of the whole completed pyramid and of the small pyramid needed for the com- pletion. Let Ok— a and 0B = ^ and OH=i where H is the required cen- troid of the frustum. Then the volumes of the large whole pyramid and small completing one are 362 ANAL YTICAL MECHANICS [art. 373 to each other as a' and ^' we may write OF.a' = OG./5' + OYi{a'-b'), or la^=lb*+z{a'-h'). Whence Thus by the relation (2) of article 369 ^=l^=OH V-/5' (6). Of course, this may be cancelled down somewhat or other expressions substituted if desired, but this has the advantage of compactness when reckoning from the apex of the completed pyramid as origin. Fig. 169. CENTK.01D OF A Frustum. Fig. 170. Centroid of Pierced Circle. 373. Difference of two Simple Figures.— As a further illustration of the principle just used, let us now determine the centroid of the difference of any two simple figures. Thus take the figure left when a circle is cut from a larger circle as shown in Fig. 170. We may write the equations of the circles x'^-\-y''=a'^ and {x—bY +y'=c', where a<^{l>-\-c). Then writing x for the abscissa of G, the centroid of the remaining surface of the pierced circle, we have irc^b+Tria' —c')x=o. -'' (7). Thus '-^ Of course, j'=o. ART. 374] PLANE STATICS OF RIGID BODIES 363 fr»i^^r°"'^^ ^^^ '^""^ "'^*^°'^ ""*y '^e applied to any figures formed irom ellipses, squares, triangles, or other combinations of simple figures. Examples— LXXII. 1. Distinguish bet\yeen centroid, centre of mass, and centre of gravity, giv- ing as Illustrations figures or bodies in which these three points are not all coincident. ^ 2. Calculate the position of the centre of mass of four equal masses at the of'tlT^'fi ^ tetrahedron, and from this establish a geometrical property 3- Establish the positions of the centroids of the surface of a triangle and the volume of a pyramid. 4. Obtain the centroid of a frustum of a pyramid in any form and reduce It to the form, distance from centroid of larger base of area A towards the centroid of the smaller base of area B is h_ _ A+2'JaM+->iB 4 A+ sTaB^B where h is the height of the frustum. Show also that the distance of the centroid from the smaller base is the above fraction altered by the transference of the coefficient 3 from the B to the A in the numerator. 5. A homogeneous cube has a pyramid cut off by a plane passing through the three corners of the cube adjacent to that original corner of the cube which forms the vertex of the pyramid. Show that the centroid of the remaining portion of the cube is on a diagonal and one-twentieth of its length from its centre. 6. 'A uniform square plate is cut in two along a straight line joining a corner to the middle point of a side. Prove that the mass centre of the larger portion coincides with that of four particles of masses 4, 6, 5, 3 situated at the corners of the original square.' (LoND. B.Sc, Pass, Applied Math., 1906, i. i.) 7. 'A triangular plate ABC of uniform thickness rests horizontally on three vertical props at A, B, and C ; show that the pressures on the props are equal.' (LoND. B.Sc, Pass, Applied Math., 1907, i. 2.) 8. ' A square board ABCD rests with its plane perpendicular to the plane of a smooth vertical wall, one corner A of the board being in contact with the wall, and another corner B tied by a string, equal in length to a side of the square, to a point in the wall. Draw carefully a diagram showing the position of equilibrium of the board, and show that the distances of the corners B, C, D from the wall are in the ratio i : 4 : 3.' (LoND. B.Sc, Pass, Applied Math., 1908, 1. 2.) 9. ' An equilateral triangular lamina ABC rests in equilibrium in a vertical plane with its sides AB, A C in contact with two smooth pegs in the same horizontal line at a distance a apart. Prove that if AD, the per- pendicular from ^_on BC, is not vertical it must make an angle 6 with the vertical where cos 6 = h '•J ^il^a and h is the length of AD.' (LoND. B.Sc, Pass, Applied Math., 1910, i. 2.) 374. Centroids by Integration. — We commence this section of the subject by the treatment of lines, passing afterwards to surfaces and then to volumes. Circular Arc, — Consider first a circular arc AB of radius a and subtending at the centre O of the circle the angle 2^. Take the axis of X through C, the middle of the arc, the origin being at O, as shown 364 ANALYTICAL MECHANICS [art. 375 in Fig. 171. Let the infinitesimal element PQ, subtending at O the angle dd, be defined by the co-ordinates *, j/ of P and the angle COP = 5. Then for P, x=acosd, and m is represented by FQ=add. Thus, applying the usual expression for the centroid G, we have x='2mxfZm, or i^—„if'^^ a ya f'^^jo isinB chord ,. / > x=a f cosOdd-v-al dd= — 3-^= radius, (i). J-p J-p ji axe That this may be put in the convenient form added in v^ords is Fig. 171. Centroid of Circular Arc. easily seen. It is also clear that G is rather nearer to C than to D, the intersection of the chord with OC. Obviously j'=o. 375. The Catenary. — Taking the directrix as the axis of x and the axis of the curve as that of j/, the catenary is expressed by y=ccosbL{xlc) and. s=cwa\\{xlc) (i). Thus ds=cosh{xlc)dx (2). Hence for a portion j the working rule gives for the abscissa x= I xds-h- 1 ds, Jo Jo sx=: j X cosh {xlc)dx=c\ xd{s\nh (x/c)} = [ex sinh (x/c) — r" cosh (x/c)]^ — .xcsinh {x/c)—c' cosh {xlc)+c^- sx=xs—(y+c' x^x-c{y-c)ls (3): or Hence and For the ordinate, the rule gives similarly ^y= \ yds=c I co%h^{xlc)dx ART. 376] PLANE STATICS OF RIGID BODIES 36s Sxic Thus 4 L2 - ys , ex 2 2 +2+«-=^'''y;c 2 =V-- or, y=\(y-Vcxls) (4). For any other curve the difficulty is only that of finding the expressions in terms of x for ds, the element of length. 376. Centroids of Surfaces. — Passing now to the centroids of surfaces, we begin with plane surfaces, taking first of all a Circular Sector. — Let the origin of co-ordinates be at the centre of the circle of radius a, the axis of x bisecting the sector of angle 2/3, so that j'=o as shown in Fig. 172. Take as element the concentric Fig. 172. Centroid of Sector. circular strip PQ of radius x and width dx. Then its centroid g is distant («sin/3)//3 from O and its area is 2^xdx. Thus by the ordinary working rule we have for the abscissa of the centroid of the sector __ I ^^^.Pj/J^i/x-^ I 2l3xdx, or Whence sin/8 /■« ,,./"", .T=— g^l xdx^ I xdx. P h Jo linyf Thus sin^ a' a' _2 sin^ „ chord ,. OG=l X radius. (■)• Another mode of arriving at the same result is to take as elements 366 ANALYTICAL MECHANICS [art. 377 triangles with their common vertices at O and their bases infinitesimal portions of the arc ACB. Then their centroids lie along a concentric arc between OA and OB and of radius two-thirds that of the circle. Hence the centroid of the sector is that of this circular arc, and so is (chord/arc) x f radius, agreeing with that, stated above. 377. Segment of a Circle. — Let us now find the centroid G of the segment ACBD subtend- ing an angle 2/S of the circle of radius a, as shown in Fig. 173. In the segment take the ele- ment PQ parallel to the base ADB, and let the axis of x bisect the segment in DC so that y—o. Let the abscissa of P and Q be x, the width of the strip dx, and denote by the angle XOP. Then we have x=acos 6, dx:= —a sin 6 dd, and PQ=2a sin 6, the limits of in- tegration for the segment being 6=/S and d=o. Hence, applying the working rule, we find -f = - 2a= r sin'6l cos 61//^-=- - 2a' r sin''6l//e =af'sm'dd{sm 6)^ f ^^^S^d9 Fig. 173. Centroid of Segment of A Circle. rsin'gT . re sin 2e ~\\ So, finally, we obtain as the abscissa of the centroid sin»|8 -h ^— sin^cos/8 = OG (^)- As an alternative method we may derive the above value by con- sidering the sector OACB as made up of the triangle OAB and the segment ADBC. Thus for the respective areas and abscissae of centroids we have Areas Sector. Abscissae OH = Sa sin/8 Triangle. a" sin § cos ^. OF=|acosi8. Segment. a\P-saiPcosP). OG=.x. ART. 378] PLANE STATICS OF RIGID BODIES 367 Thus, applying the relation for the area and abscissae of parts and the whole, we have "■''^^a ^-5^ = or sin /3 cos /8f « cos /3+a^{P — sin (3 cos fi)x. Whence |a° sin^(i-cos'/3) _ sin"/? wnence *" a«(^_sin^cos/3) -^""(/G-sin/JcosyS) ('''>' as found before in (2). As checks on the results of the present and preceding article we may note that for a semicircle, which may be regarded either as a sector or a segment, for which fi=ir/2, both (i) and (2) reduce to :J=4«/3jr. For y8=o or very small we easily find for the sector ;c=|a, as should be the case. But for the very small segment where obviously X approaches a in value, the right side of (2) becomes indeterminate of the form 0/0. So here, applying the method of the differential calculus, . we must repeatedly differentiate the numerator and denominator, after each such differentiation putting /3 = o to test if the quotient is then determinate. Thus, if the trigonometrical part of (2) is written ^(j8)~^-sini8cos^ /'"(/?)_ 6 cos"/?- 2 1 sin^j8cosj8 4>"'(/3)~ 4C0s^j8-4sin'^ we find So that /"'(°)_3 Then on inserting this value on the right side of (2) we find that for ^=0, x=a, as should be the case. Another way of evaluating the indeterminate form is, of course, available by expanding the functions in terms of /3. Thus we may write sin'j8 _ ^ 2 ^^ ^_sir^-^_^^^^_4|!...>) f/^= as found before. 3 378. Parallel Portion of any Plane Area. — Let us now derive general formulae for the centroid of a portion of a plane area cut off by the axis of x and two parallel ordinates, the fourth boundary being any curve expressed hy y—fix) say, as shown by AB6a in Fig. 174. Take in the area any infinitesimal element PQ?^ bounded by ordinates at x and x+dx. Then the area of this strip is ydx, and its centroid is, in the limit, given by x andji-Za, where j/ is the value of /P found from the equation of the curve AB. Hence, considering the area in question as made up of these strips, and applying the usual relation, 368 ANALYTICAL MECHANICS [ARTS. 379-380 we find for the co-ordinates of the centroid the following formulae, in which a and b are the limiting abscissae Oa and Ob respectively : — x= I xydx-T- I ydx (i), y-^li^/dx^l^dx (2). Of course, the values of y in terms of x must be inserted before the integrals can be evaluated. Fig. 174. Centroid of Portion of Plane Area. 379. General Sectorial Area. — Referring again to Fig. 1 74, let us now take the sectorial area AOB bounded by the same curve AB. It will now be convenient to use polar co-ordinates for part of the work. We shall accordingly regard AB as given by r=F{6), the limiting values XOA and XOB of 6 being respectively a and /3. We now take as infinitesimal element POQ, where OP=r, XOP = e, and FOQ=de. Thus the area of the element is ^r'dd, and the cartesian co-ordinates of its centroid are (|r cos 6, |r sin 6). Hence, applying the usual rule, we find the following formulae for centroid in cartesian co-ordinates : — = fd^cos e)yde-^ r^r'de Jb Jb (3)> y=r{lr sin e)^r''de^ Tirade (4). Here the values of the r's must be inserted in terms of 6 from the equation of the curve AB before the integrals can be evaluated. 380. Plane Areas by Double Integration. — Let us now considef a plane area bounded by any two ordinates x=a and x=b and any two curves )'=4>{x) and y—i'{x), as shown by ABCD in Fig. 175. This might be treated by single integrals as in article 378, but, to illus- trate another method sometimes preferable, we will here adopt double integration. We now divide the area into elementary strips parallel to both axes of co-ordinates. Thus the strip P/^Q is bounded by the abscissae x and x-\-dx. The strip R/-.fS is similarly bounded by the ordinates^ axi&y-\-dy. Hence the infinitesimal rectangle uv, common ART. 381] PLANE STATICS OF RIGID BODIES 369 to both strips, and which now forms an element of the area ABCD, has area dxdy and, in the limit, its centroid has co-ordinates x a.ndy. Thus, applying the usual rule and inserting the limits of each integra- tion, we have the following formulae :— ^— I I xdydx-i- j j dydx ■''' JifiM Jb J^ix) y= I j ydydx^ j I dydx Jb J^jiix) Jb Jm: J^PW (5), (6). Since the limits of ji' involve X, it should be noted that the integration of y must be taken first. After this integration is effected we have the expres- sions that might have been obtained in the single integral method. The advantage of the double integral lies in its power to deal with a lamina occupying the area in question and varying in surface density with distance from the co-or- dinate axes. Thus, if the density were a-xy, we should simply have a-xy introduced in each of the above four integrals in (s) and (6). Fig. 175. Centroid of Plane Area by Double Integration.

{d) and r=4'{d), as shown by ABCD in Fig. 176. We take as the infinitesimal element of the double integration the figure si bounded by the radii at angles and 9+dO and by the concentric circles of radii ;- and r-\-dr. Then the ele- ment has area rdOdr and, in the limit, its centroid has co- ordinates (r cos Q, r sin &). Thus, applying the usual rule and inserting the limits of in- tegration, we derive the foUow- FiG. 176. Plane Areas by Double ing formulae for the centroid Polar Integration. of the area : — 3A 370 Analytical MECHANICS [art. 382 /■I MS) /-a MB) x=\ \ r' COS OdrdO-^ rdrdB . . (7), n*(e) /-a /■*(«) ^ , ^ r-sinddrde^l / r^rf Fig. 177. Centroids of Surfaces of Revolution. y=i?'f/^- For the surface generated by revolution of the same curve AB about OY, the abscissae of the centroid are obviously ds , ("■ ds ^ and x—o. Case I. — For a, cylinder with its axis as the axis of .v, the_yis the con- stant radius and cancels out, and ds/dx=i, so that from (i) we have a-l> ~ 2 ^^''' as evidently should be the case. Case II. — For a sphere, let the equation be x^-\-y'-=c^. Then dyldx:= —xfy zxid yds^cdx, i.e. the surface of each ring element has the same area as that of the corresponding ring of the circumscribing cylinder. Thus here also, for a spherical zone or cap, we have, as for the cylinder, •*=^- (4)- Case III. — For a cone of semi-vertical angle u, and vertex at the origin we havej=ictana and ds=5Qca.dx. Hence (i) reduces to X— j x'dx^ I xdx—% ,_^, .... (5). Or, for the complete cone, (6). Fig. 178. Centroid of Conical Surface. 372 ANALYTICAL MECHANICS [arts. 383-384 383. Conical Surface by Projection.— Let us now illustrate the method of projection by applying it to find the centroid of a portion of the surface of a right cone on a circular base, as shown by ABC in Fig. 178. The angle between an ele- ment dS of the conical surface and dH., its projection on the base, is that between any.gen- erator and the base. Hence it is the complement of a if a is the semi-vertical angle of the cone. Thus dIildS^= sin o. And the axis of the cone being that of z, it follows from the orthogonal projection on the base that dSaad dH have the same co-ordinates in x and y, and accordingly the like equalities hold for the cen- troids of the conical surface 5 and its projection 11. Or, in other words, the projection of the centroid of any portion of the surface of a right circu- lar cone on a plane perpendicular to the axis is the centroid of the projection of that surface. For the z co-ordinate of the centroid of S we have, by application of the usual rule and the relations between S and IT, the following expressions : — __fzdS_fzdTl_ V ^~/ds~Jdn:~u where V is volume of the prism included between the conical surface S, its projection 11, and the lines of projection parallel to the axis. Where, as in the figure, 11 is the triangle AOB and 5 is the corre- sponding conical triangle ACB, it is easily seen that ^=iOC (8), which result holds also for the whole conical surface. Projections of spherical surfaces on a diametral plane or on the circumscribing cylinder are sometimes useful. 384. Centroid of a Solid of Revolution. — Suppose a curve AB to rotate about the axis of x, and let it be required to find the centroid of the volume thus generated. Take in the curve AB, Fig. 179, the point P of co-ordinates (x, y) and the adjacent point Q of abscissa X -f- dx. Then, as the curve revolves about OX, the element PQ will sweep out a volume ny^dx whose centroid, in the limit, has the abscissa x simply. Hence, by the rule, we find for the abscissa of the centroid (7). ART. 384] PLANE STA TICS OF RIGID BODIES x= I TTxy^dx^ I tti'Va- . . . Also by symmetry we obviously have jp=s=o '. . . . 373 ■ (2)- (3)- O b pf]' a. X Fig. 179. Centroid of Solid of Revolution. Thus, for 2l parallel slice of a sphere of radius c, we have j'*=r— a'°, and (i) becomes {a-iy-^{a'-d') For a hemisphere, a = c and b—o, and this reduces to x—\c (4). For a solid spherical sector of semi-vertical angle ^ we might use the above general method, but each integral in (i) would split into two. Thus from x^^o to c cos /S we should \\z.y&y^x tan ;8, while from .v=f cos /8 to ^=f we should have j'^/— a;^. Hence on this plan the work would be somewhat long. We may therefore with advantage adopt a device founded on the known centroids of the pyramid and spherical cap. For we may consider the spherical sector as composed of a number of pyramids with their common vertices meeting at the centre of the sphere and their bases making up the spherical surface of the sector. Then, since the centroid of each such pyramid is at \c for the centre of the sphere, the whole sector is replaceable, for our purpose, by a spherical cap of radius \c and semi-vertical angle /3 like that of the sector. But, by what has been found for a spherical zonal surface or cap, the centroid of this cap is distant from the centre by the arithmetic mean of the limiting distances of the zone or cap. These distances are respectively \c and \c cos ^. Hence for the abscissa of the centroid we have .f=| is zero, then the system will not begin to move in that mode of displace- ment. In this way all the possible displacements are examined, and if -v^^/ IS zero for each and every one, the given position is one of equilibnum.' 388. Concrete Example of Virtual Work.— The principle of virtual work was made by Lagrange the basis of his Mkanique Analytique, and he made a brilliant attempt to give a general proof of it. Routh, how- ever, states that 'no satisfactory method has yet been found by which the principle for a system of bodies can be deduced directly from the elementary axioms of statics.' The full treatment of the subject is accordingly regarded as beyond the scope of the present text-book, but the following simple example of a rigid bar resting on smooth inclines at its ends will serve to illustrate the principle, and afford an insight into its meaning and application. In Fig. 181 the bar touches smooth inclined surfaces at Aj and A^, its centre of mass being G. We ac- cordingly have forces Pi, Pj acting normally to the inclines at Ai and A3, also the weight Ps^=Mg say acting vertically downwards at G. Now, if any motion of the bar is supposed to occur with the restric- tion that it remains in contact with the surfaces, it is clear that, on ac- count of the smoothness, (^,=0 = dp^. If we take the axis of z ver- tically downwards we may denote dpi by dz. We accordingly have, by (r) of article 387, ^dz Fig Virtual Work for a Rigid Body. dW=Mgdz=Mgjds (3), where ds is an element of the path s described by G when the contacts Ai and As move on their surfaces. Hence by (2) the condition for equilibrium here becomes ds' (4). Or, in words, subject to contact between the rigid bar and its smooth supports, the centre of mass G of the bar can describe a certain surface, 6' say. Then, for equilibrium of the bar, the point G must occupy in the surface S a point at which the tangential plane to .S is horizontal. Many examples on equilibrium occurring presently and later may be dealt with by virtual work; but some are done quicker without it, especially if friction is present. Discretion must be exercised as to which method should be adopted for each case. The principle of virtual work is specially useful where there are pairs of equal and 378 ANALYTICAL MECHANICS [art. 389 opposite forces with the same virtual displacements as at the junction or contact of parts of the system or forces normal to all possible dis- placements, for these contribute nothing to dW, and can therefore be omitted. 389, Lever : Wheel and Axle. — Let us now examine those simple machines which include some rigid parts, taking first the lever which consists of a rigid bar, straight A C B _ or bent, movable about a fixed "a. k. b 1/^ ~ ^^'^ called the fulcrum. It is acted on by at least two forces, P and W say, besides that of the fulcrum. The parts of the lever between the fulcrum and the points of application of the other two forces are called the 'vv arms of the lever, which we Fig. 182. The Lever. may denote by a and b (Fig. 182). Let the force P act at an angle a with the arm Q,h.=a and W sX an angle ^ with the arm CB=(J. Then, for equilibrium, we have by moments Pa%ma=iWb%\n^ (i). By virtual work we should obtain for equilibrium Pdp+JVdw=o (2). But if the displacements are derived from a rotation dd of the lever, we see that dp = add sin a a.nddw=—ddO sin P (3). Hence by use of (3) equation (2) reduces to (i). Of course, for a=5r/2=/3, equation (i) reduces to Pa= Wb (4). Tke Wheel and Axle is only a modification of the lever in a form allowing of continuous motion through large angles, since the arms a and b of the lever are replaced by the radii of the wheel and axle respectively. Other modifications of the lever occur by placing B (Fig. 182) between C and A or by placing A between B and C. The ratio of W to P, called the mechanical advantage, for any such lever is easily seen to be always W _ a%\na , . 'P~bsinl3 ^^'' Or, in words, the magnitudes of the forces are inversely as the per- pendiculars from the fulcrum on their lines of action. The Differential Wheel and Axle winds the end on a thick part of the axle of radius b, while it unwinds it from a thin part of the axle of radius c, the loop of cord passing round a pulley supporting the weight. Hence Pa= W\{b—c) where a is the radius of the wheel. Weston's Pulley Blocks is a compact tackle on the above principle, in which a=b. ARTS. 390-391] PLANE STATICS OF RIGID BODIES 379 390. Efficiency of a Simple Machine. — In dealing with the lever we have supposed that the only resistance which opposes its motion is the force W, which it is used to overcome by exertion of the force P, which may be called the effort. In that case, as we have noticed, the work done against the resistance is exactly equal to that done by the effort. Indeed, it was this equality which, on the principle of virtual work, gave the relation between PFand F. But in any actual lever or other simple machine (notably in the case of screws) there is always some frictional resistance to be overcome in addition to the main resistance for which the machine is used. And, since the total work done against all resistances can but equal that done by the effort which drives the machine, the useful work done will now fall short of that total. The proper fraction expressing this ratio is called the efficiency. ThuSjin any actual displacement of the machine (whether lever or other machine), let the effort P, the resistance Q, and the frictional or other wasteful resistance R have displacements in their directions of dp, —dq, and —dr respectively. Then, by the principle of virtual work, and taking this actual displacement as the virtual one, we find dW=Pdp^Q{-dq)+R{-dr) = o,\ .. or Pdp=Qdq+Rdr f • ■ ■ ^''• Hence the efficiency rj is given by Qdq Rdr ,. ^=yi-'-pdp ^'^- Or, in words, the efficiency of a machine is the ratio of the useful work done by it to that done by the effort on it when the machine receives any small actual displacement. 391. The Screw. — A screw thread may be cut in a cylinder by a tool moving parallel to the axis with uniform speed while the cylinder rotates uniformly. If the edges of the tool are respectively parallel and perpendicular to the axis, the screw is said to be square-threaded ; if inclined, it is said to be V-threaded. For simplicity's sake we shall here confine attention to square-threaded screws. Consider such a screw mounted so as to be capable of rotation about its axis, while any motion of the screw parallel to the axis is prevented. Also, on this screw, suppose there is a nut (or piece with internal screw fitting on the other) capable of motion parallel to the axis while rotation of the nut is prevented. Let a torque act on the screw of magnitude G, which we may suppose due to a force P acting perpendicular to an arm a so that G=Pa. ^ , . Let the radius of the screw be b and its pitch, or advance of the thread axially per revolution, be q. Then if we develop the thread by unrolling it from the cylinder into a plane, we see that the angle between the thread and the base of the cylinder is given by a where tan a=ql2iTb. We may note here that although b and a are each variable quantities, differing from point to point along a radius, the pitch q is perfectly 38o ANALYTICAL MECHANICS [art. 391 Fig. 183. Screw developed into a Plane. definite. Hence if a and b are involved in any equation they must be interpreted as the mean values of angle and radius, the expression in- volving them being approximate only. From the symmetry of the screw it is obvious that we may develop the cylindrical thread into an incline of angle a, and treat the prob- lem as one of two dimensions. The horizontal force Pajb acting at the radius of the screw thus pro- duces the vertical force Q on the nut, as shown in Fig. 183. Let the normal re- action between the screw and nut threads be R and the coeffici- ent of friction between them be ju.=tan /S, then the tangential reaction may be anything up to ±/t^. If we suppose the torque acting on the screw is on the point of prevailing, then the frictional force acts upwards on the portion of screw thread AB, as shown in the figure. Thus, for equilibrium, we have from the diagram by resolving hori- zontally and vertically, Pajb = fij? cosa-|-i¥sina, and Q=Jicosa—iJi.P sina. Whence Pa— Qb tan (a +P) (i), where ft is the angle of friction. Hence the ratio of Q to P, usually called the mechanical ad- vantage, is P b tan(a-f-/3) ^"^i- But, since it is the essential property of a screw to produce or over- come an axial resistance, the effort being a torque about that axis, it would seem preferable to quote as the mechanical advantage the ratio of Q to G. We then have G 3tan(a-|-;8) ■ ■ • ^3;- The efficiency of the screw pair {i.e. nut and screw) is QbB tan a/7'a^. Qb tan o tana (4), tan (a-f /3) ■ • • • when P is on the point of prevailing. If Q is on the point of prevail- ing the sign of ft must be reversed. We then find »;=tan {a.-ft)lx.xa a. ART. 391] PLANE STATICS OF RIGID BODIES 381 If the friction is negligible /3 vanishes, the efficiency is, of course, unity, and the mechanical advantage is given by QlP=2ivalq,\ or QIG=2iTlq / (5). as would be obtained at once by the principle of virtual work. If for a given /3, a is at our disposal, we may choose it so as to make ij a maximum. Thus, differentiating t; to u, by (4), we find that, for a maximum efficiency, tan 2a := cot ^, or a = — — -^ (6). 42 ' As to the efficiency of any actual mechanism involving a screw and nut, it should be noted that it will always be lower than the expression (4), for that allows for friction only at the surfaces of the screw pair itself {i.e. the helical surfaces of the screw threads), but there must be friction also at the devices which prevent axial motion of the screw and rotation of the nut. Hence equation (4) is to be regarded as giving an ideal efficiency which is approached when the friction of parts other than the screw threads are almost negligible. To take these other frictions into account, referring to Fig. 183, we should need to introduce a horizontal resistance increasing P along AC, and also vertical resistances along the side of DE, thus reducing the available portion of Q. Both these new terms would reduce both the mechanical advantage and the efficiency. For approximate allow- ances for these quantities and tables of efficiency when /;(.=o'i5, see Goodman's Mechanics Applied to Engineerng, pp. 239-240 (London, 1908). Examples — LXXV. 1. State in words and by equations the principle of virtual work as applied to rigid bodies. 2. What do you mean by the terms mechanical advantage and efficiency when applied to simple machines ? Give an actual illustration, and find for it the value of each of the ratios mentioned. 3. ' Determine the mechanical advantage of a screw lifting jack, neglecting friction. ' Calculate the tension in a stay which is tightened up by a force of P pounds acting on a lever a inches long, which turns a double screw, composed of a right-handed screw of m threads to the inch and a left- handed screw of n threads to the inch.' (LoND. B.A. AND B.Sc, Pass, Mixed Math., 1902, i. 4.) 4. ' Enunciate the principle of virtual velocities, and mention the class of mechanical problem to which it is applicable. ' Prove that if a horse is assimilated to an articulated parallelogram, the horizontal tractive force is I -h (tan a - tan 6) of his weight, when the legs make an angle a with a horizontal road and the traces slope downwards at angle 5.' (LOND. B.Sc, Pass, Mixed Math., 1902, 11. i.) 5. ' State the laws of friction between solid bodies, and describe the experi- mental verification. 382 ANALYTICAL MECHANICS [art. 391 ' Prove that in lifting a body of weight W with tongs of weight W, they must if vertical be grasped with a force W\ W .. W / at a distance of their length from the hinge, fi denoting the coefficient of friction of the body and / of the hand on the surface of the tongs, both surfaces of contact being on the point of slipping.' (LoND. B.A. AND B.Sc, Pass, Mixed Math., 1903, i. 2.) 6. ' Mention the class of statical problem to which the principle of virtual velocities is suitable for application. ' Prove that the virtual work of a couple is the product of its moment and the angle in radians through which it works, and prove that the balancing couples on the rods AC, BD of a jointed quadrilateral ACDB pivoted at A and B are as ACICE to BDIDE, where E is the point of intersection oi AC and BD^ (LoND. B.Sc, Pass, Mixed Math., 1903, 11. i.) 7. ' Enunciate the principle of virtual work for the equilibrium of any given system of connected bodies. ''A BCD is a square formed by four equal uniform bars, each of weight W, freely jointed together at A, B, C, D ; a strut of negligible weight connects the joints B and D, so as to preserve the square figure when the system is suspended vertically from A. Show by virtual work that the pressure in the strut = 2 PF.' (Lond. B.Sc, Pass, Mixed Math., 1904, 11. i.) 8. ' A ladder AB rests on the ground at A and against a vertical wall at B. If AB is inclined to the vertical at an angle less than the angle of friction between the ladder and the ground, show geometrically that no load, however great, suspended from any point on the ladder will cause it to slip.' (Lond. B.Sc, Pass, Applied Math., 1905, i. 3.) 9. ' An effort P is applied at the end of an arm of length a to overcome a load JV placed on top of a rough screw press. The radius of the cylinder is r, the inclination of the thread of the screw to the horizon is i, and X is the angle of friction. Prove that ' If the radius of the cylinder is 2 inches, the effort arm 12 inches, the coefficient of friction o"i, and there are 8 threads to the inch, find P. Will this screw reverse if P is withdrawn ?' (Lond. B.Sc, Pass, Applied Math., 1905, i. 4.) 10. ' Prove that under a certain condition the altitude of the centre of gravity of a system of bodies in equilibrium is stationary for small dis- placements. 'Illustrate this by the case of a bar (not uniform) resting with its extremities on two smooth inclined planes which face one another.' (Lond. B.Sc, Pass, Applied Math., 1906, i. 6.) 11. 'A system of forces (Xj, Fj), (X^, Y^^), . .. act at the points {x^, jj), (■^2) ^'2)1 • • • respectively of a plane lamina. If the lamina receives a small displacement such that the component displacements of the origin are (a, 0) and the angle of rotation of the lamina is a>, prove analytically that the total work of the forces is a.2{X) + 13.2( Y) + a.^x Y -yX). ' Deduce the principle of virtual work.' (Lond. B.Sc, Pass, Applied Math., 1906, 1. 10.) ART. 392] PLANE STATICS OF RIGID BODIES 383 12. 'State the principle of work as applied to a machine working uniformly against resistance. 'Apply it in the case of a screw press which is (i) smooth, (2) rough, assuming all requisite data.' (LOND. B.Sc, Pass, Applied Math., 1907, i. 5.) 13. 'A rod of weight w and length 2/ can revolve freely in a vertical plane about one end which is fixed. At a vertical height h above the fixed end is a smooth peg over which passes a string, one end of which is attached to a smooth light ring which slides freely along the rod while the other end carries a weight P. Apply the principle of virtual work to find the position of equilibrium of the rod, explaining fully the argu- ment on which the equation used is based.' (LoND. B.Sc, Pass, Applied Math., 1910, i. 5.) 392, Stability of Equilibrium. — Let us suppose a body to be in equilibrium in any position A under the action of any forces. Let the body be successively placed at rest in each of any two adjacent positions B and C on opposite sides of A. Then the type of the equilibrium at A may be defined as follows : — (i) Let the body remain at rest at B and at Cj its equilibrium at A is then said to be neutral. (2) Let the body start moving towards A from both B and C ; its equilibrium at A is then said to be stable. (3) Let the body when at B start moving away from A and when at C start moving either from or to A ; its equilibrium at A is then said to be unstable, for in either case it will finally move away from A. Let us now find the analytical conditions to which these various types of equilibrium correspond. Suppose the body to be under the action of forces like gravity or the elastic reaction of a frictionless spring of constant properties. We then have the relations T-\- r= constant, dT-\-dV=Q . . . (i), where J" and F denote respectively the kinetic and potential energies of the body. But when a body starts to move from rest or increases any speed it has, Tis increasing; thus, by (i), Fmust be decreasing. In other words, a body moves spontaneously so as to dimirtish its potential energy. Thus for neutral equilibrium, since no motion occurs after a dis- placement, we have dT—o = dVo\ F= a constant, F„ say . (2). For stable equilibrium, since motion towards A occurs from either side, the potential energy is a minimum there, V^ say, and its increase in the neighbourhood is expressed by an even power of the displacement {x say) multiplied by a positive coefficient. Or, in symbols, V=V,+a,x' ....... (3). For unstable equilibrium, if the motion after either displacement is always from A, we have by similar reasoning an even power of x associated with a negative coefficient, or V=Fo-a,x- (4). 384 ANALYTICAL MECHANICS [ART. 392 While for unstable equilibrium with motion from A at one side, to A from the other and then through A and finally away, we have an odd power of X involved, or V=V,±a,x'' (5). NBUTRAL STABLE UNSTABLE CAB X Displacements Fig. 184. Potential Energy Graphs, showing Stability of Equilibrium. These results are collected in Table xiv. and illustrated by graphs of the potential energy in Fig. 184. Table XIV. Stability of Equilibrium. Types of Equilibrium AT Position A. Increase of Potential Energy at x from A = V- y„ = Neutral. Stable. Unstable. Zero. "~" u^x , or +a,x\ We may note here that the virtual work for any imagined friction- less displacement is numerically equal to the corresponding change of potential energy. Hence the principle of virtual work as a criterion of equilibrium is equivalent to the statement that for equilibrium of any type the first power of the displacement must vanish in the expression for dW=^Pdp, the higher powers being regarded as negligibly small. We now see that the determination of the type of equilibrium requires retention of those higher powers and an examination of their indices and coefficients. Or, looking at the graphs for the potential energy Fin Fig. 184, we may say that equilibrium of any type at A requires the slope to be 'zero there, whereas the fype of equilibrium at A depends upon the curva- ture there or its rate of change. But the slope, curvature, etc., near A depend respectively on the ART, 393] PLANE STATICS OF RIGID BODIES 385 first, second, and higher powers of x in the expansion for V, which may be written generally y= K+aiX+a:,x'+asx'+ (6). Thus the various views of the subject are harmonised. Referring again to the graphs for the potential energy V, it is seen that the positions of stable and unstable equilibrium may occur at minima and maxima; thus, in the absence of cusps and points of inflection in the curve, these two types may be expected to occur alter- nately in the equilibrium of a body. Simple examples of such alterna- tion occur in the case of a loaded sphere rolling on a table and in that of a rod revolving about a horizontal or inclined axis near one end. By bearing in mind the graphs of Fig. 184 we can often assert immediately whether the equilibrium of a body or system is stable or unstable. 393. The Balance. — The ordinary balance affords a good example of the lever, of equilibrium, of stability, and also of sensitiveness or ratio of inclination to difference of loads. Leaving to practical treatises the details of construction and manipulation, the essentials of a delicate balance for our purpose are indicated in Fig. 185. Fig. 185. The Balance. In this diagram the beam is suspended by a knife edge at S and, from other knife edges at A and B, hang the similar scale pans E and F. Let the points A and B be joined by a straight Hne, and upon it from S let fall the perpendicular SC, and produce to G the centre of mass of the beam. Denote the arms of the balance CA and CB by a and b respectively. Let SC = f and %Qi—h. Also let the beam have weight W, the scale pans each have weight w, and when weights P and ^are in the pans E and F, let the balance be in equilibrium with AB at inclination B to the horizontal as shown. Then, since the forces at A and B due to scale pans and contents 2B 386 ANAL YTICA L MECHA NICS [art. 393 are vertical, their arms are the horizontal projections of SA and SB or of SC+CA and SC+CB respectively. Thus, taking moments about S, we easily obtain the equation , (P+K/)(a cos 6-\-c sin Q)-\- Wh sin e={Q-^iv){b cos 6—c sin &) (i). A good balance in correct adjustment possesses three important properties, viz. the true zero position for equality of loads P and Q, stability, and sensitiveness, which we will notice in this order. True Zero. — When the loads in the pans are equal the line AB should be horizontal and the pointer indicate the centre of the scale. For this we must have ^=0 for Q=.p, On reference to (2) we see that this is obtained by making ^=a, which is aimed at in the adjustment of the balance. Stability of the equilibrium is obviously assured by pro viding that G is below S, for then G describes round S the lower part of a circle, and we have equation (3) of article 392 fulfilled. When the zero position of the balance is disturbed, the loads being equal, it is obvious that oscillations will occur. And, by the methods of Chapter xiii., article 258, we may write for the period of the oscillations T=2-K^I^[Wh (3), where K is the moment of inertia of the beam, pans, and load ; these being suspended at A and B, but without rotation, being reckoned as particles of same masses placed at A and B. Hence to reduce the period so as to facilitate quick working, we should have to diminish K, and therefore the arms a, but ipcrease h. It is no use increasing W, for that equally increases K. Sensitiveness. — Reverting to equation (2), we see that on putting ^=a as needed for the true zero, this may then be written tan 6 __ a , , Q^~{P+Q+2w)c+ Wh ^"^'^ which then expresses the sensitiveness which may be measured by the ratio of d or tan 6 to the difference of the loads. If we choose, this may be put in the form of a differential coeflficient. Thus writing Q=P-\- dQ, and considering the increment (/(tan ff) to correspond to it, we have for this increment sec^ddd = d6 nearly. Hence (4) becomes d0_ a dQ~ 2{P+w)c+ Wh ^S;- Thus by either (4) or (5) the sensitiveness for a given load may be increased by (i) increasing a ; (ii) decreasing c, or, making it negative ; (iii) decreasing h ; (iv) decreasing ^and w. ART. 394] PLANE STATICS OF RIGID BODIES 387 It will be noticed that some of the ways of securing great sensi- tiveness clash with some of those for securing a small period. Hence any actual good balance is one presenting the kind of compromise most suitable for a certain purpose in view. By making c negative, that is, bringing the point of suspension S below the line AB of the beam, the sensitiveness may be very greatly increased. It is seen that the sensitiveness as expressed by (4) and (5) varies with the load P-\- Q or 2F, although c and h are there supposed con- stant. But in any actual balance the beam is not perfecdy rigid. Accordingly these very small quantities c and // may appreciably change with the load, and thus cause an additional change in the sensitiveness. To approximately allow for this we may write these quantities as linear functions of the load. Thus let c=c,+(fF a.ndi h = h^+h'P . . . . (6). Then, substituting in (5), we have as a first approximation for the sensitiveness of a balance with an elastic beam the equation d9 a dQ 2{P-\-w){c,^ 'i"- - each fjure ^Lk 5 V "'^^ ^""^ '"'^" '" 'he other (see article 398). But this method suffers from the objection that lines may part y coh^c de Thus in F,g. 187 /'and Q do not coincide with 1, and ^ Tay be TveS ''[''"'• ■/"' ''J" ^'■^- ^'^^' e- andv?'had been given end ?n ^H '^ ^^"^^"' ""Z' '" ^'g- ^^7 ^and Q would have been line iv W ^ . coincident with P^Q. Again, if we indicate a line by letters at each end in one figure, it is impossible to indicate Frame Diagram. Fig. 187. Force Polygons. the corresponding line in the reciprocal figure by the same letters at each end on account of the method of construction, which brings different lines requiring different letters to meet at any given point. But since by condition (ii) of article 395 the faces of one figure corre.spond to \!ns. points in its reciprocal, these^airw in one figure and corresponding/«'«/j- in the reciprocal may bear the same letters. This constitutes Bow's notatiofi (1873), which is adopted in the small letters <7i5ir(/ of Figs. 186 and 187, and will often be used in what follows for a frame diagram and its force polygon. Thus the bars of the frame in Fig. 186 may be called ad, bd, and cd respectively. And the forces in them are represented to scale by the lengths in Fig. 187 of the lines ad, bd, and cd. In the force polygon the letters here stand at the ends of any line called by them, whereas in the frame diagram the letters stand on the faces divided by the 390 ANALYTICAL MECHANICS [art. 397 line in question. But no difficulty arises from this, the line intended being in each case quite definite. Again, a point in the frame diagram is called by the letters of the faces meeting there, acd say ; and, in the force polygon, the same letters occur at the corners of the polygon, which represent the forces ca, ad, and dc by which that point acd is in equilibrium. If desired, for further distinction, the letters may be capitals in one figure and small in the other. 397. Construction of Force Polygon and its Interpretation. — Referring still to Figs. i86 and 187, let us suppose we have given (i) the frame, (ii) the magnitude and direction of the force R applied at abd, and (iii) the fact that the point acd rests on a roller so that the force P must be vertical. Further, let it be required to find the forces P and Q and the forces exerted by the members or bars of the frame at their points of junction or joints. We cannot determine P and Q directly by the force polygon of Fig. 187, for it is evident that P may be drawn vertically of any magnitude we please, and Q then follows accordingly. We may therefore, by reference to Fig. 186, take moments of R and of P about the point bed, and equate their magnitudes. This determines P, so Q follows as the vector which joins R and P in Fig. 187. Then, by the method of lettering which we have adopted, the vector Q in Fig. 187 is to be lettered be, since in Fig. 186 (2 divides the faces b and c ; next, P must be lettered ca ; then R is already lettered ab as it should be. We next draw through the corners a, b, c of this triangle, in Fig. 187, lines parallel to the bars ad, bd, and cd in Fig. 186. At first these parallel lines may be produced both ways from the corners till it is seen where they are likely to meet in the point d. The figures may then be looked over and checked to make sure all is right. It may be noted here that in the force polygon the point c stands at the junction of P and Q, just as in the frame diagram the face e intervenes between P and Q, and so on all round. In Fig. 187 arrow heads are placed along the lines representing P, Q, and R just as they are in Fig. 186; this is done because these directions are all quite definite. But along the lines ad, bd, and cd'm Fig. 187 no arrow heads are shown, because each of these lines represents oppositely directed but numerically equal forces, when in turn it forms a side of a different polygon. Thus, if we consider the force polygon acd, the clue to the way round is afforded by the force P, and we have the forces ca, ad, and dc respectively. ]3ut these are the forces which maintain equilibrium at the point acd of Fig. 186. Hence the member cd is pulling to the right at acd; it is consequently in tension, or is a tie, which fact is shown in Fig. 186 by leaving it as a thin line. The bar ad, on the other hand, is seen to be pushing at the point acd ; it is therefore in compression, and is called a strut, and to express this in the diagram it is shown by a thick line. Thus, as the investigation of the forces in the frame proceeds by the construction and interpretation of the force diagram (or stress diagram as it is often called), the nature of the opposite forces (or ART. 397] PLANE STATICS OF RIGID B'ODIES 391 stresses) exerted by each bar upon the joints at its ends is shown by this thickening of the lines, where necessary, to denote struts in the . frame. This double use of the lines in the force polygons for opposite forces IS seen if we now take the figure bed. Here the force Q gives the clue as to direction, and it follows that we are to write the forces be, ed, and db. But this shows that the member ed is pulling to the left at the point bed, whereas the same member was before found to be pulling to the right at the point acd. This shows, as before con- cluded, that it is a tie, or in tension. It is also desirable to point out that the external forces applied to the frame should be denoted on the frame diagram, as in .Fig. 186, by external lines so as not to eut into the spaees in the frame itself. We have thus, in this very simple case, found the supports or re- actions P, Q applied to the frame under a given load E, and also the magnitudes and natures of the stresses in each member of the frame. Examples — LXXVII. 1. 'Alight framework of freely jointed rods in the form of aright-angled isosceles triangle is suspended from the right angle. Weights w and 2w are suspended from the other two joints. Determine the stresses in the rods.' (LoND. B.Sc, Pass, Applied Math., 1907, i. 3.) 2. 'Points A, B, C, D are taken on a straight line, such that AB = \BC= CD. On AB, BC, CD and on the same side of these are described equi- lateral triangles AEB, BFC, CGD. EFn-nd EG are joined. The com- pleted figure represents a system of freely jointed light rods in a vertical plane with AD horizontal and lowest. Supports are placed at A and D, and weights of 4 and 6 tons are hung on at B and C respectively. Draw a force diagram for the system. Thence determine the stresses in the rods which meet at F, indicating in each case which are tensions and which are thrusts.' (LoND. B.Sc, Pass, Applied Math., 1907, i. 6.) 3. 'AD ISO. straight line trisected at B and C, and BCEE is a square. Let AF, AB, BE, FE, BE, BD, ED be rods forming a freely jointed framework, and let this framework be supported at A and D so that AD '\% horizontal and BF vertically upwards. Find, by graphical construction, the thrusts or pulls in all the rods due to a weight W placed at E! (LoND. B.Sc, Pass, Applied Math., 1908, i. 4.) 4. ' Three equal light rods AB, AC, AD, each of length a loosely jointed at A, have their other ends joined by three strings each of length b, and rest with B, C, i? on a smooth horizontal plane so as to form a tnpod. From A is suspended a load W. Show how to find the tensions m the strings by a graphical construction or otherwise.' (LoND. B.Sc, Pass, Applied Math., 1909, i. 5-) i; 'A BCD, PBCQ are squares on opposite sides of BC, and E is the centre of the latter square. A framework is made of weightless rods^^, AD DC BC DB BE, CA" freely jointed to one another. A is freely pivo'ted to a 'smooth vertical wall, and D presses against the wall below A A weight Wis hung from E. Calculate the stresses m all the ^ojjs > (LOND. B.Sc, Pass, Applied Math., 1910, l 7.) 392 ANALYTICAL MECHANICS [ART. 398 398. Funicular Polygon for Resultant of Ooplanar Forces.— In the example dealt with in articles 396-397 we had only a single force i? applied above the frame, and found by calculation the corresponding values of the reactions or supports. But usually there are a number of forces or loads applied on the upper side of the frame, and in this and other cases it is often desirable to determine graphically the magnitude, direction, and Ihie of action of the resultant of such a system of coplanar forces. This may be done by what is termed \h& funicular polygon or link polygon. The method of drawing such a polygon and the proof of its pro- FiG. 188. Set of Coplanar Forces AND Funicular Polygon. perties may be seen from the following example : — Let the set of forces P, Q, S, T be given as shown in Fig. 188 and their resultant required. The force polygon pqst in Fig. 189 determines the magnitude and direction of the resultant R which are equal to those of r, but the line of action of R is still to be found. Fir. 189. Force Polygons from which TO DERIVE Funicular Polygon. Construction. — To determine this proceed as follows : — First, in the diagram, Fig. 189, take any convenient point O, called the pole, afld join it to the corners of the force polygon by the lines u, I, m, n, V. ART. 398a:] PLANE STATICS OF RIGID BODIES 393 Second, in the line of action of i^in Fig. 188 take any point A, and from it draw the Hnes Z/and L parallel to te and / in Fig. 189. From the intersection of L with Q draw M parallel to vi. Continue in this way, drawing next N and finally V intersecting U at K, thus com- pleting the funicular polygon whose sides are ULMNV. Finally, through K, in Fig. 188, draw R equal in magnitude in direction to r, in Fig. 189. Then K shall be a point on the line of action of the resultant, which is thus truly represented by R. Proof. — Referring to Fig. 189, we see that P is represented by/, and is equivalent to the components u and /taken in the directions of the arrows. Thus P in Fig. 189 may be replaced by forces of these magnitudes along U and L meeting at A on the line of action of P. Again, Q is represented by q or its components — / and tn, or by forces of these magnitudes along L and M meeting on the line of action of Q. Thus the force along L is cancelled, being taken in turn positively and negatively. In like manner, as we proceed along the funicular polygon ULMNV, all the intermediate forces /, m, and n are cancelled, and only those along U and J' equal to u and v are left. But their resultant is r=R, which must act through their inter- section K as shown. This line of thought may be expressed symbolically as follows, it being understood that when the small letters are used the correspond- ing forces are represented as to magnitude and direction only, the large letters denoting in addition the correct line of action. The sign ^ over the + shows that the addition is vectorial. ^ /*. /^ P+Q+S+T=p+g+s^-t ^{u+l) + {-l->rm) + {-m^-n) + {-n+v) =u+v=r = (« along U) + {v along T) = R through K. It may easily be seen that, by shifting the point A, we change to a second funicular polygon with sides respectively parallel to those of the first. Whereas, if we shift O, we change to a second funicular polygon with sides in general not parallel to those of the first. But, in either case, K remains on the same line of action of R, as shown in Fig. i8gA. 398a. Link Polygons with Different Poles.— We may now estab- lish the following important theorem on link polygons :— Enunciations.— 2%^ corresponding sides of any two Itnk polygons pr a given system of forces intersect on a straight line, which is parallel to that joining the pole of the two funiculars. . Referring to Figs. 189A and 189B, the systein of forces is deno ed in the former by PQST ^^■it.^ resultant R, and in the latter by the force polygon/, q, s, t closed by r. 394 ANALYTICAL MECHANICS [art. 398a .^0 Fig. 189A. Set of Forces and Two Link Polygons. Fig. 189B. Force Polygon and Two Poles for Link Polygons. ART. 399] PLANE STATICS OF RIGID BODIES 395 In Fig. iSqb the poles O and O' are taken, and the two sets of rays t(, I, m, n, V and u', /', m', n', v' drawn. Corresponding to these we have in Fig. 189A the two funicular polygons U, L, M, N, F and V, L', M', TV', v. Let the intersections of the sides UU', LL', MM', NN', and VV be A, A, B, C, and D respectively. Then shall ABCD in Fig. 1 89 A be a right line parallel to 00' (in Fig. 189B). Proof by a Statical Method.— At the intersection of Z and M in Fig. 1 89 A, let the force Q act as shown. Also at the intersection of L' and M' let the force — Q act. Hence, with the component forces along L, M, Z', and M', we shall have two sets of forces in equilibrium, three in each set, giving in all six forces in equilibrium, viz. + Q, —I along Z, -{-m along j1/\ , . — (2, +/' along Z', —»?' along ^'/ ' ' ' ^^'" But since the two Q's are equal and opposite along the same line, they are in equiUbrium, and consequently the other four are in equili- brium. Hence any pair of these four will equilibrate the remaining pair. Thus — /along Z and +/' along Z', each applied at A, will equilibrate —m' along M' and -\-m along M, each apphed at B. Accordingly each pair must represent a force along AB. But by Fig. 189B — /and +/' have a resultant of magnitude and direction O'O, whereas — w' and -\-m have a resultant of magnitude and direc- tion 00'. Therefore AB is parallel to 00'. And in the same way the like relation may be established for ACD. 399. GrapMcal Conditions of Equilibrium. — We may now enun- ciate from the graphical standpoint the conditions of equilibrium of a rigid body under the action of coplanar forces. We recall that, as stated analytically in (i) and (2) of article 367, these conditions are equivalent to (i) Resultants of forces parallel to x or y each equals zero, (ii) Resultant torque in plane of .ri' equals zero. Referring to Figs. 188 and 189 of article 398, we see that if five forces were given, viz. P, Q, S, T, and one of the magnitude of R but opposite in direction, then the force polygon would show that the resultant force is zero, thus fulfilling the first condition of equilibriurn. But if this reversed force R' say did not pass through the point K in Fig. 188, but was parallel to the R there shown and at a perpen- dicular distance r from it, the whole system would be equivalent to a couple of magnitude R'r. Moreover, on the line of action of this force R', the sides U and V of the funicular polygon would not intersect, but there would be two points where these sides f/ and r would respectively meet the line of action of R'. This is de- scribed by stating that the funicular polygon would not be closed. But to make the resultant couple zero it must be dosed; that is, U 396 ANALYTICAL MECHANICS [art. 400 and V must meet in the same point K on the line of action of the reversed R. Thus the formal graphical conditions for equilibrium of a rigid body or system under coplanar forces are the following two : — (i) The force polygon must be closed ; and (ii) The funicular polygon must be closed. And these clearly correspond to the equations (i) and (2) of article 367 each to each. Fig. 190. Coplanar Forces, Reactions, and Funicular Polygon. 400. Beactions determined by Funicular or Link Polygon. — We have seen that the single resultant of a system of coplanar forces may be completely deter- mined by the use of the funicu- lar polygon, and that the equi- librium of a rigid body requires that the force polygon and the funicular polygon should each be closed. We now pass to a third application of the funicular polygon, in which it is used to determine each of the two re- actions at the supports of a rigid bar or frame which is loaded by any set of coplanar forces. The case in question is illustrated by Fig 190, where P^, P^, P^, Fig. 191. Force Polygons for Funicular. ART. 401] PLANE STATICS OF RIGID BODIES 397 Pi, Pi denote the forces which constitute the load on some frame (not shown in the diagram) ; the conditions as to the reactions at the supports being tliat they occur at the points A and B, and that the reaction at A is vertical because the frame rests on a roller there. For the graphical determination of this problem we may proceed as follows : — Of the force polygon draw the sides /j, /„ /a, /j, /i (Fig. 191), and join the corners to a convenient point O chosen as pole by the lines u., q, n, m, I, u^. We have now to find a suitable point in Fig. 190 at which we may begin drawing the sides of the funicular polygon parallel to the lines meeting at O. We cannot begin at A, because that would leave it impossible to decide where a given side of the funicular not passing through B would intersect the reaction R«,, since its direction is not at first known. We therefore commence the funicular polygon at B, the only point known in the reaction R.^. We then draw through B the lines U^ and Q parallel respectively to u. and q in Fig. 191. From the intersection of (2 with P^, produced if necessary, we draw N parallel to «, and in like manner M, L, and U, meeting the known line of action of the reaction ^1 of the support at A. We can then join this point of intersection to B by the line V, and parallel to V we draw v through O in Fig. 191 ; this cuts the vertical r^ and enables us to complete the force polygon by drawing r^. Then the reactions sought are represented by R^ and R^ acting at A and'& in Fig. 190 and parallel and equal to r^ and r, in Fig. 191 each to each. It may be noticed that though U., has been drawn in Fig. 190 it is not required, since the forces it should join intersect at B. 401. Roof with Asymmetrical Load. — Let us now suppose the set of coplanar forces just dealt with to be the asymmetrical load due to wind on a specified roof principal (or roof truss) as shown in Fig. 192. We may then accept the reactions at the supports as found by the funicular polygon in article 400, and proceed to find the stresses in the members of the frame by the force polygons of Fig. 193 on the principles explained in articles 396 and 397. We may conveniently deal with the points of the frame in the order of the numbers shown in Fig. 192. And the various applied forces or stresses in the bars at each point may be taken in the order indicated by the letters after each number in the following scheme : — r. bade, 2. C(^, 3- f'^^'g^ 4. efgh^, 5. k hml , 6. hgam, 7. nlma. Thus it is well to begin at either i (or 7), because as there are only 398 ANAL YTICAL MECHANICS [art. 401 two unknown stresses here, the polygon is determinate. The point 2 is taken next in preference to 3, because the former has only two unknown stresses, while the latter has three until that in #has been determined by dealing with point 2. Tb o- FiG. 192. Roof with Wind Loads. Fig. 193. Force Polygons for Roof. In the above scheme, where a /azV of letters is underlined it denotes that it has just been found that the bar in question is a strut, i.e. is under compression. In carrying out the work, the student at this stage may well thicken the line which represents the strut in the frame diagram. ART. 402] PLANE STATICS OF RIGID BODIES 399 402. Evaluation of Stresses apparently Indeterminate.— Let us now consider a frame, of a form often adopted for roofs, in which at a Fig. 194. Frame and Loads needing Special Device for Stresses. Fig. 195. Force Polygons for above Frame. certain stage the methods hitherto adopted for the stresses needs supplementing by some other device. Thus, let the frame AMBL of 400 ANALYTICAL MECHANICS [art. 402 Fig. 194 be given with the loads there shown, and let it be required to find the reactions and the stresses in every member of the frame. We attack this problem as in articles 400 and 401, findiiig the reactions by the funicular polygon and then proceeding with the stresses at each point. Suppose we begin at B (marked also i in Fig. 194), taking next in order the points 2 and 3. There is no difficulty so far, as each point had only two unknown forces, so each of the corresponding force polygons was determinate. But a difficulty arises as to the next point to be taken. For point 4 has the three unknowns vu, ur, ra, if taken before 5. While point 5 has the three unknowns et, tu, uv, if taken before 4. There are various ways of proceeding at this juncture. Perhaps for our purpose the most convenient is that known as the Method of Sections. Goodman, in his Mechanics Applied to Engineering, p. 506 (London, 1908), states that this method, usually ascribed to Ritter, is really due to Rankine. Adopting it in the present case, we take a section of the frame along, say, the central vertical line KLMN (Fig. 194), and consider the equilibrium of the right half of the frame under (i) the loads applied to that half, (ii) the reaction at B, and (iii) the stress in the member ra at M. Now only the last-named of these is unknown. Hence, taking moments about L and equating to zero their algebraic sum, we deter- mine by computation the magnitude and nature of the stress in ra. There are then only two unknowns at point 4, -which is accordingly dealt with in the usual way. Thus, vu being found, point 5 has only two unknowns, and is therefore determinate. The other points may then follow in the order as numbered, and give no further difficulty. The whole procedure may now be summarised as follows, the references being to the drawing of the force polygons in Fig. 195 : — 1. abcxa, 2. cdwxc, 3. axwva. By method of sections find ra., 4. ravur, 5. detuvwd, 6. efste, 7. tsrut, 8- fmsf, 9- g¥s.g> 10. qporq, 11. arena, 12. phimnop, 13. mijlm, 14. mlantii, 15. Ijkal. As before, the pairs of letters in this scheme underlined are those referring to members just found to be in compression. At each step ART. 403] PLANE STATICS OF RIGID BODIES 46J the^y should accordingly be thickened in Fig. 194 to indicate a as .nnH ^' ,^"".*^°"^^ ^ere that the loads shown in the roofs hitherto re.K/c?,'^ /y/T^'^''V''P?°'^'^ '° "^^ usually derived from loads really distributed between those joints. They are replaced by equivalent torces at the joints, because we are not here concerned with the bending ot the members which are imagined rigid. . In the examples already noticed the reactions were usually needed as a preliminary to the determination of the stresses, because there were more than two unknowns where the loads were applied. But suppose the load is practically a single force, as in the case of some forms of crane for lifting heavy weights. And let it be applied at a point m which only two members meet. We may then begin at that point in the evaluation of the stresses, the reactions at the supports being determined by the force polygons simply, without any recourse to the funicular polygon. Examples— LXX VI II. 1. ' Show how to find the magnitude and line of action of the resultant of a number of coplanar forces by means of a vector and a link polygon. ' If two different poles O and O' be taken for the link polygon, prove that the intersections of corresponding links all lie on a straight line parallel to 00'.' (LoND. B.Sc, Pass, Applied Math., igog, i. 2.) 2. ' Translate and explain the following passage :— Die Grosse und Richtung der Resultante eines ebenen Kraftesystemes ergiebt sich duich die Schlusshnie des Kraftepolygones. Es geniigt aber zur Festlegung der Resultante schon die Kentniss eines einzigen Punktes auf ihrer Aktionslinie. Zur graphischen Bestimmung eines solchen wird ein Seilpolygon gezeichnet. Somit setzt sich die Resultante des ganzen Kraftesystemes aus den beiden Kraften zusammen, welche sich in den aussersten Seiten des Seilpolygones ergeben, und geht, durch deren Schnittpunkt hindurch.' (LoND. B.Sc, Pass, Applied Math., 1908, i. 5.) 3. ' Three concurrent forces are in equilibrium ; show ab initio that any funicular triangle of the system is closed. 'Also prove tliat if two funiculars be drawn, the intersections of corre- sponding sides are coUinear.' (LOND. B.Sc, Pass, Applied Math., 1906, i. 3.) 4. Draw a set of coplanar forces acting on a frame or beam, and find by the stress and link polygons the reactions. 5. For a French truss roof under asymmetrical loads find the reactioils at the ends and stresses in all the members. (See Fig. 194.) 403. Reactions at Joints. — In a jointed arrangement of rigid bars, or bars and cords, it is often a matter of interest and importance to find the reaction at some of the joints. In attacking such problems the work may frequently be shortened by keeping clearly in mind two principles pointed out by Routh as to the direction of these reactions. These may be stated as follows : — (i) Let the body be hinged at two points A and B, and let it be acted on by no other forces except the reactions at A and B. Since 2 c 402 ANALYTICAL MECHANICS [art. 403 4 W " 1 1 -k % ' T r 1 1 2c 1 sy X > 1 t [ ^ /^\ \ 1 V ^ A- <- 2b — » F G, 196. Reaction at Camp Stool. Joint of the body is in equilibrium under these two reactions, they must act along the straight line joining the hinges and be equal and opposite. (ii) Let the body and the external forces be both symmetrical about some straight line through the hinge. Then the action and reaction between the two bars must be symmetricallysituated. But they are also equal and opposite. Hence, to fulfil both conditions, the action and reaction must each be perpendicular to the line of symmetry. The first of these principles as to direction of reactions has already been illustrated in deal- ing with roofs by graphical methods, and will be useful again presently. Let us now illustrate the second principle by two examples of symmetrical problems. Take first the case of a camp stool on a smooth floor loaded by a smooth box of weight 4 W as shown in Fig. 196. Each side of the stool consists of two inclined bars as illustrated, jointed at their centre C, whose action and reaction are sought. Let the tension in the cross straps at the top be T. Then the bars and forces being all symmetrical about a vertical line through the hinge, the re- actions there are horizontal, and may accordingly be denoted by X simply, no vertical component Fbeing needed. Call the height of the frame 21 and its width 2b, and consider a single in- chned bar AB. Then the load at B and the reaction at A are each vertical and of magnitude W. Hence, taking moments about B, we find W2b^Xc, or X=2 Wbjc . (i). Then, resolving horizontally, T^X=2JV6/c . . . (2). As a second example of a frame and forces all symmetrical consider now the set of steps which in its simplified form makes the A shown in Fig. 197, the floor being supposed smooth. Let the load at the top be 2 JV, and let the reaction there and the tension of the cord below be required. Denote by 2c the spread AB of the legs, let the top C be at a height a above the ground and 6 above Reaction at Top op Steps. ART. 404] PLANE ST A TICS OF RIGID BODIES 403 the cord, whose tension is T. Then, by the principle of symmetry, the reaction at C is horizontal, and will be denoted by X. Also, because the floor is smooth, the reactions at A and B are each vertical of value Ji say Resolving vertically shows that i?= IV ■ resolving horizontally for either leg (AC say) gives T=X. Then finally, taking moments about A, we have A'a=rFf+7i;a-/;). * Whence X=lVc/d=T . ... (,\ thus giving the reactions sought. ' 404. Separation of Bars.— To find the reactions at joints it is sometimes a convenience to suppose a bar separated from the others, and then represent the reactions it bears, and so determine them. This method will be illustrated by the framework of seven members shown in Fig. 198, one of the side pieces AG being considered detached to show the forces more clearly, as represented on a larger scale in Fig. 199. + W 2a 2 W + Z W Fig. 198. Reactions at Joints A AND G. Fig. 199. Bar AG separated FROM Frame. The frame AEBCFDGH is suspended at E, the middle point of AB, and has a weight 2 tV attached at F, the middle point of CD. Let it be required to find the reactions on the ends of one of the inclined bars, AG say. We cannot treat this by the symmetrical principle, because the line of symmetry does not pass through either of our points. We may, however, apply the other principle of article 403 to show that the action S exerted at G by the bar GC must be along GC, since this bar has forces at its ends only. The same principle shows us that at A the action due to the bar AB is not solely along it, for there is the force 2 IV a.t E which is not a joint, AB and CD being stiff bars with joints at their ends only. We thus obtain the view of the forces on AG which is represented in Fig. 199, viz. that A has the force W vertically upwards and the thrust X horizontally to the left, while G has the tension 7' horizontally to the right and S obliquely 404 ANAL YTICAL MECHANICS [art. 405 downwards in the direction of the bar GC in Fig. 198, making an angle, 6 say, with the vertical. Let AC = 2a, AB — GH = 2^, the lengths of each of the four inclined bars being c. Then sin Q^bjc and cos 6=alc. We may now consider the equilibrium of the bar AG in the usual way. Thus, resolving vertically gives JV= S cos = Sa/(; resolving horizontally, X+Sbjc—T; and, taking moments about G, Xa= IVb. Hence we have S=JFc/a,X=m>/a, and T= 2 lV6/a . . . (4), showing that the resultant of JF and ^is along GA as should be. Compounding 5 and 7" to give the resultant J?, we find that .^=6', and is along AG . ... (5), as obviously should be the case, and so provides another check. These relations might also have been obtained graphically. 405. Reactions inside Bodies. — We now pass to the consideration of the reactions between two adjacent portions of the same continu- ous body by which the equilibrium of each of those portions is main- tained in spite of certain impressed forces or systems of forces. We here confine our attention to the determination of these forces of reaction and their variation with the circumstances of the case, because at present we are dealing only with bodies supposed rigid. The discussion of any changes in size or shape of the body con- sequent upon the operation of these forces forms part of the theory of elasticity and is accordingly deferred to Chapter xxi., which is devoted to bodies exhibiting elastic properties. We start with the following simple example : — Let a horizontal beam be subject to a vertical force /*! at Xi, and consider the reactions 2X and 2 Y which must act horizontally and vertically at x to maintain in equiHbrium the portion of the beam between these two points (see Fig. 200). We suppose the beam to be 1 „ held in equilibrium by some forces ' applied at or beyond x, but with the Fig. 200. Reactions in a Beam, exact nature of these forces we are not now concerned. Then, for the equilibrium of the part of the beam under considera- tion, resolving vertically and horizontally, and taking moments about the point of co-ordinates {x, o), we find 2F=-/', (,) 2Z=o and 2(- J(»^^^(.v_.v,) .... (,). The signs of SZand 2 F must be reversed if we wish to express the actions of this portion of the beam upon that portion beyond x. The vertical components of this action and reaction, acting parallel to the vertical section at x, constitute what is called the shearing stress there, its magnitude being force per unit area. We are here concerned with X\i&patr of total forces of this stress, and shall denote them by S, which refers to the magnitude of either force, the positive sign being ARTS. 406-408] PLANE STATICS OF RIGID BODIES 405 used for the present case, in which the positive force is experienced on the positive side of the section in question. The horizontal components have a distribution as yet undeter- mined, but have a definite moment as given by (2), and their resultant is equivalent to a pure couple, the possible force being zero. They con- stitute what is called \h& benditig motnent ?it the. section under considera- tion. This moment will be denoted by M, the positive sign being used when the parts near the section in question tend to become concave towards the positive direction oiy. 406. Two or more Forces. — Suppose that to the same beam, in addition to Pi, other forces F^, F„ etc., are now applied at a\, .\\, etc., each such abscissa being less than x; and let it be required to express the reactions at .v. Then, by the same reasoning which led to the former equations for the single force, we find P -\-P„-\-P -\-, . ,^='2,p^S . . . . l-d and '2{~Xy) = P,{x-x,)+P,{x-x,)+P,{x-x.;)-\-&\.c. = M (4).' As before, 2X=o. These expressions for the shearing forces and the bending moments show that there is a simple relation between them. For, on differentiating (4) with respect to x, we have '^^=P,+P,+P,+ . . .=-^P^S (S). Or, in words, though the absolute value of the bending moment depends upon the positions as well as the magnitudes of the forces, its rate of change along the beam at any place depends only on the sum of all the forces up to that place, which is also the shearing force at the place in question. 407. Distributed Load. — It is obvious from (3), and the reasoning which led to it, that, if the forces instead of being few and finite are very many and correspondingly small so as to become practically a con- tinuous or distributed load, still the sum of all the forces up to a point expresses the shearing forces at that point, and we also see that this stress will then be a continuous function of the abscissa. Hence, for this case, we may again differentiate to x, operating upon equation (5). We thus find d'M dS ^ /.. -,-:2==77,= <2say (6), ax ax where Q clearly denotes the total impressed force per unit length at X. 408. Diagrams for Bending Moments and Shearing Forces in Beams. — It is now desirable to consider how these reactions in beams may be graphically represented. For the diagrams so formed illustrate the relations just developed, and also lend themselves to the solution of problems. , ,, , , , Take first the case of a horizontal beam held at or beyond x and with vertical forces F,, I',, F, acting at a„ x,, x, respectively. And 4o6 ANALYTICAL MECHANICS [art. 408 let us draw diagrams in which the abscissae represent lengths along the beam, the ordinates representing in one diagram the bending moments i?/"and in the other the shearing forces 5. The beam and these diagrams are shown in Figs. 201, 202, and 203. ^ a Xl X3 sc X p. Pz Pj W/M Fig. 201. Beam with Three Forces. Fig. 203. Shearing Force Diagram. In the bending moment diagram the inclinations of the line are denoted by ^1, <^3, and <^3 after the applications of the forces /'j, P^, and P,, respectively. And, since at the various parts in this diagram the values of dMjdx are the tangents of the corresponding angles, we have the general relations \.an = dM/dx^^F=S (7). We may also represent the increase, M — Mi, of M between two ART. 409] PLANE STATICS OF RIGID BODIES 407 points, Xi and x say, as the areas of the shear diagram over that range. For from (7) we have M-M,= r Sdx (8). Jxi But, in using either (7) or (8), care is necessary as to the scales used in the horizontal and vertical directions of the diagrams. Thus, if the horizontal scale of the bending moment diagram were i inch to the foot, and the vertical scale i inch to 10 foot-tons weight, a line inclined at 45° would have a tangent 10 when properly interpreted instead of unity, as in ordinary geometry, where the scales are equal in all directions. The comparison of Figs. 202 and 203 shows instructively what is expressed by equations (6) and (7), namely, that the slope of the bending moment diagram always equals the shearing force, and there- fore any change in this slope equals the increase in the ordinates of the shear diagram at the place in question. These changes in slope occur abruptly because the loads are discrete forces. The next example better illustrates equation (.6). 409. Uniformly loaded Cantilever and Beam. — Let us now take the case of a cantilever (or beam projecting from a wall into which it is built) loaded uniformly along its length /, the forces being downwards Fig. 204. Uniformly loaded Cantilever, M Fig. 205. Shear Diagram for above, Y Fig. 206. Bending Moment Diagram for above. 4o8 ANALYTICAL MECHANICS [art. 409 and of distribution whose numerical value is 7 per unit length. Let us draw the cantilever projecting to the right and take the origin of co-ordinates at its free end with j; upwards, as shown in Figs. 204, 205 and 206; giving the beam itself and the two diagrams. Then (2= -7, and by (6) 5= -7;^ . . . (9). But by (5) dM=Sdx— — ixdx. Whence rx M— — ']\ xdx= — -]x^\2=Qx^l2 . . (10). Thus, since our abscissae are all negative ^ is by (9) positive, as shown in Fig. 205 ; and M\5 by (10) negative for any values of ^, and is accordingly shown negative in Fig. 206. We see from (10) that the bending moment diagram is part of the parabola -^y=-Qy (II). Hence its vertex is at the origin, its axis is vertical, and its branches extend downwards. If the length of the cantilever is /, we see that the maximum shear- ing force at the root, where x=- —I, is given by — Ql=- + 7/. Also the bending moment at the root, by (10) or (11), on substitution of —/for X, is seen to be — 7/72- Uniform Load. m^ 21 '/, Fig. 207. Uniformly loaded Beam. Fig. 209. Bending Moment Diagram. ART. 410] PLANE STATICS OF RIGID BODIES 409 We may now very simply pass to the case of a beam supported at each end with clear span of 2/ and a downward load of 7 per unit length, I.e. as before (2= -7- The diagrams for this case are given in Figs. 207-209, and need little if any further explanation. It may be noted that although the shear and bending moment diagrams seem simply extended to twice the width by producing their hnes, still in the case of the latter (Fig. 209) the ordinates are all different, the maximum moment being now at the place of zero shear instead of both vanishing together as before. In choosing the scale of ordinates for the bending moment diagram it is neither necessary nor in general convenient to represent by the same height a moment M=Sx i, as was used in the shear diagram to represent the shearing force S. To avoid making the ordinates in the bending moment diagram unduly large, 5 X one-third or half the length of the beam may be represented by the same ordinate as 5 itself in the shear diagram. See also the next article. 410. Bending Moment Diagram a Particular Link Polygon. — It is now desirable to note that the bending moment diagram as hitherto drawn on a horizontal base is a particular form of the link polygon for vertical forces. To illustrate this, consider the case of a beam supported at the ends and loaded at three points as shown in Fig. 210. From the data there shown a force diagram is drawn as in Fig. 211, in which the polygon closes to a single line BAG since all the forces are vertical. Now let some point O be taken as pole and the rays OB, OWj, OWj, OW3 be drawn, and consider their slopes. Obviously, as we pass from ray to ray, the change of tangent of the inclination is, on the scales chosen, equal to the corresponding forces interposed between those rays. Hence, if any one ray has the right slope for the corre- sponding part of the bending moment diagram, they all have the right slopes. Suppose ^1 is the reaction at the support at the origin in Fig. 210, and in Fig. 211 take AB=^i, and let AO be perpendicular to AB. Then, if the bending moment diagram begins with a slope parallel to OB in Fig. 211, it is evidently right, on the understanding that the scale of ordinates is such that a bending moment equal to (OA X AB) is represented by AB simply. Then this first slope being right, the others will be right if drawn parallel to OW^, 0W„ and OW, respectively, the junctions of the slopes corresponding to the points of application of the loads ^^^, Vt\, and W,. Thus i?i being found, by calculation or a preliminary link polygon with any pole O', and the pole O being taken as shown, the shear diagram may be appropriately drawn, as in Fig. 212, level with the force polygon, the bending moment diagram being at foot as in Fig. 213. If J?, is determined by a link polygon with any pole O', that link polygon may be retained as the bending moment diagram. But if O' were not level with A in Fig. 211, as O is, then the bending moment diagram so obtained would be on a s/oJ>tng base parallel to O'A. Ordinates could still be measured on it, as on the other, but it would 4IO ANALYTICAL MECHANICS [art. 410 Fig. 210. Beam WITH Irregular Loads. (^^Tb&M^t. Fig. 2X1. Force Diagram and Rays FOR SAME. Fig. 213. Bending Moment Diagram as Link Polygon. ART. 470] PLANE STATICS OF RIGID BODIES 4,1 not be possible to apply to it the equations written for rectangular co-ordinates and suppose them to be still valid as rectangular co-ordinate equations. The relation of the scales of the ordinates in the shear and bending moment diagrams has been shown to depend upon the pole distance OA. A full statement as to all the scales may be put as follows : — In all diagrams let i inch of abscissae represent a feet, and in the shear diagram let i inch of ordinate represent a force of b lbs. wt. ; then, in the bending moment diagram, r inch of ordinate shall repre- sent a bending moment of abc ft. lbs. wt. where c is the pole distance OA, and the slopes of the bending moment diagram are parallel to the corresponding rays from O. For further information on these beam diagrams, such works as Professor A. Morley's Stretigth of Materials or D. A. Low's Applied Mechanics may be consulted. Examples — LXXIX. 1. State how to find the reactions at joints of a frame, illustrating your answer by a numerical example. 2. Explain the nature of the reaction between the two parts of a beam, divided by an imaginary cross section, when one or more forces are applied at places beyond this section. 3. Explain the terms shearing forces and bending moment as applied to a beam under load, and obtain relations between the above quantities and the load per unit length. 4. Make shear and bending moment diagrams for a beam fixed in a wall and loaded at two or more points. 5. Discuss the stresses in a beam resting on two supports and loaded uniformly throughout its length. Draw the diagrams for the shear and for the bending moments, and obtain the equations of the curves. 6. ■ .V bar AB is in equilibrium under the action of given forces ; show how to find the resultant action exerted over the cross section at any point P of the bar by the portion BP on the portion PA. ' In particular, if AB is a uniform bar resting on two supports at A and i) in a horizontal line, what is the action between the two halves of the bar ? Explain the result.' (LoND. B.Sc, Pass, Applied Math., 1905, \. 2.) 7. ' A framework ABCD is formed of four similar uniform heavy rods freely jointed at their extremities. The rod AB is of length 2), and {z—c) respectively. But the expressions for the components of R do not contain x, y, and z. Hence the principal force R is the same in magnitude and direction whatever base is chosen. Let the components of the couple G for the new base be Z', M', and N'. We then find L=^{Z{y-b)-Y{z-c)\ = L- JFb+ Vc \ M'=^X{z-c)-Z(x-a)]=M-l7c+IVa\ . . (7). iV'=2{ V(x-a)-X(y-b))=W- Fa+ Ub ] We accordingly see that the magnitude and axis of the principal couple are in general different for different bases. 412. Conditions of Ectuilibrium. — For equilibrium it is evident that the resultants of the system of forces must vanish, i.e. for the reduction already effected i?=oand G=o .... (8). But these involve the vanishing of each of the six components of the forces, giving as the conditions C/=o, F=o, W=o (9), and L=o,M=o, N=o . . . (10). Suppose the base is shifted to O', the principal force and couple being nowi? and G'. We then have (9) as before, but (ro) replaced by Z'=o, J/'=o, andi\^' = o (11). But by (7) we see that (11) reduces to (10) when (9) is fulfilled. Hence, for any base whatever, the conditions of (9) and (10) are sufficient. It is not, however, necessary that the axes should be at right angles. It may easily be seen that any oblique axes that are non- coplanar will serve. It may be noted here that the six conditions of equihbrium correspond to the absence of linear and angular accelerations of the six possible modes open to a rigid body or in the six degrees of freedom possessed by it. 413. Components of a Force. — As seen from Table xv. of article 410, it is sometimes convenient, y»r the sake of the possible addition, to consider as the six components of a force P the expressions X, y, Z, Zy- Vz, Xz-Zx, and Vx-Xy . . . (12), instead of regarding, as usual, its magnitude and the equations of its line of action. ART. 414] SOLID STATICS OF RIGID BODIES 415 To represent on this plan the line of action of the force apart from Its magnitude, we may temporarily write for the magnitude unity. Then, if {x,y, z) are the co-ordinates of a point on the line and (/, m, n) are its direction cosines, the six components of this unit force (or co-ordinates of the line) are /, m, n; X=iny — mz, ii = lz — fix, v = mx — h . (13), From which it is evident we have the relation /X+mfi+nvzizo . . (14). If the force along this line be jP instead of unity, it follows that its six components are PhPm,P,t; P\Psx,Pv . . . .(15). To compound several such forces we obviously have, using the former notation, U=1.{PI), V=^Pm),W^1{Pn) . . . (16). L = ^P\),M=l{Pi.), N^^Pv) .... (17). But it is clear that the relation LU+MV+NlV^o . . (18), corresponding to (14), is fwf now necessarily true. It may be shown later that when (18) holds then either (i) R=o, (ii) G=o, or (iii) the couple can be made zero by shifting R. (See ends of articles 415 and 420.) 414. Moment of a Force about a Line. — It was stated at the end of article 411 that L, M,-Na.x& called the moments of the forces about the axes. Let us now examine how this terra agrees with the definition of the moment of a vector with respect to a point. The latter was defined as the product of the vector and the perpendicular upon it from the point (article 25a). The same definition can obviously be extended to the moment of a vector about a line, if that line passes through the point previously named and is perpendicular to the plane containing the point and the vector. Thus in dealing with coplanar vectors we might speak indifferently of their moments about given points in their common plane, or about axes through those points and perpendicular to this common plane. But,, when the vectors and the axes are inclined, further definition is needed. Thus, referring to Fig. 214 of article 411, what is the moment about OZ of the force P of components X, V, and Z applied at the point {x,y, z) ? Let the moment of P about OZ be the algebraic sum of the moments of its components about the same axis, and con- sider as zero the moment of a vector about a parallel line. Thus the moment of Z about OZ vanishes, and the moments of Fand Xare seen to be + Yx and —Xy respectively. For these components lie on the plane parallel to X'OY and distant z from it, and this plane is intersected at right angles at (o, o, z) by the axis OZ. Let two inclined vectors P^ and P^ be compounded and the moments be taken about OZ of the separate vectors and of their resultant. Then in each of the three cases the Z components give no moment and, as already seen for the plane case, the moment of the resultant equals 4i6 ANALYTICAL MECHANICS [art. 414 the algebraic sum of the moments of the components. Hence adding the moments on this plan will correctly give that of their resultant. And what holds for the axis OZ will apply to any other axis. Consider again the single force P, and let it make the angle ^ with OZ ; then the resultant of X and Y will be the component of P parallel to the xy plane, and will be P^va )p ; hence yx—Xy=Ppsin(l> (19). We may thus enunciate generally as follows : — Definition. — The product Pp sin ^ is the moment about any straigh line AB of the vector P localised in a line inclined at the angle <\> to AB, the shortest distance between the lines being/. The usual relation between rotation and translation is to be observed in fixing the sign of the above product. If the line in which the vector is localised is called CD, it is obvious that the quantity / sin ^ has reference to those lines only, and that the numerical value of the moment is the same however P acts along one line, the other being the axis of the moment. Thus the product/ sin <^ may be called the moment of either line about the other, or their mutual moment. Thus the principal couple G, of a system of forces, may be seen to be the algebraic sum of the moments of all those forces about the line which is the axis of G. A more formal proof of this is as follows : — If we keep to the sam origin, R is independent of the direction of the axes, for it is the resultant of all the forces as if transferred to the origin. But if, by a new set of axes with the same origin, we could obtain a different couple G" say, then R and G' would be equivalent to R and G, and therefore R, G, —R, and —G" would form a system in equilibrium. But this is impossible unless G"=G and their axes were coincident or parallel. That is, both in magnitude and direction G is indepen- dent of the direction of the axes if the origin remains fixed. Thus, since the direction of the axes is arbitrary, we may let the axis of X coincide with that of G ; then M=o, JV=o, and G and Z are identical. Hence G is the algebraic sum of the moments of all the forces with respect to the straight line which is the axis of G. Examples — LXXX. 1. Show how to reduce any system of forces acting on a rigid body to a single force and single couple. 2. If, in the reduction of the former question, the base is changed, show that the principal force is unaltered but that the principal couple usually is altered. 3. State what is meant by the six components of a system of forces, and express them in a second form. ARTS. 415-416] SOLID STATICS OF RIGID BODIES 417 4. Explain carefully what is meant by the moment of a force about a line, and show that the principal couple of a set of forces is the algebraic sum of the moments of all the forces about that line which is the axis of the couple. 415. Conditions for a Single Resultant. — Having seen how to reduce any system of forces acting on a rigid body to a force and a couple, let us now examine the conditions for these to give a single resultant, one force only. Obviously one set of conditions would be ^±oand(?=o"l , , or U, V, and WnoX. &\\ zero, hut L = M—N=oi • ■ V^oj- But a second set of conditions may be found. For, if both G and R are finite but at right angles, as in article 364, they may be reduced to an equal force shifted parallel to itself, as there shown. The condition that G and H are at right angles is obviously that the cosine of the angle between them shall vanish. But cos 9=l>^-\-mu.+nv H' G'^ Ji' G^ Ji ' G' Thus, the second set of conditions is analytically expressed by LU+MV-^NW=o\ ,. where U, ?^ and /^^F do not all vanish/ ' ' . ■ . ^^21;. When (21) is satisfied the reduction to a single force proceeds as in article 364. Thus the plane of the couple G is made to contain Ji and to consist of a force opposite but numerically equal to Ji and applied at the same point, and another force equal to R in magnitude and direction but distant from it by the arm GjR. Hence the couple is absorbed, R at the original point is annulled, and we have left only the equal force R transferred parallel to itself through the distance GfR. We have now arrived at the proof alluded to after equation (18) at the end of article 413. For, in (18) or the identical equation of (21), (i) if we have U= V= W=o, then R=o; (ii) if L=M—N=o, then G=o; and (iii) if neither (i) nor (ii) is fulfilled, i? and G are each finite but at right angles, and G may be absorbed by the shifting of R, as just seen. 416. Line of Action of Single Resultant. — Supposing there is a single resultant for a system of forces acting on a rigid body, let it be required to determine the equations of its line of action. We may conveniently find these by shifting the origin to O', whose co-ordinates are (a, b, c), writing the values for the corresponding moments Z', M', and N', and then equating them to zero. For the single resultant force acts through that new origin for which the couple vanishes. Thus, quoting (7) in article 411a, we have L'=L-Wb+Vc=o . . ■ (22). M'=:M-Uc-{-lia = o . . . . .(23). JV'=N'-Fa+Ub=o (24). 2 D 4i8 ANALYTICAL MECHANICS [art. 417 But, since there is a single resultant, we also have, as in (21), LU^MV-\-NW=o (25). And this shows that the former three equations are not all inde- pendent. For, if we eliminate c between (22) and (23), we obtain LU^-MV->rW{Va-Ub) = o (26), and (25) and (26) give (24). We are thus confined to two of the three equations of (22) to (24), say (22) and (23), and these give two relations for the three unknowns a, b, c, which accordingly may have an infinite number of values subject to these conditions. That is, (22) and (23) do not determine some one definite point O' of fixed co-ordinates (a, b, c), but define a locus of O'. Hence, writing the current co-ordinates x, j, and z instead of a, b, and c, we have L- Wy+ Vz=o\ , . M-Uz^Wx=o] • • • • • (27;. which are the equations of a straight line any point in which is an origin O' for which the couple G vanishes. In other words, these equations are those of the line of action of the single resultant force R to which the system was reducible. They may be thrown into the form x-\-MlW _ y-L\W _ z UjR ~ VjR ~ W/R ^^^^' showing that the line has direction cosines [//R, V/R, and W/R, as we know should be the case, and also that it intersects the xy plane at the point whose co-ordinates are {—MjW, Z/ W). Alii. Reductioa to Two Forces.— We have already seen that any system of forces acting on a rigid body may be reduced to a force and a couple, which, under certain conditions, may be a couple only, or even a single force only. But we may now notice that though these further reductions are particular, we may in all other cases reduce the system to two forces. We may also make one of them act at an assigned point and give to the other an assigned value. Thus, suppose the system has been reduced to the force R and couple G for the base or origin O. We may then consider the couple to be composed of any two unlike parallel forces F and —i^ distant G^/i^ apart, -7^ acting at O, and both in the plane through O perpen- dicular to the axis of G. These two forces may, of course, have any orientation we choose provided they are in the above plane We may then compound R and -F at the origin, giving the resultant S say. Thus the system has been reduced to two forces 6" and F. Further since the origin may be anywhere we choose, one of the forces S may be made to pass through any assigned point. Also, since the magnitude of i^was arbitrary, we may choose it to have any assigned value, though, of course, this choice modifies the magnitude and direction of the 5 thereafter determined ARTS. 418-419] SOLID STATICS OF RIGID BODIES 419 418. EoLuilibrium of a Body with One or Two Points Fixed.— Take first the case of a rigid body acted on by any applied forces and with one point fixed, which we will choose as the origin of co-ordinates. Then these forces will produce on the fixed point a pressure whose components may be denoted by X\ Y', and Z'. Hence, using the former notation, we have U-X' = o, V- F=o, W-Z' = o .... (i). L — o, M=o, N—o . (2). Thus, (i) gives the pressure components at the fixed point, and (2) gives the three conditions of equilibrium, viz. that the moments of the applied forces shall vanish with respect to any three rectangular axes meeting in the fixed point. It is easily seen that these three conditions correspond to the three degrees of freedom possessed by the body. Take now the case in which tivo points are fixed. Let the axis of z pass through both points, their distances from the origin being z' and z", and the components of pressures being X', V, Z', and X", Y", Z" . Then we have U-X'-X"=Q,V-Y'-Y"=o (3). W-Z'-Z"^o (4). L-\-Y'z'-^Y"z=o,M-X'^-X"z=o (5). N=o (6). Hence the four equations (3) and (5) determine the components X', Y, X", Y" of the pressures on the fixed points, while (4) gives the sum of Z' and Z", it being impossible to discriminate between them for an ideally rigid body. Finally, (6) gives the sole condition of equilibrium, viz. the vanish- ing of the moment of the applied forces about the line through the two fixed points, which obviously corresponds to the sole degree of freedom left to the body. 419. Equilibrium of a Body with Three Points on a Smooth Plane. — Let the plane be that of xy, the co-ordinates of the points of contact being (*',/, o), («", /, o), {x"',y"', o), the corresponding pressures of the body on the points bemg Z\ Z", and Z". Then the equations are , , U=o, V=o (?)• lV-Z'-Z"-Z"'=o . . (8). z-zy-z"/-z"'y"^o\ / ) M+z'x'+z"x"+z"'x"'=oj ■ • ^^' N^o .... (10). Hence, of these six equations, the three contained in (8) and (9) determine the pressures Z', Z", Z" exerted by the body on the plane; while the other three numbered (7) and (10) form the conditions of equilibrium corresponding to the three degrees of freedom still left to the body. 420 ANALYTICAL MECHANICS [art. 420 EXAMPLES^LXXXI. 1. If a set of forces applied to a rigid body have the components U, V, and fF parallel to the co-ordinate axes and the moments L, M, and TV about them, show that they all reduce to a single force if LU+MV+NIV = and L, M, and TV' are not all zero. 2. When a general system of forces may be reduced to a single resultant force, find the equation to the line of action of that resultant. 3. Show that 'any system of forces in three dimensional space may be reduced to two forces if not to one. 4. State the conditions of equilibrium of a rigid body with no points fixed, and show to what these reduce if two points are fixed. 5. If a rigid body is constrained to have three points in contact with a plane, obtain the pressures on these points under any set of applied forces, and. find also the conditions of equilibrium of the body. 420. Reduction of Forces to a Wrencli. — We have already seen (article 411) how to reduce a set of forces to their six components Z, M, JV, U, V, W a.nA thence to a couple G and force £. It is now desirable to note how they may be still further reduced to a single force and a couple in a plane perpendicular to {i.e. with axis parallel to) the direction of the force. Such a combination is known as a wrench, this use of the word being due to Sir R. Ball. Let the axis of G be inclined at the angle 6 to the direction of S as shown in Fig. 215. Resolve G into its rectangular components OQ = Gcos^and OS=GsinS, with axes along and perpendicular to the direction of i?. Further, let the component couple OS be repre- sented by forces —RaXO and Ji' at O', each parallel and nu- merically equal to £. Then OO' is obviously perpendicular to the plane of R and the axis of G, and its length is {G sin 6)1 R. Further, R and —R each applied to O annul each other, so we are left with R at O', equal and parallel to the original i?, together with the couple OQ=G^ cos 6, with its axis parallel to the original R. Since the axis of a couple may be regarded as shifted to any parallel line, we have thus a single force R at O' and a couple, G cos ^=r say, with axis coincident with the line of action of this force. We have accordingly reduced the forces to a wrench, as was desired. The axis O'R', to which the force R has been transferred, is called Foinsofs Central Axis. It is obviously constructed as the line through O' parallel to^?, where 00' is perpendicular both to R and to the axis of G, of length 00'= (G sin 6)1 R, and of direction such that the couple G sin 6 would carry O' in the direction of R'. Fig. 215. Reduction to Wrench. ART. 420] SOLID STATICS OF RIGID BODIES 421 And the wrench, to which we have reduced all the forces, is the forced along this central axis, together with the couple V=Gcosd about the central axis. Since the cosine of the angle between two lines is the sum of the products of their corresponding direction cosines, we have for the angle 6 between R and G IiV = RGco5e=LU+MV-\-NW=I . . . . (i), where / is an invariant for the given forces. For, however the base is changed, U, V, and ff remain unaltered. And if L, M, and Ware changed to Z', M\ and N' by shifting the base, it may be seen (from equation (7) of article 4na) that the above sum of three products is not changed thereby. Thus for /=o we have either R=o or r=o, which is another proof of the statement at the end of article 413. To obtain the equations of the central axis, take any point (x, y, z) on it and form the expressions Z', yl/', N' for the moments of the forces about parallel axes through this new origin (see equation (7) of article 41 la). Then, since the central axis is parallel to R, these moments are proportional to U, V, W, the components of R. The symboUc ex- pression of these relations gives the equations desired for the central axis, viz. L- Wy+ Vz _M- U z+ Wx _ N- Vx+ Uy ^ I ,^. U ~ V ' W R'- ' ^ '' The final expression on the right is obtained by multiplying the' three on the left by U°; V\ and W respectively, adding, and remem- bering (i). If U, r, and IV all vanish, (2) fails to give the equations sought. But in this case R is zero, and any straight line parallel to the axis of G is the central axis. We may now show that the couple of the wrench has the minimum value for the principal couple. Let any base be chosen, and denote by G' and Z', M', N' the corresponding principal couple and its components. Then we have already seen that L'U-k-M'V+N'W^LU+MV+NW^I . . (3)- We may also write R'G'={u"-+ r-+ w'){r'+M"+ir'-). And, by using (3), this may be thrown into the form R'G" = {N- V-M' ivy + {L' iv-N'uy + {M'U-L'VY- + I' .... • (4). But as both R and / are invariants, the minimum value of G' obviously corresponds to the vanishing of each of the three bracketed squares of (4)- Hence by ( i ) R'GL^=^r-^R'r-,orG'^^.^V (5)- 452 ANALYTICAL MECHANICS [arts. 42* -432 421. Tripod by Virtual Work.— Let us now apply the principle of virtual work to the solution of a few problems of equilibrium of bodies or systems in three dimensions. Take first the case of a weight W hangmg D from the freely jointed summit of a tripod ABCD, /^ Fig. 216, whose legs have each the length a, / l\ and whose feet touch a smooth horizontal plane / I \ at the corners of an equilateral triangle of side / 11 \ s, and are there bound by a string whose tension / ^j\^^ r is to be found. /,^---'^jr. /U All being symmetrical, it is evident that the A^^^^lII^^y V/ vertical from D meets the triangle ABC at the B point F, where AF is 2/3 of AE and F is the FIG. 216. Tripod by "^i^dle point of BC. Thus AE=x J3/2 and Virtual Work. AF=s/ J 3. Denote by A the height FD, then we have /i'-+sy5=a' (i). Hence, on differentiating, we obtain ^Mk+sds—o ... . . (2). But, by the principle of virtual work, we have lVd/i+T{3ds) = o ........ (3). Thus the comparison of coefficients in (2) and (3) gives W/3A = 3T/s, lV^'^^g^'^/3 ' ■ ■ ^^'' Pile of Four Equal Spheres. — Let us now imagine three equal smooth spheres in contact on a smooth plane and encompassed by a fine thread in contact with them in the horizontal plane through their centres, the thread having no tension until a fourth equal smooth sphere of weight W\s, placed on the other three and in contact with them all. It is required to"determine the tension T' of the thread. This problem, so different in the form of the bodies composing the system, is easily seen to be essentially the same as the tripod just treated, except that now a=s. Thus (4) yields for this case r _ I 422. Bifilar Suspension by Virtual Work. — Let us now consider the relation between the couple G about a vertical axis and the angle of twist which it can maintain in a bifilar suspension, whose equal threads have length a, the distance between their fixed points of sup- port being 2b, and the mass they carry being M. The bifilar suspen- sion is shown in Figs. 217-219, the full lines representing the displaced position and the dotted lines the equilibrium position. When the bar of the bifilar is turned through the angle B about a vertical axis, let its depth below the fixed points CD be decreased from ART. 422] SOLID STATICS OF RIGID BODIES 423 a to h. Next imagine the virtual displacements dd and dh, then we nave the relation Gde+Mgdh = o (i), which states for this case the principle of virtual work. It is accord- ingly the only mechanical relation needed, the others required being Fig. 217. Side Elevation. Fig. 218. Edge Elevation. (C) D 26 C D % I I t I I a In !___i___5-, Fig. 219. Plan of the Bar. F{B) Three Views of Bifilar Suspension. 424 ANALYTICAL MECHANICS [art. 422 obtainable from the geometrical conditions. In deriving these it is convenient to use the angle with the vertical assumed by the threads in their displaced positions. It should be noted that this angle is not seen at its true value in either of the elevations, for in neither case IS its plane coincident with that of the diagram. In the figures, AB denotes the equiUbrium position of the bar, EF its position if raised {a—h) parallel to itself, PQ its position in the actual displacement in which it turns through the angle 6 about the vertical through its centre and at the same time rises vertically by (a—h), the threads swinging till at the angle <^ with the vertical in oblique planes. Then by the plan, Fig. 219, we have EP = FQ = 2^ sine/2, and by it and the elevations we see that EP=FQ=flsin^. Hence a sin <^= 2^ sin 5/2 (2). Again, by the elevations, we have h^a cos 4' (3), which completes the required geometrical equations. We must now eliminate <^ between (2) and (3), differentiate the equation so obtained, and use the result in (i). We thus find in turn h''=a'-a' sinV=fl'-4'^' sin" 6 j 2, n n and 2hdh=—i,V 2 sin — cos — {\dB), Ai. ^° sin Q or dh= . do . . . (4). And this in (1) gives the exaci result ^^_^ JfgP sine ., a/ a" —46^ sin' 2 r.i«S 2 Whence, when 26 sin — is small compared with a, we have the 2 approximate result Q^JMpl (g^ For the work done in twisting the bifilar through the angle /8 from the equilibrium position, we have from (6) the approximate value W Jo ('Gde=^i-cosP). . . (7). Jo "^ ART. 422] SOLID ST A TICS OF RIGID BODIES 425 Examples— LXXXII. 1. Reduce any system of forces to a wrench, and show that the couple then involved is the minimum foi' the system in question. 2. Apply the principle of Virtual Work to obtain the couple required to main- tain any specified angular displacement of a bifilar suspension. 3. A horizontal bar of radius of gyration k about a central vertical axis is suspended by two parallel threads each of length a and at a distance ib apart. Show that if slightly disturbed the bar will oscillate in the period given by T= -7-27r'V— ■ b y g 4. ' (i) A bar AB, of weight IV, is guided by rings at its ends so that A can move on a smooth horizontal rail OX, and 5 on a smooth vertical rail O y. Employ the principle of Virtual Work to evaluate the horizontal force at A necessary to maintain equilibrium when the angle OAB is 6. ' (ii) A nut of weight IV is mounted on a fixed smooth screw of pitch /, whose axis is inclined to the vertical at an angle a ; what couple is required to keep the nut from moving ? ' (LoND. B.Sc, Pass, Applied Math., 1908, l 7.) 5. 'Three equal spheres are lying in contact on a horizontal plane and are held together by a string. A cube of weight W is placed with one diagonal vertical so that its lower faces touch the spheres, and the cube is supported in this position by the spheres ; show that the tension in the string is (LoND. B.Sc, Pass, Applied Math., 1908, i. 8.) 426 ANAL YTICAL MECHANICS [art. 423 PART v.— HYDROMECHANICS CHAPTER XIX HYDROSTATICS 423, Natures of Fluids, Liquids, and Gases. — In this, the fifth part of the present work, we consider the rest and motion of fluids, forms of matter offering only very small resistances to changes of shape how- ever large, provided only that time enough is allowed in which those changes may occur. This distinguishes them from rigid bodies which are supposed to retain their exact shape under all circumstances, and from elastic solids each of which exhibits a certain nearly constant ratio between its almost instantaneous changes of shape and the external actions to which those changes are ascribed. If the resistance offered by a fluid to even a sudden change of shape is quite negligible, the fluid is said to be very mobile or of negligible viscosity. This last term denotes a property really possessed by all fluids and which, in the case of sufficiently sluggish fluids, needs taking into account according to a quantitative definition. It is obvious that in the present chapter on the statics of fluids we are not concerned with viscosity, for we examine the equilibrium state after sufficient time has been allowed for all motions to cease. And in the next chapter on the motions of fluids we shall, for simplicity's sake, exclude all notions of viscosity (as being negligible) except where it is expressly introduced. We may now subdivide fluids, discriminating between liquids and gases. Liquids are fluids whose volume per unit mass are practically independent of the pressures to which they are subjected ; in particular, the specific volume is finite when the pressure is almost at zero. In other words, the density of a given liquid is practically constant and its specific volume always finite. Gases are fluids whose volumes per unit mass may become as large as we please by our suitably diminish- ing the pressure to which they are subjected. In other words, the density of any gas is a fairly simple function of the pressure such that its specific volume has no finite limit. We may accordingly sum up the chief points of distinction by the serai-popular remark that a solid body has both size and shape, a given mass of liquid has size only, while for a given mass of gas there is neither shape nor size. Hence a solid body requires no vessel to hold it, a liquid requires no lid to the vessel, but a gas needs both vessel and lid, or it would expand to fill all the exterior space open to it. Arts. 424-425] tJYDRdSTATtCS 427 Z Fig. 220. Hydrostatic Pres- sure Independent of Direction. In the case of gases near the change of state called liquefaction, the relations between pressure and volume are often much more compli- cated than the ideal simple forms which ordinarily represent them with sufficient accuracy. The fluid is then called a vapour. But with the details of these complexities we are not here concerned. They must be studied in physical rather than in mechanical text-books. 424. Hydrostatic Pressure Independent of Direction.— In a fluid at rest take a small triangular pyramid OABC bounded by the three rectangular co-ordinate planes and a base ABC of area A whose perpendicular distance from the origin has length ^ and direction cosines /, ot, n. And consider the equilibrium of this pyramid under the forces X, Y, Z per unit mass parallel to the axes, the normal pressures P, Q, R on the mutually rect- angular faces meeting at O, and N on the oblique base, as shown in Fig. 220. It should be noticed that whether the fluid has appreciable viscosity or not, these pressures must be normal since the fluid is at rest. Then, by the geometry of the figure, the areas of the three mutually rect- angular faces are respectively /A, wA, «A (since their inclinations with the base are those between the axes and the normal h) ; also the mass of fluid in the pyramid is Jp,^A where p denotes the density. Hence, taking components parallel to the axes, we have Pl^-~Nl^+\phx^=o (0 and two similar equations. But when, to make the pressures F, Q,R, and A^all act at the same point O, the pyramid is indefinitely reduced in size, its volume ^hl\, being of the third order of small linear quantities, vanishes in com- parison with A, which is of the second order only. Hence, each equation reduces to its first two terms, and the set simplifies to P=N\ Q=N\ . Or in words, the hydrostatic pressure of a fluid in equilibrium on any small surface through a given point acts normally to that surface but is otherwise independent of its aspect. Or again, the normal of the surface is the direction of the pressure, its magnitude rematmng the same for all orientations of that surface. 425 Fundamental Equations.— Consider a small parallelepiped of edges a, /?, and y parallel to the co-ordinate axes, and let it be occu- D ed by fluW of density p in equilibrium under the pressures and the forces^ Y and Z per unit mass. Let +p be the positive pressure (2). 428 ANALYTICAL MECHANICS [art. 426 on the /8y face at the negative end of the element and — /' the corresponding pressure on the opposite face. Then, taking all components parallel to the x axis, we have the condition of equi- librium j>Py-f'j3y+pafiyX=o ... (3). 'Rui p' =p-{-adpldx, axiA accordingly /—/'=— o^/rfj;. Thus (3) and the two similar relations for the other axes yield the set ^-1= ^-1= (4). These are the fundamental equations of hydrostatics, and show that the space rate of increase of pressure in any direction equals the applied force per unit volume in the same direction. 426. Pressure and Depth in a Liquid. — For the case of a liquid of constant density p in equilibrium under gravity only let us take the z axis vertically upwards. Then we have X^Y=o,Z=-g . (5). And these, put in (4) of the last article, give dp dp . d^=dy^^ (^>' ^"'^ fz^-P^ (7). Equation (6) shows that the pressure is constant at all points in any horizontal plane. It therefore agrees with the common statement that ' water finds its own level ' ; or the famihar experiment of Pascal's vases, in which the liquid reaches the same level in any limbs of a complicated vessel of communicating parts of bulbous and other forms. Since the density p is supposed constant for our liquid under all moderate pressures, equation (7) yields / dp=-pg\ dz, or p-po=Pg{z<,—z) = pgh (8), where /» is the pressure at some standard height z„ (say the free surface of the liquid), h being the depth of z below 0„. If either p or g vary, or both, (8) would need modifying by keeping these variables under the sign of integration and dealing with them as functions of z. Since the pressure, or force per unit area, is quite independent of the area, the ratio between forces on given movable surfaces in contact with the same liquid may be magnified as much as we please by correspondingly magnifying the ratio of the areas of those surfaces. ART. 427] HYDROSTATICS 429 Thus if the same pressure/ is exerted by a liquid on the plunger of a pump of radius a and on the ram of radius 3 in a hydraulic press, the corresponding forces being P and Q, we have P\Q.=pa-lpF- = a^lb'' (9). Hence with alb one hundredth, we have PjQ. one ten thousandth. Or one pound weight on the plunger gives a force of 10,000 lbs. wt. (or nearly four and a half tons weight) on the ram. The fact that, by means of an intervening fluid, a small force can be made to balance a much larger one is often referred to as the ' hydrostatic paradox.' The above expression given in (9) is, of course, easily obtained on the principle of virtual work by the kinematical relation between the two corresponding displacements possible to the plunger and ram connected by the liquid, supposed incompressible. The principle of this article as contained in equation (8) has many applications scientific, technical, and familiar. Thus the method of determination of densities of liquids by balancing one against another in a U-tube, the use of siphons and pumps, may be mentioned here, but call for no detailed treatment. 427. Resultant Force on a Submerged axes of -v and y in the upper surface of the liquid and that of z vertically downwards as shown in Fig. 221, ABCD being the area inclined at an angle B with the vertical. Take a point E in the plane at a depth FE=5 below the free surface of the liquid, and through E take the horizontal line BD in the plane. Take also in the plane a parallel line at a depth z-\-dz so as to cut off a slice of the plane of width dzjcos 6. Then, if the length of BD is denoted by /, we have, as the force due to the liquid pressure on one side of this element, the expressions Plane Area. — Take the Fig. 221. Forces on Plane Area. Ids -.6 (i). the pressure /o being that on the free surface of the liquid. To obtain the resultant force on the whole surface due to the liquid we have simply to integrate (i), because since the area in question is plane, all the forces of the liquid pressures on it are parallel. We thus find P=[(po+Pg^)~-0=AS+PgSA . . . (2). where c and a are the limiting depths of the plane ABCD, S its area, and h the depth, JG, of its centroid G below the level of the free surface of the liquid of density p. . . „ i, • , Where />„ is zero or negligible the expression for H obviously 430 ANALYTICAL MECHANICS [art. 428 reduces to its last term, which gives the pressure due to the liquid only. It is worth noting that the magnitude of this term expresses the weight of the column of liquid that would stand vertically on the area ABCD if rotated into the horizontal plane through its centroid; the direction of R is, however, normal to the plane in its actual position as shown. The vertical component of R is evidently R^vcvd=pi^S^va.Q-\-pgSh^vi\Q (2a), showing its value to be that of the weight of the vertical columns of fluid standing on the area in question in its actual inclined position. Examples— LXXXI 1 1. 1. Discuss carefully the distinction between solids and fluids and that between liquids and gases. 2. Show that the hydrostatic pressure of a fluid at a point is independent of the orientation of the surface on which it presses. 3. Obtain the fundamental equations of a fluid in equilibrium under speci- fied forces, and apply them to the state of a fluid at rest under the action of gravity only. 4. Find the relation between pressure and depth from the free surface of a heavy liquid. 5. Obtain an expression for the resultant force on a plane area submerged in a heavy liquid and also one for the vertical component of this force. 6. ' Prove that the average thrust per unit area of a liquid on a plane area immersed vertically is equal to the pressure intensity at the centroid of the area. ' The water upon one side of a dock gate 15 feet wide is 10 feet deep and upon the other side is 20 feet. Taking the gate as rectangular, find the resultant thrust of the water and its line of action.' (LoND. B.Sc, Pass, Applied Math., 1905, in. g.) 7. ' If at any point P in a perfect fluid a small plane area is imagined as separating fluid on one side from fluid on the other side of the area, and if the direction of the force exerted over this area by the one part of the fluid on the other is always normal to the area, whatever be the aspect of the area at P, prove that the magnitude of the force is constant for all positions of the area. ' A canal lock gate is 12 feet broad, and the depths of the water at oppo- site sides of the gate are 16 and 10 feet ; find, in tons weight, the magnitude of the resultant water pressure on the gate, assuming that a cubic foot of water has a mass of 62"5 lbs.' (LoND. B.A., Pass, Applied Math., 1906, 11. i.) 428. Centre of Pressure. — We have just found the resultant force of the liquid pressures on a plane area and know its direction. We have now to find a point on the line of its action. This point, if taken in the plane itself, is called the centre of pressure. Obviously its position sideways, or in the horizontal direction, is simply that of the centroid of a lamina of the same shape as the plane area in question, but of surface density proportional to the depth below the free surface of the liquid. We are therefore concerned now simply with the depth z of the centre of pressure below the liquid surface. To find this we consider moments of the horizontal components of the forces about the surface of the liquid, which we still take as the }cy plane, referring ART. 429] HYDROSTATICS 431 again to Fig. 221. Draw any vertical plane whose intersection with ABCD is a horizontal line, then .S cos 6 is the area of the projection of ABCD on this vertical plane, and our horizontal components will be perpendicular to this vertical plane. We shall also need symbols for the radii of gyration of this projection about horizontal lines in the plane of projection. When this line or axis is in the free surface of ■ the liquid, or xy plane, let .^be the radius of gyration, k denoting the corresponding value for the parallel axis through the centroid of the projection. Then, writing H for the horizontal component of pressure on the whole surface and M ior its moment about the xy plane, we find and Jl—pgi zldz=pgShcosd .... M=pg{\^ldz-pgSK'' cos e . . . Thus, for the depth of the centre of pressure, we have M A" ' h )i'+k- -.h+- (3), (4). (5). is here The pressure, if any, on the free surface of the liquid supposed negligible. We accordingly find that the depths of the centroid and of the centre of pressure of a plane area below the free surface of the liquid are related to each other like the corresponding distances of the centroid and centre of oscillation of a physical pendulum from the axis of sus- pension (article 258). 429. Resultant Force on a Closed Surface.— Consider any closed surface 5 described in a liquid at rest in equilibrium as shown in Fig. 222, in which the axes of x and y are taken horizontally in the free surface of the liquid and that of z vertically downwards. Take, in the volume enclosed by S, any prism AB with axis horizontal ^nd of an infinitesimal cross section. Let the pressure at this level be/, and denote by dS the area of the inter- section of the prism with the surface ^ at A. Then the force on this base of the prism due to the liquid pressure is pdS, and, if its normal is inclined 6 with the horizontal, the horizontal component of this force is pdS cos 6. But ^6' cos 6 is the area, da- say, of the normal cross section of the prism, so the horizontal component of either end force is the symmetrical expres- cinruiz/o- Hence the horizontal component of the inward forces on 5 at A is opposed by one of equal magnitude at B so tha the resultant horizontal force on this prism is zero. And 1. fhis appS to any horizontal prism parallel to the axes of x or y, or Fig. 222. Resultant on a Closed Surface. 432 ANALYTICAL MECHANICS [art. 429 oblique to them, and at any level, it is clear that the horizontal com- ponent of the resultant force on the entire closed surface S vanishes. Consider now, in the closed surface S, any vertical prism CD, and produce it to E in the free surface of the liquid. Then, by what we found at the end of article 427, the vertical components of the forces due to the liquid at C and D are each equal to the weights of the columns of liquid which could stand on C and D. Hence, taking these forces inwards on the closed surface S, we see that they are in opposite senses, and have for their resultant a vertically upward force equal to the weight of the column of the liquid CD extending between the lower and upper limits of S along the vertical prism in question. And this is true whether there is one liquid only, or two liquids, or more round the surface 6'. Accordingly, the resultant of all the inward forces on the closed surface S, due to the pressure of the surrounding liquid (or liquids) at rest in equilibrium, is a vertically upward force, equal to the weight of the liquid (or liquids) occupying the interior of S, and acting through the centre of gravity of the liquid (or liquids) so contained. The centre of gravity of the liquid (or liquids) thus contained by the closed surface S is called the centre of buoyancy of 5 under the circumstances in question. Corollary i. — If the closed surface 5 be that of a solid body intro- duced into the liquid, it is evident that the body will be subject to the resultant force just described, which is now the weight of liquid (or liquids) displaced by the body. Usually some other force is required to keep the solid in equilibrium beside that of the liquid pressures. This may be supplied by a thread attached above and from which the body hangs, if it is denser than the liquid. If, on the other hand, the liquid is denser than the body, then the latter may be tethered down by a thread attached below, or the body may be held by a sinker or cage of denser material suspended from above. Such devices are used when finding the densities of bodies by the hydrostatic balance. Thus, a body first in air and then in water is balanced each time by weights in air, and the difference in grams gives its volume in c.c. (nearly). Then the density is found as the quotient, mass divided by volume. Further, by taking into account the weight of the air displaced by bodies and by the weights of the balance, the weighings may be reduced to vacuo when extreme accuracy is desired. Corollary 2.—\i the closed surface 5 were at a great depth in an incompressible liquid of small density, we might have a considerable pressure, but differing only very slightly at the upper and lower limits of S, the weight of the liquid displaced by S being correspondingly small. Hence, if the depth of 5 below the free surface were continually increased while the density of the liquid were proportionally diminished, we might maintain the pressure finite while the weight of liquid dis- placed by 5 and the difference of pressures at top and bottom of S each vanished. We may accordingly state that the resultant of any uniform normal pressure on any closed surface is zero. ARTS. 430-431] HYDROSTA TICS 433 "t M^£=2-- H Fig. 223. Floating Body. Corollary 3.— Knowing the resultant force for a closed surface and lor a plane surface we can deduce that for an unclosed surface, if closable by a plane; e.g. the curved surface of a cone. 430. Floating Bodies.— Let a body float in equilibrium partly immersed in one Hquid, the upper portion of the body being in an- other liquid or fluid, all in equilibrium as in Fig. 223. Then the body is in equihbrium under the action of two resultant forces : — (i) its own weight IV acting vertically downwards through G, the centre of gravity of the body, and (2) the force R equal to the weight of fluids dis- placed and acting vertically upwards through the centre of buoyancy H. Hence we have ,^=^ (I). also G and H are in the same vertical line (2), as the conditions for equilibrium: for these ensure that final resultant forces and torques are each zero. If the upper fluid be air and the lower any liquid, it is often near enough for practical requirements to neglect the part of R contributed by the air. We then obtain the special forms of (i) and (2) known as the Principle of Archimedes. The hydrometers of fixed or variable immersion are scientific devices for obtaining the densities of solid and liquids by application of this principle. 431. Stability of Floating Bodies: Metacentre. — Having seen what are the conditions for the equilibrium of floating bodies, it is now desirable to examine the stability of that equilibrium and the circum- stances on which it depends. Any shift of a floating body can be regarded as made up of translations and rotations. The only translation with which we can be concerned here is a vertical one, and for this the equilibrium is obviously stable. We turn therefore to the question of the stability of a floating body when slightly tilted but without change of the volume V of liquid displaced, which may be called its displacement. Let this tilt occur in the plane of the diagram Fig. 224, which may be regarded as a cross-sectional elevation of a boat. Fig. 225 shows a sectional plan of the same boat at the water-line, AKBL being called the surface of flotation, whose area we shall denote by S. Instead of drawing the boat twice, in the equilibrium and tilted positions, it is shown once only, viz. upright, the water-level being shown horizontal by a full line AB, corresponding to the boat's equilibrium position, and again by the broken line A'B', inclined dB, corresponding to the boat's tilted position. Thus, since the volume displaced by the boat is to be the same in each position, the wedge of 2 E 434 ANALYTICAL MECHANICS [art, 431 d,e Fig. 224. Cross- Sectional Elevation OF Boat. Fig. 225, Sectional Plan of Boat. ART. 43 1 ] HYDROS TA TICS 435 immersion FEB' must be equal in volume to the wedge of emersion FAA'. This circumstance serves to define the position of F where the wedges meet. For, take the origin at F, and let the positive direction of X be from F towards B, and at the distance FP=x take a small vertical prism of horizontal cross section dS and extending from P to P', the height {x tan dd) of the wedge of immersion at the place. Then the volume of this prism will be tan dO.xdS. And, when we pass to the wedge of emersion, x will be negative and make the volume negative. Hence, by integrating the above expression over the whole surface of flotation (AKBL, Fig. 225) we obtain the addition to the displaced volume F" caused by the tilting. But this addition is zero, accordingly we have o = tan dd 'I xdS^iandO.Sx (i), where x is the distance of the centroid of .S from F. We thus see that '^■ = ° (2), or the centroid of 5 is on the line through F perpendicular to the plane of Fig. 224. In other words, the wedges of immersion and emersion meet on a line passing through the centroid of the surfaces of flotation. We shall let F represent this centroid in each figure. Let H and H' be the respective centres of buoyancy in the equilibrium and slightly tilted positions, and let the verticals through them in each case meet in M, then M is called the metacentre of the body for the type of tilt in question, i.e., in the present case, for rolling. The metacentre must be located to determine the behaviour of the body for the tilts under consideration, and its position depends only on the form of that part of the body which is immersed. But to determine the stability of the body, when loaded so as to sink to the given mark, a knowledge of the position of G, the centre of gravity of the floating body, is also required. For, in the tilted position, we evidently have a force equal to the weight W of the body acting down through G parallel to MH', and an equal force Vpg acting up through H' along H'M, and due to the displacement of a volume V oi liquid of density p. These forces form the restoring or righting couple, if G is below M ; the couple vanishes if G coincides with M, while the couple tilts the body farther if G is above M. To locate M we may conveniently find HM. To do this consider the body in the tilted position and, about the axis of tilt through F, take the moments of the buoyancy due to the liquid displaced. We may write two expressions for this moment, one regarding the total force Vpg acting through H', and another regarding the volume displaced in the tilted position as made up of (i) the original yolume in the equilibrium position, (ii) minus the wedge of emersion FAA', (iii) plus the wedge of immersion FBB'. But the wedge of emersion has a negative moment about F, so taking this wedge away adds a term to the moments, just as adding the wedge of immersion does (since its 436 ANALYTICAL MECHANICS [art. 432 moment is positive). We thus obtain the following equation of moments : — - yp^HFsmdd+pglSinddjx'-dS+pgtsLndelx'-dS^ Vpg{HM-llF)smde Whole displacement Wedge of Wedge of Whole displacement, in equilibrium position. emersion. immersion. in tilted position. or tan dd 'rx°-dS=VHM.sinde (3). Let us now denote by / the moment of inertia of the surface of flotation AKBL about the axis of tilt KFL. Then fz= I x'-dS (4). Hence using this in (3), and in the limit sinking the distinction between tan dd and sin d6, we obtain HM=//F (5). This distance HM is sometimes called the metacentric height, but it is to be noted that GM is the height upon which the stabiUty depends. So it is perhaps safer to avoid the vague phrase meta- centric height where any confusion might occur. If HG is known to be less than the value given by (5) for HM, or if HG can be adjusted so as to be less than HM, then G falls below M, and stability is attained. Since for any initial infinitesimal tilt H cannot vary in height, it is evident that M is the intersection of consecutive normals to the locus of H, which locus is called the surface of buoyancy. In other words, the metacentre for any plane of tilt is the centre of curvature of the cor- responding section at H of the surface of buoyancy. 432. Practical Determination of Metacentric Height. — On ships the height GM may be found conveniently by shifting a weight across the deck, or, what is practically the same thing, alternately filling with water two boats at opposite sides of the deck. The consequent inclina- tion is found by observing the shift of a plumb bob on a string of known length. With the movable weight at one side, let the floating body be at rest in the symmetrical position ; then its centroid is at G, Fig. 224. Call the total weight W, and let the inclination dB be pro- duced by shifting the weight i/fF across through a distance a. Then the centroid of the whole body floating must have shifted from G to some point G' in MH', in order that the weight acting at G' and the equal buoyancy at H' shall act along the same straight line. Hence the change of moment of the floating body's weight may be regarded in either of two ways : — (i) dW.a cos dO ; (ii) f^GM.sin dO. Hence, remembering that the cosine of a small angle may be written unity and its sine assimilated to its circular measure, we have _., adW ,,, ^^-IVdQ (^>- ART. 43^] HYDROSTATICS 437 As an illustration of this method, we may take the following numerical example : — Let J^= 5000 tons weight, dW^^o „ „ a = 50 feet, dd= 1/20. Then GM = -S°X2o ^ ^^^^ 5000 X 1/20 It should be borne in mind that the theoretical treatment obtains the height HM of the metacentre above the centre of buoyancy, whereas the practical method determines the height GM of the metacentre above the centre of gravity of the ship as then loaded. Thus the former result holds for the given floating body every time it is sunk to the place in question, while the latter result varies also according to the arrangement of the loading, but is the vital criterion of stability for that loading. Examples— LXXXIV. 1. Define centre of pressure of a plane area submerged in a heavy liquid, and obtain an expression which locates this point. 2. Show that if a closed surface be described in a liquid at rest under gravity, the resultant of the inward pressures on this surface is numeri- cally equal to the weight of the liquid inside that surface and acts upwards along the same line as that weight. If such a surface is that of a solid, what follows ? 3. Obtain an expression for the height of the metacentre above the centre of buoyancy of a floating body. 4. ' State and prove the principle of buoyancy. 'A solid cone is floating in water with its axis vertical and vertex down- wards. To cause it to sink until 3/4 of its axis is immersed requires a load of 50 grams on its base ; and to cause 4/5 of the axis to be immersed requires a load of 96 grams. Show that the specific gravity of the body is very nearly o'324.' (LOND. B.A., Pass, Applied Math., 1906, n. 3.) 5. 'Two liquids which have different densities and do not mix are poured into a vessel. Prove that their surface of separation is a horizontal plane. ' A solid cylinder of specific gravity 07 floats with its axis vertical in a vessel containing two liquids whose specific gravities are o-6 and 0-9, the cylinder being completely submerged ; how much of its axis is in the upper fluid ? 'Explain how the upper fluid contributes towards the upward force on the body.' (Lond. B.A., Pass, Applied Math., 1906, 11. 4.) 6. ' Show how the depth of the centre of pressure on any given plane area in a liquid is calculated. ' A circular area of radius r whose plane is vertical has its highest point in the surface of water, and its centre of pressure is at a depth rl\ below the centre of the circle. Prove this by considering the separate equilibrium of the hemisphere of water standing on the given circular area, having given that the centre of gravity of a homogeneous hemisphere is 3^/8 from the centre.' (LOND. B.Sc, Pass, Applied Math., 1906, in. 8.) 438 ANALYTICAL MECHANICS [art. 433 7. 'Define a metacentre, and establish the formula for its position in the case of a solid of revolution floating with its axis vertical. ' A solid right circular cone, of specific gravity s, floats in water with its axis vertical and vertex downwards. If r is the radius of the base and h the height, find the condition for stability.' (LoND. B.Sc, Pass, Applied Math., 1907, in. 7.) 8. ' Prove that the resultant of the pressure on a body immersed in a fluid is an upward force equal to the weight of the fluid displaced. ' A thin rectangular board of specific gravity a- is hinged along one of its shorter edges to the flat bottom of a tank. Find the position assumed by the board when water is poured in toji depth h, and prove that the vertical position is attained when h is v'o-Jiimes the length of the longer edge of the board.' (LoND. B.Sc, Pass, Applied Math., 1908, iii. 9.) 9. ' Prove that if a ship be displaced through a small angle about a longi- tudinal axis in the section of flotation, then, approximately, Righting Couple = Displacement x metacentric height x sin (angle of heel). ' A weight of 10 tons is shifted through 22 feet across the deck of a ship of 7000 tons. The bob of a pendulum suspended from a height of 70 feet above the deck is found to move 5^ inches across the deck at the same time. Calculate the metacentric height of the ship.' (LoND. B.Sc, Pass, Applied Math., 1909, in. 8.) 10. 'Find formulae giving the position of the centre of pressure on a rect- angular lamina, submerged with two sides horizontal. 'ABCD is a square of side 2a, and P, Q, R are the middle points oi DA, AB, EC. A lamina in the form of the pentagon PQRCD is submerged in water so that Q is in the surface, DC is horizontal, and the plane of the lamina makes an angle, 5 with the vertical. Compare the total pressures on the portions PQR and PRCD ; find the positions of the corresponding centres of pressure. {N.B. — The atmospheric pressure is to be neglected).' (Lond. B.Sc, Pass, Applied Math., 1910, in. 6.) 11. 'Discuss the relation between the stability of a floating body and its metacentric height. ' The section of a barge by the plane of flotation is a rectangle of 40 feet by 10 feet. Compare the couples required to produce a given small angular displacement about the fore and aft line and about a perpendicular horizontal line amidships.' (LoND. B.Sc, Pass, Applied Math., 1910, in. 8.) 433. Heights by Barometer. — In article 426 we treated the relation between pressure and height in a liquid of constant density. We now turn to the problem of the form that the relation assumes for a fluid whose density is proportional to the pressure. Taking as before the axes of x and y horizontal and that of z verti- cally upwards, we may quote equations (6) and (7) from article 426. dp_dp , , and %r-ps (2)- For a fluid like air we may now write the approximate relation expressing the behaviour of an ideal gas, viz. PIp=M, or p=plR6 .... (3) *^T. 434] HYDROSTATICS 439 where i? is a constant for the gas in question and 6 is its absolute temperature. Then, putting (3) in (2), and integrating from the height z. at which the pressure is /„, we have I J IW' (4). Hence, provided g and 6 can be treated as constants, we find loge/-loge/o= ~^q{z-z,), -^^=f^^{f) (5). Now using (3), and inserting numerical values for air at 0° C. and standard pressure, we have i?:=4 = -76Xi3:6x.^ pv 0001293X273 ^ ' Hence, by (6) in (5), and transforming to common logarithms, we obtain finally ._.„= 76_xiy6xi /A\ 0-001293x273 ^ ^ \p) ^" Since the height of the standard mercury barometer has been written 76 cm. and the density of mercury has been put at i3'6 gm./c.c, etc., etc., the difference of heights z—z^ will be expressed in centi- metres. It may be noted that for the /„ and p in (7), since they form a pure ratio, any units may be used provided they are the same for each. Indeed, the values of p at each level could be put in (5) or (7) instead of the/'s, since the /)'s are proportional to the/'j. The fall of the mercury on ascending is roughly of the order i cm. in 11,000 cm. or i inch per 1000 feet. If the temperature varies considerably in a given ascent, equation (7) may be applied to separate sections of the ascent, the mean temperature being used for each such section. Hence a barometer and thermometer give the data for a determina- tion of the heights ascended on a mountain or in the air. For other refinements as to convective equilibrium and variation of g the student may consult Webster's Dynamics of Particles and of Rigid, Elastic, and Fluid Bodies, p. 466 (Leipzig, 1904). 434. Pressure on Curved Membrane of Uniform Tension. — As a preliminary to a brief treatment of surface tension or capillary pheno- mena, it will be convenient to find here the relation between the difference of pressures on the two sides of a membrane or surface, its radii of curvature, and its tension or force per unit length, supposed the same in every direction. The method followed is that used in the writer's Text-Book of Sound (pp. 262-263, 1908), and consists of equating the work done by the pressures for a small imaginary displacement of an element of the 440 ANALYTICAL MECHANICS [ART. 435 surface to that done against the tensions. It is, in fact, an application of the principle of Virtual Work. The first expression for the work is obviously the normal force into normal displacement or excess of normal pressure on concave side into volume described by the element of surface. The second expression for the work is the tension of the surface into the increment of area acquired by the element in its normal displacement. Let the plane of xy be tangential to the surface. And, at the point of contact, take as the element an infinitesimal rectangle of sides a. and /? parallel to the axes of X and y, the corresponding radii of curvatures of the surface being r^ and r„. Then as we shall suppose these radii large compared with the sides of our element, its normal displacement at each point may be written dz. Hence the work considered as pressure into volume is dW=po.fidz (i), where p is the excess of the pressure on the concave side over that on the convex side. To obtain the increment of the area of the surface in consequence of the normal displacement, we need an expression for the increase of each side of the rectangle. Thus a a-\-da. da. Ti r^-\-dz dz' since each of these expressions is the circular measure of the angle subtended by the side of length a at the centre of its curvature. Hence da.= — ^z and similarly (^)- But by hydrostatic considerations (see equation (8) of article 426) we have , , n\V>^V>n Fig. 226. Capillary Ascent. 442 ANALYTICAL hfECHANICS [art. 437 Thus (9) in (8) gives iTco^^—pghr (10). This shows that for a given liquid hr is a constant, or hoz ijr. It also expresses the product T cos in terms of readably observable quantities. The value of ^, the angle of contact, varies with the liquid and solid concerned, and must be experimentally determined for any given pair of substances. Since water wets glass, the value of <^ is zero for water in a glass tube. Thus, considering also p practically unity for water, (10) simplifies to the approximate equation for water in a glass tube. 2T=gkr ... (11). It should be noticed that in dealing with the meniscus we were concerned only with the radius of the tube at that place. Further, equation (9), giving the relation between pressures and difference of levels, is independent of the radius altogether. Hence the tube may be of any section we please provided it is circular and of radius r at the meniscus ABC and the equations (8), (9), and (10) are still valid. It may, however, be observed that if the tube have a bulbous portion just below ABC, the liquid would not spontaneously ascend past that bulb but would need forcing or sucking up to ABC, and would then remain in equilibrium there. It is usually near enough to measure the height h to the middle point of the meniscus as shown in the figure. If a liquid like mercury is used which does not wet glass and whose meniscus is convex upwards in equilibrium, the values of cos and of h become negative. In other words, the ascent is changed into a depression. 437. Ascent between Plates. — If we use parallel vertical plates dipping into the liquid a distance a apart instead of a tube, equation (8) of article 436 is replaced by (j>^—p)la = 2lTcos (iia), a horizontal length / along the faces of the plates being considered. Our previous equation (9) still holds, viz. Hence, we obtain 2 T cos (j>=pg/ia .... ... (12), showing that Aa is constant, or Aoci/a. Thus, if two vertical plates are used very nearly touching at one vertical edge and quite so at the other, the form of the upper surface of the liquid is an equilateral hyperbola with the free liquid surface and the line of contact of the plates as asymptotes. Examples— LXXXV. 1. Establish the law that in an atmosphere of uniform temperature the pressures diminish in a geometrical progression as the corresponding heights increase in an arithmetical progression. 2. If the tabular density of air is taken as 0-00129 gii- per c.c, show that at ART. 438] HYDROSTATICS 443 15° C. a fall of the barometer from 76 to 75 cm. corresponds to an ascent of 11,080 cm., and a fall from 30 to 29 inches to an ascent of 940 feet. Show that the resultant force of uniform normal pressures P all over the convex surface of a hemisphere of radius a is na'P perpendicular to the base. Find the relation between the excess pressure on the concave side of a membrane, film, or other flexible sheet, and its curvature and tension. A long cylindrical glass tube of i cm. internal bore has an excess pressure inside of 50 atmospheres, show that the circumferential tension is 25 X 10" dynes per cm. nearly. 6. Obtain formulae for the surface tension of a soap bubble of given size and excess internal pressure, also for the tension per square inch in a spherical steel shell subjected to high internal pressure. Derive formulae for the capillary ascent of liquids in tubes and between plates. A rectangular frame of opposite wires of length c connected at their ends by threads is dipped in soap solution, and one wire is fixed horizontally the other supporting a weight W; the vertical height between the wires is then found to be b, and the distance apart of the threads midway between the wires is a. Show that the surface tension T of the soap solution is given by T=^W- '-" 5- 7- c' + i -Let us now consider the 438. Equilibrium Form of Large Drop. form of a large drop of a liquid upon a horizontal plate which it does not wet, say mercury on clean glass. Take the origin of co-ordi- nates at the centre of its upper sur- face, the axis of z being vertically downwards and that of x in the plane of the diagram, see Fig. 227, which shows a central vertical sec- tion of the drop resting on the sur- face cC. Let the pressure outside the drop be /„, practically the same at all parts of it, the pressure inside it at the level z being /. Then, for the point P on the surface at this level, we have from article 426, equa- tion (8), and article 434, equation (4), the following expressions— p-po=Pgz ... . . (i), . . . . (2). Fig. 227. Equilibrium of Large Drop. Whence (3)> T" being the surface tension of the liquid air interface and r^, r„_ the radii of curvature at P. Let r^ denote the radius on a plane perpendi- cular to that of the diagram, and passing (of course) through PK, the 444 ANALYTICAL MECHANICS [ART. 438 normal at P, to the surface of the drop. Then, as the drop is supposed large, we see that r^ will be large, and therefore its reciprocal may be neglected in comparison to that of r^, which refers to the plane of the diagram. Also, if ^ denote the angle between OX and the tangent at P, and s is distance along the curve, we have 1 d^ d\!' dx . d^ , > — = -/-=-/-•- =cosi/'-/- . . (4). ^1 ds dx ds dx But, in (3) and (4), we have still three variables, 0, i/-, and x. Let us therefore eliminate the latter by use of the relation -^ = '^"^ (5). Then (4) and (5) in (3) give or pg I zdz— t\ sin ^ di^. Jo Jo Whence, on integrating, pgz'' ^ 2 T(i- cos tf) (6). At the point H, where the vertical is tangential to the curve, let the depth below the summit be A. Then, since tf=ir/z at H, equation (6) gives PgA' = 2T (7), a formula which may be used in the experimental determination of T without the angle (j> of contact being known. Let the total thickness of the drop be c ; then, as ^ at the depth c is equal to , equation (6) gives Pg£' = 2 T{i- cos ) (8). And from this ^ may be experimentally determined when T is known from (7). What is here only approximate for a drop of finite size (and therefore of finite curvature in plan) would be rigidly true for the shape of the liquid surface near a plane plate dipped into the liquid. As shown in Fig. 227, it would be right for a liquid not wetting the plate, the surface, convex upwards, being accordingly depressed near the plate, as mercury is near glass. If the curve of Fig. 227 were inverted by rotation through 1 80° about OX, it would then correctly represent the liquid surface, concave upwards, and raised above the general free surface OX, because it was near a plate which that liquid wets, as for water and glass. In either case the curve would be valid from O to P, H, etc., according to the inclination of the plate dipping into the liquid and the angle of contact of that liquid with it. Further, the investigation made and shown for a drop on the upper surface of a plate applies also to an air bubble blown in liquid on the under surface of a plate. ARTS. 439-440] HYDROSTATICS 445 439. Relative Equilibrium of a Liquid in an Accelerated Vessel. — The problems of liquids in motion belong, of course, to the next chapter, but the relative equilibrium of liquids in vessels whose acceleration is uniform or follows a very simple law may be noticed here. We have seen in articles 226-229 l^o^ gravity appears to be changed in magnitude and direction by the acceleration of the chamber in which a plumb bob or pendulum is hung. And, since the free equilibrium surface of a liquid is at right angles to gravity, we may at once deduce the form of a liquid surface in relative equilibrium in an accelerated vessel. The relative or disturbed value of gravity ^ at an angle ^ say with the vertical is a normal to the surface of relative equilibrium of the liquid in this chamber or vessel. This surface is accordingly inclined at the angle ^ with the horizon- tal, the vertical plane which cuts it at the steepest angle being that which contains the acceleration of the vessel. Thus, in Fig. 228, let the vessel ABCD have horizontal acceleration Yb in the plane of the diagram. Then we may re- gard this acceleration as vectorially sub- tracted from the acceleration P^ due to gravity, thus leaving the effective or relative gravity P^' at an angle i/- with the vertical and in the plane of the diagram. Thus the relative equilibrium surface of the liquid . is A'PB' at the angle f with the horizontal in the plane of the diagram. The quantities concerned are obviously connected by the relations X.d.n^=—blg . Fig. 228. Liquid in Accelerated Vessel. (2). Fig. 229. Surface of Rotat- ing Liquid. If there is liquid in a chamber mov- ing with acceleration down an incline we should have horizontal and vertical accelerations, say b and a respectively. Then they could be both vectorially subtracted from the vertical g, leaving the effective ^, which is a normal to the surface of the liquid. (See articles 226-228.) 440. Uniform Rotation about a Vertical Axis : Liquid Surface a Para- bola. — Let the vessel ABCD rotate uniformly at speed w about the vertical axis OY, Fig. 229; and let it be re- quired to find the form of the liquid surface when it has settled to a steady state and is rotating like a rigid solid. Let AOPB represent this form, and consider the point P of co- 446 ANALYTICAL MECHANICS [art. 440 ordinates x and y. At this point the acceleration of the liquid is obviously expressed by , ^ ^=-=/„, u=v=o, and y = w. Thus the velocity w of outflow is given by putting these values in (21) and substituting for £ from (23), viz. Ho Po O"^ . . w"- = 2gh (24), which is the approximate formula known as Torricelli's Theorem. It is seen that the velocity thus determined is the same as that of a body falling from the level of the free liquid surface. The same result might have been obtained by considerations of energy. 446. Vena (3ontracta. — We must now examine the approximate relations used in obtaining Torricelli's theorem and note how, for strictness, they need modification. In the first place, the stream lines above the orifice must be of a converging character. Further, this convergence continues for a short distance outside the orifice, as may easily be observed. Hence, although at the surface of the issuing jet the pressure of the liquid may be atmospheric only, ivithin the jet itself the pressure is rather greater than atmospheric. Accordingly, the velocity here is less than that applicable to the surface as given by Torricelli's theorem, which would in consequence fail to give a correct estimate of the quantity of liquid issuing from a given orifice. The exact behaviour of the liquid near the orifice presents great theoretic diflficulties which have been only partially overcome. But experiments have shown that at a little distance from a simple orifice the stream lines have all become parallel, and at this place, called the vena contractu, the cross section of the jet has a minimum value. Thus, over the cross section of the vena contracta, we may take the velocity as uniform and the pressure practically atmospheric. Hence, if we know experimentally the position and size of the vena contracta for a given orifice, we may make an amended calculation of the total outflow. For a circular hole in a thin plate the ratio of the area S' of the vena contracta to that 6' of the orifice is approximately SIS=o-62 (25). This ratio is called the coefficient of contraction. The distance of the vena contracta from the circular orifice may be taken as between 0-39 and 0-5 of its diameter. For practical details relating to a variety of openings the student may refer to Goodman's Mechanics Applied to Engineering (pp. 548- 563, 1908), 454 ANALYTICAL MECHANICS [aRTS. 447-448 447. Liquid Rotating as Eigid Solid.— Suppose a mass of liquid of constant density to rotate as a rigid solid with velocity w about the axis of s, which is taken vertically upwards. Then we have X=o, F=o, Z^-g] • ^'^''• The equation of continuity is then satisfied, and these equations (26) put in (7) of article 442 give I dp — to X=: , p ax I dp p dy I dp °=-^--pT.] (^7)- Multiplying these in order by dx, dy, and dz, adding, and integrating, we find y(*'+/)-^2=|+C (28). Taking the origin at the surface, where the pressure is/,, the con- stant of integration is C=-P,Ip (29)- Hence the equation of the surface is expressed by b>^{x''+y-) = 2gz (30), which is a paraboloid of revolution with latus rectum 2^/(0'', as found by other considerations in article 440. Examples— LXXXVII. 1. Derive the equation of continuity for a compressible and for an incom- pressible fluid. 2. Obtain the equations of motion for a fluid. 3. State the various relations which hold under different circumstances between the pressure and density of a fluid. 4. Assuming Euler's fundamental hydrodynamical equations, obtain the equation of steady motion along a stream line in the form ds p ds 5. Obtain an equation which expresses Bernoulli's theorem, and interpret each term. Can the equation be put in another useful form ? 6. Show that, with the assumption of certain approximate relations, the out- flow velocity of a liquid is that due to a free fall through the pressure head. 7. How is the result of question 6 affected by the phenomenon known as the vena contract a ? 448. Angular Velocities of Elements of a Fluid. — We have hitherto expressed the motion of a fluid in terms of its linear velocity components. Let us now introduce expressions for the angular velocities of its elements and determine the relation between the two. ART. 449] HYDR OKINE TICS Take the 455 d/3. B m .c edees dx dv dTlS^' ^' ) l°"l ''°™^' °^ ^ parallelepiped of eages dx, dy, dz, and consider its change of shape parallel to the xy plane in consequence of the p.H'inei to me linear velocity components u, v, ru of the fluid occupying it (see Fig. 231)- Thus in time dt let the face PACE change to the size and shape PA'C'B', in which the corner originally at P is shown as if brought back there to simplify the figure. Let the angles APA' and BPB' be da and d/S. Then the mean angular displacement about OZ of the element of fluid in time dt may be represented by ^(da—dfi). C J — Z Fig. 231. Angular Velocities of A Fluid. v)dt=—-dxdt, dx But dadx=AM={v' where v' is the velocity parallel to the y axis at A We thus have da. = ^rdt, dx and similarly dliJ-^dt. dy Hence, for the angle of mean rotation in time dt about the 2 axis, we have da. — dp .{dv du\ ,^ ,., Thus, denoting by ^ and y\ the angular velocities about the axes of X ^x\A y respectively, and obtaining the corresponding expressions by symmetry, we find . dw dv " 2-q-. dy du dw dz dx ._ dv du dx dy ^ (0. expressions which give the component angular velocities in terms of the linear velocities as required. 449. Velocity Potential. — When the three angular velocity com- ponents all vanish, the fluid is devoid of rotation of its elements, and its linear velocity components are then derivable as the space differentials of a function <^ of the space co-ordinates and time, and called the velocity potential. deb d4> d4> /,\ We then have ,,= _^, z,=:_-, z.= __ - . . (2), and o = ^=V = C (3)- 456 ANALYTICAL MECHANICS [art. 450 Thus the velocity potential stands in the same relation to the linear velocity as the gravitational potential does to the field or attraction. The introduction of the velocity potential accordingly effects a similar simplification in certain cases. 450. Angular Accelerations of Elements of a Liquid. — To obtain the angular accelerations of the elements of a liquid, we must intro- duce the expressions for the angular velocities into the equations of motion. We may conveniently do this as follows : — By differentiation of the square of the resultant linear velocity and its components, we have .do" . d , , , „ , „. du ' dv , dw , . Now take the right side of (4) from the left side of the first of the equations of motion (7) of article 442, and we obtain du (du dv\ (du dw\ ^ i dp ,dg^ , . Tt-^'"Kdy-Txr'"\Jz-Txr^—iTx-^-Tx ■ ■ ^5)- Then, introducing in this the angular velocities from (i) of article 448, and writing the other two equations by symmetry, we find dv . , ^ ., \ dp ,dq^ dt ' ^ ^ ^' p dy ^dy dw , / * \ .T 1 dp ,dq^ (6). Now differentiate with respect to y the third of equations (6), and from the result subtract the second differentiated with respect to z. Then the last two terms on the right disappear for constant p, and the first on the left, interpreted by (i), gives 2d^ldt. We accordingly have l+|<"l-"')-5«-"f)=i(f-f ) ■ ■ <"■ Now, considering the liquid to be incompressible, we may apply the special equation of continuity, (3a) of article 441, viz. a+|+*=» W- But using (i), which defines the rotations, we find by differentiation that Tx-^^^dE-° (9). Then, by aid of (8) and (9), we may transform (7) into the following : — '^f , <^£ , '^i I '^$ fdu du .du , (dZ dY\ , , We have hitherto used ^, »;, and f to apply to whatever portions of liquid occupied a given position at a specified instant. Let us now change to some particular poirtio.n of liquid which at time / is at (x,y, z) AI^T. 451] HYDROKINETICS 457 and is followed in imagination in its motion. Then, using as before JJint as the symbol of particle differentiation, we have D^ d^ d^^ d^ ^ d^ and similar equations for j; and f just as we had for u, v, and w. See (6) of article 442. Hence, using (11) in (10), it becomes the first of the following set, the others being written by symmetry : — nt ^dx'^'^dy'^^dz'^^Kdy dz I Di) dv , dv . Jv , . (dX dZ\ , . wr^+'^d^+^d^+Ki^-dx )\ ■ ■ ■ ('^)- Dt~^dx'^'^dy'^^Tz^\dx~'dy). Now let the forces be derivable from a potential Fso that ^ dV .^ dV ^ dV , . ^=-^'^=-^'^=-^ (^3), then the component torques due to these forces are all zero, as seen by the terms in round brackets at the right in (12). If, in addition, we have also at any instant ^ = r, = f=o (14), then the other angular accelerative terms on the right of (12) vanish. Hence, for the case under consideration, no rotation being present in our ideal liquid, it appears that none can be acquired} It should be noted that viscosity is supposed quite absent here. Its presence would change this conclusion. 451. Coaxial Circular Currents. — Let us now consider the case of the flow of an incompressible liquid in coaxial circles round the axis of z, which is taken vertically upwards. Within the cylinder of radius /■„, let the value of the angular velocity of the elements be given by f = f„, a constant. But, outside this cylinder of radius /■„, let f=o. It is required to find the velocities of the particles and the velocity potential, if any. If u) is the angular velocity about OZ at r, the radius vector, we have for the velocity and force components X=o, F=o, Z=-g] ■ ■ ■ ■ ^'^'■ Then, remembering that r-=x"-+y' and dr/dy=y/r, we find by partial differentiation du du) dr _ _ y d(n — =— w— j^-^— ^ ^^ dv , d , , Thus, within the cylinder r=^r^ we have rdii> f-^". + P-^=o (18). And, on integrating, log (w-f„)+log r"=log ^, or <« = &+;:5 (i9)> where ^ is the integration constant. But, to avoid an infinite value of (u at the axis where ^= o, we see that the integration constant A must be zero. Hence, for the whole interior of the cylinder of radius r^, we have '« = fo (20). which means that this rotates as a rigid solid with angular velocity fo, and that each part of it is therefore rotating at the same angular velocity as specified at the outset. We now deal with the outer part, for which the rotation of the individual parts is zero. Then, from (17), we have r dti) , (o^ r- = o . (^i)- 2 ar ^ ' Whence, proceeding as before, we find (0 = -^ . . . (22). To avoid a discontinuity of velocities, we will make the ui's for the inner cylinder and beyond it agree for the value r=r^. Thus fo=-2-. OX B=i^r% (23). Accordingly the outer linear velocities in the circles are given by '"'■=^r (24)- And this expresses the grading of the velocities of flow with radius, which correspond to the absence of rotalion of the parts. We see that in this region the angular momentum about the axis of z per unit mass is constant, for it is given by ""■' = fo'-o (25). Let us now find the velocity potential <^ of this outer region. We evidently have from (24) Tr=°^''^-^e=''''--T ■ ■ •(^^)- Thus, by the first of these 4> does not contain r, and by the second we have 'i'^-iA^+C^C-i.rlt&n-'ijIx) (27). ^^'^- 452] HYDROKiNETICS 459 For the inner cylinder of radius r„ there is no velocity potential, since the condition for it, that g, r,, f should all vanish, is not fulfilled. Ur, we may note direct from (15) that no velocity potential can be assigned to this motion. The free surfaces of a liquid rotating like this under gravity are easily found. For the inner cylinder it is, of course, the paraboloid we have dealt with before in (30) of article 447, viz. from r=o to r=r^. ^g^=a{r-->i) (28), the origin being now at the level of the hquid at r=r^. While for the outer part, using (24) for tar, whose square is pro- portional to the change of level, we find the hyperboloid, for r>r„, 2g^=Q(K-rt/r') (29). Thus, the depression of the centre below the surface for r=oo is expressed by '^=Qf^ols: ' . . . (30), the level at r=r^ being equidistant from the levels at the centre and at infinite radius. 452, Steady Flow past Cylinder. — Let us now consider a plane problem of the steady flow of an incompressible liquid parallel to the circular base of a right cylinder of infinite length, which forms the obstacle in the path of the stream, whose velocity far away from this disturbance is «,), parallel to the x axis, that of z being the axis of the cylinder of radius c. Then everywhere we have w—o, and far away from the cylinder v=o also, but near the cylinder v is finite, and both it and u are variables, being functions of r, where r'=x'+y'. Since the liquid is incompressible the simple form of the equation ' of continuity applies, and, using the velocity potential cp, we may put (3a) of article 441 in the form d-4> d-4> d-cl> _ /,) -d^--^df^az'-° ^ '' But for our case, since w = o, this reduces to ^+?^ = o (^)- ax' ay' This equation, together with the condition as to the undisturbed velocity, is satisfied by =-u^{i-\-c'lr') ■ ■ -^ (7)- Thus, differentiating for u and », we obtain d / , ^\ 2U„<^X^ , , "=-Tx=''V+v) ?^ • • • • ^^)' d 2Unc'xy I ■■ and .= -|=_^L^ (9). Hence, for the resultant velocity q at the surface of the cylinder, we have Then, introducing this value of q in equation (21) of article 444, we find for the normal pressure on the cylindrical surface ^=p„(^-^.-^) (,x). The pressure is accordingly just as great on the hinder part of the cylinder, where the liquid is flowing away from it, as on the fore part, where the liquid is meeting it. Hence the cylinder has no resultant force from this t'deat liquid flowing past it. Or, if the liquid is conceived as at rest at all parts far from the cylinder, which is moving through it at speed «„, then the cylinder would, under the ideal conditions assumed, experience no resistance from the liquid. The differences between this state of things and those obtaining in any actual experiment are due to the presence of viscosity in all actual liquids and of friction between the solid and the liquid, called skin friction. Further, one cannot in an experiment obtain the ideal geometric relations here supposed. 453. Water Waves. — Of the many kinds of waves possible in liquids, theory recognises three chief classes : — (i) Long waves, or tidal waves, in which the motions of the particles are chiefly horizontal, and are equally great on the surface and below, where the bottom is level. (2) Ripples, or very small disturbances on the surface, in which the restoring forces called into play are due to the surface tension, the liquid skin tending to flatten itself and assume the form of minimum area. (3) The commonest class of all, and those most noticeable, which are variously called oscillatory waves, surface waves, and gravity waves. Their oscillatory character they have in common with the ripples, from which, however, they are distinguished by the term gravity, which refers to the nature of their restoring forces. The term surface dis- ART. 453] HYDROKINETICS ,. 401 tinguishes thetn from the tidal waves, in that, unlike the tidal waves these are produced by surface disturbances and are confined to a region near the surface, the amplitude of the disturbance diminishing rapidly as we descend. "naimig In the present article we now confine attention to this commonest class of water waves, which we shall suppose to be propagated horizon- tally along the axis oix, that of z being taken vertically upwards from the level bottom of the vessel. We have accordingly no motion parallel to the ji- axis, hence the equation of continuity reduces to the two terms du dw _ \ dx dz~ \ We next suppose the wave to be of the simplest harmonic form and test the legitimacy of that assumption. Thus let =Fcosk{x—at) (2), where F\s a function of z only. Then, differentiating (2) and putting in (i), we obtain -f-i— —/i''Fco5k{x—at), and 2"=^°^ (3). Whence F=At^'-\-Be'^' (4). Now for the bottom of the liquid, where z = o, we must have w=o=d4>ld2. But, by (2) and (4), ^^k{Ae'''-Be-'"')cQsk{x-ai) (5). Hence for z = o one of two alternatives must be chosen. Thus (i) we may write B=A in (5) and (4), which reduces (2) to the form =A{e'"'+e-'"')cos.k{x-at) (6). Or (ii) if the depth is great we might write £=0 and put A small, so that A into the factor e*^ is considerable at the surface, although practically zero long before the bottom is reached. To obtain a, the velocity of propagation of the waves, we must use the equations of motion (7) of article 442. In these, if we suppose the amplitudes we deal with are small, we may neglect the products udujdx, etc., which reduces the left side of each equation to its first term. Hence, introducing the velocity potential, these become _d_/d^\__dj' idfi dx\dt)~ dx p dx' and two similar ones for y and z. Hence multiplying them by 462 ANALYTICAL MECHANICS [ART. 453 dx, dy, dz respectively, adding, remembering the differentiations are partial, and integrating, we obtain- -'>'+i=^ ....... (7), gz being written for V, since it is supposed due to gravity only. Further, since we suppose the amplitudes to be small, we may consider the pressure everywhere independent of the time. Thus differentiating (7), we find -S+4- w- Hence (8) becomes We have to obtain these terms from (6) and substitute them in (10). Thus -^=—k''a''A{e'"+e-'''')coskix-at) . . (11), and g-^=gkA{^'—e'''")cosk{x—at). . . . (12). Whence, equating the sum to zero, we have /^a'(«*2+«-*^)=^(«*^— «-*«) (13). Thus to obtain the velocity of propagation of the waves at any height above the bottom of the liquid, we put the corresponding value oi z=h say in (13) and find We may now fitly introduce the wave length A, and so transform (14) by aid of the relation derived from (6), etc., viz. kX=2Tr, ox k = 2ir/X. (15)- Thus (14) may be written (?) <■«)• If k is very great compared to A, then the exponential fraction in (14), involving e^''''l\ reduces to unity, and we have a'—-~=— nearly (17). « 27r ^ " We thus see that the velocity of propagation in each case depends on the wave length, varying directly as its square root ; thus there is in water waves an analogy to the phenomena of optical dispersion. Further, again referring to the exponential fraction in (14), we see that for small values of A sl reduction in A involves a reduction in a. That is, the waves advance more slowly in shallower water. If the depth of the water is great, we may now take the alternative a"-=^tanh( 27r ART. 454] HYDROKINETICS 463 potlntiaf ■ '^"^ ^ """'' ^^ '"°- ^' ''^^^ ™'^ f°^ th« ^^'°"ty Thus for the velocities at {x, z) we have and v=-^^ = -kAe^'zo%k{x-ai)\ ' ' ' '^ ' Hence the path of the particles at {x, z) is a circle, described with angular velocity ak and of radius r, such that rak=g = kAe''', 1 or /-=_gfe=— ^2tzM I (20)- a Thus the radii of these circles diminish in geometrical progres- sion as depth below the surface increases in arithmetical progression. And, at the depth of one wave length only, the ratio of diminution is ^^^ or 535 : I nearly. Thus, as Tait mentions, an Atlantic roller of 40 feet high from trough to crest (if such occur) and 300 feet long would produce a disturbance from the mean position of only half an inch or less at a depth of 300 feet. For aerial waves, works on physics may be consulted, as, e.g., the writer's Text-Book on Sound (Macmillan, 1908), Chapter iv., articles 119-122, 125-127, 130-139, 145-146, and 149-155- 454. Steady Flow of Viscous Liquid through a Narrow Cylinder. — We have hitherto usually ignored viscosity but will now in conclusion deal with one very simple case in which viscosity is paramount. It is the problem of the steady flow under pressure and gravity (or the former only) of a viscous liquid through a narrow cylindrical tube. We suppose the velocity of the liquid to be zero at the wall of the tube and to increase to a maximum which is reached only at the centre. Thus, the liquid is sliding in concentric cylindrical layers, each inner one urging the outer one along, which in turn is retarded by the next outer layer, which moves slower than itself. The magnitude of this effect for a given liquid is expressed by the coefficient of viscosity q, which may be defined by dA = +ri-~-dzdx (i), where dA is the force parallel to the x axis on the elementary area dzdx, when the velocity u in that direction changes as we pass along the axis oiy. To fix our ideas, let us consider the arrangement shown in Fig. 232, in which liquid, kept at a constant depth c in the upper vessel, flows through the fine tube of radius a and length / set obliquely so that the difference of levels of its ends is h. In the tube, consider an elementary cylindrical layer of liquid, of radii r and r+dr, in steady flow, the velocity u parallel to the axis 464 ANALYTICAL MECHANICS [art. 454 and down the slope being a function of r only and the velocity extremely small. Then the motion is executed and maintained con- FiG. 232. Steady Viscous Flow. stant by a zero force, which is the resultant of five separate forces falling into three groups, viz. (i) That due to gravity, which is • 2i:rdrlp = zirpghrdr (2). (3), (2) That due to pressures at ends {{po+cpg)—po}2TTrdr=2Trpgcrdr where /o is the atmospheric pressure. (3) That due to the tangential viscous forces on its inner and outer surfaces. The former is y—r)-r-2Tvrl\, and acts downwards, that is, posi- tively, dujdr being negative. The latter is numerically greater by the differential of the former, whose variable part is rdufdr. Hence their difference, the former minus the latter, is given by ''-^''Tr{''fy (4), and acts upwards. Now, equating the sum of these expressions to zero, we find or where 2irpg{c+h)rdr+ 27rh^dU^=o, frdr+rtd{r^ = o . . f=pg(c+h)ll (5)> (6), (2= fV(2=- Jo ART. 454] HYDROKINETICS ^65 Integrating (5) we obtain fr' du ^ 2 dr ' or f^'^*' 1 ^ C , "■^ -^+W« = — ^r . (7). Hence, integrating again, we have ■^ — !-';«= Clog /-+Z) (8). 4 Now for r=o, if « is not infinite, we have 6'=o. Again for r=a, «=o by hypothesis, so D=fa:l^. Thus (8) becomes u=f{a'-r-)l^-n ....... (9). To find the volume Q of liquid discharged per second, we have Hence integrating we obtain "W ^'°^^ or, putting in from (6) the value of_/j we find ^ 8/7; ^' These relations may be used for a practical determination of the viscosity coefficient y\ at any given temperature. Of course, if the tube be vertical h becomes /, while if it be placed horizontally h vanishes. In the present chapter much help has been derived from Professor G. Jager's excellent brief treatise on Theoretical Physics (Sammlung Goschen, Leipzig, igo6). Examples— LXXXVIII. 1. Obtain expressions for the angular velocities of the elements of a fluid. If the angular velocity for any element is zero, does that element necessarily behave like a rigid solid? Take some actual example of motion, and indicate how the element behaves although rotation is absent. 2. Assuming the fundamental equations, derive expressions for the angular accelerations of the elements of a liquid. If at any instant, in an ideal fluid under forces derived from a potential, the angular velocities of the elements are all absent, what follows ? Prove your assertion. ^. All the liquid in a certain region is in coaxial circular motion about a vertical axis, the portions inside a certain coaxial cylinder having linear velocities directly as the radii and the portions outside that cylinder having linear velocities inversely as the radii, there bemg no discon- tinuity of velocity anywhere. Show that the free surface is a paraboloid within the cylinder mentioned and a hyperboloid outside it. 2 G 466 ANALYTICAL MECHANICS [ART. 454 Also show that the depression of the centre is double that of the boundary of the cylinder, both being reckoned from the free surface at infinity. Prove that the liquid outside the cylinder has no rotation of its elements, but is not like a rigid body, while that within is like a rigid body and has rotation of its elements. 4. Obtain the flow of an ideal liquid past an infinitely long right circular cylinder in planes perpendicular to its axis. Plot curves showing the stream lines, and mark the speeds at several points on one line. 5. Derive the differential equation of motion for gravity waves, and solve it, showing that the motion of the particles is in circles which diminish as we descend. 6. Obtain an expression for the velocity of propagation of gravity waves, and point out how it varies with depth and wave length. 7. Derive a general formula for the steady flow of a viscous liquid through a capillary tube, and adapt it to the cases where this tube is (i) horizontal and (ii) vertical. 8. For very small vibratory disturbances of a compressible fluid of negligible weight obtain the equations of motion in the form where j is defined by p = pg{i +s). ART. 455] STATICS OF ELASTIC SOLIDS 467 PART VI.— ELASTICITY CHAPTER XXI STATICS OF ELASTIC SOLIDS 455. Nature of Elastic Bodies. — Elasticity may be regarded as that property of matter in virtue of which a body (i) resists forces tending to change its bulk or form or both ; (ii) requires the continuance of those forces for the unimpaired maintenance of those changes ; and (iii) recovers its original bulk and form when those forces are removed. Adopting the word strain to express the change in volume or form or both combined, and the word stress to express that combination of forces associated with the strain, we may say that the theory of elasticity is that branch of mechanics which discusses the, mutual relations of stress and strain. But before proceeding to this theory, even to that elementary degree compatible with the plan of this work, we must note the general significance of certain other terms applied to bodies in this connection, and also give more formal and quantitative definitions of stress and elasticity. If, on the removal of the stress, the strain entirely disappears, the body is said to be perfectly elastic. If, however, on the removal of the stress some part of the strain remains, that part is called the permanent set, and the body in question is said to be imperfectly elastic for such stresses. The condition at which a marked permanent set occurs is called the elastic limit of a material. Very small stresses and strains are found to be practically proportional. In other words, they satisfy Hooke's law : ut tensio sic vis. The place at which this simple pro- portionality ceases to hold is called the proportionality point on the graph of stresses and strains. It is usually near the elastic limit. On the removal of a stress, if practically none of the strain disappears, the body is said to heplastic; and if only a small part of the strain disappears, it is said to be ductile. If a stress, maintained constant, causes in a body a strain which increases continually with the time, that material is said to be viscous. ^^'hen a continuous alteration in form is produced only by stresses exceeding a certain value, the material is called a solid, however soft it may be. But if the very smallest stresses, when continued long enough, cause a constantly increasing change of form, the material must be regarded as a viscous fluid, however hard it may be. (Maxwell's Heat, p. 303, London, 1894.) 468 ANALYTICAL MECHANICS [art. 456 ' A body is called homogeneous when any two equal, similar parts of it, with corresponding lines parallel and turned towards the same parts, are undistinguishable from one another by any difference in quality ' (Kelvin and Tail's NaturalPhilosophy, Part ir. p. 2 1 6, Cambridge, 1 8go). If we push our scrutiny to the utmost conceivable limit perhaps no material would survive the test and be held as homogeneous. But, in the theory of elasticity, glass, continuous crystals, india-rubber, and fluids are usually considered as homogeneous. 'The substance of a homogeneous solid is called isotropic when a spherical portion of it, tested by any physical agency, exhibits no difference in quality however it is turned. Or, which amounts to the same, a cubical portion cut from any position in an isotropic body exhibits the same qualities relatively to each pair of parallel faces ' (ibid. p. 217). In what follows we shall restrict our attention to materials which are both homogeneous and isotropic. 456. Stress and its Relation to Strain. — When the terms stress and strain were introduced into the theory of elasticity in 1854 by Rankine, he used stress to denote the equilibrating set of forces which represents the elastic reaction of a strained body. In the following year Kelvin adopted the term stress, but used it for the numerically equal but opposite set of forces, defining as follows : — '^Definition. — A stress is an equilibrating application of force to a body. ' Corollary. — The stress on any part of a body in equilibrium will thus signify the force which it experiences from the matter touching that part all round, whether entirely homogeneous with itself, or only so across a part of its bounding surface. ' Definition. — A strain is any definite alteration of form or dimensions experienced by a solid ' {Encyclopaedia Britannica, ninth edition, vol. vii. p. 819). It is well to note here that Kelvin's use of the term stress brought the mechanics of elasticity into line with that previously developed for particles and rigid bodies. For just as Force = Mass X Acceleration and Torque = Moment of Inertia X Angular Acceleration, so Stress — Elasticity x Strain. Or, in symbols, the three analogous relations may be written F=Ma\ and S=zKs (2), °' ^'^Sjs (3), where S denotes stress in force per unit area, s strain as fractional change, and K the elastic constant or elasticity concerned, it being understood that the stresses and strains are kept below the elastic limit. Thus (2) and (3) each give the symbolic quantitative definition of elasticity. But although the meaning attached by Kelvin to the word stress ART. 457] STATICS OF ELASTIC SOLIDS 469 was probably the one whose need to be named was then most keenly felt, experience showed that a word was required for another closely allied conception, and for this also the same word stress became generally adopted. We must accordingly be prepared to meet the word in its more modern usage, which is indicated in the following passages or quasi-definitions. When there is no tendency to relative motion between the parts of a body, so that if cuts are made in it there is neither opening, closing, nor sliding anywhere, the body is said to be in a state of ease. When, however, on that test being made, it is found that there is tendency to relative motion as shown by opening, closing, or sliding, the body is said to have been in a state of stress. Stress is the pair of forces constituting the mutual interaction in or across a plane, or the entire action and reaction, of which force is one- half. Stress is force per unit area in or across a plane. Some writers reserve the word stress for the interior of a body and use load and reactions for the weight and supports applied externally. Whereas by Kelvin's definition the load and reactions would constitute the stress on the whole body, and the force per unit area in any direction at any place in it would be strictly a stress component on an element there. 457. We may further illustrate the different uses of the word stress by diagrams. Thus, in Fig. 233. eight forces are shown, being four pairs at the planes i to 4. Then on Kelvin's original definition and strict usage A and H constitute the stress on the whole body included be- tween the planes i and 4, A and D the stress on the part i to 2, and so forth. On the more modern method, A and H would be called the load and reaction, C and D, or either alone, the stress at the plane 2, and so on. In either case all the stresses are called compressive, because they are normal and tend to compress or crush the substance. Also, in either usage the value of the stress is the quotient g force per unit area. Again, in Fig. 234 is shown a cube about to be subjected to a shearing stress, i.e. one which produces the strain called a simple shear. Then, in the modern usage, any one of the four tangential forces B, E , D, or D', reckoned per unit area, would be called a shearmg stress, because it is a tangential force or a force in the plane and not normal to it or inclined. On Kelvin's definition the whole set of four forces is required to constitute an 'equilibrating application Fig. 233. Compressive Stresses. B Fig. 234. Shearing Stress. 470 ANALYTICAL MECHANICS [art. 458 of force ' ; and so nothing less is, by that definition, a shearing stress, the separate forces being stress components merely. It should be noted also that, in this respect, Kelvin's definition is strictly logical. For any one force only, say B, would produce linear acceleration simply. A pair of forces, on opposite faces, as B and B', constitute a couple, and would accordingly, by themselves, produce angular acceleration. There is no necessary production of strain until we have completed the set of forces B, B', D, D', which, on a rigid body, would be an equilibrating set, the strain then follows if the body is elastic. But however the word stress is used qualitatively to apply to one force, two, or four, both usages are in accord quantitatively, as the various forces in question are equal and any one is the gauge of the rest. We may thus without confusion use the term stress in the various recognised senses, as found convenient, generally leaving the context to show the exact meaning intended in each case. 458. General Homogeneous Stress and its Components. — Let us now consider a general application of forces to a body, simplify it till it forms a homogeneous equilibrating system or stress, and then specify that stress by its rectangular components. Further, let us take as oiir element which is the subject of this stress a cube, which may be the whole of the body or may be in the interior of a larger body from which it is separated in imagination only. Whatever the forces applied to the faces may be, we can denote them by their rectangular components, which are respectively normal to the face and parallel to its edges. We have thus three possible components for each face, and therefore eighteen in all for the six faces of the cube. But to reduce these to an equilibrating system we first make the normal components on opposite faces numerically equal. This reduces the total number of differing values from eighteen to fifteen. We next make the parallel tangential components on opposite faces equal and opposite so as to form a couple. This reduces the fifteen different components to nine, as there are two tangential forces along each of the three pairs of opposite faces. But there is a still further reduction of these nine different com- ponents to six; for, referring to Fig. 234, not only must £'=B and jy=D as already just provided for, but also B must equal D. For B=B'=D=D' measures the simple shearing stress corresponding to a simple shear in the plane of that djagram. Hence the most general application offerees to the faces of a cube, when reduced to an equilibrating system or stress, become a pair of equal and opposite normal forces on a pair of opposite faces, a set of four equal tangential forces parallel to the edges of these faces and applied on the other four faces as in Fig. 234; and then a similar set for each of the other two pairs of opposite faces. Thus to specify a stress we require only six different components in all, three normal and three tangential. ART. 458] STATICS OF ELASTIC SOLIDS 471 The subject may also be treated analytically. Thus, take the stage at which the parallel forces on opposite faces have been made equal, so that we have three forces to specify on each of three adjacent faces, or nine in all. Then we may denote them by the following scheme of symbols : — ^xj X,i, XA Kr- V... V. L Z. 'XI -Jj/, (4), y, ^-z J in which the large letters denote the direction of the forces and the subscripts give the normals to the faces on which those forces act. Following now a notation analogous to that used for strains in Table v. of article 179, we may express these nine components by the first letters of the alphabet : — A, £, c^ d,e,f\ Comparing these two schemes, we see that the normal components are denoted by A, E, I, the first three vowels, corresponding to the three elongations, which were previously denoted by a, e, i. We also see that our reduction from the nine to the final six com- ponents will consist in the equations (5)- or the equivalent set Yx = Xy, Zx = Xz, Zy= Y2 (6), D = B,G=C,H=F (7). Hence our set of six components which specify the stress are A, B, C^ (8), B,E,F\ C,F, I] the full nine being retained to show the equalities. These correspond to the general pure strain shown in Table v. of article 179. The relation of these six com- ponents of a general stress will per- haps be best understood by reference to a diagram such as that in Fig. 235. One force of each pair of equal ones is shown on the near face of the cube, the equal and opposite one on the hidden face being understood. All the forces must be imagined dis- tributed equally over the face to which it is applied, the capital letters must be taken as denoting the values of the quotients force per unit area, and the positive directions of the components on the cube are those indicated in the figure. Six Components of Fig. 235. General Stress. 472 ANALYTICAL MECHANICS [art. 459 Examples— LXXXIX. Give an account of the nature of elastic bodies, carefully distinguishing between the various special terms used in this connection. Explain carefully the various meanings which have been assigned to the word stress, indicating how we may still use the word quantitatively without fear of confusion. Illustrate your answer by concrete examples and sketches. Consider the most general set of forces acting on the faces of a cube, and show how, to represent the most general homogeneous stress, the eighteen components reduce to six. With the notation of article 458, specify a uniform hydrostatic stress and two shearing stresses, showing by a diagram how the forces act. Also prove that six components are sufficient to specify the most general homogeneous stress. If the application of force were such as to require more than six quan- tities to specify it, what would happen to the body ? 459. Stress across any Plane. — Let -the stress to which a material is subjected be specified by its six components A, E, /, F, C, B, referred to the co-ordinate planes, and take any plane cutting the axes at U, V, W, the direction cosines of its normal being /, m, n. It is required to find, on UVW, the stress N with direction cosines A, /M, V and components P, Q, and Ji parallel to the axes (see Fig. 236). Let the area of the plane tri- angle UVW be A, then those of its projections OVW, OWU, and OUV will be respectively /A, wA, and i^A. Then, using the con- ditions of equilibrium for the portion of material OUVW, we have by resolving parallel to the axes PA-AlA+BmA+ CnA, and two similar equations. Thus, cancelling out the A's all through, we find for the components and resultant F=A/+Bm+Cn=Arn Q=B/+£m+Fn=Nfji.\ . Ji=a+Fm+In=JVu I Fig. 236. Stress at given Plane. (0 also and _A P Hi. Q V I ''IV To embody this result in soid S=Ax'+£y+Iz' + 2(Pyz+Czx+Bxy) = K and take on its surface a point defined by the co-ordinates x=lr, y=mr, z = tir a geometrical form consider the ellip- (3), (4), (5). ART. 459] STATICS OF ELASTIC SOLIDS 473 Now, differentiating (3), we find ^-=2{AA-\.£y^C,) ~ = 2{Cx+Fy+Iz) Thus, remembering (4) and (i), we have dS dS dS_ dx'- dy '■ dz~ ■^'■'' • • • • (6),i which shows that the perpendicular (of lengthy say) from the origin to the tangent plane at x, y, z has the same direction cosines as the resul- tant stress 77 across the plane UVW, whose normal has the same direc- tion as the radius vector r to the point just named. For, from (5) and (i) we ma)' write MrX^zAx+By^ Cz] JVrtx. = £x+£y + Fzl /,) and JVrv^Cx+Jy+Is J ^^' Multiplying these equations by .v, y, and ,■: respectively, adding, and using (3) and (6), we obtain JVr(Xx+fiy+i'z) = IC=JVrp (8), oi- N=K\pr (9).' Thus, ' For any fully specified state of stress in a solid, a quadric surface may always be determined, which shall represent the stress graphically in the following manner : — ' To find the direction and the amount per unit area of the force acting across any plane in the solid, draw a radius perpendicular to this plane from the centre of the quadric to its surface. The required force will be equal' (ox proportional, unless K=i) 'to the reciprocal of the product of the length of this radius into the perpendicular from the centre to the tangent plane at the extremity of the radius, and will be perpendicular to this tangent plane. ' From this it follows that for any stress whatever there are three determinate planes at right angles to one another such that the force acting in the solid across each of them is precisely perpendicular to it. These planes are called the principal or normal planes of the stress ; the forces upon them, per unit area, its principal or normal tractions ; and the lines perpendicular to them its principal or normal axes, or simply its axes. The three principal semi-diameters of the quadric surface are equal (or proportional) to the reciprocals of the square roots of the principal tractions. If, however, in any case each of the three principal tractions is negative, it will be convenient to reckon 1 The student not familiar with tliis may easily verify the corresponding relations for an ellipse, its tangent, and the perpendicular upon it from the centre. 474 ANALYTICAL MECHANICS [arts. 460-461 (10). them rather as pressures ; the reciprocals of the square roots of which will be the semi-axes of a real stress ellipsoid representing the distribu- tion of force in the manner explained above, with pressure substituted throughout for traction' (Kelvin and Tait's Natural Philosophy, Part II. pp. 207-208, Cambridge, 1890). 460. Composition of Stresses. — The composition of stresses may, of course, be effected by the algebraic addition of like components. Thus, if several stresses are defined by A^, E^, /i, Fx, C^, B^, and the same with subscripts 2, 3, etc., the resultant stress has components l.A=A,+A,+ ...\ l.E=E,+E,+ .. 2/=A+/,+ ... ^F=F,+F,+ ... 2C=C, + C-1- .. 2B=B, + £,+ . . If all the stresses to be compounded have their principal axes coincident, each one reduces to its three principal tractions A, E, and /, and the resultant is, of course, given by the three terms "EA, XE, and 2/. Finally, if the strains to be compounded are each a simple traction, and all at right angles, we have them represented by A, o, o; o, E, o; and o, o, I, and their resultant by A, E, I. 461. Two Aspects of Shearing Stress. — In the chapter on strains we considered a simple shear at first (article 165) as a combination of an elongation and an equal rectangular contraction. We afterwards saw (articles 170-174) that it could be viewed as a sliding of planes at 45° to the above elongation and contraction, the amount of the shear, or slides per unit distance apart, being double the elongation. In referring to shearing stresses in the present chapter we have taken first the tangential-force aspect of the matter which corresponds to the sliding of the planes, which in the strain was con- sidered second. We now note the con- version of the other form into this, and shall find that here no doubling nor halving occurs in passing from one aspect io the other. Thus let the shearing stress be specified by its principal tractions o, E, —E, and take a plane at angles of 45° with the axes of y and z, so that its normal lies in their plane and between their positive directions, as shown in Fig. 237. Then, using the results in (i) and (2) of article 459 for the stress on Fig. 237. Two Aspects of Shearing Stress. this inclined plane, we have Q=EI J2 =NiJ. F=-EI ■j2=Nv (II). ART. 462] STATICS OF ELASTIC SOLIDS 475 Whence N=E, k=Q, fi^ij j2=: — v . . . (12). Thus, showing that the stress on the plane indined at 45° to the axes of traction and pressure has a tangential force of equal value per tmit area and directed as shown in the figure. This result can also be found easily by elementary considerations without the general formulae of article 459. 462. Elasticities and their Relations. — Let us now take symbols for the chief elasticities, find expressions for them, and establish relations between them. Let the traction ^ along the axis of x produce in an isotropic material the strain (a, — /, — «'), or (a, —a-a, —a-a) where o- is called Poisson's ratio. The elasticity involving change of size only is called the volume elasticity or bulk modulus. It is measured by the quotient hydrostatic pressure divided \iy fractional ditnitmtion of volume, or uniform normal tractions divided by fractional increase of volume. Thus denoting this elasticity by k, and using equations (2) of article (164) and (11) of 169, we find 8 id 3a(i-2qKb, i^Wb ^")' a formula which is often used in the laboratory exercise of finding q for wood and metals. For symmetrical cross sections it is clear that the neutral surface is central. Passing from the case of a clamped bar to that of one drooping b at the middle under a central load W, the distance between the supports being Z, we see that /=Z/2a.ndJ?=WJ2 (12). Thus writing these values in (n), it transforms into JFL' 1= ifiKb (13), this arrangement being often more convenient than the former, to which (ii) apphes. 465. Twisted Cylinder. — Let us now consider a right circular cylinder of radius a and length / with axis along the axis of z, the base in the xy plane held still while the opposite end is rotated through the angle Q in its own plane, QU being called the twist and denoted by t. It is required to determine the stress needed to maintain this twist. It is almost self-evident that the stress in question will consist of equal and opposite couples in the planes of the ends of the cylinder respectively. And further, it is obvious that the strain will consist of a progressive angular displacement of planes parallel to xy, increasing uniformly with z from zero at the base to 6 at the other end where z=/, none of these planes being them- selves distorted. A little reflection shows that the mutual mteraction of any elements of the material meeting at these planes is the Fig. 240. Twisted Cylinder. ART. 465] STATICS OF ELASTIC SOLIDS 481 undiminished handing on of a force parallel to the xy plane and perpendicular to the radius, such interaction being imposed by the apphcation at the ends of the cylinder of the forces constituting the stress, which is a pair of equal but opposite couples about the axis. Hence, referring to Fig. 240, we may write the strain as follows :— x' — x^=—Tyz\ y'-y=--{-Txz\ (i). z'—z=o J Whence, with our ordinary notation, we may write a=e^i=o \ f=TX, c=~Ty, b=o} ^'^'• Thus, for the stress components, we have A=E=I=o \ F—mx, C= — nTy,B=oj ■ • • ■ \6h where n is the rigidity of the material. All these equations show that the strain with which we have here to do is not of the type called homogeneous, for the displacements now involve theproducts of the variables instead of being a linear function of them. The components F and C give a resultant (refer, if necessary, to Fig. 235, at end of article 458) tangential to circles about the axis of z and of value expressed by T=nTr (4), as shown at the top of the cylinder in Fig. 240. Hence for any ring of radii r and r^dr the force would be 2irr(rnr)dr, and its moment about the axis of z would be r times that value. Thus, for the whole area, the torque G is given by 7=2irnTi r^, '^'-=— — (s). ""'^y T=^ ^^^• An alternative method of obtaining this relation is given in the writer's Text-Book of Sound, pp. 128-129, and is there followed by a description of some methods for the determination of the rigidity of a material. , t ■ If the rod is variable in radius we might write r (as a function of z) instead of a. In this case the twist, angle per unit length, will also vary along the axis from point to point. We may accordingly write dBldz in place of t. Making these substitutions in (5), we find dd=^-^-^A (M. Tzn r 2 H 482 ANALYTICAL MECHANICS [art. 466 466. Elongation of Helical Spring. — To deal rigorously with the problems of a helical spring is beyond the scope of this work, but by the simple device used in Perry's Applied Mechanics, pp. 628-629 (London, 1898), we may calculate ap- proximately the small elongations of a close helical spring of circular wire. In Fig. 241 is shown the spring, made of wire of radius r bent into coils of radius R to the centre of the wire. It is fixed at A, and at the lower end B in the axis a vertical force P is applied depressing that point by the amount p say. Let the weight of the spring be negligible in comparison with P. At any point Q in the spring take a section by a plane through the axis AB. Then, since the spring is a close one, this section is practically a normal section of the wire. Consider the equilibrium of the portion BQ. The force P down the axis is balanced by some forces at the section at Q. These must be equivalent to a force P vertically upwards and a torque of magnitude PR acting on the upper end Q of the portion BQ of the spring. And since Q is any point, this force and torque remain constant throughout the purely helical portions of the spring. The vertical force P \% 3, shear component of the stress, and its effect may be neglected in com- parison with the torsion. The torque PR applied throughout the spring will produce the corresponding uniform twist as found in equation (6) of article 465. And this twist will give the depression / of the point B proper to its final angle and the arm BC=.^. Thus, writing L for the length of the wire bent into the helical form and 6 for the total angle due to the twist, we have from (6) PR -Knr' Re P'IG. 241. ElON GATION OF HeLI CAL Spring, and p Whence, eliminating G between these equations p 2LR' (8). (9), where N is the number of turns in the helix, its slope being accounted negligible. For rigorous theories of helical springs the interested student may consult Perry {ibid. pp. 633-638) and Gray, Physics, vol. i. pp. 600- 609 (London, 1901). ART. 466] STATICS OF ELASTIC SOLIDS 483 Examples — XCI. I. Obtain expressions for the work per unit volume when an elastic body is strained from a state of ease to a specified finite strain. What does this general expression reduce to in the case of a wire stretched by an amount/, the final force being Pf 3. Prove that to bend a bar to radius r needs the application of bending moments of the value gKjr, where g is the Young's modulus of the bar and K the moment of inertia of the cross section about its intersection with the neutral plane. 3. Show that if a straight uniform bar is fixed at one end and very slightly bent by a perpendicular force at the other, its equation may be written ' y = Ax'^-Bx^, and state the values of the constants A and B. 4. Find the inclinations at the ends of a beam loaded in the middle and supported at the ends, also the droop in the middle. 5. A beam of clear length L between its supports has a load W uniformly distributed along it ; show that the droop in the middle is 5 WL^/^S^glC, where g is the Young's modulus of the material and IC the moment of inertia of the cross section of the beam about the line where the neutral surface cuts it. 6. Treat the problem of the pure torsion of a right circular cylinder of radius a and length /, and show that the couple per radian of total angular displacement of one end relatively to the other is ■irna*l2l, where n is the rigidity of the material. 7. Assuming the result of the previous example, show that the torsional oscillations of a cylinder or other body, of moment of inertia / about a central vertical axis, wh en suspen ded by a wire fixed at its top, may be expressed by T = 2Tr ^illinna'^, whence the rigidity of the wire is given by n = iirlIJ7'^a'^. 8. Investigate the pure torsion of a frustum of a cone whose bases have the slightly differing radii a and b, the length of the frustum being /, and show that the total angle 6 through which one base is turned relatively to the other by the opposite couples of magnitude G is given by 2£/ a^ + ab + b"- Q A filament of circular cross section fluctuates several times between the radii- a and b, all parts of it being conical without any intervening cylindrical portions ; show that, if one end is fixed, the couple per radian displacement of the other end is expressed by G _ ■K^ncfib^ T" iV ' where F is the volume of the filament. ,. „ j r • , 10 Show that for a close helical spring of radius R, made of circular wire of length L and of radius r, the relation between load and axial eloneation is that between load and displacement when the load is applied to an arm of length R on the end of the straight wire of length Z and radius r of the same material as the spring. 1 1 A uniform beam is placed horizontal y on two end supports, and the in- tervening portion then bears a uniformly distributed load. Taking he origin at centre of the beam, the axis of ^horizontally parallel to the orilinal position of the beam before loading and the axis oi y vertically upwards, show that the shearing force, the bending moment ^heW/of the beam, and its ordinate are proportional respectively to he/r5, second, third, ^^d fourth powers of the abscissa .r. 484 ANALYTICAL MECHANICS Examples — XCII. : Chikfly Kinematics. 1. 'Show how to connect linear, angular, and areal velocity, period, and revolutions per second in uniform motion in a circle ; and calculate the angular velocity of the hands of a clock. 'Determine the revolutions per second of a bullet fired with velocity 2000 feet per second from a rifle, the grooves of the rifling making one turn in 10 inches.' (LoND. B.A. AND B.Sc, Pass, Mixed Math., 1902, i. 5.) 2. ' Investigate the simple harmonic vibration of a weight suspemled by a vertical spiral spring, and determine the length of the simple pendulum which will synchronise. 'Prove that a train on a perfectly straight railway, not curved to the radius of the earth, will oscillate if unresisted on each side of the position of equilibrium in the same period as a grazing satellite, about one-seven- teenth of a day, or i h. 25 m.' (LoND. B.A. AND B.Sc, Pass, Mixed Math., 1902, i. 9.) 3. 'Write down formulas connecting angular velocity, linear velocity in feet/second, revolutions/minute, and period of revolution in seconds. 'Prove that a bicycle geared up to D inches requires 336 SjD revolu- tions/minute of the pedals for a speed of 5 miles/hour.' (LoND. B.A. AND B.Sc, Pass, Mixed Math., 1903, i. 5.) 4. ' Explain the theory, units, and notation of the formulas (1) Pt = —-; (11) Ps = ----; (ui) T/=2— . ' Supposing that one m in is the steepest incline a train can crawl up with uniform velocity, and one in n is the steepest incline on which the brakes can hold the train, prove that the quickest run up an incline of one in p from one station to stop at the next, a distance of a feet, can be made in V \m 71 J \in p )\ n p ) ■ seconds. ' Calculate for ;« = 50, ;? = 5, / = 100, a = 5280.' (LoND. B.A. AND B.Sc, Pass, Mixed Math., 1903, i. 6.) 5. 'Calculate the velocity at any point in a centrifugal railway and the thrust on the rails of a car, where its C. G. describes a vertical circle of radius a feet, due to entering at the lowest point from an incline with a fall of h feet vertical. ' Prove that h should not fall short of 5^/2, and the car should be strong enough to sustain the weight sixfold.' (Lond. B.A. .'VND B.Sc, Pass, Mixed Math., 1903, i. 8.) 6. ' Show that an angular velocity . (LoND. B.Sc, Pass, Applied Math., 1905, 11. 6.) EXAMPLES— CHIEFLY KINEMATICS 485 7. ' One end 5 of a rod AB describes a circle, while the other end A is con- strained to move along a line which passes through the centre C of the circle. Prove that the ratio of the speeds of A and B is equal to the ratio of AM io AJV, where BJV is the perpendicular from B upon AC and AxM is that from A upon CB. ' If AB (or AB produced) meets the circle again in B', show that the rate of increase of AB' is to the speed of A as 2AC\s to AB.' (LoND. B.Sc, Pass, Applied Math., 1906, 11. 6.) 8. ' If a particle is projected from 0, under the action of gravity, at an eleva- tion a, with a velocity due to a height /;, show that the equation of the parabola described with reference to horizontal and vertical axes at is where m = tan a. Find the greatest height attained.' (LoND. B.A., Pass, Applied Math., 1907, i. 6.) 9. ' A particle describes an ellipse under a centre of force in a focus which produces an acceleration fijr^ ; prove the formula for the velocity ^-K^-T> where a is the major semi-axis of the ellipse. 'The maximum velocity of the earth in its orbit is 30,000 metres per second, and the minimum velocity is 29,200 metres per second ; deduce the eccentricity of the earth's orbit.' (LoND. B.A., Pass, Applied Math., 1907, i. 9.) 10. ' Obtain a formula for the velocity at any point of an elliptic orbit described under a central force to one of the foci. ' The greatest and the least velocities in such an orbit being 1 10 ft. per sec. and 90 ft. per sec. respectively, and the periodic time being 20 min., calculate the eccentricity and (approximately) the length of the major axis.' . (LoND. B.Sc, Pass, Applied Math., 1907, 11. 5.) II 'A particle of unit mass moves in a straight line from rest under a constant accelerating force ^ and a retarding force ^Z'"'. Show that after describ- ing a distance x its velocity is given by Z/2 = ^(l-«-2*"^).' (LoND. B.Sc, Pass, Applied M.ath., 1908, 11. 7-) I-' 'The arms AC, CB of a wire bent at right angles slide upon two fixed circles in a plane. Show that the locus of the instantaneous centre m space is a circle, and that its locus in the body is a circle of double the radius of the space centrode.' „ ,r..r„ Tnr,s m -, 'i (LoND. B.Sc, Pass, Applied M.4.TH., 1908, in. 2.) n 'A particle is projected at right angles to the line joining it to a centre ^' of force attracting according to the law of the inverse square with a :?s:^^S^iKSS^:S^^;S^istS^^ timetakentodescribetliemajoi.^^^^^^^^^^^ ''■ '?^Po tin";: P/TSeSe'^i'o^S^^^^^^ circles of radii ., and at the centre of the circles such that ^ ^^^'^-(toND^fetSTppirD MATH., 19IO, n. 2.) 486 ANAL YTICAL MECHANICS 15. 'If (r, 6) be the polar co-ordinates of a point P moving in any manner in a plane, show that the accelerations of P along and perpendicular to the radius vector are r=r6\ —-^(r^e). r at ' 'A smooth horizontal tube OA of length a is movable about a vertical axis OB through the extremity O. A particle placed at the extremity A is suddenly projected towards O with velocity aa>, while at the same time the tube is made to revolve about OB with angular velocity a. Show that the particle will have travelled half-way down the tube after a time — loge2, and will not reach in any finite time.' (LoND. B.Sc, Pass, Applied Math., 1910, 11. 3.) 16. 'Translate : — ' Les organes qui permettent de faire passer le mouvement, de la pifece menante k la pihce menee, se nomment m^canismes. Les mecanismes peuvent Itre classes en deux categories ; la premiere categorie comprend les syst^mes articules dans lesquels les angles varient, les distances des articulations restant constants. La deuxifeme categorie comprend les systfemes constitues par une piece P anim^e d'un mouvement M qui, par contact continu, imprime un mouvement JVT k une autre pifece P.^ (LoND. B.Sc, Pass, Applied Math., 1910, 11. 10.) 1 7. ' Prove that the orbit described by a particle under a force which tends to a fixed centre, and varies inversely as the square of the distance from the centre, is a conic. ' Prove that, if P is the particle, 5 the centre of force, and N the foot of the perpendicular from P on the axis of the conic, the velocity of N is greatest when it coincides with 5.' (LoND. B.Sc, Pass, Applied Math., 1910, in. i.) 18. ' Investigate the motion of a heavy particle allowed to fall from rest in a medium which offers a resistance proportional to the velocity. ' Prove that the velocity constantly increases, but tends to a finite limit.' (LoND. B.Sc, Pass, Applied Math., 1910, in. 4.) 19. 'Translate the following passage : — 'Das Planetensystem erleidet allerdings im Laufe der Jahrtausende bedeutende Umgestaltungen. Doch treffen sie diejenigen beiden Elemente nicht, welche man mit voUstem Recht als die hauptsach- lichsten bezeichnen kann, namlich die mittleren Entfernungen der Planeten von der Sonne und ihre Umlaufszeiten um dieselbe. Hat man die letzteren als Mittel von hunderten oder tausenden von beobachteten Umlaufen bestimmt, so ist man sicher, dass dieses Mittel die unveran- derliche Umlaufszeit genau darstellt.' (LoND. B.Sc, Pass, Applied Math., 1910, in. 10.) 20. ' Prove that, at any point of the path of a moving particle, the normal component of the acceleration is v'^R, where v is the velocity and R the curvature at the point. Deduce its expression in terms of x, % and /.' (Board of Education, Theo. Mech., Solids, Stage 3, 1908, 47.) 21. 'What is meant by the motion of a lamina in its own plane? ' Show that a lamina can be moved in its own plane from any one posi- tion to another by a rotation round a point in that plane. Point out any case that presents an apparent exception. ' The motion of a body at any instant can be represented by two angular velocities round parallel axes ; find a simpler mode of representing the motion.' (Board of Education, Theo. Mech., Solids, Stage 3, 1908, 48.) EX A MPLES— CHIEFL Y PAR TICLE KINE TICS 487 22. ' Show that two equal and opposite rotations, effected successively round two parallel axes A and B, are equivalent to a single motion of translation. ' Illustrate your answer with reference to a carefully drawn diagram of the following case : — Suppose that AB is a side of a square and that the square is made to turn in its own plane round A through 60°, and then, in the opposite direction, round B through an angle of 60° ; also, show the direction of the translation.' (Board of Education, Theo. Mech., Solids, Stages, 1909, 46.) 23. 'A motor starts from rest to go a distance .s', and the acceleration of its velocity at time /is k{i ), r being the time taken to acquire the T maximum \elocity, which after that time is maintained. ' If rbe the whole time for the distance, and T^ that of the same motor with a flying start, prove that r=3(r-ri), (Board of Education, Theo. Mech., Solids, Stage 3, 1909, 47-) 24. 'Suppose that a square ABCD undergoes a very small elongation parallel to the side AD caused by a uniform tension T; also that it undergoes a very small compression parallel to the side AB caused by a uniform pressure T. Show that the square is under a shear in the direction A C, and find its magnitude.' (Board of Education, Theo. Mech., Honours, 1909, 62.) 25. 'Explain what is meant by a rotation and by an axis of rotation. Con- sider a body which turns round an axis, and a straight line in the body which is in a plane at right angles to the axis of rotation. Show that, at the end of the motion, the line makes with its initial position an angle equal to that through which the body has been turned. 'AB, AC, AP are three lines forming a solid angle at A, and AP is such that a rod coinciding with AB can be brought into the position ^C by a rotation round A P. Supposing that AB and ^C are fixed find the locus of AP. Find also the relation between the angle of rotation and the angle BA C (Board of Education, Theo. Mech., Solids, Stage 3, 1910, 48.) Examples— XCIII. : Chiefly Particle Kinetics. I ' Pro^•e that a jet of water of delivery P lb. per second, and ^■elocity 7/ feet per second, impinging on a plane pallet fixed perpendicular to its direction, will exert a thrust Pvjg pounds. r„„„^p „f „ ..-heel ' If such a series of pallets are mounted on the circumference of a «heei moving with velocity u, the horse-power given out will be a maximum i'^.V-oo^-vhen « = .73; and half the energy of the jet is then utilised ^Y *e -heel.' ^^^ ^^^^^^ , ,.) resistance, assumed constant.'^ ^^^^ ^^^^^ ^^^^^^ ^^^^^_^ ,g„,_ , 3.) j88 ANALYTICAL MECHANICS 3. ' A particle m describes in succession the sides of a regular polygon, each with the same constant velocity v. Prove that in order that it may do this an impulse of amount tnvHjr must be applied to it at each corner, where / is the time of describing a side and r is the radius of the circle circumscribing the polygon.' (LOND. B.A. AND B.Sc, Pass, Mixed Math., 1904, i. 5.) 4. 'A force whose magnitude is at each instant completely given acts always in a given right line ; show how to draw a diagram representing (a) the work done by the force in a given interval ; {b) the whole impulse of the force in this interval. ' A ball whose mass is 4 ounces strikes normally a fixed plane with a velocity of 50 f./s. ; the coefficient of restitution is 2/5, and the time of contact is 1/256 second; taking ^=32, what is the mean pressure between the Ijall and the plane in lbs. weight ? 'What is, approximately, \hc greatest ^ressurel' (LoND. B.Sc, Pass, Mixed Math., 1904, 11. 3.) 5. 'At the top, B, of a rough plane inclined to the horizon tan "'3/4 is fixed a pulley ; a uniform chain having a mass of 5 lbs. per foot lies on the plane along the line of greatest slope and passes over the pulley at B. If the coefficient of friction is 1/2, and 20 feet of chain lie on the plane, find the amount of work done against friction when the free end of the chain is pulled until {a) 10 feet of chain have come over ; (*) 20 „ „ „ (Lond. B.Sc, Pass, Applied Math., 1905, i. 7.) 6. ' A particle makes small abrasions on a horizontal straight line under the influence of a spring of negligible mass attached to a fixed point. Dis- cuss the motion, and prove that for different particles the time of oscillation varies as the square root of the mass attached.' (Lond. B.Sc, Pass, Applied Math., 1905, in. i.) 7. ' Assuming that the attraction of the earth is directed towards the centre, and is constant at all points of its surface, prove that the deviation of the direction of gravity from the radius due to the rotation of the earth at a place in latitude X varies as sin iX, and that its maximum value is 6' approximately.' (Lond. B.Sc, Pass, Applied Math., 1905, iii. 2.) 8. ' Prove that the work done by an impulse on a particle in the direction of its line of motion is measured by the product of the impulse and the mean of the initial and final velocfties of the particle. ' Hence find the loss of kinetic energy in the direct impact of two given inelastic spheres.' (Lond. B.Sc, Pass, Applied Math., 1905, in. 3.) 9. ' Discuss the motion of a heavy particle making complete revolutions within a smooth circular tube which is fixed in a vertical plane. ' If the speed at the lowest point is n times the speed at the highest point, prove that the pressure of the particle on the tube, when the particle is moving vertically, bears to the pressure at the lowest point the ratio of 2{n^+ i) : s»2- I.' (Lond. B.Sc, Pass, Applied Math., 1905, in. 4.) 10. 'A mass M strikes a mass vi which is at rest with velocity v ; prove that the velocity communicated to m is always less than 2.v. ' A hammer whose head weighs 3 lbs. strikes a steel ball weighing 5 ozs. with a velocity of 50 f.s. ; find the velocity communicated to the latter, assuming that there is no loss of energy in the impact.' (Lond. B.A., Pass, Applied Math., 1906, i. 6.) EXAMPLES- CHIEFLY PARTICLE KINETICS 489 n. A number of equal particles of mass m are connected by strings, each of length <7, so as to form a regular polygon of n sides, and the whole revolves about the centre, the velocity of each particle being v. Find the tension in each string.' , . (LoND. B.A., Pass, Applied Math., 1906, i. 8.) 12. Define simple harmonic motion, and find the velocity in any given phase, in terms of the period and the amplitude. 'A mass of 10 lbs. is executing a S. H. motion of amplitude 3 ins., with a frequency of 2i complete vibrations per second. Find (in foot-lbs.) its maximum kinetic energy.' (LoND. B.A., Pass, Applied Math., 1906, \. 9.) 13. 'Find in gravitation units the force which must act along the normal on a particle of weight Ti' which is moving in a curved path. In what sense along the normal must this force act ? ' A uniform chain AB of length / and weight w per unit length rests on a smooth horizontal table. If the chain revolves freely round A, which is fixed, with uniform angular velocity m, find the tension at a point distant X from A.' (LOND. B.Sc, Pass, Applied Math., 1906, in. i.) 14. 'What is meant by a work diagram? A force acts in a fixed right line OA, its point of application being P ; the magnitude of the force is directly proportional to OP, and has the value F when OP = a ; draw the diagram representing the work done by the force in displacing P from the given position B to the position C along OA. ' A mass of 500 lbs. moves down 100 feet of a rough plane inclined at sin~'o'o5 to the horizon, frictional resistance being 15 lbs. weight. By a direct application of the principle of work and energy, find the velocity of the body when it reaches the foot of the incline.' (Lond. B.A., Pass, Applied Math., 1907, i. 3.) 15. ' If a particle of weight w is moving in a plane curved path whose radius of curvature at a given point is p, prove that there must act on the particle a force equal to •wv'^lg-p along the normal towards the concave side of the path, v being the velocity of the particle at the point. ' If a particle of weight w is suspended from the roof of a railway carriage which is mo\'ing with a constant speed v, prove that in the position in which the particle is at rest relatively to the carriage the tension of the cord is t£/(l+W«/^p2)l/2.. (LoND. B.Sc, Pass, Applied Math., 1907, iii. i.) 16. 'A particle revolves in a circle about a spherical body attracting accord- ing to the law of the inverse square. Show that, if Tbe the time of a complete revolution, the mass of the attracting body is given m astronomical units by 477 V/r^ r being the radius of the orbit described. ' Given that the time of revolution of the Earth about the Sun is approxi- mately 13 times that of the Moon about the Earth, and the Sun's distance is 400 times the Moon's distance, compare the masses of the Sun and Earth.' (Lond. B.Sc, Pass, Applied Math., 1908, 11. 5.) 17. 'A particle acted upon bv gravity is projected with velocity v, and at an inclination a in a uniform medium of which the resistance varies as the velocity ; find the altitude of the particle at a given time, and show it is a maximum at time -^log(i + — I'sina), where k is the resistance per unit mass experienced by the body when it is moving with unit velocity.' (Board OF Education, Theo. Mech., Solids, Stage 3, 1909, 45.) .490 ANALYTICAL MECHANICS i8. 'A heavy particle hangs from a point O by a string of length a. It is projected horizontally with velocity v such that ■v'' = {7.-\r ■Ji)ga. ' Show that the string becomes slack when it has described an angle and that the subsequent path of the particle passes through O.' (Board of Education, Theo. Mech., Solids, Stage 3, 1909, 48.) 19. 'A heavy particle is placed at the highest point of a smooth vertical circular disc ; it is connected by an inextensible string with an equally heavy particle which is at the extremity of a horizontal radius. If motion be allowed to ensue, prove that the upper particle will leave the disc when at an angular distance from the highest point given by the equation 2 cos 6 = 6+ i. ' At that instant find the tension of the string and the velocity of the system. Find also an approximate value of 6.' (Board of Education, Theo. Mech., Solids, Stage 3, 1910, 41.) 20. 'Two particles, whose masses are given, are connected by an inex- tensible string, and are projected in any way, but so as to move in a vertical plane. Define their motion, and find the tension of the string at any instant during the motion.' (Board of Education, Theo. Mech., Honours, 1910, 65.) Examples — XCIV. : Chiefly Rigid Dynamics. 1. 'Write down the radius of gyration of a homogeneous billiard ball, and prove that it will roll down a plane slope a with acceleration fifsin a. ' Prove that if rolled horizontally on the plane with velocity V, the ball will proceed to describe a parabola with latus rectum 'j* l^'/g' sin a. (Galileo's experiment.)' (Lond. B.Sc, Pass, Mixed Math., 1902, 11. 9.) 2. ' Investigate the harmonic vibration of the balance wheel of a chronometer of moment of inertia / (Ib.-ft.^), and prove that if the wheel svnngs through N° from rest to rest in T seconds, the maximum couple exerted by the balance spring must be nHN/i8ogT^ (lb. -ft.).' (Lond. B.Sc, Pass, Mixed Math., 1902, 11. 10.) 3. ' Define the centre of oscillation of a compound pendulum, and show that it is convertible with the centre of suspension. 'A uniform solid rectangular parallelepiped has its edges 6, 9, and 12 inches long ; what is the least time in which it can oscillate about a horizontal axis, and how must the axis be fixed in the body?' (Lond. B.Sc, Pass, Mixed M.\th., 1904, n. 9.) 4. 'A uniform rod turns in a vertical plane about a point distant one-third of its length from one end. Find the centre of percussion. ' Determine also the impulse on the axis when the rod receives a given horizontal impulse at a point distant c from the lower end.' (Lond. B.Sc, Pass, Applied Math., 1905, in. 7.) 5. ' A fly-wheel is movable in smooth bearings, about a horizontal axis, and is set in motion by a descending mass of weight' P, which hangs from one end of a cord coiled round the axle of radius r. This mass is found to descend through k feet in n seconds ; prove that the moment of inertia of the wheel about its axis is (' ,» •)'■''■• (Lond. B.Sc, Pass, Applied Math., 1906, in. 3.) ' If /" is in lis. weiglit, the answer is in lbs.-ft.2 EXAMPLES-CHIEFL Y RIGID t> YNAMICS 491 6. ' A uniform rod of length / and weight w is held at an inclination a to the vertical with its lower end in contact with a smooth horizontal plane, and IS then let fall. Find its angular velocity and its pressure on the plane, when it is inclined at d to the vertical.' ._, . ^ , (LOND. B.Sc, Pass, Applied Math., 1907, III. 9.) 7. Obtain a formula for the period of the oscillations of a compound pendulum. 'A uniform rod of length za is swinging as a pendulum about one of its ends, its greatest angular deviation from the downward vertical being a. At an instant when the rod is vertical its fixed end is suddenly released ; find how far the centre of the rod descends before the rod is again vertical.' (LoND. B.Sc, Pass, Applied Math., 1908, in. 5.) 8. ' Define the principal axes of a rigid body. ' If the body be a plane lamina, explain why one of the principal axes at any point must be at right angles to the plane. ' If the lamina be an equilateral triangle of uniform density, find the principal axes at one of the angular points. Find also the principal moments of inertia at that point.' (Board of Education, Theo. Mech., Solids, Stage 3, 1908, 49.) 9. ' Find the period of a complete double oscillation of a compound pen- dulum when the angular swing is small. How must the pendulum be suspended to make the period a minimum ? ' (Board of EDUCATiobf, Theo. Mech., Solids, Stage 3, 1908, 50.) 10. 'Find an expression for the kinetic energy of a body moving, in any given way, in one plane. '' AB is an inclined plane and BC is the horizontal plane, through the lowest point B^ and both planes are smooth. A uniform rod is placed on AB with one end at B, and is allowed to slide down. Find its angular velocity just before its upper end leaves the inclined plane.' (Board of Education, Theo. Mech., Solids, Stage 3, 1908, 51.) 11. 'A uniform rod can turn freely round one end. It is held in its highest position, and is then allowed to fall. On reaching its lowest position it encounters a fixed obstacle at its lower end. There is no rebound. ' Find the impact on the obstacle and that on the point of suspen- sion. ' The rod is 12 ft. long and weighs 10 lbs. ; compare the impacts with the inomentum of a body which weighs 5 lbs. and has a velocity of 8 ft. a second.' (Board of Education, Theo. Mech., Solids, Stage 3, 1909, 49-) 12. 'A uniform beam AB of length ia hangs vertically from the end A, which lies in a smooth horizontal groove. ' If the end A be projected with velocity tc along the groove, show that the middle point of the beam will move with a velocity whose horizontal component is where k is the radius of gyration of the beam about its middle point. 'Find the equation which determines the angular velocity of the beam ""' (Board OF Education, Theo. Mech., Solids, Stage 3, 1909, S'-) n 'Discuss the sensibility of a balance with equal arms. ^ 'Show how to find the period of a small oscillation and, for a given load. Drove that it varies directly as the square root of the sensibility. prove tnat K^^^^ ^^ Education, Theo. Mech., Honours, 1908, 62.) 492 ANALYTICAL MECHANICS 14. 'A uniform rod AB lies on a smooth horizontal plane ; C is a fixed point vertically over B ; a thread carries a weight P, and after passing over C is fastened to B ; the end B can move freely in a vertical guide or groove coinciding with the vertical line BC. The weights of P and of the rod are equal. If P is allowed to fall, find {a) the angular velocity of the rod in any assigned position, (b) the direction of the motion of the centre of gravity, {c) the reaction of the groove on the end B.' (Board of Education, Theo. Mech., Honours, igo8, 68.) 15. 'A beam of mass M and length a rotates about one extremity on a smooth horizontal plane, there being no forces except the resistance of the atmosphere. If the retarding effect of the resistance on a small element of the beam be equal to A times the square of its velocity, show that in time t the angular velocity will be reduced from O to a, where (Board of Education, Theo. Mech., Honours, 1909, 66.) 16. 'A uniform rod is placed very nearly upright on a smooth horizontal floor and against a smooth vertical wall, and it slides down in a plane at right angles to both wall and floor. Find its position at the instant of its leaving the wall ; find also how it is moving at that instant, and the pressure on the floor.' (Board of Education, Theo. Mech., Honours, 1909, 67.) 17. 'A cone 8 inches high, radius of base 4 inches, weighs 5 lbs. Deter- termine its moment of inertia about an axis through its centre of gravity parallel to its base.' (Board of Education, Theo. Mech., Solids, Stage 3, 1910, 47.) 18. ' In a compound pendulum form the equations of motion to determine the horizontal and vertical components, X and Y, of the force acting at the point of suspension at any instant. ' Let W be the weight, r the distance of the centre of gravity from the point of suspension, and k the radius of gyration about the same point. Suppose that the pendulum is allowed to fall from the position in which the centre of gravity and the point of suspension are in the same horizontal line ; show that, when the pendulum is inclined at an angle of 45° to the vertical, x-Y={.-g)w: (Board of Education, Theo. Mech., Solids, Stage 3, 1910, 50.) 19. 'Find, in the motion of a ballistic pendulum, the relation between the centre of percussion and the axis of spontaneous rotation.' (Board of Education, Theo. Mech., Solids, Stage 3, 1910, 51.) 20. ' Show how to find the moment of inertia of a hemisphere (of uniform density) about an axis drawn through its centre of gravity and parallel to its base, assuming that the moment of inertia of a sphere about a diameter is 2mr^/s. 'A hemisphere rests with its curved surface in contact with a rough horizontal plane. It is slightly disturbed from its position of equilibrium ; find the time of a small oscillation.' (Board of Education, Theo. Mech., Solids, Stage 3, 1910, 52.) EXAMPLES-CHIEFL V STA TICS 493 Examples— XCV. : Chiefly Statics. 1. ' Define the angle of friction 0, and by means of it determine graphi- cally where jamming begins when a drawer is pulled out by a handle to one side. 'Prove that a sash window of height a, counterbalanced by weights, cannot be raised or lowered by a vertical force unless it is applied within a middle distance a cot (^ ; and prove that if the cord of a counter- balance breaks, the window will fall unless the width is greater than a cot 0.' (LoND. B.A. AND B.Sc, Pass, Mixed Math., 1902, i. 2.) 2. 'Discuss the conditions of equilibrium of three forces. Determine graphically the stress in each bar of a jointed triangular framework, strained by three forces acting at the angles.' (LoND. B.A. and B.Sc, Pass, Mixed Math., 1903, i. i.) 3. ' Determine the force to be exerted by the hand at the end of each arm a feet long of a copying press to set up a thrust of P pounds, the screw being smooth and cut with n threads to the foot. 'Sketch in plan the forces which act on the press and the man to maintain equilibrium.' (LoND. B.A. AND B.Sc, Pass, Mixed Math., 1903, i. 4.) 4. ' A uniform bar is bent into the shape of a V with equal arms, and hangs freely from one end. Prove that a plumbline suspended from this end will cut the lower arm at a distance of one-third its length from the angle.' (LoND. B.A. AND B.Sc, Pass, Mixed Math., 1904, i. 9.) 5. ^ABCD is a rectangle; AB=i2 inches, AD = &; aX A, B, C, D are placed particles whose masses are proportional to 8, 10, 6, 16 respec- tively. Find the position of the centre of mass {a) by the theorem of mass moments ; (b) by means of funicular polygons.' (LOND. B.Sc, Pass, Applied Math., 1905, i. 5.) 6. ' A uniform square plate is divided into two portions by a straight line joining a corner to the middle point of a side. Prove that the line joining the mass centres of the two portions is perpendicular to the dividing line.' (LoND. B.A., Pass, Applied Math., 1906, \. i.) 7. ' Four equal light bars are jointed freely so as to form a rhombus ABCD, and the corners A, C are connected by a light chain. The whole hangs from A, which is uppermost ; and two equal weights W'are suspended from B and D. Find (graphically or otherwise) the tension m the cham.' (LoND. B.A, Pass, Applied Math., 1906, i. 4.) 8. 'A uniform ladder, of length / and weight W, rests with its foot on the ground (rough) and its upper end against a smooth wall, the mclma- tion to the vertical being a. A force F is applied horizontally to the ladder at a point distant c from the foot so as to make the foot approach the wall. Prove that P must exceed JV, — (u + itana), where M is the coefficient of friction at the foot.' . . '^ (LOND. B.Sc, Pass, Applied Math., 1906, i. 2.) 9. 'Assuming the position of the centre of gravity of a triangular pyramid deduce the position of the centre of gravity of a homogeneous solid right circular cone. 494 ANALYTICAL MECHANICS ' The radii of the bases of a frustum of such a cone are 6 feet and 3 feet, and the thickness of the frustum is 9 feet ; find the position of its centre of gravity.' (LoND. B.A., Pass, Applied Math., 1907, i. 7.) 10. 'A uniform bar, AB, rests with its end A on a. rough horizontal plane, for which the angle of friction is X ; the bar is to be kept at a given inclination to the horizon by means of a cord attached to B. Ejdiibit in the figure the extreme directions of the cord which will allow of equilibrium.' (LoND. B.Sc, Pass, Applied Math., 1907, i. 4.) 11. 'Es ist zu beweisen dass die an einem starren Kdrper angreifenden Krafte, falls sie in einer Ebene liegen, im allgemeinen einer einzigen resultierenden Kraft statisch Equivalent sind, welche auch in ein Poinsot'sches Kraftepaar iibergeben kann. Wie setzt man die Krafte graphisch zusammen?' (LoND. B.Sc, Pass, Applied Math., 1907, i. 10.) 12. 'Two equally rough pegs A and B are a distance 2a apart in a straight line inclined at an angle 6 to the vertical. A rod passes over the peg A and under the peg B, and is just kept from sliding down by friction at the pegs. Prove that the centre of gravity of the -rod must be at a distance from the upper peg A equal to a(cot6/fi.- i), %1'here /t is the coefficient of friction between the rod and the pegs.' (LoND. B.Sc, Pass, Applied Math., 1908, i. 6.) 13. ' Show how to reduce a system of coplanar forces to a single force and a couple. ' Forces of magnitudes i, 2, 3, 4, 5, 5 act in the order named round the sides of a regular hexagon, the senses of the forces in adjacent sides being either both towards, or both away from, the corresponding vertex. Find the single force at the centre of the hexagon and the couple to which the system reduces.' (LoND. B.Sc, Pass, Applied Math., 1909, i. 3.) 14. 'A heavy uniform bar rests with one end on the ground and the other end against the vertical face of a rectangular block, which also rests on the ground. The vertical plane through the bar is perpendicular to the given face, and passes through the centre of gravity of the block. The coefficient of friction for the contact of the bar, both with the ground and with the block, is n, and the weight of the block is four times that of the bar. Show that if the bar is on the point of slipping and the block on the point of overturning, the ratio of the length of the bar to the width of the block is (i +/i^)(2 + 3;j.^)//»(i - /i^).' (LoND. B.Sc, Pass, Applied Math., 1910, i. 4.) 15. A sphere is composed of a solid homogeneous hemisphere and a very thin hemispherical shell of equal mass. Show that the composite body cannot rest in equilibrium upon a rough plane if its inclination with the horizontal exceeds the angle whose sine is 0*0625. 16. '■ ABCDE is a frame of five equal bars, kept in the form of a regular pentagon by two bars AC, AD. The frame is hung up by the point A, and carries equal weights ( W) at B and E. Find the stresses in the bars, putting out of the question the weight of the bars and the friction of the joints. ' Explain how the results would be altered if the weights were hung at C and D instead of at B and E.^ (Board of Education, Theo. Mech., Solids, Stage 3, 1908, 41.) EXAMPLES-CHIEFL V A TTRACTIONS 495 17. 'Show that any system of forces, acting on a rigid body, can be replaced by a single resulting force acting at any chosen point and a couple. ' Give further information concerning this force. 'Define Poinsot's central axis, and show how to construct it for any given system of forces.' (Board of Education, Theo. Mech., Solids, Stage 3, 1908, 43.) 18. 'A particle of weight ui rests against the circumference of a circular plate, whose plane is vertical ; a cord attached to the particle passes over a pulley placed vertically above the highest point of the circle at a distance from the circle equal to the radius and carries a weight/ ; show that the particle will rest at an angular distance 6 from the highest point of the circle where cosd = {^/4-p^jw''). ' Prove also that the pressure on the circle is independent of the magni- tude of/.' (Board of Education, Theo. Mech., Honours, 1908, 61.) 19. ' The end of a cylinder is pressed against a rough plane by a force which is equally distributed over the area of contact. The cylinder could move freely round its axis were it not for the friction. Find the force applied along a tangent of the base which will just make the cylinder turn.' (Board of Education, Theo. Mech., Solids, Stage 3, 1909, 44.) 20. ' A system of concurrent forces in space are given in magnitude and line of action. ' Show how from the plans and elevations of these forces to draw the plan and elevation of the resultant force. 'A horizontal triangle has sides 10', 10', and 5' respectively; from the three corners hang three ropes, each 6' long ; they are joined at their extremities and support a weight of i cwt. ; find the pull on each rope.' _ ^ (Board of Education, Theo. Mech., Honours, 1910, 61.) Examples — XCVI. : Chiefly Attractions. I. 'Find the attraction at any point in the substance of a solid sphere of given uniform density. 'A thick shell of uniform density is bounded by spherical surfaces which are not concentric ; prove that the attraction in the internal cavity is uniform in magnitude and direction.' (LOND. B.Sc, Pass, Applied Math., 1905, 11. 9.) 2 ' D emontrer que lorsqu'on passe au travers d'une couche de density super- ficielle T I'intensite de la force d'attraction dans la direction perpen- diculaire \ la surface recoit un accroissement subit de 41770-.' (LoND. B.Sc, Pass, Applied Math., 1907, 11. 9-) 7. ' Show that the attraction of a solid homogeneous sphere at any point outside it is the same as if its mass were collected at the centre. ' Prove that if the earth were homogeneous throughout, the decrease in gravitational attraction as one rose through a certain height in a balloon would be approximately twice the decrease as one descended an equal depth into a mine.' ^^^^ ^^^^^ ^^^^^^^ ^^^^^ ^^^^_ „ ^^ 4. In a homogeneous sphere, of radius a and density p, is drilled to the centre a cylindrical hole of very small radius 6. Show that the attrac- tTon of Se sphere on a plug' of same density filling the hole is 5. 'Ssrigate the attraction of a thin horoogeneous circular plate of radius 496 ANALYTICAL MECHANICS a at a point which is at a perpendicular distance c from the centre of the plate. ' Remark upon the cases : — (1) a infinite, c finite ; (2) a finite, c infinitesimal.' (Board of Education, Theo. Mech., Solids, Stage 3, 1908, 44.) 6. ' Define the potential of a system of attracting or repelling masses at any point. 'Mass, attracting according to the law of nature, is uniformly distri- buted on the circumference of a circle. Prove that the chord of contact of tangents drawn from an external point divides the mass into two parts having equal potentials at the point.' (Board of Education, Theo. Mech., Solids, Stage 3, 1908, 46.) 7. 'Define the potential of a single particle and of a given distribution of matter at an assigned point, and state what is its physical meaning in the case of gravity. ' Find the potential of a spherical shell of uniform density at an assigned external point.' (Board of Education, Theo. Mech., Honours, 1909, 61.) 8. ' Consider a cylindrical surface of given length and radius and of uniform surface density ; find the attraction at one end, /", of the axis.' (Board of Education, Theo. Mech:, Solids, Stage 3, 1910, 42.) 9. ' Find the attraction of a plane, of uniform surface density and of in- definite extent, at a point outside it. 'Let AB he a. diameter of a spherical surface of uniform surface density. Show how to draw a plane at right angles to AB, which will divide the surface into two parts, such that their attractions at A shall be equal.' (Board of Education, Theo. Mech., Honours, 1910, 63.) Examples — XCVII. : Chiefly Hydrostatics. 1. ' Show how to determine the specific gravity (S. G.) of a solid or liquid by the Hydrostatic Balance, and investigate the formula. ' Prove that a brass pound weight made to equilibrate the standard pound of platinum in air is too great by the fraction (4-4)/('-4)' ~ where A, B, P denote the S. G. of air, brass, and platinum.' (Lond. B.Sc, Pass, Mixed Math., 1902, 11. 3.) 2. ' State the principle of Pascal for the transmission of fluid pressure. ' A rectangular area is immersed vertically in water with one side hori- zontal at a depth of 9 feet and the opposite side at a depth of ij feet ; show that the centre of pressure is 3 inches below the middle point of the rectangle.' (Lond. B.Sc, Pass, Mixed Math., 1904, 11. 7.) 3. ''ABC is a triangular area immersed vertically in water with C in the surface and AB horizontal ; show how to divide the area by a horizontal line PQ into two portions on which the pressures are equal, P and Q being points in yiC and DC respectively. 'If ^ is the length of the perpendicular from C on AB, prove that the height above AB of the centre of pressure on the area APQB in the above case js -|-(3x4'«-4).' (Lond. B.A., Pass, Applied Math., 1906, 11. 2.) EXAMPLES— CHIEFL V HYDROSTA TICS 497 4. ' Obtain an expression for the total pressure exerted by water on a plane area occupying any position. 'A rectangular area A BCD has the side AB'm the surface of water, the side AD (10 feet long) being vertical and submerged ; divide the area by horizontal lines into three parts on each of which the pressure is the same.' (LOND. B.Sc, Pass, Applied Math., 1906, iii. 7.) 5. ' A hollow cone consisting of a curved surface closed by a circular base, both made of thin sheet metal, is 12 inches high and has a radius of 5 inches. The cone is to rest completely submerged in water with its vertex fixed and its axis horizontal ; find the necessary weight of metal per square inch of surface, assuming that a cubic foot of water has a mass of 1000 ounces.' (LoND. B.Sc, Pass, Applied Math., 1906, in. 9.) 6. 'A rectangular tank is divided into two compartments by a vertical diaphragm. The two compartments are then filled to heights /;, k with liquids of densities p, o- respectively. Show how to choose /; and k so that the resultant of the pressures on the diaphragm shall reduce to a couple, and find the magnitude of this couple per unit breadth of the tank.' (LoND. B.A., Pass, Applied Math., 1907, 11. i.) 7. ' State the conditions for equilibrium and for stability of a body floating freely in water. 'A solid sphere of radius r, weight ]V, and specific gravity s lies in the bottom of a cylindrical vessel of radius a and height //, which contains water just up to the top. Prove that the work required to raise the sphere just clear of the water is (LoND. B.Sc, Pass, Applied Math., 1907, in. 5.) 8. 'A rectangular area is immersed vertically in water with one side hori- zontal, and at a depth .r, the opposite side being at a depth h+x; show that the distance of the centre of pressure from the upper side is // 3-r+_2/^. 3 2.r + h ' Show that if the area is divided by a horizontal line into two parts on which the water pressures are the same, the depth of this line below the surface of the water must be {x"- ■vhx + \h^f '-.' (LoND. B.Sc, Pass, Applied Math., 1907, iii. 6.) Q. -A thin hollow vessel in the shape of a paraboloid of revolution floats in water with its axis vertical and vertex downwards. If the weight of the vessel itself be neglected, find approximately to what height it must be filled with mercury (specific gravity 13-6) in order that its vertex may be 18 inches below the free surface of the water, lb incnes oeio ^^^^^^ ^^^^ p^^^^ Applied M.«h., 1908, iii. 10.) 10 ' State and prove the principle of Archimedes. fl„,^:„„ 'A vessel contains two liquids that do not mix, and a cyhnder floating .^th axis vertical ; the lighter liquid is 5 - /-p, and '^^P gr is o 8 the so er of the heavier liquid is ri5, and li m of the height ottne cyUndef is above the upper liquid. If the sp. gr. of the cyhnder ^ o 75, what is its height?' ^ g^^ p^^^^^ ^pp^j^^ ^j^^^_^ ,g^^ l„. 7.) ■'■ SiiiiSillSii 2 1 498 ANALYTICAL MECHANICS the imprisoned air is found to occupy half the length of the pipe. Find the specific gravity of the liquid, assuming the water barometer to stand at 33 feet.' (LoND. B.Sc, Pass, Applied Math., 1909, iii. 9.) 12. 'A circular hole, of radius a, in the plane vertical side of a cistern is exactly filled by a solid wooden sphere of specific gravity o-, which is free to turn about a fixed smooth horizontal axle which is diametral to both the circle and the sphere. The cistern is filled with water of density w to a height h, greater than a, above the centre of the sphere. Evaluate the action between the sphere and the axle. ' What couple, if any, is required to keep the sphere from rotating ? ' (LoND. B.Sc, Pass, Applied Math., 1910, iii. 7.) 13. A casting is to be made by running molten metal (of density 1/4 lb. to the cubic inch) into a sand mould consisting of top and bottom parts. The casting is a rectangular table 12 feet by 5 feet with legs 3 feet high. The pattern is moulded face down and legs up, the joint between top and bottom parts being at the junction of the legs and the rectangular face or table. Show that, to hold the top part of the mould down at the instant of casting, a distributed load of nearly 35 tons is needed, with almost an additional ton for every inch of accidental excess height in the legs when pouring the metal in. 14. 'Explain a general method for determining the position of the centre of pressure of a plane surface immersed in any manner in a fluid. ' A plate in the form of a quadrant of a circle is immersed vertically in water with one edge in the surface. Find the distance of the centre of pressure from the horizontal and vertical edges respectively.' (Board of Education, Theo. Mech., Fluids, Stage 3, 1908, 43.) 15. 'A cylindrical vessel, radius a, contains water, and a cylindrical body (of the same height), whose radius is b, is lowered into the vessel until it stands upright on the bottom ; none of the water is spilt. Show that the ratio of the increase of the potential energy of the water to its original potential energy is U^j^c? - b^.' (Board of Education, Theo. Mech., Fluids, Stage 3, 1908, 47.) 16. 'Define the capillary curve. Find an expression for its radius of curvature at any point of its length.' (Board of Education, Theo. Mech., Fluids, Stage 3, 1908, 52.) 17. 'Find the centre of pressure in the case of an equilateral triangle, com- pletely immersed with one edge horizontal, and its plane inclined at a given angle to the horizon. 'A vessel has the form of a regular tetrahedron. It is filled with water and made watertight. It is held with one face horizontal. Find the resultant pressures on the several faces {a) when the vertex is below the horizontal face, {b) when it is above that face.' (Board of Education, Theo. Mech., Fluids, Stage 3, 1909, 44.) 18. ' Find the least value of the specific gravity of a cube of uniform density that the conditions of equilibrium may be satisfied when it has no more than one angular point out of water. Also show that, when its specific gravity exceeds that value, the parts above water of the edges meeting in that point are equal. ' Consider the case when the specific gravity is 47 -f 48.' (Board of Education, Theo. Mech., Honours, 1909, 70.) 19. ' Investigate the height to which a liquid rises in a capillary tube of any cross section. ' Show that it rises as high in a cylindrical tube of diameter d as in one having for cross section a square of side d.' (Board of Education, Theo. Mech., Fluids, Stage 3, 1910, 50.) EXAMPLES-CHIEFL Y HYDROKINETICS 499 "■ Xtt'it"-^/f '"'' *^ ^^"'^'^ ^"■^^^"' ^^°^ *^' ^' h-^h' ^ 'What assumptions have been made in obtaining this result? What is k ? Determme it m foot-second units from the data height of barometer = 30 in., specific gravity of mercury = 13-596, ^=32-2, , „ specific gravity of air = o-ooi 3.' (Board of Education, Theo. Mech., Fluids, Stage 3, 1910, 52.) Examples— XCVIII. : Chiefly Hydrokinetics. 1. 'Discuss the flow of a fluid in a tube of non-uniform cross section, showing how the pressure varies. ' Will the fluid tend to push the tube before it ?' ^ (Board of Education, Theo. Mech., Honoitrs, 1908, 72.) 2. If water be flowing steadily through an inclined pipe of varyinsr section show that ' -— + "^— + ^ = constant, ig iu ' where at any section 2i = velocity in feet per second, ^ = pressure in lb. per sq. ft., ■zc = weight of a cubic foot of water, ^• = height of section in feet above a fixed horizontal plane. ' Show that for steady motion u cannot exceed a certain limiting value.' (Board of Education, Theo. Mech., Honours, 1909, 72.) 3. 'State the physical fact expressed by the "equation of continuity " in the motion of fluids. ' Establish that equation in the form ^Kp dKpv _ and show how to express it in terms of rectangular co-ordinates.' (Board of Education, Theo. Mech., Honours, 1910, 69.) 4. 'A vessel of water discharges through a large pipe of variable section. If the flow be steady, investigate the relation between velocity, pressure, and height above datum level at any section of the pipe. ' If the vessel supply a stream running in a channel of any size, either closed or open, and is undisturbed by frictional resistances, show that the total energy of all parts of the water is the same.' (Board of Education, Theo. Mech., Honours, 1910, 70.) 'A vessel symmetrical about a vertical axis contains a gas. If it be rotated uniformly about the axis, investigate the pressure at any point of the gas, assuming it to move in relative equilibrium with the vessel. ' Apply to the case in which the vessel, a cylinder of radius a and height k, contains a weight W of gas.' (Board of Education, Theo. Mech., Honours, 1910, 71.) Examples — XCIX. : Chiefly Elasticity. I. ' State the law^(Hook;e's) connecting the tension of an elastic cord with its extension. ' A uniform india-rubber cord has a length of 26 inches under a tension of 2^ pounds weight, and a length of 20 under a tension of I pound 500 ANALYTICAL MECHANICS weight ; calculate the amount of work done in stretching it from its natural length to a length of 30 inches, and draw a work diagram.' (LOND. B.Sc, Pass, Applied Math., 1907, i. 8.) 2. ' Prove that the potential energy stored up in a stretched elastic string is half the product of the tension and the extension. 'A uniform bar 6 feet long, weighing 20 lbs., lies on perfectly rough horizontal ground. Evaluate the work done in raising one end from the ground to a height of 3 feet, by means of a vertical string attached to that end, (i) when the string is inelastic ; (ii) when the string is elastic, 2 feet long, and such that its length would be doubled by a pull of 30 lbs. weight.' (LoND. B.Sc, Pass, Applied Math., 1908, i. 10.) 3. ' A beam length a, width b, depth d, coefficient of elasticity E, is held by one end in a horizontal position so as to be bent simply by its own weight, which is w per unit of length ; show that the curvature at a distance x from the fixed end is (twia - xf '~EbcP ' Find also the deflection of the other end of the beam.' (Board of Education, Theo. Mech., Honours, 1908, 65.) 4. ' A rod, naturally straight, is slightly bent by forces at right angles to its length in one plane. Show that the radius of curvature at any point of its length is given by the formula £'yiA'*-f (Bending Moment), and explain the notation. ' A rod of uniform cross section is supported horizontally on three points, viz. one at each end and one in the middle, so that there is no droop at the middle. Show that the greatest droops are very nearly at one-fifth of the length from each end. (\/33 = 57446.) ' (Board of Education, Theo. Mech., Honours, 1908, 66.) 5. 'An iron bar, 2 inches in diameter, for which Young's modulus is 29,000,000, is bent into an arc of a circle 400 feet in diameter. Find the maximum stress at any point of the transverse section. ' Show further that if the stress be limited to 4 tons per square inch, the diameter of the circle must not be less than 540 feet.' (Board of Education, Theo. Mech., Honours, 1909, 63.) 6. ' A uniform horizontal beam is supported in any manner. Establish the equation ~dx^' EI of the deflection curve. ' If the beam be supported at the ends and centre, show that there is no bending moment at points which are at a distance from an end equal to 3/8 of the length of the beam.' (Board of Education, Theo. Mech., Honours, 1909, 64.) 7. ' A uniformly loaded beam is supported at the ends, the supports being in the same horizontal line and propped in the middle. ' If the centre prop is at the same level as the supports, find the points of zero bending moment. ' If now the centre prop be raised a distance equal to 1/4 of the deflection of the centre when the prop is wholly removed, prove that it sustains a pressure equal to 25/32 of the whole load.' (Board of Education, Theo Mech., Honours, 1910, 62.) EXA MPLES—MISCELLANEO US Examples — C. : Miscellaneous. 501 1. State an analogy, pointed out by Poinsot, between kinematics and statics, and give a dynamical illustration which links up the apparently ^ conflicting elements m the analogy. if vv :i 2. ' Determine the position of equilibrium of a body movable about a fixed point having one or more spherical cavities in «hich a sphere or some liquid can roll about. 'Prove that a tilting basin, of thin metal in the form of a segment of a spherical surface, movable about a diameter of the base, will not upset when water is poured into it, if the weight of the basin exceeds (radius of base/height of basin)^ - i times the weight of water.' (LOND. B'A. AND B.Sc, Pass, Mixed Math., 1902, L 3.) 3. Prove that if fi^tons is conxeyed s feet in t seconds, being moved from rest by a force of P, tons up to velocity v feet per second, and then brought to rest by a force of P^ tons, " ^ ' ig p,+p^' ^^' g p,+p,' (3)/ i2s/ir 11 \ ' With a coefficient of adhesion fx, a motor car actuated and braked on the hind axle can get up a speed v in x feet, or be brought to rest again in _y feet, given by 7'- /I /l\ V-/ 1 // \ a denoting" the distance between the axles and A the height of the C. G. (midway between the wheels) from the ground.' (LoND. B.A. AND B.Sc, Pass, Mixed Math., 1902, i. 6.) 4. ' Prove that the line joining the centre of gravity and the centre of buoyancy of a floating body is vertical in a position of equilibrium, and normal to the curve or surface of buoyancy ; and show how the stability is determined. ' Find where a cylindrical wooden pile of given S. G. will begin to leave the vertical position when lowered into water by a chain fastened to the top.' (LoND. B.Sc, Pass, Mixed Math., 1902, il 4.) 5- ' On the experimental law that the resistance of similar steamers is pro- portional to the wetted surface and the square of the velocity, prove that if a 6-foot model of o'oi ton displacement run at a speed of 2 knots in an experimental tank experiences a resistance of 0-2 pound, a similar 600-foot 10,000-ton steamer at 20 knots would experience a resistance equivalent to an incline of one in 112, and require over 12,000 effective horse-power. 'Show that for the Atlantic passage an increase of 1% in speed re- quires 6% increase in displacement tonnage and 7% increase in horse- power.' (LoND. B.Sc, Pass, Mixed Math., 1902, 11. j.) 6. ' Prove that the bursting rim velocity of a circular wire ring is ^gA, where // is the breaking length of the wire when straight and hung up vprtif 3.11 V 'Determine the greatest velocity in miles an hour possible with wheel tires of density i cwt./ft.=' and tensile strength 2 cwt./in.'^ ' (LoND. B.Sc, Pass, Mixed Math., 1902, 11. 7.) 502 ANALYTICAL MECHANICS 7. ' Prove that as the C. G. of a part of a body of weight P is moved about in a curve, the C. G. of the whole body of weight W will describe a similar curve, reduced in the linear scale of P\.oW. 'A rectangular block of stone is slung by two equal parallel chains fastened to points in its upper face, and suspended from the ends of a uniform beam supported by a fulcrum. Prove that the block can be tilted from the horizontal to any desired inclination 6 by moving the fulcrum support through a distance {i-PllV)hta.n6, where h is the depth of the C. G. of the block below the upper face, IV the total weight, and P that of the beam and chains.' (LOND. B.A. AND B.Sc., Pass, Mixed Math., 1903, i. 3.) 8. 'Prove that if a body is resisted by a force "proportional to its displace- ment, the body is in a position of stable equilibrium, about which it will perform harmonic oscillation in an invariable period ; and mention some familiar illustrations. ' Prove that the free oscillation of the mercury of a barometer in a U tube of uniform section will synchronise with a pendulum of half the length of the mercury column.' (LoND. B.A. and B.Sc, Pass, Mixed Math., 1903, i. 7.) 9. ' Prove that if a chain ABC fastened at A is led over a pulley B, so as to rest on a smooth inclined plane BC, the part AB will assume a catenarj' curve in which the tension at any point P will be the same as at Q at the same. level on BC; and deduce the analytical properties of the catenary. ' Prove that if the plane is rough the length of the chain BC may be altered without affecting AB, within the limits .5C sin a cos ^ cosec (a + <^), a denoting the slope of BC and (^ its limiting angle of friction.' (LoND. B.Sc, Pass, Mixed Math., 1903, 11. 2.) 10. ' Define the metacentre ; and prove that the metacentric height of a ship of futons displacement is mbPj IV, Ma. moment of bP foot-tons obtained by moving P tons transversely b feet gives the ship a heel of one in m. ' Assuming the experimental laws that the normal pressure of the wind and the tangential friction of the water per square foot are proportional to the square of the velocity, prove that a ship and its model are heeled over to the same angle by a wind proportional to the square root of the length or the sixth root of the tonnage, and move through the water at a proportional speed.' (LOND. B.Sc, Pass, Mixed Math., 1903, 11. 3.) 11. 'On the experimental law of the last question, prove that steamers geometrically similar, propelled at speed proportional to the sixth root of the tonnage, will experience the same resistance expressed in pounds per ton or equivalent incline, and will burn the same coal per ton-mile ; but the horse-power (H.P.) per ton and the period of revolution of the screw will be proportional to the speed, and the steam pressure to the square of the speed. 'Taking the H.P./ton as 1/16 (speed in knots), prove that this impjies a resistance of about 20 lbs. per ton, or an equivalent incline of one in 112; and with a coal consumption of 2 lbs. per H.P. -hour, the coal capacity required for a voyage of 3000 miles is about one-sixth of the tonnage. ' Calculate lor a steamer of 26,000 tons at a speed of 23-5 knots.' (LoND. B.Sc, Pass, Mixed Math., 1903, n. 4.) EXAMPLES— MISCELLANEOUS 503 12. ' Determine the motion of a circular hoop of radius a feet, whirling in a vertical plane on a round horizontal stick, if released when the centre is moving with velocity V f./s. at an angle a with the horizon ; and prove that It will make 27ra/K revs. /sec. in the air. ' Prove that the tension in the hoop will be the weight of a length V^lg feet of the rim (the tension length).' (LoND. B.Sc, Pass, Mixed Math., 1903, 11. 6.) 13. Prove that about the diameter of a homogeneous sphere (radius of gyration)2 = o-i(diameter)2 ; and show how this may be verified experimentally by allowing the sphere to roll down a slit of uniform breadth in an inclined plane. 'Explain why the cushion of a billiard table is made to receive the impact at a height 07 of the diameter of the billiard ball.' (LoND. B.Sc, Pass, Mixed Math., 1903, 11. 7.) 14. 'Investigate the torsional vibration of a body suspended by a vertical wire in which the restoring couple is proportional to the angle turned through from the position of equilibrium. ' Prove that the angular velocity is 27r/(period) times the G. M. of the angular distance in radians from the two ends of the swing.' (LoND. B.Sc, Pass, Mixed Math., 1903, 11. 8.) 15. 'A rectangular tray with vertical sides is made of thin sheet metal. If its length be 3 feet, its breadth 2 feet, and its depth 4 inches, find the height of the centre of gravity above its base.' ' Also find approximately how much the centre of gravity will be raised if the tray is filled with water ; the density of the metal being eight times that of water, and its thickness one-sixtegnth of an inch.' (LOND. B.A. AND B.Sc, Pass, Mixed M.-vth., 1904, i. 8.) 16. ' A light bar AB can turn freely about the end A^ which is fixed, and is supported in a horizontal position by a string CB, C being a fixed point vertically above A. If a weight W\>t. suspended from any point P of the bar, find geometrically the direction and magnitude of the reaction at A ; also the tension of the string. Work out numerically the tension of the string when W= 10 lbs., AB= 18 in., AP=\i in., ^C=9 in.' (Lond. B.A. AND B.Sc, Pass, Mixed Math., 1904, i. 10.) 17. 'If two systems of mass lying in a plane have the same centre of mass and the same radii of gyration about three different lines in the plane, prove that they have the same radii of gyration about every line. ' Hence prove that when calculating the radius of gyration of a triangular area (or uniform plate), we may replace the triangle by three equal particles at the middle points of the sides.' (Lond. B.Sc, Pass, Mixed Math., 1904, 11. 10.) 18. 'An unclosed curved surface of given shape is immersed in a given position in water ; show how to find the vertical component of the pressure exerted over one side of the surface. 'A hollow cone of inner radius 4 feet and inner height 10 feet, and not closed by a base, is placed with its rim on a horizontal plane ; the cone is filled with water through a small hole at the vertex and the water does not flow out. Find the force, in tons weight, with which the water tends to lift the cone, assuming the mass of water per cubic foot to be ^^ ^ ' ^' (Lond. B.Sc, Pass, Mixed Math., 1904, 11. 6.) IQ 'AB BC, and CD are three bars in a horizontal plane, freely jomted at ^' B aAd C\ and movable about fixed pins at A and D. At a g'ven point P in 5Cis applied perpendicularly to BC a given force F ; and the bars are^o be kept in a given configuration by means of a couple applied to die bar AB Assuming all necessary data, calculate the magnitude of 504 ANALYTICAL MECHANICS this couple, and exhibit the hnes of action of the stresses at the points ABC D' (LoND. B.Sc, Pass, Applied Math., 1905-, I. 8.) 20. 'A rod of length / and weight W rests against a rough horizontal rail with its lower end on a smooth horizontal plane. The height of the rail above this plane is h, and the angle of friction between the rail and the rod is X. Find the greatest possible inclination of the rod to the vertical, and the corresponding pressure on the horizontal plane. [Assume that /z> J/ sin X.].' (LOND. B.A, Pass, Applied Math., 1906, \. 3.) Examples — CI. : Miscellaneous. 1. ' Show that a uniform rod, mass m, is kinetically equivalent to three particles rigidly connected and situated one at each end of the rod and one at its middle point, the masses of the particles being \m, \m, ^m. 'A rod consists of two parts of equal length which are uniform but of different densities. Find three particles which situated respectively at the ends and the middle point of the rod form a system kinetically equivalent to the rod. If M and M' are the masses of the two portions of the rod, prove that this is always possible with actual particles, if M/M' lies between 5 and 1/5.' (LoND. B.Sc, Pass, Applied Math., 1905, in. 5.) 2. ' Prove that the moment of momentum of a rigid body moving in two dimensions about an axis through the mass centre perpendicular to the plane of motion is Mk'^a. 'A uniform heavy sphere, whose mass is i lb. and radius 3 inches, is suspended by a wire from a fixed point, and the torsion couple of the wire is proportional to the angle through which the sphere is turned from the position of equilibrium. If the period of an oscillation is 2 sees., find the couple which will hold the sphere in equilibrium in a position in which it is turned through four right angles from the equilibrium position.' (Lond. B.Sc, Pass, Applied Math., 1905, iii. 8.) 3. 'Eine zylindische Taucherglocke von der Hohe a wird in Wasser getaucht, bis ihre oberste Spitze in einer Tiefe b unter der Wasserflache ist. Bestimmen Sie wie weit die Luft komprimirt wird, wenn das Wasserbarometer auf h steht. ' Wenn die Glocke mit einer gleichmassigen Geschwindigkeit v sinkt, so bestimmen Sie die Geschwindigkeit, mit welcher man Luft mit atmospharischem Druck in die Glocke pumpen muss, um kein Wasser in dieselbe zu bekommen.' (Lond. B.Sc , Pass, Applied Math., 1905, iii. 10.) 4. ' Four bars are loosely jointed so as to form a plane quadrilateral ABCD ; and it is in equilibrium under the action of four forces P, g, R^ S, applied to the joints A, £, C, D respectively. The lines of action of P, Q meet in E, and those of R, S meet in F. Prove that EE pro- duced will pass through the intersection oi AD, BC J . _ (Lond. B.Sc, Pass, Applied Math., 1906, i. 4.) 5- If a particle is acted upon by a force always directed towards a fixed pomt and varying inversely as the square of the distance, obtain the conditions to be satisfied when the particle is initially projected so that the path of the particle may be (i) a circle, (2) a parabola.' (Lond. B.Sc, Pass, Applied Math., 1906, in. 2.) EXAMPLES— MISCELLANEOUS 505 6. ' Write down the general equation for the transformation of a given mass of gas whose volume, temperature, and intensity of pressure are altered. ' Calculate the height to which the water will rise in a cylindrical diving bell, 1 2 feet high, when its top is lowered to a depth of 60 feet, being gi\en that the temperature of air at the surface is 80° F., height of water barometer at surface 33 feet, and temperature of water 40° F.' (LoND. B.Sc, Pass, Applied Math., 1906, iii. 10.) 7. ' State the necessary and sufficient conditions of equilibrium to be satisfied by a system of forces acting in one plane on a rigid body. '' AB is a uniform ladder 37 feet long resting at A on the ground, where the coefficient of friction is 1/2, and at B against a vertical wall, where the coefficient is 5/12; the distance of A from the wall is 12 feet; a horizontal force is applied at A to move the ladder towards the wall ; find the magnitude of the requisite force in terms of the weight, \\\ of the ladder.' (LOND. B.A., Pass, Applied Math., 1907, i. 1.) 8. 'Two uniform bars, AB and yJC, are rigidly attached together at A, so that the angle BA C is 1 20°, and are freely movable in a vertical plane about A, which is fixed ; find the inclination of ^^ to the horizon in the position of equilibrium, being given length of AB = 6 feet, mass of ^.S=iolbs., length of^C=5, mass of^C=8.' (LoND. B.A., Pass, Applied Math., 1907, i. 3.) 9. 'What is meant by unavailable energy? A mass B placed on a smooth horizontal table is connected by a light slack cord with a mass Q lymg on the ground. Show that, if B is projected along the table with a velocity V, the energy available for raising Q is B PV P + Q' 2^' and write down the energy which has become unavailable.' (LOND. B.A., Pass, Applied Math., 1907, i. 4.) 10. 'Give a proof that the time of a small semi-oscillation of a simple pendulum is ir v//f. • 1 ^ j • 'If the bob of the pendulum is drawn out from the vertical to a devia- tion a such that the tension in the vertical position exceeds the weight of the bob by i/io of the weight, show that a must be 18 1 1 . (LoND. B.A., Pass, Applied Math., 1907, i. 5-) II 'Two perfectly elastic particles impinge directly on each other; prove that the product of the sum of their masses and the amount of energy transferred from one to the other is equal to the product of their total momentum and the momentum transferred. , , , ^ . 'S he ratio of the momentum transferred to the total momentum in thL case of two imperfectly elastic spheres of masses 6 and 10 with co- efficLnt of restitution 3/4, when the smaller one moving with velocity 8"ges'dfrectly on the larger one moving with velocity 3 m the same sense.' ^^^^ ^^^ p^^^^ Applied Math., 1907, i- 8.) mass.' ^^^^ g^ p^gg_ Applied Math., i9o7> i- i°-) «. ^^rlcUtc: nf a U tube of uniform bore and open to ''■ tratmoT/hfr-t ^ ouferlimb'the increase of pressure in the inner 5o6 ANAL YTICAL MECHANICS limb being read by the depression there of the surface of the liquid of density A occupying the lower portions of each limb. A special gauge has the upper parts, of each limb of the U tube, enlarged in cross-sectional area to n times that of the lower parts, and is charged with two liquids as follows :— A liquid of density p meets the atmosphere in the large upper portion of the outer limb, and extends round the bend of the U, finishing at some level P in the thin tube of the inner limb. Above P in the inner limb there extends a second liquid of density o-, not mixing with the former, but reaching to the upper enlarged portion of the tube where it is exposed to the pressure to be determined. The gauge is read by the position of the interface P of the two liquids. Show that the ratio of the sensibility of this special gauge to that of the ordinary one first mentioned may be expressed by /)(«+ \)-a(n- i) 14. 'Two reservoirs, 10,000 square feet in area, with vertical walls, contain the one salt and the other fresh water. They are both filled to the same absolute level. If the reservoirs are connected by a pipe 50 feet below the free surface of either, find how much salt water has passed into the fresh water reservoir before equilibrium is established, the specific gravity of the salt water being i'026, assuming that any salt water which enters the fresh water reservoir sinks below the level of the pipe.' (LoND. B.A,, Pass, Applied Math., 1907, 11. 2.) 15. ' State the conditions of equilibrium of bodies wholly or partly immersed in a fluid. ' A cylinder of radius r floats in liquid of density o- inside a cylindrical vessel of radius a. Show that if a mass W\it placed on the floating cylinder, it will sink by an amount na\r' i>' (LoND. B.A., Pass, Applied Math., 1907, 11. 4.) 16. 'A thin rod of length 2a and density a floats in a sloping position in a liquid of density p, the end out of the liquid being suspended by a string. Find the proportion of the length of the rod immersed, and show that it is independent of the inclination of the rod. Find also the tension of the string.' (LoND. B.A., Pass, Applied Math., 1907. 11. 5.) 17. 'Prove that, if r, 6 be the polar co-ordinates of a point moving in a plane, the radial component of acceleration is ^ - rff^. 'A particle P of unit mass, free to slide on a straight smooth wire, is attracted towards a point O of the wire with a force p.. OP. If the wire be made to revolve about O in a horizontal plane with constant angular velocity a, show that the mo tion of P on the wire is a simple harmonic motion of period 2irl sip- a>\ and that when a^ = fi/2 the path of P in space is a circle.' (LoND. B.Sc, Pass, Applied Math , 1907, 11. 4.) 18. 'Two equal particles are connected by an inextensible string of length a + b, which passes through a hole in a smooth table, so that one particle hangs freely and the other is on the table. At an instant when the system is at rest, and a length a of the string is horizontal, the particle on the table is sudd enly made to move perpendicularly to the string with velocity ^'4^c. Employ the principles of energy and angular EXAMPLES— MISCELLANEOUS 507 momentum to show that the hanging particle will not rise to the level of the table unless b is a rectangular lamina of uniform density. Find the moment of inertia, about the side AB, of each of the triangles into which it is divided by the diagonal BD.' (Board of Education, Theo. Mech., Fluids, Stage 3, 1910, 41.) 9. ' Find the position of the centre of pressure of a plane area immersed in water, supposing the plane to be vertical. 'Apply the method to the case in which the area is the part of a parabola cut off by the focal chord at right angles to the axis, the chord being vertical and just immersed.' (Board of Education, Theo. Mech., Fluids, Stage 3, 1910, 43.) 10. ' Prove that if a volume be cut off a solid body by a plane section and an equal volume be supposed cut off by any other plane section, making a small angle with the first plane, the two planes will intersect in a line passing through the centre of gravity of the first plane. ' Explain the importance of this theorem in the question of the stability of a floating body.' (Board of Education, Theo. Mech., Fluids, Stage 3, 1910, 45.) 11. ' A body floating in stable equilibrium is slightly disturbed, and makes small vertical oscillations. Find the time of an oscillation. ' Apply the method to the case of a cone of given height, whose vertical angle is 60° and specific gravity s/8. {N.B. — \/5 = i7i. It may be as- sumed that the cone will float in stable equilibrium with its vertex downwards.) ' (Board of Education, Theo. Mech., Fluids, Stage 3, 1910, 46.) 12. 'Describe the hydrometer of variable immersion, and explain how the specific gravity of a liquid may be found by means of it. 'A hydrometer floats with 8 inches of the stem above the surface of a liquid whose specific gravity is 0-95 ; also it floats with 2 inches of the stem above the surface of a liquid whose specific gravity is 075. Find the specific gravity of the liquid in which it is found to float with 5-5 inches of the stem above the surface.' (Board of Education, Theo. Mech., Fluids, Stage 3, 19 10, 47.) 13. 'A spherical surface of radius a and of small thickness c contains gas at a given pressure p. Investigate the magnitude of the tension per unit area of the section of the material at any point. ' A thin india-rubber ball contains air. Show that the tension per unit area of the material varies as the absolute temperature.' (Board of Education, Theo. Mech., Fluids, Stage 3, 1910, 51.) 14. 'A body is in motion about a fixed point O, and three rectangular axes, fixed in the body, are drawn through O. The angtilar velocities of the body about these axes respectively are known. Find equations for determining the motion of the body with reference to three rectangular axes drawn through O and fixed in space.' (Board of Education, Theo. Mech., Honours, 1910, 64.) 15. 'In the motion of a rigid body in space of two dimensions establish the mdependence of the motions of the centre of gravity of the body and of the body relative to the centre of gravity. ' Examine the motion of a heavy rod sliding between a smooth vertical EXAMPLES— MISCELLANEOUS 515 plane and a smooth horizontal plane in a plane perpendicular to ' When will the rod leave the vertical plane ? ' (Board of Education, Theo. Mech., Honours, 1910, 66.) 10. ^atate the mechanical principle called the conservation of areas. ^ Explam the application of the principle to the following case :— A thm but heavy cylindrical shell is fitted with a solid cylinder or core, supposed to be perfectly smooth. The compound body is placed on a rough mclmed plane, and is allowed to roll down it. Find how the body is movmg when it has rolled down a given length of the plane. Also compare the results that would be obtained :— (i) If the core had not been put in. (2) If the core had adhered firmly to the shell, the masses of shell and core being equal.' (Board of Education, Theo. Mech., Honours, 1910, 67.) Examples — CV. : Miscellaneous. 1. 'Define the centre of gravity of a body and establish the formula i^mz) -r 2ot for the distance from a plane of the centre of gravity of a number of particles. ' Find the position of the centre of gravity of the part of a solid sphere which lies between a diametral plane and a parallel plane whose dis- tance from the former is half the radius of the sphere.' (Lond. B.Sc, App. Math., Subsidiary to Hons. Physics, 1909, i. 2.) 2. ' State the principle of \'irtual Work. In what respects is the method of Virtual Work preferable to that of resolving and taking moments as a means of solving problems ? ' A circular cylinder of radius a is fixed with its axis horizontal and its curved surface in contact with a vertical wall. A disc of radius b {b>a) rests on the cylinder and against the wall, the plane of the disc being parallel to the axis of the cylinder. If in the position of equilibrium the plane of the disc makes an angle a with the horizon, prove that « = ^cosa(l +sina).' (Lond. B.Sc, App. Math., Subsy. to Hons. Physics, 1909, i. 4.) 3. ' Define (i) potential, (ii) tube of force, and prove the fundamental pro- perty of a tube of force. ' If the lines of force are arcs of circles whose centres are at a fixed point O and whose planes pass through a fixed line OA, compare the intensity of the force at two points in free space which lie on the same line of force.' (Lond. B.Sc, App. Math., Subsv. to Hons. Physics, 1909, i. 6.) 4. ' Find the attraction of a thin uniform circular plate at any point of the line through its centre at right angles to its plane. ' Find the force of attraction exerted by a solid uniform hemisphere of density p and radius a upon a particle of mass ;// placed at the centre of the plane face.' (Lond. B.Sc, App. Math., Subsy. to Hons. Physics, 1909, i. 7-) 5. ' Determine the potential of a straight uniform rod at any pomt, and deduce the law of attraction of an infinite rod extendmg to infinity m both directions. , . , • , 1. ' Prove that if two such infinite rods intersect, their plane is cut by the equipotential surfaces in hyperbolas.' (Lond. B.Sc, App. Math., Subsy. to Hons. Physics, 1909, i. 8.) 6. ' Find by any method the centre of pressure of a circular area, wholly immersed in water. Si6 ANALYTICAL MECHANICS ' Prove that, if a is the radius of the circle and A is the height of the water barometer, the centre of pressure lies on or within a concentric circle of radius ia^-r {a + /[).' (LOND. B.Sc, Apr Math., Subsy. to Hons. Physics, 1909, i. 9.) 7. ' If a particle is subjected to two simple harmonic motions of the same period in perpendicular directions, show that the resulting motion is in general elliptic. ' If the motions have the same period and amplitude, differ in phase by one quarter of a period, and are in directions inclined to one another at 60°, find the resulting motion.' (LoND. B.Sc, App. Math., Subsy. to Hons. Physics, 1909, 11. i.) 8. 'Find expressions for the tangential and normal accelerations of a particle moving in a plane curve. If the motion is such that the tangential and normal accelerations are always equal, and v^, v^ are the velocities at any two points of the path, show that where 6 is the angle turned through between the points.' (LoND. B.Sc, App. Math., Subsy. to Hons. Physics, 1909, 11. 2.) 9. ' A particle of unit mass moves in a straight line under the action of a force kx directed towards a fixed point in the straight line, where x is the distance of the particle at any time from the fixed point and k is con- stant. The resistance to motion at any time is / times the velocity of the particle where / is constant. Write down the equation of motion of the particle, and show that if P cosec 6 from the centre of the plane section.' (Calcutta B.A. and B.Sc, Honours, Math., 1909, v. 4.) 5. ' Assuming the general equations of equilibrium for a flexible inextensible string deduce the equation of the curve formed by a string hanging under the action of gravity, its extremities being attached to fixed points in the same horizontal plane. 'Show that the tension at any point of the catenary is equal to the weight of a portion of the string whose length is equal to the ordinate of the point.' (Calcutta B.A. and B.Sc, Honours, Math., 1909, v. 5.) 6. 'Define momentum, kinetic energy, potential energy, power. 'An engine of 350 horse-power, whose weight is 20 tons, is attached to a train weighing 130 tons, and pulls it up an incline of i in 300 at a rate of 40 miles an hour. Find the resistance per ton due to friction, etc' (Calcutta B.A. and B.Sc, Honours, Math., 1909, v. 6.) 7. ' A point is moving in a curve with velocity v ; show that its acceleration inwards along the normal is w^Pj where p is the radius of curvature of the curve at the point. 'A particle is projected, from the lowest point inside a smooth elliptic cylinder whose major axis is vertical, in a direction perpendicular to the generators. Show that the particle will go completely round the cylinder if the velocity of projection is greater than Vs - (^•<^Si where la is the major axis and e is the eccentricity of the cylinder.' (Calcutta B.A. and B.Sc, Honour.s, Math., 1909, v. 7.) 8. 'Define the hodograph of the path of a particle, and show that the velocity in the hodograph is equal to the acceleration in the path. By means of the hodograph determine the acceleration of a point describing a circle with constant speed. ' A heavy particle slides down a smooth cycloid with its axis vertical and its vertex uppermost. If the particle start from rest at the vertex, prove that the hodograph consists of the quadrant of a circle and a straight line touching the circle.' (Calcutta B.A. and B.Sc, Honours, Math., 1909, v. 8.) 9. ' Discuss the motion of a particle of mass ot, attached by a weightless elastic string of length / to a fixed point, which has been allowed to fall from a point vertically below the fixed point and at a distance h from it {h nitni^itti + u,_ tan a) (Calcutta B.A. and B.Sc, Honours, Math., 1909, v. 1 1.) Examples — CVIII. : Miscellaneous. ' Show that for a beam supported horizontally and loaded in any manner the ordinates of the funicular polygon represent to some scale the Bending Moment for the beam, at any point, and show how to find this scale. ' Draw to scale the Shearing Force and Bending Moment diagram for a uniform beam AF, 20 feet long, under the following conditions : — It is supported at one end A and an intermediate point B, where AB is 15 feet, while weights of 10, 15, and 20 cwts. are attached at points C, /?, and E where AC, AD, and AE equal 10, 12, and r6 feet respectively ; in addition, there is a uniform load of half a cwt. per foot run over EF. The weight of the beam may be neglected.' (Calcutta B.E., Statics and Dynamics, 1909, 4.) ' A mass M draws up another M' on the wheel and axle ; if a is the radius of the wheel and ci that of the axle, find the motion and the tensions of the strings, assuming \i to be the mass of the revolving body and k its radius of gyration about axis of revolution.' (Calcutta B.E., Statics and Dynamics, 1909, 5.) ' Show that, assuming the earth to be a homogeneous sphere, it would be necessary for the rotatory motion to be about seventeen times as fast as it is at present, if the centrifugal force was to exactly neutralise the action of gravity. The earth's radius equals 20,880,000 feet, and the acceleration due to gravity is 32'2 feet per sec. per sec' (Calcutta B.E., Statics and Dynamics, 1909, 6.) ' A particle of mass m is acted on by a force varying inversely as the square of the distance of the centre of mass from a fixed point ; show that the particle will describe an ellipse round the fixed point, that equal areas will be traced out about the centre of force in equal times, and that the square of the periodic time of revolution bears a constant ratio to the cube of major axis of the ellipse described.' (Calcutta B.E., Statics and Dynamics, 1909, 7.) ' Show that for a particle moving in a plane the radial and transversal accelerations with reference to any axes fixed in the plane can be expressed in the form ' Use this result to investigate the motion of two particles of masses m and ;;/, connected together by a light string of length / and revolving in a smooth tube in a horizontal plane at constant angular velocity. (Calcutta B.E., Statics and Dynamics, 1909, 8.) 520 ANALYTICAL MECHANICS 6. ' State the fundamental notion that distinguishes " solids " from "fluids," and show that in a fluid the pressure at any point is the same in all directions. 'Differentiate between "Total Pressure," "Centre of Pressure," and "Pressure at a Point."' (Calcutta B.E., Hydrostatics, 1909, 2.) 7. ' Investigate the necessary conditions of equilibrium for a body floating freely in a liquid, pointing out the use and meaning of the terms "metacentric height," "plane of flotation," and "surface of buoyancy." 'A rectangular block of wood of given volume and square in section floats in a homogeneous liquid. Find the least ratio of breadth to height that the block may just float upright.' (Calcutta B.E., Hydrostatics, 1909, 3.) MATHEMATICAL TABLES (reproduced here by kind permission of the controller of h.m. stationery office.) {A copy of these Tables will be supplied to each candidate at the Examinations in Practical Plane and Solid Geometry, Machine Constructiott and Drawing (Stage 3 and Honours'), Building Construction (Stage 3 and Honours), Naval Architecture, Practical Mathematics, Ajf lied Mechanics, and Heat Engines.) USEFUL CONSTANTS. I Inch = 25 -40 millimetres. I Gallon = -1604 cubic foot= 10 lb. of water at 63° F. I Knot = 6080 feet per hour= i Nautical mile per hour. Weight of I lb. in London = 445,000 djTies. One pound avoirdupois = 7000 grains = 453'6 grammes. I Cubic foot of water weighs 62-3 lb. I Cubic foot of air at 0° C. and i atmosphere, weighs -0807 lb. I Cubic foot of Hydrogen at 0° C. and i atmosphere, weighs -00559 lb. I Foot-pound = I •3562 X lo^ ergs. I Horse-power-hour = 33000 x 60 foot-pounds. I Electrical unit = 1000 watt-hours. . „ , , TT ■ f 774 ft--lb. = I Fah. unit. Joule's Equivalent to suit Regnaulfs H, is jj^^, ft.-lb.= i Cent. „ I Horse-power = 33000 foot-pounds per minute = 746 watts. \'olts X ampferes = watts. I Atmosphere= 147 lb. per square inch= 3116 lb. per square foot = 76o mm. of mercury = id's dynes per sq. cm. nearly. A column of water 23 feet high corresponds to a pressure of i lb. per sq. inch. Absolute temp., t = 6° C. + 273° or 6° F. + 460°. Regnault-s H =606-5 + -305 ^° C = 1082 + -305 6 F. ^„1.0646 = 479. R C logio/ = 6-ioo7-7-72' where log,oB = 3i8i2, logioC. = 5'o882. ^ ,■ j p is in pounds per sq. inch, t is absolute temperature Centigrade. V is the volume in cubic feet per pound of steam. ?H:^::S;^nt^:: Napierian logarithms, multiply by 2-3026. The base of the Napierian logarithms is ^ = 2-7183- The value of^ at London = 32-i82 feet per sec. per sec. 522 ANALYTICAL MECHANICS Logarithms. 1 2 3 4 5 6 7 8 9 12 3 4 5 8 7 8 9 10 oooo 0043 0086 0128 0170 0212 0253 0294 0334 0374 4 9 13 17 4 8 12 16 21 20 26 30 34 38 24 28 32 37 u 0414 0453 0492 0531 0369 0607 0645 0682 0719 0755 4 8 12 13 4 7 II 15 19 19 23 27 31 35 22 26 30 33 12 13 0792 0828 0864 0899 0934 0969 1004 1038 1072 1 106 3 7 II 14 3 7 10 14 18 17 21 25 28 32 20 24 27 31 1 139 "73 1206 1239 1^71 1303 1333 1367 1399 1430 3 7 10 13 3 7 10 12 16 16 20 23 26 30 19 22 23 29 14 I46I 1492 1323 1553 1584 1614 1644 1673 1703 1732 3 6 9 12 3 6 9 12 13 15 18 21 24 28 17 20 23 26 15 1761 1790 j8i8 1847 1873 1903 1931 1959 1987 2014 3 6 9 jl 3 5 8 II 14 14 17 20 23 26 16 19 22 25 16 2041 2068 2095 2122 2148 2173 2201 2227 2233 2279 3 5 8 II 3 5 8 lo 14 13 16 19 22 24 13 18 21 23 17 2304 2330 2355 2380 2405 2430 2433 2480 2304 2329 3 5 8 10 2 5 7 10 13 12 13 18 20 23 13 17 19 22 18 2553 2577 260X 2623 2648 2672 2693 2718 2742 2765 2379 2579 12 II 14 16 19 21 14 16 18 21 19 2788 2810 2833 2836 2878 2900 2923 2945 2967 2989 2479 2468 II II 13 16 18 20 13 15 17 19 20 3010 3032 3034 3075 3096 3118 3139 3160 3181 3201 2468 II 13 13 17 19 21 22 23 24 3222 3424 3617 3802 3243 3444 3636 3820 3263 3464 3838 3284 3483 3674 3836 3304 3502 3692 3874 3324 3522 3711 3892 3343 3541 3729 3909 3365 3360 3747 3927 3385 3579 3766 3945 3404 3398 3784 3962 2468 2468 2467 2437 10 10 9 9 12 14 16 x8 12 14 13 17 II 13 15 17 II 12 14 16 25 3979 3997 4014 4031 4048 4063 4082 4099 4116 4133 2357 9 10 12 14 15 26 27 28 29 4150 4314 4472 4624 4166 4330 4487 4639 4183 4346 430a 4654 4200 4362 4518 4669 4216 4378 4683 4232 4393 4548 4698 4249 4409 4564 4713 4265 4425 4379 4728 4281 4440 4394 4742 4298 4436 4609 4757 2337 2356 2336 1346 8 8 8 7 10 II 13 15 9 II 13 14 9 II 12 14 9 10 12 13 30 4771 4786 4800 4814 4829 4843 4857 4871 4886 4900 1346 7 9 10 II 13 31 32 33 34 4914 3051 5185 3315 4928 5063 3198 5328 4942 5079 52H 5340 4953 509a 5224 5353 4969 3103 5237 5366 49S3 3119 5250 5378 4997 5132 5263 5391 301 1 3145 5276 5403 5024 5159 5289 3416 5038 5172 5302 3428 1346 1343 1345 1345 7 7 6 6 8 10 II 12 8 9 II 12 8 9 10 12 8 9 10 II 35 5441 3453 5465 5478 5490 3302 3514 5327 5539 5551 1243 6 7 9 10 II 36 37 38 39 5563 5682 5798 591 1 5573 3694 3809 5922 5587 5703 5821 5933 5599 5717 5832 5944 5611 3729 3843 5955 5623 5740 3853 3966 5633 5752 5866 5977 5647 5763 5877 3988 5658 5775 3888 5999 5670 3786 5899 6010 1243 12 3 3 1235 12 3 4 6 6 6 5 7 8 ID II 7 8 9 10 7 8 9 10 7 8 9 10 40 6021 6031 6042 6053 6064 6073 6085 6096 6107 6117 1234 5 6-8 9 10 41 42 43 44 6128 6232 fi335 6435 6138 6243 6343 6444 6149 6253 6353 6454 6160 6263 6464 6170 6274 6375 6474 6180 6284 6385 6484 6191 6294 6395 6493 6201 6304 6405 6503 6212 6314 6415 6513 6222 6325 6425 6522 1234 1234 1234 1234 5 5 5 5 6789 6789 6789 6789 45 6532 6342 6551 6561 6571 6580 6590 6599 6609 6618 1234 5 6789 46 47 48 49 6628 6721 6812 6go2 6637 6730 6821 6911 6646 6830 6920 6656 6749 6839 6928 6665 6758 6848 6937 6673 6767 6837 6946 6684 6776 6866 6935 6693 6785 6875 6964 6702 6794 6884 6972 6712 6803 1234 1234 1234 1234 3 3 4 4 6778 5678 5678 60 6990 6998 7007 7016 7024 7033 7042 7050 7039 7067 1233 4 5678 The copyright of that portion of the above table which gives the logarithms of numbers from 1000 to 2000 is the property of Messrs. Macmillan & Company, Limited, who, however, have authorised the use of the form in any reprint published (or educational purposes. MATHEMATICAL TABLES Logarithms. 523 1 2 3 4 5 6 7 8 9 12 3 4 5 6 7 8 9 61 62 53 64 7076 7160 7243 73a4 7084 7168 7251 7332 7093 7177 7259 7340 7101 7185 7267 7348 7no 7193 7275 7356 71J8 7202 7284 7364 7126 72IO 7292 7372 7135 72l8 7300 7380 7143 7226 7308 7388 7132 7235 7316 7396 1233 1223 12 2 3 1223 4 4 4 4 5678 5677 5667 5667 55 7404 7412 7419 7427 7433 7443 7431 7459 7466 7474 1223 4 5567 56 67 58 59 748a 7559 7634 7709 7490 7566 7642 7716 7497 7574 7649 7723 7505 7582 7657 7731 7513 7589 7664 7738 7320 7397 7672 7745 7328 7604 7679 7752 7336 7512 7686 7760 7343 7619 7694 7767 7551 7627 7701 7774 1223 12 2 3 I I 2 3 1123 4 4 4 4 5367 5567 4367 4567 60 7782 7789 7796 7803 7810 7818 7823 7832 7839 7846 I I 2 3 4 4566 61 62 63 64 7853 7924 8062 7860 7931 8000 8069 7868 7938 8007 8075 7875 7945 8014 8082 7882 7952 8021 8089 7889 7959 8028 8096 7896 7966 8033 8102 7903 7973 8041 8109 7910 7980 8048 8ll6 7917 7987 8055 8122 I I 2 3 I I 2 3 I I 2 3 I I 2 3 4 3 3 3 4366 4566 4556 4556 65 8129 8136 8142 8149 8156 8162 8169 8176 8182 8189 I I 2 3 3 4356 66 87 68 69 ?'?5 8261 8325 8388 8202 8267 8331 8395 8209 8274 8338 8401 8215 8280 8344 8407 8222 8287 8351 8414 8228 8293 8357 8420 8233 8299 8363 8426 8241 830S 8370 8432 8248 8312 8376 8439 8254 f3J9 8382 8445 I I 2 3 I I 2 3 I I 2 3 112 2 3 3 3 3 4356 4356 4456 4456 70 8451 8457 8463 8470 8476 8482 8488 8494 8500 8306 112 2 3 4456 71 72 73 74 8513 8633 S692 8519 8579 8639 8698 5525 8585 8643 8704 8531 8591 8651 8710 8537 8597 8657 8716 8603 8663 8722 8549 8609 8669 8727 8555 8615 8675 8733 8561 8621 8681 8739 8567 8627 8686 8745 112 2 I I 2 2 112 2 112 2 3 3 3 3 4 4 5 5 4 4 5 5 4 4 5 5 4 4 5 5 76 8731 8756 8762 8768 8774 8779 8785 8791 8797 8802 112 2 3 3 4 5 5 76 7? 78 79 8808 8865 8921 S976 8814 8871 8927 8982 8820 8876 8932 8987 8823 8882 8938 8993 8831 8887 8837 8893 8949 9004 8842 8899 S934 9009 8848 8904 8960 9015 8854 Sgio S965 go20 8839 8915 8971 9023 112 2 112 2 112 2 112 2 3 3 3 3 3 4 3 3 3 4 4 5 3 4 4 5 3 4 4 5 80 9031 9036 9042 9047 9053 9058 9063 9069 9074 9079 11^2 3 3 4 4 5 81 82 83 84 9083 9138 9191 9243 9090 9143 9196 9248- 9096 9149 9201 9253 9101 9154 9206 9258 9106 9159 9212 9263 9112 9165 9217 9269 9117 9170 9222 9^74 9 122 9175 9227 9279 912S 9180 9232 92S4 9133 9186 9238 9289 112 2 112 2 I I 2 2 112 2 3 3 3 3 3 4 4 5 3 4 4 5 3 4 4 5 3 4 4 5 85 9294 9299 9304 9309 9315 9320 9323 9330 9335 9340 112 2 3 3 4 4 5 86 87 88 89 9345 9395 9443 9494 9350 9400 9450 9499 9355 9405 9453 9504 9360 9410 9460 9509 9365 9415 9465 9513 9370 9420 9469 9518 9375 9425 9474 9323 9380 9430 9479 9528 9383 9435 9484 9533 9390 9440 9489 9538 I I 2 2 113 0112 I I 2 3 2 2 2 3 4 4 3 3 3 4 4 3 3 4 4 3 3 4 4 90 9542 9547 9552 9557 9562 9566 9571 9576 9381 9586 Oil" 2 3 3 4 4 91 92 93 94 9590 9638 9685 9731 9595 9643 9689 9736 9600 9647 9694 9741 9605 9652 9699 9745 9609 9657 9703 9750 9614 9661 9708 9734 9619 9666 9713 9759 9624 9671 9717 9763 9628 9675 9722 9768 9633 9680 9727 9773 0112 I I 2 112 0112 2 2 2 2 3 3 4 4 3 3 4 4 3 3 4 4 3 3 4 4 95 9777 9782 9786 9791 9795 9800 9805 9809 9814 9818 0112 2 3 3 4 4 96 97 98 99 9823 9868 9912 9956 9827 9872 9917 9961 9832 9877 9921 9965 9li? 9926 9969 9841 9886 9930 9974 9845 9890 9934 9978 9850 9894 9939 9983 9S54 9899 9943 9987 9859 9903 9948 9991 9863 9908 9932 9996 0112 I I 2 I I 2 0112 2 2 2 2 3 3 4 4 3 3 4 4 3 3 4 4 3 3 3 4 ^Ml __— J ' ■*""" 524 ANALYTICAL MECHANICS Antilogarithms. - — 1 2 3 4 5 6 7 8 9 12 3 4 6 6 7 8 9 ■00 1000 1002 1005 1007 1009 1012 1014 10x6 1019 1021 X X I X 2 2 2 •01 •OS •OS •04 1023 1047 1073 1096 1026 1050 1074 1099 1028 1052 1076 1 102 1030 1054 1079 1104 1033 1057 1081 1 107 1033 1059 1084 1 109 1038 1062 1086 1112 1040 1064 1089 1114 X042 1067 109X 1117 1045 X069 1094 XX19 X X X X X I I I I I X X X X 2 2 2 X 2 2 2 X 2 2 2 2 2 2 2 •06 XI22 1125 II27 1130 1132 "35 1138 XX40 "43 X146 X X I I 2 2 2 2 •08 •07 •08 ■09 II48 "75 1202 1230 1151 1178 1205 1233 II53 1180 I208 1236 1156 1183 1211 1239 1159 1186 12 13 1242 Il6x 1189 X2l6 1245 X164 1191 12 19 1247 1x67 1194 X222 X250 X169 X197 1225 1253 X172 "99 1227 1256 X X X X X X I I X I X X X X X I 2 2 2 2 2 2 2 2 2223 2223 •10 1259 1262 1265 1268 1271 1274 1276 1279 X282 X285 X X X X 2223 •11 •12 •13 ■14 1288 1318 1349 1380 1291 1321 I3S2 1384 1294 1324 1355 1387 1297 1327 1358 1390 1300 1330 1361 1393 1303 1334 1365 1396 1306 1337 1368 1400 1309 1340 I371 1403 13x2 1343 X374 X406 1315 1346 1377 1409 I X X 1 X X X X X X X X 2 2 2 2 2223 2223 2233 2233 •15 1413 1416 I419 1422 1426 1429 1432 X435 1439 X442 I X I 2 2233 •16 •17 •IS •19 1445 1479 1514 1549 1449 1483 1517 1552 1521 1556 1455 1489 1524 1560 1459 1493 1528 1563 1462 1496 1531 1567 1466 1500 1535 1570 1469 1503 1538 1574 X472 1507 1542 1578 1476 1510 1545 1581 X X X X X I I I I X X I 2 2 2 2 2233 2233 2233 2333 •20 1585 1589 1592 1596 1600 1603 1607 1611 1614 16x8 X X X 2 2333 ■21 •22 •23 •24 1622 1660 1698 1738 1626 1663 1702 1742 1629 1667 1706 1746 1633 1671 1710 1750 1637 1675 1714 1754 1641 1679 1718 1758 1644 1683 1722 1762 1648 1687 1726 1766 1652 1690 5730 1770 1656 1694 1734 1774 I I 2 I I 2 X X 2 X X 2 2 2 2 2 2333 2333 2334 3 3 3 4 •28 1778 1782 1786 1791 1795 1799 1803 1807 18" x8l6 1X2 2 2334 •26 •27 •28 ■29 1820 1862 1905 1950 1824 1866 1910 1954 1S28 1871 I914 1959 1832 1875 1919 1963 1837 1879 1923 1968 1841 1884 1928 1972 1845 1888 1932 1977 1849 1892 1936 1982 1854 1897 1941 X986 1858 19OX 1945 X991 01x2 X I 2 X X 2 0X12 2 2 2 2 3 3 3 4 3 3 3 4 3 3 4 4 3 3 4 4 •30 1995 2000 2004 2009 2014 2018 2023 2028 2032 2037 I I 2 2 3 3 4 4 •31 •32 ■33 ■34 2042 2089 2138 2i88 2046 2094 2143 2193 2051 2099 2148 2198 2056 2104 2153 2203 2061 2109 2158 2208 2065 2113 2163 2213 2070 2Il8 2168 2218 2075 2x23 2173 2223 2080 2128 2x78 2228 2084 2133 2x83 2234 I I 2 1X2 I I 2 1 I 2 2 2 2 2 3 3 3 4 4 3 3 4 4 3 3 4 4 3 4 4 5 ■35 2239 2244 2249 2254 2259 2265 2270 2275 2280 2286 I I 2 2 3 3 4 4 5 ■36 •37 •38 •39 2291 2344 2399 2453 2296 2350 2404 2460 2301 2355 2410 2466 2307 2360 2415 2472 2312 2366 2421 2477 2317 2371 2427 24S3 2323 2377 2432 2489 2328 2382 2438 2495 2333 2388 2443 2500 2339 2393 2449 2506 I I 2 2 X I 2 2 1X22 1X22 3 3 3 3 3 4 4 5 3 4 4 5 3 4 4 5 3 4 5 5 •40 2512 2518 2523 2529 2535 2541 2547 2553 2559 2564 T X 3 2 3 4 4 5 5 ■41 -42 ■43 ■44 2570 2630 2692 2754 2576 2636 2698 2761 2382 2642 2704 2767 2588 2649 2710 2773 2594 2655 2716 2780 2600 2661 2723 27B6 2606 2667 2729 2793 2612 2673 2735 2799 2618 2679 2742 2805 2624 2685 2748 2812 X X 2 2 1X22 1x23 X X 2 3 3 3 3 3 4 4 5 5 4456 4456 4456 ■45 2818 2825 2831 2838 2844 2851 2858 2864 287X 2877 1X23 3 4556 •46 •47 ■48 •49 2884 2951 3020 3090 2891 2958 3027 3097 2897 2965 3034 3105 2904 2972 3041 3112 29II 2979 3048 3II9 2917 2985 3055 3126 2924 2992 3062 3133 2931 2999 3069 3141 2938 3006 3076 3148 2944 3013 3083 3155 X X 2 3 I I 2 3 I I 2 3 1x23 3 3 4 4 4556 * ' 1 « 4566 MATHEMATICAL TABLES Antiloga?iithms. 525 1 2 3 4 6 6 7 , 8 9 12 3 4 5 6 7 8 9 ■50 3162 3170 3177 3184 3192 3199 3206 3214 3221 3228 I I 2 3 4 4367 •51 ■52 •63 •54 3236 3311 3388 3+67 3243 3319 3396 3475 3251 3327 3404 3483 3258 3334 3412 3491 3266 3342 3420 3499 3273 3350 3428 3508 3281 3357 3436 3516 3289 3365 3443 3324 3296 3373 3451 3532 3304 3381 3459 3540 1223 1223 1223 1223 4 4 4 4 5567 5567 5667 5667 ■56 3548 3556 3565 3573 3581 3589 3597 3606 3614 3622 1223 4 5677 ■56 ■57 •68 •69 3631 3715 3802 3890 3639 3724 38H 3899 3648 3733 3819 3908 3656 3741 3828 3917 3664 3750 3837 3926 3673 3846 3936 3681 3767 3835 3945 3690 3864 3954 3698 3784 3873 3963 3707 3882 3972 1233 1233 1234 1234 4 4 4 5 5678 5678 5678 5678 ■80 3981 3990 3999 4009 4018 4027 4036 4046 4055 4064 1234 5 6678 ■61 •62 ■63 ■64 4074 4169 4266 4365 4083 4178 4276 4375 4093 4188 4285 4385 4102 4198 4295 4395 4111 4207 4305 4406 4121 4217 4315 4416 4130 4227 4325 4426 4140 4236 4333 4436 4150 4246 4345 4446 4159 4256 4355 4457 1234 1234 1234 1234 5 5 5 5 6789 6789 6789 6789 •65 4467 4477 4487 4498 4508 4519 4529 4539 4550 4360 1234 5 6789 •66 ■67 •68 •69 4571 4677 4786 4898 4581 4688 4797 4909 4592 4920 4603 4710 4819 4932 4613 4721 4831 4943 4624 4732 4842 4955 4634 4742 4966 4645 4733 4864 4977 4656 4764 4875 4989 4667 4775 4887 5000 1234 1234 1234 1235 5 5 6 6 6 7 9 10 7 8 9 10 7 8 9 10 7 8 9 10 •70 5012 5023 5035 5047 5058 5070 5082 5093 5105 5117 1245 6 7 8 9 II ■71 •72 ■73 ■74 5129 5248 5370 5495 5140 5260 5383 5508 5152 5272 5395 5521 5164 5284 5408 5534 5176 5297 5420 5546 5188 5309 5433 5559 5200 5321 5445 5572 5212 5333 5458 5583 5224 5346 5470 5598 5236 5358 5483 5610 12 4 5 1245 1345 1345 6 6 6 6 7 8 JO II 7 9 10 II 8 9 10 11 8 9 10 12 •76 5623 5636 5649 5662 5675 5689 5702 5715 5728 5741 1345 7 8 9 10 12 •76 •77 ■78 •79 5754 5888 6026 6:66 5768 5902 6039 6180 5781 5916 6053 6194 5794 5929 6067 6209 5808 6081 6223 5821 5957 6095 6237 5834 5970 6109 6252 5848 5984 6124 6266 3861 3998 6138 6281 5875 6012 6152 6295 1345 1345 1346 1346 7 7 7 7 8 9 II 12 8 10 II 12 8 10 II 13 9 10 II 13 '80 6310 6324 6339 6353 6368 6383 6397 6412 6427 6442 1346 7 9 10 12 13 •81 •82 •83 ■84 6457 6607 6761 691S 6471 6622 6776 6934 6486 6637 6792 6950 6501 6633 6S08 6966 6516 6668 6823 6982 6531 6683 6839 6998 6546 6699 6855 7015 6561 6714 6871 7031 6577 6730 68«7 7047 6592 6745 6902 7063 2356 2356 2356 2356 8 8 8 8 9 II 12 14 9 II 12 14 9 " 13 14 10 n 13 13 ■86 7079 7096 7112 7129 7145 7161 7178 7194 721J 7228 2357 8 10 12 13 15 ■86 •87 •88 •89 7244 7413 7586 7762 7261 7430 7603 7780 .7278 7447 7621 7798 7295 7464 7638 7816 7311 7482 7656 7834 7328 7499 7674 7852 7345 7516 7691 7870 7362 7534 7709 7889 7379 7551 7727 7907 7396 7568 7745 7925 2357 2357 2457 2457 8 9 9 9 10 12 13 15 10 12 14 i5 11 12 14 16 II 13 14 16 •90 7943 7962 7980 7998 8017 8033 8054 8072 8091 8110 2467 9 II 13 15 17 ■91 •92 •93 ■94 8128 8318 8511 8710 8147 8337 8531 8730 8166 8356 8551 8750 8185 8375 8570 8770 8204 8395 8590 8790 8222 8414 8610 8810 8241 8630 8831 8260 8453 8650 8851 8279 8472 8670 8872 8299 8492 S690 8892 2468 2468 2468 2468 9 10 10 10 11 13 15 17 12 14 13 17 12 14 16 18 12 14 16 18 •96 8913 8933 8954 8974 8995 9016 9036 9057 9078 9099 2468 10 12 15 17 19 ■96 •97 ■98 ■99 9120 9333 9550 977a 9141 9354 9572 9795 9162 9376 9594 9817 9183 9397 96x6 9840 9204 9419 9638 9863 9226 9441 9661 9886 9247 9462 9683 9908 9268 9484 9705 9931 9290 9506 9727 9954 9311 9528 9750 9977 2468 2479 2479 2579 II II II 11 13 15 17 19 13 15 17 20 13 16 18 20 14 16 18 20 526 ANALYTICAL MECHANICS Angle. Chord. Sine. Tangent. Co- tangent. Cosine. Degrees. Radians. 0° c» I 1-414 1-5708 90° I 2 3 + •0175 •0349 ■0524 •0698 ■017 ■035 •052 -070 •0175 ■0349 •0698 ■0175 ■0349 •0524 •0699 57-2900 28-6363 I9-08II 14-3007 .9998 -9994 •9986 -9976 1-402 1-389 1-377 1-364 1-5533 1-5359 1-5184 1-5010 89 88 87 86 5 ■0873 •087 •0872 •0875 II-430I -9962 I-35I 1-4835 85 6 I 9 •1047 •1222 ■1396 ■1571 •105 •122 ■140 •157 •1045 •1219 •1392 •1564 •1051 •1228 •1405 •1584 9^5144 8-1443 7-1154 6-3138 •9945 •9925 •9903 •9877 1-338 1-325 I-3I2 1-299 1-4661 1-4486 1-4312 1-4137 83 82 81 10 •1745 •174 •1736 ■1763 5-6713 •9848 1-286 1-3963 80 II 12 13 I* •1920 ■Z094 •2269 ■2443 ■192 '209 •226 •244 •1908 ■2079 •2250 •2419 •1944 •2126 •2309 •2493 51446 4-7046 4^33I5 4-0108 •9816 •9781 •9744 •9703 1-272 1-259 1-245 1-231 1-3788 1-3614 1-3439 1-3265 79 78 77 76 15 •2618 •261 •2588 •2679 3-7321 ■9659 1-218 1-3090 75 16 18 19 •2793 •2967 •3142 •3316 •278 •296 •313 ■330 •2756 •2924 •3090 •3256 •2867 •3057 •3249 ■3443 3-4874 3-2709 3-0777 29042 •9613 •9563 •95 1 1 •9455 1-204 l-lgo 1-176 1-161 1-2915 1-2741 1-2566 1-2392 74 73 72 71 20 •3491 •347 •3420 •3640 2-7475 •9397 1-147 1-2217 70 21 22 23 24 ■3665 ■3840 •4014 ■4189 •364 •382 •399 ■416 •3584 ■3746 •3907 ■4067 ■3839 •4040 •4245 •4452 2-6051 2-4751 2-3559 2-2460 ■9336 •9272 •9205 •9135 1-133 1-118 1-104 1-089 1-2043 1-1S68 1-1694 1-1519 67 66 25 •4363 •433 •4226 •4663 2-1445 •9063 1-075 1-1345 65 26 27 28 29 •4538 ■4712 •4887 ■5061 •450 •467 ■484 •501 •4384 ■4540 •4695 •4848 •4877 •5095 •5317 •5543 2-0503 1-9626 1-8807 1-8040 •8988 •8910 •8829 ■8746 I -060 1-045 1-030 I-0I5 1-1170 1-0996 1-0821 1-0647 64 62 61 30 ■5236 ■518 •5000 •5774 f732l ■8660 I -000 1-0472 60 31 32 33 34 •54" ■5585 •5760 ■5934 •534 •551 •568 •585 ■5150 •5299 •5446 •5592 ■6009 •6249 ••6494 •6745 I'6643 1-6003 1-539? 1-4826 •8572 •8480 •8387 •8290 •985 •970 •954 •939 1-0297 I-OI23 •9948 •9774 % % 35 •6109 •601 •5736 •7002 1-4281 •8192 •923 •9599 55 36 37 38 39 •6283 ■6807 •618 •635 •651 ■668 •5878 •6018 •6157 •6293 •7265 •7536 ■7813 ■8098 1-3764 1-3270 1-2799 1-2349 •8090 ■7986 •7880 •7771 •908 •892 ■877 •861 •9425 •9250 •9076 •8901 54 53 52 51 40 •6981 •684 ■6428 •8391 1-1918 •7660 ■854 -8727 50 41 42 43 44 ■7156 •7330 ■7679 •700 •717 •733 ■749 •6561 •6691 •6820 ■6947 •8693 •9004 ■9325 •9657 1-1504 1-1106 1-0724 1-0355 •7547 •7431 •7314 •7193 •829 •813 •797 •781 1=52 •8378 •8203 •8029 49 48 45° •7854 •755 ■7071 1*0000 I -0000 •7071 ■765 •7854 45° Cosine. Co-tangent. Tangent. Sine. Chord. Radians. Degrees. Angle. INDEX The Numbers refer to the Articles. Accelerated chamber, pendulum in, 226-229. rotation about fixed axis, 236- 237«. Acceleration, ig. diminished : — proportionally to speed, 37-38 ; proportionally to square of speed, 39-43. inversely as distance squared, 34-36. proportional to displacement, 29-30. varying with displacement and speed, 45. Accelerations of point in any cur\-e, 112. radial and transversal, 74-75. tangential and normal, 69. Addition of vectors, 15. Air bubble under plate, 438. Algebraic formulae, 4. Amount of shear, 165, 170, 172. Amplitude, 29. Analogous relations between linear and angular momenta, etc., 293. Analogy between precessional and circular motions, 120. Poinsofs, between statics and kinematics, 366. Angle of friction, 221. Angular acceleration for steady pre- cession, 124. accelerations about moving axes, 123-124; of fluid elements, 45°- , . acceleration, uniform, 92. and areal velocities, 73. and linear velocities com- pounded, 97. momenta, 237 ; general expres- sions for, 290. momentum is a vector, 291 ; of Angiilar momentum — continued. rigid body in plane motion, 266- 267. Angular velocities, composition of, 25(5. velocities of fluid elements, 448. velocities about intersecting axes, 118. — — velocity and angular accelera- tion compounded, 119-120. Application of force occurs at a sur- face or throughout a volume, 212. Archimedes' principle, 430. Areal velocities and angular, 73. velocity constant under central acceleration, 76-78. Asymmetrical load on roof, 401. Atwoods machine, 223-224 ; allow- ing for inertia of pulley, 257. Attractions : — graphical representa- tion of, 348 ; in cavity, 349 ; of cylinder, 343 ; of discs, 340-342 ; of filaments, 336-339, 342 ; of par- ticles, 334 ; of shell, 346 ; of solid sphere, 347 ; of spherical shell, 344-345 ; of thick shell, 347. Axodes, body and space, 95. Balance, the, 393. Ball, Sir R., on wrenches, 420. Ballistic pendulum, 273. Bar, bent, 464. Barometer, heights by, 433. Bar pendulum, 258-261. Base point in most general displace- ment of rigid body, 126. Beam, bent, 464. Beams, bending moments and shear- ing forces in, 405-410. Belts, 136. Bending moment diagram is a link polygon, 410. 6-27 ANALYTICAL MECHANICS Bending moments and shearing forces in beams, 405-410. Bent bar, 464. Bernoulli's theorem, 444. Bifilar suspension by virtual work, 422. Binormal, 112. Body centrode, 95, loi. Bow's notation for graphical statics, 396. Boyle's law, 202. Boys, C. v., on constant of gravita- tion, 350. Brachistochrone, 61-62 ; cycloid is a, 64-65 ; equation of, 63. Brennan, Louis, mono-rail car, 305. Bubbles and films, 435. Bucket descending from roller, 255- 256. Bulk modulus, 462. Buoyancy, centre of, 429 ; surface of, 431- Calculus formulae, 7-8. Cantilever, 409. Capillary ascent, 436-437. Catenary, 324-331 ; approximations to, 329 ; elementary properties of, 328 ; elementary relations for, 325 ; equations of, 326-327 ; para- meter of, 331. Causal relations not assumed, 212. Central acceleration : — formula, 81 ; involves constant areal velocity, 76-78 ; proportional to radius, 66- 68 ; under natural law inversely as distance squared, 83-90(2. axis in general displacement of rigid body, 130. Centre of mass, 369-384 ; for bodies with varying densities, 385 ; in- dependence of translation of and rotation about, 307-309 ; of rigid body in plane motion, 264-269. of percussion, 270. of pressure, 428. Centres of oscillation and suspension convertible, 259. Centrifugal reactions and torques, 306. Centrodes, body and space, 95, loi. Centroids defined, 251?; determina- tion of, 369 ; elementary examples of, 370-373; by integration, 374- 386 ; of lines, 374-375 ; of plane surfaces, 376-381 ; of surfaces of The Numbers refer to the Articles. C en troids — contin ued. revolution, 382-383 ; of solids of revolution, 384. Chain, falling, 231 ; uncoiling, 232. kinematic, 139. Change points, 150. Circle, motion in vertical, 57. Circular cord, loads for, 333. disc, moment of inertia of, 245. motion, uniform, 71. Closed surface, resultant force on due to liquids, 429. Conical circular currents, 451. Coefficient of viscosity, 454. Composition and resolution of forces, 314- Composition of : — angular velocities, 25^, 98, 118 ; angular velocity and angular acceleration, 11 9- 120; col- linear simple harmonic motions, 31-33; displacements, 15; linear and angular velocities, 97 ; pure strains and rotations, 184 ; rectan- gular vibrations, 67-68 ; strains, 166-169 ; stresses, 460 ; vectors, 15 ; velocities and accelerations, 23- Compound pendulum, 258-261 ; variation of period with axis, 260 ; treatment by energy, 261. Conditions of equilibrium, 315 ; 367- 368. Cone of friction, 221. Conical motion, 107. pendulum, 72. Conic, natural orbit is a, 85-86. Connected particles on rough in- clines, 225. Conservation of energy, 212. Constraints and degrees of freedom, 26. Contact, angle of, 436, 438. Continuity, equation of, 441. Convertibility of centres of oscillation and suspension, 259. Co-ordinates, cartesian and polar, 10. Coplanar forces, resultant of, 364. motions, 100-102. Cord in circle, loads for, 333. equilibrium of, 322-325. Coulomb on friction, 201. Couples, 363. Crab tree, H., on spinning top, 298, 305- Crank and lever, 148. INDEX 529 The N%t77ibers refer to the Articles. Criterion for rotation in linkworks, Elliptic lamina, moment 152. Curvature, i^c. Curved membrane in tension, 434. Cycloidal oscillations, 59. pendulum, 60. Cycloid, as brachistochrone, 64-65 ; as involute of cycloid, 60; intrinsic equation of, 58 ; motion in vertical, 58-60. Cylinder, attraction of, 343. : steady flow past, 452. twisted, 465. Cylindrical motion, 106. Damped vibration, 45. Dead points, 150. Deformability and rigidity in link- ages, 144. Deformable figures and systems, 134; Density and pressure, 443. Description of phenomena, mechan- ics a, 212. Differential equations, 9. wheel and axle, 389. Dilatation, a uniform, 165. Dimensions of units, 12. Discs, attractions of, 340-342. Disc, moment of inertia of, 245. Displacement, 13. ' Displacement ' of floating body, 431, 432- of rigid body, 126-130. Distributed loads on beams, 407, 409-410. Double-crank linkwork, 149. Drop, form of large, 438. Ductile bodies, 455. Dunkerley, S., on mechanism, 144, 163. Durley, Ji. J., on kinematics of machines, i6i. Efficiency of a simple machine, 390-391- Elastic bodies, nature of, 455 ; per- fectly and imperfectly, 455. limit, 455. Elasticities and their relations, 462. Elasticity, 202 ; defined, 456. Electrons, masses of at high speeds, 206. Elevation of exterior rail, 229. Ellipsoid, strain, 174, 180-181 ; shear, 182-183. of inertia of, 245a. orbit, 85-86 ; velocity, period and focal acceleration in, 89-90. Elongation of helical spring, 466. Energy : — conservation of, 212 ; kin- etic, 212 ; lost by impact, 219; of plane motion of rigid body, 268 ; potential, 215 ; transformation of, 215. Encyclopaedia Bntanntca, 456. Engineering, on gyroscopes, 305. Enunciations, etc., of mechanical bases : — Newton's, i8g-i()i iMach's, 194 ; Pearson's, 197 ; Love's treat- treatment, 198 ; set of brief, 209. Epoch, 29. Equations of: — continuity for fluids, 441 ; motion for axes of fixed directions, 310; motion for fluids, 442. Equilibrium: — conditions of, 315; conditions of under coplanar forces, 367-368 ; conditions of under general forces, 412 ; neutral, 392 ; of body with fixed points, 418 ; of body with points on plane, 419 ; stable, 392 ; unstable, 392. Equinoxes, precession of, 305. Equipotential surfaces, 356. Ether, 206. Enter's dynamical equations, 296, 306. theorem, 11 3- 11 4. Evaluation of stresses apparently indeterminate, 402. Evans and Main, 188. Exterior rail, elevation of, 229. Falls of : — growing raindrop, 233 ; hailstone, 39-40 ; mist, 37-38. Field and potential, 354. Filaments, attractions of, 336-339 ; moment of inertia of, 243. Finite angular displacements, 116- 117. arcs, simple pendulum in, 54-56. Fixation of point in rotating body, 272a. Floating bodies, 430 ; stability of, 431-432. Flotation, surface of, 431. Flow past cylinder, 452. Fluids," liquids, and gases, natures of, 423- Flux, 352. 2 L 53° ANALYTICAL MECHANICS The Numbers refer to the Articles. Focal acceleration, for any conic, 91 ; in elliptic orbit, 90. Forced vibration, 46-48. Force on submerged plane, 427. polygons, 396-410. Formulae, algebra, 4 ; trigonometry, 5 ; co-ordinate geometry, 6 ; cal- culus, 7, 8 ; diflferential equa- tion, 9. Foucaulfs pendulum, in. Freedom and constraint, 26. Free or natural vibration, 48. Frictional couple for axle, 279. Friction, angle of, 221 ; cone of, 221 ; coeflficient of, 221 ; in At- wood's machine, 224 ; laws of, 201. Fundamental equations of hydro- statics, 425. Funicular polygons, 398-401, 410 ; closed for equilibrium, 399 ; for resultant of forces, 398 ; reactions found by, 400 ; with different poles, 398a. Galileo, 187, 196. Gases, fluids, and liquids, natures of, 423- Gauche polygon of vectors, 25. Gauss' theorem, 352. General displacement of rigid body, 126-130. motion of rigid body, 131 ; with one point fixed, 125. Geometrical clamp, 26. formulae, 6. Goodman on efficiencies of screws, 391 ; friction, 201 ; method of sections, 402 ; ' vena contracta,' 446. Graphical composition of coUinear simple harmonic motions, 32-33. method for moments of inertia of laminae, 254. statics, 394-410. treatment for rise and fall of shot, 44. Graphs, displacement, 20 ; speed, 21 ; other forms of, 22. Gravitational field, 335. Gravitation, 200 ; constant of, 335. Gravity waves, 453. Gray, A., on compound pendulum, 259 ; on helical springs, 466. Greenhill, Sir G., on analogous re- lations in mechanics, 293 ; on I Greenhill, Sir G. — r.ontinucd. ] ballistic pendulums, 273 ; on gyro- \ scopes, 298, 305. Growing raindrop, fall of, 233. Gyration, radius of, 237. Gyroscopic motions, examples of, 305- Gyroscopes, 287-289. Hailstone, fall of, 39-40. j%r/'j cell, 155-156. Hayward^s equations, 311. Head, pressure, 444 ; velocity, 444. Heights by barometer, 433. Helical spring, 466. Henrici and Turner an linkages, 144. Herpolhode, 312. Hertz, H., 211. Heterogeneous strains, 185. Higher pairing, examples of, 163. High falls, 34-36. Hodograph, 70 ; for natural orbits, 84. ' Hole, slot, and plane,' 26. Homogeneous bodies, 455. strains, 164, 173-184 ; nine con- stants of, 175. stress and its components, 458. Hookis law, 202. Huyghens, 187, 196. Hydrometers, 430. Hydrostatics, fundamental equations of, 425- . Hydrostatic pressure independent of direction, 424. Hyperbolic orbit and hodograph, 87. Impact, direct, 217; oblique, 218; loss of energy in, 219 ; of molecule, 220. Impulse, 212. Impulsive torque, 237. Inclined plane, 316. Incline, motion down, 52. Incompressible fluid as 'link,' 138. Independence of translation of centre of mass of rigid body and rotation about it, 269, 307-309. Inertia, 186 ; at high speeds, 206. moment of, defined, 237. Inextensibility of cord, 137. Initial conditions, for orbit, 88 ; for simple harmonic motion, 30. Instantaneous axis of rotation, 95. centres of rotation, 95, loi ; for linkages, 145. Invariants of forces in reduction to wrench, 420. Inversions of linkages, 143. Isotropic bodies, 455. Jager, G., on hydrokinetics, 454. Jeans, J. H., on change of latitude, 312. Joints, reactions at, 403-404. Kelvin, Lord, on strain, 164. Kelvin and Tait on homogeneous bodies, 455 ; New/on' s laws, 192 ; strains, 164, 173 ; stress on given plane, 459 ; time, 204. Kennedy, A. B. M^., 161. Kepler's laws, qoa, 200. Kinetic energy, 212; of rigid body in plane motion, 268 ; in rotation, 299. Kinematic chain, 139. Kirchhoff, 198. Krigar-Menzel, on constant of gravi- tation, 350. Lagrange's ' Me'canique Analytique,' 388. Lagrange on '\''irtual Work, 388. Lamffs ' Hydrodynamics,' 444. Laminae, moments of inertia of, elliptical, 245a ; graphical method for, 254 ; rectangular, 244 ; trian- gular, 248-252. Lamina theorem for moments of inertia, 239. Laplace's theorem, 357. Large drop, form of^ 438. Latitude, change of, 312. Lazy-tongs, 157. Level of liquid disturbed by accelera- tions, 439-440- Lever, 389. Limiting speed, 37-42. Linear and angular velocities com- pounded, 97. momentum and acceleration of rigid body in plane motion, 265. Lines of force, 351. Linkages, plane, 143-161. quadric, 146-152. Link polygons, 398-401, 410 ; closed for equilibrium, 399 ; for resultant of forces, 398 ; reactions found by, 400 ; with different poles, 398(2. INDEX The Numbers refer to the Articles. S31 Links and their relative motion, 139- 141. Linkworks, velocity ratios for, 147, 159-160. Liquid in accelerated vessel, 439-440. force on plane in, 427. — ; — rotating as rigid solid, 447. Liquids, fluids, and gases, natures of, 423; Localisation of vectors, 16. Lodge, Sir Oliver, on axioms, 199 ; on dynamics, 2 1 1 ; on Kepler's laws, 200. Logarithmic decrement, 45. spiral, 45. Longitudinal mass of electron, 206. Long waves, 453. Lorentz, H. A., masses of electrons at high speeds, 206. Loss of energy at impact, 219. Love, A. E. H., on Newton, 198. Low, D. A., on beams, 410. Lubricated surfaces, friction of, 201. jMach, Ernst, on Newton's enuncia- tions, 193-196, 205, 207, 211. Main, Evans and, i88. Mass, at high speeds, 206 ; brought into equations, 212 ; defined, 209 ; longitudinal and transverse, 206 ; need for, 186 ; quantities usually proportional to, 207. Maxwell, J. C, on reciprocal figures, 395 ; on time, 204. Mechanical advantage, 389-391. Mechanisms, subdivisions o^ 134. Meniscus, 436. Metacentre, 431-432. Method of sections for evaluation of stresses apparently indeterminate, 402. Miuchin and Dale aa. catenary, 328. Minimal velocities for top to spin and to sleep, 304. Mist, fall of, 37-38. Moduli of elasticity, 202. Molecule, impact of, 220. Moment of couple, 363. of a force about a line, 414. of inertia defined, 237. of inertia table or pendulum, 263. of momentum, 237. Moments of inertia, table of, 253. of inertia, theorems on, 238- 241 ; evaluation of, 242-252. 532 ANALYTICAL MECHANICS Moments, definition and theorem, 25a. Momentum, angular, 237 ; linear, 212. Morley, A., on beams, 410 ; on mo- ments of inertia found graphically, 254. Motion, equations of, for fluids, 442. Motions of rigid body with one point fixed, 125, 286-306. Moving axes, fixed in body, 295- 296 ; torques about, 294. axis, rotation about a, 121. Multiplied cord, 135, 318. Mutual moment of two lines, 414. Natural law, orbits under, 83-90^. ' Nature ' on gyrostats, 305. Neutral equilibrium, 392. surface of bent bar, 464. Newtonian constant of gravitation, 335, 350- NewtorHs prmciples, 188; defini- tions, 189; axioms or laws of motion, 190 ; corollaries, 191 ; dis- ciples and critics, 192. Nine constants of homogeneous strain, 175. Normal and tangential accelerations, 69. Nutation, 301 ; velocity of, 303. Oblique axis theorem for moments of inertia, 241. Orbit, curvature of, 80 ; differential equation of, 79 ; under natural law, 83-goa ; velocity in, 82. Oscillation, centre of, 259. Oscillations, rolling, 285. Oscillatory waves, 453. Outflow velocity, 445. Pairing of links, lower, 140 ; higher, 141- Pantograph, 153. Pappu^ theorems, 386. Parabolic cord has uniform hori- zontal load, 332. orbit, 85-86. Parallel axes theorem for moments of inertia, 238. cranks, 149, 157. forces, resultant of, 362. motion, 151. Parallelepiped, moment of inertia of, 244. Parameter of catenary, 331. The Numbers refer to the Articles. Particles on rough inclines, 225. Pearson, Karl, on Newton, etc., 197, 211. Peaucellier's cell, 154. Pendulum, ballistic, 273 ; compound, 258-261 ; conical, 72 ; cycloidal, 60 ; in accelerated chamber, 226- 229 ; rigid, 258-261 ; simple, 53- 57; spherical, 109-111; torsional, 262. Percussion, centre of, 270. Period and velocity in elliptic orbit, 89-90. of vibration, 30. Permanent set, 455. Perry, J., on elongation of springs, 466 ; on gyrostats, 305. Physical conceptions introduced, 186. pendulum, 258-261. Pile of spheres by virtual work, 421. Pitch of screw, 130. Plane linkages, 143-161. motion of rigid body, 264-269 ; analysed, 96. Planetary orbit is a conic, 85-86. Plastic bodies, 455. Plates, air bubble under, 438 ; ascent of liquid between, 437. Poincare, H., 211. Poi7isofs analogy between statics and kinematics, 366 ; central axis, 420; reduction of forces, 411. Poissoiis ratio, 165, 462 ; theorem, 357- Polar diagrams for linkwork velocity ratios, 147, 160. Polhode, 312. Potential, and field, 354 ; and work, 355 ; energy, 215 ; graphic repre- sentation of, 361 ; introduced, 353 ; of disc, 358 ; of solid sphere, 360 ; of spherical shell, 359 ; velocity, 449. Poyntitig, J. H., on constant of gravitation, 350. Precessional motion, 120, 122. velocities at limiting inclina- tions, 302. Precession, angular acceleration for steady, 124 ; conical without torque, 298 ; maintenance of, 287-289 ; of equinoxes, 305 ; of top, 297 ; starting of, 300-301. Pressure and density, 443 ; at any depth in a liquid, 426 ; centre of Pressure and density— con/imied. 428 ; head, 444 ; on curved stretched membrane, 434 ; pairing of links, 142. Principal axes and planes of stress, 459- Principia,' Newtoris, 188, 196, 211. Prism, moment of inertia of, 247. Projectile, 49-51. Proportionality point, 455. Pulley or multiplied cord, 318. Pure strain, 164 ; along co-ordinate axes, 179 ; conditions for, 177 ; derived from homogeneous strain, 178. QUADRic linkages, 146-152. Quoit, motions of, 312. Radial and transversal velocities and accelerations, 74-75. Radius of gyration, 23.7. Rankifie, W. J. M., on method of sections, 402 ; on stress and strain, 164. Range of projectile, 49 ; on incline, 5?- Rational mechanics, 198. Reactions, at joints, 403-404 ; inside bodies, 405-410. and torques, centrifugal, 306. Reciprocal figures, 395. Rectangular vibrations, composition of, 67-68. Reduction of forces to a wrench, 420. to two forces, 417. Relative character of mechanics, 203. Repose, angle of, 221. Resolution of forces, 314; of strains, 166-169. Resultant, angular velocity and mo- mentum not about same axes, 293 ; conditions for single, 415; force on closed surface in liquid, 429 ; line of action of single, 416 ; of forces in solid space, 411; of parallel forces, 362 ; work is al- gebraic sum of component works, 320. Reuleaux on mechanisms, 139, 161. Richarz, F., on constant of gravita- tion, 350. Right-handed system of co-ordmates, 94- INDEX The Numbers refer to the Articles. 533 Rigid body introduced, 92. Rigidity, 462 ; and deformability of linkages, 144. Ripples, 453. Rise and fall of shot, 41-44. Ritter, method of sections, 402. Rodriguez co-ordinates, 115; theo- rem, 1 16- 1 17. Roller and bucket problem, 255-256. Rolling and sliding, down incline, 278 ; on level, 280-282 ; up incline, 283-284. cones, 122, 125 ; contact, 141 ; down incline, 274-277 ; motion, 95, loi ; oscillations, 285. Roof, snow slipping on, 234 ; with asymmetrical load, 401. Rotating liquid, 447, 451. vessel has liquid surface a para- boloid, 440. Rotational energy, general expres- sions for, 299. Rotation, about a moving axis, 121 ; and translation compounded, 94 ; and translation of rigid body, in- dependence of, 269 ; in homo- geneous strains, 176 ; in linkworks, graphical criterion for, 152 ; of orbit of spherical pendulum, iii ; of rigid body under uniform ac- celeration, 236-2373 ; simple har- monic, 93. Rotations and translations compared, 92. of elements of fluid, 448, 450. Rough incline, motion on, 222. Rouih, E. y., on ballistic pendulum, 273 ; centrifugal reactions, 306 ; Gauss' theorem, 352 ; rule for moments of inertia, 253 ; screws, 133 ; theorem of six constants, 241. Russell, Bertram, 211. Sag of telegraph wires, 330. Scalars, 14. Schlick, Otto, on steadying vessels, 305- Scope of mechanics, i, 2. Screw pairs, 162. Screws, 130, 391. Sections, method of, for stresses in roof, etc., 402. Sectorial velocity, T2I- Sensitiveness of the balance, 392. Separation of bars for reactions, 404. 534 ANALYTICAL MECHANICS Shear, amount of, 165, 170, 172 ; ellipsoid, 182-183 ; simple, 165. viewed as slidings, 1 70-171. Shearing forces and bending mo- ments in beams, 405-410. stress, two aspects of, 461. Shell, attractions of, 344-346. Shot, rise and fall of, 41-44. Simple harmonic motion, 29-30. machines, 389-391. pendulum, 53-57. Simultaneous elliptic orbits, 90a. Single resultant, conditions for, 415 ; line of action of, 416. Skin friction, 452. Sleeping top, 304. Slider crank chain, 158. other inversions of, 161. Sliding contact, 141. ' and rolling down incline, 278 ; on level, 280-282 ; up incline, 283- 284. Slip of snow on slope, 234. Snow slipping on slope, 234. Soap bubbles, 435. Solid angles, 351. co-ordinates, 103-104. sphere, attraction of, 347, 360 ; moment of inertia of, 246 ; poten- tial of, 360. Space centrode, 95, loi. Speed, 18. Spencer, Herbert, 211. Spherical motion, 108. pendulum, 109-111. shell, attractions of, 344-346. moment of inertia of, 246. Spinning contact, 141. Spiral spring, 464. Spontaneous rotation, axis of, 270. Spring, helical, 466. Stable equilibrium, 392. Stability of equilibrium, 392. of the balance, 392. Statics, graphical, 394-410. Steady flow, of viscous liquid, 454 ; past cylinder, 452 ; under gravity, 444. precession, angular acceleration for, 124. Straight line, linkworks to draw a, 154-156. Strain, definition of, 164; ellipsoid, 174, 180-181. Strains, composition and resolution of, 166-169. The Nuvibers refer to the Articles. Strains, homogeneous, 164, 173-184. Stream line, 444. Stress, 202, 456-457 ; across any plane, 459 ; and strain, work of, 463. homogeneous, 458. polygons, 396-410. Stresses, composition of, 460. Subdivisions of mechanics, i. Successive finite angular displace- ments, 1 1 6- 1 17. Surface of liquid disturbed by accel- erations, 439-440. waves, 453. Tackle or purchase, 318. Tait, Kelvin and, 192 ; on strains, 164, 173 ; on time, 204. P. G., on gravitation, 200 ; on motions of quoit, 312. Tangential and normal accelerations," 69. Tarleton, Wilhainson and, on screws, 133- Telegraph wires, sag and excess length in, 330. Tension pairing of links, 142. Terminal speed, 37-42. Three connected masses, motion of, 230. Tidal waves, 453. Time and motion, 17. measurement of, 204. Todhunter on virtual work, 321. Top, steady precession of, 297. TorricelWs theorem, 445. Torque, 237. Torques, centrifugal, 306. Torsional pendulum, 262. Torsion of cylinder, 465. Tortuosity, 112. Tower, Beauchanip, on friction, 201. Trajectory of projectile, 50. Transformations of energy, 215. Translation and rotation of rigid body, independence of, 269. Translations and rotations compared, 92. Transversal and radial velocities and accelerations, 74-75. Transverse mass of electron, 206. Trap door, fall of, 271-272. Triangular lamina, moments of in- ertia of, 248-252. Trigonometrical formulae, 5. Tripod by virtual work, 421. Tubes of force, 351. INDEX 535 Turner, HenricianA, on linkages, 144. Twisted cylinder, 465. Twisting velocity, resultant, 1 33. Twists, 130. Tw:o forces, system reduced to, 417. Typical pure strains, 165. Uniform acceleration, 27-28. angular acceleration, 92. circular motion, 71. Uniformly loaded beams, 409-410. Units, II, 12 ; and systems of units, 213-214. Unstable equilibrium, 392. Vapour, 423. Vectorial polygons, 24. Vectors, 14. Velocity, 18 ; and period in elliptic orbit, 89-90 ; head, 444 ; of any point of rigid body, 132 ; potential, 449 ; ratios for slider crank chain, 159-160; ratios of link works, 147, 159-160. * Vena contracta,' 446. Vertical circle, motion in, 57. Vibrations, note on, 235. Virtual work, bifilar suspension by, 422 ; for particles, 321 ; for rigid bodies, 387-388 ; pile of spheres by, 421 ; pressure on membrane by, 434; tripod by, 421 ; zero for equilibrium, 321. The Numbers refer to the Articles. Viscosity, qualitative mention of, 423 ; determination of, 454. Viscous fluids, 455. Volume, change in strain, 164. elasticity, 462. Water waves, 453. IVat/'s parallel motion, 151. Waves, 453. Webster, A. G., on general motions of a rigid body, 312 ; on heights by barometer, 433. Wedge, 317. Wedges of immersion and emersion, 431- Weston's pulley blocks, 389. Wheel and axle, 389. Williamson and Tarleton, on screws, 133- Wind loads on roof, 401. Work, 212 ; in oblique displacements, 216; of given force and displace- ment, 319; of stress and strain, 463; virtual, 321, 387-388, 421- 422, 434. Worthington, A. M., on gyroscopes, 298, 305. Wrench, forces reduced to, 420. Young's modulus, 462. Ziwet, Alexander, 211. Printed by T. and A. Constable, Printers to His Majesty at the Edinburgh University Press EXAMPLES— MISCELLANEO US 511 vacuum of a cylindrical barometer tube ? If the height of the barometer be 30 inches, the length of the Torricellian vacuum 5 inches, and the cross section of the tube i square inch, calculate the effect, on the column of mercury, of the insertion of 1/4 cubic inch of air into the vacuum, the absolute temperature having changed during the experi- ment in the ratio of 48 to 49.' (Board of Education, Theo, Mech., Fluids, Stage 3, 1908, 50.) 7. ' State the laws of the ascent and depression of liquid in capillary tubes. 'Show in a diagram the state of equilibrium when a capillary tube is immersed vertically in a liquid (a) when the liquid wets the tube ; \b) when it does not do so. ' In each case, show by lines with arrow heads, supplemented by explana- tions, how the forces act which maintain the liquid inside the tube in equilibrium.' (Board of Education, Theo. Mech., Fluids, Stage 3, 1908, 51.) 8. ' In a suspension bridge the roadway, weighing w lb. per foot of the length, is upheld by a uniform chain suspended from two points in a horizontal line. Neglecting the weight of the chain in comparison with that of the roadway, prove that the tension at a point is equal to Nw where N is the length of the normal at the point between the curve and the axis of symmetry.' (Board of Education, Theo. Mech., Honours, 1908, 63.) 9. ' A heavy string, of weight w per unit of length and of length 2/, is sus- pended from two points which are in a horizontal line and ia apart, so as to form a catenary. A circular disc of radius b and weight W is placed so as to rest symmetrically on the string and in the same vertical plane. Form an equation for the determination of the length of the string which is in contact with the disc in the position of equilibrium.' (Board of Education, Theo. Mech., Honours, 1908, 64.) 10. ' Find the centre of gravity of a circular arc of uniform density. 'A uniform ' flexible rope is wrapped round a cylinder, whose axis is horizontal, and the length of the rope equals the circumference of the cylinder. Its free end is at the end of a horizontal diameter. The cylinder is turned through a right angle, so that the free end falls through a distance equal to a quarter of the circumference. Find the work done by gravity.' (Board of Education, Theo. Mech., Honours, 1908, 67.) 11. 'Draw a triangle ABC with BC vertical, and suppose it to represent a frame of three weightless bars. A weight J*^is hung from^, and BC is kept vertical by being fastened to a wall at two given points, X and Y. Find the stresses in AB and AC, and the forces at X and Y caused by W: (Board of Education, Theo. Mech., Solids, Stage 3, 1909, 41.) 12. ' Show that a force, acting on a rigid body in an assigned direction along a given line, can be replaced by an equal force, acting in the same direction along a parallel line, and a couple. 'Let AB AC, AD be three edges of a cube, and consider AD and two other edges parallel to AB and ^ C respectively, which neither intersect AD nor one another. Suppose that equal forces, P, act along these edges respectively in the directions AB, AC, AD. Show how to reduce them to a resultant force and a couple. Also, show how to draw the line along which the resultant acts when the plane of the couple is at ria-ht angles to the direction of the resultant' (Board of Education, Theo. Mech., Solids, Stage 3, 1909, 42.) 13. 'A ladder, whose centre of gravity is half-way between its ends, rests