^Mimi§i CORNELL UNIVERSITY LIBRARY CORNELL UNIVERSITY LIBRARY 3 1924 059 509 897 Cornell University Library 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/cu31924059509897 ANALYTICAL MECHANICS FOR STUDENTS OF PHYSICS AND ENGINEERING HAROUTUNE M. gADOURIAN, M.A., Ph.D. Instructor of Physics in the Sheffield Scientific School of Yale University NEW YORK D. VAN NOSTRAND COMPANY 25 Paek Place 1913 t.v. CoPTBiaHT, 1913, BT D. VAN NOSTRAND COMPANY Stanbope iprees F. H. GILSON COMPANY BOSTON, U.S.A. PREFACE. The following work is based upon a course of lectures and recitations which the author has given, during the last few years, to the Junior class of the Electrical Engineering Department of the Sheffield Scientific School. It has been the author's aim to present the subject in such a manner as to enable the student to acquire a firm grasp of the fundamental principles of Mechanics and to apply them to problems with the minimum amount of mental effort. In other words economy of thought is the goal at which the author has aimed. It should not be understood, however,, that the author has been led by the tendency toward reducing: text-books to collections of rules, mnemonic forms, and formu- lae. Rules and drill methods tend toward the exclusion of reasoning rather than toward efficiency in thinking. The following features of the treatment of the subject may be noted: In order to make the book suitable for the purposes of more than one class of students more special topics are discussed than any one class will probably take up. But these are so arranged as to permit the omission of one or more without breaking the logical continuity of the subject.. In deciding on the order of the topics discussed two more or less confUcting factors have been kept before the eye, i.e., to make the treatment logical, yet to introduce as few new concepts at a time as possible. It is to secure the second of these ends, for instance, that the historical order of the development of mechanics is followed by discussing equilibrium before motion. This arrangement not only iv PREFACE grades the path of the student by leading him from the easier to the more difficult dynamical ideas, but it also gives him time to acquire proficiency in the use of his mathe- matical tools. As a result of the severe criticisms of Newton's laws of motion by such men as Heinrich Hertz, Ernst Mach, and Karl Pearson, authors of recent text-books on Mechanics have shown a tendency to give either a new set. of laws or none at all. There is no doubt that a subject like Mechanics should start, as in the case of Thermodynamics, with a few simple laws and the entire structure of the science should be based upon them. In the present work the following law is made the basis of the entire subject: To every action there is an equal and opposite reaction, or, the sum of all the actions to which a body or a part of a body is subject at any instant vanishes. Four concepts are associated with the term action, namely, the concepts of force, torque, linear kinetic reaction, and ■angular kinetic reaction. These are introduced one at a time and in connection with the application of the law to a certain class of problems. Force is introduced with the ■equilibrium of a particle (pp. 15, 16), torque with the equihb- rium of a rigid body (pp. 35, 39, 40), linear kinetic reaction with the motion of a particle (pp. 100-106), angular kinetic reaction with the motion of a rigid body (pp. 218-221). Thus by introducing the concepts of linear and angular reactions and by extending the meaning of the term action to include these reactions, the fundamental principle of Mechanics is put in the form of a single law, which is equiva- lent to Newton's laws of motion and which has the advan- tages of the point of view involved in D'Alembert's principle. This law has the directness and simpUcity of Newton's third law, so that the beginner can easily understand it and apply it to simple problems of equilibrium, and yet it admits of wider interpretation and apphcation with the growth of PREFACE V the student's knowledge. By making this law the central idea of the entire subject and by gradually extending its interpretation the treatment is made uniform, coherent, and progressive. While appeal is made to the student's experience in in- troducing the principles of the conservation of dynamical energy and of the conservation of momentum they are shown to be direct consequences of the law of action and reaction. The equivalence and the alternative character of the conservation principles and of the law of action and reaction are emphasized by working out a number of prob- lems by the appUcation of both the law and the principles. The two types of motion, i.e., motion of translation and motion of rotation, are treated not only in the same general manner, but are developed along almost parallel Unes. The simpler types of motion which are generally treated under Kinematics are given in the present work as problems in Dynamics. The author believes that the practice of divesting the physical character of the motion from the simpler types and reducing them to problems in integration is imfortunate. On account of their freedom from mathe- matical difficulties the simpler tj^jes of motion are particu- larly well adapted to illustrate the principles of dynamics. In order to differentiate between vectors and their magni- tudes the former are printed in the Gothic type. In conclusion the author wishes to express his obligations to Mr. Leigh Page for reading the manuscript and to Dr, David D. Leib for reading the proofs and to both for many valuable suggestions. H. M. Dadourian. Yale University, January, 1913. TABLE OF CONTENTS. PAQB Table of notations xi Introduction 1 Chaptek I. Addition and Resolution of Vectors. Scalar and vector magnitudes 3 Addition of two vectors 4 Difference of two vectors 7 Resolution of vectors into components 8 Resultant of any number of vectors 9 Chapter II. Equilibrium of a Particle. Particle 14 Force 14 The law of action and reaction 15 Condition of equilibrium of a particle 16 Sliding friction 21 Resultant of a system of forces 25 Chapter III. Equilibrium of a Rigid Body. Rigid body 31 Theorems on the motion of rigid bodies 31 Linear and angular action 35 The law of action and reaction 39 Conditions of equiUbrium of a rigid body 40 Resultant of a system of forces 47 Friction on journals and pivots 51 Rolling friction 56 Chapter IV. Equilibrium of Flexible Cords. Simplification of problems 61 Suspension bridge problem 62 The catenary 63 Friction belts 67 vii viii TABLE OF CONTENTS Chapter V. Motion. paob Analysis of motion 73 Fundamental magnitudes 74 Velocity 77 Angular velocity 86 Acceleration 89 Angular acceleration 97 Chapter VI. Motion op a Particle. Kinetic reaction 100 The law of action and reaction 101 Force equation 106 Difference between mass and weight 109 Motion of a particle under a constant force 113 Motion of a particle imder a variable force 125 Chapter VII. Center op Mass and Moment op Inertia. Center of mass 140 Moment of inertia 152 Chapter VIII. Work. Work done by a force 164 Work done by a torque 169 Hooke's law 172 Virtual work 180 Chapter IX. Energy. Results of work 185 Kinetic energy of translation 186 Kinetic energy of rotation 188 Power 191 Potential energy 193 Conservation of energy 196 Chapter X. Fields op Force and Newtonian Potential. Fields of force 203 Degradation of energy 203 Force experienced by a particle in a field of force 204 Torque experienced by a body in a field of force 205 TABLE OF CONTENTS ix PAGE New conditions of equilibrium 207 Stability of equilibrium 208 Newtonian potential 211 Field intensity 212 Chapter XI. Uniplanab Motion of a Rigid Body. Angular kinetic reaction 218 The law of action and reaction 218 Experimental definition of moment of inertia 220 Torque equation 221 Comparison 222 Torque and energy methods 223 Motion about a fixed axis 224 Motion about instantaneous axes 228 Chapter XII. Impulse and Momentum. Impulse 238 Momentum 239 Conservation of momentum 241 Motion of center of mass 242 Central collision 244 Impact 249 EflSciency of a blow 251 Motion where mass varies .' 253 Oblique impact 258 Chapteb XIII. Angular Impulse and Angular Momentum. Angular impulse 265 Angular momentum 266 Conservation of angular momentum 268 Ballistic pendulum 271 Motion relative to the center of mass 272 Reaction of axis 276 Center of percussion 279 Chapter XIV. Motion of a Particle in a Central Field of Force. Central field of force 283 Equations of motion 283 Motion of two gravitating particles 287 Types of orbits 290 Mass of a planet which has a sateUite 293 Kepler's laws 294 X TABLE OF CONTENTS Chapter XV. Periodic Motion. ^^^^ Simple harmonic motion 297 Composition of simple harmonic motions of equal period 303 EUiptic harmonic motion 306 Pendulums 308 Damped harmonic motion 320 Vibrations about a position of equilibrium 325 Appendix A. Mathematical Fohmxtl.*!. Algebraic relations 337 Trigonometric relations 338 Exponential and hyperboUc relations 339 Appendix B. Mathematical Tables. Logarithms of numbers 343 Natural trigonometric functions 345 Exponential functions 347 Index 349 TABLE OF NOTATIONS. a = radius, length, constant. b = radius, length, constant. c = const., center of mass. d = distance. e = nap. base, coef. of restitu- tion. / = acceleration. ■ g = grav. acceleration. h = vertical height, constant. i = V^. k = constant. I = length, direction cosine. m = mass, direction cosine. n = number, direction cosine. o = origin, center. ■p — pressure, page. q = variable magnitude, r = radius, radius vector. « = strain, length of curve. t = time. u = velocity. V = velocity, volume. V) = weight. X = variable magnitude. y = variable magnitude, z = variable magnitude. A = area, point, constant. B = point, constant. C = point, constant. D = distance. E = total energy. F = force, frictional force. G = moment of force or torque. H = height, force derived from po- tential, angular impulse. / = moment of inertia. K = radius of gyration, constant. L = length, linear impulse. M = mass. N = normal component of force. = origin, point. P = period, point, power. Q — point. R = total reaction, resultant force. S = stress. T = tensile force, kinetic energy. U = potential energy. V — velocity, potential. W = weight, work. X = a>component of force. Y = j/-component of force. Z = 2-component of force. X, /3, T, S = constant angles. 6, ,'l' — variable angles. tt> = angle of friction. M = coef. of friction. ;x = modulus of elasticity, -y = angular acceleration. oi = angular velocity. £ = a small quantity, or angle. <7 = surface density. T = volume density. p = hnear density, radius of curvature. "S" = " sum of aU the . . . s." " = " = "is identical with." tt _1_ )} = "approaches." tC ii )) = "equals approximately." "<" = " is smaller than." "<" = "is very small compared with." " V, tl = "is greater than." "»" = "is very large compared with." |n = n! = l-2'3. . . . n. Xll TABLE OF NOTATIONS lb. = pound, the unit of weight, pd. = mass of a body which weighs one pound, in. = inch, ft. = foot. H.P. = horse power. S.H.M. = simple harmonic motion. gm. = gram, kg. = kilogram, mm. = millimeter, cm. = centimeter, m. = meter, km. = kilometer. Letters in gothie type denote vector magnitudes. A dot on a letter indicates that the latter is differentiated with respect to- time. A letter with a bar above it denotes an average magnitude. The letter " i " when used as a subscript denotes " any one of . . . ,"' thus " Fi " stands for " any one of Fi, Fj, F,, etc." ANALYTICAL MECHANICS INTRODUCTION. 1. Scope and Aim of Mechanics. — Mechanics is the science of motion. It has a twofold object : First, to describe the motions of bodies and to interpret them by means of a few laws and principles, which are gen- eraUzations derived from observation and experience. Second, to predict the motion of bodies for all times when the circmnstances of the motion for any one instant are given, in addition to the special laws which govern the motion. The present tendency in science is toward regarding all physical phenomena as manifestations of motion. Comph- cated and apparently dissimilar phenomena are being ex- plained by the interactions and motions of electrons, atoms, molecules, cells, and other particles. The kinetic theory of heat, the wave theories of sound and hght, and the electron theory of electricity are examples which illustrate the tend- ency toward a mechanical interpretation of the physical universe. This tendency not only emphasizes the fundamental im- portance of the science of mechanics to other physical sciences and engineering but it also broadens the aim of the science and makes the dynamical interpretation of all physi- cal phenomena its ultimate object. The aim of elementary mechanics is, however, very modest and its scope is Umited to the discussion of the simplest cases of motion and equiHbrium which occur in nature. 1 2 ANALYTICAL MECHANICS 2. Divisions of Mechanics. — It is customary to divide Mechanics into Kinematics and Dynamics. The fonner treats of the time and space relations of the motions of bodies without regard to the interactions which cause them. In other words, Kinematics is the geometry of motion. In Dynamics, on the other hand, motion and eqmUbrium are treated as the results of interactions between bodies; conse- quently not only time and space enter into dynamical discus- sions, but also mass, the third element of motion. Dynamics in its turn is divided into Statics and Kinetics. Statics is the mechanics of bodies in equilibrium, while Kinetics is the mechanics of bodies in motion. Chapters II, III, and IV of the present work are devoted to problems in statics, while the rest of the book, with the exception of Chapters I, V, and VII, is given to discussions of problems in kinetics. The subject matter of Chapters I and VII is essentially of a mathematical nature. In the former the addition and resolution of vectors are discussed, while in the latter the Calculus is apphed to finding centers of mass and moments of inertia. Chapter V is devoted mainly to kinematical problems. CHAPTER I. ADDITION AND RESOLUTION OF VECTORS. 3. Scalar and Vector Magnitudes. — Physical magnitudes may be divided into two classes according to whether they have the property of orientation or not. Magnitudes which have direction are called vectors, while those which do not have this property are called scalar s. Displacement, veloc- ity, acceleration, force, torque, and momentum are vector magnitudes. Mass, density, work, energy, and time are scalars. 4. Graphical Representation of Vectors. — Vectors are rep- resented by directed Hnes or arrows. The length of the directed line represents the magnitude of the vector, while its direction coin- cides with that of the vector. For brevity the directed lines as well as the physical quantities which they represent are called vectors. The head and the tail of the directed line ^- ^ are called, respectively, the terminus and the origin of the vector. In Fig. 1, for instance, P is the origin and Q the terminus of the vector a. 5. Notation. — Vectors wiU be denoted by letters printed in Gothic type, while their magnitudes will be represented by the same letters printed in itahc type. Thus in Fig. 1 the vector PQ is denoted by a, but if it is desired to represent the length PQ without regard to its orientation a is used. 6. Equal Vectors. — Two vectors are said to be equal if they have the same length and the same direction. It follows, therefore, that the value of a vector is not changed when it 3 4 ANALYTICAL MECHANICS is moved about without changing its direction and magni- tude. 7. Addition of Two Vectors. — Let the vectors a and b, Fig. 2, represent two displacements, then their sum is another vector, c, which is equivalent to the given vectors. In order to find c let us apply to a particle the operations indi- cated by a and b. Each vector displaces the particle along its direction through a distance equal to its length. There- FiG. 2. fore applying a to the particle at P, Fig. 3, the particle is brought to the point Q. Then applying the operation indi- cated by b the particle is brought to the point R. There- fore the result of the two operations is a displacement from P to R. But this is equivalent to a single operation repre- sented by the vector c, which has P for its origin and R for its terminus. Therefore c is called the sum, or the resultant, of a and b. This fact is denoted by the following vector equation, a + b = c. (I) 8. Order of Addition. — The order of addition does not affect the result. If in Fig. 3 the order of the operations indicated by a and b is reversed the particle moves from P to Q' and then to R. Thus the path of the particle is changed but not the resultant displacement. 9. Simultaneous Operation of Two Vectors. — The operations indicated by a and b may be performed simultaneously without affecting the final result. In order to illustrate ^i^i-'ixiuiN AJND RESOLUTION OF VECTORS 5 the simultaneous operation of two vectors suppose the particle to be a bead on the wire AB, Fig. 4. Move the wu-e, keeping it parallel to itself, until each of its particles is given a displacement represented by b. Simultaneously with the motion of the wire move the bead along the wire giving it a displacement equal to a. At the end of R 1:^^ Z^f^ r' / / -f — / — -^ A ^ ' ■ a ' > ^ B Fig. 4. these operations the bead arrives at the point R. If both the wire and the bead are moved at constant rates the resultant vector c represents not only the resulting dis- placement but also the path of the particle. 10. Rules for Adding Two Vectors. — The results of the last three paragraphs furnish us with the following methods for adding two vectors graphically. Triangle Method. — - Move one of the vectors, without changing its direction, until its origin falls upon the terminus of the other vector, then complete the triangle by drawing a vector the origin of which coincides with that of the first vector. The new vector is the resultant of the given vectors. Parallelogram Method. — Move one of the vectors until its origin falls on that of the other vector, complete the parallelo- gram, and then draw a vector which has the common origin of the given vectors for its origin and which forms a diagonal of the parallelogram. The new vector is the resultant of the given vectors. 11. Analytical Expression for the Resultant of Two Vectors. — Let a and b, Fig. 5, be two vectors and c their result- 6 ANALYTICAL MECHANICS ant. Then, solving the triangle formed by these vectors, we obtain c2 = a2 + 62+2a&cos0 (II) b sin and tan d = a + b cos (III) where a, b, and c are the magnitudes of a, b, and c, respec- tively, while and 6 are the angles b and c make with a. Equation (II) determines the magnitude and equation (III) the direction of c. Fig. 5. (a) (b) A - / jtL^ a (e) Fig. 6. Special Cases, (a) If a and b have the same direction, as in Fig. 6a, then 0=0. Therefore and c^ = a'' + b^+2ab, :. c = a + b, tan 6 = 0, :. 6 = 0. Thus c has the same direction as a and b, while its magni- tude equals the arithmetical sum of their magnitudes. (b) When a and b are oppositely directed, as in Fig. 6b, = X. Therefore c2 = 0^ + 62- 2 aft, .-. c = a-b, and tan 5 = 0, :. e = 0. ADDITION AND RESOLUTION OF VECTORS 7 Thus the magnitude of c equals the algebraic sum of the magnitudes of a and b, while its direction is the same as that of the larger of the two. It is evident that if the magnitudes of a and b are equal c vanishes. Therefore two vectors of equal magnitude and opposite directions are the negatives of each other. In other words, when the direc- tion of a vector is reversed its sign is changed. (c) When a and b are at right angles to each other, as in Fig. 6c, <^ = ^- Therefore C2= 0^+62 and tane = -- a 12. Difference of Two Vectors. — Subtraction is equivalent to the addition of a negative quantity. Therefore, to subtract b from a we add — b to a. Thus we have the following rule for subtracting one vector from an- other. In order to subtract one vector from another reverse the one to he subtracted and add it to the other vector. It is evident from Fig. 7 that the sum and the difference of two vectors form the diagonals of the parallelogram determined by them. ILLUSTRATIVE EXAMPLES. A particle is displaced 10 cm. N. 30° E., then 10 cm. E. Find the resulting displacement. Representing the displacements and their resultant by the vectors a, b, and c, Fig. 8, we obtain ANALYTICAL MECHANICS e2 = a'' + b' + 2ab cos (j> = (10cm.)'+(10cm.)2 + 2xl0cm. Xl0cin.cos(60°) = 300 cm. 2 c = 10 Vs cm. = 17.3 cm.* bsin4> tan 9 = a + 6 cos (t> 10 cm. sin (60°) 10 cm. + 10 cm. cos (60°) = iV3. .-. e = 30°. o Therefore the resxiltant displacement is about 17.3 cm. along the direction N. 60° E. Fig. 8. PROBLEMS. 1. A vector which points East has a length of 16 cm., and another vector which points Southeast is 25 cm. long. Find the direction and the magnitude of their sum. 2. Find the direction and the magnitude of the difference of the vectors of the last problem. 3. The sum of two vectors is perpendicular to their difference. Show that the vectors are equal in magnitude. 4. The sum and the difference of two vectors have equal magnitudes. iShow that the vectors are at right angles to each other. 13. Resolution of Vectors into Compo- y nents. — The projection of a vector upon a line is called the component of the vector along that line. The vectors a^ and a^ in rig. 9, for instance, are the components of a along the x-axis and the 2/-axis, respec- tively. The following relations are evident from the figure and do not need further o" explanation. Fig. 9. * The sjrmbol " = " will be used to denote approximate equality. There- fore" = " should be read " equals approximately," or "equals about," or "equals nearly." See table of notations, p. x. ADDITION AND RESOLUTION OF VECTORS 9 a — Sx ~r a-K) ax= a cos 8, tty — a sin 0, a = ^aJ^+Uy^, ax When a has components along all three axes of a rectangular system, Fig. 10, the following equations express the vector in terms of its components. Y tan 5: (IV) (V) (VI) (VII) Fig. 10. a ^ ai -p a^/ -p- a^. ax= a cos ai, ay= a cos a2, a^= a cosaa. (IV) (V) a = VaJ+Oy^+a?, (VI') where ai, a-i, and az are the angles a makes with the coordi- nate axes. 14. Resultant of Any Number of Vectors. Graphical Methods. — The resultant of a number of vectors a, b, c, etc., may be obtained by either of the foUov/ing methods. First : move b, without changing either its direction or its magnitude, until its origin falls on the terminus of a, then 10 ANALYTICAL MECHANICS 'X^v^' FlQ. 11. move c until its origin falls on the terminus of b, and so on until all the vectors are joined. This gives, in general, an open polygon. Then the resultant is obtained by drawing a vector which closes the polygon and which has its origin at the origin of a. The validity of this method will be seen from Fig. 11, where r repre- sents the resultant vector. Evi- dently the resultant vanishes when the given vectors form a closed polygon. Second: draw a system of rectangular coordinate axes; resolve each vector into components along the axes; add the components along each axis geometrically, beginning at the origin. This gives the components of the required vector. Then draw the rectangular parallelopiped determined by these components. The resultant is a vector which has the origin of the axes for its origin and forms a diagonal of the parallelopiped.* This method is based upon the following analytical method. 15. Analytical Method. — Expressing the given vectors and their resultant in terms of their rectangular components, we have a = Si ~r &» ~r 3-2,' b = bi+ by + b^, (1) r = Tx + r„ + r,. Substituting from (1) in the vector equation r=a+b+c+ • • • (2) and collecting the terms we obtain ri+ra-f r, = (ax+bx+ • • •)+(&„+ b„+ • • •) + (a,+ b,+ - • •). (3) But since the directions of the coordinate axes are indepen- * When the given vectors are in the same plane the parallelopiped reduces to a rectangle. ADDITION AND RESOLUTION OF VECTORS 11 dent, the components of r along any one of the axes must equal the sum of the corresponding components of the given vectors. Therefore (3) can be spht into the following three separate equations. rx= ax+b:.+ Cx+ • • • , r„=a^+b„+Cj,+ • • • , (4) _ r, = a, + b, + c^ + • • ■ . It was shown in § 11 that when two vectors are parallel the algebraic sum of their magnitudes equals the magni- FiG. 12. tude of their resultant. This result may be extended to any number of parallel vectors. Therefore we can put the vector equations of (4) into the following algebraic forms: rx = ax+bx + Cx+ • • • r„ = ay+by+Cy+ ■ • ■ r, = a^ + b^ + c,+ --- Equations (5) determine r through the following relations: (5) V; rx' + ry'^ + r.. 2 2 ) T cos ai = — > r cos q;2 = — ' r cos as = (6) (7) where ai, a2, and as are the angles r makes with the axes. 16. Multiplication and Division of a Vector by a Scalar. — When a vector is multiplied or divided by a scalar the result is a vector which has the same direction as the original vec- tor. If, in the equation b = ma, m be a scalar then b has the same direction as a but its magnitude is m times that of a. 12 ANALYTICAL MECHANICS ILLUSTRATIVE EXAMPLE. A man walks 3 miles N. 30° E., then one mile E., then 3 miles S. 45° E., then 4 miles S., then one mile N. 30° W. Find his final position. Representing the displacements by vectors we obtain the graphical solution given in Fig. 13, where r represents the resultant displacement. Fig. 13. In order to find r analytically we first determine its components. Thus r^ = [3 cos (60°) + cos (0°) + 2 cos (-45°) + 4 cos (-90°) + cos (120°)] miles = (2 + V^) miles = 3.41 miles. Ty = [3 sin (60°) + sin (0°) + 2 sin (-45°) + 4 sin (-90°) + sin (120°)] miles = (2V3-\/2-4) mUes = —1.95 mUes. = 3.93 mUes. The direction of r is given by the following relation. , . »•«.. -1.95 *^^^ = r. = ^ir' .-. 9 = -37°.l. Therefore the final position of the man is about 3.93 miles S. 52°.9 E. from his starting point. ADDITION AND RESOLUTION OF VECTORS 13 PROBLEMS. 1. The resultant of two vectors which are at right angles to each other is twice the smaller of the two. The magnitude of the smaller vector is a; find the magnitude of the, greater vector. 2. In the preceding problem find the resultant vector. 3. Find analytically the sum of three equal vectors which point in the following directions — East, N. 30° W., and S. 30° W. 4. In the preceding problem make use of the first graphical method. 5. In problem 3 make use of the second graphical method. 6. A vector which is 15 cm. long points N. 30° E. Find its compo- nents in the following directions. (a) N. 30° W. (c) W. (e) S. 60° E. (b) N. 60° E. (d) S. 30° W. (f) E. 7. A vector a is in the xy-p\a.ne. If 3 is added to a^ and 4 to a„ the direction of the vector is not changed but its magnitude becomes a^ + Oy. Find the magnitude and direction of a. 8. Three vectors a, b, and c he in the xy-pla.ne. Find their resultants analytically, taking the magnitudes of their components from the follow- ing tables : a^ ay 6x hy Ci Cy (1) 6, 9, t5. 2, 0, 10. (2) -3, 7, 5, 0, 6, -8. (3) 0, -10, 8, 5, 3, -2. (4) 2, 0, -6, 4, 0, 8. 9. In the preceding problem make use of the second graphical method. 10. Straight horizontal tunnels in a mine connect the points Pi, P2, Pa, and P4, in the given order. The length of each tunnel and the angle it makes with the meridian are given in the following tables. Find the lengths and directions of the tunnels which have to be dug in order to connect Pi with P3 and P4 directly. P1P2 = 200 feet, and makes 30° with the meridian. P2P3 = 100 feet, and makes 120° with the meridian. P3P4 = 400 feet, and makes 300° with the meridian. 11. Work out the preceding problem by the first graphical method. 12. Work out problem 10 by the second graphical method. 13. Find the direction and magnitude of the force experienced by an electrical charge of five units placed at one vertex of an equilateral tri- angle due to two unlike charges of 10 imits each placed at the other vertices. The sides of the triangle are 2 cm. CHAPTER II. EQUILIBRIUM OF A PARTICLE. ACTION AND REACTION. FORCE. 17. Particle. — A body whose dimensions are negligible is called a -particle. In a problem any body may be considered as a particle so long as it does not tend to rotate. Even when the body rotates it may be considered as a particle if its rotation does not enter into the problem. For instance, in discussing the motion of the earth in its orbit the earth is considered as a particle, because its rotation about its axis does not enter into the discussion. 18. Degrees of Freedom. — The number of independent ways in which a body can move is called the number of degrees of freedom of its motion. It equals the number of coordinates which are necessary in order to specify completely the posi- tion of the body. A free particle can move in three inde- pendent directions, that is, along the three axes of a system of rectangular coordinates, therefore it has three degrees of freedom. When the particle is constrained to move in a plane its freedom is reduced to two degrees, because it can move only in two independent directions. When it is con- strained to move in a straight line it has only one degree of freedom. 19. Force. — While considering the motion or the equilibrium of a body our attention is claimed not only by that body but also by others which act upon it. In order to insure concentration of attention problems in Dynamics are sim- plified in the following manner. All bodies are eliminated, except the one the motion of which is being discussed, and their actions upon the latter are represented by certain vec- tor magnitudes known as forces. As an illustration consider 14 EQUILIBRIUM OF A PARTICLE 15 the equilibrium of the shaded part of the rope in Fig. 14a. The shaded part is acted upon by the adjoining sections of the rope. Therefore we consider the shaded part alone and represent the actions of the adjoining parts "F ^, by the forces - F and F, as shown Fig 14 in Fig. 14b. 20. Definition of Force. — Force is a vector magnitude which represents the action of one body upon another. The interac- tion between two bodies takes place across an area, while the forces which represent them are supposed to be applied at one point. Therefore the introduction of the idea of force presupposes the simplification of dynamical problems which is obtained by considering bodies as single particles, or as a system of particles. 21. Internal Force. — A force which represents the action of one part of a body upon another part of the same body is called an internal force. 22. External Force. — ^A force which represents the action of one body upon another body is called an external force. 23. Unit Force. — The engineering unit of force among English speaking people is the pound. The pound is the weight, in London, of a certain piece of platinum kept by the British government. 24. The Law of Action and Reaction. — The fundamental law of Mechanics is known as the law of action and reaction. Newton (1692-1727), who was the first to -formulate it, put the law in the following form. "To every action there is an equal and opposite reaction, or the mutual actions of two bodies are equal and oppositely directed." Let us apply this law to the interaction between a book and the hand in which you hold it. Your hand presses upward upon the book in order to keep it from falling, 16 ANALYTICAL MECHANICS while the book presses downward upon your hand. The law states that the action of your hand equals the reaction of the book and is in the opposite direction. The book reacts upon your hand because the earth attracts it. When your hand and the earth are the only bodies which act upon the book, the action of your hand equals and is opposite to the action of the earth. In other words the sum of the two actions is nil. Generalizing from this simple illustra- tion we can put the law into the following form: To every action there is an equal and opposite reaction, or the sum of all the actions to which a body or a part of a body is subject at any instant vanishes : SA = 0.* (A) 25. Condition for the Equilibrium of a Particle. — The condi- tion of equilibrium of a particle is obtained by replacing the term "action" by the term "force" in the last form of the fundamental law and then stating it in the form of a condi- tion. Thus — in order that a particle he in equilibrium the sum of all the forces which act upon it must vanish. In other words if Fi, F2, F3, . . . , F„ are the forces which act upon a particle, then the vector equation F1+F2+F3+ • • • +F„=0 (I) must be satisfied in order that the particle be in equilib- rium. Equation (I) is equivalent to stating that when the forces are added graphically they form a closed polygon. But when the sum of a number of vectors vanishes the sum of their components also vanishes. Therefore we must have Xi+X2+ • • • +x„=o, Y1+Y2+ • • • +Y„ = 0, (IF) Zi + Z2 + • ■ ■ + Z„ = 0, where Xj, Y^, and Z; are the components of F;.* Since the vectors in each of the equations of (II') are parallel we can * See table of notations. EQUILIBRIUM OF A PARTICLE 17 write them as algebraic equations. Therefore we have the foUowrag equations for the analytical form of the condition of equilibrium of a particle. SX = Zi + Z2+- • ■ +X„=0,* 2F=Fi+F2+- • • +F„ = 0, (II) 2Z = Zi + Z2 + • • • + Z„ = 0. The condition of equilibrium may, therefore, be stated in the following form. In order that a -particle he in equilibrium the algebraic sum of the components of the forces along each of the axes of a rec- tangular system of coordinates must vanish. The following" rules will be helpful in working out prob- lems on the equilibrium of a particle. First. Represent the particle by a point and the action of each body which acts upon it by a properly chosen force-vector. Be sure that all the bodies which act upon the particle are thus represented. Second. Set the sums of the components of the forces along properly chosen axes equal to zero. Third. If there are not enough equations to determine the unknown quantities, obtain others from the geo- metrical connections of the problem. Fourth. Solve these equations for the required quantities. Fifth. Discuss the results. ILLUSTRATIVE EXAMPLES. 1. A particle suspended by a string is pulled aside by a horizontal force until the string makes an angle a with the vertical. Find the tensile force in the string and the magnitude of the horizontal force in terms of the weight of the particle. The particle is acted upon by three bodies, namely, the earth, the string, and the body which exerts the horizontal force. Therefore, we * The relation SX = Xi + X2 + • • • + Xn is not an equation. It merely states that SZ is identical with and is an abbreviation for Xi + X2 + • ■ • +X„. 18 ANALYTICAL MECHANICS represent the actions of these bodies by three force-vectors, W, T, and F, Fig. 15, and then apply the conditions of equilibrium. Setting equal to zero the sums of the components of the y forces along the x- and 2/-axes, we get •LX^F-Tam.a = Q. (a) SF s -Tr+rcosa = 0. (b) Solving equations (a) and (b) we have W and T =■ cos a F =Tsm.a = W tan a. Discussion. — When o; = 0, T=W and F = Q. When a ■■ 2' r = 00 and Fig. 15. F=. Therefore no finite horizontal force can make the string perfectly hori- zontal. 2. A uniform bar, of weight W and length a, is suspended in a horizontal position by two strings of equal length I. The lower ends of the strings are fastened to the ends of the bar and the upper ends to a peg. Find the tensile force in the strings. The bar is acted upon by three bodies, namely the earth and the two strings. We represent their actions by the forces W, Ti, and T2, Fig. 16a. The tensile forces of the strings act at the ends of the bar. On the other hand the weight is distributed all along the rod. But we may consider it as acting at the middle point, as in Fig. 16a, or we may replace the rod W by two particles of weight — each, as shown in Fig. 16b. In the last case Ji the rigidity of the bar which prevents its ends from coming together is represented by the forces F and — F. Considering each particle separately and setting equal to zero the sums of the components of the forces along the axes, we obtain SZsTiCosa-F =0, SF ^ Ti sin a - ^ = 0, for the first particle, and 2Zs -r2cosQ;-l-F = 0, W 2F = Tj sin a - ^ = 0, EQUILIBRIUM OF A PARTICLE for the second particle. It follows from these equations that W 19 2 sia a I W. Fig. 16. Discussion. — The tensile force of the strings increases indefinitely as their total length approaches that of the bar. On the other hand as the length of the strings becomes very large compared with that of the W bar the tensile force approaches — as a limit. The problem can be solved also by considering the forces acting on the peg, as shown in Fig. 16b. PROBLEMS. 1. Show that when a particle is in equilibrium under the action of two forces the forces must he in the same straight line. 2. Show that when a particle is in equUibrium under the action of three forces the forces he in the same plane. 20 ANALYTICAL MECHANICS 3. Find the horizontal force which will keep in equiUbrium a weight of 150 pounds on a smooth inclined plane which makes 60° with the horizon. 4. A ring of weight W is suspended by means of a string of length I, the ends of which are attached to two points on the same horizontal line. Find the tensile force of the string if the distance between its ends is d. Also discuss the Umiting cases in which I approaches d or becomes very large compared with it. 5. A body of weight W is suspended by two strings of lengths k and h. The upper end of each string is attached to a fixed point in the same horizontal line. Find the tensile forces in the strings if the distance between the two points is d. 6. A weight is suspended by four equal strings, the upper ends of which are attached to the vertices of a horizontal square. Find the tensUe forces in the strings. 7. A particle is in equilibrium on a smooth inclined plane under the action of two equal forces, the one acting along the plane upwards and the other horizontally. Find the inchnation of .the plane. 8. Apply the conditions of equilibrium to find the magnitude and direction of the resultant of a number of forces acting upon a particle. 9. Two spheres of equal radius and equal weight are in equilibrium in a smooth hemispherical bowl; find the reactions between the two spheres and between the spheres and the bowl. 10. The ends of a string, 60 cm. long, are fastened to two points in the same horizontal line and at a distance of 40 cm. apart; two weights are hung from points in the string 25 cm. and 20 cm. from the ends. Find the ratio of the weights if the part of the string between them is hori- zontal. 11. A single triangular truss of 24 feet span and 5 feet depth supports a load of 3 tons at the apex. Find the forces acting on the rafters and the tie rod. 12. A particle of weight W can be kept in equilibrium upon a smooth inchned plane by a force Fi acting horizontally; it can also be kept in equilibrium by a force F2 acting parallel to the plane. Express W in terms of Fi and Fn. 13. In the following arrangements of puUeys find the relation between F and W. EQUILIBRIUM OF A PARTICLE mm vMmmm. 21 SLIDING FRICTION. 26. Frictional Force. — Consider the forces acting upon a body which is in equihbrium on a rough inclined plane, Fig. 17. The body is acted upon by twoi forces, namely, its weight, W, and the reaction of the plane, R. The reaction of the plane is th6 result of two distinct and independent forces. One of these, N, is perpendicular to the plane and is called the normal reaction. The other, F, is along the plane and is called the frictional force. The normal reaction is due to the rigidity of the plane. It re- sists the tendency of the body to go through the plane. The frictional force is due to the roughness of the contact between the body and the plane. It prevents the body from sliding down the plane. 27. Angle of Friction. — As we increase the angle of elevation of the inclined plane a certain definite angle will be reached Fig. 17. 22 ANALYTICAL MECHANICS when the equilibrium is disturbed and the body begins to slide down the plane. This angle is called the angle of friction. This definition for the angle of friction does not hold when the body is acted upon by other forces besides its weight and the reaction of the plane. The following definition, however, is vahd under all circumstances: The angle of friction equals the angle which the total reaction makes with the normal to the surface of contact when the body is on the point of motion. 28. Coefficient of Friction. — Denoting the angle of friction by , we obtain F = R sin (j>, N = R cos (j). Therefore F = Ntan^ = MiV, (III) where fi = tan ^ and is called the coefficient of friction. The angle of friction and consequently the coefficient of friction are constants which depend upon the surfaces in contact. The last four equations hold true only when the body is on the point of motion. 29. Static and Kinetic Friction. — The friction which comes into play is called static friction if the body is at rest and kinetic friction if it is in motion. 30. Laws of Friction. — The foUoAving statements, which are generalizations derived from experimental results, bring out the important properties of friction. They hold true within certain limits and are only approximately true even within these limits. 1. Frictional forces come into play only when a body is urged to move. 2. Frictional forces always act in a direction opposite to that in which the body is urged to move. 3. Frictional force is proportional to the normal reaction, EQUILIBRIUM OF A PARTICLE 23 4. Frictional force is independent of the area of contact. 5. The static frictional force which comes into play is not greater than that which is necessary to keep the body in equUibrium. 6. Kinetic friction is smaller than static friction. Laws 1 to 4 hold true for both static and kinetic friction. The coefficient of friction between two bodies depends upon the condition of surfaces in contact. Therefore the value of M is not a. perfectly definite constant for a given pair of sub- stances in contact. The values given in the following table are averages of values obtained by several experimenters. Condition of surfaces in contact. Coefficient of friction. Static. Kinetic. Wood on wood Wood on wood Wood on wood Heavy rope on wood Heavy rope on wood Cast iron on cast iron Cast iron on cast iron Cast iron on oak Leather on oast iron Dry Wet Polished and greased Dry i Wet Dry Greased Wet .50 .68 .35 .60 .80 .24 .15 .65 .30 .36 .25 .12 .40 .35 .18 .13 ILLUSTRATIVE EXAMPLES. 1. A body which is on a rough horizontal floor can be brought to the point of motion by a force which makes an angle a with the floor. Find the reaction of the floor and the coefiicient of friction. The body is acted upon by three forces, Fig. 18, P, the given force, W, the weight of the body, R, the reaction of the floor. Replacing R by its components F and N, and applying the conditions of equilibrium, we obtain SZ = P cos a - F = 0, 27 = P sin a + iV - 17 = 0. Therefore F = P cos a, iV = TF - P sin a, 24 and ANALYTICAL MECHANICS = VP'' + W^-2PWsma. But since the body is on the point of motion the relation F = fiN holds. Therefore _ Z = P coso: '^^ N~ PT-Psina' Discussion. — (a) When a = 0, R = VP" + W and m = ^.■ W (b) When a = ^, R = P-W = 0, therefore P = TF, and ai is mdeter- minate. (c) When P = 0, fj. = 0, a,nd R = W. Yl Fig. 19. 2. A body which rests upon a rough inclined plane is brought to the point of motion up the inclined plane by a horizontal force. Find /i and R. The body is acted upon by three forces, Fig. 19, P, the horizontal force, W,the weight, R, the reaction of the plane. Replacing R by its components F and N , and taking the axes along and at right angles to the plane, we obtain SX s P cos a - P - W sin a = 0, S Y = -P sm a + TV - TT cos a = 0. Therefore P = P cos a — TF sin a, A'^ = P sin a + TY cos a, R = VF^ + m = VP^ + lY", EQUILIBRIUM OF A PARTICLE 25 J F P cos a — W sia a N P sm a+W cos a Discussion. — (a) When a = 0, fi = —, and R = W V/t^ + !• (b) When P = 0, /* = —tan a; therefore a = —^, that is, the indined plane must be tipped in the opposite direction and must be given an angle of elevation equal to the angle of friction in order that motion may take place towards the positive direction of the a;-axis. PROBLEMS. 1. A body which weighs 100 pounds is barely started to move on a rough horizontal plane by a force of 150 pounds acting in a direction making 30° with the horizon. Find R and /*. 2. A body placed on a rough inclined plane barely starts to mov& when acted upon by a force equal to the weight of the body. Find the coeflScient of friction, (a) when the force is normal to the plane; (b) when it is parallel to the plane. 3. A horizontal force equal to the weight of the body has to be applied in order to just start a body into motion on a horizontal floor. Find the coefiScient of friction. 4. A weight W rests on a rough inchned plane, which makes an angle a with the horizon. Find the smallest force which wUl move the weight if the coefficient of friction is jj.. 5. How would you determine experimentally the coefiicient of friction between two bodies? 6. A weight of 75 pounds rests on a rough horizontal floor. Find the magnitude of the least horizontal force which wiU move the body if the coefficient of friction is 0.4; also find the reaction of the plane. 7. A particle of weight W is in equflibrium on an inclined plane under the action of a force F, which makes the magnitude of the normal pres- sure equal W. The coefficient of friction is /* and the angle of elevation of the inchned plane is a. Find the magnitude and direction of the force. 8. An insect starts from the highest point of a sphere and crawls down. Where wiU it begin to slide if the coefficient of friction between the insect and the sphere is f ? 9. The greatest force, which can keep a particle at rest, acting along an inchned plane, equals twice the least force. Find the coefficient of friction. The angle of elevation of the plane is a. 31. Resultant of a System of Forces. — The resultant of a nxunber of forces which act upon a particle is a force which 26 ANALYTICAL MECHANICS is equivalent to the given forces. There are two criteria by which this equivalence may be tested. First : The resultant force will give the particle the same motion, when applied to it, as that imparted by the given system of forces. We cannot use this test just now because we have not yet .studied motion. Second: When the resultant force is re- versed and applied to the particle simultaneously with the given forces the particle remains in equilibrium. According to the second criterion, therefore, the resultant, R, of the forces Fi, F2, . . . , F„, must satisfy the equation -R + (Fi+F2 + • • • +FJ = 0,1 or R = Fi+F2+ ■ • • +F„. (IV) Splitting the last equation into three algebraic equations, we obtain X = Xi + X2+ ■ ■ ■ +X„, F = 7i+F2+- • • +F„, (IV) Z =Zi+Z2 + - ■ ■ +Z„, Tvhere X.-, Y^, and Z; are the components of Fi. The magnitude of R is given by the relation R = Vx^+Y^+Z^, (V) while the direction is obtained from the following expressions for its direction cosines. X Y Z cos ai = — ) C0Sa2 = — ' COS 0:3 = — -• (VI') K R R Special Case. — ^When the forces lie in the xy-phne the 2:-component of each force equals zero. Therefore we have R = Vx^+ F^ (V) y and tan0 = -:, (VI) -A, where 6 is the angle R makes with the a;-axis. EQUILIBRIUM OF A PARTICLE 27 PROBLEMS. 1. Three men puU on a ring. The first man pulls with a force of 50 pounds toward the N. 30° W. The second man pulls toward the S. 45° E. with a force of 75 pounds, and the third man pulls with a force of 100 pounds toward the west. Determine the magnitude and direction of the resultant force. 2. Show that the resultant of two forces acting upon a particle lies in the plane of the given forces. ' 3. Show that the line of action of the resultant of two forces lies within the angle made by the forces. 4. Find the direction and magnitude of the resultant of three equal forces which act along the axes of a rectangular system of coordinates. GENERAL PROBLEMS. 1. A particles is in equilibrium under the action of the forces P, Q, and R. Prove that P ^ Q ^ R sin (Q, R) sin (P, R) sin (P, Q) ' where (Q, R), etc., denote the angles between Q and R, etc. 2. Two particles of weights Wi and W2 rest upon a smooth sphere of radius a. The particles are attached to the ends of a string of length I, which passes over a smooth peg vertically above the center of the sphere. If A is the distance between the peg and the center of the sphere, find (1) the position of equilibrium of the particles, (2) the tensile force in the string, and (3) the reaction of the sphere. 3. The lengths of the mast and the boom of a derrick are a and b respectively. Supposing the hinges at the lower end of the boom and the pulley at the upper end to be smooth, find the angle the boom makes with the vertical when a weight W is suspended in equilibrium. 4. Find the tensile force in the chain and the compression in the boom of the preceding problem. 5. Two rings of weights Wi and Wj are held on a smooth circular wire in a vertical plane by means of a string subtending an angle 2 a at the center. Show that the incUnation of the string to the horizon is given by t-^^=Tf:^Tr/ana. 6. A bridge, Fig. (a), of 60-foot span and 40-foot width has two queen- post trusses 9 feet deep. Each truss is divided into three equal parts by two 28 ANALYTICAL MECHANICS posts. What are the stresses in the different parts of the trusses when there is a load of 150 pounds per square foot of floor space? k 20— *|< ^20 ^ 20—^ ^STJ^^- Fig. (a). Fig. (b). 7. Find the force in one of the members of the truss of figure (b) . C . A weight rests upon a smooth inclined plane, supported by two equal strings the upper ends of which are fastened to two points of the plane in the same horizontal line. Find the tensile force in the strings and the reaction of the plane. 9. In the preceding problem suppose the planf^ to be rough. 10. A particle is suspended by a string which passes through a smooth ring fastened to the highest point of a circular wire in a vertical plane. The other end of the string is attached to a smooth bead which is movable on the wire. Find the position of equUibrium supposing the bead and tlie suspended body to have equal weights. 11. A particle is in equilibrium on a rough inclined plane under the action of a force which acts along the plane. If the least magnitude of the f ^rce when the inchnation of the plane is a equals the greatest magnitude when it is a^, show that 4> — -^-r — -, where is the angle of friction. 12. Two weights Wx and PFz rest upon a rough inclined plane, con- nected by a string which passes through a smooth pulley in the plane. riaJ the greatest inclination the plane can be given without disturbing t,he equilibrium. 13. Two equal weights, which are connected by a string, rest upon a r jugh inclined plane. If the direction of the string is along the steepest ijlope of the plane and if the coefficients of friction are \x\ and /i2, find the greatest inclination the plane can be given without disturbing the equi- librium. 14. In the preceding problem find the tensile force in the string. 1j. One end of a uniform rod rests upon a rough peg, while the other end is connected, by means of a string, to a point in the horizontal plane which contains the peg. When the rod is just on the point of motion it EQUILIBRIUM OF A PARTICLE 29 is perpendicular to the string. Show that 21 = na, where I is the length of the string, a that of the rod, and n the coefficient of friction. 16. A particle resting upon an inclined plane is at the point of motion under the action of the force F, which acts downward along the plane. If the angle of elevation of the plane is changed from a^ to az and the direction of the force reversed the particle will barely start to move up the plane. Express n in terms of ai and a^- 17. A string, which passes over the vertex of a rough double inclined plane, supports two weights. Show that the plane must be tilted through an angle equal to twice the angle of friction, in order to bring it from the position at which the particles will begin to move in one direction to the position at which they will begin to move in the opposite direction. 18. Three equal spheres are placed on a smooth horizontal plane and are kept together by a string, which surrounds them in the plane of their centers. If a fourth equal sphere is placed on top of these, prove that the W tensile force in the string is — — , where W is the weight of each sphere. 3v6 19. Three equal hemispheres rest with their bases upon a rough hori- zontal plane and are in contact with one another. What is the least value of fi which will enable them to support a smooth sphere of the same radius and material? 20. If the center of gravity of a rod is at a distance a from one end and b from the other, find the least value of fi which will allow it to rest in aU positions upon a rough horizontal ground and against a rough vertical wall. 21. A string, which is slung over two smooth pegs at the same level, supports two bodies of equal weight W at the ends, and a weight W at the middle by means of a smooth ring through which it passes. Find the position of equilibrium of the middle weight. CHAPTER III. EQUILIBRIUM OF RIGID BODIES. TRANSLATION AND ROTATION. 32. Rigid Body. — There are problems in which bodies cannot be treated as single particles. In such cases they are considered to be made up of a great number of discrete par- ticles. A body is said to be rigid if the distances between its particles remain unchanged whatever the forces to which it may be subjected. There are no bodies which are strictly rigid. All bodies are deformed more or less under the action of forces. But in most problems discussed in this book ordi- nary solids may be treated as rigid bodies. 33. Motion of a Rigid Body. — A rigid body may have two distinct types of motion. When the body moves so that its particles describe straight paths it is said to have a motion of trans- lation. Evidently the paths of the particles are parallel, Fig. 20. If the particles of the body describe circular paths it is said to have a motion of rotation. The planes of the circles are parallel, while their centers lie on a straight line per- pendicular to these planes, which is called the axis of rotation. The motion of a flywheel is a well-known example of motion of rotation. Suppose A, Fig. 21, to be a rigid body which is brought from the position A to the position A' by a motion of rotation about an axis through the point perpendicular to the plane of the paper, then the paths of its particles 30 Fig. 20. EQUILIBRIUM OF RIGID BODIES 31 Fig. 21. are arcs of circles whose planes are parallel to the plane of the paper and whose centers lie on the axis of rotation. 34. Uniplanar Motion. — When a rigid body moves so that each of its particles re- mains at a constant distance from a fixed plane the motion is said to be uniplanar. The fixed plane is called the guide plane. 35. Theorem I. — Uniplanar motion of a rigid body consists of a succession of infinitesimal rotational displacements. Suppose the rigid body A, Fig. 22, to describe a uniplanar motion parallel to the plane of the paper and let A and A' be any two positions occupied by the body. Then it may be brought from A to 4' by a rotational displacement about an axis the position of which may be found in the fol^ lowing manner. Let P and Q be the positions of any two particles of the body in a plane parallel to the plane of the paper when the body is at the position A, and P' and Q' be the positions of the same particles when the body occupies the po- sition A'. Then the desired axis is perpendicular to the plane of the paper and passes through the point of intersection of the perpendicular bisectors of the lines PP' and QQ', drawn in the plane determined by these lines. ' Therefore the body can be brought from any position A to any other position A' by a single rotational displacement. The actual motion between A and A' will be, in general, Fig. 22. 32 ANALYTICAL MECHANICS quite different from the simple rotation by which we accom- phshed the passage of the body from one of these positions to the other. But the result, which we have just obtained, is true not only for positions which are separated by finite distances but also for positions which are infinitely near each other. Therefore by giving the body infinitesimal' rotational displacements about properly chosen axes it may be made to assume all the positions which it occupies during its actual motion. 36. Instantaneous Axis. — As the body is made to occupy the various positions of its actual motion the axis of rota- tion moves at right angles to itseK and generates a cylin- der whose elements are perpendicular to the guide plane. The elements of the cyHnder are called instantaneous axes, because each acts as the axis of rotation at the instant when the body occupies a certain position. The curve of inter- section of the cylinder and the guide plane is called the centrode. The motion of a cylinder which rolls in a larger cylinder is a simple example of uniplanar motion. In this case the common element of contact is the instantaneous axis. As the cylinder rolls different elements of the fixed cylinder become the axis of rotation. Motion of translation and motion of rotation are special cases of imiplanar motion. In motion of translation the axis of rotation is infinitely far from the moving body. In rotation the cyUnder formed by the instantaneous axes reduces to a single line, i.e., the axis of rotation. 37. Theorem II. — Rotation about any axis is equivalent to a rotation through the same angle about a parallel axis and a translation in a direction perpendicular to it. The truth of this theorem will be seen from Fig. 23, where the rigid body A is brought from the position A to the posi- tion A' by a single rotation about an axis through the point perpendicular to the plane of the paper. This displace- EQUILIBRIUM OF RIGID BODIES 33 ment may be produced also by rotating the body to the position A" and then translating it to the position A'. Fig. 23. PROBLEMS. 1. Show that in theorem II the order of the rotation and of the trans- lation may be changed. 2. Show that the converse of theorem II is true. 38. Theorem III. — The most general displacement of a rigid body can he obtained by a single translation and a single rotation. Let A and A' be any two positions occupied by the rigid body and P and P' be the corresponding positions of any one Fig. 24. of its particles. Then the body may be brought from A to A' by giving it a motion of translation which will bring the particle from P to P' and then rotating the body about a properly chosen axis through P'. A special case of this 34 ANALYTICAL MECHANICS theorem is illustrated in Fig. 24, where the direction of the translation is perpendicular to the axis of rotation. 39. Theorem TV. ^- The most general displacement of a rigid body can be obtained by a displacement similar to that of a screw in its nut, that is, by a rotation about an axis and a translation along it. This theorem states that the axis of rotation of the last theorem can be so chosen that the translation is along the axis of rotation. In theorem III let PP', Fig. 25, be the path of any point of the body described during the transla- tion and BB be the line about which the body is rotated. Draw CC through P parallel to BB and drop the perpendicular P'P" upon CC. The displace- ment may be accomplished now in the following three stages. First: translate the body along the line CC until the point which was at P arrives at P'-'. Second : translate the body along P"P' until the point arrives at P'. Third: rota'te the body about BB until it comes to the desired position. But by theorem II the last two operations can be accomplished by a single rota- tion about CC. Therefore the desired displacement can be ob- tained by a translation along and a rotation about the line CC. Evidently the last theorem holds for infinitesimal dis- placements as well as for finite displacements; therefore however compUcated the motion of a rigid body it can be reproduced by a succession of infinitesimal screw-displace- ments, each displacement taking the body from one position which it has occupied during the motion to another position infinitely near it. Thus at every instant of its motion the rigid body is displaced like a screw in its nut. In general Fig. 25. EQUILIBRIUM OF RIGID BODIES 35 the pitch and the direction of the axis of the screw-motion change from instant to instant. In the case of the motion of a screw in its nut these do not change. Translation and rotation are special cases of screw- motion. When the pitch of a screw is made smaller and smaller it advances less and less during each revolution. Therefore if the pitch is made to vanish the screw does not advance at all when it is rotated. Thus rotation is a special case of screw-motion in which the pitch is zero. On the other hand as the pitch of the screw is made greater and greater the screw advances more and more during each revolution. Therefore at the limit when the pitch is in- finitely great the motion of the screw becomes a motion of translation. Thus translation is a special case of screw- motion in which the pitch is infinitely great. LINEAR AND ANGULAR ACTION. TORQUE. 40. Two Types of Action.— We have seen that a rigid body may have two different and independent types of motion, namely, motion of translation and motion of rotation. These motions are the results of two independent and entirely different kinds of actions to which a rigid body is capable of being subjected. We will differentiate between these two types of action by adding the adjectives "linear" and "angular" to the term "action." Thus the action which tends to produce translation will be called linear action and that which tends to produce rotation, angular action. 41. Torque. — The vector magnitude which represents the angular action of one body upon another is called torque. 42. Couple. — Although a single force is not capable of giving a rigid body a motion of pure rotation, two or more external forces will do it when properly applied. The simplest system of forces which is capable of producing rotation is known as a couple. It consists of two equal and opposite forces which are not in the same Une, Fig. 26. 36 ANALYTICAL MECHANICS It is evident from Fig. 26 that a couple is capable of giving a rigid body a motion of rotation. But this is not enough to show that the effect produced by a couple is the same as that produced by a torque. We must show also that the couple is not capable of producing a motion of translation. Consider the rigid body A, Fig. 27, which is acted upon by a couple. Suppose the couple did tend to Fig. 26. Fig. 27. produce a translation in a direction BB'. Then pass through the body a smooth bar of rectangular cross-section in the direction of the supposed motion, so that the body is free to move along the bar but not free to rotate. When this constraint is imposed upon the rigid body it behaves like a particle and therefore cannot be given a motion by two equal and opposite forces. But since any motion in the direction BB' is not affected by the presence of the bar, the assumption that the couple produces a motion of trans- lation along BB' must be wrong. Hence we see that when the bar is taken out the motion due to the couple will be one of pure rotation. 43. Measure of Torque. — When a rigid body is in equi- librium under the action of two couples" it is always found that the product of one of the forces of one couple by the distance apart of the forces of the same couple equals the EQUILIBRIUM OF RIGID BODIES 37 corresponding product for the other couple. In order, for instance, that the rigid body A, Fig. 28, be in equilibrium, we must have FD = F'D'. Therefore the product FD is the measure of the torque of the couple formed by the forces F and — F, the lines of action of which are separated by the distance D. Thus denoting the torque of a couple by G, we have G = FD. The distance D is called the arm of the couple and the plane of the forces the plane of the couple. 44. Unit Torque. — The torque of a couple whose forces are one pound each and whose arm is one foot is the unit of torque. The symbol for the unit torque is the lb. ft. 45. Vector Representation of Torque. — Torque is a vector magnitude and is represented by a vector which is perpen- FiQ. 28. (I) Fig. 29. dicular to the plane of the couple. The vector points away from the observer when the couple tends to rotate the body in the clockwise direction and points towards the observer when it tends to rotate the body in the counterclockwise direction, Fig. 29. In the first case the torque is considered to be negative and in the second case positive. 38 ANALYTICAL MECHANICS 46. Equal Couples. — Two cou- ples are equal when the vectors which represent their torques are equal in magnitude and have the same direction. The three couples in Fig. 30 are equal if d = G2 = G3. Resultant of two couples is a third couple, whose torque is the vector sum of the torques of the given couples. . FiQ. 30. PROBLEMS. 1. Find the direction and magnitude of the resultant torque of three couples of equal magnitude the forces of which act along the edges of the bases of a right prism. The bases of the prism are equilateral triangles. 2. In the preceding problem let the forces have a magnitude of 15 pounds eaph, the length of the prism be 2 feet and the sides of the bases 10 inches. 3. In problem 1 suppose the prism to have hexagonal bases. 4. In problem 2 suppose the prism to be hexagonal. 6. A right circular cone, of weight W and angle 2 a, is placed in a circular hole of radius r, cut in a horizontal table. Assuming the coeffi- cient of friction between the cone and the table to be ii, find the least torque necessary to rotate the former about its axis. 47. Moment of a Force. — The most common method of giving a rigid body a motion of rotation is to put an axle through it and to apply to it a force which acts in a plane per- pendicular to the axle. The rotation is produced by the couple formed by the applied force and the reaction of the axle. The torque due to the couple equals the product of the applied force by the shortest distance from the axle to the line of action of the force. It is often more con- FiQ. 3L EQUILIBRIUM OF RIGID BODIES 39 venient to disregard the reaction of the axle. When this is done the torque of the couple is called the moment of the force applied. Therefore the moment of a force about an axis equals the product of the force by its lever-arm. The lever- arm of a force is the shortest distance between the axis and the line of action of the force. In Fig. 31 the moment of F about the axis through the point and perpendicular to the plane of the paper is G = Fd, (II) where d is the lever-arm. PROBLEMS. 1. Prove that the moment of a force about an axis equals the moment of its component which lies in a plane perpendicular to the axis. 2. Prove that the sum of the moments of the forces of a couple about any axis perpendicular to the plane of the couple is constant and equals the torque of the couple. 48. Degrees of Freedom of a Rigid Body. — A rigid body may have a motion of translation along each of the axes of a rectangular system of coordinates and at the same time it can have a motion of rotation about each of these axes. Therefore a rigid body has six degrees of freedom, three of translation and three of rotation. When one point in it is constrained to move in a plane the number of degrees of freedom is reduced to five. When the point is constrained to move in a straight line the niunber becomes four. When the point is fixed the body has only the three degrees of freedom of rotation. If two points are fixed the body can only rotate about the line joining the two points. There- fore its freedom is reduced to one degree. When a third point, which is not in the line determined by the other two, is fixed the body cannot move at all, that is, it has no freedom of motion. 49. The Law of Action and Reaction. — The law from which the conditions of equilibrium of a particle were obtained is a 40 ANALYTICAL MECHANICS universal law applicable to all bodies under all conditions; therefore it is applicable to rigid bodies as well as to single particles. But since rigid bodies may be subject to two distinct types of action the law may be stated in the fol- lowing form. The sum of all the linear and angular actions to which a body or a part of body is subject at any instant vanishes: S(A, + AJ = 0. (A') But since the two types of action are independent of each other the sum of each type must vanish when the combined sum vanishes. Therefore we can split the law into the fol- lowing two sections. To every linear action there is an equal and opposite linear reaction, or, the sum of all the linear actions to which a body or a part of body is subject at any instant vanishes: SA, = 0. (At) To every angular action there is an equal and opposite angular reaction, or, the sum of all the angular actions to which a body or a part of body is subject at any in- stant vanishes: SA„ = 0. (AJ 50. Conditions of Equilibrium of a Rigid Body. — If we replace the term "linear action" in the first section of the law by the word "force" and the term " angular action " in the second section of the law by the word "torque" we obtain the two conditions which must be satisfied in order that a rigid body be in equilibrium. Thus, in order that a rigid body be in equilibrium the following conditions must be satisfied. First. The sum of all the forces acting upon the rigid body must vanish, that is, if Fi, Fa, . . . F„ denote all the forces acting upon the body then the vector equation F1+F2+ • • • +F„=0 (III) must be satisfied. (V) EQUILIBRIUM OF RIGID BODIES 41 Second. The sum of all the torques acting upon the rigid body must vanish, that is, if Gi, G2, . . . G„ denote all the* torques acting upon the body then the vector equation G1 + G2+ • • • +G„=0 (IV) must be satisfied. The following forms of the statement of these two condi- tions are better adapted for analysis. First. The algebraic sum of the components of all the forces along each of the axes of a rectangular system of coordinates must vanish, that is, 2X = Xi + Z2+ • • • +Z„=0, S7=Fi+F2+- ■ • +F„=0, SZ = Zi + Z2 + • • • + Z„ = 0. Second. The algebraic sum of the components of all the torques about each of the axes of a system of rectangular coor- dinates must vanish, that is, TiGy ^ Gy -\- Gy + ' ' " + (t„ =0, SG,^GJ + GJ'+ ■ ■ ■ +GT^ = 0. 51. Coplanar Forces. — If two or more forces act in the same plane they are said to be coplanar. If a system of coplanar forces act in the a;y-plane then the conditions of equilibrium reduce to the following equations: SZ^Zi + X2+ • • • +X„=0,' 2F^Fi+F2+- ■ ■ +F„=0, SG,=Fidi +F2d2+- ■ ■ + FA = 0, (VI) where di, c^, . . . , d„ are the lever-arms of the forces Fi, F2, . . . F„, respectively, about any axis which is perpen- dicular to the plane of the forces. The 2-components of the forces and the x- and y-components of the moments vanish identically. Consequently they need not be con- sidered. (vio (V) 42 ANALYTICAL MECHANICS 52. Transmissibility of Force. — A force which acts upon a rigid body may be considered to be appUed to any particle of the body which lies on the line of action of the force. In order to prove this statement consider the rigid body A, Fig. 32, which is in equiUbrium under the action of the two Fig. 32. equal and opposite forces F and — F. Now suppose we change the point of application of F, without changing either its direction or its line of appUcation. Evidently the equilibrium is not disturbed, because by moving F in its line of action we neither changed the sum of the forces nor the sum of their moments about any axis. Therefore the line of action of a force is of importance and not its point of application. 53. Internal Forces. — Internal forces do not affect the equi- librium of a rigid body. This is a direct consequence of the law of "action and reaction." Since by definition the in- ternal forces are due to the interaction between the particles of the system these forces exist in equal and opposite pairs, therefore mutually annul each other. ILLUSTRATIVE EXAMPLES. 1. A uniform beam rests with its lower end on smooth horizontal ground and its upper end against a smooth vertical wall. The beam is held from shpping by means of a string which connects the foot of the beam with the foot of the wall. Find the tensile force in the string and the reactions at the ends of the beam. There are four forces acting upon the beam, i.e., the two reactions, Ri and Rs, the tensile force T and the weight W. Since both the ground and the wall are supposed to be smooth, Ri is normal to the ground, and Rj EQUILIBRIUM OF RIGID BODIES 43 to the wall. Therefore denoting the lengths of the beam and the string by I and a, respectively, we have 2Z s 7?2 - r = 0, 27 ^ Bi - W= 0, 'Z,Go'= -Rilsaia+w\cosa=Q, where liGj denotes the sum of the mo- ments of the forces about an axis through the poLat 0' perpendicular to the sy-plane. Solving the last three equations we have and it2 2 cot a 2 a y/P- a" T _W_ 2 a VI'- a' Fig. 33. Discussion. — It should be noticed that in taking the moments the axis was chosen through the point 0' in order to eliminate the moments of as many forces as possible and thus to obtain a simple equation. The reaction Ri is independent of the angular position of the beam and equals the weight W. On the other hand R2 and T vary with a. When a ■ both R2 and T vanish. As a is diminished from - to 0, Rz and T increase indefinitely. 2. A ladder rests on rough horizontal ground and against a rough vertical wall. The coefficient of friction between the ladder and the ground is the same as that between the ladder and the wall. Fiad the smallest angle the ladder can make with the horizon without slipping. There are three forces acting on the ladder, i.e., its own weight W and the two reactions Ri and R2. Replacing Ri and R2 by their components and writing the equations of equilibrium we obtain SZ ^ Fi - iVa = 0, 27 s iVi + F2 - TT = 0, XGo' = F2I cos a + N2I sin a — PT - cos a = 0, where a is the required angle. We have further 5. 44 ANALYTICAL MECHANICS Solving these we get Fi = — ^^ W ' 1 + M^ ' N2- l + fi^ 1 W, vT+7^ w, l + M' TF, :W. Zt2 = tan a = 1 + m' 1 -m' 2/* Tf, Fig. 34. Discussion. — The last expression gives the value of a. for a given value of fj,. When fi = 1, a = 0, therefore in this case the ladder wiU be in equilibrium at any angle between and - with the ground. £1 Evidently this is true for any value of /i greater than unity. 3. Find the smallest force which, when applied at the center of a carriage wheel of radius a, wUl drag it over an obstacle. The forces acting on the wheel are : its weight W, the required force F, and the reaction R. Since the first two meet at the center of the wheel, the direction of R must pass through the center also. Take the coordinate axes along and at right angles to R, as shown in Fig. 35, and let F make an angle % with the a;-axis. Then the equations of equiUbrium become 2Z s 2?- cos 9 - i? + TT cos a = 0, SF =^sin0-PFsina = O, SG<,'= PT-asin a-F&m.Q-a = 0. From either of the last two equations we get F = ^,W. sm Q Since W and a are fixed F can be changed only by changing 9. Therefore the minimum value of F is given by the maximum value of sin Q, i.e., 6 = -, which makes /^ = Tf sin a. EQUILIBRIUM OF RIGID BODIES 45 From the figure we obtain cos a = > a therefore sin a = - VA (2 a — h), and F=— Vh{2a-h). Fig. 35. Since cos 5 = the first equation of equilibrium gives R = W cos a a Discussion. — It will be observed that the first two of the equations of equihbrium are sufficient to solve the problem. When h is zero, F = and R = W. On the other hand when h = a, F = WajidR = 0. PROBLEMS. 1. Prove that the true weight of a body is the geometric mean between the apparent weights obtained by weighing it in both pans of a false balance. 2. A uniform bar weighing 10 pounds is supported at the ends. A weight 'of 25 pounds is suspended from a point 20 cm. from one end. Find the pressure at the supports if the length of the bar is 50 cm. 3. A uniform rod which rests on a rough horizontal floor and against a smooth vertical wall is on the point of slipping. Find the reactions at the two ends of the rod. 46 ANALYTICAL MECHANICS 4. A body is suspended from the middle of a uniform rod wliich passes over two fixed supports 6 feet apart. In moving the body 6 inches nearer to one of the supports the pressure on the support increases by 100 pounds. What is the weight of the body if 5 pounds is the weight of the rod? 5. A uniform rod of length a and weight W is suspended by two strings having lengths k and h. The lower ends of the strings are attached to the ends of the rod, while the upper ends are tied to a peg. Find the tensile force in the strings. 6. A safety valve consists of a cylinder with a plunger attached to a uniform bar hinged at one end. The plunger has a diameter of J inch and is attached to the bar at a distance of 1 inch from the hinge. The bar is 2 feet long and weighs 1 pound. How far from the hinge must a shde-weight of 2 pounds be set if the steam is to blow off at 120 pounds per square inch? 7. The two legs of a stepladder are hinged at the top and connected at the middle by a string of negUgible mass. Find the tensile force in the string and the pressure on the hinges when the ladder stands on a smooth plane. The weight of the ladder is W, the length of its legs I, and the length of the string a. 8. A uniform rod rests on two smooth inclined planes making angles of oil and ai with the horizon. Find the angle which the rod makes with the horizon and the pressure 9n the planes. 9. A rectangular block is placed on a rough incUned plane whose in- chnation is gradually increased. If the block begins to shde and to turn about its lowest edge simultaneously find the coefficient of friction. 10. A uniform rod rests with one end against a rough vertical wall and the other end connected to a point in the wall by a string of equal length. Show that the smallest angle which the string can make with the wall is tan~^ I - 1 • 11. A uniform rod is suspended by a string wliich is attached to the ends and is slung over a smooth peg. Show that in equiUbrium the rod is either horizontal or vertical. 12. A ladder 25 feet long and weighing 50 pounds rests against a vertical wall making 30° with it. How high can a man weighing 150 pounds climb up the ladder before it begins to slip? The coefficient of friction is 0.5 at both ends of the ladder. 13. A rod of negligible weight rests wholly inside a smooth hemispheri- cal bowl of radius r. A weight W is clamped on to the rod at a point whose distances from the ends are a and h. Show that the equilibrium EQUILIBRIUM OF RIGID BODIES a — b position of the rod is given by sin d 47 , where 6 is the angle it 2 Vr^ - ah makes with the plane of the brim of the bowl which is horizontal. 14. Prove that when a rigid body is in equilibrium under the action of three forces their lines of action He in the same plane and intersect at the same point. 15. Find the forces which tend to compress or extend the different members of the following cranes. 1,750 .lbs, 16. Supposing the weights of the following figures to be in equilibrium find their relative magnitudes. The circles which are tangent to other circles represent gears. 64. Resultant of a System of Forces Acting upon a Rigid Body. — ^We have already shown that the most general displacement of a rigid body consists of a translation along, and a rotation about, a certain liae. Therefore such a displacement can be prevented by a single force opposed to the translation and a single torque opposed to the rotation. Thus a single force and a single torque can be found which will keep a rigid body in equilibrium against the action of any system of forces. X = Xi + Z2+ • • ■ +x„, F=Fi+F2+ • • ■ +F„, R = VX2+ F^, Y tan S = —> 48 ANALYTICAL MECHANICS The resultant of a system of forces consists, therefore, of a single force and a single torque which, when reversed, will keep the rigid body in equilibrium against the action of the given system of forces. 55. Resultant of Coplanar Forces Acting upon a Rigid Body. — Let Fi, F2, . . . F„ denote the given forces and let the xy- plane be their plane of action. Then, if R, X, and Y denote the resultant force and its components, respectively, we have (VII) (VIII) and tane = -^j (IX) where the terms in the right-hand members of the first two equations are the components of the given forces, and 6 is the angle R makes with the a;-axis. On the other hand if G,, denotes the resultant torque and di, d2, . . . , d„ denote the distances of the origin from the Unes of action of the forces, then G, = Fidi + FA + • • ■ +FJ„. (X) If we represent this torque by the moment of the resultant force about the s-axis, then RD^Fidi + F2d^+ ■ ■ ■ +FJ„ I,{Fd) or D = R (XI) gives the distance of the line of action of the resultant force from the origin. ILLUSTRATIVE EXAMPLE. Find the resultant of the six forces acting along the sides of the hexa- gon of Fig. 36. Taking the sum of the components along the x and y directions, we have EQUILIBRIUM OF RIGID BODIES 49 X = 2F + 3F Gosl - 2F cos'^ - F - 2F cos'^ + F cos- O O o o Y = 0-3Fsm~-2Fsml + + 2F8m^ + FsiD.'^ = -fVs. .: B = VF^ + 3F^ = 2F and tan 8 = — Vs. Therefore the resultant force has a magnitude 2 F and makes an angle of — 60° with the x-axis. Taking the moments about an axis through the center of the hexa- gon, we obtain PiQ- 36. BD = {2F + 3F + 2F + F + 2F + F)a = 11 Fa, therefore D = 5.5 a, where a is the distance of the center from the lines of action of the forces. 56. Resultant of a System of Parallel Forces. — Let R be the resultant of the parallel forces Fi, Fa, . . . , F„, which act upon a rigid body. Then, since the forces are parallel, the resultant force equals the algebraic sum of the given forces. Thus B=Fi + F2+ • ■ • +F„, and RD = Fidi + F2d2+ • ■ • +i^„d„. Now take the z-axis parallel to the forces and let Xi and t/,- denote the distances of F^ from the 2/2-plane and the xz- plane, respectively. Then the last equation may be split into two parts, one of which gives the moments about the a;-axis and the other about the y-axis. Thus, Rx = F].Xi + FiX2+ ■ ■ ■ +F„x„,} Ry = F12/1 + F22/2 + • ■ • + F„y^, S ^^^^^ where x and y are the coordinates of the point in the xy- plane through which the resultant force passes. In other r^ 50 ANALYTICAL MECHANICS words, {x,y) is the point of application of the resultant force. The resultant force is evidently parallel to the given forces. The last two equations may be written in the following forms x = R R (XIII) ILLUSTRATIVE EXAMPLE. Find the resultant of two parallel forces which act upon a rigid body in the same direction. Let the y-axis be parallel to the forces. Then R = Fi + Fi, - FiXi + F1X2 and ^1 Fi F1 + F2 X2 — X Fig. 37. X — Xi But since Xi — x and x — Xi are the distances of F2 and Fi from R,"we have El = 4? F2 di' or Fidi = Fidi. Therefore the distances of the resultant from the given forces are in- versely proportional to the magnitudes of the latter. PROBLEMS. 1. Find the resultant force and the resultant torque due to the forces P, 2 P, 4 P and 2 P which act along the sides of a square, taken in order. 2. Three forces are represented in magnitude and line of action by the sides of an equilateral triangle. Find the resultant force, taking the directions of one of the forces opposite to that of the other two. 3. The lines of action of three forces form a right isosceles triangle of sides a, a, and a •\/2. The magnitudes of the forces are proportional to the sides of the triangle. Find the resultant force. 4. The sum of the moments of a system of coplanar forces about any three points, which are not in the same straight line, are the same. Show that the system is equivalent to a couple. EQUILIBRIUM OF RIGID BODIES 51 6. Three forces are represented in magnitude, direction, and line of action by the sides of a triangle taken in order; prove that their resultant is a couple the torque of which equals, numerically, twice the area of the triangle. 6. Three forces act along the sides of an equilateral triangle; find the condition which will make their resultant pass through the center of the triangle. FRICTION ON JOURNALS AND PIVOTS. 57. Friction on Journal Bearings. — If the horizontal shaft of Fig. 38 fits perfectly in its bearings the friction which comes into play is a sliding friction, therefore the laws of sliding friction may be assumed to hold good. The most important of these laws is : the frictional force which comes into play is proportional to the normal reaction, that is, in the relation n is independent of N. We will assume therefore that this law holds at each point of the surface of contact and thus reduce the problem under discussion to one of sliding fric- tion. There is an important difference, however, between the problem under discussion and the problems on friction which we have already discussed. In the present problem the normal reaction is not the same at all the points of the surfaces in contact. We must apply, therefore, the laws of friction to small elements of surfaces of contact over which the normal reaction may be considered to be constant. Let the element of surface be a strip, along the length of the shaft, which subtends an angle de at the axis of the shaft. Further let dN be the normal reaction over this element of surface, and dF be the corresponding frictional force; then we have dF=' ndN = up -l- add, where p is the normal reaction per unit area or the pressure, a is the radius of the shaft, and I the length of the bearing. 52 ANALYTICAL MECHANICS Therefore the total frictional force and the total frictional torque are, respectively, and G = fjiuH f'p de. In order to carry out the integral of the foregoing expressions we have to make some assumption with regard to the nature Fig. 38. of dependence of p upon d. But whatever the relation between p and d it is obvious that the sum, over all the sur- faces of contact, of the vertical component of the normal reaction must equal the load which rests upon the bear- ings. If P denotes this load, then p must satisfy the condi- tion i/O psind • dA = al I psind de, where A is the total area of contact. ILLUSTRATIVE EXAMPLE. The normal pressure on the bearings is given by the relation p=po sin fl; find the total frictional force and the total frictional torque. EQUILIBRIUM OF RIGID BODIES 53 Substituting the given value of p in the expression for F we obtain F = luilpQ \ saiddB Jo = 2 juoZpo. In order to determine po in terms of the total load on the bearings we make p satisfy the condition P = aZr'psined9. ^0 Substituting the given value of p in the right-hand member of the pre- ceduig equation we have = aZpo \i Jo sin'' Odd TToZpo - 2 ' Va 2P =s — ■ iral F IT G IT Therefore and It will be observed that the total frictional force varies with the load and is independent of the radius and of the length of the bearing; in other words it is independent of the area of contact. PROBLEMS. 1. Supposing the normal pressure to be the same at every point of the surfaces of contact, derive the expressions for the total frictional force and the resisting torque due to friction. 2. Supposing the vertical component of the total reaction at every point of the surfaces of contact to be constant, derive the expressions for the total frictional force and the resisting torque due to friction. 3. Derive expressions for the total frictional force and the resisting torque upon the assumption that the normal pressure is given by the relation p = po sin^ d. 58. Friction on Pivots. — The problem of friction on pivots also is a problem of sliding friction. The feature 54 ANALYTICAL MECHANICS \ which distinguishes the pivot from the journal bearing is this : in the former the lever arm of the frictional force varies from point to point, while in the latter it is constant and equals the radius of the shaft. Let dN be the normal reaction upon dA, an ele- ment of area at the base of the flat-end pivot of Fig. 39; then if dF denotes the corresponding frictional force, we have dF = p.dN = ij,p dA, where p is the normal pres- sure. Evidently p is con- stant ; therefore we can write F= ixp j dA ira^fip. The expression for the resisting torque due to the friction is obtained as follows : G r-dF -f Jo = j r • updA n2ir riip . rde -dr _ = TTUP I i = f TraVp where P is the total load on the pivot. r^dr EQUILIBRIUM OF RIGID BODIES PROBLEMS. 55 1. Derive an expression for the resisting torque due to friction in the coUar-bearing pivot of the adjoining figure. 2. Supposing the normal pressure to be constant, derive an expression for the resisting torque due to' friction in the conical pivot of the adjoining figure. 3. In the preceding problem suppose the vertical component of the normal pressure to be constant. 4. In problem 2 suppose the horizontal component of the normal pressure to be constant. 1' ..-'i-i 6. Taking the normal pressure to be constant derive an expression for tho resisting torque, due to friction in the spherical pivot of the adjoin- ing figure. 56 ANALYTICAL MECHANICS 6. Prove that the resisting torque due to friction is greater for a hollow pivot than for a solid pivot, provided that the load and the load per unit area are the same in both cases. 7. Show that the resisting torque due to friction for a hemispherical pivot is about 2.35 times as large as that for a flat end pivot. ROLLING FRICTION. 59. Coefficient of Rolling Friction. — Consider a cylinder, Fig. 40, which is in equilibrium on a rough horizontal plane under the action of a force S- y In addition to this force the cylinder is acted upon by its weight and by the reaction of the plane. Applying the con- ditions of equilibrium we ob- tain 2F= -W + N=0, ^Go = ND-Sd=0, where F and N are the com- ponents of R, the reaction of the plane, while D and d are, respectively, the distances of the points of apphcation of R and S from the point 0, about which the moments are taken. These equations give us = Vs^+VP, (1) and I>=4d. (2) Fig. 40. W If the cyUnder is just on the point of motion F=(.N, S and consequently M = W (3) EQUILIBRIUM OF RIGID BODIES 57 Combining (2) and (3), we obtain D = ixd. (XIV) The distance D is called the coefficient of roiling friction. Equation (XIV) states, therefore, that the coefficient of the rolling friction equals the coefficient of the sliding friction times the distance of the point of contact from the fine of action of the force which urges the body to roll. 60. Friction Couple. — It is evident from the above equa- tions that a change in the value of d does not affect the values of N and F, consequently it does not change the value of /i. This is as it should be, since, according to the laws of sliding friction, /i depends only upon the natm-e of the surfaces in contact. A change in d, however, changes the value of D; in other words, it changes the point of application of R. When d = 0, that is, when S is apphed at the point of con- tact, D = 0, in which case the body is urged to slide only. But when d is not zero the force S not only urges the body to sHde but also to roll; therefore, in addition to the resist- ing force F, a resisting torque comes into play. This torque, which is due to the couple formed by N and W, is called the friction couple. PROBLEMS. 1. A gig is so constructed that when the shafts are horizontal the center of gravity of the gig is over the axle of the wheels. The gig rests on perfectly rough horizontal ground. Find the least force which, act- ing at the ends of the shafts, will just move the gig. 2. Find the smallest force which, acting tangentially at the rim of a flywheel, wiU rotate it. The weight and the radius of the flyTvheel, the radius of the shaft, and the coefficient of friction between the shaft and its bearings are supposed to be known. 3. A flywheel of 500 pounds weight is brought to the point of rotation by a weight of 10 pounds suspended by means of a string wound around its rim. Find the coefficient of friction between the axle and its bearings. The diameters of the wheel and the axle are 10 feet and 8 inches, respectively. 58 ANALYTICAL MECHANICS 4. A wheel of radius a and weight W stands on rough horizontal ground. If ix is the coefficient of friction between the wheel and the ground find the smallest weight which must be suspended at one end of the horizontal diameter in order to move the wheel. GENERAL PROBLEMS. 1. A table of negUgible weight has three legs, the feet forming an ■equilateral triangle. Find the proportion of the weight carried by the legs when a particle is placed on the table. 2. A rectangular board is supported in a vertical position by two smooth pegs in a vertical waU. Show that if one of the diagonals is parallel to the line joining the pegs the other diagonal is vertical. 3. A uniform rod rests with its two ends on smooth inclined planes making angles a and j3 with the horizon. Where must a weight equal to that of the rod be clamped in order that the rod may rest horizontally? 4. A uniform ladder rests against a rough vertical wall. Show that the least angle it can make with the horizontal floor on which it rests is given by tan 6 = — , where ix and ix' are the coefficients of friction for the floor and the wall, respectively. 5. A uniform rod is suspended by two equal strings attached to the «nds. In position of equiUbrium the strings are parallel and the bar is horizontal. Find the torque which wiH turn the bar, about a vertical axis, through an angle d and keep it in equilibrium at that position. 6. The line of hinges of a door makes an angle a with the vertical. Find the resultant torque when the door makes an angle /8 with its equi- librium position. 7. The lines of action of four forces form a quadrilateral. If the magnitude of the forces are a, b, c, d times the sides of the quadrilateral find the conditions of equilibrium. 8. A force acts at the middle point of each side of a plane polygon. Each force is proportional to the length of the side it acts upon and is perpendicular to it. Prove that the polygon will be in equilibriimi if all the forces are directed towards the inside of the polygon. 9. A force acts at each vertex of a plane convex polygon in a direc- tion parallel to one of the sides forming the vertex. Show that if the forces are proportional to the sides to which they are parallel and if their directions are in a cyclic order their resultant is a couple. 10. A uniform chain of length I hangs over a rough horizontal cylinder of radius a. Find the length of the portions which hang vertically whea EQUILIBRIUM OF RIGID BODIES 59 the chain is on the point of motion under its own weight, (1) when a is neghgible compared with I, (2) when it is not negligible compared with I. 11. Two equal weights are attached to the extremities of a string which hangs over a rough horizontal cylinder. Find the least amount by which either weight must be increased in order to start the system to move. The weight of the string is negligible. 12. Three cylindrical pegs of equal radius and roughness are placed at the vertices of a vertical equilateral triangle the two lower corners of which are in the same horizontal line. A string of negligible weight is attached to two weights and slung over the pegs. Find the ratio of the weights if they are on the point of motion. 13. A sphere laid upon a rough inchned plane of inclination a is on the point of sliding. Show that the coefficient of friction is j tan a. 14. A uniform ring of weight W hangs on a rough peg. A bead of weight w is fixed on the ring. Show that if the coefficient of friction W between the ring and the peg is greater than — — the ring wiU VW' + 2wW be in equilibrium whatever the position of the bead with respect to the peg. 16. A uniform rod is ia equHibrimn with its extremities on the interior of a rough vertical hoop. Find the limiting position of the rod. 16. A weight W is suspended from the middle of a cord whose ends are attached to two rings on a horizontal pole. If w be the weight of each ring, n the coefficient of friction, and I the length of the cord, find the greatest distance apart between the rings compatible with equihbrium. CHAPTER IV. EQUILIBRIUM OF FLEXIBLE CORDS. 61. Simplification of Problems. — The simplest phenome- non in nature is the result of innumerable actions and reactions. The consideration of all the factors which con- tribute to any natural phenomenon would require unlimited analytical power. Fortunately the factors which enter into dynamical problems are not all of equal importance. Often the influence of one or two predonunate, so that the rest can be neglected without an appreciable departure from the actual problem. Any one who attempts to solve a physical problem must recognize this fact and use it~to advantage by repre- senting the actual problem by an ideal one which has only the important characteristics of the former. This was done in the last two chapters in which bodies were treated as single particles and rigid bodies, and the problems were thereby simplified without changing their character. The same procedure will be followed in discussing the equilibrium of flexible cords, such as belts, chains, and ropes. These bodies will be represented by an ideal cord of negli- gible cross-section and of perfect flexibiUty. The solution of the ideahzed problems gives us a close enough approxima- tion for practical purposes. If, however, closer approxima- tion is desired smaller factors, such as the effects of thickness and imperfect flexibility, may be taken into account. 62. Flexibility. — A cord is said to be perfectly fleodhle if it offers no resistance to bending; in other words, in a perfectly flexible cord there are no internal forces which act in a direction perpendicular to its length. 60 EQUILIBRIUM OF FLEXIBLE CORDS 61 63. Suspension Bridge Problem. — The following are the important features of a suspension bridge which should be considered in order to simpUfy the problem: 1. The weights of the cables and of the chains are small compared with that of the road-bed. 2. The road-bed is practically horizontal. 3. The distribution of weight in the road-bed may be considered to be uniform. We can, therefore, obtain a sufficiently close approxima- tion if we consider an ideal bridge in which the cable and the chains have no weight and the distribution of weight in the road-bed is uniform in the horizontal direction. With these simplifications consider the forces acting upon that part of the cable which is between the lowest point and any point P, Fig. 41. Fig. 41. The forces are : The tensile force To, which acts horizon- tally at 0. The tensile force T, which acts along the tangent to the curve at P. The weight of that part of the bridge which is between and P. If w be the weight per unit length of the road-bed and x denotes the length, O'F', then the third force becomes wx. Therefore the conditions of equiUbrium give 'LX= -To + r cos 9=0; 27= -wx+Tsm e=0; rcos0=ro. (1) Tsmd= wx. (2) 62 ANALYTICAL MECHANICS It is evident from equation (1) that the horizontal compo- nent of the tensile force is constant and equals Tq. Squaring equations (1) and (2) and adding we get T^ = To^ + wV. (3) Thus we see that the smallest value of T corresponds to X = and equals To, while its greatest value corresponds to the greatest value of x. If D denotes the span of the bridge then the greatest value of T, or the tensile force of the cable at the piers, is t^ = \/t, In order to find the equation of the curve which the cable assumes we eUminate T between equations (1) and (2). This gives w tan e=-—x. (4) ■to Substituting -~ for tan B and integrating we get where c is the constant of integration. But with tlie axes we have chosen, y=Q when a;=0, therefore c = 0. Thus the equation of the curve is ^^2To ' ^^^ which is the equation of a parabola. Dip of the Cable. — Let H be the height of the piers above the lowest point of the cable. Then iorx = —,y=H, therefore fi=^.i>'. (6) It is evident from the last equation that the greater the tension the less is the sag. EQUILIBRIUM OF FLEXIBLE CORDS 63 Peoblem. a bridge is supported by two suspension cables. The bridge has a weight of 1.5 tons per horizontal foot and has a span of 400 feet. Supposing the dip of the bridge to be 50 feet find the values of the tensile force at the lowest and highest points of the cable. 64. Eqiiilibrium of a Uniform Flexible Cord which is Sus- pended from Its Ends. — The problem is to determine the natm-e of the curve which a ^^ perfectly uniform and flexi- ble cable will assume when suspended from two points. Let AOB, Fig. 42, be the curve. Consider the equilibrium of that part of the cable which is between the lowest point and any other point P. The part of the cable which is imder consideration is acted upon by the following three forces: The tensile force at the point 0, To. The tensile force at the point P, T. The weight of the cable between the points and P. Since the cable is perfectly flexible To and T are tangent to the curve. Therefore we have EX= -To + Tcose=0, or Tcose = To, (1) HY^ —ws+Tsin e = 0, or Tsmd = ws, (2) where w is the weight per unit length of the cable and s is the length of OP. Squaring equations (1) and (2) and adding we obtain T^=To^ + w^s\ (3) Eliminating T between equations (1) and (2) we get s= — tan 6, w (4) which is the intrinsic equation of the curve. 64 ANALYTICAL MECHANICS In order to express equation (4) in terms of rectangular coordinates we replace tan dhy -r and obtain ax w dx (5) But ds^ = dx^ + dy^, therefore eliminating dx between this equation and equation (5) and separating the variables dy = sds V^^+a} (6) and then integrating To ' = VsH-^ + c, where a = — and c is the constant of integration. w Let the x-axis be so chosen that when s = 0, y = a, then c = 0. Therefore y = Vs^ + a^, or s = V^/^— a\ (7) Differentiating equation (7), squaring and replacing ds^ by (dx^+dy^) we have dx^ + dy^ = Solving for dx, dx= — ady y-' — a^ dy ady (8) where i= ^ — 1. iVa^-y"^ Integrating equation (8) we get — = cos-i- +c'. a a -il EQUILIBRIUM OF FLEXIBLE CORDS 65 But y = a, when a; = 0, therefore c' = 0. Thus we get ice 2/ = a cos— , (9) a = a cosh -, (10) = f(e^+e"^l (11) = ~W'-\-e ^« , (12) 2'w which are different forms of the equation of a catenary. Discussion. — Expanding equation (12) by Maclaurin's Theoremf we obtaia ' = a . 2 Va/ ^ 24 \a) (13) In the neighborhood of the lowest point of the cable the value of X is small, therefore in equation (13) we can neglect all the terms which contain powers of x higher than the second. Thus the equation y^a + f^ (14) represents, approximately, the curve in the neighborhood of the lowest point. It will be observed that (14) is the equation of a parabola. This result would be expected since the curve is practically straight in the neighborhood of and consequently the horizontal distribution of mass is very nearly constant, which is the important feature of the Suspension Bridge problem. The nature of those parts of the curve which are removed from the lowest point may be studied by supposing x to be X large. Then since e " becomes negligible equation (11) re- duces to 2/=|e^, (15) * See Appendix Avin. t See Appendix Av. 66 ANALYTICAL MECHANICS The curve, Fig. 43, defined by equation (15) is called an exponential curve. It has an interesting property, namely, its ordinate is doubled every time a constant value P is added to its abscissa. This constant is called the half-value period of the curve. The value of P may be determined in the following manner. By the definition of P and from equa- tion (15) we have 2 y = -e Dividing equation (16) by equation (15) we get p 2 = e°, or P = aloge2. Length of Cable. — In order to find the length in terms of the span eliminate y between equations (7) and (11). This gives (X x\ &■ —e °-] x+i» (16) = a; + - 1 3 a^ ^2 ■ ha^^ (17) (18) 2.3 a' 2.3-4 where the right member of equation (18) is obtained by expanding the right-hand member of equation (17) by Maclaurin's Theorem. If D and L denote the span and the length of the cable, respectively, we have & = \L when x = | D. Therefore substituting these values of s and x in equation (18) and replacing a by its value we obtain EQUILIBRIUM OF FLEXIBLE CORDS 67 When the cable is stretched tight To is large compared with w. Therefore the higher terms of the series may be ne- glected and equation (19) be put in the following approxi- mate form. 1 2 Hence the increase in length due to sagging is — 7^I>', 24 To approximately. PROBLEMS. 1. A perfectly flexible cord hangs over two smooth pegs, with its ends hanging freely, while its central part hangs in the form of a catenary. If the two pegs are on the same level and at a distance D apart, show that the total length of the string must not be less than De, in order that equilibrium shall be possible, where e is the natural logarithmic base. 2. In the preceding problem show that the ends of the cord will be on the a;-axis. 3. Supposing that a telegraph wire cannot sustain more than the weight of one mile of its own length, find the least and the greatest sag allowable in a line where there are 20 poles to the mile. 4. Find the actual length of the wire per mile of the line in the pre- ceding problem. 5. The width of a river is measured by stretching a tape over it. The middle point of the tape touches the surface of the water while the ends are at a height H from the surface. If the tape reads S, show that the width of the river is approximately 1/ — 6. Show that the cost of wire and posts of a telegraph line is mini- mum if the cost of the posts is twice that of the additional length of wire required by sagging. The posts are supposed to be evenly spaced and large in number. 7. A uniform cable which weighs 100 tons is suspended between two points, 500 feet apart, in the same horizontal line. The lowest point of the cable is 40 feet below the points of support. Find the smallest and the greatest values of the tensile force. 8. In the preceding problem find the length of the cable. 65. Friction Belts. — The flexible cord AB, Fig. 44, is in equilibrium under the action of three forces, namely, To 68 ANALYTICAL MECHANICS and T, which are apphed at the ends of the cord, and the reaction of the rough surface of C, with which it is in con- tact. It is desired to find the relation between To and T when the cord is just on the point of motion towards To. Fig. 44. Consider the equilibrium of an element of that part of the cord which is in contact with the surface. The element is acted upon by the following three forces: The tensile force in the cord to the right of the element. The tensile force in the cord to the left of the element. The reaction of the surface. Let the tensile force to the left of the element be denoted by T, then the tensile force to the right may be denoted by T + dJ. On the other hand if R denotes the reaction of the surface per unit length of the cord, the reaction on the element is R ds, where ds is the length of the element. We will, as usual, replace R by its frictional component F and its normal component N. EQUILIBRIUM OF FLEXIBLE CORDS 69 Taking the axes along the tangent and the normal through the middle point of the element and applying the conditions of equilibrium we obtain SX^ (r+ dT) cos^ - r cos f - F ( - ds)* = 0, I.Y=Nds-Tsm^-{T+dT)sm^ = 0, or drcos^+i^ds = 0, and i\rds- 2 rsin^-drsin- = 0, 2 2 ' where dQ is the angle between the two tensile forces which act at the ends of the element. But since the cord is sup- posed to be perfectly flexible the tensile forces are tangent to the surface of contact. Therefore 6 is the angle between the tangents, and consequently the angle between the nor- mals, at the ends of the element. As an angle becomes indefinitely small its cosine approaches unity and its sine approaches the angle itself,t therefore we can make the substitutions dfl , , . de de cos -- = 1 and sm — = — - 2 2 2 in the last two equations, and obtain dT+Fds = 0, (1) and Nds-Tde + idTde=0. (2) Neglecting the differential of the second order in equation (2) and then eliminating ds between equations (1) and (2) we get ^=-^d0=-f.d9, (3) where fi is the coefficient of friction. Integrating the last * The negative sign in i*" ( — ds) indicates the fact that F and ds are measured in opposite directions, t See Appendix Avi. 70 ANALYTICAL MECHANICS equation and passing from the logarithmic to the exponen- tial form, we have T = ce-*^, where c is the constant of integration. If d is measured from the normal to the surface at the point where the right-hand side of the cord leaves contact we obtain the initial condition, T =Ta when 6=0, which determines c. Applying this con- dition to the last equation we have T=Tae-''\ (4) Discussion. — Equation (4) gives the relation between the values of the tensile force at any two points of the cord. It must be observed that 6 is measured in the same direction as F; in other words, opposite the direction towards which the cord is urged to move. Therefore T or To has the larger value according to whether 6 is positive or negative. As a concrete example suppose a weight W to be suspended from the right- hand end of the cord and to be held in equilibrium by a force F applied at the left-hand end. If F is just large enough to prevent W from falling then the cord wiU be on the point of moving to the right, therefore 6 is measured in the counter-clockwise T direction and is positive. In this F = We -,,) In case F is just large enough to start W to move up, then 6 is measured in the clockwise direction and is nega- tive. Therefore F = Wei^. The value of T drops very rapidly with the increase of d. This fact is made clear by drawing the graph ^ ° ^^ 3P of equation (4), Fig. 45. The graph ^'^- 45. may be constructed easily by making use of the half-value period of the cxu-ve. If P denotes the period, then, by definition, the ordinate is reduced to one-half its value every time P is added to 6* We have therefore * The difference between this definition of P and the one given in the pre- ceding section is accounted for by the difference in the signs of the exponents in equation (4) and in equation (14) of the preceding section. EQUILIBRIUM OF FLEXIBLE CORDS 71 or Dividing equation (4) by the last equation we get 1, P=il0ge2 (5) Thus if S = nP, then by equations (4) and (5) T = — 2"' (6) Therefore taking 0.53 for hemp rope on oak and -2ir, we obtain n = 4.76 and 2" = 27.3. Hence in this case To is 27.3 times as great as T. Application to Belts. — The tensile force on one side of a belt which transmits power is greater than that on the other side. The relation between the tensile forces on the two sides of the belt is given by equation (4). Thus if Ti denotes the tensile force on the driving side and T^ that on, the slack side, then Ti = Tie-i^ or Ti = Tae'^- (4'> The difference between Ti and Ti is the effective force which drives the pulley. Denoting the effective force by F, we have F = Ti-Ti = Ti (1 - e-'^) (7) = Ti (e"" - 1). We have neglected the cross-section of the cord in the solution of the foregoing problem. Therefore the results which we have obtained are applicable to actual problems only when the cross-section of the cord is negligible com- pared with that of the solid with which it is in contact. 72 ANALYTICAL MECHANICS PROBLEMS. 1. A weight of 5 tons is to be raised from the hold of a ship by means of a rope which takes 3 J turns around the drum of a steam windlass. If li = 0.25 what force must a man exert at the other end of the rope? 2. By puUing with a force of 200 pounds a man just keeps from surg- ing a rope, which takes 2.5 turns around a post. Find the tensile force at the other end of the rope. ^ = 0.2. 3. A weight W is suspended by a rope which makes IJ turns around a clamped pulley and goes to the hand of a workman. If /* = 0.2, find the force the man has to apply in order (a) to support the weight, (b) to raise it. 4. Two men, each of whom can exert a puU of 250 pounds, can sup- port a weight by means of a rope which takes 2 turns around a post. On the other hand, one of the men can support it alone if the rope makes 2.5 turns. Find the weight. 5. In order to prevent surging a sailor has to exert a force of 150 pounds at the end of a hawser, which is used to keep the stern of a boat at rest while the bow is being turned by the engines. Find the pull exerted by the boat upon the hawser under the following conditions: [Hint. — Make use of equations (5) and (6).] (a) e = l M = 0.2. (g) e = 2ir, fl = 0.1. (b) 6=1 M = 0.5. (h) 4 fl = 0.4. (c)e = l, M = 0.5. (i) ^- 2' M = 0.5. (d) e = w, fi = 0.4. (J) = 3x, fl = 0.3. (e) e = '-f, jx = 0.3. (k) fl 13x fl = 0.4. (f) e = ^, M = 0.2. (1) ^- 2' M = 0.5. 6. A belt has to transmit an effective force of 500 pounds. Find the tensile force on both sides of the belt, under the following conditions: (a) e = 135°, fl = 0.5. (e) 6 = 165°, fi = 0.2. (b) e = 135°, fl = 0.4. (f) e = 180°, n = 0.3. (c) e = 150°, fl = 0.3. (g) e = 180°, M = 0.5. (d) e = 165°, M = 0.5. (h) e = 195°, n = 0.4. 7. In the preceding problem find the width of the belt, supposing the permissible safe tensile force to be 50 pounds per inch of its width. CHAPTER V. MOTION. FUNDAMENTAL MAGNITUDES. 66. Analysis of Motion. — The conception of motion neces- sarily involves four ideas, namely, the ideas of (a) A body which moves. (b) A second body with respect to which it moves. (c) A distance which it covers. (d) An interval of time during which the distance is covered. 67. Relativity of Motion. Reference System. — The first important inference to be drawn from the foregoing analysis is the fact that motion presupposes at least two bodies, namely, the body which is supposed to move and the body to which the motion is referred. The words "motion" and "rest" become meaningless when applied to a single particle with no other body for reference. Whenever we think or talk about the motion of a particle we refer its motion, consciously or unconsciously, to other bodies. The body to which motion is referred is called a reference system. The choice of a particular body as a reference system is a question of convenience. If a man walks in a crowded car fast enough to disconunode its occupants he will be blamed, not because he is moving at the rate of, say, 20 miles per hour with respect to the ground, but because he is moving at the rate of 4 miles per hour with respect to the car. In this case the car should be taken as the reference system, and not the ground. On the other hand if the man wants to leave the moving car, it is of great importance for him to 73 74 ANALYTICAL MECHANICS consider the velocity with which he is going to land. In this case, therefore, the surface of the earth should be taken as the reference system. 68. Fundamental Magnitudes. — The first two of the four conceptions into which we analyzed motion are similar; therefore three distinct conceptions are associated with motion. The first of these is the idea of body, or of matter; the second is that of distance, and the third is that of time. Distance and time are terms which are too familiar to be made clearer by definitions, therefore we will not attempt to define them. In their efforts to reduce natural phenomena to their simplest terms scientists have come to the conclusion that all physical phenomena are the result of motion. It is the main object of science to describe the compUcated phe- nomena of nature in terms of motion, in other words, to express all physical magnitudes in terms of the three magni- tudes involved in motion. Therefore time, mass, and length are called fundamental magnitudes and all others derived magnitudes. 69. Fundamental Units. — The units of time, length, and mass are called fundamental units, while those of other magnitudes are called derived units. 70. The Unit of Time is gsAoo part of the mean solar day, and is called the second. 71. The Unit of Length is the centimeter, which is ^-^q part of the standard meter. The latter is the distance at 0° C. between two parallel lines drawn upon a certain platinum- iridium bar in the possession of the French government. 72. Mass.— The choice of the units of time and length is comparatively easy. We associate only one property with each of these quantities, therefore in choosing a unit all we have to do is to decide upon its size. Matter, on the other hand, has a great number of properties, such as volume, shape, temperature, weight, mass, elasticity, etc. We com- MOTION 75 pare and identify different bodies by means of these proper- ties. In selecting one of these properties to represent the body in our study of motion we must see that the property fulfills two conditions : that it is intimately related to motion and that it is constant. Weight is often used to represent a body in its motion. So far as bodies on (the earth are concerned weight is intimately connected /with motion, but it is not constant. Besides, when bodies are far from the earth, weight does not have a definite meaning. Therefore weight does not satisfy the foregoing conditions. The property which serves the purpose best is known as mass. It is intimately connected with motion and is constant.* The nature of this property will be discussed in the next chapter. Therefore we will con- tent ourselves by defining mass as that property with which bodies are represented in discussions of their motion. 73. Unit of Mass. — The unit of mass is the gram, which is 1^5-0 part of the mass of the standard kilogram. The latter is the mass of a piece of platinum in the possession of the French Government. 74. Dimensions. — The fundamental magnitudes enter into the composition of one derived magnitude in a manner dif- ferent from the way they enter into that of a second. Length alone enters into the composition of an area, while velocity contains both length and time, and all three of the funda- mental magnitudes combine in work and momentum. The expression which gives the manner in which time, length, and mass combine to form a derived magnitude is called the dimensional formula of that magnitude. Thus the dimen- sional formulae for area, velocity, and momentum are, respec- tively, [A] = [L'], [V] = [LT-'], and [H] = [MLT-'], where M, L, and T represent length, mass, and time. The exponent of each letter is called the dimension of the de- * Cf. §101. 76 ANALYTICAL MECHANICS rived magnitude in the fundamental magnitude which the letter represents. Thus area has two dimensions in length and zero dimension in both time and mass, while momen- tvun. has one dimension in mass, one dimension in length, and minus one dimension in time. 75. Homogeneity of Equations. — Magnitudes of different dimensions can neither be added nor subtracted. There- fore in a true equation the sum of the magnitudes of one kind which are on the left of the equation sign equals the sum of the magnitudes of the same kind which are on the right. When all the terms of an equation have the same dimensions the equation is said to be homogeneous. 76. Systems of Units. — The C.G.S. System is used in most of the civilized countries and by scientists all over the world. In this system the centimeter, the gram, and the second are the fundamental units. English-speaking people use another system, known as the British gravitational system, in which weight, length, and time are the fundamental magnitudes and the pound, the foot, and the second are the fundamental units. Thus the unit of time is the same in both systems. The following equations give the relation between the centimeter and the inch with an error of less than one-tenth of one per cent. 1 in. = 2.54 cms. 1 cm. = 0.3937 in. The relation between the mass of a body which weighs one pound and the gram is given by the following equations with an error of less than one-tenth of one per cent. 1 kg. = 2.205 pds. 1 pd. = 453.6 gms., where kg. is the abbreviation for the kilogram, or 1000 gms., while pd. denotes the mass of a body which weighs one pound in London and is often called pound-mass. Denoting MOTION 77 the pound (weight) by its usual abbreviation we have 2.205 lbs. = the weight of 1000 gms.* VELOCITY. 77. Displacement. — When the position of a particle with respect to a reference system is slightly changed it is said to have been displaced, and the vector s, Fig. 46, which has its origin at the initial position and its terminus at the final position, is called a displacement. 78. Velocity. — If a particle un- dergoes equal displacements in equal intervals of time, however - small these intervals, it is said to " pj^ ^g have a constant velocity. In this particular case the velocity equals, numerically, the distance covered per second. When, therefore, a distance s is covered in an interval of time t, the velocity is given by s By equal displacements are meant displacements equal in magnitude and the same in direction. Therefore con- stant velocity means a velocity which is constant in direc- tion as well as in magnitude. The magnitude of velocity without regard to its direction is called speed. In general, bodies not only cover unequal distances in equal intervals of time, but also change their directions of motion. Therefore we need a definition of velocity hke the following, which is perfectly general. The velocity of a particle at any point of its path equals, in magnitude, the time rate at which it describes that part of the path which is in the immediate neighborhood of the point and has the direction of the tangent at that point. * For the relation between mass and weight see p. 109. 78 ANALYTICAL MECHANICS In order to express this definition of velocity in analytical language, consider a particle describing a curved path with a changing speed. The most natural way of determining the speed at a point P, Fig. 47, is to observe the interval of time which the particle takes to pass two points, Pi and Pi, which are equidistant from P, y then to divide the distance P1P2 by that interval of time. This gives the average speed from Pi to P2, which may or may not equal the actual speed at P- If, however, we take the points Pi and P2 nearer to P we obtain an average speed which is, in general, nearer the speed at P, because there is less chance for large variations. If we take Pi and P2 nearer and nearer the average speed approaches more and more to the value at P- Therefore the limiting value of the ratio speed at P- In other words ds o Fig. 47. P1P2 • J.I. — -^ IS the = 6* dt (I) is the analytical definition of speed. Therefore the velocity is a vector which has s for its magnitude and which is tan- gent to the path at the point considered, that is, V = s. (10 * The Differential Calculus was invented by Newton and Leibnitz inde- pendently. Newton adopted a notation in which the derivative of a variable s with respect to another variable is denoted by s. This notation is not con- venient when derivatives are taken with respect to several variables. The notation introduced by Leibnitz is more convenient and is the notation which is generally adopted. Newton's notation, however, is often used to denote differentiation with respect to time. On account of the compactness of s ds compared with j:, we will denote differentiations with respect to time by Newton's notation whenever compactness of expression is desired. MOTION 79 79. Dimensions and Units of Velocity. — The dimensions of velocity are [LT~^]. The C.G.S. unit of velocity is the centimeter per second, — ■'. The British unit of velocity is the foot per second, A sec. 80. Rectangular Components of Velocity. — Let v, Fig. 48, Fig. 48. denote the velocity at P, then the magnitude of its compo- nent along the x-axis is Vx= V cos 6 ds = — cos e dt ds cos d dt dx dt Vy _dy _ dt dz y, V ■> dt (11) Similarly and Equations (II) state that the component of the velocity of & particle along any line equals the velocity of the projec- tion of the particle upon that line, in other words, the ve- 80 ANALYTICAL MECHANICS locity along any direction equals the rate at which distance is covered along that direction. The velocity and its components evidently fulfill the relation ._ V=VvJ+Vy^+V,\ (III) When, as in the case of Fig. 48, the particle moves in the xy-p\ane, 2 = 0, therefore v=Vvj+Vy\ (mo The direction of v, ia this case, is given by y tan 6 = where 6 is the angle v makes with the a;-axis. (IV) ILLUSTRATIVE EXAMPLE. Find the path, the velocity, and the components of the velocity of a particle which moves so that its position at any instant is given by the following equations : X = at, (a) 2/=-iff<^ (b) Eliminating t between (a) and (b), we obtain ^2= - — y, for the equation of the path, therefore the path is a parabola, Fig. 49. To find the component -velocities we differentiate (a) and (b) with respect to the time. This gives i: = a, y = -gt- .: V = Va'^ + gt^- Discussion. — The horizontal compo- nent of the velocity is directed to the right and is constant, while the vertical com- ponent is directed downwards and increases at a constant rate. We will see later that these equations Fig. 49. represent the motion of a body which is projected horizontally from an elevated position. MOTION 81 PROBLEMS. 1. Find the path and the velocity of a particle which moves so that its position at any instant is given by the following pairs of equations : (a) X = at, (b) X = at, (c) X = at, (d) X = a sin uit, (e) X = a sin wi, (f ) X = a sin uit, (g) X = ae*', y = ht. y = at- lgt\ y = b cos oit. y = U. y = a cos cot. y = b sin oit. y = ae~*'. 2. Prove the relation v = Vx^ + y^ + z^- 81. Radial and Transverse Components of Velocity, — The magnitude of the velocity along the radius vector is, accord- ing to the results of the preceding section, dr (1) The expression for the velocity at right angles to r is ob- tained by considering the motion of the projection of the particle along a perpendicular y to r. When the particle moves through ds, its projection moves through r de, Fig. 50, therefore the required velocity is rde ' dt de . ='dr''' (2) Fig. 50. The components Vr and Vj, may be expressed in terms of x and y by differentiating the equations of transformation and fl=tan-i- X (3) (4) 82 ANALYTICAL MECHANICS with respect to the time. Differentiating (3) we obtain dr Vr = dt _x dx ydy r dt r dt = xcose + ysiD.e. Differentiating (4) we get de '^='dt _^ xy-yx x^ + y^ = y cos d — X sin t These components satisfy the relation (5) (6) (7) ILLUSTRATIVE EXAMPLE. A particle describes the motion defined by the equations X = a cos kt, (a) and y = a sin kt. (b) Find the equation of the path, the velocity at any instant, and the com- ponents of the latter. Squaring and adding (a) and (b) we eliminate t and obtain x' + : o^ for the equation of the path. Differentiating (a), we have . _ dx = —ka sin kt = -ky. Differentiating (b), we obtain dy y = dt = ka cos kt = kx. Fig. 51. MOTION 83 Therefore v = y/i? + jr2 = ka. Thus the particle describes a circle with a constant speed ka. The direc- tion of the velocity at any instant is given by the relation tan = ^ z = -l. x The components w, and Wp may be obtained at once by remembering, (1) that the radius vector is constant: e.g., r = 0, (2) that it is always normal to the path: e.g., rdB = ds. Therefore dr dt 0, and dd v„= r— = dt ds dt : V = ka. 82. Velocity of a Particle Relative to Another Particle in Motion. — Consider the motion of a particle Pi, Fig. 52, with Fig. 52. respect to a particle P2, when both are in motion relative to the system of axes XOY. Let the system of axes X'PiY' have P2 for its origin and move with its axes parallel to those of the system XOY. Further let (xi, yi) and (xz, 2/2) be the positions, and vi and va the velocities of Pi and P2 with respect to XOY. Then 84 ANALYTICAL MECHANICS if (x'j y') denotes the position and v' the velocity of Pi with respect to X'P^Y', we get X' = Xi— Xi, y' = yi- 2/2. Differentiating the last two equations with respect to the time X' = Xi— Xi, y' = yi- 2/2. Therefore v' = x' + y' = (xi + yi) — (x2 + ya) = vi-V2. (V) Equation (V) states that the velocity of a particle with respect to another particle is obtained by subtracting the velocity of the first from that of the second. ILLUSTRATIVE EXAMPLE. Two particles move in the circumference of a circle with constant speeds of v and 2 v. Find their relative velocities. Let the slower one be chosen as the reference particle, and let the angle PtOPi, Fig. 53, be denoted by 6. Then the velocity of Pi relative to P2 is But Vi = 2y and v% = v, therefore v' = \/4y2 -2w.2!).cose + w=i Fia. 53. Discussion. — Whenever Pj passes P2 the value of fl is a multiple of 2 TT, therefore cos 9 = I and v' = v. When the particles occupy the ends MOTION 85 of a diameter cos 5 = — 1, therefore «/ = 3v. When they are separated by an angle which is an odd multiple of - , cos 9 = 0; therefore Vi=v Vs. PROBLEMS. 1. An automobUe is moving at the rate of 30 miles an hour in a direc- tion at right angles to a train which is making 40 mUes an hour; find the velocity of the automobile with respect to the train. 2. Two trains pass each other on parallel tracks, in opposite directions. A passenger in one of the trains observes that it takes the other train 4 seconds to pass him. What is the length of the other train if the veloci- ties of the two trains are 50 and 40 miles per hour? 3. A man of height h walks on a level street away from an electric lamp of height B.. If the velocity of the man is v, find the velocity of the end of his shadow (a) with respect to the ground and (6) with respect to the man. 4. Two particles move, in opposite directions, on the circumference of the same circle with the same constant speed. Find an expression for their relative velocity and see what this expression becomes at special positions of the particles. 5. A train is moving due north at the rate of 50 miles an hour. The wind is blowing from the southeast with a velocity of 20 mUes an hour. Find the apparent direction and magnitude of the wind to a man on the train. 6. The wind seems to blow from the north to an automobile party travehng westward at the rate of 15 mUes an hour. On doubling the speed of the automobUe the wind appears to come from the northwest. Find the actual direction and magnitude of the velocity of the wind. 7. Find the velocity of a particle moving on the circumference of a circle with uniform speed relative to another particle moving with equal speed in a diameter of the circle. 8. Express the speed of a mile a minute in the C.G.S. units. 9. Express the C.G.S. unit of velocity in mUes per hour. 10. Prove that x^ -j- ^^ = j.2 + ^2^2 11. Prove analyticaUy that Vx = Vr COS d — Vp sin 6, Vy = Vr sin d -\- Vp cos 6. 12. Prove graphically that Vr = Vx COS 6 + Vy sin 6, Vv = Vy cos 6 — Vx sin 6. 86 ANALYTICAL MECHANICS ANGULAR VELOCITY. 83. Angular Displacement. — When the motion of a particle is referred to an axis, then the angle which the axial plane, i.e., the plane determined by the particle and the axis, describes, is called an angular displacement. Angular dis- placement is a vector magnitude which is represented by a vector drawn along the axis; as in the case of the vector representation of a torque. The directional rela- tions are the same; that is, the vector points towards the observer and is considered as positive when the rotation is counter-clockwise. It points away from the observer and is negative when the rotation is clockwise. The relation between the linear displacement of a particle and its angular displacement about an axis may be found from a consideration of Fig. 54: Fig. 54. _ ds COS r ■where ds is the linear displacement of the particle P, dd is the corresponding displacement about an axis through the point perpendicular to the plane of the paper, and <^ is the angle ds makes with the normal to the axial plane. When r is constant is zero, and the particle describes a circle, in which case the last equation becomes dd= — r or e = MOTION 87 84. Unit Angle. — In the last equation 6=1 when s = r; therefore the angle which is subtended at the center of a circle by an arc equal to the radius is the unit of angle. This unit is called the radian. Angles and angular dis- placements have no dimensions. Why? 85. Angular Velocity. — The conception of angular velocity is similar to that of linear velocity. It is the time rate at which the axial plane sweeps over an angle. When con- stant it is numerically equal to the angle swept over per second. If we denote the angular velocity by 2 a ' de 03 == — dt = X^ 2 a dt v_ ~ 2a' Thus the angular velocity about is independent of the position of the particle and equals one-half the angular velocity about the center. PROBLEMS. 1. The radius of the earth is 4000 miles and that of its orbit 93 million miles. Compare the angular velocities of a point on the equator with respect to the sun at midday and midnight. 2. In what latitude is a buUet, which is projected east with a velocity of 1320 feet per second, at rest relatively to the earth's axis; the radius being taken as 4000 miles? MOTION 89 3. A belt passes over a pulley which has a diameter of 30 inches and which makes 200 revolutions per minute. Find the linear speed of the belt and the angular speed of the pulley. 4. The wheels of a bicycle, which are 75 cm. in diameter, make 5000 revolutions in "65 minutes. Find the speed of the rider; the angular speed of the wheels about their axles; the relative velocity of the highest point of each wheel with respect to the center. 5. A point moves with a constant velocity v. Find its angular veloc- ity about a fixed point whose distance from the path is a. 6. A railroad runs due west in latitude X. Find the velocity of the: train if it always keeps the sun directly south of it. 7. Find the expression for the angular velocity of any point on the- rim of a wheel of radius a, moving with a velocity v; the wheel is supposed to be roUing without shpping. Discuss the values of the velocity for special points. 8. In the preceding problem find the relative velocity of any point on the rim with respect to the center of the wheel, and the velocity of the; center with respect to the point of contact with the ground. 9. The end of a vector describes a circle at a constant rate. If the origin is outside the circle find the velocity along and at right angles to the vector. Discuss the values for interesting special positions. 10. In the preceding problem derive an expression for the angular velocity of the vector and discuss it. ACCELERATION. 86. Acceleration. — When the velocity of a particle changes it is said to have ah acceleration. The change may be in the magnitude of the velocity, in the direction, or in both; further it may be positive or negative. Therefore the term acceleration includes retardation as well as increase in ve- locity. Retardation is negative acceleration. If the particle moves in a straight path with a velocity which increases or diminishes at a constant rate its accel- eration equals, numerically, the change in the velocity per second and is said to be constant: X V2 - Vl ^~ t ' where f is the acceleration and vi and vz are the velocities at the beginning and at the end of the interval of time t.. 90 ANALYTICAL MECHANICS Since vi and V2 are in the same line, their difference will be a vector in the same direction. Therefore in this particular case the acceleration is constant not only in magnitude but also in direction. The following definition of acceleration is general and holds true whatever the manner in which the velocity changes. The magnitude of the acceleration of a particle at any point of its path equals the time rate at which its velocity changes at the instant it occupies that point. The analytical expression for this definition may be obtained by a reasoning similar to that employed in deriving the analyt- ical definition of velocity. Sup- pose it is required to find the acceleration at P (Fig. 56). Let vi and V2 denote the veloc- ities at two neighboring points Pi and Pi. Then the ratio f = V2- vi t gives the average rate at which the velocity changes during the interval of time t, which it takes the particle to move from Px to Pi. Therefore f is the average acceleration for that interval of time. In general this average acceleration will not be the same as the acceleration at P. But by taking Pi and P2 nearer and nearer to P the difference between the average acceleration and the required acceleration may be made as small as desired. Therefore at the limit when Pi, P, and P2 become successive positions of the particle, the aver- age acceleration becomes identical with the acceleration at P, and the last equation takes the form f=| = v. (VIII) MOTION 91 It must be remembered that dv is the vector difference of the velocities at the beginning and at the end of the interval of time dt; therefore f is a vector magnitude with a direction which is, in general, different from that of the velocity. 87. Dimensions and Units of Acceleration. — The dimen- sions of acceleration are [LT~^]. The unit of acceleration is a unit change in the velocity per second. Therefore the C.G.S. unit is — - — '- or '-. Thus if the velocity of a sec. sec.^ cm particle increases by an amount of one — '- during each sec. second it has a unit acceleration. The engineering unit of acceleration isthe foot per sec. per sec, — '—• sec.^ PROBLEMS. 1. Express the engineering unit of acceleration in terms of the C.G.S. unit. 2. Taking the value of the gravitational acceleration to be 980 • — '- > c J •* , . ft. , miles ^°- find its value m and -; — —■ sec.^ hr.-' 3. A train moving at the rate of 30 kilometers per hour is brought to rest in two minutes. Find the average acceleration and express it in , - cm. ft. J km. terms of -< ; and r^- sec.^ sec.^ hr.^ 88. Components of Acceleration along Rectangular Axes. — Suppose a particle to describe a path in the xy-plane. Then if vi and V2 be the velocities at two neighboring points, we can write dv = V2 — vi = (x2+ y2)-(xi+ yi) = dx+dy. , dv dx I dy dt dt dt But since f = f ^ + f „, ^'^^'' dt^ dt 92 ANALYTICAL MECHANICS The last equation cannot be true unless , dx ^'~ dt and ^"^ ~dl' Therefore the component of the acceleration along a fixed Une equals the time rate of change of the component of the velocity along that line. It follows from the last two equations that : f _dx _ d^ _ ■■ ^^~ dt~ dt ~^' f -dy_d^_.. ^'~ dt" dt ^" (IX) The magnitude and the direction of the acceleration are given by the following equations: /= VWW, (X) tan6>=^, (XI> X where is the angle f makes with the a;-axis. 89. Tangential and Normal Components of Acceleration. — The tangential component of the acceleration at P (Fig. 57) equals the rate at which the velocity increases along the di- rection of the tangent at P. In order to find this rate we consider the velocities at two neighboring points Pi and Pa. Let vi and V2 be the velocities at these points and ei and e^ the angles which vi and V2 make with the tangent at P. Then the change in the velocity along the tangent at P, while the particle moves from Pi to P^, is V2 COS 62 — Vi cos ei. Dividing this by the corresponding interval of time we ob- tain the average rate at which the velocity increases from. MOTION 93 Pi to Pi along the tangent at P. Therefore the average tangential acceleration is VjGOS €2 — Vi COS ei /.= t This average approaches the actual tangential acceleration at P as Pi and P2 are made to approach P as a limit. But N Fig. 57. as these points approach P the angles ei and 62 approach zero as a limit and their cosines approach unity, the tangential acceleration at P is Therefore /'='S'[*f^] dv d^ dt^ s. By similar reasoning we obtain I- Vj sm €2 — V\ sm ei t for the average normal acceleration between Pi and P2. The actual normal acceleration at P is the limiting value of this expression as Pi and P2 approach P. But as these points approach P, Vi and Vi approach v, the velocity at P, while 94 ANALYTICAL MECHANICS sin €i and sin a approach ei and €2,* respectively. Therefore the normal acceleration at P is = limit — V- = -^^=-^^. (XII) where fl = «i+e2 is the total change in the direction of the velocity in going from Pi to P2. Since the direction of the velocity coincides with that of the tangent, d is the rate at which the directions of the tangent and the normal change. But the rate at which the normal changes its direction equals the angular velocity of the particle about the center of curvature. Therefore if p denotes the radius of curvature at P, we have ! = ,-■ [byvii] and ^"^~T ^^^^^^ The negative sign in (XIII) shows that/„ and p are measured in opposite directions. Since p is measured from the center of curvature, /„ must be directed towards the center of cur- vature. Therefore the total acceleration is always directed towards the concave side of the path. The following are the principal results obtained in this section and the conclusions to be drawn from them. (a) The magnitude of the tangential acceleration is v; fr=V. (b) The normal acceleration is directed towards the center of curvature and has — for its magnitude,- J" p * See Appendix Avi. MOTION 95 (c) The magnitude of the total acceleration is given by / t^ the relation / = y {i^ + _. (d) The total acceleration is directed towards the concave side of the path and makes an angle with the tangent which is defined by tan^ = ^ = --,. , (e) When the path is straight, that is, when p = oo , the normal acceleration is nil; therefore in this case the total acceleration is identical with the tangential acceleration. (/) When the path is circular and the speed constant, then p = r, the radius of the circle, and v = 0; therefore 90. Radial and Transverse Components of Acceleration. — Let P (Fig. 58) be any point of the path at which the acceleration of the particle is to be considered. Take two neighboring points Pi and Pi, and let vi and va be the veloci- ties at these points. Then the change in the radial velocity in going from Pi to P2 is obtained by subtracting the radial component of vi from that of va. Replace vi and V2 by their components along and at right angles to n and rj, respectively, and denote these components by Vr„ Vp, and Vrj, Vp^; then it will be seen from the figure that {Vr^ cos 62 - Vp^ sin ea) - (Vr, cos ei + Vp^ sin ei) is the total change in the radial velocity. Therefore the radial component of the acceleration is f _y ■ , pr, cos €2 — Vp, sin 62 — Vt, cos ei — Vp, sin ei "| ' (=0 L t J where t is the time taken by the particle to go from Pi to P2. But as the points Pi and P2 approach P as a limit, the follow- ing substitutions become permissible. 96 ANALYTICAL MECHANICS COS «i = COS 62 = 1*, sin ei = ei, sin 62 = 62. Vr, - Vry = AVr, Vp^ = Vp, = Vp, €1 + 62 = dd. Making these substitutions in the expression for/r, we obtain /.= limitr ^'^--'^-'^7^''^^^ + ^'^1 (=0 L t J _dvr _ dd " dt ~^^dt dv_ /dey (XIV) df \dtl' where 6 is the angle r makes with the a;-axis. Fig. 58. By similar reasoning we obtain the following expressions for the transverse acceleration, that is, the component of the acceleration along a perpendicular to the radius vector. * See Appendix Avi. MOTION 97 J = limit P""' ^^ *' "^ '^''' ^"^ ^ ~ ( ~ '^'•- ^^'^ ^1 +^'" ^"^ ^i) ] " «=o L < J = limit h^^^ + ^^^+^'^^'-'^^'^l _ ds dvp ~'''dt'^ dt dr . d , . = ''dt+dt^'''^ = J|(rM, (XV) where we have du> "^'di^'' -df^-^- (XVI) If the angular velocity of a body increases uniformly '- in one second the body is said to have a unit angular .1 sec. 98 ANALYTICAL MECHANICS acceleration. Therefore the unit is the — -^. The dimen- sec. sions of angular acceleration are given by [T"^]. ILLUSTRATIVE EXAMPLE. A particle moves so that the coordinates of its position at any instant are given by the equations X = a cos kt, y = a sin kt. Find the acceleration and its components. In a previous illustrative example, p. 82, it was shown that these equations represent uniform circular motion, with the following data: V — ka,
  • R Fig. 66. Therefore the total velocity at any instant is v= VxM-^ = '^vo^ —2vogs,m.a.-t + gVl and makes an angle with the horizon defined by X Vo sm a gt (5) (6) Integrating equations (3) and (4) we obtain X= i;oCOS a •<+ C3, and y= vosin a 't — ^gP+Ci. But when t= 0, x = y= 0, therefore 03= Ci = 0, and con- sequently a; = j'o cos a • t, (7) y= Vosina-t—i gt\ (8) 122 ANALYTICAL MECHANICS It is interesting to note that the motions in the two directions are independent. The gravitational acceleration does not affect the constant velocity along the x-axis, while the mo- tion along the 2/-axis is the same as if the body were projected vertically with a velocity Vq sin a. The projectile virtually rises a distance of Vot sin a on account of its initial vertical velocity, and falls a distance J gt^ on account of the gravita- tional acceleration. The Path. — The equation of the path may be obtained by eliminating t between equations (7) and (8). This gives y = xtsina--—r^—r-x', (9) 2 2;o^COS^a which is the equation of a parabola. The Time of Flight. — When the projectile strikes the ground its y-coordinate is zero. Therefore substituting zero ior y in equation (8) we get for the time of flight y _ 2 t>o sin « _ ,jQ, g ' The Range. — The range, or the total horizontal distance tjovered by the projectile, is found by replacing t in equation (7) by the value of T, or by letting y = in equation (9). By -either method we obtain „ 2 Vq'^ sin a cos a K = g Vq'^ sin 2 a g (11) "Since Vq and g are constants the value of R depends upon a. It is evident from equation (11) that R is maximum when sin 2 a = 1, or when « = -. The maximum range is, there- fore, ^•» = 7- (12) MOTION OF A PARTICLE 123 In actual practice the angle of elevation which gives the maximum range is smaller on account of the resistance of the air. The Highest Point. — At the highest point y = 0. There- fore substituting this value of y in equation (4) we obtain — or -T for the time taken to reach the highest point. Substituting this value of the time in equation (8) we get for the maximum elevation H = Vo^ sin^ a. 2g (13) The Range for a Sloping Ground. — Let /3 be the angle which the ground makes with the horizon. Then the range is the distance OP, Fig. 67, where P is the point where the projectile strikes the sloping ground. The equa- tion of the line OP is y = x tan /3. (14) T Fig. 67. Eliminating y between equations (14) and (9) we obtain the a;-co6rdinate of the point, 2 Vq^ cos^ a (tan a — tan p) But Therefore 9 Xp = R' cos ft where R' = OP. „, 2 Vo'^ cos a . . -V R = —^ TT- Sm (a-0) g cos^ |3 _ ;;o^ sin (2 a — /3) — sin j3 ~ g cos^/S (15) 124 ANALYTICAL MECHANICS Thus for a given value of /3, B' is maximum when sin (2 a -$) = 1, that is, when a = - +-: ' 4 2 "• g cos^iS ^(1 + sin/3) ^ When |3 = equations (15) and (16) reduce to equations (12) and (13), as they should. PROBLEMS. ft 1. A body is projected horizontally with a speed of 105 — ^from a cliff S6C. 365 feet high. Find the magnitude and direction of the velocity at the time it reaches the ground. ft 2. The muzzle velocity of a gun is 3000 — '-. Find the area it covers S6C. if it is mounted on top of a hill 500 feet above the surrounding plain. 3. A shot fired horizontally from the top of a tower strikes the ground at a distance d from the base of the tower, with a velocity the vertical component of which equals the initial velocity of the shot. Find the height of the tower. 4. A buUet is projected at an angular elevation of 45° with a velocity of 400 ^^ . At the highest point of its flight the buUet goes through a sec. target 6 cm. thick and strikes the ground at a distance of 1200 m. from the place where it was projected. Find the average resisting force offered by the target. 6. After shding 200 m. down a slope of 30° a ski-jumper leaves the ground making 45° with the horizon and lands further down the same sloping ground. Supposing the coefficient of friction to be 0.05 and neglecting the resistance due to air, find (a) the speed with which he left the ground, (b) the speed with which he landed. 6. In the preceding problem find the leap measured along the ground. 7. A man can make 6 feet in the high jump. How many feet could he make if he were on the moon? The gravitational acceleration on the moon • KQ ft. IS 5.3 — - . sec .2 8. A particle sUdes down a chord of a vertical circle and then falls on a horizontal plane h feet below the lower end of the chord. Find the chord which will give the greatest possible range on the plane. MOTION OF A PARTICLE 125 9. Show that a rifle wiU shoot three tunes as high when its angle of elevation is 60° as when it is 30°, but will carry the same distance along a horizontal plane. 10. An emery wheel bursts into small pieces while making 100 revo- lutions per second. If the radius of the wheel is 10 cm. find farthest dis- tance reached by any of its pieces. MOTION OF A PARTICLE UNDER THE ACTION OF A VARY- ING FORCE. 110. I. Uniform Circular Motion. — Consider the motion of a particle projected into a circular tube, Fig. 68, the inner surface of which is perfectly smooth. Let to be the mass of the particle, v its speed of pro- jection and r the radius of the circle formed by the tube. The radius of the cross-section of the tube is supposed to be neg- ligible. Suppose the particle to be acted upon by no forces ex- cept the reaction of the inner surface of the tube. Then, since the surface is smooth, the reac- tion is normal and consequently there is no force along the tangent to the path. Therefore the force equations for the tangential and the normal directions give m- = 0, (1> Fig. 68. and m— = — N, r (2> where N represents the reaction of the walls of the tube upon the particle. It is evident from equation (1) that the mag- nitude of the velocity is constant. Therefore the particle describes the circular path with a constant speed. Equation (2) states that the normal reaction of the surface equals the 126 ANALYTICAL MECHANICS normal kinetic reaction and is oppositely directed. There- - fore the reaction of the surface is directed towards the center of the circular path. Equations (1) and (2) are independent of the special method used to keep the particle in a circular path. If, for instance, the particle were connected to the center by means of an iaextensible string and then projected in a direction perpendicular to the string the results would have been the same. The force which constrains the particle to move ia the circular path is called the central force. This force may be the reaction of a surface, the tension of a string, or the pull of a center of attraction. In order to emphasize the fact that this force is directed towards the center it is often denoted by Fc. Since the subscript makes clear the fact that the force is directed towards the center, we can drop the negative sign from equation (2), and write ^c=^ (3) r = -^ ' (4) where P is the time of one revolution. PROBLEMS. 1. A particle of mass mi, which describes a circle on a perfectly smooth horizontal table, is connected with another particle of mass rrit which hangs freely; the string which connects the two particles passes through a smooth hole in the table. Find the condition necessary to keep m^ at rest. 2. Find the smallest horizontal velocity with which a body must be projected at the equator in order that the body may become a satellite. Find the period of revolution. 3. A locomotive weighing 125 tons moves in a curve of 600 feet radius, with a velocity of 20 miles per hour. Find the lateral pressure on the rails if they are on the same level. 4. Derive the expression for the period of a conical pendulum. 5. A number of particles of different masses are suspended from the same point by means of strings of different lengths. Show that when MOTION OF A PARTICLE 127 the bodies are given the same angular velocity about a vertical axis through the point of suspension the particles will lie in the same hori- zontal plane. 6. If the masses in the preceding problem are equal how will the tensile force vary with the length of the strings? 7. Supposing the earth to be spherical discuss the variation in the weight of a body due to the rotation of the earth about its axis. 8. The moon describes a circular path around the earth once in every 27 days, 7 hours, and 43 minutes; find the acceleration at the center of the moon due to the attraction of the earth. Take 240,000 miles for the radius of the moon's orbit. 9. If the earth rotated fast enough to make the weights of bodies vanish at the equator show that the plumb Une at any latitude would become parallel to the axis of the earth. 10. In the preceding problem what would the length of the day be? 11. How much would the weight of a body be increased at latitude 30° if the earth stopped rotatiilg? 12. A particle suspended from a fixed point by a string of length I is projected horizontally with a speed Vllg; show that the string wUl become slack when the particle has risen to a height f I. 13. How much should the outer rail of a railroad track be raised at a curve in order that there be no lateral pressure on the rails when a train makes the curve at the rate of a mile a minute? The radius of the curve is 1500 feet and the distance between the tracks is 4 feet SJ inches. 14. Prove that a locomotive will upset if it takes a curve with a speed greater thani/^, on tracks the outer rails of which are not raised, where g denotes the gravitational acceleration, r the radius of the curve, a the distance between the rails, and h the height of the center of mass of the locomotive above the tracks. 16. Show that if there is no lateral pressure on the outer rails, while a car takes a curve, the relation tan 9 = — gr is satisfied, where 9 is the angle the floor of the car makes with the hori- zon, V is the speed of the car, and r the radius of the curve. 111. n. Bodies Falling from Great Distances. — When the distance from which a body falls is not negligible compared with the radius of the earth the gravitational acceleration cannot be considered as constant during the fall. Therefore 128 ANALYTICAL MECHANICS the variation of the gravitational attraction must be taken, into account. According to the law of gravitational attrac- tion the force between two gravitating spherical bodies is of the following form: P- -^"TT' (1) where m and m' are the masses of the spheres, r is the dis- tance between their centers, and 7 is a positive constant. The negative sign indicates the fact that r is measured in a direction opposed to that in which F acts. When the grav- itating bodies are the earth and a body which is small com- pared with the earth 7 = ^ , where M is the mass of the ^ M earth, a its radius, and g the gravitational acceleration on the surface of the earth. In order to show this observe that when the body is on the surface of the earth, that is, when r = a, the force is —mg, the weight of the body. Therefore replacing in equation (1) F by —mg and to' by M and solving for 7 we obtain 7 = ^- (2) ^ M ^ ' Substituting in equation (1) this value of 7 we get ^=-™^ (3) for the force which acts upon a body of mass m during its fall towards the earth. Therefore the force equation is Dropping to from both sides of equation (4) and writing v — dr for — we obtain at dv qa^ ;— = — 2 — . dr T^ V^=-'-r- (5) MOTION OF A PARTICLE 129 ■Separating the variables and integrating we have v^ = -^^ + c. Now suppose the body starts to fall from a distance r' from the center, then v=0 when r=r' and c = — ^. Therefore r ^^=2?a^(;-^^) (6) ^ives the velocity at a distance r from the center. When the body falls from an infinite distance r' = (Xi and the Telocity at any distance is v^ = a\l^. (7) Therefore the velocity with which it will reach the surface of the earth is 1^00= ^2ga „ miles 7- (8) sec. If the body starts to fall from a distance above the surface equal to the radius of the earth, then in equation (6) r' = 2r. Therefore , miles = 4.95- sec. Therefore about seventy-one per cent of the velocity attained in falling from an infinite distance is developed in the last 4000 miles. PROBLEMS. 1. A meteorite falls to the earth. Supposing it to start from infinity find the time it takes to travel the last 4000 miles. 2. A particle is attracted towards a fixed point by a force which varies inversely as the cube of the distance of the particle from the fixed point. Find the time it wiU take the particle to fall to the point if it starts from ■& distance d. 130 ANALYTICAL MECHANICS 3. Discuss the motions of a particle which is repelled from a fixed point with a force which varies directly as the distance of the particle from the fixed point. 112. III. Motion of a Particle in a Resisting Medium. — As a concrete example of motion in a resisting medium consider the motion of falling bodies, taking the resistance of the atmosphere into account. At any instant of the motion two forces act on the body, i.e., the weight of the body and the resistance of the air. Denoting the resisting force by F we get m— = mg-F (1) for the force equation. In order to be able to integrate the last equation we must make an assumption as to the nature of F. Case I. Resistance Proportional to the Velocity. — Suppose F to be proportional to the velocity, then F = kiv, where ki is a positive constant. Substituting in the force equation this expression for F we obtain dv , m — = mg- kiVi, dv , g-kv, (2) or ki where k = ~. Rearranging the last equation m Integrating dt pi dv = -kdt. log(.-l) kt+c, or v — ? = e" . e"*'. MOTION OF A PARTICLE 131 Let V = Vo when i = 0, then e° = fo — f • Therefore V — 9 or (--!)-' -!)• v=l + {vo (3) Limiting Velocity. — The last equation has a simple in- terpretation which comes out clearly by plotting the time as abscissa and the velocity as ordinate. There are four special cases which depend upon the following values of the initial velocity : (a) Vo= 0, (c).o=f. (b) vo < |, (d) .o>f. Curves (a), (b), (c), and (d) of Fig. 69 represent these cases. It is evident from these curves that whatever its initial value the velocity tends to the same limiting value |, called the Km- iting velocity. In the third case the velocity remains constant, as shown by the horizontal line (c), because the resisting force ex- actly balances the moving force. Integrating equation (3) we get vpk-g , k" k' Fig. 69. S=U '■ + c. Let s = when t = 0, then c = vpk-g Therefore -1^+^ (1- (4) 132 ANALYTICAL MECHANICS If we plot the last equation for the four different cases of the initial velocity we obtain the curves of Fig. 70. It is evident from these curves that a very short time after the beginning of the motion the distance covered increases at a constant rate, as would be ex- pected from the meaning of the lim- iting velocity. Case II. Resistance Propok- TIONAL TO THE SQUARE OF THE VE- LOCITY. — The assumption that resist- a,nce varies as the velocity holds only for slowly moving bodies. It is found that for projectiles whose velocities He under 1000 feet per second and over 1500 feet per second the resistance varies, approxi- naately, as the square of the velocity, while between these values it varies as the cube and even higher powers of the velocity. The experimental data on the subject are not enough to find a law of variation which holds in all cases. If we assume the resistance to vary as the square of the velocity, then the force equation for a falling body becomes Fig. 70. TO — =mg— kiv^, or dv Jt = g- kv^, (5) h where k= — = constant. In order to integrate the last equa- TO tion we replace ;^ by y-r- and rearrange the terms so that we get vdv = — kds. MOTION OF A PARTICLE 133 Integrating we have or 1)2 _ f = e<= . e-2 *», k Let V = Vo when s = 0, then e' = v^^ — ^- Therefore a; Therefore the limiting velocity isy f. In other words, for large values of s the distance traversed approximately equals ^i' PROBLEMS. 1. A man finds that the resistance of the air to a body moving at the rate of 20 -. — '■ equals 1000 dynes per square centimeter of the resisting sur- hr. face. If 600 — '■ is the hmit of the velocity with which he can safely land, find the smallest parachute with which he can safely descend from any height. The man and his parachute have a mass of 75 kg. Take the re- sistance to be proportional to the velocity. 2. In the preceding problem take the resistance to be proportional to the square of the velocity. 3. Discuss the equation of motion of a boat in still water, after the man who was rowing ships his oars. Suppose the resistance to be pro- portional to the velocity. 4. A particle is projected with a velocity v in a resisting medium and is acted upon by no other force except that due to the resistance of the me- dium. Show that, (a) the particle will describe a finite distance in an in- finite time when the resistance is proportional to the velocity; (b) it will describe infinite distance in infinite time when the resistance is propor- tional to the square of the velocity. 6. A bullet is projected vertically upwards with a velocity Wo- Sup- posing the resistance of the air to be proportional to the square of the 134 ANALYTICAL MECHANICS velocity of the bullet, find the expression for the highest point reached. Also find the time of upward flight. 6. In the preceding problem suppose the resistance to be proportional to the velocity. 113. IV. Simple Harmonic Motion. — The motion of a par- ticle is called simple harmonic when the particle is acted upon by a force which is always directed towards a fixed point and the magnitude of which is proportional to the distance of the particle from the same point. Fig. 71. Let m be the mass of the particle, the Hne A A' its path. Fig. 71, and the fixed point. Then denoting the dis- placement, i.e., the distance of the particle from the fixed point, by x we obtain F=- k% dv or TO dt — k^x, (1) for the force equation. It is evident that equation (1) is nothing more or less than the analytical expression for the foregoing definition of simple harmonic motion. The fac- tor k^ is a constant. The negative sign in equation (1) indicates the fact that the force and the acceleration are directed towards the fixed point, while x is measured from MOTION OF A PARTICLE 135 it. Since the motion is along the x-axis the velocity has doc no components along the other axes, consequently ^= -yr- Therefore equation (1) may be written in the form (2) KM Ju 70 or -372 = - w% d^ where u^ = — • Multiplying both sides of equation (2) by dx ^-rrdt and integrating (dxV _ \dt) - or v= Vc^ — coV, where c^ is the constant of integration. Let v =Vo when x = 0, then c" = Vq^. Therefore v= Vvo^-coV- (3> In order to find the second integral of equation (2) rewrite, equation (3) in the form at Separating the variables in the last equation and integrating we have sm 1 — = ci)i + c , or x= — sin (wi + c') Ct) = a sin (wf + c'), where c' is the constant of integration and a=— . Let CO x=0 when t=Q, then c' = 0, and a; = a sin ut. (4) 136 ANALYTICAL MECHANICS When equation (4) is plotted with the time as abscissa and the displacement as ordinate the well-known sine curve is obtained, Fig. 71. It is evident both from equation (4) and from the curve that the maximum value of x is equal to a. This value of the dis- placement is called the amplitude. The minimum value of a; is a displacement equal to the ampHtude in the negative direction. Therefore the particle oscillates between the points A and A'. The displacement equals the positive value of the amplitude every time sin cot equals unity, that is, when wt assumes the values -, — , -~, etc. In other Ji Ji Ji words, the particle occupies the extreme point A at the instants when t has the values-—, — ^, — ^, etc. Therefore 2 a) 2 a) 2 a) the particle returns to the same point after a lapse of time 2 T «equal to — . This interval of time is called the period of O) the motion and is denoted by P. Thus P = — • (5) CO PROBLEMS. 1. A particle which moves in a straight groove is acted upon by a force ■which is directed towards a fixed point outside the groove, and which varies as the distance of the particle from the fixed point. Show that the motion is harmonic. 2. Within the earth the gravitational attraction varies as the dis- tance from the center. Find the greatest value of the velocity which a .body would attain in falling into a hole, the bottom of which is at the >center of the earth. 3. Show that when a particle describes a uniform circular motion, its projection upon a diameter describes a harmonic motion. GENERAL PROBLEMS. 1. The speed of a train which moves with constant acceleration is -doubled in a distance of 3 kilometers. It travels the next l-h kilometers in one minute. Find the acceleration and the initial velocity. MOTION OF A PARTICLE 137 2. Show that the ratio of the distances described by a falling body during the (n — l)th and the nth seconds is - — ^ • 2n+ 1 3. A juggler keeps three balls going in one hand, so that at any in- stant two are in the air and one in his hand. Find the time during which, a ball stays in his hand; each ball rises to a height h. 4. Find the shortest time in which a mass m can be raised to a height k by means of a rope which can bear a tension T. 5. A train passes another on a parallel track. When the two loco- motives are abreast one of the trains has a velocity of 20 miles per hour ft and an acceleration of 3 — '- , while the other has a velocity of 40 miles sec.'' ft per hour and an acceleration of 1 — '— . How soon will they be abreast sec.'' again, and how far will they have gone in the meantime? 6. A mass of 1 kg. is hanging from a spring balance in an elevator. After the elevator starts the balance reads 1100 gm. Assuming the. acceleration of the elevator to be constant, find the distance moved in 5 seconds. 7. A smooth inclined plane of mass m and inclination a stands with its base on a smooth horizontal plane. What horizontal force must be applied to the plane in order that a particle placed on the plane simulta- neously with the beginning of action of the force may be in contact with the plane yet fall vertically down as if the inclined plane were not there?' 8. The puU of a train exceeds the resisting forces by 0.02 of the weight, of the train. When the brakes are on fuU the resisting forces equal 0.1 of the weight. Find the least time in which the train can travel between two stopping stations 5 rmles apart, the tracks being level. 9. Give a construction for finding the line of quickest descent from a point to a circle in the same vertical plane. 10. A mass Wi falling vertically draws a mass rrh up a smooth inchned plane which makes an angle of 30° with the horizon. The masses are connected with a string which passes over a small smooth puUey at the top of the plane. Find the ratio of the masses which will make the acceleration t ■ 4 11. A particle is projected up an inchned plane which makes an angle a with the horizon. If Ti is the time of ascent, T2 the time of descent,, and the angle of friction, show that (TiV ^ sin (a - ) \tJ sin(a:+<^)' 138 ANALYTICAL MECHANICS 12. The time of descent along straight lines from a point on a verti- cal circle to the center and to the lowest point is the same. Find the position of the point. 13. A uniform cord of mass m and length I passes over a smooth peg and hangs vertically. If it slides freely, show that the tension of the ^ X (1/ ~ xl ■ 1 cord equals ^ -, when the length on one side is x. Ir 14. In an Atwood's machine experiment the sum of the two moving masses is m. Find their values if in t seconds they move through a dis- tance h. 15. Given the height h of an incUned plane, show that its length must be — ^, in order that mi, descending vertically, shall draw m^ up the plane in the least possible time. 16. A gun points at a target suspended from a balloon. Show that if the target be dropped at the instant the gun is discharged, the bullet wiU hit the target if the latter is within the bullet's range. 17. Find the position where a particle sliding along the outside of a smooth vertical circle wiU leave the circle. 18. A particle falls towards a fixed point under the action of a force which equals •yr~^, where 7 is a constant and r is the distance of the parti- cle from the fixed point. Show that starting from a distance a the particle will arrive at the fixed point with an infinite velocity in the time — — . V37 19. A particle falls towards a fixed point under the attraction of a force which varies with some power of the distance of the particle from the cen- ter of attraction. Find the law of force, supposing the velocity acquired by the particle in falling from an infinite distance to a distance o from the center to be equal to the velocity acquired in falling from rest from a dis- tance o to a distance 7- 4 20. A particle is projected toward a center of attraction with a velocity equal to the velocity it would have acquired had it fallen from an infinite distance to the position of projection. Supposing the force of attraction to be yr^", where 7 is a constant and r is the distance of the particle from the center of attraction, show that the time taken to cover the distance between the point of projection and the center of attraction is n+lV 27 MOTION OF A PARTICLE 139 21. From the following data show that the velocity with which a body- has to be projected from the moon in order to reach the earth is about 1.5 miles per sec. Both the earth and the moon are supposed to be at rest. The mass of the moon is A of that of the earth. The radii of the earth and the moon are 4000 miles and 1100 miles, respectively. The distance between the earth and the moon is 240,000 miles. 22. Taking the data of the preceding problem, show that if the earth and the moon were reduced to rest they would meet, under their mutual attraction, in about 4.5 days. CHAPTER VII. CENTER OF MASS AND MOMENT OF INERTIA. CENTER OF MASS. There are two useful conceptions, known as center of mass and moment of inertia, which greatly simplify dis- cussions of the motion of rigid bodies. It is, therefore, desirable to become familiar with these conceptions before taking up the motion of rigid bodies. 114. Definition of Center of Mass. — The center of mass of a system of equal particles is their average position; in other words, it is that point whose distance from any fixed plane is the average of the distances of all the particles of the system. Let Xi, X2, Xt, . . . Xn denote the distances of the particles of a system from the 2/2-plane ; then, by the above definition, the distance of the center of mass from the same plane is -. Xi+X2+X3-\- ■ ■ ■ +X„ n = -2x. n When the particles have different masses their distances must be weighted, that is, the distance of each particle must be multipUed by the mass of the particle before taking the average. In this case the distance of the center of mass from the 2/2-plane is defined by the following equation : (mi+m2+ • • • + mj X = TOiXi + m2X2 + • • • +m„x„, 140 CENTER OP MASS AND MOMENT OF INERTIA 141 or similarly and -_ ^mx - _ 'Lmy ^~ Sto' - Hmz 2jm (I') Evidently x, y, and z are the coordinates of the center of mass. ILLUSTRATIVE EXAMPLES. 1. Find the center of mass of two particles of masses m and nm, which are separated by a distance a. Taking the origin, of the axes at the particle which has the mass m, Fig. 72, and taking as the a;-axis the line which joins the two particles we get + nma 2m X Fig. 72. 2. Find the center of mass of three particles of masses m, 2 m, and 3 m, which are at the vertices of an equilateral triangle of sides o. Choosing the axes as shown in Fig. 73 we have - ^ + 2 ?»a + 3 ?wa cos 60 ° m -\- 2 m + 3 m V = A a, + + 3 mo sin 60° 2=0. 6m 142 ANALYTICAL MECHANICS 115. Center of Mass of Continuous Bodies. — When the par- ticles form a continuous body we can replace the summation signs of equation (I') by integration signs and obtain the following expressions for the coordinates of the center of mass: I xdm* x = 2 = dm I y dm dm nm I z dm Jo dm (I) where m is the mass of the body. ILLUSTRATIVE EXAMPLES. 1. Find the center of mass of the parabolic lamina bounded by the curves y^ = 2px and x = a, Fig. 74. Obviously the center, of mass lies on the a;-axis. Therefore we need to Y Y Fig. 74. * In general if 2/ is a function of x then the average value of y between the limits xi and x, is given by the relation: y = — = — f'y dx. Xi — Xi Jxi -CENTER OF MASS AND MOMENT OF INERTIA 143 detennine x only. Taking a strip of width Ax for the element of mass we have dm = (r2ydx = 2 cr -\/2 px dx, where o- is the mass per miit area. Therefore substituting this expression of dm in equation (I) and changing the limits of integration we obtain J»ffi > x V 2 px dx X = °- 2(7 I V 2 pa; da; Jn ■ j^xidx x^dx ^ 3a 5 ' 2. Find the center of mass of the lamina bounded by the curves y^ = 4:ax and y = bx, Fig. 75. Let dx dy be the area of the element of mass, then dm = a dx dy. Therefore substituting in equation (I) and introducing the proper limits of integration we obtain 4£ 40 4Jr y ia Xx ^y^'^ Jo Jh. '^y^- 4a 40 r (2 Vm — bx) X dx C \2ax — — xAdx 40 4£ ~ C (2 Vox — 6x) ia; C (2 V^ — 6a;) dx 8a ^ 2a 562' &■ 144 ANALYTICAL MECHANICS 3. Find the center of mass of a semicircular lamina. Selecting the coordinates and the element of mass as shown in Fig. 7& we have din = (T'pdd-dp, Jj y 'crpdpdff -^0 2/ = Jf a-pdpdd rrp^smddpdB Jo Jo rcpdpdB Jo Jo 4a x = 0. PROBLEMS. Fmd the center of mass of the lamina boimded by the following curves : (1) y = mx, y = — mx, and y = a. (2) y = asmx,y = 0, x = 0, and x = w. (3) y^ = ax and x' = by. (4) x^ + 2/2 = o^, a; = 0, and 2/ = 0. (5) bV + ahf = aV, x = 0,a,ndy = 0. (6) r = a(l + cose). (7) r = a,e = 0, and 8 = 6. (8) r = a, r = b, = 0, and 6 =^- 116. Center of Mass of a Homogeneous Solid of Revolution. — Let Fig. 77 represent any solid obtained by revolving a plane curve about the x-axis. Then the center of mass lies on the axis of revolution. The position of the center of mass is found most conveniently when the element of mass is a thin slice obtained by two transverse sections. The expression for the mass of such an element is dm = T • iry'^ • dx, where t is the density of the solid, y the radius of the slice,, and dx its thickness. CENTER OF MASS AND MOMENT OF INERTIA 145 Y Fig. 77. ILLUSTRATIVE EXAMPLE Find the center of mass of a parab- y oloid of revolution obtained by revolv- ing about the a;-axis that part of the parabola y^ = 2px which lies between the lines x = and x = a. dm = Tiry^ dx = TTT 2 px dx; 2irTp ) x^dx — ^0 .*. X = 27rTp i xdx Jo = f a. PROBLEMS. Find the center of mass of the homogeneous solid of revolution gener- ated by revolving about the s-axis the area bounded by (1) y = -rx, X = h, and y = 0. h (2) x^ = iay, X = 0, and y = a. (3) a;' + 2/2 = a^, and a; = 0. (4) bV + a^ = a^b^, and a; = 0. (5) y = smx, X = 0, and x = -■ Li (6) x^^y'^ = a}, x'^ + y'^ = V, and a; = 0. Fig. 78. 146 ANALYTICAL MECHANICS 117. Center of Mass of Filaments. — The transverse dimen- sions of a filament are supposed to be negligible; therefore it can be treated as a geometrical curve. Taking a piece of length ds as the element of mass and denoting the mass per unit length by X we have dm = X ds. ILLUSTRATIVE EXAMPLE. Find the center of mass of a semicircular filament. (a) Taking x^ + y^ = a^ to be tlie equation of the circle we get dm = \ds = X = Xa Mt] dx dx Va' - z^ dx x' J^ dx x—== Va^ - ; dx Va^ - a;2 Vc? sm- 2a Fig. 79. (b) Referrmg the circle to polar coordinates we have r = a for its equation. Therefore dm = \ds = \adB. p x\add p acosOdd f \ade r 2a CENTER OF MASS AND MOMENT OF INERTIA 147 PROBLEMS. Find the center of mass of a uniform wire bent into the following curves: (1) An arc of a circle subtending an angle 2 9 at the center. (2) y = a sin x, between x = and x = tt. (3) 2/^ = 4 ax, between a; = and x = 2a. (4) The cycloid x = a (d — sin 6), 2/ = a (1 — cos 6), between two successive cusps. (5) Half of the cardioid r = a (1 + cos 6). 118. Center of Mass of a Body of Any Shape and Distribution of Mass. — The illustrative examples of the last few pages are worked out by special methods in order to bring out the fact that in a great number of problems the ease with which the center of mass may be determined depends upon the choice of the element of mass. The following general ex- pressions for an element of mass may be used whatever the shape of the body or the distribution of its mass : (a) When the bounding siirfaces of the body are given ia the Cartesian coordinates the mass of an infinitesimal cube is taken as the ele- ment of mass : dm = T ' dxdy dz. (b) When the bounding surfaces of the body are given in spherical coordi- nates the element of mass is chosen as shown in Fig. 80. In this case the fol- lowing is the expression for the element of mass : dm = T ■ p do • dp • p d4 sin 6 = Tp^ sin eded(j) dp. (c) When the density, t, varies from point to point it is expressed in terms of the coordinates and substituted in the expression for dm. Fig. 80. 148 ANALYTICAL MECHANICS ILLUSTRATIVE EXAMPLES. 1. Find the center of mass of an octant of a homogeneous sphere, (a) Suppose the bounding surfaces to be x^ + y^ + z^ = a^, X = 0, y = 0, and 2 = 0. Then the Umits of integration are X = and x = a, y = and y = Va^ — x'^, g = and z = ^/a' — x^ — y\ Therefore j ( xdxdydz I i dxdy dz a Jq J 3a and by symmetry y = z = -^• 8 (b) Suppose the equations of the bounding surfaces to be given in spheri- cal coordinates, then we have r = a, e = |, = 0, and = | • The limits of integration are r = and r = a, e = and 61 = -, 2 Fig. 81. <^ = and (^ = |. Therefore r C C r^ sva^e cos d>drdBdd> *^o Jq Jq TT IT J^ J^ ("r'' sin ddrdBd^ 3a [x = rsinOcosi^l. CENTER OF MASS AND MOMENT OF INERTIA 149 2. Find the center of mass of a right circular cone whose density varies inversely as the square of the distance from the apex, the distance being measured along the axis. dm = T ' iry^ • dx _ Ti aV = '!^!^dx 'dx where n is the density at a unit distance from the apex. There- fore Tjira' xdx TiTTO' /•* dx Fig. 82. PROBLEMS. 1. Find the center of mass of a right circular cone, the density of which varies inversely as the distance from, the vertex. 2. Find the center of mass of a circular plate, the density of which varies as the distance from a point on the circumference. 3. Find the center of mass of a cylinder, the density of which varies with the nth power of the distance from one base. 4. Find the center of mass of a quadrant of an ellipsoid. 6. Find the center of mass of a hemisphere, the density of which varies as the distance from the center. 119. Center of Mass of a Number of Bodies. — Let Wi, m2, etc., be the masses and Xi, x^, etc., be the a;-co6rdinates of the centers of mass of the individual bodies. Then if x denotes the x-co6rdinate of the center of mass of the entire system we can write xdm 150 ANALYTICAL MECHANICS xdm+ • • I xdm + dm+ I dm + Jo TOlXi + TO2X2 + • • • TOi + m2 + Sot Similarly 2 = Sot Therefore the mass of each body may be considered as being concentrated at its center of mass. ILLUSTRATIVE EXAMPLE. Find the center of mass of the plate indicated by the shaded part of Fig. 83. (a) Suppose the plate to be separated into two parts by the dotted line. Then the coordinates of the center of mass of the lower part are b J - b — a xi = - and 2/1 = — — On the other hand the coordinates of the center of mass of the upper part are X2 ■■ b — a , - 26 — a -r— and 2/2= — Therefore the coordinates of the center of mass of the entire plate have the following values : 6 6 , b — a rrii + nh (Tb{b-a)-+(7a{b-a) -~ ab {b—a)+a- {b—a) a 6" + 06 - g' 2 (a + 6) mi- tt , 2b — a : — \-m2 — —. — y=. mi + mi ab{b-a)^-^+Mb-a)^^ 0-6(6— o)+o-a(6— a) 6' + ab - g' 2 (a + 6) CENTER OF MASS AND MOMENT OF INERTIA 151 (b) Suppose the square OA to represent a plate of positive mass and the square O'A to represent a plate of negative mass. Then if the two plates have the same thickness and density the positive and the negative masses aimul each other in the square O'A. Therefore the two square plates form a system which is equivalent to the actual plate represented by the shaded area of the figure. Hence the center of mass of the square plates is also the center of mass of the given plate. The masses of the square plates are o-fe^ and — aa^, while the coordi- nates of their centers of mass are ^' = ^'=2 ™<^ x"=y"= — ^ Therefore the coordinates of the center of mass of the two are 2"*" ^~''"'^ — 2 — a; = « = _ 6' + a6 — o" 2 (a + 6) ' which are identical with those obtained by the first method. PROBLEMS. 1. Find the center of mass of the homogeneous plates indicated by the following figures: 2. A sphere of radius 6 has a spherical cavity of radius a. Find the center of mass if the distance between the centers is c. 3. A right cone is cut from a right circular cylinder of the same base and altitude. Find the center of mass of the remaining solid. 4. A right cone is cut from a hemisphere of the same base and altitude. Find the center of mass of the remaining solid. 5. A right circular cone is cut from another right circular cone of tha same base but of greater altitude. Find the center of mass of the remain- ing solid. 6. A right circular cone is cut from the paraboloid of revolution gener- 152 ANALYTICAL MECHANICS ated by revolving about the s-axis the area bounded by y^ = 2 px and X = a. Find the center of mass of the remaining solid if the parab- oloid and the cone have the same base and vertex. MOMENT OF INERTIA. 120. Definition of Moment of Inertia. — The moment of inertia of a body about an axis equals the sum of the products of the masses of the particles of the body by the square of their distances from the axis.* Thus if dm de- notes an element of mass of the body and r its distance from the axis then the following is the analytical state- ment of the definition of moment of inertia: r^ dm. (II) The integration which is involved in equation (II) is often simplified by a proper choice of the element of mass. The choice depends upon the bounding surfaces of the body and the position of the axis ; therefore there is no general rule by which the most convenient element of mass may be selected. There is one important point, however, which the student should always keep in mind in selecting the element of mass, namely, the distances of the various parts of the element of mass from the axis must not differ by more than infinitesimal lengths. ILLUSTRATIVE EXAMPLES. 1. Find the moment of inertia of a rectangular lamina about one of its sides. Y Suppose the lamina to lie in the xy- plane. Further suppose the side with respect to which the moment of inertia is to be found to lie in the a;-axis. Then the most convenient element of mass is dm dy FiQ. 84. X a strip which is parallel to the x-axis. ^ Let a be the length (Fig. 84), 6 the width, and *^2' ^^^ P = a- r^dm {p'^-p^siii^dcos^(l>)dm a TT IT = ^"^^P j P* (sin 9 -sio^e CQa^) dB d dp CENTER OF MASS AND MOMENT OF INERTIA 163 126. Routh's Rule. — The following is a useful rule for re- membering the moments of inertia of certain types of bodies : Moment of inertia with respect to any axis of symmetry _ sum of the squares of the perpendicular semi-axes 3, If., 5 The denominator of the right-hand member is 3, 4, or 5 ac- cording as the body is rectangular, elliptical, or ellipsoidal. The following illustrate Routh's rule. Rectangular lamina; about axis perpendicular to its plane: J 4"^4 a^ + V Circular lamina; about axis perpendicular to its plane: Elliptical lamina; about axis perpendicular to its plane: r g' + ft' 4 Rectangular parallelopiped; about axis perpendicular to one of its sides : r 4:'^ 4: a^ + 5' Circular cylinder; about longitudinal axis: Sphere; about a diameter: J a^ + g' 2 „ 1 = m — - — = - ma\ 5 5 Ellipsoid; about one of its axes: r a' + b' I = m — i — • CHAPTER VIII. WORK. 127. Work. — The mechanical result produced by the ac- tion of a force in displacing a particle may be considered to be proportional to the interval of time during which the force acts or to the distance through which it moves. In other words, we can take either the time or the displacement as the standard of measure. The effect measured when the time is taken as the standard is different from that which is obtained when the displacement is made the standard. The first effect is called impulse. It will be discussed in a later chapter. The second is called work, the subject of this chapter. 128. Measure of Work. — When a force moves a body it is said to do work. The amount of work done equals the product of the force by the distance through which the body is displaced along the line of action of the force. In this defi- nition the force is considered to be constant. When it is vari- able the definition holds for infinitesimal displacements, since during the time taken by an infinitesimal displacement the force may be considered as constant. Therefore if the par- ticle P, Fig. 92, is displaced through ds, imder the action of the force F, the work done is dW = F 'ds cos a, where a is the angle between the direc- tions of the force and the displacement. ^ When the displacement is finite the '*'' work done equals the sum of the amounts of work done in 164 WORK 165 successive infinitesimal displacements. Therefore the work done in any displacement is given by the integral W = I F cos a ds. (I) When the path of the particle is curved the direction of ds coincides with that of the tangent to the curve. Therefore F cos a is the tangential component of the force. In other words the tangential component of force does all the work. Hence W^f'Prds. (I') The normal component does no work because the particle is not displaced along it. Special Cases. Case I. — When the force is constant, both in direction and in magnitude, it can be taken out of the integrand. Therefore W = F j cos a ds. The last integral equals the projection of the path upon the direction of the force. Therefore the product of the force by the projection of the path upon the hne of action of the force equals the work done. Case II. — When the force is constant and the path is straight then the angle between the force and the displace- ment is constant. Therefore W= Fcosa \ ds = Fs cos a. Case III. — When the force is not only constant but is also parallel to the path, then a = 0. Therefore W = Fs. Case IV. — When the force is at right angles to the dis- placement a = ? , and cos a = 0. Hence W = 0. Therefore the force does no work unless it has a component along the 166 ANALYTICAL MECHANICS path. In this case the motion of the particle is not due to the force in question. 129. Work Done Against the Gravitational Force. — These special cases may be illustrated by considering the work done in raising a body from a lower to a higher level against the gravitational attraction of the earth. Consider the work done in taking a particle from A to B, along each of the three paths shown in Fig. 93. (a) Suppose the particle to be taken from A to C and then to B; the work done in taking it from A to C comes under Case IV. The direction of motion is at right angles to that of the gravitational force, therefore no work is done against it. The work done in taking the particle from C to B comes under Case III; the force and the motion are in the same direction. Therefore the work done is W = mgh, where h is the vertical height of B above A. (b) Suppose the particle to be taken along the straight line AB. This comes under Case II. The angle between the force and the direction of motion constant. Therefore W = mgl cos a, where I is the length of the line AB. But since I cos a = h, the work done is "" fiq. 93 the same as in (a), that is, mgh. (c) Suppose the particle to be taken along the curve AB. This comes under Case I. Then W = mg I cos a ds Jo = mgj' = mgh. % dh [since ds cos a = dh] ■■ WORK 167 Therefore the work done against the gravitational force in taking a body from one position to another depends only upon the vertical height through which the body is raised and not upon the path. 130. Dimensions and Units of Work. — Work is a scalar magnitude and has for its dimensions [ML^T~^]. The C.G.S. unit of work is the erg. It equals the work done by a force of one dyne in displacing a particle through a distance of one centimeter, measured along its line of action. It is firm cm symbolized by - — '- — —■ . The erg is a small unit, therefore sec.'' a larger unit called the joule is also used. 1 joule = 10^ ergs. The British unit of work is the foot-pound (ft.-lb.). It is, the work done against the gravitational attraction of the earth in lifting one poimd through a vertical distance of one foot. Since the work done in lifting bodies is mgh, we can express the foot-poimd in terms of the fundamental imits, thus 1 ft. lb. = 1 pd. X 32.2-^ X 1 ft. sec' sec' where pd. represents the pound-mass. 131. Work Done by Components of Force. — The work done by a force F in giving a particle a displacement ds is F cos d ds, where 6 is the angle between F and ds. Let X, Y, and Z be the rectangular components of F, then the direction of F X Y Z is defined by its direction cosines — , — , and -. Therefore p' p ' p if I, m, and n denote the direction cosines of ds, we get cosd=l- + m- +TO- * See Appendix Aiv. 168 ANALYTICAL MECHANICS and F cos eds= (IX + mY + nZ) ds = X dx + Y dy + Z dr, where dx, dy, and dz are the components of ds along the axes. Thus the total work done in a finite displacement is given by W = f F cose ds Jo = rxdx+ f''Ydy+ f'z dz. (II) Jo Jo Jo Equation (II) states that the work done by a force equals the sum of the amounts of work done by its components. PROBLEMS. 1. Find the number of foot-pounds in one Joule. :2. Find the number of ergs in one foot-pound. :3. Find the work done in dragging a weight w up an inclined plane ■-of length I, height h, and coefficient of friction fi. 4. A body of 100 kg. mass is dragged up, then down, an inclined plane. Compare the work done in the two cases if the length of the plane is 15 m., ±he height 5 m., and the coefficient of friction 0.5. 5. What is the work done in winding a uniform chain which hangs from .•a horizontal cylinder? The chain is 25 m. long, and has a mass of 125 kg. 6. A body has to be dragged from a point at the base of a conical hill to a point diametrically opposite. Show that, if the angle which the sides of the hill make with the horizon equals the angle of friction, the work done in dragging the body over the hill is less than in dragging it around the base. 7. A steam hammer falls vertically from a height of 3 feet under the action of its own weight and of a force of 2000 pounds due to steam pres- sure. At the end of its fall it makes a dent of 1 inch depth in an iron plate. Find the total amount of work done in making the dent. The hammer weighs 1000 pounds. 8. In the preceding problem find the average resisting force. 9. A locomotive which is capable of exerting a draw-bar pull of 1.5 tons is coupled to a train of six cars. The locomotive and the tender weigh 50 tons. The cars weigh 15 tons each. Find the time it takes the loco- motive to impart to the train a velocity of 60 miles per hour and the work done under the following conditions. WORK 169 (a) Horizontal tracks and no resistance. (b) Horizontal tracks and a resistance of 12 pounds per ton. (c) Down a grade of 1 in 200 with no resistance. (d) Same as in (c) but with a resistance of 12 pounds per ton. (e) Up a grade of 1 in 200 with no resistance. (f) Same as in (b) but with a resistance of 12 pounds per ton. 10. A mass of 5 pds. is at the bottom of a vertical shaft which reaches the center of the earth. How much work will have to be done in order to bring it to the surface? The weight of a body varies, within the earth, directly as its distance from the center. Take 4000 miles to be the depth of the shaft. 11. Express the result of the last problem in joules. 132. Work Done by a Torque. — Suppose the rigid body A, Fig. 94, to be given an angular displacement dd about a fixed axis through the point 0, per- pendicular to the plane of the paper. The displacement may- be considered to be] due to a single force which forms a couple with the reaction of the axis, or it may be considered to be due to small forces acting upon every element of mass of the body. Taking the latter view, let dF be the resultant force* acting upon the element of mass dm. Then since dm can move only at right angles to the line r, which joins it to the axis, dF must be perpendicular to r. When the body is given an angular displacement dd, dm is displaced through ds, therefore the work done by dF is d^W = dF -ds = dF -rde = dG de, * d F is the resultant of the external forces which act directly on dm, and of the internal forces which are due to the connection of dm with the rest of the body. Fig. 94. 170 ANALYTICAL MECHANICS where dG is the moment of dF about the axis. Thus the total work done by all the forces acting upon all the elements in producing the angular displacement de is dW= f dG de -*x ff {de is the same for every element of mass.) = Gde, where G is the sum of the moments about the axis of all the forces acting upon the elements of the body, i.e., the result- ant torque about the axis. The work done in giving the body a finite angular displacement is, therefore, W= f^Gde (III) Jo = Gd [when G is constant]. Therefore work done by a constant torque in producing an angular displacement equals the product of the torque by the angle. PROBLEMS. 1. A weight of 10 tons is to be raised by a jackscrew. The pitch of the screw is i inch and the length of the bar which is used to turn the nut on the screw is 2 feet long. Supposing the work done by the torque to be expended entirely against gravitational forces, find the force which must be applied at the end of the bar. 2. A ball, which is suspended by a string of negligible mass, is pulled aside until the string makes an angle B with a vertical line. Show that the work done is the same whether it is supposed to have been done in raising the ball against the action of gravitational forces, or in rotating the ball and the string, as a whole, about a horizontal axis through the point of suspension, against the action- of the torque. 3. In the preceding problem take the following data and calculate, by both methods, the amount of work done. Weight of ball = 12 ounces, length of string = 3 feet, and 6 = 60°. 4. The torque which has to be applied to the ends of a rod varies directly with the angle through which it is twisted; derive an expression for the work done in turning one end of the rod with respect to the other end through an angle 0. WORK 171 6. In the preceding problem suppose one end of the rod to be fixed, while the other end is firmly attached to the middle of another rod perpen- dicular to it. A torqne of 10 pounds-foot is necessary in order to keep the second rod in a position turned through 15° about the axis of the first rod. How much work must be done in order to produce an angular deflection of 45° ? 6. If in problem 5 the torque is due to a couple the forces of which are applied at points 4 inches from the axis of rotation, find the forces appUed and show that the work done by the forces equals the work done by the torque. 7. Making the following assumption with regard to the normal pressure at the bearings, obtain an expression for the work done in giving a fly- wheel an angular displacement. (Hint. — For this and the following problems consult §§57 and 58.) (a) Normal pressure is constant. (b) Vertical component of the total reaction is constant. (c) Normal pressure is a sine function of the angular position; the latter being measured from the horizontal plane through the axis of the shaft. (d) Normal pressure varies as the square of the sine of the angular posi- tion. 8. Find the work done in giving a S.ywheel a complete rotation. The following data are given. The flywheel weighs 5 tons, the diameter of the shaft is 10 inches, the coefficient of friction in the journal bearings is 0.05, and the normal pressure on the bearings satisfies one of the foUow- ing conditions: (a) Normal pressure is constant. (b) Vertical component of the total reaction is constant. (c) Normal pressure varies as the sine of the angular position, measured from the horizontal plane through the axis of the shaft. (d) Normal pressure varies as the square of the angular position. 9. Derive an expression for the work done in giving an angular dis- placement to a load which is supported by a flat-end pivot. 10. The rotating parts of a water turbine which weigh 50 tons are sup- ported by a flat-end pivot. The diameter of the shaft is 10 inches and the coefficient of friction is 0.03. Find the work lost per revolution. 11. Supposing the normal pressure to be constant derive an expression for the work done in giving a loaded spherical pivot an angular displace- ment about its axis. 12. Supposing the normal pressure to be constant derive an expression for the work done in giving a .loaded conical pivot an angular displacement. 172 ANALYTICAL MECHANICS 13. A vertical shaft carries a load of 10 tons. Find the work lost per revolution if the shaft is 8 inches in diameter and has a flat-end bearing; the coefficient of friction being 0.01. 14. Derive an expression for the work done in giving a collar-bearing pivot an angular displacement. 15. A vertical shaft carries a load of 5 tons. Find the work lost per revolution if the shaft is supported by a coUar-bearing pivot which has an inner diameter of 6 inches and an outer diameter of 8 inches. The coefficient of friction is 0.1. HOOKE'S LAW. 133. Stress. — When a body is acted upon by external forces which tend to change its shape and thus give rise to forces between its contiguous elements, the body is said to be under stress. The measure of stress is force per unit area: S=j> (IV) where ;S denotes the stress, F the force, and A the area on which the latter acts. 134. Pressure, Tension, and Shear. — Stresses which occur in bodies are often of a complex nature, but they may be resolved into three component stresses of simple character. These are called pressure, tension, and shear. Pressure tends to compress, tension to extend, and shear to distort bodies. Shearing stress is the result of a compressive stress com- bined with a tensile stress at right angles. A special case of shear, which comes into play within a shaft when the latter is twisted, is called torsion. 135. Strain. — Strain is the deformation produced by stress. The measure of strain is the percentage deformation. For instance, if the deformation consists of a change in length the strain equals the ratio of the increase in length, to the original length : / where s denotes the stress, I the increase in length, and L the original length. WORK 173 136. Hooke's Law. — The relation which connects a stress with the strain which it produces is known as Hooke's law. It states that stress is proportional to strain: S=\s, (VI) where X is the constant of proportionaUty, and is called the modulus of elasticity. 137. Elastic Limit. — Hooke's law holds true so long as stress is small enough to leave no appreciable permanent def- ormation. In other words, Hooke's law holds true strictly only while the body under consideration behaves like a per- fectly elastic body under the action of the given stresses. All bodies are more or less imperfectly elastic ; that is, stresses always leave bodies with permanent strains. Therefore at the best Hooke's law is approximately true when applied to material bodies. The approximation, however, is close enough for practical purposes so long as the permanent def- ormation is negligible compared with the total deformation produced by the stress. If a considerable portion of the deformation becomes permanent the body under stress is said to have reached its elastic limit, when Hooke's law does not give a close enough approximation and consequently cannot be used. 138. Young's Modulus. — The modulus of elasticity of a body which is being stretched is called Young's modulus. Let the body be an elastic string, a wire, or a rod, and let A be the area of its cross-section, L its natural or unstretched length, and I the increase in length due to stretching. Then we have Therefore or S-- = — and A I '=1 F A '' -A' F L \ = ^ A'l' 174 ANALYTICAL MECHANICS Thus Young's modulus of a substance equals, numerically, the force necessary to stretch a uniform rod of unit cross- section, which is made of the given substance, to double its length. During the process of stretching Hooke's law is, of course, supposed to hold. 139. Work Done in Stretching an Elastic String, — Let L denote the natural length of the string and x its length at any instant of the process of stretching. Then the work done in increasing the length by dx is dW = Tdx = AS dx, where T is the tensile force, S the tension, and A the area of the cross-section of the string. But by Hooke's law, S=\s. In this case s = X— L Therefore dW=A-\ X- ' -dx and W = j- / {x-L)dx — A_72 ~2L^' Fig. 95, where X' = A\, and I is the total in- crease in length. Thus the work done varies as the square of the increase in length. Plotting I as abscissa and W as ordinate we obtain a parabola. Fig. 95. 140. Work Done in Compressing Fluids. — Let C, Fig. 96, be a cyUnder which contains a compressible fluid and which is provided with a piston. When the piston is displaced toward the left work is done against the force with which the fluid presses upon the piston. If dx denotes the dis- WORK 175 placement and F the force on the piston then the work dW = — Fdx = — pAdx , = — pdv, where p is the pressure in the fluid, A the area of the piston, and dv the change in the volume of the fluid. Therefore the total work done in compressing the fluid from a volume Vi to a volmne V2 is W=- \pdv. (1) When the law connecting p and v is given the work done in compressing or expanding a fluid can be found by carry- ing out the integration indicated in equation (1). During expansion, however, the displacement has the same direction as the force which causes the expansion ; therefore the sign before the integral is positive. 141. Representation of the Work Done in the PV-Dia- gram. — When the volume of the expanding fluid is plotted as abscissa and the pressure as ordinate, a curve is obtained, which repre- sents, graphically, the law connecting p and v. Such a representation is called a PV-diagram. It is evident from equation (1) that the area bounded by the curve, the «j-axis, and the two vertical lines whose equations are v = Vi and V = Vi, represents the work done in compressing the fluid from Vi to %• 142. Isothermal Compression of a Gas. — If a gas is com- pressed without changing its temperature the compression Fig. 96. 176 ANALYTICAL MECHANICS is called isothermal, in which case the relation between p and V is given by Boyle's law, i.e., pv = k. (2) Substituting in equation (1) the value of p given by equation (2) we obtain = A;log^- (3) 143. Adiabatic Compression of a Gas. — If no exchange of heat is allowed between the gas and other bodies while the former is being compressed the compression is called adiabatic. The law which connects p and w in an adiabatic compression or expansion of a gas may be expressed by the relation pv = k, (4) where y and k are constants for a given gas. Substituting in equation (1) the value of p, which is given by equation (4), we obtain w^-kT^ 7-1 - 7-1 ■ ^^'' 144. Modulus of Elasticity of a Gas. — Let -dv denote the change in volume due to an increase in the pressure of a gas by an amount dp. Then the stress is dp and the strain ^^ . V Therefore by Hooke's law dp = ^Z± V x.-.|. (6> WORK 177 The modulus of elasticity X is not a definite constant for a given gas, because the value of -j^ depends upon the tem- dv perature and the amount of heat of the gas. Therefore the state of a gas for which -f^ is calculated should be stated in dv order that the value of X may have any meaning at all. There are two states for which X is calculated, namely, the isothermal and the adiabatic states. 145. Isothermal Elasticity. — When the compression is iso- thermal 'pv =^ k and f = -\- dv v^ .: \=^=p. (7) V Therefore the isothermal elasticity of a gas niunerically equals the pressure. 146. Adiabatic Elasticity. — When the gas is compressed adiabatically pv^ = k and ^ = - kyv-y-^ dv = — ypv~\ ■■• X= yp. (8) 147. Torsional Rigidity of a Shaft. — Suppose the upper end of the cylinder of Fig. 97 to be rotated about the axis of the cyUnder through an angle 6, while the lower end is fixed, and consider the stresses and the strains in the cylin- der. It is evident that the strain is nil at the axis and in- creases uniformly with the distance from the axis. Further the strain is nil at the lower base and increases uniformly with the distance from it. Since Hooke's law holds these statements are true with regard to the stress in the cylinder. 178 ANALYTICAL MECHANICS Let dF denote the force acting on the area, on the upper base, of a ring of radius r and width dr, then the stress equals ^^ , . But if e is the angle of twist at the upper 2-kt -dr base and I the length of the cylinder, then the strain equals Mil y- Therefore by Hooke's law 6 dF re 2 irrdr In this case X is called modulus of shearing elasticity or, simply, shear modulus. Solv- ing the preceding equation for dF we get dF=^er'dr. V Therefore the torque acting upon the area of the ring is dG = r- . dF _ 2 7rX , I 2 7rx„ r G = I - dr^ dr. a r^dr -^21^' (9) Fig. 97. where G is the total torque applied at the upper end and a the radius of the cyhnder. Thus the torque necessary to produce a given angle of twist varies directly as the fourth power of the radius and inversely as the length. On the other hand for a given shaft the torque varies directly with the angle of twist. The torsional rigidity of the shaft is defined as the torque necessary to produce a unit angular twist ; therefore = X Tra" 21 (10) WORK 179 It will be observed that the torsional rigidity of a solid shaft varies directly as the fourth power of the radius and inversely as the length. 148. Work Done in Twisting a Rod. — Work done by a torque is obtained by substituting the expression for the torque in the work equation. Thus w= rode. =X' ^TtJ}'^' [byeq.O).] = hkd^ (11) where k = X^ • 21 PROBLEMS. 1. What are the dimensions of stress, strain, and modulus of elasticity? 2. A steel rod of J-inch radius is found to stretch 0.004 inch in 10 inches of its length when a load of 10,000 pounds is gradually applied. Find the Young's modulus of the rod. 3. The Young's modulus of a brass wire is 10.8 X 10"^^^. Find cm.2 the load (in pounds) necessary in order to produce an elongation of 0.5 mm . in 1 meter. The diameter of the wire is 1 millimeter. 4. The modulus of shearing elasticity of a steel shaft is 1 1 X 10* pounds per square inch. What force acting at the end of a lever 30 inches long win twist asunder the shaft if it is 0.5 inch in diameter? 5. A brass rod, 4 feet long and 1 .5 inches in diameter, is twisted through an angle of 9° by a force of 1500 pounds acting 6 inches from the axis of the rod. If on removal of the stress the bar recovers its original position, calculate the modulus of shearing elasticity of the rod. 6. Taking the data of the preceding problem find the force necessary to give an angle of twist of 2° to a rod 15 inches long, 0.5 inch in diameter. 7. An elastic string of natural length I is stretched to twice its length when it supports a weight W. The ends of the string are connected to two points at the same level and a distance d apart, while the weight W is attached to the middle of the string. Find the position of equihbrium of the weight. 8. A spider hangs from the ceiling by a thread which is stretched by the weight of the spider to twice its natural length. Show that the work 180 ANALYTICAL MECHANICS done by the spider in climbing to the ceiling equals | mgh, where m is the mass of the spider and h its distance from the ceiling. 9. The outer end of a flat spiral spring is fixed, while the inner end is attached to the center of a bar 20 cm. long, in such a way that the bar is parallel to the plane of the spring. Two forces of 500 dynes each applied at the ends of the bar, at right angles to the bar and parallel to the plane of the spring, can keep the bar turned through an angle of - radians. What torque must be applied in order to keep the bar in position after giving it three turns? 10. In the preceding problem find the work done in giving the bar three turns. What portion of the total work is done in the last turn? 11. Prove that the following is the expression for the torsional rigidity of a hoUow shaft : Z I •where 6 is the inner radius of the shaft, while the other letters represent the same magnitudes as in § 147. 12. Derive expressions for the saving of material and loss of rigidity due to making a shaft of a given external diameter hollow. 13. Find the value of the quotient of the inner to the outer radius which wiU make the quotient of the saving of material to the loss of rigid- ity a maxiilium. 14. The weight and the length of a shaft are fixed; find the ratio of the inner to the outer diameter which wiU make the rigidity of the shaft a maximum. 15. The torsional moment which a shaft has to withstand and the length are fixed; find the ratio of the inner to the outer diameter which will make the weight of the shaft a minimum. VIRTUAL WORK. 149. Principle of Virtual Work. — The concept of work enables us to formulate a principle, called the principle of virtual work, which can be applied to equilibrium problems to great advantage. In order to derive this principle consider a particle which is in equilibrium. Evidently the resultant force acting upon the particle is nil and remains nil so long as the particle is in the equilibrium position. But when the particle is given a WORK 181 small displacement, the resultant force assumes a value dif- ferent from zero. If the displacement is small enough, so that the departure from equilibrium position and conse- quently the resultant force remains small, the displacement is called a virtual displacement and the work done by the resultant force virtual work. We will call virtual force the small resultant force, which is called into play by the virtual displacement. Let Fi, F2, etc., be the forces under the action of which the particle is in equilibrium. When the particle is given a virtual displacement ds, these forces are changed, in general, in magnitude and direction so that a virtual force dF acts upon the particle during the displacement. Then the virtual work is dF ■ds=Fi-dsi+Fi-ds2 + ■ ■ • , (VII) where dsi, ds2, etc., are the displacements of the particle along the forces Fi, Fi, etc., due to the virtual displacement ds. But since the left-hand member of the last equation is an infinitesimal of the second order while the terms of the right-hand member are infinitesimals of the first order we can neglect the left-hand member and write Fx dsi + F2 ds2 -J- • • • + = 0. (VIII) Equation (VIII) states: when a particle which is in equi- librium is given a virtual displacement the total amount of work done by the forces acting upon the particle vanishes. This is the principle of virtual work. The principle of virtual work is apphcable not only to particles, but also to any system which is in equihbrium. If the system is acted upon by torques as well as forces, then the sum of the work done by the virtual torques and the virtual forces vanishes : Fidsi + F2ds2+ ■ • ■ +Giddi + G2de2+ • • • = 0. (IX) 182 ANALYTICAL MECHANICS ILLUSTRATIVE EXAMPLES. 1. Supposing the weights in Fig. 98 to be in equilibrium and the con- tacts to be smooth, find the relation between the two weights. If Wi is given a virtual displace- ment towards the left along the in- clined plane, then the virtual work is -T ds + Wi' dssin a + N -0 = 0, or T = Wi sin a. But T = W2. Therefore Wi = Wi sin a. Fig. 98. 2. Two uniform rods of equal weight W and equal length a are jointed at one end and placed, as shown in Fig. 99, in a vertical plane on a smooth horizontal table. A string of length I joins the middle points of the rods. Find the tension of the string. The following forces act upon each rod — the weight of the rod, the pull of the string, the reaction at the joint, and the reaction of the table. Suppose a slight displacement to be given to the system by pressing down- ward at the joint. The work done by the force which produced the dis- placement equals the sum of the work done by the other forces which act upon the rods during the displace- ment. But since both the force ap- plied and the displacement produced are very small their product is negli- gible. Therefore the sum of the work done by all the other forces is zero. The reactions at the ends of the rods do not contribute to the virtual work because each of the reactions is perpendicular to the corresponding surface of contact along which the displacement takes place. Therefore the weights and the tensile force of the string contribute all the virtual work. If dl and dh denote, respectively, the increase in length of the string and the distance through which the centers of mass of the rods are lowered during the virtual displacement the virtual work takes the form 2(T-~ + Wdh] = 0. Fig. 99. ^ + Wdhy- But from the figure I ■■ and dh =— a sin 6 di obtain a sin 5, and h = acos 6. Therefore dl = a cos 6 dB Making these substitutions and simplifying we r = 2prtan0. WORK 183 / 3. Find the mechanical advantage of the jack-screw. Let p be the pitch of the screw, I the length of the lever arm, F applied and P the force derived. Then since at any instant the system is supposed to be in equilibrium the virtual work, due to a small displacement, must vanish. Let dd denote a small angular displacement and dh the corresponding rise of the screw. Then if G denotes the torque applied the virtual work takes the form Gdd-Pdh = 0. the force ButG = F -l and dh = ^ p. Therefore Fide = ^. Fig. 100. Hence the mechanical advantage, which is the quotient of the force de- rived to the force applied, is P _2irl F p PROBLEMS. 1. By the application of the principle of virtual work derive the ex- pression for the mechanical advantage of (a) the lever; (b) the wheel and axle; (c) the hydraulic press; (d) the pulley (a) of problem 13 on page 21; (e) the pulley (b) of problem 13 on page 21; (f) the pulley (c) of problem 13 on page 21; (g) the pulley (d) of problem 13 on page 21. 2. Apply the principle of virtual work to (a) illustrative problem 1 on page 17; (b) problem 4 on page 20; (c) problem 6 on page 46; (d) problem 16 on page 47. 3. Four rods of equal weight W are freely jointed so as to form a square. The system is suspended vertically from one of the joints. A string of negligible weight connects two of the joints so as to keep the square shape of the system. Find the tension of the string. 184 ANALYTICAL MECHANICS 4. An elastic band of weight W and natural length Z is slipped over a smooth circular cone the axis of which is vertical. The force necessary to stretch the string to double its natural length equals X. Find the position of equilibrium and the tension of the string. 6. Find the mechanical advantage of the following machines. 4 ^ gw V J (a) CHAPTER IX. ENERGY. 150. Results of Work. — Consider the work done by the engine of a train in pulling it upgrade. The work done may be divided into three parts: (a) Work done against frictional forces. (b) Work done against gravitational forces. (c) Work done against the kinetic reaction. The result of work done against frictional forces is heat. The amount of heat generated is proportional to the amount of work done. The heat may be utilized, at least theoreti- cally, to do work. Thus a part, if not all, of the original work may be recovered. The apparent result of the work done against the gravi- tational forces is the elevation of the train to a higher level.. The work done may be recovered by letting the train come down to its former level and thereby do work. Therefore the work done against gravitational forces may be considered to be stored up. The apparent result of the work done against the kinetic reaction in accelerating the train is an increase in the ve- locity of the train. The work done may be recovered by- letting the train overcome a force, which tends to reduce the velocity of the train to its original value. Therefore in this case also the work done may be said to have been stored up. In fact in all three cases the work done is stored up. In the first case, however, work is not available as readily as in the other two cases. In order to convert 185 186 ANALYTICAL MECHANICS heat into work special means, such as heat engines, etc., have to be used, which do not belong to the domain of ordinary mechanics; therefore work done against frictional forces is considered as lost. On the other hand work which is done against gravitational forces or against kinetic reac- tions is directly available for mechanical work. 151. Energy. Potential, Kinetic, and Heat Energy. — Energy- may be defined as work which is stored up. Work stored up in overcoming kinetic reactions is called kinetic energy. Work stored up while overcoming nonfrictional forces, such as gravitational forces, is called potential energy. Work done while overcoming* frictional forces is called heat energy. 152. Transformation of Energy. — ^ Potential, kinetic, and -Iheat energy are different (at least apparently*) forms of the ;same physical entity, i.e., energy. Energy may be changed from any one of these forms into any other form. Whenever isuch a change takes place energy is said to be transformed. Transformation of energy is always accompanied by work. In fact the process of doing work is that of transformation of energy. The amount of energy transformed equals the amount of work done. The units and dimensions of energy are the same as those of work. KINETIC ENERGY. 153. Kinetic Energy of a Particle. — By definition kinetic energy equals the work done against the kinetic reaction in giving the particle its velocity. Since there is no mo- tipn along the normal to the path of the particle no work is ■done against the normal component of the kinetic reaction. Therefore we need only consider the work done against the tangential component. * Recent developments in physical sciences tend to show that differences between different forms of energy are only apparent and that all forms of energy are, in the last analysis, kinetic. ENERGY 187 Denoting the kinetic energy by T and putting the defini- tion into analytical language we obtain dv\ rds , = m I -^dv Jo at = i mv''. (I) Therefore the kinetic energy of -a particle equals one-half the prod- uct of the mass by the square of its velocity. Since both m and v" are positive, kinetic energy must be a positive magnitude. The kinetic energy of a system of particles, therefore, equals the arithmetic sums of the kinetic energies of the individual particles. Thus T=\i:mv^. (II) When all the particles of the system have the same velocity r = i Mv\ where M is the total mass of the system. 154. Work Done in Increasing the Velocity of a Particle. — If the velocity of a particle is increased from Vo to v then the work done against the kinetic reaction equals the increase in the kinetic energy of the particle. This will be seen from the following analysis : = m I V = \ mv^ — \ mvo = T- To. dv (III) * The first negative sign indicates the fact that T is the work done against and not by the kinetic reactions. The second negative sign belongs to the kinetic reaction as it was explained in Chapter VI. 188 ANALYTICAL MECHANICS PROBLEMS. 1. Show that the dimensions of work and kinetic energy are the same. 2. A body of 50 gm. mass starts from the top of an inclined plane 10 m. long, and arrives at the bottom with a velocity of 300 — '-. Find the aver- S6C. age frictional force. The angle of elevation of the plane is 30°. 3. A body of 100 gm. mass, which is projected up an incHned plane, cm arrives at the top of the plane with a velocity of 150 — ^. Find the velocity of projection, supposing the frictional force to be constant and equal to 10,000 dynes. The length of the plane is 5 inches, and the angle of elevation is 30°. ft 4. A bullet enters a plank with a velocity of 1500 — '-, and leaves it ft with a velocity of 1350 — - ■ How many such planks can the bullet sec. penetrate? 5. In the preceding problem find the average resisting force which the planks offer. The bullet weighs J ounce. 6. A catapult, which consists of an elastic string 15 cm. long, with ita ends tied to the prongs of a forked piece of wood, is used to throw stones. What velocity will it give to a stone of 5 gm. mass when stretched to twice its natural length. The modulus of elasticity of the string is 2 pounds. 7. The kinetic energy acquired by a weight of 750 pounds in faUing through a distance of 4 feet is to be absorbed by a helical spring, 5 inches long. Find the modulus of elasticity of the spring so that it wiU not be compressed more than 1 inch. 8. Having a given size and shape, how will the penetrative power of a bullet depend (a) on its weight, and (b) on its velocity. The resisting force is supposed to be constant. 155. Kinetic Energy of a Rigid Body Rotating About a Fixed Axis. — Suppose the rigid body A, Fig. 101, to rotate about an axis through the point 0, at right angles to the plane of the paper. Consider the kinetic energy of an element of mass dm at a distance r from the axis. If v denotes the velocity of the element and dT its kinetic energy, then dT=^v^ dm = I r^oi^ dm, [v = rcc] ENERGY 18& where w is the angular velocity of the body, total kinetic energy of the rotating body is Therefore the .' — a iCO . ^dm- (IV) Fig. 101. where / is the moment of inertia of the body about the axis of rotation. Comparing the expression for the kinetic energy of rotation with the expression for the kinetic energy of translation we observe that moment of inertia plays the same role in motion of rotation as mass, the Unear inertia,, plays in motion of translation. The expression for the kinetic energy of a rotating body may be put in a Uttle different form by substituting for / its value in terms of the moment of inertia about a parallel axis through the center of mass. Thus = hmvj^+hlc' (V> where v^ is the velocity of the center of mass. We have thus divided the kinetic energy into two parts — (a) kinetic energy due to the motion of translation of the body as a whole with the velocity of the center of mass, (b) kinetic energy due to the rotation of the body about an axis through the center of mass. 156. Work Done in Increasing the Angular Velocity of a Rigid Body. — It was shown in § 154 that the work done against the kinetic reaction of a particle equals the increase in the kinetic energy of the particle. Therefore the work done against the kinetic reaction of any number of particles is the 190 ANALYTICAL MECHANICS sum of the increments in the kinetic energies of the individ- ual particles. Therefore W =2 (I mv^ — i mvo^). When the particles form a continuous system, we can replace the particles by elements of mass and the sxmamation sign by the integration sign. Thus nm W = f (^v^ dm — J Vo^ dm) Jo = 2 / {^^'^^ dm — r^ojo^ dm) r^ dm. — J coo^ / ^^ dm, = §/a.2-i7'oo^ (VI) where oo and w are the initial and the final values of the angular velocity of the body. Therefore in this case also work done equals the increase in the kinetic energy. PROBLEMS. 1. The flywheel of a metal punch is 4 feet in outside diameter and weighs 500 pounds. What must be its initial velocity in order that the punch may exert a force of 25 tons, through a distance of 1 inch, without reducing the speed of the flywheel by more than 25 per cent? Neglect the effect of the shaft, and consider the flywheel to be a disk. 2. The power of a 15-ton car was shut off and the brakes were put on at a time when the car was making 50 miles an hour. On each of the 8 wheels a normal brake-shoe force of 5000 pounds was applied. Find the distance covered by the car before coming to rest. The diameters of the wheels are 30 inches, the tracks are horizontal, and the coefficient of fric- tion for the contact between the shoes and the wheels is 0.2. 3. A 100-ton locomotive making a mile a minute is to be stopped withm 500 yards. What brake-shoe force must be applied? The diameters of the wheels are 6 feet. The coefficient of friction is 0.3. 4. Find the amount of heat which would be generated if the rotation of the earth about its axis were stopped. The mean density of the earth = 5.5 -^^j the radius = 4000 miles; 1 calorie = 4.2 (10)' ergs. ENERGY 191 6. How many cubic miles of ice could be melted by the heat computed in the preceding problem? The latent heat of ice is 80 calories per gram. 6. The winder of a spinning top is a helical spring, which is set in a cylindrical piece 1 inch in diameter. When the winder is hooked to the top and twisted through ir radians a force of 1 pound has to be appUed to the cyhnder tangentiaUy in order to keep it from untwisting itself. After the spring is given a twist of 2^ turns the top is released. Find the kinetic energy the top would acquire if there were no frictional forces. 7. In the preceding problem find the angular velocity of the top suppos- ing it to consist of a circular plate of 2 inch radius, and of i pound weight. 8. If the top of the preceding problem turns for 2 minutes before stop- ping, find the mean torque due to friction and resistance; also find the total number of revolutions made. 9. A top is given a motion of rotation by pulling at a string wound around it. Derive an expression for the energy communicated, (a) when the force apphed to the string is constant; (b) when it varies directly with the length of the string which is unwound. POWER. 157. Power. — Power is the rate at which work is done. When put into the language of calculus this definition be- comes P = ^. (VII) Power is a scalar quantity and has the dimensions [ML^T~^]. The C.G.S. unit of power is the erg per second. This unit is too small for engineering purposes; therefore two larger units are adopted, which are called the watt and the kilowatt. The following relations define these units: lwatt = V^^ 1 sec. = 107 ergs sec. 1 kilowatt = 10' watts = lOio^ISS sec. 192 ANALYTICAL MECHANICS The British unit of power is the horse power, defined by the following equation : 1 H.P. = 33,000 -^*" ^^^' = 550 nun. ft. lbs. sec. PROBLEMS. 1. Show that 1 horse power equals about 746 watts. 2. The engine of a train, which weighs 150 tons, is of 200 horse power> Find the maximum speed the train can attain on a level track if there is a. constant resisting force of 15 pounds per ton. 3. The diameter of the cylinder of a steam engine is 9 inches, and its- length 10 inches; the mean effective pressure per square inch is 90 pounds, and the number of revolutions per minute is 100. Find the indicated horse power. 4. Each of the 2 cylinders of a locomotive is 16 inches in diameter^ the length of the crank is 9 inches, the diameter of the driving wheels is 6 feet, the velocity of the train is 40 miles per hour, and the mean effective pressure is 75 pounds per square inch. Find the power developed. 5. A train weighing 125 tons moves at the rate of 50 miles an hour, along a horizontal road. Find the power, in kilowatts, transformed by the motors of the electric engine which puUs the train. The resistance is 10 pounds per ton. 6. Find the horse power developed by an engine which moves a train at the rate of 30 miles an hour up an incline of 1 in 300. The train weighs 120 tons and there is a resistance of 15 pounds per ton. 7. A belt traveling at the rate of 45 feet per second transmits 100 horse power. What is the difference in tension of the tight and the slack sides of the belt. The width of the belt is 20 inches. 8. A 150-horse-power steam engine has a piston 18 inches in diameter which makes 100 strokes per minute. Find the mean effective pressure of the steam in the cylinder. The length of the stroke is 24 inches. 9. The average flow over the Niagara Falls is 10,000 cubic meters per second. The average height is 160 feet. Find the power, in kilowatts, which could be generated if all the energy were utilized. 10. A fire engine pumps water with a velocity of 125 — through a sec. " nozzle 1 inch in diameter. Find the horse power of the engine required to drive the pump, if the efficiency of the pump is 75 per cent and the ENERGY 193 nozzle is 15 feet above the surface of the reservoir which supplies the water. 11. Find the power of a machine gun which projects 600 bullets per minute with a muzzle velocity of 500 — - and angular velocity of 600 tt ra- sec. ■dians per second. The bullets are cylinders 0.9 cm. in diameter and 15 gm. mass. 12. A shaft transmits 50 horse power and makes 150 revolutions per -minute. Express the torque transmitted in pounds-foot and dynes-cm. 13. An electric motor develops 25 kilowatts at 900 revolutions per minute. Find the torque on the rotating armature due to the field mag- nets. Neglect friction. 14. Find the power of a clock which has a maximum run of 8 days. "The weight which moves the works has a mass of 10 kg. At its highest position the weight is 15 inches above its lowest position. 15. A twin-screw steamer has engines of 20,000 horse power and when working at fuU power the engines make 75 revolutions per minute. Find the torque transmitted by the shaft of each screw. 16. The pitch of the screw propeller of a ship is 25 feet. The power transformed by the propeller is 15,000 horse power, when the ship makes :20 knots. Assuming that there is a slip of 10 per cent at the propeller iscrew and that the efficiency is 0.75, find the torque transmitted by the •shaft, also the thrust on the bearings. 17. A feed pump delivers water into a boiler at the rate of 20 lbs. an hoiu-. If the pressure in the boiler is 150 lbs. per square inch above the atmospheric pressure, find the effective horse power of the pump. POTENTIAL ENERGY. 158. Configuration. ^ The arrangement of the parts of a •system is called the configuration of the system. The system which consists of this book and the earth, for instance, is in •one configuration when the book is on the desk and in an- other configuration when it is on the floor. During the trans- fer of the book from the floor to the desk the system passes, continuously, through infinite number of configurations, be- cause the book occupies infinite number of different positions relative to the earth. 159. Conservative Forces. — If the work done in bringing & system from one configuration to another configuration is 194 ANALYTICAL MECHANICS independent of the manner in which the change of configura- tion takes place, the forces acting upon the system are said to be conservative forces. Gravitational forces are examples of conservative forces. This is evident from the result of § 129 where it was shown that the work done against gravitational forces in taking a body from one point to another is inde- pendent of the path along which the body is carried. 160. Dissipative Forces. — Forces which are not conserva- tive are called dissipative or nonconservative forces. All fric- tional and resisting forces are of this type. 161. Potential Energy. — The potential energy of a system in any configuration equals the work- done against the con- servative forces which act upon the system, in bringing it' from a standard configuration to the configuration in ques- tion. For instance, if the unstretched state of an elastic string is taken to be its standard configuration, then the potential energy of the string at any stretched state equals the work done in producing the extension. The potential energy of this book when on the table equals the work done in raising, it from the floor to the table, provided the book is considered to.be at the standard configuration when it is on the floor. The selection of the standard configuration is quite arbi- trary and is a matter of convenience only. It is evident from the definition of potential energy that its value is zero at the standard configuration. Comparing the definitions of potential energy and of con- servative forces we see that the potential energy at any given configuration is independent of the manner in which the system is brought from the standard configuration. This is equivalent to stating that the potential energy of a system depends upon its configuration. But coordinates define the configuration of a system; therefore potential energy is a function of the coordinates. If the sea level is taken as the standard configuration, i.e., the position of zero potential energy, then the potential ENERGY 195 energy of a body, due to gravitational forces, is a function of the vertical height of the body above the sea level; in fact it equals mgh, where mg is the weight of the body and h its height above the sea level. 162. Difference of Potential Energy. — The difference be- tween the potential energy of a system in two different con- figurations equals the work done in taking the system from the configuration of lower potential energy to that of higher potential energy. Let the point A, Fig. 102, represent the standard con- figuration and the points B and C represent two other con- FiQ. 102. figm-ations. Then if Ub and Uc denote the potential energies at B and C respectively, then by definition Ub=Wab, Uc=Wac, where Wab and Wac are equal, respectively, to the work done in going from A to B and from A to C. Therefore Uc-Ub= Wac - Wab = Wbc, (VIII) where Wbc equals the work done in taking the system from B to C. Thus the work done against conservative forces acting upon a system equals the increase in the potential energy of the system. 163. Isolated System. — A system which is not acted upon by external forces is called an isolated system. An isolated system neither gives energy to external bodies nor receives energy from them. This is an immediate result of the defi- nition of an isolated system, because exchange of energy 196 ANALYTICAL MECHANICS presupposes work by or against external forces, which in its turn presupposes interaction with external bodies. But since no external forces are supposed to act upon the system, there cannot be interaction with external bodies or exchange of energy. 164. The Principle of the Conservation of Energy. — One of the greatest achievements of the nineteenth century was the recognition and the experimental verification of the great generaUzation known as the principle of the conservation of energy, which states that the total amount of energy of an isolated system is constant. By means of the interaction of the different parts of an isolated system the various forms of its energy may be changed into other forms, and the distribution of the energy within the system may be altered, but the total amount of energy remains constant. In other words, energy may be transformed or transferred but cannot be annihilated or created. 165. Djmamical Energy. — -Kinetic and potential forms of energy are called dynamical energy. The distinction be- tween dynamical and nondynamical energy, such as heat energy, chemical energy, etc., is a matter of convenience. Heat energy may be treated as kinetic energy, but in order to do that molecules and their individual motions have to be taken into account. On the other hand chemical energy may be treated as potential energy if molecular and atomic forces can be taken into account. It is to avoid the compUcations of the molecular structure of bodies that these forms of energy are considered as nondynamical. 166. Conservation of Dynamical Energy. — When all the forces acting within an isolated system are conservative the interchange of energy is confined to the potential and kinetic forms of the energy of the system. Therefore apply- ing the general principle of the conservation of energy we see that in such a system the sum of the djmamical energy remains constant, that is, T+U = const. (IX) ENERGY 197 If To and Uo denote the initial values of T and U, then the last relation gives T+U=T,+ Uo and T-Tf,= -(U -Uo). (X) Therefore if only conservative forces act between the various parts of an isolated system, the sum of the potential and kinetic energies of the system remains constant, in other words, the gain in the kinetic energy equals the loss in the poten- tial energy. Equation (X) will be called the energy equation. 167. Conservation of D]maniical Energy and the Law of Action and Reaction. — The principle of the conservation of dynam- ical energy may be obtained from the Law of Action and Reaction. In order to prove this statement consider an isolated conservative system. Suppose the configuration of the system to have changed under the action of its inter- nal forces. Let Uo and U be the potential energies in the initial and final configurations, respectively. Then the change in the potential energy is {U- Uo). During the change in the configuration of the system the positions and the velocities of the particles, which form the system, undergo changes. Therefore let Sp and s denote the positions, and vo and v the velocities of any particle in the initial and final configurations of the system. Further let F denote the resultant force which acts upon the particle. Then the change in the potential energy of the system due to the displacement of the particle from Sq to s is J Fr ds* where F^ is the tangential component of the force. The normal component contributes nothing to the work. There- * Potential energy is, by definition, the work done by external forces against internal forces. Therefore when the change in potential energy is obtained by computing the work done by internal forces the result is the negative of the change in the potential energy. Hence the negative sign. 198 ANALYTICAL MECHANICS fore the total change in the potential energy of the system equals the sum of the work done, during the rearrangement, on all the particles of the system; i.e., (t7-C/o) = -2 fPrds, where the smnmation covers all the particles of the system. Therefore substituting the expression for F„ which was obtained by applying the law of action and reaction to the motion of particles, we obtain (f7_ ?7o) = -2 Tm^-ds V dv = — 2 (I mv^ — J mvo^) = -{T-T,), where To and T are, respectively, the values of the total kinetic energy of the system in the initial and in the final configurations. Rearranging the' terms of the last equation we get C7+ r= C/o+ Ta= const. which is the principle of the conservation of dynamical energy. Therefore the principle of the conservation of dynamical energy and the law of force are not independent of each other but form two different aspects of the same universal principle. ILLUSTRATIVE EXAMPLE. Talcing into account the variation of the gravitational attraction with the distance of a body from the center of the earth, find the potential energy of a body with respect to the surface of the earth. Outside the earth the weight of a body varies inversely as the square of its distance from the center of the earth. Therefore denoting this variable weight by F we have P = — ENERGY 199 where A; is a constant and r is the distance of the body from the center of the earth. But at the surface of the earth the weight of the body is —mg, therefore F = —mg when r = a, where a is the radius of the earth. Therefore making these substitu- tions in the last equation we obtain k —mg = -;, or k = — mga^. Therefore and F = - mga' U ■■ CPdr ^ a (---)■ ■■ — mga — mga Fig. 103. Discussion. — Plotting the potential energy as abscissa and the height above the surface of the earth as ordinate we obtain the curve of Fig. 103, where the circle represents the earth. When r = a, U = 0; as it should. When r= cc , U = mga. Therefore m^a is the maximimi value of the potential energy. In the figure this is evident from the fact that the curve approaches asymptotically to the Une U = mga. When r=2a,U= —^- Therefore at a height of about. 4000 nules the potential energy equals half its maximum value. It win be seen from the following analysis that for small heights the- potential energy may be considered to increase linearly with h, where h is, the height above the surface of the earth: V = mga^l ) \a rl il ^-] \a a + hj = mga' = mga a + h = mgh, when h X r. ™™' But F = -y — r- • dr r' dr Therefore V = ym j — = -T^- (VI) The negative sign indicates the fact that when a particle is brought to the field of another attracting particle work will be done by the particle and not by the agent which brings it. Therefore the potential due to a material particle, as we have defined it, is everywhere negative, except at infinity where it is zero. In case of electrical and magnetic masses potential is defined as the work done in bringing a unit posi- tive charge, or unit positive pole, from infinity. Therefore 212 ANALYTICAL MECHANICS the potentials due to a negative charge and a negative pole are negative, while the potentials due to a positive charge and a positive pole are positive. 179. Potential Due to Any Distribution of Mass. — When the field of force or the potential field is due to a number of par- ticles (material, electrical, or magnetic), the potential at a point equals the algebraic sum of the potentials due to the various particles. Thus if Wi, rrh, rrh, etc., be the masses of the particles and n, 7-2, n, etc., their distances from the point considered, then the potential at the point is = — 7ZI — r + 7 h (VII) When the field is due to a continuous distribution of mass the last equation may be put in the form of an integral. Thus V=-yf"^- (VII') 180. Intensity of the Field. — The intensity at any point of a potential field, or a field of force, is defined as the force experienced by a unit mass when placed at that point. Let H denote the intensity at a point. Then, if F is the force experienced by a mass to' when placed at that point, we have, by definition, H = -^. (VIII) TO ^ and H=—, TO m'\ds/ ds \m'l FIELDS OF FORCE AND NEWTONIAN POTENTIAL 213 Similarly and = dV ds Hx = dV dx Hy = dV dy H.= dV dz (IX) Therefore the component, along any direction, of the intensity at any point equals the rate at which the potential diminishes at that point as one moves along the given direction. ILLUSTRATIVE EXAMPLES. 1. Find the expressions for poten- tial and intensity at a point due to a spherical shell. Let P, Fig. 107, be the point and R its distance from the center of the shell. Then taking a zone for the element of mass, as shown in the figure, we get Fig. 107. dm = tr '2 ira sin. 6 • aa and r = V{R -acosey + a" sin^^ = Va'' + B^-2aRcose. Therefore 'dm = -Tr2.a^/;^^ sinddd ^ + R^-2aRcosd = - 32l|2™ [(a2 + R^-2aR cos 0)i]„ R = - ^^^^^|^[(o2 + 2aR + R^)i - ia^-2aR + R^)h There are two different cases which have to be considered separately. 214 ANALYTICAL MECHANICS (a) Point Outside the Sphere. — In this case B >a. Therefore the expression for the potential may be put in the form ■.-y r47ro^ R m = -^¥- Therefore outside the shell the potential is the same as if the mass of the shell were concentrated at its center. (b) Point Within the Sphere. — In this case Ra. = when R < a. Therefore the shell attracts a particle which is outside with the same force as if all of its mass were concen- trated at its center. On the other hand the shell exerts no force on a particle which is within the shell. The dis- tribution of V and H in the field are represented graph- ically in Fig. 108, where curve (I) represents the po- tential and (II) the intensity. 2. Find the expressions for the potential and the intensity due to a solid spherical mass. FIELDS OF FORCE AND NEWTONIAN POTENTIAL 215 There are two cases- which have to be considered separately. (a) Point Outside the Sphere. — Consider the sphere to be made of concentric shells of thickness dp. Then, since the point is outside every one of these shells the potential due to any one of the shells is, according to the results of the last problem, jTr dm dV^-y^, where dm is the mass of the shell and R the distance of the point from the center. Hence the potential due to all the shells in the sphere is V = — f" ^ Jo R m where m is the mass of the sphere. Therefore the potential at a point outside of a sphere is the same as that due to a particle of equal mass placed at the center. (b) Point Within the Sphere, -t- In this case we divide the sphere into two parts by means of a concentric spherical surface which passes through the point. Then the potential due to that part of the sphere which is within the spherical surface is obtained by the result of case (a). Thus if mi denotes the mass of this part of the sphere and Vi its potential, then T^i = - 7 ^ = - -^TryrR^. In order to find the potential due to the rest of the sphere suppose it to be divided into a great number of concentric spherical shells. Then since every one of the shells contaius the point the potential due to any one of them is dVi = — y — = — 4^TryTp dp, P where dm is the mass, p the radius, and dp the thickness of the shell. Therefore the potential due to all the shells having radii between R and a is '^2 = — 4x7T f pdp = - 2T7yT {a^ - R'^ 216 ANALYTICAL MECHANICS Therefore the potential due to the entire sphere is 7 = Fi + 72 = — ym 2 a? When R is plotted as abscissa and V as ordinate the distribution of the potential is given by a curve similar to (I) of Fig. 109. Now consider the intensity at a point in the field of the sphere. (a) Point Outside the Sphere. H =- V H dV dR Fig. 109. = -T R' Therefore the distribution of the field intensity outside of the sphere is the same as that due to a particle placed at the center, (b) Point Within the Sphehe. "-~dR Therefore within the sphere the distribution of the field intensity obeys the harmonic law; i.e., the intensity varies directly as the distance from the center. In Fig. 109, curve (II) gives the distribution of the intensity of the field. PROBLEMS. 1. Find the potential and the field intensity due to a hollow sphere at a point (1) outside, (2) within the hollow part, and (3) in the solid part of the sphere. 2. Find the potential and the field intensity due to a circular disk of negligible thickness at a point on its axis. 3. Find the potential and the field intensity due to a straight wire of length I and mass m at a point on the axis of the wire. The cross-section of the wire is negligible. FIELDS OF FORCE AND NEWTONIAN POTENTIAL 217 4. Find the potential and tlie field intensity due to a straight circu- lar rod at a point on its axis. 5. Show that problems 2 and 3 are special cases of problem 4. 6. Find the magnetic potential and the field intensity due to a cylin- drical magnet at a point on its axis; suppose the magnetism to be dis- tributed at the ends only. 7. Find the potential and the field intensity due to two spherical charges at a point equidistant from centers of the two charges. 8. Find the potential and the field intensity due to a right cone at a point on its axis. 9. A uniform sohd sphere is cut in two by a diametral plane. Show 3 7)1 that the gravitational force between the two parts wiU be — 7 — , where lb a^ m is the mass of the sphere, a the radius, and y the gravitational constant. 10. Show that if any two points on the surface of the earth were joined by a straight and smooth tunnel a particle would traverse it in about 42.5 minutes. 11. Two spheres of masses m and m' attract each other with a force, F =y — — , where 7 is a constant and r is the distance between the centers.. Taking the configuration when the spheres are in contact to be that of zero potential energy, find their potential energy when the centers are separated by a distance D. The radii of the spheres are a and &. 12. In the preceding problem suppose the spheres to repel each other with the same law of force and take the configuration when the spheres are separated by an infinite distance to be that of zero potential energy. 13. Find the potential due to a small magnet at a point whose distance- is large compared with the length of the magnet. 14. In the preceding problem find the components of the intensity of the field along and at right angles to the line joining the point to the magnet. Also find the total intensity and its direction. CHAPTER XI. UNIPLANAR MOTION OF A RIGID BODY. 181. Angular Kinetic Reaction. — It will be remembered that in considering the equihbrium of rigid bodies the Law of Action and Reaction was divided into the following two sections : To every linear action there is an equal and opposite linear reaction, or, the sum of all the linear actions to which a body or a part of a body is subject at any instant vanishes. SA, = 0. (AO To every angular action there is an equal and opposite angular reaction, or, the sum of all the angular actions to which a body or a part of a body is subject at any instant vanishes. 2A„ = 0. (A„) In Chapter VI the first section of the law was applied to particles in motion; but in order to do this the meaning of the terms "linear action" and "hnear reaction" was enlarged so as to include linear kinetic reactions as well as forces. In the present chapter the second section of the law will be applied to the motion of rigid bodies; but before doing this we must introduce another form of kinetic re- action, which we will call angular kinetic reaction. If we replace in the second section of the law the terms "angu- lar action" and "angular reaction" by the terms "torque" and "angular kinetic reaction," we obtain the following form which is directly applicable to problems of rotation : The sum of all the torques acting upon a rigid body tplus the angular kinetic reaction equals zero, or the 218 UNIPLANAR MOTION OF A RIGID BODY 219 resultant torque equals and is oppositely directed to the angular kinetic reaction. Resultant torque = —(angular kinetic reaction). (I) In order to understand the nature of the angular kinetic reaction consider the following experiment: If we try to rotate a flywheel, which is free to move about a horizontal axis, by pulling at one end of a string which is wound around the axle. Fig. 110, we find that the greater the angular veloc- ity which we want to impart in a given interval of time the harder we must pull at the string. But since the pull of the string and the reaction of the bearings form a couple and since the increase in the angular velocity per unit time means angular acceleration, we conclude that a torque must be applied to the flyTvheel in order to impart to it an angular acceleration, and that the greater the acceleration desired the greater must the torque be. Evidently the torque which we apply to the flywheel expends itself in overcoming certain reactions. The resisting torque due to the friction between the axle and its bearings and between the surface of the flywheel and the surrounding air must be overcome. But if we gradually diminish this resisting torque by reducing the friction we observe that the torque which must be applied, in order to give the flywheel a certain angular acceleration, tends towards a constant value different from zero. In other words even if all the resisting torques due to friction were eliminated we would have to apply a torque of definite magnitude in order to give the flywheel a desired angular acceleration; that is, the flywheel resists torques which im- part to it an angular acceleration. This resistance to angular acceleration is the angular kinetic reaction. 220 ANALYTICAL MECHANICS 182. Experimental Definition of Moment of Inertia. — If in the experiment of the preceding section all frictional forces and torques are eliminated and then torques of different magnitudes are applied to the flywheel, it will be found that the torques are proportional to the angular accelerations pro- duced; that is, if Gi, G2, etc., denote the torques obtained by multiplying the pull of the string by the radius of the axle and 71, 72, etc., the corresponding angular acceleration, then we shall find that the following relations hold : ^ = ^=^=. . . =1, (II) Tl 72 73 where / is a constant which depends only upon the rotating, system. In fact, as will be shown in § 186, it is nothing more or less than the moment of inertia of the rotating system. We have, therefore, the following definition for the moment of inertia of a body, in addition to the analytical definition given in Chapter VII : The moment of inertia of a body about a given axis is a con- stant of the body, relative to the given axis, which equals the quotient of the torque applied by the angular acceleration ob- tained; both being referred to the given axis* 183. Measure of Angular Kinetic Reaction. — It is evident from equation (II) that Gi, G2, etc., which measure the angu- lar kinetic reactions of the flyTvheel for the accelerations 71, 72, etc., are proportional to these accelerations. Therefore the angular kinetic reaction of a body varies directly with the angular acceleration imparted. If, on the other hand, a number of bodies of different moments of inertia are given the same angular acceleration, it is found that the kinetic re- actions are proportional to the moments of inertia; that is, 1 = 1 .. (HI) * Note the striking similarity between this definition of moment of inertia and the definition of mass given in § 94. UNIPLANAR MOTION OF A RIGID BODY 221 where 7 is the common angular acceleration. Therefore the angular kinetic reaction varies directly as the product of the moment of inertia by the angular acceleration, angular kinetic reaction = kly, where h is the constant of proportionaUty. When all the magnitudes involved in the last equation are measured in the same system of units k becomes unity. Introducing this simplification in the last equation and putting it into vector notation we have angular kinetic reaction = — 7Y. (IV) The negative sign indicates the fact that the direction of the angular kinetic reaction is opposed to that of the angular acceleration. 184. Torque Equation. — Combining equations (I) and (IV) and denoting the resultant torque by G we obtain G = 7Y, = 7(0. (V) The last equation, which will be called the torque equation, states that the resultant torque about any axis equals the product of the moment of inertia by the angular accelera- tion and has the same direction as the angular acceleration. 185. The Two Definitions of Moment of Inertia. — In order to show that the constant, 7, of equation (II) and the moment of inertia defined by equation (II) of page 152 are the same magnitude, consider the motion of the rigid body A, Fig. Ill, about a fixed axis through the point 0, perpendicular to the plane of the paper. Let d? be the resultant force acting upon an element of mass dm, that is, the vector sum of the forces due to external fields of force and the forces due to the connection of dm with the rest of the body. Then dF = dm — dt is the force equation for the element of mass. 222 ANALYTICAL MECHANICS The linear acceleration varies from point to point, but the angular acceleration is the same for all the elements. Therefore the discussion of the problem becomes simpler if we replace the Hnear acceleration by the angular acceleration. This may be done by taking the mo- ments of the forces about the axis. Since dm can move only in a direction perpendicular to the line r, the resultant force dV must be perpendicular to r. Therefore the magnitude of the moment dG, due to dV, is dG=rdF Fig. 111. rdm-r- dt = r^dm dw ~dt' (v = rui). Therefore the resultant torque acting upon the body, or the sum of the moments due to the forces acting upon all the particles of the body, is du G r^ dm - , dt = CO / r^ dm. But by equation (V) G = lia. Therefore 7= / r^dm, Jo which is the definition of the moment of inertia given in Chapter VII. 186. Comparison. — There is a perfect analogy between motion of pure translation and motion of pure rotation. This is clearly brought out m the following lists of the magnitudes involved in the two types of motion: UNIPLANAR MOTION OF A RIGID BODY 223 Magnitudes involved in motion of translation. Their analogues in motion of rotation. s, linear displacement. 6, angular displacement. V, linear velocity. o 2W ' Wo — OOJo Vs = 2 ' Vo — acao UNIPLANAR MOTION OF A RIGID BODY 235 Therefore after the time fe the hoop will roU along towards the right with a Unear velocity vi less than Vo, and with an angular velocity w greater than too- Case ///. — Suppose the initial rotation to be counter-clockwise. In this case we obtain dv ^ ^^,,,\ (5'") (8") (9") where 4 is the time when sliding ceases. There are three special cases to be considered : (a) When Vo > acoo, «3 is positive, and consequently the hoop goes on roUing towards the right. (b) When vo = aoio, vs = 0, and consequently att = t, the hoop comes to rest. (c) When Vo < oojo, Vz is negative. Therefore at the instant t = ts the hoop begins to roll backwards. PROBLEMS. 1. Discuss the motion of the following bodies rolling down an inclined plane without sUpping : (a) A hollow cylinder of mass m and inner and outer radii n and raj respectively. (b) A hoop of mass m and radius r. (c) A sphere of mass m and radius r. (d) A hollow sphere of mass m and inner and outer radii r^ and r2, respectively. (e) A spherical shell of negligible thickness of mass m and of radius r. (f) Compare the times of descent in (c) and (e). 2. A sphere is projected, without initial rotation, up a perfectly rough inclined plane. Discuss the motion. 3. A wheel which is rotating about its own axis is placed on a per- fectly rough inclined plane. Discuss the motion up the plane. 236 ANALYTICAL MECHANICS 4. The return trough of a bowling alley is 50 feet long and has a slope of 1 foot in 20 feet. Supposing the contact to be perfectly rough find the time a ball will take to return. The sides of the trough are perpendicular to each other. 5. In the adjoining figure the largest cir- cle represents a solid disk wheel, which roUs along a rough horizontal table under the action of a falling body. The left-hand end of the string is spliced and connected to two smooth rings on the axle of the wheel. The puUey over which the string passes is smooth. Discuss the motion. 6. In the preceding problem suppose the pulley to be rough and to rotate about its axis. 7. Same as problem 5 except that the wheel rolls up an inclined plane. 8. In the preceding problem suppose the pulley to rotate. 9. Same problem as the third illustrative example, p. 231, except that the cylinder is hoUow and has a negUgible thickness. , 10. Same as the preceding problem, but the cylinder roUs up an inchned plane. 11. How can you tell a solid sphere from a hollow one which has exactly the same diameter and mass? 12. Two men of different weights coast down a hill on exactly similar bicycles. Which will reach the bottom of the hiU first, the lighter or the heavier man? 13. A thin spherical shell of perfectly smooth inner surface is filled with water and allowed to roll down an inchned plane. Discuss the motion. 14. A hollow cylinder of negUgible thickness and perfectly smooth inner surface is filled with water and allowed to roll down an inclined plane. Discuss the motion. GENERAL PROBLEMS. 1. A sphere of radius a starts from the top of a fixed sphere of radius 6 and roUs down. If there is no sliding find the position at which they will separate. 2. Two masses mi and nh are suspended by means of strings which are wound around a wheel and its axle, respectively. The wheel and axle are rigidly connected and are free to rotate about a horizontal axis. Dis- cuss the motion. UNIPLANAR MOTION OF A RIGID BODY 237 (a) When Mi and M2, the masses of the wheel and axle, are negligible; (b) When they are not negligible. 3. In the Atwood machine problem show that if the pulley is not rough enough the acceleration of the two moving masses is Trr— g ^ ^ _ _ ^ M + mei"' ^ where n is the coefficient of friction. Hint. — If r and T' are the tensile forces in the string on the two sides, 4. Same as the third illustrative problem, but the pulley P is supposed to rotate. 5. In the preceding problem suppose the cyUnder to roll up an incUned plane. 6. A tape of negUgible mass and thickness is wound around the middle of a cyhnder. The free end of the tape is attached to a fixed point and then the cylinder is allowed to faU. Show that the cylinder falls with an acceleration of f 3 and the tensile force of the tape is \ W, where W is the weight of the cylinder. 7. In the preceding problem the fixed point is on an inclined plane and the cylinder roUs down the plane. 8. Discuss the motion of a log which moves along its length down an incUned plane, upon two rollers, which stay horizontal. 9. A uniform rod is allowed to fall from a position where its lower end is in contact with a rough plane and it makes an angle a with the horizon. Show that when it becomes horizontal its angular velocity is y -^ sin a, where I is the length of the rod. 10. Discuss the motion of a cylinder down an inclined plane, supposing the contact to be imperfectly rough, so that the cylinder both slides and roUs. 11. In the preceding problem suppose the cylinder to be hollow. CHAPTER XII. IMPULSE AND MOMENTUM. 190. Impulse. — It was stated at the beginning of Chapter VIII that when a force acts upon a body two entirely dif- ferent mechanical results are produced which are called work and impulse. The former is the result of the action of force in space. The latter is the result of the action of force in time. We have already discussed work. Impulse is the subject of the present chapter. 191. Measure of Impulse. — If a force which is constant both in direction and magnitude acts upon a particle the impulse which it imparts to the particle equals the product of the force by the time during which it acts. Since time is a scalar while force is a vector, impulse is a vector which has the same direction as the force. If L denotes the im- pulse which a constant force F imparts in the interval of time t, we can write L=F.i. (10 When the force is variable in magnitude or in direction, or in both, we must consider the impulses imparted in infini- tesimal intervals of time and add them up. Thus dL = Fdt and L= ffcit. (I) Jo Substituting in the last equation tov for F we have L = / mv dt dw, -J mv - TOVo, (II) 238 IMPULSE AND MOMENTUM 239 where Vo and v are the velocities at the instants t = and t= t, respectively. If Vo and v are parallel, equation (II) may be written in the form L = mv — mvo. (II') 192. Momentum. — The vector magnitude mv is called mo- mentum. Therefore the momentmn of a particle equals the product of the mass by the velocity and has the same direc- tion as the latter. Equation (II) states, therefore, that iw- pulse equals the vector change in momentum. PROBLEM. Show that the component of the impulse along any direction equals the change in the component of the momentum along the same direction, that is, X dt = mx — mxa, etc. D 193. Dimensions and Units. — Substituting the dimensions of force and time in the definition of impulse and those of mass and velocity in the definition of momentiim, we obtain [MLT-^-] for the dimensions of both. The C.G.S. unit of soon cm both impulse and.momentimi is the - — '- '-. The British sec. unit is the pound-second. Force and Momentum. — Let F denote the resultant of all the forces acting upon a particle of mass m. Then we have F= TOV = |(-v), (III) which states that the resultant force experienced by a particle equals the time rate of change of the momentum of the particle. In order to extend this result to a system of particles let F denote the resultant of all the external forces acting upon the system. Further let F^ be the resultant of all the forces acting upon any one of the particles. Evidently F. is the resultant of two sets of forces, namely, those which are 240 ANALYTICAL MECHANICS external and those which are internal to the system. Let F/ denote the resultant of the external forces acting on the particle, and F/' denote the resultant of the internal forces acting upon it, due to its connection with the rest of the system. Then F,= F/+F/'. But since F is the resultant of all the external forces acting upon all the particles of the system we have F=2F/ =2(F,-F/') =2F,-2F/'. The second sum of the left-hand member is the sum of the internal forces and is nil, because the internal forces come in pairs which mutually annul each other. Therefore F = SF, (IV'> = 2tov (IV)' = |(2mv). (V> These are results which are worth noting. Equation (IV) states that the resultant external force acting upon a system equals and is opposite to the vector sum (or the resultant) of the kinetic reactions of all the particles of the system. Equation (V) states that the resultant external force acting upon a system equals the time rate of change of the resultant momentum of the system. PROBLEMS. (1) Show that the component, along any direction, of the resultant force acting upon a particle equals the rate at which the corresponding component of its momentum changes, that is, X = — (mx), etc. (2) Show that the component, along any direction, of the resultant external force acting upon a system equals the rate at which the corre- IMPULSE AND MOMENTUM 241 sponding component of the resultant momentum of the system changes, that is, Z = — (2ma;), etc. at 194. The Principle of the Conservation of Momentum. — When the resultant external force is zero equation (V) gives |(2»v)-0 or S (mv) = const. (VI) Therefore when the sum of the external forces acting upon a system vanishes the resultant momentum of the system remains constant, both in direction and magnitude. This is the prin- ciple of the conservation of momentum. The momenta of the various parts of an isolated system may and, in general, do change, but the vector sum of the momenta of all the particles of the system cannot change either in direction or in magnitude. PROBLEM. Show that if the component, along any direction, of the resultant external force vanishes, the corresponding component of the resultant momentum of the system remains constant, that is, Xmx = const., when X = 0. 195. Momentum of a System. — ^The magnitude of the x-component of the resultant momentum of a system may be put in the following forms: 21 mi = — (Smx) = — (Mx) [by equation (I'), P- 141] = MxA Similarly Umif = My, [ (VH') 2ml = Mz, I 242 ANALYTICAL MECHANICS where M is the total mass and x, y, and i are the coordi- nates of the center of mass of the system. Combining the last three equations in a single vector equation we obtain 2mv = Mv, (VII) which states that the resultant momentum of a system equals the product of the total mass of the system by the velocity of its center of mass. 196. Motion of the Center of Mass of a System. — Com- bining equations (V) and (VII) we get F = Mv, (VIII) which states that the resultant external force acting upon a system equals the product of the total mass of the system by the acceleration of its center of mass. But equation (VIII) is the force equation for a particle of mass M, which is acted upon by a force F. Therefore the center of mass of a system moves as if the entire mass of the system were concentrated at that point and all the forces acting upon the system were ap- plied to the resulting particle. PROBLEM. Show that when the component, along any direction, of the resultant force acting upon a system vanishes the corresponding component of the velocity of the center of mass remains constant, that is, X = const., when X = 0. ILLUSTRATIVE PROBLEM. A bullet penetrates a fixed plate to a depth d. How far would it penetrate if the plate were free to move in the direction of motion of the buUet? Let F be the mean resisting force which the plate offers to the motion of the bullet. When tjie plate is fixed all the energy of the bullet is ex- pended in doing work against this force. Therefore we have Fd=i mv^, (1) where m is the mass and v the velocity of the bullet. When the target is free to move part of the energy of the buDet is expended in giving the IMPULSE AND MOMENTUM 243 target and the bullet a common velocity v'. Therefore if d' be the new depth of penetration we have Fd' = J mw2 - i (m + M) v"", (2) where M is the mass of the target. Eliminating F between equations (1) and (2) we get d'=[i_??i±M(^y]d. (3) But by the conservation of momentum we have WW = (m + M) v'. (4) Therefore eliminating the velocities between equations (3) and (4) we get d'=l^d. (5) M + m It is evident from equation (5) that when the target is free, but very large compared with the bullet, the depth penetrated is about the same as when it is fixed. PROBLEMS. 1. A particle which weighs 2 ounces describes a circle of L5 feet radius on a smooth horizontal table. If it makes one complete revolution in every 3 seconds find the magnitude and direction of the impulse imparted by the force, which keeps the particle in the circle, (a) in one-quarter of a revolution; (b) in one-half of a revolution; (c) in three-quarters of a revolution; (d) in one complete revolution. 2. Find the expression for the impulse imparted to a particle in de- scribing an arc of a circle with uniform speed. 3. Considering the rate of change of the momentum of a particle which describes a uniform circular motion derive the expression for the central force. 4. If we neglect the resistance of the 'air to the motion of a projectile what can we state with regard to the components of the momentum in the horizontal and vertical directions? 5. A train which weighs 100 tons runs due south at the rate of one mile a minute. Find the lateral force on the western rails due to the rotation of the earth, while the train passes the line of 30° latitude. 6. At what latitude will the force of the preceding problem be a maxi- mum? Determine its amount. 244 ANALYTICAL MECHANICS 7. Two trains, weighing 150 tons each and moving towards each other at the rate of 40 miles an hour, coUide. Find the average force which comes into play if the collision lasts 1.5 seconds. 8. A body explodes while at rest and flies to pieces. If at any instant after the explosion the different parts of the body are suddenly connected, wiU it move? 9. A shell of mass m explodes at the highest point of its flight and breaks into two parts, the one n times the other. Find the velocity of one piece if the other is brought to rest for an instant by the explosion. The velocity of the shell at the instant of explosion is v. 10. In the preceding problem will the motion of the center of mass of the entire shell be affected by the explosion? Answer this question on the assumption (a) that there is no air resistance, (b) that there is an air resistance. 11. A man walks from one end to the other of a plank placed on a smooth horizontal plane. Show that the plank is displaced a distance M+m ' where M and m are the masses of the man and of the plank, respectively, and I is the length of the plank. 12. A shell, which weighs 150 pounds, strikes an armor plate with a velocity of 2000 feet per second and emerges on the other side with a velocity of 500 feet per second. Supposing the resisting force to be uni- form, find its magnitude and show that the impulse produced by it equals the change in the momentum of the shell while plowing through the plate. The plate is 10 inches thick. COLLISION AND IMPACT. 197. Central Collision. — If two bodies collide while moving along the line which joins their centers of mass the collision is said to be central. In order to fix our ideas suppose the colliding bodies to be spheres, then Fig. 118 represents roughly the state of affairs during the collision. For a short interval of time after the spheres come into contact their centers approach each other and a little deformation takes place in the neighborhood of the point of contact at the end of which the centers of the spheres are, just for an instant, at rest with respect to one another, and are moving with a IMPULSE AND MOMENTUM 245 common velocity. Then the deformed parts of the spheres begin to regain, at least partially, their original forms and cause the spheres to separate. The process of colUsion may, therefore, be divided into two parts. The first part lasts from the initial contact at t= until the instant when the centers of the spheres are nearest together at t = ti. The second part be- gins at < = ti and lasts until the spheres separate at t = ti'. The impulse imparted to each body during the first part of the collision is called the impulse of compression, while that im- parted during the second part is called the impulse of restitution. Let mi and TOz be the masses of the coUiding bodies, vi and V2 be their velocities just before and vi' and V2' just after the collision, and let v be their common velocity at the instant of maximum compression, that is, when the distance between the centers of mass is shortest. Further, let L and L' denote the impulses of compression and of restitution, respectively. Then we have L= I 'Fdt = mi{v— Vi) = — nhiv— V2), I^' = I F dt= m-i, (vi — v) = —nh {v^ — v). The foregoing relations follow at once from the definition of impulse and from the fact that the colhding bodies form a system which is not acted upon by external forces, and consequently the sum of their momenta remains constant during the collision. Fia. 118. 246 ANALYTICAL MECHANICS 198. Coefficient of Restitution. — It is found by experiment that the ratio of the impulse of restitution to the impulse of compression depends only upon the nature of the bodies in coUision. The ratio, therefore, is a constant of the sub- stances in collision. This constant is called the coefficient of restitution, and is generally denoted by the letter e. Thus e=-^ (IX) ;;/— ;; V — Vi Vi — V V — Eliminating v we obtain e=-^^i^^^. (X') Vi- Vi But {vi - Vi) and {— Vi' + v^) are the velocities of the first body relative to the second, just before and just after the collision. Denoting them by V and V', respectively, we obtain _ relative velocity after impact relative velocity before impact 199. Resiliency. — When two bodies rebound after col- lision they are said to have resiliency, and the contact is called elastic contact. The coefficient of restitution is a measure of the resiliency of the colliding bodies. When e = 1 the resiliency of the colliding bodies is perfect and the contact is said to be perfectly elastic. The coefficient of restitution cannot have a value greater than unity, as will be seen from a consideration of the trans- formation of energy which takes place during colUsion. At the beginning of the collision the bodies have a certain amount of kinetic energy which depends upon their relative IMPULSE AND MOMENTUM 247 velocity at that instant. During the compression part of the collision a fraction of their energy is transformed into potential energy of compression and the rest into heat energy. During the restitution a fraction of the potential energy is transformed into kinetic energy and the rest into heat energy. Thus, in general, the kinetic energy at the end of the collision is less than that at the beginning. Therefore the relative velocity at the end of the collision is less than that at the beginning. Thus the coefficient of restitution is, in general, less than imity. If none of the energy, which is due to the relative motion of the colUding bodies, is lost in the form of heat, it is all transformed into potential energy during the compression and back into kinetic energy during the resti- tution. In this case the relative velocity at the end of the collision equals that at the beginning, which makes the coefficient of restitution imity.* The relative velocity at the end of the collision may be made greater than that at the beginning by having explosives at the point of con- tact. But this does not come in the definition of the coeffi- cient of restitution. Therefore unity is the highest value of e. When all the kinetic energy is transformed into heat during the colUsion the bodies have no relative velocity after the coUision. In this case the contact is called 'perfectly inelastic. Evidently e is zero when the contact is perfectly inelastic. Therefore the value of e lies between zero and imity. The values of the coefficient of restitution are 0.95 for glass on glass, 0.81 for ivory on ivory, and 0.15 for lead on lead. 200. Loss of Kinetic Energy of Colliding Bodies. — The kinetic energy of a system equals the kinetic energy due to the linear motion of the system with the velocity of its * In working out problems in which the contact is perfectly elastic instead of introducing the coefficient of restitution make use of the principle of the conservation of energy. The conservation of dynamical energy holds only when the contact is perfectly elastic. But the conservation of momentum and the conservation (general) of energy are true under all circumstances. 248 ANALYTICAL MECHANICS center of mass plus the kinetic energy of its parts due to their motion relative to the center of mass. Collision does not affect the motion of the center of mass of the system formed of the colhding bodies, because the forces which arise during the collision are internal forces. Therefore that part of its kinetic energy which is due to the motion of its center of mass does not suffer any loss. The loss occurs in that part of the energy which is due to the motion of the parts of the system with respect to the center of mass. Referring all the velocities to the center of mass and denoting the loss of kinetic energy by Ti, we have where Vi and V2 are the velocities just before and Vi and v^' the velocities just after the collision. "We can eUminate Vi and V2' from this expression for Ti by means of the principle of the conservation of momentum and the definition of e. According to the former m^Vi + vwi = miVi + TThv-^ and by (X') «i' - 2^2'= -e {vx — V2). Eliminating V2' between the last two equations we have rrii + nh ' The following changes in the expression of T, are effected by means of the last three equations. Ti = \ m^W - vl^) + i TO2 («^2^ - v^^) = i mi (di - v(){vx + v^) + I TO2 (?;2 - v^){v2 + v^) = I TOi (wi - ?;i')(fi - 1^2 + v^ - V2') = imi {vi - vi')(vi - ViXl - e) = *STkC^ -•*)'('-«'>• (^) When the colliding bodies are perfectly elastic then e = 1 IMPULSE AND MOMENTUM 249^ and Tj = ; on the other hand if the bodies are perfectly- inelastic, 6=0: therefore, Ti = h — i— 2— (vi — V2)^. TOi + ma 201. Impact. — When the mass of one of the colliding bodies is very large- compared with that of the other the velocity of the former with respect to the center of mass of the colliding system does not change appreciably during the collision. In such a case the body with the greater mass is considered to be fixed and the collision is called an impact. The impact of a falling body when it strikes the ground is a case in point. The velocities of the larger mass before and after the collision, as well as the common velocity at the instant of maximum compression, are negligible. Therefore making these changes in the expressions for L, L', e, arid Ti and, dropping the subscripts we obtain L = mv, L' = — mv, v' e=-' (X") V and Ti=^mi)^(l- e^), (XI') where m is the mass of the impinging body, while v and v' are its velocities just before and just after impact, respectively. ILLUSTRATIVE EXAMPLE. A ball which is thrown vertically down from a height h rises to the point of projection after impinging against a horizontal floor. Find the ve- locity of projection and the loss in energy. Let Vo be the velocity of projection, then the velocities just before and just after the impact are V = V«o^ + 2gh and v' = y/igh, respectively. But w' = &>. _ Therefore 250 ANALYTICAL MECHANICS Discussion. — The energy lost during the impact equals the kinetic energy of projection, as would be expected from the conservation of energy. When e = 1, Wo = and Ti = 0. In other words when the ball is per- fectly elastic it will rise to the height from which it is dropped. The entire kinetic energy is transformed, during the impact, into potential energy and back to kinetic energy without any loss. When e = 0, Vo = / + V dm, _ dv , dm d , X = ^(mv), which is the same equation as (III), except that in (III) m was considered to be constant, while here it is considered as a variable. If dm has an initial velocity u, then the change in the momentum of dm is (v — u) dm. Therefore _ dy , , .dm 254 ANALYTICAL MECHANICS ILLUSTRATIVE EXAMPLES. 1. A jet of water strikes a concave vessel with a velocity of 80 feet per second and then leaves it with a velocity which has the same magnitude as the velocity of impact but makes an angle of 120° with it. If the diameter of the jet is 1 inch find the force necessary to hold the concave vessel in position. The force experienced by the vessel equals the rate at which it receives momentum. Suppose the vessel to be symmetrical with respect to the axis of the jet, as in Fig. 119, then by symmetry there can be no resultant force on the vessel in a direction perpendicular to the axis of the jet. Therefore we need to consider only the change in. momentum along the axis. Let m be the mass of water delivered by the jet in the time t, V the velocity of impact, and a the change in the direction of flow. Then the force is F = mv — mv cos a t V (1 — cos a) t W gt WiA I WiAv^ V (1 — cos a) w (1 — cos a) (1 — cos a), where A is the area of the cross-section of the jet and Wi is the weight of a cubic foot of water. Replacing the various magnitudes by their nu- merical values we obtain . ^^■^T3X-(^4^0'x(«»£.)'x(^ + ^"-60°) 32 ft. sec.^ = 102.3 lb. Discussion. — It is evident from the general expression of F that its value depends upon a and varies between zero for a = and ^ ^'^"^ foj 9 a = IT. w hen a = „ , /' = — — - • 2 g IMPULSE AND MOMENTUM 255 2. A uniform chain is hung from its upper end so that its lower end just touches an inelastic horizontal table, and then it is allowed to faU. Find the force which the table will experience at any instant during the fall of the chain. The force is partly due to the weight of that part of the chain which is on the table at the instant considered and partly due to the rate at which the table is receiving momentum. Let x be the height of the upper end of the chain above the table, I the total length, and p the mass per unit length. Then pg (I — x) is the weight of that part of the chain which is on the table. On the other hand the momentum which the table receives in the interval of time dt is pvdt-v. Therefore the rate at which it re- ceives momentum is pv^, where v is the velocity of that part of the chain which is above the ground. This velocity is the same as that of the upper end of the chain, therefore v= V2g{l-x). Hence the total force is F = p(l-x)g + p-2gil-x) = 3p{l-x)g. Discussion. — When x = I, that is, at the beginning of the motion, the force is zero. When x = 0, that is, at the end of the motion, it is 3 pig, or three times the weight of the chain. As soon as the entire chain comes to rest on the table the force equals the weight of the chain. 3. A spherical raindrop, descending by virtue of its weight, receives continuously, by precipitation of vapor, an accession of mass proportional to the surface. Find the velocity at any instant. The external force acting upon the drop at any instant equals the rate at which its momentum changes, therefore mg = ^^ (mv), (1) where m, the mass of the drop, is variable. Since the accession of mass is proportional to the surface the rate of change of radius of the drop will be constant. Let a be the radius of the drop when it begins to fall, r its radius at any later instant, and k the rate at which r increases. Then at any instant m = TyTrr' 256 ANALYTICAL MECHANICS where r is the density of water. Substituting this expression for m in. equation (1) = {a + UY^ + Zia-rhtYkv at _f_3k_ ^^ f-l^dt -l the integral of which is y = e "+*' \ fs^ ""''*' d< + c • .-. V = e-"°- '" + ^') r rffe"°B <" + *') d< + c~\ = (o + AiO-'fsi J(a + kty dt + c] = (o + A;0-=|_S [aH + — ^ 1 g— + -^1 + cj ^ (a + fcO-'fl (4 a3< + 6 a%f + 4 akH' + /b^i^) + cj . Let V = when t = 0; then c = 0. 4 (o + fct; 4 V r- "^ r2 "^ r'/ ^gt 4 a=' + 6 a'fct + 4 akV + fc^t^ "4 (a + kty PROBLEMS. 1. Find the pressure upon the canvas roof of a tent produced by a. shower. The following data are given — the raindrops have a velocity of ft 50 — - at right angles to the roof: the intensity of the shower is such as sec. to produce a deposit of 0.2 inch per hour; 1 cubic foot of water weighs 62.5 pounds. 2. Find the pressure on horizontal ground due to the impact of a column of water which falls vertically from a height of 500 feet. cm 3. Water flowing through a pipe at the rate of 100 — '■ is brought to sec. dv * Equation (2) is of the form — + Pi/ = Q, which is the typical linear equation, with the integral y = e'^^^^V CQbJ^^ dx + cl. IMPULSE AND MOMENTUM 257 rest in 0.1 second by closing a valve at the lower end. Find the in- crease of pressure produced near the valve in both the C.G.S. and the British units. The length of the pipe is 500 meters. 4. A jet of water strikes a blade of a turbine normally. If the velocity of the jet is 150 feet per second, find the pressure it exerts on the blade, (a) when the blade is fixed; (b) when it has a velocity of 50 feet per second along the jet. 5. Figure 120a represents a horizontal trough with smooth vertical walls. The stream is supposed to have the same speed, 6 miles per hour, in all three parts of the trough. The stream in C is one-third of that in B. Find the force on the wall BC. The cross-section of the stream in A is 5 feet by 3 feet. 6. In the preceding problem suppose the branch C to be closed. Fig. 120. 7. A stream of water flowing in a horizontal direction is divided into two equal streams, as shown in Fig. 120b. Supposing the velocity of the water to remain unchanged derive an expression for the force on the obstacle, and discuss it for special values of 6. 8. In the preceding problem suppose the velocity of the stream to be 5 miles per hour, its cross-section before it is divided to be 4 feet by 2 feet, and d = 120°. 9. In the preceding problem take d = t. 10. In (8) take 9=^- 11. In(8)take0 = 27r. 12. A machine gun delivers 500 bullets per minute with a velocity of 1800 feet per second. If the bullets weigh 0.5 ounce each find the average force on the carriage of the gun. 13. A train scoops up 1500 pounds of water into the tender from a trough 500 yards long while making 50 miles per hour. Find the added resistance to the motion of the train. 258 ANALYTICAL MECHANICS 204. Oblique Impact of a Particle upon a Fixed Plane. Case I. Smooth Contact. — Let Vt and v^, Fig. 121, be the compo- nents of the velocity along the plane and along the normal, respectively, just before the impact; and let w/ and v„' be the corresponding components just after the impact. Since the plane is smooth, no hori- zontal forces arise during the impact; hence the horizontal component of the momentum remains constant. Therefore or mvt = mvt Vt = vi So far as the vertical compo- nent is concerned the impact is direct; therefore ev„ Fig. 121. Denoting by. a and j3 the angles which the resultant velocity makes with the normal just before and just after the im- pact we obtain tan a = —> v^ tan /3 = -^• :. tan a. = e tan ^. (XIV) Discussion. — When the contact is perfectly elastic e = 1; therefore the angle of incidence equals the angle of reflection as in the case of the reflection of light. In this case the magnitude of the velocity is not changed by the impact, as is to be expected from the conservation of energy. When the contact is imperfectly elastic the angle of reflection Ues between - and the angle of incidence, while the normal component of the velocity and consequently the magnitude of the total velocity is di- minished. When the contact is perfectly inelastic e = 0, and since a is not zero ^ must be | in order that e tan ,8 may have a finite value. There- fore in this case the particle slides along the plane after the collision. IMPULSE AND MOMENTUM 259 205. Case II. Rough Contact. — When the plane is rough frictional forces come into play and change the tangential component of the momentum. Let F be the tangential force due to friction, N the normal force, and fi the coefficient of friction; then we have XT Ndt= -mv„, Jt Ndt = mv„'; Jrr pT Fdt= \ fiN dt= — ymv^, Jo pT fT' Lt = \ Fdt= I fiN dt == - efiinv„. Jt Jt But Lt+ L/ = mvt - rnVf - Therefore m {v/ — Vt) = — mn {1 + e) v„ and v/ = Vt - nil+e) v„. Substituting this value of v/ in the expression for tan /3, which is obtained from Fig. 121, we get tan/3=^ = ^'-^^^+^)'^" - Vn ev„ Ehminating Vt between the last equation and the relation tan a = — we obtain Vn etanj3 = tana -m(1 +e). (XV) Discussion. — When ^ = 0, equation (XV) reduces to equation (XIV). When M = °° » tan /3 = — ooor/3 = — -; therefore the particle shdes along the plane towards the left. When e = and tan a.> n, tan j8 = oo and )3 = ^; therefore the particle shdes along the plane towards the right. 260 ANALYTICAL MECHANICS When e = and tan a < M, tan i3 = - 00 and /3 = - I ; therefore the particle is reflected towards the left and slides along the plane. PROBLEMS. 1. A perfectly elastic ball impinges obliquely on another ball at rest. Prove that their masses are equal if, after unpact, the balls move at right angles. 2. A billiard ball strikes simultaneously two bilhard balls at rest, and comes to rest. Show that the coefficient of restitution is |. 3. A particle slides down a smooth inchned plane and then rebounds from a horizontal plane. Find the range of the first rebound. 4. A bullet strikes a target at 45° and rebounds at the same angle. Prove that e = ~^ , where ju is the coefficient of friction. 1 + M 5. Four smooth rods, which form a square, are fixed on a smooth horizontal plane. A particle which is projected from one corner of the square strikes an adjacent corner after three reflections; show that e (1 + e) tan a ■ 1 + e (1 + e) ■where a is the angle the initial velocity makes with the rod joining the two corners and e is the coefficient of restitution. 6. In the preceding problem discuss the values of a for special values of e. 7. Derive an expression for the percentage of energy lost during oblique impact (a) when the contact is smooth; (b) when the contact is rough. 8. Two biUiard balls which are in contact are struck, simultaneously, by a third ball moving with a velocity v, in a direction perpendicular to the line of centers of the first two. Supposing the table to be perfectly smooth find the velocity of each baU after impact. 9. In the preceding problem obtain the expression for the loss of energy and find its value for the following special cases. The balls weigh 6 ounces each. (a) w = 16 feet per second, e = 0.8. (b) y = 20 feet per second, e = 0.5. 10. A ball impinges against another ball which has twice as large a mass and is at rest. The smaller ball has a velocity of 60 feet per second in a direction which makes 135° with the line of centers. Find the veloci- ties after impact; e = 0.5. IMPULSE AND MOMENTUM 261 GENERAL PROBLEMS. 1. A gun is free to move on smooth horizontal tracks. Show that the loss of energy due to recoil is -rr"; — E, where M and m are the masses of the gun and the projectile respectively, and E is the kinetic energy which is transmitted to the gun and projectile. 2. In the preceding problem compare the velocities of the projectile when the gun is fixed and when free to move. Also show that the actual ]\£ -\- vt angle a' at which the projectile leaves the gun is given by tan a' = — — — tan a, where a is the angle which the gun makes with the horizon. 3. A man stands on a plank of mass m, which is on a perfectly smooth horizontal plane. He jumps upon another plank of the same mass, then back upon the first plank. Find the ratio of the velocities of the two planks if the mass of the man is M. i. A stream of water delivering 1000 gallons per minute, at a velocity ft of 20 — '-, strikes a plane (1) normally, (2) at an angle of 30°. Find the sec. force exerted on the plane. 5. A uniform chain is held coiled up close to the edge of a smooth table, with one end hanging over the edge. Discuss the motion of the chain when it is allowed to fall, supposing the part hanging over the edge to be very small at the start of the motion. 6. In the preceding problem show that the acceleration is constant if the density of the chain varies as the distance from that end of the chain which is in motion. 7. A mass of snow begins to slide down a regular slope, accumulating more snow as it moves along, thus forming an avalanche. Supposing the path cleared to be of uniform depth and width, show that the acceleration of the avalanche is constant. 8. A baU falls on a floor from a height h and rebounds each time verti- cally. (a) Show that T = \^^\M, where T is the total time taken by the ball to come to rest. Find the value of T ioi h = 25 feet and e = 0.5. (b) Show that H = \^~h, where H is the total distance described. Find the value of H ior h = 25 feet and e = 0.5. 262 ANALYTICAL MECHANICS 9. A shell explodes at the highest point of its path and breaks up into two parts, the centers of mass of which he in the line of motion. Find the velocity of the pieces just after explosion, taking m for the mass of the shell, n for the ratio of the masses of the pieces, v for the velocity of the shell just before explosion, and E for the energy imparted to the pieces by the explosion. 10. A particle sUdes down a smooth incUned plane which is itseK free to move on a smooth horizontal plane. Discuss the motions of the particle and of the plane. 11. After falhng freely through a height h a particle of mass m begins to pull up a greater mass M, by means of a string which passes over a smooth pulley. Find the distance through which it will hft M. 12. A smooth inclined plane which is free to move on a smooth hori- zontal plane is so moved that a particle placed on the inchned plane remains at rest. Discuss the motion of the plane. 13. A disk and a hoop slide along a smooth horizontal plane with the same velocity v, then begin to roU. up the same rough inclined plane. How high will each rise? 14. A ball, moving with a velocity v, coUides directly with a ball at rest. The second ball in its turn collides with a third baU at rest. If the masses of the first and last ball are TOi and OTj, respectively, show that the velocity acquired by the third ball is greatest when the mass of the second ball satisfies the relation m^ = VotiWj. 15. Find the maximum velocity acquired by the third ball of the pre- ceding problem. 16. A billiard ball, moving at right angles to a cushion, impinges directly on an equal ball at rest at a distance d from the cushion. Show 2 c^ that they will meet again at a distance — - — d from the cushion. 1 + e 17. A ball is projected from the middle point of one side of a biUiard table, so that it strikes an adjacent side first, then the middle of the opposite side. Show that if I is the length of the adjacent side, the ball strikes the adjacent side at a point — — from the corner it makes with the opposite side. 18. A simple pendulum hanging vertically has its bob in contact with a vertical wall. The bob is pulled away from the wall and then it is let go. If e is the coefficient of restitution find the time it will take the pendulum to come to rest. 19. A particle strikes a smooth horizontal plane with a velocity v, IMPULSE AND MOMENTUM 263 making an angle a with the plane, and rebounds time after time. Prove that (a) r= 2^sin^_ ^^^ ^^ ■— • gii-e) g{i-e) where T is the total time of flight after the first impact, R the total range, and e the coefficient of restitution. 20. In the precediag problem find the values of T and R for the fol- lowing special cases : (a) V = 500 meters per second, a = 30°, e = 0.5. (b) V = 500 meters per second, a = 90°, e = 0.9. 21. A particle is projected horizontally from the top of a smooth inclined plane. Derive an expression for the time at the end of which the particle stops rebounding and slides down the plane. Compute its value for the following special cases: (a) Vo = 500 feet per second, a = 45°, e = 0.5. (b) Vo = 500 feet per second, a = 30°, e = 0.3. 22. In the preceding problem find the distance the particle moves along the plane before it stops rebounding. 23. In problem 21 find the velocity of the particle at the instant it. stops rebounding. 24. A bead slides down a smooth circular wire, which is in a vertical plane, and strikes a similar bead at the lowest point of the wire. If during the collision the first bead comes to rest, show that the second bead will rise to a height e% and on its return wUl follow the first bead to a height e* (1 — e)^ h, where h is the height from which the first bead falls. 25. Two equal spheres, which are in contact, move in a direction per- pendicular to their line of centers and impinge simultaneously on a third equal sphere which is at rest. Supposing the contacts to be perfectly smooth and elastic find the velocity of each sphere after the coUision. 26. A bullet hits and instantly kills a bird, while passing the highest point of its trajectory. Supposing the bullet to stay imbedded in the bird, and the bird to have been at rest when shot, find the distance between the place of firing and the point where the bird strikes the ground. 27. Two particles of masses mi and nh are conn,ected by an inextensible string of negligible mass. The second particle is placed on a smooth hori- zontal table while the first is allowed to fall from the edge of the table. When the falling particle reaches a distance h from the top of the table the string becomes tight. Find the velocity with which the second particle begins to move. 264 ANALYTICAL MECHANICS 28. A uniform chain lies in a heap close to the edge of a horizontal table. One end of the chain is displaced from the edge of the table so that it begins to fall. Show that when the last portion of the chain leaves the table the chain will have a velocity of y-^, where I is the length of the chain. 29. A uniform plank is placed along the steepest slope of a smooth inclined plane. Show that if a man runs down the plank making its length in the time given by 2M a f = M + m gsca.a the plank remains stationary during his motion. 30. A number of coins of equal mass are placed in a row on a smooth horizontal plane, each coin being in contact with its two neighbors. A similar coin is projected along the line of the coins with a given velocity. Find the velocity with which the last coin will start to move. 31. A baU of mass m, which is at rest on a smooth horizontal plane, is tied by means of a string to a fixed point at the same height as the center of the ball. A second ball of equal radius but of mass m' is pro- jected along the plane with a velocity v which makes an angle a with the .string. The second ball collides with the first centrally and gives it a velocity u. Show that m' sin a (1 + e) w = i , . ' V. m + m sm'' a CHAPTER XIII. ANGULAR IMPULSE AND ANGULAR MOMENTUM. 206. Angular Impulse. — The mechanical results produced by a torque may be measured in two ways. If the torque is considered to act through an angle the result measured is the work done by the torque; on the other hand if the torque is considered to act during an interval of time the result measured is called angular impulse. The angular impulse which a constant torque imparts to a. body in an interval' of time equals the product of the torque by the interval of time. If H denotes the angular impulse, G the torque, and t the time of action, then H = G . «. (I') When a vector is multiphed by a scalar the product is a vector which has the same direction as the original vector. Therefore H is a vector and has the same direction as G. When torque is not constant angular impulse equals the vector sum of infinitesimal impulses imparted during infini- tesimal intervals of time. Therefore Gdt I a>dt I I da, (I) where in equation (II) we find that both angu- lar impulse and angular momentum have the dimensions [ML^r-i]. The units are also the same for both. The C.G.S. unit is ^— ^ '- and the British unit is ft. lb. sec. sec. ANGULAR IMPULSE AND ANGULAR. MOMENTUM 267 210. Torque and Angular Momentum. — When the moment of inertia of a body remains constant under the action of a torque we have dt G =/- =l<^"'- (III) Therefore torque equals the time rate of change of momentum. The following analysis proves that the last statement is true when the moment of inertia varies with the time as well as when it remains constant. Let A, Fig. 122, represent a body, or a system of bodies, which is acted upon by one or more external torques. For the sake of simpUcity suppose the planes of the torques to be parallel to the plane of the paper, and the axis of rotation to pass through the point and. to be perpendicular to the plane of the paper. Let dF be the resultant force acting upon an element of mass dm. Then the moment of dF about the axis of rotation equals the product of r, the distance of dm from the axis, by dFj,, the component of dF perpendicular to r. Therefore dG = r. dF„ Fig. 122. = r = r ■dmfp • dm • - — (r^co) r dt d [p. 97] = — (r^ dm ' u) . dt 268 ANALYTICAL MECHANICS Therefore the resultant external torque acting upon the body is "-ii-f?""") or ° = |(^")' '^^"^ where I is supposed to vary with the time. Equation (III) is the general form of torque equation, of which equation (V) of Chapter XI is a special case. Introducing this expression of G in the definition for angular impulse we obtain H= T'cdf = Jo) - Zocoo, (IV) where 7o and wo denote the moment of inertia and the angular acceleration at the instant t = 0, and / and a those a.t t = t. Equation (IV) is a generahzation of equation (II). It states that angular impulse equals the change in the angular momen- tum under all circumstances. 211. The Principle of the Conservation of Angular Momen- tum. — When the resultant external torque acting upon a body or system of bodies vanishes, it follows from equation (III) that and consequently Im = const. (V) Therefore if the resultant of all the external torques acting upon a system vanishes, the angular momentum of the system ANGULAR IMPULSE AND ANGULAR MOMENTUM 269 remains constant, in direction as well as in magnitude. This is the principle of the conservation of angular momentum. ILLUSTRATIVE EXAMPLE. Discuss the effect of a shrinkage in the radius of the earth upon the length of the day. Let P and P' be the lengths of the day when the radius of the earth is a and a', respectively. Further, let oj and co' be the corresponding values of the angular velocity of the earth about its axis. Then ^ = ^- (1) But since the earth is not supposed to be acted upon by any external torques its angular momentum remains constant. Therefore 7a) = /'«'. (2) From equations (1) and (2) we obtain P' P r " I or P-P' P a'2 and SP P a + a 9LM a (3) where SP and da denote the diminutions in the length of the day and the radius, respectively. When 8a is small a' is very nearly equal to a, there- fore equation (3) may be written in the form Therefore the percentage diminution in the length of the day is twice as large as the percentage diminution in the radius. Hence when the radius is diminished by 1 mile the length of the day is diminished by about 43 seconds. PROBLEMS. 1. How do the oceanic currents from the polar regions affect the length of the day? 2. A uniform rod of negligible diameter falls from a vertical position with its lower end on a perfectly smooth horizontal plane. What is the path of its middle point? 270 ANALYTICAL MECHANICS 3. While passing through the tail of a comet an amount of dust of mass TO settles uniformly upon the surface of the earth. Find the conse- quent change in the length of the day. 4. In the preceding problem find the torque due to the addition of mass. Suppose the passage to take n days and the rate at which mass is acquired to be constant. 6. A particle revolves, on a smooth horizontal plane, about a peg, to which it is attached by means of a string of negligible mass. The string winds around the peg as the particle rotates. Discuss the motion of the particle. 6. A mouse is made to run around the edge of a horizontal circular table which is free to rotate about a vertical axis through the center. Find the velocity of the mouse relative to the table which will give the latter 20 revolutions per minute? The table weighs 2 pounds and has a diameter of 18 inches; the mouse weighs 5 ounces. 7. In the preceding problem find the velocity of the mouse with re- spect to the ground. 8. A cylindrical vessel of radius a is filled with a liquid, closed tight, and made to rotate with a constant angular velocity coo about its geomet- rical axis, which is vertical. Suppose the frictional forces between the inner surface of the vessel and the liquid and between the molecules of the liquid to be small, yet enough to transmit the motion to the liquid if the rotation is kept up for a long time. After each particle of water at- tains an angular velocity about the axis given by the relation oj = coor the torque which kept the angular velocity constant is stopped and the liquid is suddenly frozen. What will be the angular velocity of the system if (a) The mass of the vessel is neghgible. (b) The mass is not negligible but the thickness is. Take the ends into account. (c) Neither the mass nor the thickness of the cylinder is negligible. Do not take the ends into account. (d) In (c) take the ends into account. 9. In the preceding problem suppose the distribution of the angular velocity of the liquid about the axis just before it is frozen to be given a~ r by the relation cu = cooe ^ , where r is the distance from the axis. ANGULAR IMPULSE AND ANGULAR MOMENTUM 271 APPLICATION TO SPECIAL PROBLEMS. 212. Ballistic Pendulum. — A ballistic pendulum is a heavy- target which is used to determine the velocity of projectiles. The target, which is suspended from a horizontal axis, is given an angular displacement when it receives the projec- tile. Considering the target and the bullet which is projected into it as an isolated system we apply the principles of the conservation of energy and of the conservation of angular momentiun. Just before the bullet hits the target the angular momentum of the system about the axis is that due to the velocity of the bullet and equals I' -, where /' is the moment of inertia of the bullet about the axis, v is its velocity, and b is its distance from the axis just before it hits the target, Fig. 123. The bullet is supposed to hit the target normally, when the latter is in the equilibriumposition, and to be imbedded in it. The angular momentum just after the buUet hits the target is (/ + /') w, where I is the moment of inertia of the target and co its initial angular velocity. Then, by the conservation of the angular momentum, we have Fig. 123. (/ + /')'o. 1,1 + 1' V= — ;i— CO. (1) If we suppose the energy lost during the impact to be negli- gible the kinetic energy of rotation just after the bullet hits the target equals the potential energy of the system at its position of maximum angular displacement. Therefore I (7 -t- 70 0,2 = (M + m) ga (1 - cos a), (2) where M and m are the masses of the target and of the bullet, respectively, a is the distance of the center of mass of the system from the axis, and a is the maximmn angular dis- 272 ANALYTICAL MECHANICS placement. Eliminating oi between equations (1) and (2) we obtain V = j-,V2 ga (I + r)iM + m){l- cos a). (3) The moment of inertia of the target may be determined by ob- serving the period of oscillation when it is used as a pendulum. It will be shown later* that if P denotes the period then P=2xv/-/±^J (4) ^ (M + m)ga Eliminating (J+7') between equations (3) and (4) we get tt/' V^ _ Pabg (M+to) , / I — cos a Pabg(M+m) . ,a ^' ^^^2- (5) But in practice m is very small compared with M, the bullet is small enough to be considered as a small particle, and a is small; therefore we can neglect m in the numerator, substi- tute to6^ for I', and replace sin ^ by ^. When these simph- fications are introduced into equation (5) we get PagM , 213. Motion Relative to the Center of Mass. — Suppose a rigid body to have a uniplanar motion. Let M be the mass of the body, I its moment of inertia with respect to an axis perpendicular to the plane of the motion, 7, its moment of inertia about a parallel axis through the center of mass, and a the distance between the two axes. Then the angular momentum about the first axis is where v is the veiocity of the center of mass. In the right hand member of the last equation the first term represents * Page 309. ANGULAR IMPULSE AND ANGULAR MOMENTUM 273 the angular momentum of the body due to the motion of its particles relative to the center of mass, while the second term represents the angular momentum of the body due to the motion of its particles mth the center of mass. The second term depends upon the position of the center of mass relative to the axis of rotation. The first term does not at all depend upon this position. It depends upon the distri- bution of the particles of the body about the center of mass. The two terms are, therefore, independent; that is, if the center of mass of a body is suddenly fixed the angular mo- mentiun of the body due to the motion of its particles about the center of mass is not at all affected. On the other hand if the motion about the center of mass is destroyed the angu- lar momentum about a given axis due to the motion of the particles of the body with the center of mass is not changed. In other words motion about the center of mass and motion with the center of mass are distinct and independent* As an illustration of this important fact consider two disks, Fig. 124, of equal mass, radius, and thickness, which have equal and opposite angular velocities about a common axle, and which move with the axle in a direction perpendicular to it. Suppose each of the disks to have two similarly placed holes, as shown in the figure, so that they can be made one soUd piece by dropping a pin in each pair of holes when they are in Une. If the rotational motion is stopped by dropping the pins into the holes, the motion of the axle goes on as if nothing had happened. On the other hand if the motion with the axle is changed or even stopped, the rotations of the disks about the axle are not at all disturbed. Pig. 124. This result holds true for all bodies and systems, whether rigid or not. 274 ANALYTICAL MECHANICS ILLUSTRATIVE EXAMPLE. A uniform circular hoop rotates about a peg on a perfectly smooth horizontal plane; find the angular velocity of the hoop if the peg is sud- denly removed and simultaneously another peg is introduced, about which it begins to rotate. Let and 0', Fig. 125, be the positions of the first and second peg, respectively. The circle in continuous hne may be considered to repre- sent the position of the hoop just before it stops rotating about and just after it begins to rotate about 0'. The only force which comes into play when the hoop strikes the peg 0' passes through 0', hence it pro- duces no effect upon the angular momentum about 0'. Therefore the angular momentum about 0' just after the hoop strikes the peg equals the angular momentum just before. The angular momentum after the hoop begins to rotate about 0' is H'o' = lu' = 2 maW, where H'^' is the angular momentum and J the angular velocity about the 0', m the mass, and a the radius of the hoop. The angular momentum about 0' just before the hoop begins to rotate about 0' equals the angular momentum of the hoop due to the motion of the hoop about its geometrical axis plus its angular momentum due to its motion with its center of mass. Therefore H^ = 1 ci^ + mv • a cos a = ma^u + TOa^oj cos a = ma^o> (1 -|- cos a), where to is the angular velocity about the peg 0, and a the angle which the arc 00' subtends at the center of the hoop. But since H'o' = Ho; 2 ma'^ia' = ma^co (1-|- cos a) and ^, ^1-t-cosa or ti' = 1 -|- cos a ANGULAR IMPULSE AND ANGULAR MOMENTUM 275 where v is the hnear velocity of the center of the hoop while the latter rotates about 0, and v' the velocity afterwards. Discussion. — When a = 0, that is, when the two pegs coincide, «' = o) and v' = v, as they should. When a = ^, «'= -, v'= -• When 2 2^ a = TT, w' = and v' = 0, that is, the hoop comes to rest. PROBLEMS. 1. A rod of negUgible transverse dimensions and length I is moving on a smooth horizontal plane in a direction perpendicular to its length. Show that if it strikes an obstacle at a distance a from its center it will have an angular velocity equal to -rr-, where v is its linear velocity before meeting the obstacle. 2. A uniform circular plate is turning about its geometrical axis on a smooth horizontal plane. Suddenly one of the elements of its lateral sur- face is fixed. Show that the angular velocity after fixing the element equals -, where w is the angular velocity before fixing it. o 3. A circular plate which is rotating about an element of its lateral surface is made to rotate about another element by suddenly fixing the second and freeing the first. Show that co' = to, where oi and o ci>' are the values of the angular velocity of the plate before and after fixing the second element, and a is the angular separation of the two elements when measured at the center of the disk. 4. Three particles of equal mass are attached to the vertices of an equilateral triangular frame of neghgible mass. Show that if one of the vertices is fixed while the frame is rotating about an axis through the center of the triangle perpendicular to its plane the angular velocity is not changed. 5. A square plate is moving on a smooth horizontal plane with a veloc- ity V at right angles to two of its sides. Find the velocity with which it will rotate if (a) one of its corners is suddenly fixed; (b) the middle point of one of its sides is fixed. 6. A uniform rod of negligible transverse dimensions is rotating about a transverse axis through one end. Find the angular uelocity with which it win rotate if the axis is suddenly removed and simultaneously a parallel axis is introduced through the center of mass of the rod. 7. An equilateral triangular plate is rotating about an axis through 276 ANALYTICAL MECHANICS one of the vertices perpendicular to the plane of the plate. Find the resulting angular velocity due to a sudden removal of the axis and a simul- taneous introduction of a parallel axis through the center of mass. 8. In the preceding problem, suppose the new axis to pass through one of the other two vertices. 214. Reaction of the Axis of Rotation.— Suppose B, Fig. 126, to be a rigid body free to rotate about a fixed axis through the point 0, perpendicular to the plane of the figure. If an ex- ternal force F is applied to the body a part of its action is, in gen- eral, transmitted to the axis of rotation. This results in the re- action, R, of the axis, which we will investigate. For the sake of simplicity suppose F to lie in the plane which passes through the center of mass, c, perpendicular to the axis. Since F and R are supposed to be the only external forces acting upon the body, then by equation (VIII) of p. 242 mv = F + R, (1) where ^ is the acceleration of the center of mass. If F„ and F,. denote the components of F along and at right angles to the line Oc, respectively, and P and Q the components of R along the same directions, equation (1) may be resolved into the following component-equations : mf^ = F^ + P, (2) mf^ = F^+Q, (3) where f„ and T, are the components of v. But since the path of the center of mass is a circle 7.2 a and fr ANGULAR IMPULSE AND ANGULAR MOMENTUM 277 where a is the distance of the center of mass from the axis and CO the angular velocity of the body. Making these sub- stitutions in equations (2) and (3) and solving for P and Q we obtain P=-F,+ maui^, (VII) Q=- Ft + mail. (VIII) The magnitude and the direction of R are given by the relations and t&r\.=^, where is the angle R makes with the line Oc. ILLUSTRATIVE EXAMPLE. A uniform rod, which is free to rotate about a horizontal axis through one end, falls from a horizontal position. Find the reaction of the axis at any instant of its fall. Evidently F„= — mg cos 0. Ft = — mg sin d. The negative sign in the first equation is due to the fact that in equa- tion (VII) Fn is supposed to be directed towards the axis, while mg cos 9 is directed away from the axis. The negative sign in the second equation is due to the fact that 6 is measured in the counter-clockwise direction, while ?ng sin d points in the opposite direction. Substituting these values of F„ and Ft in equations (VII) and (VIII), we obtain P = mg cos 6 + macii^, Q = mg sin 6 + maoi. But by the conservation of energy 1 7a)2 = mga cos B, where a is one-half the length of the rod. Therefore „ 2 mga /.So „ o)^ = — :f— cos 9 = —^ cos d I 2a and (Ji = — — ^ sin 9. 4a 278 ANALYTICAL MECHANICS Making these substitutions P = \mg cos 0. Q = Img sin d. 11 = m Vl + 99cos2 0. 4 tan = -h tan B. Discussion. — The reaction and its direction are independent of the length of the rod. When 6 = 0, Q = and /? = P = f wfir. In other words at the instant when the rod passes the lowest point the force on the axis is i times as large as the force when rod hangs at rest. When ^ = o' P = and E = Q = \mg. If the rod is held in a horizontal position by- supporting the free end the reaction of the axis is | mg. But as soon as the support is removed from the free end the reaction on the axis is changed from ^ mg io \ mg. PROBLEMS. 1. A uniform rod which is free to rotate about a horizontal axis falls from the position of unstable equilibrium. Find the reaction of the axis. 2. In the preceding problem find the position where the horizontal component of the reaction is a maximum. 3. A uniform rod which is free to rotate about a horizontal axis falls from a horizontal position. Show that the horizontal component of the reaction is greatest when the rod makes 45° with the vertical. 4. A cube rotates about a horizontal axis which coincides with one of its edges. If at the highest position it barely completes the revolution, show that P = ' W and Q = — — W, where W is the weight of ^ 4 the cube. 6. A cube which is free to rotate about a horizontal axis through one of its edges starts to fall when its center is at the same level as the axis of rotation. Find the reaction of the axis. 6. Show that if the body of § 214 is a particle connected to the axis with a massless rod the reaction perpendicular to the rod vanishes. 7. Consider the reactions of the axis when the latter passes through the center of mass of the rigid body. 8. A circular plate is free to rotate about a horizontal axis which forms one of the elements of its cyUndrical surface. The plate is let fall from the position when its center of mass is vertically above the axis. De- termine the reaction of the axis at 9 = ^ and at = 0. ANGULAR IMPULSE AND ANGULAR MOMENTUM 279 9. A hoop barely completes rotations about a horizontal axis which passes through its rim and is perpendicular to its plane. Determine the reaction of the axis at the lowest and the highest positions. 10. A uniform rod which rotates about a horizontal axis through one end has four times as much kinetic energy as it has potential energy at the instant it passes the highest point. Find the reaction of the axis when the rod is (a) at the highest position; (b) horizontal; (c) at the lowest position. 215. Impulsive Reaction of an Axis. Center of Percussion. — If a rigid body which is free to rotate about a fixed sixis is so struck that no impulse is imparted to the axis during the blow, any point of the Une of action of the blow is called a center of percussion for that axis. It is evident that if the axis be removed and the blow applied at a center of percus- sion which corresponds to the removed axis, the body will rotate as if the axis were not removed. The axis about which a free rigid body rotates when it is given a blow is called the axis of spontaneous rotation. Suppose the rigid body of Fig. 127 to be free to rotate about an axis through perpendicular to the plane of the figure. For the sake of simplicity suppose the blow to be appUed in such a direction that it tends to produce rotation only about the given axis. Let L denote the liaear impulse of the blow and L' the impulse given to the body by the reaction of the axis of rotation. Then by the conservation of Hnear momentum the linear momentum of the body must be equal to the impulse given to it by the blow and by the reaction of the axis. Therefore mv = L + L', (1) Fig. 127. 280 ANALYTICAL MECHANICS where m is the mass of the body and v the velocity of its center of mass. But by the conservation of angular momen- tum the angular momentum of the body about the axis after the blow must equal that of the blow itself. Therefore Jco = Lh, (2) where I is the moment of inertia of the body, co its angular velocity and h the distance of the line of action of the blow from the axis. Eliminating L between equations (1) and (2) and solving for L' we obtain L' = mv-^, (3) (ma — -] CO, (IX) = [ma where a is the distance of the center of mass from the axis of rotation. Equation (IX) gives the impulse produced by the reaction of the axis. If the blow is appUed at a center of percussion L' = 0. Therefore ma —7 = and b= (X) ma ^ ' PROBLEMS. 1. A square plate is moving on a smooth horizontal plane with two of its sides parallel to the direction of motion. Find the angular velocity with which it will rotate, also the impulsive reaction of the axis (a) if one of the corners is fixed; (b) if the middle point of one of the sides is fixed. 2. An equilateral triangular plate is moving on a smooth horizontal plane in a direction perpendicular to one of its sides. Find the resulting angular velocity, also the impulse given by the axis, (a) if one of its corners is fixed; (b) if the middle point of one of its sides is fixed. 3. A hoop is moving on a smooth horizontal plane with its axis perpen- ANGULAR IMPULSE AND ANGULAR MOMENTUM 281 dicular to the plane. Suppose a point on it to be fixed and find expres- sions for tlie resulting angular velocity and impulse imparted. Discuss the expressions for special positions of the fixed point. 4. While a circular plate is moving on a smooth horizontal plane one of the elements of its lateral surface is fixed. Find expressions for the resulting angular velocity and the impulse given by the axis. Discuss the results for special positions of the axis of rotation. 5. A imiform rod Hes on a smooth horizontal plane. Where must a blow be struck so that it rotates about one end? 6. In the preceding problem can the rod be made to rotate about its middle point by a single blow? 7. A circular plate which lies on a smooth horizontal plane is struck so that it rotates about one of the elements of its lateral surface as an axis. Find the position where the blow is apphed. 8. Find the center of percussion of a hoop which is free to rotate about an axis perpendicular to its plane. 9. How must a triangular plate, placed on a smooth horizontal plane j be struck so that it may rotate about one of its vertices? GENERAL PROBLEMS. 1. Two particles of equal mass are connected by a string of length ? and of negUgible mass and placed on a smooth horizontal table so that one of the particles is near an edge of the table and the string is stretched at right angles to the edge. The particle near the edge is given a small displacement so that it begins to fall. Show that the interval of time between the instant at which the second particle leaves the table and the instant at which the string occupies a horizontal position is given by 2 V„ 2. A uniform bar of negligible cross-section, which is rotating on a smooth horizontal plane about a vertical axis, strikes an obstacle and begins to rotate in the opposite direction. If L and U denote the impulses given by the collision to the axis and the obstacle, respectively, co and co' the angular velocities of the bar before and after the collision, I the length and m the mas? of the bar, and a the distance of the obstacle from the axis, show that (a) 0)' = eco; (b)L=m(H-e)^-i4^a,; (c)L' = m(l + e)g^a,. 282 ANALYTICAL MECHANICS 3. A circular table is perfectly free to rotate about a vertical axis through its center. Show that if a man walks completely around the edge of the table the latter turns through an angle of „.. , ' — • 2 ir, where m and M are the masses of the table and of the man, respectively. 4. A circular plate is rotating about its axis, which is vertical, with an angular velocity oi and is moving on a smooth horizontal plane with a linear velocity v. Find the angular velocity it will have if one of the elements of its lateral surface is suddenly fixed, and determine the impulse given to the axis of rotation. Discuss the results for special positions of the fixed axis. 5. A uniform rod strikes at one end against an obstacle while falhng transversely. Show that the impulse which the obstacle receives wiU be one-half that which it would have received if the other end of the rod had struck an obstacle simultaneously with the first. 6. A particle is projected into a tube which is bent to form a circle and is lying on a smooth horizontal table. If the inner surface of the tube is perfectly smooth, show that the center of mass of the two moves in the direction of projection of the particle with a velocity of rrw, while m + M the particle and the center of the tube describe circles about it with an angular velocity - , where M is the mass and a the radius of the tube, while m is the mass and v the velocity of projection of the particle. 7. Find the direction and point of application which an impulse must have in order to make a sphere rotate about a tangent. 8. A uniform rod which is rotating on a smooth horizontal plane about a pivot through its middle point breaks into two equal parts. Determine the subsequent motion. 9. A uniform rod rotates on a smooth horizontal plane about a pivot. What will be the motion when the pivot breaks? 10. A uniform rod falls from a position where its lower end is in con- tact with a rough horizontal plane with which it maJies an angle a. Show that when it becomes horizontal its angular velocity is y ^ " , where I is the length of the rod. 11. Show that in problem (10) the angular velocity wiU be the same when the horizontal plane is smooth. 12. A uniform rod which lies on a smooth horizontal plane is struck at one end, transversely. Show that the energy imparted equals i of the energy which would have been given to the bar by the same blow if the other end of the bar were fixed. CHAPTER XIV. MOTION OF A PARTICLE IN A CENTRAL FIELD OF FORCE. 216. Central Field of Force. — A region is called a central field of force when the intensity of the field at every point of the region is directed toward a fixed point. The fixed point is called the center of the field. The force which a particle experiences when placed in a central field of force is called a central force. 217. Equations of Motions. — Consider the motion of a par- ticle which is projected into a central field of force. It is evident from symmetry that the path will lie in the plane determined by the center of the field and the direction of projection. The expressions for the radial and transverse components of the acceleration are, according to the results of §90, , _ dv fdey ^'~dt' ''[dt)' When the center of the field is chosen as the origin the force acts along the radius vector. Therefore the transverse ac- celeration vanishes. Suppose the force and the acceleration to be functions of the distance of the particle from the cen- ter, then the last two equations become -di^-Adt)=-^^'^' ^'^ |(rM = 0. (II) 283 284 ANALYTICAL MECHANICS where — / (r) is the total acceleration. The negative sign in the right-hand member of equation (I) indicates the fact that the acceleration is directed toward the center, while the radius vector is measured in the opposite direction. Equations (I) and (II) are the differential equations of the motion of a particle in a central field of force. 218. General Properties of Motion in a Central Field. — Integrating equation (II) we get r^co = h, (III) where h is a, constant. The following properties, which are direct consequences of equation (III), are common to all motions in central fields of force. (1) The radius vector sweeps over equal areas in equal intervals of time. When the radius vector turns through an angle dd it sweeps over an area equal to ^r-rdd; therefore the rate at which the area is described equals -r^ — - = -r^co = -h= constant. 2 at 2 2 (2) The angular velocity of the particle varies inversely as the square of the distance of the particle from the center of force. This is evident from equation (III). (3) The hnear velocity of the particle varies inversely as the length of the perpendicular which is dropped upon the direction of the velocity from the center of force. It was shown on page 87 that VGOSd) CO = ^1 r where v is the linear velocity and the angle which the velocity makes with a Une perpendicular to the radius vector. Let p denote the length of the perpendicular dropped from MOTION OF A PARTICLE 285 the center of force upon the direction of the velocity; then it is evident from Fig. 128 that COS0 -P Substituting this value of cos in the preceding equation we obtain pv or v = (IV) Fig. 128. rUo _ h P ~ P (4) The angular momentmn of the particle with respect to the center remains constant. This result is obtained at once by multiplying both sides of equation (III) by to, the mass of the particle. Thus TOr^o) = mh, but mr'^w = loi. Therefore Iw = mh = constant. 219. Equation of the Orbit. — The general equation of the orbit is found by eliminating t between equations (I) and (III) . The analytical reasoning which follows does not need further explanation: dt dr de dr de'dt "dd h dr [by (III)] = -h- G) d i ,du dd 286 ANALYTICAL MECHANICS where u = -. Therefore r (Pr ^ _i^(^ dd df ~ dd^' dt dd^ dV dd Substituting this value of — and the value of -^ , which may be obtained from equation (III), in equation (I), we have de^ ^ hV ^^ for the equation of the orbit. When the law of force is given / (r) is known and the orbit is determined by equation (V) . On the other hand if the orbit is given equation (V) deter- mines the law of force. Thus, if F denotes the force, F = — mf (r) = — mhV I ILLUSTRATIVE EXAMPLE. A particle describes a circle in a / r. ^^^ \ central field of force. Determine the / r^ — \ law of force if the center of the field ^ ^ lies on the path. X Taking the center as the origin, \ / Fig. 129, and the diameter through \ / the origin as the axis and referring \. / the circle to polar coordinates we ob- \ ^ tain Fig. 129 r = 2acos0, (!') 9 a nna f> (1) for the equation of the orbit. Differentiating the last equation dhjiX 1 \ de^ \a cos' Q 2 a cos 5/ ■ u. (2) MOTION OF A PARTICLE 287 Substituting in equation (VI) from equations (1) and (2) we get r, _ _ 8a%hn fo\ Therefore the force varies inversely as the fifth power of the distance from the center of force. The negative sign in the second member of equation (3) shows that the force is directed towards the origin; in other words, it is an attractive force. PROBLEMS. 1. Show that if a particle describes the reciprocal spiral rd = a in a, central field of force, the force is attractive and varies inversely as the cube of the distance from the origin, which is the center of attraction. 2. Show that if a particle describes the logarithmic spiral r = e"* in a central field of force, the expression for the force is F = ^ '-■ 3. A particle moves in a central field of force where the force is away from the center and is proportional to the distance. Show that the orbit is a hyperbola. 4. Show that in the preceding problem the radius vector sweeps over equal areas in equal intervals of time. 6. A particle describes an elhpse in a field of force the center of which is at the center of the elhpse. Show that the force varies directly as the distance and is directed towards the center. 6. In the preceding problem show that the radius vector sweeps over equal areas in equal intervals of time. 7. A particle describes an ellipse in a field of force, the center of which is at one focus. Show that the force is towards the center of force, and is inversely proportional to the square of the distance. 220. Motion of Two Gravitating Particles. — Suppose two particles of masses m and M to move under the action of their mutual gravitational attraction, as in the case of the sun and the earth or the earth and the moon. Then if r is the distance between the centers and y the gravitational con- stant the mutual force of attraction is mM In order to fix our ideas let M be the mass of the sun and m the mass of the earth. Then the sun gives the earth 288 ANALYTICAL MECHANICS an acceleration — y -z , while the earth imparts to the sun an acceleration equal to t — • Suppose at any instant we impart to both the sun and the earth a velocity equal and opposite to that of the sun 71% and apply an acceleration — -. This will bring the sun to rest and keep it at rest, without altering the motion of the earth relative to the sun.* This reduces the problem of the motion of the earth to that of a particle moving in a central field of force where the acceleration is ., . Mm M+m or f(r) = ^. and i^ = -.?' where n= y {M + m). (VII) (VIII) Substituting from equation (VIII) in equation (V) we obtain for the equation of the orbit. Let u' = u— ^, then the equation of the orbit takes the form de^ + u'=0. (2) In order to integrate equation (2) let v = 2^, dd * The acceleration of a particle relative to another moving particle is found by adding the negative of the acceleration of the second particle to that of the first. MOTION OF A PARTICLE 289 then ^ = dv^dv_ de^ dd du' ' and i,^ + u'=0. du' Separating the variable and integrating i;2 = A2 - u'\ Integrating again cos~i — = e + 5 Jx or u' = A cos (9+ 8). Let u' ^ A whenO =0, then 5 = 0. Therefore u'= A cos e (3) is a solution of equation (2). Substituting the value of u' in equation (3), (4) (5) (6) Equation (5) is the well-known equation of a conic section. Therefore the orbit is a conic section with an eccentricity equal to • The expression for the velocity at any point of the orbit flv rid may be obtained by substituting the values of —and — . which M = 7^ + A cos a,nd replacing u by its value ^ ep 1 — e cos e , h^A where e= ■, p= - 1 A 290 ANALYTICAL MECHANICS may be obtained from equations (5) and (III), respectively, in equation (3) of page 82. Thus -(I)'-'" (I) = ^,sm'«+^. (7) Eliminating sin^ 6 between equations (5) and (7) we get i;2=^(e2_l) + 2l' = -^,(.^-1)+^, (8) where k= — = const. ep Therefore v''-— = ^Ae''-\) = const. (9) 221. Conditions which Determine the T3rpe of the Orbit. — Suppose a gravitating body to be projected into the field of another gravitating body, which acts as the center of force; then the type of the orbit is determined by the initial con- ditions, that is, the magnitude and the direction of the velocity of projection and the distance of the particle from the center of force at the instant of projection. Substituting the initial values of v and r in equation (9) and rearranging we obtain the following expression for the eccentricity : W-v} («> 5^= 1 + The character of the orbit is determined by the value of the factor in the parentheses of equation (10). When it van- ishes e is one, therefore the orbit is a parabola; when it is negative e is less than one, therefore the orbit is an ellipse; and when it is positive e is greater than one, therefore the orbit is a hyperbola. We have, therefore, the following criteria : MOTION OF A PARTICLE 291 2k Case I. The orbit is a parabola, when Vo^ = *'o 2k Case II. The orbit is an elUpse, when Vo^< 2k Case III. The orbit is a hyperbola, when Vo^> The general expression for the velocity, which is given by equation (9), may be put ia the following special forms: 2k I. v^ = — , when the orbit is a parabola. II. v^= kl j, when the orbit is an eUipse. III. v^= kl- + -j, when the orbit is a hyperbola. The quantity a is the length of the semi-transverse axis. 222. Velocity from Infinity. — The velocity which the par- ticle acquires in falling towards the center from a point infinitely distant from the center is called the velocity from infinity. This velocity may be computed from the energy equation. Thus \mv^= I Fdr 'Jar, m =/ — dr 1^ 2 Therefore iim r 2m. r But the last equation is identical with the relation which gives the velocity of a particle moving in a parabolic path, therefore if a particle describes a parabolic orbit its velocity at any point of its orbit is equal to the velocity it would have acqiiired if it had started from infinity and arrived at that 292 ANALYTICAL MECHANICS point of the field of force. This fact enables us to state the conditions which determine the type of the orbit in the fol- lowing forms : I. When the velocity of projection equals the velocity from infinity the orbit is a parabola. II. When the velocity of projection is less than the velocity from infinity the orbit is an ellipse. III. When the velocity of projection is greater than the velocity from infinity the orbit is a hyperbola. Thus if a comet starts from rest at an infinite distance from the sun and falls towards the sun its orbit wUl be a parabola. If it is projected towards the sun from an infinite distance its orbit will be a hyperbola. If it falls from rest, starting from a finite distance, its orbit will be an eUipse. 223. Period of Revolution. — From equation (III) we have 7 9 n OjU h = r^a = r^-r- at .-. hdt = r •rde=2 dA, where dA is the area swept over by the radius vector in the time dt. Therefore when the orbit is an ellipse the period of revolution is p dt hJo 2Trab irab dA (irab = area of elUpse) where a and b are the semi-major axis and semi-minor axis of the ellipse, respectively. But by equations (6) h = Vep'.fi b^ /h^ and by the properties of the ellipse ep =—, therefore h=\~^, 0, ~ a MOTION OF A PARTICLE 293 Substituting the last expression for h in that for P we obtain 2 7ral Vy {M + to) It will be noted that the period of revolution depends upon the major axis but not upon the minor axis of the orbit. The results obtained in discussing the motion of two gravi- tating particles are as they appear to an observer who is lo- cated on one of the bodies. The form and size of the orbit, the period of revolution, etc., will be the same whether the observer is located on one or on the other of the two bodies. For instance, to an obseryer on the moon the earth describes an orbit which is exactly similar to the orbit which the moon appears to describe to an observer on the earth. 224. Mass of a Planet which has a Satellite. — In order to fix our ideas let the earth be the planet. Then, since the ac- celeration due to the sun is practically the same on the moon as it is on the earth, the period of revolution of the moon around the earth is the same as if they were not in the gravi- tational field of the sun. Therefore the period of the moon around the earth is _, 2 wa'l v-y {m + m') while that of the earth around the sun is . 2 7raf ^yiM + m) where M, m, and m' are the masses of the sun, of the earth, and of the moon, respectively, a is the semi-major axis of the earth's orbit, and a' that of the moon's orbit. Squaring these equations and dividing one by the other m+m' ^ /P\^ /a/y M + m [p'l '\a/ ' 294 ANALYTICAL MECHANICS Since m! is negKgible compared with m, and m compared with M, the last equation may be written in the form m M =©•(!')■. which gives the ratio of the mass of the planet to that of the sun. 225. Kepler's Laws. — In estabhshing the truth of the law of gravitation Newton showed that the same law which makes the apple fall to the ground keeps the moon in its orbit. Then he extended the application of the law to the other members of the solar system by accounting for the empirical laws which Kepler (1571-1630) had formulated from the observations of Tycho Brahe (1546-1601). The fol- lowing are the usual forms in which Kepler's laws are stated. 1. Each planet describes an ellipse in which the sun occu- pies one focus. 2. The radius vector describes equal areas in equal inter- vals of time. 3. The square of the period of any planet is proportional to the cube of the major axis of its orbit. The first law is, as we have seen, a direct consequence of the inverse square law. The second law follows from equation (III), which holds good for all bodies moving in central fields of force. The third law amounts to stating that the masses of the planets are negUgible compared with the mass of the sun. For if TO, a, and P refer to one planet and to', a', and P' to another planet, then V7 (M + to) Vt, (M + to') Therefore ©' = ©' M +m MOTION OF A PARTICLE 295 Evidently when m and m' are negligible compared with M ©'=©' which is of Kepler's third law. PROBLEMS. 1. The gravitational acceleration at the surface of the earth is about 980 cm. /sec.'' Calculate the mass and the average density of the earth, taking 6.4 X 10* cin. for the mean radius, and supposing it to attract as if all its mass were concentrated at its center. 2. The periods of revolution of the earth and of the moon are, roughly, 365i and 27| days. Find the mass of the moon in tons. Take 6.0 X 10" gm. for the mass of the earth. 3. The periods of revolution of the earth and of the moon are 365i and 27| days, respectively, and the semi-major axes of their orbits are, approximately, 9.5 X 10^ and 2.4 x 10* miles. Find the ratio of the mass of the sun to that of the earth. 4. Taking the period of the moon to be 27 i days, and the radius of its orbit to be 3.85 X 10^" cm., show that the acceleration of the moon, due to the attraction of the earth, is equal to what would be expected from the gravitational law. Assume the gravitational acceleration at the surface of the earth, that is, at a point 6.4 X 10* cm. away from the cm center, to be 980 — '-• sec.'' 6. Show that if the earth were suddenly stopped in its orbit it would fall into the sun in about 62.5 days. 6. Show that if a body is projected from the earth with a velocity of 7 miles per second it may leave the solar system. GENERAL PROBLEMS. 1. Find the expression for the central force under which a particle describes the orbit r" = o" cos nB and consider the special cases when (a) n = J, (c) n = 1, (e) n = 2. (b)n = -i (d)n = 2, 2. A particle moves in a central field of force with a velocity which is inversely proportional to the distance from the center of the field. Show that the orbit is a logarithmic spiral. 296 ANALYTICAL MECHANICS ' 3. A gun can project a shot to a height of — , where R is the radius of the earth. Taking the variation of the gravitational force with altitude, show that the eun can command — ; of the earth's surface. 4. A particle is projected into a smooth horizontal circular groove. The particle is attracted towards a point in the radius which joins the position of projection with the center, with a force equal to 'y Show that in order that the particle may be able to make complete revolutions the ini- tial velocity must not be less than . ., , where a is the radius of the a^ — b^ groove and 6 the distance of the center of force from the center of the groove. 5. A comet describing a parabolic orbit about the sun coUides with a body of equal mass at rest. Show that the center of mass of the two describes a circle about the sun as center. 6. Prove that the least velocity with which a body must be projected from the north pole so as to hit the surface of the earth at the equator is about 4i miles per second, and that the angle of elevation is 22°. 5. 7. A particle moves in the common field of two fixed centers of force of equal intensity. The particle is attracted towards one of the centers with a force which varies as its distance from that center, and repelled from the other center according to the same law. Show that the orbit is a parabola. 8. A particle moves in a field in which the force is repulsive and varies inversely as the square of the distance from the center of force. Show that the orbit is a hyperbola. 9. In the preceding problem show that the radius vector sweeps over equal areas in equal intervals of time. CHAPTER XV. PERIODIC MOTION. 226. Simple Harmonic Motion. — When a particle moves in a straight line under the action of a force which is directed towards a fixed point and the magnitude of which varies directly as the distance of the particle from the fixed point, the motion is said to be simple harmonic. Let 0, Fig. 130, be the fixed point, to, the mass of the par- ticle, and X its distance from 0; then the foregoing definition gives F=-kx, (I') i 'i- 1 * ■ O mX. m Fig. 130. where h is the constant of proportionality. The negative sign in the right-hand member of the equation (I') accounts for the fact that F is directed towards the fixed point, while X is measured in the opposite direction. Substituting this expression for F in the force equation we get mf^=-kx, (I) or — = — co^x, (I") k dv dv where a>^ = — - Substituting u— for — in equation (I") and integrating we have t;2 = C^ — uV. Let v=Vq when x=Q, then c = va. Therefore v = Vvo^-oi'^x'^. (II) 297 298 ANALYTICAL MECHANICS Putting equation (II) in the form and integrating we obtain |-"\/5--' sin~^ — = ut+b, Vo or x = —sm(cot+8) 03 = osin(a><+5), (III) where 5 is the constant of integration and a = — • 03 227. Displacement. — The distance, x, of the particle from the fixed point is called the displacement. 228. Amplitude. — The maximum displacement is called the ampUtude. It is evident from equation (III) that the ampUtude equals a. 229. Phase. — The particle is said to be in the same phase at two different instants, if the displacement and the velocity at the one instant equal, respectively, the displacement and the velocity at the other instant. 230. Period. — The time which elapses between two suc- cessive instants at which the particle is in the same phase is called the period of the motion. In order to find the period we will make use of the definition of a periodic function.* It is evident from equation (III) that x is a periodic function of t; therefore we can write x = a sin [coi + 5] = asm[co(f+P) + 5]. * If any variable a; is a periodic function of any other variable t and if the dependence oi xont is given by the relation x = f (t), then the function satis- fies the following condition: f(t) = f{t+nP), where P is the period and n any positive or negative integer. As an illustra- tion consider the function x = sin 0. This function evidently satisfies the relation sin 9 = sin (9 + n • 2 ir). Therefore 2 tt is the period. PERIODIC MOTION 299 But since sin e is a periodic function of B with a period of 2 t, we have sin 8 = sin (e+2 7r), (IV) 231. Frequency. — The number of complete vibrations which the particle makes per second is called the frequency of the vibration. If n denotes the frequency, then 1 0) ther efore x = asin[a,(« + P) + 5] = a sin [wi + 5 + 2 tt] and consequently P=^^ n = (V) 232. Time-distance Diagram. — Suppose the particle to describe the vertical line AA', Fig. 131, the middle point of Fig. 131. which is the fixed point. Then 0A = OA' = a = the amph- tude. The relation between the position of the particle and the time may be visualized by plotting equation (III) with X as ordinate and t as abscissa. This gives the well-known sine curve. A mental picture of the motion of the particle may be formed by supposing that the particle under consideration is a projection of another particle which moves in a circle of radius a with a constant speed. The second particle and its path may be called the auxiliary particle and the auxiliary circle, respectively. 300 ANALYTICAL MECHANICS (III') 233. Common Forms of Equation (III). — The following are the typical forms in which equation (III) is written: x = a sin {u>t + 5) = a sin oj (i + A)) = a SLXi\--^t-\- s] = asm.-^{t+to) = a cos (coi+ 5') = a cos co (<+ W) = acos(-^i+ 5'] = aGOS-~{t-\-to'), ■K P where 5' = - — 5 and to' = to. 234. Epoch. — The constants to and to' are called epochs, and d and 5' are called epoch angles. The meanings of these constants will be seen from Fig. 131. 235. Velocity. — The following expressions for the velocity of the particle may be obtained either from equation (II) or from equation (III) : V = Vvo^ — la^x^ = CO Va^ — x^ = aw cos {ii>t-\- 5) ' = aw sin ( coi + 5 + (I) Is the displacement - time curve (II) is the velocity - time curve Fig. 132. It is evident from these expressions that the velocity is a simple harmonic function of the time, that it has the same PERIODIC MOTION 301 period as the displacement, and that it differs in phase from the latter by -, as shown in Fig. 132. 236. Energy of the Particle. — The following do not need further explanation. U Jo dx T^^mv^ 2 -X' ■'-^(a^-x^) 2 TT^a^TO -^^sm^ — (i + fo).(VII) E=T+U = 2 TT^a^m cos^y(«+«. (VI) (VIII) Thus the total energy of the particle is constant and equals the maximum values of the potential and kinetic energies. The total energy varies, evidently, directly as the square of the amplitude and inversely as the square of the period. In Fig. 133, T, U, and V are plotted as ordinates and the time as abscissa, with phase relations which correspond to the curves of Fig. 132. (I) is the U and t Curve. (II) is the T and. t Curve. CIII)is the E and t Curve. Fig. 133. 237. Average Value of the Potential Energy. — Since U may be considered as a function of either x or t, we will find its average value with respect to both variables. Taking and 302 ANALYTICAL MECHANICS a as the limits of x and the corresponding values of < as the limits of t we have a — 0«/o 0*^" =— rx^dx p 2a«^o v.=^£udt* Udx P 6 P P P^ Jo P - fi P PTx, 1 . 4xn* E=3U. = 2Ut. (IX) = |S PROBLEMS. 1. A particle which describes a simple harmonic motion has a period of 5 sec. and an ampUtude of 30 cm. Find its maximum velocity and its maximum acceleration. 2. When a load of mass m is suspended from a helical spring of length L and of negligible mass an extension equal to D is produced. The load is puUed down through a distance a from its position of equilibrium and then set free. Find the period and the amplitude of the vibration. Hooke's law holds true. 3. Within the earth the gravitational attraction varies as the distance from the center. Suppose there were a straight shaft from pole to pole, with no resisting medium in it. What would be the period of oscillation of a body dropped into the shaft? Suppose the earth to be a sphere with a radius of 4000 mUes. 4. In the preceding problem find the velocity with which the body would pass the center of the earth. 5. A particle describes a circle with constant speed. Show that the projection of the particle upon a straight Une describes a simple harmonic motion. 6. The pan of a heUcal spring balance is lowered 2 inches when a weight of 5 pounds is placed on it. Find the period of vibration of the balance with the weight on. * See footnote p. 142. PERIODIC MOTION 303 7. A particle which is constrained to move in a straight line is at- tracted by another particle fixed at a point outside the hne. Show that the motion of the particle is simple harmonic when the force varies as the distance between, the particles. 8. A particle of mass m describes a motion defined by the equation a; = a sin {oil + 5) . Find the average value of the following quantities, with respect to the time, for an interval of half the period : (a) displacement; (e) momentum; (b) velocity; (f) kinetic energy; (c) acceleration; (g) potential energy. (d) force; 9. In problem 8 take the averages with respect to position. 10. In problem 8 suppose the motion to be given by X = a cos CO (t + to) . 11. In problem 10 take the averages with respect to position. 12. In problem 8 suppose the motion to be given by the following equations : I. X = a sin^ ((at + S). II. X = acos^ {(at + 5). III. X = a sin (lit cos {cot + 5) . 238. Composition of Two Parallel Simple Harmonic Motions of Equal Period. Analytical Method. — Suppose xi= aisin((at + 81), (1) and X2 = 02 sin (cot + Sa), (2) to define the motions which a particle would have if acted upon, separately, by two simple harmonic forces. Then the motion which will result when the forces act simultaneously is obtained by adding equations (1) and (2). Thus X = Xi + a;2 = tti sin ((ot + Si) +. 02 sin {cot + Sj). 304 ANALYTICAL MECHANICS Expanding the right-hand member of the last equation and rearranging the terms we get X = (ai cos 81+02 cos 52) sin wt + (tti sin 3i + 02 sin ^2) cos ut = a cos 5 sin cat + a sin S cos ut = a sin {oit + 5), where a cos 5 = ai cos 5i + a^ cos §2, and a sin 5 = Oi sin Si + a^ sin 52. (3) Fig. 134. It is evident from equation (3) that the resulting motion is simple harmonic and has the same period as the component motions. Squaring the last two equations and adding we obtain the amplitude of the motion in terms of the constants of equations (1) and (2). Thus a^ = Oi^ +02^ + 2 aiQi cos (52 — 5i). (4) The phase angle of the motion is evidently defined by ai sin 5i + 02 sin 52 tan 5 = ai cos 5i + 02 cos 52 (5) 239. Graphical Method. — The graph of the resulting mo- tion may be obtained by either of the following methods: (1) Represent the given motions by displacement-time curves, then add the ordinates of these curves in order to PERIODIC MOTION 305 obtain the curve which represents the resultant motion. In Fig. 135 the curves (I) and (II) represent the component motions and curve (III) represents the resultant motion. 4- 1 1 ^1 * rw« '''\ \ V — \J VlSl) Fig. 135. (2) Draw two concentric auxiliary circles with radii equal to the ampUtudes of the component motions; draw a radius in each circle making an angle with the i-axis, equal to the phase angle of the corresponding motion; the vector sum of these radii gives the radius of the auxiliary circle for the resultant motion and the corresponding phase angle. By the help of this auxiliary circle the displacement-time curve of the resulting motion can be drawn without drawing those of the component motions. PROBLEMS. Find the resultant motion due to the superposition of two motions ■defined by the following pairs of equations : (1) Xi = tti sin cot and Xa = 02 sin ( ojJ + - j • (2) a;i = ai sin «f and 2:2 = aacosfcoi — ^j- (3) xi = Oi cos coi and X2 = aicoslut +^y (4) Xi = Ui sin cot and X2 = Oi sin (cat +S). {S) Xi = tti sin cat and X2 = Ch cos ciit. 306 ANALYTICAL MECHANICS (6) Xi = ai sin — t and X2 = 02 cos -^(t + Q. (7) xi = ai cos (lit and X2 = a2 sin (wt + d) . (8) xi = ai cos (at and % = Ch cos (oji + 6). 271- (9) xi = fli cos (ot and 2:2 = 02 cos -^ (i + = ^ when e= a; further 2 sin%os0 'd IT \mQD\ + - sm^ - + 8 ) sm^ - + = Po(l+Jsi = Po(l + |sin2a) ■ ■ ) [by (XII)] [ [when a is small higher I L terms may be neglected J when a is small sin a ^ a I 2~ 2A Therefore Po = P ('-t' (XIV) 242. Simple Pendulum. — A ball which is suspended by means of a string forms a simple pendulum when it is free to swing about a horizontal axis through the upper end of the string, provided the mass of the string is neghgible compared with that of the ball and the radius of the ball is negh- gible compared with the length of the string. If TO denotes the mass of the ball and I the length of the string then Fig. 138, * This is called an elliptic integral. t This expansion is carried out by the Binomial theorem . See Appendix A^ X See Appendix Ai. 312 * ANALYTICAL MECHANICS the moment of inertia of the pendulum equals mP. Therefore substituting this value of Z in the expressions for Po and P and replacing D by i we obtain Po=2t = 2x\/-' » n mgD l 9 (XV) -.4\[^ for the first approximation, and for the second approximation. 243. Equivalent Simple Pendulum. — A simple pendulum which has the same period as a physical pendulum is called the equivalent simple pendulum of the latter. If I denotes the -length of the equivalent simple pendulum, then P=2V"- = 2^\/^- ■•• 1 = ^-^- (XVI) For a given value of D and a given direction of the axis, K Is constant. Therefore if the direction of the axis is not changed I is a function of D alone. If we plot the last equa- tion with I as ordinate and D as abscissa we obtain a curve similar to that of Fig. 139. It is evident from the curve that the value of I is infinitely large for D = 0, but it diminishes rapidly to the minimum value as D reaches the value K. 9 As D is increased further I increases continually. It wiU be ob- 2 K served that for a given value of I greater than — there are two 9 values of D, one of which is less and the other greater than K. PERIODIC MOTION 313 The group of parallel axes about which the rigid body- oscillates with the same period forms two coaxial circular cyl- inders, Fig. 140, whose common axis passes through the cen- ^-%-^ Fig. 139. Fig. 140. ter of mass. The cyhnders which correspond to the minimum value of the period coincide and have a common radius X. PROBLEMS. 1. Find the period of the following physical pendulums : (a) A uniform rod, the transverse dimensions of which are negligible compared with the length, oscillates about a horizontal axis through one end. (b) A sphere suspended from a horizontal axis by means of a string of negligible mass. Discuss the changes in the period as the axis approaches the center of the sphere. (c) A circular flat ring oscillates about an axis which forms an element of the inner surface. (d) A door oscillates about the line of the hinges which make an angle a with the vertical. 2. A sphere of radius a oscillates back and forth in a perfectly smooth spherical bowl of radius 6. Find the period of oscillation. The sphere is supposed to have no rolhng motion. 3. What effect on the period of a pendulum would be produced by a change in the mass of the bob, or of the length of the string, or in the radius of the earth, or in the length of the day, or in the latitude of the, location? 4. A seconds pendulum loses 30 seconds per day at the summit of a. mountain. Find the height of the mountain, considering the earth to be 314 ANALYTICAL MECHANICS a sphere of 4000 miles radius and the gravitational force to vary inversely as the square of the distance from the center of the earth. 5. Given the height of a mountain above the surrounding plain and the period of a pendulum on the plain and on the top of the mountain, find a relation from which the radius of the earth can be computed. 6. Supposing the gravitational attraction within the earth to vary as the distance from the center, find the depth below the surface at which a seconds pendulum will beat 2 seconds. 7. Derive a relation between the distance of a pendulum from the center of the earth and its period. 8. A balloon ascends with a constant acceleration and reaches 400 feet in one minute. What is the rate at which the pendulum gains in the bal- loon? 9. A pendulum of length I is shortened by a small amount M. Show that it will gain about -;^ vibrations in an interval of time of n vibra- tions. n is supposed to be a large integral number. 10. How high above the surface of the earth must a seconds pendu- lum be carried in order that it may have a period of 4 seconds? 11. WhUe a train is taking a curve at the rate of 60 miles per hour a seconds pendulum hanging in the train is observed to swing at the rate of 121 oscillations in 2 minutes. Show that the radius of the curve is about a quarter of a mile. 12. Find the expressions for the least period of oscillation the following bodies can have; also determine the corresponding position of the axes. (a) Rod of negligible transverse dimensions. (d) Sohd cylinder. (b) Square plate of negUgible thickness. (e) SoUd sphere. (c) Circular plate of negligible thickness. (f) Spherical shell. 244. Determination of the Gravitational Acceleration by Means of a Reversible Pendulum. — A physical pendulum which is provided with two convenient axes of vibration is called a reversible pendulum. Let D and D', Fig. 141, denote the distances of the axes from the center of mass. Then the corresponding periods are and P'=2V^^^- ^ gD' PERIODIC MOTION 315 Eliminating K and solving for g we get D'2 _ J)2 g=4: £)'p'2 _ Dp2 (1) Reversible pendulums which are made for the purpose of determining g are so constructed that the two periods are very nearly equal. Therefore we can write P'=P + bP, [&p<:p], and obtain = 4' D' {P + 5Py - DP'' 2)'2 _ 2)2 P^ (Z)' -D) +2 PD'dP + D' (3P)2 [(Spy is neglected] = 4.^ ^^t^ p./l I 2D^ SP\ ^V + d'-d'p) ,D+D' ( _ 2D' SP\ * \ d'-d'p) (2) / The approximate expression which is given in equation (2) is better adapted for computing the value of g from experimental data than the more exact expression given in equation (1). This is due to the fact that (D' — D), which cannot be determined with a high degree of accuracy, enters into equation (1) I as a factor, while it appears only in \ the correction term of equation (2). 245. Bifilar Pendulum. — A rigid body which is suspended by means of two parallel strings, as shown in Fig. 142, is called a bifilar pendulum. X K -S- . \ ^f-D-. Fig. 141. When the body is given an angular displacement about a vertical axis through '■ See Appendix Ai. 316 ANALYTICAL MECHANICS y/////////////////M'/V'/^A g its center of mass and then left to itself it vibrates with a definite period. Let m = the mass of the rigid body,. I = the moment of inertia of the body about a vertical axis through its center of mass, I = the length of each string, 6 = the angular displacement of the bar, * sin0 = 0=-Te. Making these substitutions in the torque equation we get dt mgP'' I e, which is the equation of simple harmonic motion. Therefore D ^ mg is the period of the motion. Second Approximation. — From Fig. 143 we have Fig. 143. e T COS b^, then Va^-b^ is real, radical by c we have d = Ai[e-'-''-'^'-e-^''+''>% The character of the motion is brought out by the graph of equation (XXII), Fig. 145. The graph is easily obtained by drawing the dotted curves, which are _ plotted by considering the ° terms of the right-hand member of equation (XXII) separately, and then adding them geometrically. It is evident from the curve that the value of 6 starts at zero, increases to a maximum, and then diminishes to zero asymptotically. In this case the motion is said to be aperi- odic or dead-beat. Case III. L et a^< b^ then V^2 V - 1 = i and Vfe2 -a^= CO. Fig. 145. Then Va^ this substitution in equation (XXI) we obtain 0= Aie-°'(e'*'-e-"0 = Aie~ '* •2i sin wt * = Ae~ "* sm. oit, * See Appendix Avii. h'^ is imaginary. Let 6^ = ioi. Making (XXIII) PERIODIC MOTION 323 where A = 2 iAi. Equation (XXIII) is the integral equa- tion of harmonic motion with the additional factor e~'", which is called the damping factor. On account of this factor the amplitude of the motion continually diminishes. It is evident from equation (XXIII) that the motion is periodic and has a period P- 2£ CO V62 (XXIV) The character of the motion is brought out clearly by the dis- placement-time curve of Fig. 146. A mental picture of the Fig. 146. damped harmonic motion of a particle may be formed by con- sidering the motion of an auxiliary particle which moves in a logarithmic spiral. If the auxiliary particle describes the logarithmic spiral of the figure in the counter-clockwise di- rection, in such a way as to give the radius vector a constant angular velocity, then the motion of the projection of the auxiUary particle upon the e-axis is damped harmonic. The logarithmic spiral may be used as an auxiliary curve in drawing the graph of equation (XXIII), as the circle is used in drawing a sine cm-ve. 324 ANALYTICAL MECHANICS 249. Logarithmic Decrement. — The logarithm of the ratio of two consecutive amphtudes is constant and is called the logarithmic decrement of the motion. The amplitudes occur whenever the relation tan {(at) = — (J/ is satisfied. Let the first amplitude occur at the instant t = ti; then since the period of the tangent is x, the times of the succeeding amplitudes are given by tan (iat) = tan (wti + n-w), or by t = ty-\ ) 111 where n is a positive integer. Hence, denoting the loga- rithmic decrement by X and the nth amphtude by 0^, we have ex. X = log — ^ (by definition) 01-71 + 2 _ (i 4-^^ = log ^^"; ;|f^(-^^ + ^-) [by (XXIII)] Ae-<'"+-^') sin [a,t = 0. PERIODIC MOTION 325 250. Effect of Damping on the Period. — Substituting the values of a and 6 in the expression for the period, 2t P = Ik' X Vt-_ \2 ^P.(l+^). (XXVI) where Po is the period for the undamped motion. It is evi- dent from equation (XXVI) that the damping increases the period. VIBRATIONS .ABOUT A POSITION OF EQUILIBRIUM. 251. Lagrange's Method. — In the various pendulum prob- lems which we have discussed the vibrating body was consid- ered to be either a particle or a rigid body. These simphfi- cations were necessary because the methods we have used cannot be applied conveniently to compUcated systems. La- grange (1736-1813) introduced into Dynamics a method which can be apphed to any vibrating system. The following is a special case of his method adapted to conservative systems which have only one degree of freedom of motion. Express the potential energy of the system as a function of a properly chosen* coordinate q, so that when expanded in ascending powers of q the first power of q does not appear. Then the potential energy takes the form C7= ft+ p,q^ + ftg' + • • • , (XXVII) where /3o, ^2, etc., are constants. The constant /3o can be * It is shown in books on advanced Dynamics that such a choice is always possible. 326 ANALYTICAL MECHANICS eliminated by taking the origin as the position of zero poten- tial energy. Thus we have U=M'+m'+ ■ ■ ■ • (XXVII') But since the vibrations are supposed to be small, q remains a small quantity diu-ing the motion. Therefore the higher powers of q are neghgible compared with q^- Thus neglect- ing the higher terms we obtain the following expression for the potential energy of the system. U=hm\ (XXVIII) where ^ |8 = 182. The kinetic energy, on the other hand, takes the form ^=iag^ (XXIX) where a is a constant and g = ^ . But since the system is conservative the sum of its dynamical energy remains con- stant. Therefore E= T+U = ^aq^ + i ^q\ (XXX) Differentiating both sides of the last equation with respect to the time, ag + ;8g'=0, (XXXI) which is the differential equation of simple harmonic motion. Therefore we have g = a sin y ^ {t + ^) (XXXII) and P=2xy^. (XXXIII) Hence the main part of Lagrange's method consists of select- ing the coordinate which defines the position of the system in such a way as to make the expressions for the kinetic and potential energies of the forms U=^pq\ PERIODIC MOTION 327 ILLUSTRATIVE EXAMPLES. 1. A weight which is suspended by means of a heUcal spring vibrates in the gravitational field of the earth. Find the expression for the period, taking the mass of the spring into account. Let m = mass of the suspended body. m' = mass of the spring. p = mass per unit length of the spring, L = length of the spring before the body is suspended. D = increase in the length of the spring due to the weight of the suspended body. a = the distance through which the body is pulled down in order to start the vibration. In Fig. 147 let denote the position of equihbrium, A the lowest posi- tion, and B any position of the body. The coordinate in terms of which we want to express the energy of the system must vanish at the position of equilibriima. Therefore we wiU define the position of the suspended body in terms of its distance from the position of equilibrium. The dis- tance win be considered as positive when measured downwards. Let q denote this distance then the kinetic energy of the suspended body equals J mq'. In order to express the kinetic energy of ■ tr — p- the spring in terms of this coordinate let x denote the distance g j of an element of the spring from the point of suspension. ^S t Then the kinetic energy of the entire spring is T. 2 J. x'dm = 2^0 I^^'-P'^ - 2 ^J^ X dx 2 3 3 1 m' ., = 2-3"«- the kinetic energy of the entire system is T-ii-n+fy (1=1) 328 ANALYTICAL MECHANICS By Hooke's law the force which produces the extension of the string is a harmonic force, that is, if Q denotes the force then Q=- kg, where k is a constant. Therefore the potential energy of the system is = k C qdq = ikq\ But Q = mg when q = -D. Therefore mg = kD, or k = ^ . Making this substitution in the expression for the potential energy we obtain U = lfq. Therefore the total energy of the system is E = T+U 1/ ,m'\ Differentiating the last equation with respect to t we obtain 2r' 3J^^+I?«^- (m + f)g ^ + fg = o. which is the equation of simple harmonic motion. Therefore g = a sin / ^-yr — (t + t„) It will be observed that, as in the case of every true harmonic motion, the period is not affected by the amplitude. When the mass of the spring is negligible compared with that of the suspended weight the last two equations become 5 = osini/-^(« + to), P -y/f- Therefore in this case the length of the equivalent simple pendulum equals the stretch in the length of the spring produced by suspending the weight. 2. A particle of mass m is attached to the middle point of a stretched elastic string of natural length L, modulus of elasticity X, and of negligible PERIODIC MOTION 329' mass. Find the period with which the particle will vibrate when dis- placed along the string. Let L' be the stretched length of the string, A the area of its cross-sec- tion, g the distance of the particle from its position of equilibrium, and Tx and T-i the tensile forces of the two parts of the string. Then by Hooke's law we have A L L 2 7'2_> V2 V 2 _., L'-L- 2(7 A L L ' 2 Therefore the resultant force on the particle is n-T -T 4AX 4V where X' = AX. Hence the potential energy equals But since the kinetic energy is given by we obtain -^ = o '"^i^ + 1~ 'f for the total energy of the system. Differentiating the last equation we- get .. , 4X' „ which gives g = asiny— («-!-<„) and P =-k\ 3. A cylinder performs small oscillations inside of a fixed cylinder. Find the period of the motion, supposing the contact between the cyUn- ders to be rough enough to prevent sUding. 330 ANALYTICAL MECHANICS Let m be the mass of the vibrating cylinder and a and 6 the radii of the vibrating and the fixed cylinders, respectively. Then at any instant where T denotes the kinetic energy of the vibrating cylinder, I its mo- ment of inertia about the element of contact and w its angular velocity. But / = I ma', b '/ \ and {h-a)Q, Fig. 148. where v is the linear velocity of the axis of the moving cylinder and Q its angular velocity. Therefore T= imib-ayeK On the other hand we have the following expressions for the potential ■energy : U = mgh = mg(b — a) (1 — cos Q) ■ mg •'-"'[-(-l+s- ••■)']• ■Since 6 is supposed to remain small aU the time, it is permissible to neglect the higher terms of d in the last expression for U. Therefore we have U = imgib-a) e\ 'Thus both T and U are expressed in forms which are adapted to the apph- •cation of Lagrange's method. The total energy of the system is E= im{b-aye^ + img{b-a)e\ Differentiating the last equation with respect to the time we obtain 3{b-a)'e + 2ge = 0. Therefore e = ««-\/^ 3 (6 - a) it + to), .and P = ^y/61iiz«i. * The expansion is carried out by Maclaurin's Theorem. See Appendix Ai. PERIODIC MOTION 331 When the contact is smooth we have and U = i mg (b - a) 6'. Therefore = a sin 1 / r^— {t + «o) , \ — a and V g Thus the length of the equivalent simple penduliun is (6 — a) when the contact is smooth and — ^"-r — - when it is rough. PROBLEMS. 1. A butcher's balance is elongated 1 inch when a weight of 4 pounds is placed in the pan. If the spring of the balance weighs 5 ounces, find the error introduced by neglecting the mass of the spring in calculating the period of oscillation. 2. Find the expression for the period of vibration of mercury in a U-tahe. 3. If in the illustrative problem on p. 329 the particle divides the string in the ratio of 1 to n, show that the period is P = 2 xV — ;~' ^ • ' n' X 4. Find the period of vibration of a homogeneous hemisphere which performs small oscillations upon a horizontal plane which is rough enough to prevent sliding. 5. Find the period of vibration of a homogeneous sphere which makes smaU oscillations in a fixed rough sphere. 6. A particle of mass m is attached to a point on a smooth horizontal table by means of a spring of natural length L. If the particle is pulled so that the spring is stretched to twice its natural length and then let go, show that it wiU vibrate with a period P = 2 (tt + 2) y — , where T is the force necessary to stretch the spring to twice its natural length. The mass of the spring is negligible. 7. Two masses Wi and m^ are connected by a spring of negligible mass. The modulus of elasticity of the spring is such that when mi is fixed nii makes n vibrations per second. Show that when mi is fixed Wi makes vibrations per second. m. 332 ANALYTICAL MECHANICS 8. In the preceding problem suppose both of the particles to be free and show that they make n V ' — vibrations per second. ' mi 9. A string which connects two particles of equal mass passes through a small hole in a smooth horizontal table. One of the particles hangs vertically while the other, which is on the table at a distance D from the hole, is given a velocity y/g Z) in a direction perpendicular to the string. Show that the suspended particle will be in equiUbrium and that if it is sUghtly disturbed it will vibrate with a period of 2 x y — • 10. The piston of a cyUnder, which is in a vertical position, is in equi- Hbrium under the action of its weight and the upward pressure of the gas in the cyhnder. Show that when the cyUnder is given a small displace- ment it wUl vibrate with a period equal to 2 x V ~ > where h is the height ^ g of the piston above the base of the cylinder when the former is at its equi- librium position. Assume Boyle's law to hold. 11. In illustrative problem 2 (p. 328) take the mass of the string into account and obtain the expression for the period of vibration. 12. In problem 6 take the mass of the spring into account and obtain an expression for the period. 13. In problem 7 take the mass of the spring into account and find the expression for the period of vibrations. 14. In problem 8 take the mass of the spring into account and find the expression for the period. 16. A particle is placed at the center of a smooth horizontal table; two particles of the same mass as the first one are suspended by means of strings of neghgible mass, each of which passes over a smooth pulley at the middle point of one of the edges of the table and is attached to the first particle. The particle at the center is given a small displacement at right angles to the strings. Show that it performs small oscillations ^-il with a period of 2 ir y - where a is the distance between the two pulleys. 16. A particle rests at the center of a square table which is smooth and horizontal. Four particles are suspended by means of strings each of which passes over an edge of the table and is connected to the particle on the table. Find the period with which the system wiU vibrate when the particle which is on the table is displaced along one of the strings. The particles have equal mass. Neglect the mass of the strings. PERIODIC MOTION 333 17. A particle is in equilibrium at a point midway between two centers of attraction, which attract the particle with forces proportional to the distance. Show that if the particle is displaced toward one of the centers it win vibrate with a period of , = , where K and K' are the forces VK+ K' which a unit mass would experience when placed at a unit distance from each center of force. APPENDIX A. MATHEMATICAL FORMULA. 335 I. BINOMIAL THEOREM. (a + a;)» = a" + ^ a'-^x + ^^^~^ a^V + n(ri-l)(n-2) a"-'a;3 + »..(i+^). When a; C 1, and consequently x^, x', etc., •C a;.] (f) log(l+a;)=^-|^ + |^-|j+- • .,for-l