BOUGHT WITH THE INCOME FROM THE SAGE ENDOWMENT FUND THE GIFT OF Henrg W. Sage 1S91 A A l'dt>H6\ ^.A^.:^^.'r..i- Date Due NOV 1> ! llfiW i 'J r+*^ _ Cornell University Library TA 350.Z81 Elements of theoretical mechanics. 1924 003 999 111 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/cu31924003999111 THEORETICAL MECHANICS ^The^y^ o. Elements of Theoretical Mechanics ALEXANDER ZIWET Junior Professor of Mathematics in the University of Michigan Revised Edition of "An Elementary Treatise on Theoretical Mechanics/ Especially Designed for Students of Engineering THE MACMILLAN COMPANY LONDON: MACMILLAN & CO., Ltd. 1904 -^hl i OS A,, i^.'iHUv Copyright, 1894, 1904 By the MACMILLAN COMPANY New edition^ set up, electrotyped and printed January, 1904 Berwick & Smith, Norwood, MasB., U.S.A. PREFACE. The present edition differs considerably from the original edi- tion of 1893-94, especially in the third part. It represents essentially the required course in theoretical mechanics as given in the Engineering Department of the University of Michigan. In order to keep within the bounds of a three-hour course extending through a year and within the reach of the mathemati- cal attainments of a second or third year's college student it seemed best to confine the treatment largely to problems in one and two dimensions (except in Statics). Thus the motion of a rigid body about a fixed point had to be omitted, in spite of its importance. But rectilinear motion and rotation about a fixed axis have received more ample treatment, and at least some illus- trations of plane motion have been given. It is hardly necessary to say that the text has been carefully revised throughout, and that the exercises have in part been modified and increased in number. For the sake of completeness, certain fundamental subjects, such as simple and compound harmonic motion, the determina- tion of centroids, motion under central forces, the theory of moments of inertia and principal axes, have been retained in greater fullness than might be thought necessary in so elementary a work. Where a shorter course is required the teacher will find no difficulty in retrenching. Thus, Arts. 129-148, 155-156, 170-174, 176-178, 234-236, 239-240, 248-249,312-313, 324, 334, 382-388,392, 411-417,479-495, 509-51 1, 556-571,637- 663, 702-704 may be omitted, as well as many of the numerous applications and the more difficult exercises. VI PREFACE. The work is not a treatise on applied mechanics, the applica- tions being merely used to illustrate the general principles and to give the student an idea of the uses to which mechanics can be put. It is intended to furnish a safe and sufficient basis, on the one hand for the more advanced study of the science, on the other for the study of its more simple applications. I wish in particular that it may serve to stimulate the study of theoretical mechanics in engineering schools. At a future time, I hope to embody in a more advanced treatise, together with other matter, those por- tions of the old edition which could not find a place in the present volume. To Professor E. R. Hedrick, of the University of Missouri, who had the kindness of reading the manuscript and proofs of a large part of the work, I am greatly indebted for pertinent criti- cism and many valuable suggestions. My thanks are also due to Mr. Wm. Marshall, of the University of Michigan, who has ably assisted me in reading the proofs and eliminating errors. Alexander Ziwet. University of Michigan, January, 1904. CONTENTS. Introduction PAGE I PART I : Geometry of Motion ; Kinematics. CHAPTER I. Geometry of Motion. I. Linear Motion ; Translation and Rotation II. Plane Motion III. Spherical Motion IV. Screw Motion V. Composition and Resolution of Translations Vectors 3 7 i6 i8 CHAPTER II. Kinematics. I. Time II. Linear Kinematics: 1. Uniform rectilinear motion ; velocity 2. Variable rectilinear motion ; acceleration . 3. Applications ....... 4. Rotation ; angular velocity ; angular acceleration 29 30 34 37 49 III. Plane Kinematics : 1. Velocity; composition of velocities ; relative velocity . . 52 2. Applications 58 3. Acceleration in curvilinear motion ..... 62 4. Applications ......... 68 5. Velocities in the rigid body ...... 109 6. Applications . 117 VUl CONTENTS. PART II : Introduction to Dynamics ; Statics. CHAPTER III. Introduction to Dynamics. I. Mass ; Moments of Mass ; Centroids : 1. Mass ; density 2. Moments and centers of mass 3. Centroids of particles and lines 4. Centroids of areas . 5. Centroids of volumes II. Momentum ; Force ; Energy . CHAPTER IV. Statics. I. Forces Acting on the Same Particle II. Concurrent Forces ; Moments . HI. Parallel Forces .... IV. Theory of Couples .... V. Plane Statics : 1. The conditions of equilibrium . 2. Stability ...... 3. Jointed frames .... 4. Graphical methods .... 5. Friction ...... VI. Solid Statics : 1. The conditions of equilibrium . 2. Constraints ..... VII. The Principle of Virtual Work 129 133 137 141 153 IS9 170 '75 '?3 201 208 217 219 230 243 255 258 PART III : Kinetics. CHAPTER V. Kinetics of the Particle. I. Impulses ; Impact of Homogeneous Spheres II. Rectilinear Motion of a Particle . 275 293 CONTENTS. ix III. Free Curvilinear Motion : 1. General principles . . , . , . . .325 2. Central forces .... o ... . 337 IV. Constrained Motion: 1. Introduction ......... 362 2. Motion on a fixed curve ....... 365 3. Motion on a fixed surface . . . . . -375 CHAPTER VI. Kinetics of the Rigid Body. I. General Principles 379 II. Moments of Inertia and Principal Axes : 1. Introduction . 39^ 2. Ellipsoids of inertia ....... 399 3. Distribution of principal axes in space . . . .4'° III. Rigid Body with a Fixed Axis 416 IV. Plane Motion 450 THEORETICAL MECHANICS. INTRODUCTION. The science of theoretical mechanics has for its object the mathematical study of motion. The idea of motion is intimately related to the fundamental ideas of space, time, and mass. It will be convenient to introduce these consecutively. Thus we shall begin with a purely geomet- rical study of motion, without regard to the time consumed in the motion and to the mass of the thing moved, the moving object being considered as a mere geometrical configuration. This intro- ductory branch of mechanics may be called the Geometry of motioa. The introduction of the idea of time will then lead us to study the velocity and acceleration of geometrical configurations. This constitutes the subject-matter of Kinematics proper. The term kinematics is often used in a less restricted sense, so as to include the geometry of motion. Finally, endowing our geometrical points, lines, and other con- figurations with mass, we are led to the ideas of momentum, force, energy, etc. This part of our subject, the most comprehensive of all, has been called Dynamics, owing to the importance of the idea of force in its investigation. For the sake of convenience it is usually divided into two branches, Statics and Kinetics, In statics those cases are considered in which no change of motion is pro- duced by the acting forces, or, as it is commonly expressed, in which the forces are in equilibrium. The investigations of statics are therefore independent of the element of time. Kinetics treats of motion in the most general way. 2.] LINEAR MOTION. PART I: GEOMETRY OF MOTION ; KINEMATICS CHAPTER I. GEOMETRY OF MOTION. I. Linear Motion ; Translation and Rotation. 1. Motion consists in change of position. We begin with the simple case of a point moving in a straight line. The position of a point /" in a Hne is determined by its dis- tance 0P= X from some fixed point or origin, 0, assumed in the line, the length x being taken with the proper sign to express the sense (say forward or backward, to the right or to the left) in which it is to be measured on the line. This sense is also indi- cated by the order of the letters, so that PO = — OP, and OP+F0 = o. The position of a point in a line is thus fully determined by a single algebraical quantity or co-ordinate; viz. by its abscissa x=OP. 2. Let the point P move in the line from any initial position Pg (Fig. i) to any other position P^, and let OP^ = x^, OP^ = x^. — % -4- Fig. 1. This change of position, or displacement, is fully determined by the distance PJ'^ = x^— x^ traversed by the point. 4 GEOMETRY OF MOTION. [3. Now let this displacement P^/'j be followed by another dis- placement in the same line, from P^ to P^, in the same sense as the former, or in the opposite sense. In either case the total, or resultant, displacement Pf,P^ is the algebraic sum of the two dis- placements PJ-'-i, P1P2, which are called its components ; i. e., we have P^P^ = P^P^ + P,P^, or P^P^ + P^P^ + PJ"^ = o, whatever may be the positions of the points P^, P^, P^ in the line. This reasoning is easily extended to any number of compo- nent displacements ; that is, the resultant of any number of con- secutive displacements of a point in a line is a single displacement in the same line equal to the algebraic sum of the components. Similar considerations apply to the motion of a point in a curved line provided the displacements be always measured along the curve. 3. Let us next consider the motion of a rigid body. The term rigid body, or simply body, is used in kinematics to denote a figure of invariable size or shape, or an aggregate of points whose distances from each other remain unchanged. Examples are : a segment of a straight line, a triangle, a cube, an ellipsoid, etc. Imagine such a body brought in any manner from some initial position into any other position. This change of position is called the displaceinent of the body. We shall see (Arts. 30—37) that, even in the most general case, the displacements of three points of the body determine those of all other points, and consequently the displacement of the whole body. There are, however, two special cases of motion, translation and rotation in which the displacement of the body is fully deter- mined by the displacement of a single point ; such motions can be called linear. There are also motions determined by the dis- placements of only two points of the body ; an example of this is plane motion (see Art. 11). 4. The displacement of a rigid body is called a translation when the displacements of all of its points are parallel and equal. It is evident that in this case the displacement of any one point of the 5.] LINEAR MOTION. 5 body fully represents the displacement of the whole body. The translation of a rigid body from one position to another is there- fore measured by the segment PJ^^ of the straight line joining the initial and final positions of any point P of the body. Two or more consecutive translations of a rigid body in the same direction produce a resultant translation in the same direc- tion equal to the algebraic sum of the component translations. 5. When a rigid body has two of its points fixed, the only motion it can have is a rotation about the line joining the fixed points as axis. In a motion of rotation all points of the body excepting those on the axis describe arcs of circles whose centers lie on the axis while the points on the axis are at rest. Thus any point P, not on the fixed axis, is carried from its initial position P^ to its final position P^ along a circular arc whose center C lies on the axis. The rotation is evidently measured by the angle P^CP^ subtended at Chy this arc P^P^ (Fig. 2). Fie. 2. The position of any point P (not on the axis) fully determines the position of the whole body and is given by the angle 6 made by CP with some initial line CO, passing through C at right angles to the axis. If OCP^ = 6^, OCP^ = 6^, the angle P^CP^ =^1-^0 measures the rotation, just as (Art. 2) the distance P^P^ = x^ — x^ measures the displacement of a point, and hence (Art. 4) the trans- lation of a rigid body. Two or more consecutive rotations of a rigid body about the same axis give a resultant rotation about the same axis whose 6 GEOMETRY OF MOTION. t^' angle is the algebraic sum of the angles of the component rotations. 6. The particular case when the rigid body is a plane figure whose motion is confined to its plane deserves special mention. If one point, C, of such a figure be fixed, the figure can only have a motion of rotation about C, the center of rotation, every other point of the figure describing an arc of a circle whose center is C (Fig. 2). 7. We have seen that a translation as well as a rotation is meas- ured by a single algebraical quantity, the translation by a distance, the rotation by an angle. This is the reason why such motions may be called linear. The two fundamental forms of motion, translation and rotation, are thus seen to correspond to the two fundamental magnitudes of metrical geometry, viz., distance and angle. It is to be noticed that both for translations in the same direc- tion and for rotations about the same axis the resultant displace- ment is found by algebraic addition of the components, not only when the components are consecutive motions, but even when they are simultaneous. Thus we may imagine a point P displaced by the amount P^P^ along a straight line, while this line itself is moved along in its own direction by an amount Q^Q^. The resultant dis- placement of P is the algebraic sum Pj/'^ -\- Q Q . 8. Translations being measured by distances or leno-ths and rotations by angles, we need in mechanics a unit of length and a unit of angle. The two most important systems of measurement are the CCS (z. e. centimeter-gram-second) system and the F. P S (' foot-pound-second) system. The former is frequently called th scientific system ; it is based on the international or metric tem of weights and measures. The F. P. S. or British syste " still used in England and the United States almost universall ' engineering practice.* * For fuller information on all questions relating to standards andunuTs T Everett, Illustrations of the C.G.S. system of units, with tables of pkysi slants; London, Macmillan, 1902. II.] PLANE MOTION. 7 9. The unit of length in the C. G. S. system is the centimeter (cm.), i. e. ^i-g- of the meter. The original standard meter is a platinum bar preserved in the Palais des Archives in Paris ; two carefully compared copies, known as prototypes, are kept at the National Bureau of Standards, in Washington, D. C. The meter can be defined as the distance between two marks on the standard meter when at a temperature of o^ C. In the F. P. S. system, the unit of length is the foot, i. e. \ of the standard yard. The original British standard yard is a bronze bar preserved in London. The relation between these two fundamental units of length is, according to the United States Coast and Geodetic Survey Bulletin, No. 9, i88g, I cm. = 0.032 808 2 ft. For practical use we have the following approximate relations : I m. = 3.2808 ft, I ft. = 30.48 cm. I cm. = 0.3937 in. I in. = 2.54 cm. 10. The unit of angle is either the degree, z. e. ^i-^ of one revolution, or the radian, i. e. the angle subtended by a circular arc equal in length to the radius. If a be any angle expressed in radians, and a° , a' , a" the same angle expressed respectively in degrees, minutes, seconds, we have the relations 7r „ '^ , ■"■ ,, 180 10800 648000 or a = 0.017 453^^° = o.OOO igia' = 0.000 004 85a"- II. Plane Motion. 11. The position of a plane figure in its plane is fully deter- mined by the positions of any two of its points since every other point of the figure forms with these two points an invariable tri- angle. But the position of the figure can of course be determined GEOMETRY OF MOTION. [12. in other ways ; for instance, by the position of one point and that of a line of the figure passing through the point ; or by the posi- tion of two hnes of the figure. 12. Let us now consider the motion of a plane figure in its plane from any initial position to any other position. This displacement can be brought about in various ways. Thus, it would be suffi- cient to bring any two points A, B of the figure from their initial positions A^, B^ (Fig. 3) to their final positions A^, B^ This can be done, for instance, by first giving the whole figure a translation Fig. 3. through a distance A^A^ and then a rotation through an angle equal to the angle between AJ3^ and AJi^ ; or by such a rotation followed by the translation. Instead of A we might have selected any other point of the figure. But it is important to notice that the angle of rotation required for a given displacement is always the same, while the translation will differ according to the point selected as center. 13. This leads us to inquire whether the center of rotation can- not be so selected as to reduce the translation to zero. Now any rotation that is to bring A from A^ to A^ must have its center on the perpendicular bisector of A^A^ ; similarly for B. Hence the intersection C of the perpendicular bisectors of A^A^ and BJi^ is the only point by rotation about which both A and B can be brought from their initial to their final positions. That they I5-] PLANE MOTION. actually are so brought follows at once from the equality of the angles AJ^B^ and A^CB^ (and hence of the angles A^CA^ and B^CB^ which are corresponding angles in the equal triangles AfB^ and A^CB^. We thus have the proposition : Any displacement of an inva- riable plane figure in its plane can be brought about by a single rotation about a certain point which we may call the center of the displacement. 14. The construction of the center C given in the preceding article becomes impossible when the bisectors coincide (Fig. 4) and when they are parallel (Fig. 5). In the former case, C is readily found as the intersection of A^B^ and A^B^- In the latter, i. e. whenever A^A^ = B^B^, the center lies at infinity, and the rotation becomes a translation. Fig. 4. C Fig. 5. Any translation may therefore be regarded as a rotation about a center at infinity. 15. Let the figure F pass through a series of displacements. Each displacement has its angle and its center. If the successive positions F^, F^, ... F^ of the figure are taken each very near the preceding one, the angles of rotation will be very small, and the successive centers C^, C^, ... (7 will follow each other very closely. In the limit, i. e. when the series of finite displacements passes into a continuous motion of the figure, the centers C will form a continuous curve {c) and the successive angles of rotation approach zero while the radii of the successive arcs described by lO GEOMETRY OF MOTION. [1 6. any point pass into the normals to the continuous path of that point. The point C about which the figure rotates in any one of • its positions during the motion is now called the instantaneous center ; the locus of the centers, that is the curve {c), is called the fixed centrode, or path of the center. It is apparent that in any position of the moving figure the normals to the paths of all its points must pass through the instantaneous center, and the direction of motion of any such point is therefore at right angles to the line joining it to the center. 16. The centers C'are fixed points of the fixed plane in which the figure F moves. But in any position F^ of this figure some point C\ of i^will coin- cide with the point C^ of the fixed plane. Thus, in the case of finite dis- placements (Fig. 6), let the figure F begin its motion with a rotation of angle d^ about a point C^ of the fixed plane ; let C\ be the point of the moving figure that coin- cides during this rota- tion with Cj. The next rotation, of angle 6.^, takes place about a point C^ of the fixed plane. The point of the movino- figure that now coincides with C^ was brought into the position C^ by the preceding rotation. Its original position is therefore obtained by turning C^C^ back by an angle — Q^ into the posi- tion C^C\. The rotation of angle 0^ about C^ brings a new point O ^ of the moving figure to coincidence with the fixed center C^ ; and the original position C ^ of this point can be determined by first turning C^C^ back about C^ by an angle — ^2 into the position CJD, and then turning the broken line Fig. 6 I9-] PLANE MOTION. I I C^Cfi by a rotation of angle — Q^ about C^ back into the posi- tion C\C'^C\. Continuing this process we obtain, besides the broken line C^C^C^ . ., formed by joining the successive centers of rotation in the fixed plane, a broken line C\C\C\... in the moving figure formed by joining those points of this figure which in the course of the motion come to coincide with the fixed centers. The whole motion may be regarded as a kind of roUing of the broken line C\C'^C'^... over the broken hne C C C 17. In the case of continuous motion each of the broken lines becomes a curve, and we have actual rolling of the curve {c'), or body centrode, over the curve (c), or fixed centrode. The con- tinuous ^notion of an invariable plane figure in its plane may therefore always be produced by the rolling (without sliding) of the body centrode over the fixed centrode. The point of contact of the two curves is of course the instantaneous center. 18. It appears from the preceding articles that the continuous motion of a plane figure in its plane is fully determined if we know the center of rotation for every position of the figure. This center can be found as the intersection of the normals of the paths of any two points of the figure, so that the motion of the figure will be known if the paths of any two of its points are given. This, however, is only one out of many ways of determining plane motion by two conditions. 19. When a circle rolls (without slipping) over a straight line the path of any point on the circumference of the circle is called a cycloid, that of any other point rigidly connected with the roll- ing circle is called a trochoid. When a circle rolls over another circle three cases may be distinguished : {a) when each circle lies outside the other, the corresponding paths are called epicycloid and epitrochoid ; (b) when a smaller circle rolls over the inside of a fixed larger circle, the paths are called hypocycloid and hypo- trochoid ; (c) when a larger circle rolls over an enclosed fixed smaller circle, the paths are called pcricycloid and peritrochoid. 12 GEOMETRY OF MOTION. [20. The following examples illustrate the method of finding the centrodes and the path of any point of the moving figure in ine motion. 20. Elliptic motion : Two points of a plane figure move along two fixed lines that are at right angles to each other. Let A, B (Fig. 7) be the points moving on the lines Ox, Oy; the perpendiculars to these lines erected at A and B intersect at the in- stantaneous center C. Denoting by 2a the invariable distance of A and B, we have OC= AB = 2a for all positions of the moving figure. The fixed centrode ( c ) is therefore a circle of radius 2a described about the intersec- tion O of the fixed lines. To find the body centrode (.:') we must construct the triangle ABC for all possible positions of AB. As BCA is always a right angle, the body centrode will be a circle described on AB as diameter. Hence the whole motion can be produced by the rolling of a circle of radius a within a circle of radius 2a. The student is advised to carry out carefully the constructions indi- cated in this as well as the following problems. Thus, in the present case, draw the moving figure, i. e., the segment AB, in a number of its successive positions in each of the four quadrants, and con- struct the instantaneous center C in every case. This gives a number of points ■ of the fixed cen- trode. Then take any one position of AB and transfer to it as base Fig. 7. Fig. 8. 22.] PLANE MOTION. 1 3 all the triangles ABC previously constructed. The vertices of these triangles all lie on the body centrode. 21. To find the equation of the path of any point P of the moving figure, let this point be referred to a co-ordinate system fixed in, and moving with, the figure (Fig. 8) ; let the middle point O' oi AB be the origin, and O'A the axis O'x' , of this system. Then the co-ordi- nates x' , y of B in this moving system are connected with its co-ordi- nates X, y in the fixed system Ox, Oy by the equations a: = ( (Fig. 9). This case is readily reduced to the preceding one. For, let the circle through O, A, B intersect at B' the perpendicular erected at O to OA, and imagine AB rigidly connected to AB' ; then the points A and B will move, by Art. 21, along OA and OB as desired, pro- 14 GEOMETRY OF MOTION. [23- vided A and £' be made to move along the perpendicular lines OA and 0£'. The figure shows that, since ■:^AB'B = ii^AOB= m, the diameter of the rolling circle is 0C-= AB' = AB/sinco. Fig. 9. 23. Connecting' Rod Motion ; One point A of the figure describes a circle, while another point B moves on a straight line passing through the center O of the circle (Fig. lo). ,^0 The two centrodes are readily constructed by points for a given ratio //a, say 4, 3, or 2. If / > a the fixed centrode consists of two branches having a common asym- ptote ; the body centrode has two branches with a common tangent at A. For I— a both centrocies become circles, one of radius 2a about O, the other of radius a about the point A. For I <^a, the point B can describe only a portion of the crank circle, and the centrodes become closed curves. 24. Conchoidal Motion: A point A of the figure moves along a fixed straight line I, while a line of the figure, I' , containing the point A, always passes through a fixed point B (Fig. 11). Fig. 10. 25-J PLANE MOTION. IS The fixed point B may be regarded as a circle of infinitely small radius, which the line /' is to touch. The instantaneous center is there- fore the intersection C of the perpendiculars erected at ^ on / and at B on /'. The fixed centrode is a parabola whose vertex is B. To prove this we take the fixed line / as axis of jc, the perpendicular OB to it drawn through the fixed point .5 as axis of X. Then, putting -^OBA = (p and OB = a, we have for the co-ordinates of C ^'s- ' ' ■ X = a + y tanp, y ^ a tanp; hence x — a =y'/a, or, for B as origin and parallel axes, y = ax. The proportion _) at 10 h. 11 m., the distance .<4Z) being 46.2 miles. Considering the motion as uniform : (a) What is the velocity ? (^) What is the equation of motion ? ( and CD? (^) If after stopping 10 minutes at Z> the train goes on with the same velocity as before, when will it reach £, 20^ miles beyond £>? ( (^) sound in air, 333 meters per second. (6) Two men starting (in opposite sense) from the same point walk around a block forming a rectangle of sides a, d; if their con- stant speeds are v^, c\, when and where will they meet ? (7) How is the unit of velocity changed if the minute be adopted as unit of time, the unit of length remaining unchanged? (8) The mean distance of the sun being 92^ million miles, find the velocity of light if it takes light 16 m. 42 s. to cross the earth's orbit (a) in miles per second, (/') in centimeters per second. (9) Two trains are running on the same track at the rate of 30 and 24 miles per hour, respectively. If at a certain instant they are 1 2 miles apart, find both graphically and analytically when they will col- lide (a) if they are headed the same way ; (^) if they run in opposite directions. (10) In what latitude is a bullet shot west with a velocity of 1320 ft. per second at rest relatively to the earth's axis, the radius being taken as 4000 miles ? (11) Two trains, one 250, the-other 420 ft. long, pass each other on parallel tracks in opposite directions with equal velocity. A pas- seng-er in the .shorter train observes that it takes the longer train just 6 seconds to pass him. What is the velocity? 2. VARIABLE RECTILINEAR MOTION ; ACCELERATION. 62. If the motion be not uniform, the definition of velocity given in Art. 54 is not applicable, as it would not give any defi- nite meaning to the term. We may of course divide the whole space, or any portion of it, by the corresponding time ; and the quotient so obtained is called the mean, or average, velocity for that space or time. But its value is in general different for different portions of the path. It simply represents that constant velocity with which the space could be described in the same time in which it is actually described. 63. While we cannot speak, generally, of the velocity of a variable motion, we attach a perfectly definite meaning to the 64.] LINEAR KINEMATICS, 35 expression : the velocity of the motion at any particular point or instant. To obtain a mathematical expression for this velocity at the point P, or at the time t, let us consider a point moving in a straight line. Let P (Fig. i8) be its position at the time t, P' its position at the time ^ + A^ ; let the spaces be counted from the point as origin so that 0P= s, OP' = J + Aj. Then, by ____^ p P' Ofc s ^|< — As— A Fig. 18. Art. 62 the quotient As/A.t is the average velocity in the interval PP'. As P' approaches P, i. e., as Aj and A^ approach zero, this average velocity As/ At will approach a definite limit. This limit is called the velocity v of the point at P ; we have thus the definition As ds V = Iim r- = -J- ; ^t=<,At dt' t. e., in variable motion in a straight line the velocity at any par- ticular point of the path, or at any instant of time, is the value, at that point or time, of the first derivative of the space with respect to the time ; in other words, velocity is the time-rate of change of space. This definition includes uniform motion as a special case ; for in this case, v being constant, the equation of uniform motion (i), -^rt. 55, gives dsldt= v. 64. The definition of velocity given in the last article, ds , ^=Jt' (^) enables us to find the velocity if the space be given as a function of the time, say s =f(f) ; and conversely, if the velocity be given as a function of time or space, we find by integrating the differ- ential equation dsfdt = v an integral equation of motion s =f(^t). 36 KINEMATICS. [65. If V be given as function of t, say v = ^(i), we find from (2) (/j = vdt, and hence by integration z/^i-, (3) where s^ is the space described during the time t^^. Similarly, if v be given as a function of s, say v = ■^{s), we have from (2) dt = ds/v, and hence ^0^ 65. We have seen (Art. 56, equation (i")) that in the case of uniform motion the velocity z> ^ (s — -fo)/^, z. f., the rate of change of space with time, is constant. The simplest case of variable motion is that in which the velocity varies uniformly. In recti- linear motion, t/ie rate at which the velocity varies with the titne is called the acceleratioii ; we shall denote it by /. If the velocity vary uniformly, the acceleration is constant, and we have y= (j/ — Z'g)//, where ^ is the time during which the velocity changes from v^ to v. By reasoning analogous to that employed in Art. 63, we find for the acceleration of anv rectilinear motion at the time t dv d^s ^^lit^di^' ^s) that is, in rectilinear motion the acceleration at any point or instant is the value, at that point or instant, of the second derivative of the space with respect to the time. Negative acceleration will thus indicate a decreasing velocity. When the acceleration is constant, the motion is said to be uni- formly accelerated. 66. Conformably to the definition of acceleration, its unit is the "cm. per second per second " in the C. G. S. system, and the "foot per second per second " in the F. P. S. system. As it can rarely be convenient to use two different time units in the unit of acceleration (say, for instance, mile per hour per second), it is 68.] LINEAR KINEMATICS. 37 customary to mention the time unit but once and to speak of an acceleration of so mjiny feet per second, or cm. per second, it being understood that the otlier time unit is also the second. For the dimensions of acceleration we have (see Art. 57) J = VT-' = LT-l Denoting, as in Arts. 58, 59, the concrete value of an accelera- tion by J, its unit by J^, and similarly for length and time, we have the equation which shows that (a) the numerical value y/y^ of an acceleration varies directiy as the square of the unit of time, and inversely as the unit of length ; and ib) the unit of acceleration, J^, varies di- rectly as the unit of length, and inversely as the square of the unit of time. 67. Exercises. (i) A point moving with constant acceleration gains at the rate o\ 30 miles an hour in every minute. Express its acceleration in F. P. S. units. (2) At a place where the acceleration of gravity is g=- 9.810 meters per second, what is the value of ^ in feet per second? (3) A railroad train, 10 minutes after starting, attains a velocity of 45 miles an hour; what was its average acceleration during these 10 minutes ? (4) If the acceleration of gravity, _f = 32 feet per second, be taken as unit, what is the acceleration of the railroad train in Ex. (3) ? 3. APPLICATIONS. 68. Uniformly Accelerated Motion. As in this case the acceler- ation 7' is constant (see Art. 65), the equation of motion (5) d'^s . dv d^=J' °' -dt=J' can readily be integrated : V =jt + C. 38 KINEMATICS. [69. To determine the constant of integration C, we must know the value of the velocity at some particular moment of time. Thus, if V = v^ when t = o, we find v^= C ; hence, substituting this value for C, v-v^ =jt. (6) This equation, which agrees with the definition of/ given in Art. 65, gives the velocity at any time t. Substituting ds^dt for V and integrating again, we find s ^ v^t -\- ^jt^ + C, where the constant of integration, C, must again be determined from given "initial conditions." Thus, if we know that s — s^ when t = o, we find .f, = C ; hence ^-\= ^'0^ + y^'- (7) This equation gives the space or distance passed over in terms of the time. 69. EUminating/ between (6) and (7), we obtain the relation ■^ - -^0 = K^'o + ^y< which shows that in uniformly accelerated motion the space can be found as if it were described uniformly with the mean velocity 70. To obtain the velocity in terms of the space, we have only to eliminate i between (6) and (7) ; we find J(^^-0 =>(.-.„). (8) This relation can also be derived by eliminating dt between the differential equations v = dsjdt, dvjdt =j, which gives vdv ^jds, and integrating. The same equation (8) is also obtained directly from the fundamental equation of motion d'^sjdt''' =j by a process very frequently used in mechanics, viz., by multiplying both members of the equation by dsjdt. This makes the left-hand member the exact derivative of l(ds/dty or Jz/^, and the inte- gration can therefore be performed. 73-] LINEAR KINEMATICS. 39 71. The three equations (6), (7), (8) contain the complete solution of the problem of uniformly accelerated motion. For uniformly retarded motion, taking the direction of motion as positive, we have only to write —j for -\-j. If the spaces be counted from the position of the moving point at the time / = o, we have s,^ = o, and the equations become ^=V+i>^'. (7') K^ - ^'0') =js. (8') If in addition the initial velocity v^ be zero, the point starting from rest at the time / = o, the equations reduce to the following : V =jt, (6") s = \ji\ (7") \v' =js. (8") 72. The most important example of uniformly accelerated motion is furnished by a body falling in vacuo near the earth's surface. Assuming that the body does not rotate during its fall, its motion relative to the earth is a mere translation, and it is sufficient to consider the motion of any one point of the body. It is known from obser\'ation and experiment that under these circumstances the acceleration of a falling body is constant at any given place and equal to about 980 cm., or 32 ft., per second per second ; the value of this so-called acceleration of gravity is usually denoted by g. In the exercises on falling bodies (Art. 74) we make through- out the following simplifying assumptions : the falling body does not rotate ; the resistance of the air is neglected, or the body falls in vacuo ; the space fallen through is so small that g may be regarded as constant ; the earth is regarded as fixed. 73. The velocity v acquired by a falling body after falling from rest through a height h is found from (8") as V = 1 ^2^/2. 40 KINEMATICS. [74- This is usually called the velocity due to the height (or head) h, while is called the height (or head) due to the velocity v. 74. Exercises. (i ) A body falls from rest at a place where ^= 32.2. Find («) the velocity at the end of the fourth second ; (/') the space fallen through in 4 seconds ; (ir) the space fallen through in the iifth second. (2) If a railroad train, at the end of 2 m. 40 s. after leaving the station, has acquired a velocity of 30 miles per hour, what was its acceleration (regarded as constant) ? (3) Galileo, who first discovered the laws of falling bodies, ex- pressed them in the following form : (a) The velocities acquired at the end of the successive seconds increase as the natural numbers ; (3) the spaces described during the successive seconds increase as the odd numbers; (^) the spaces described from the beginning of the motion to the end of the successive seconds increase as the squares of the natural numbers. Prove these statements. (4) A stone dropped into the vertical shaft of a mine is heard to strike the bottom after t seconds ; find the depth of the shaft, if the velocity of sound be given = c. Assume ^'= 4 s., ;: = 332 meters, ^=980. (5) A railroad train in approaching a station makes half a mile in the first, 2000 ft. in the second, minute of its retarded motion. If the motion is z/wj/ir;;*/)' retarded : {a) When will it stop ? {l>) What is the retardation ? {c') What was the initial velocity? (^/) When will its velocity be 4 miles an hour? (6) Interpret equations (6) and (7) geometrically. (7) A body being projected vertically upwards with an initial velocity r'„, (a) how long and (3) to what height will it rise? {c) When and {a) with what velocity does it reach the starting-point ? (8) A bullet is shot vertically upwards with an initial velocity of 1200 ft. per second, {a) How high will it ascend? (^) What is its velocity at the height of 16,000 ft. ? (,r) When willit reach the ground 75-] LINEAR KINEMATICS. 41 again? (^d) With what velocity? (^) At what time is it 16,000 ft. above the ground ? Explain the meaning of the double signs wherever they occur in the answers. (9) With what velocity must a ball be thrown vertically upwards to reach a height of 100 ft. ? (10) A body is dropped from a point ^ at a height AB = h above the ground ; at the same time another body is thrown vertically upward from the point B, with an initial velocity v^. (a) When and (/5) where will they collide? (it) If they are to meet at the height •^ h, what must be the initial velocity ? (ir) If a train can attain its regular speed in 3 minutes and can be brought to rest in the same time, how much time is lost by making four stops of 2 minutes each between two stations ? (12) The barrel of a rifle is 30 in. long; the muzzle velocity is 1300 ft. /sec. ; if the motion in the barrel be uniformly accelerated, what is the acceleration and what the time ? (13) If a stone dropped from a balloon while ascending at the rate of 25 ft. /sec. reaches the ground in 6 seconds, what was the height of the balloon when the stone was dropped ? (14) If the speed of a train increases uniformly after starting for 8 minutes while the train travels 2 miles, what is the velocity acquired? (15) A train running 30 miles an hour is brought to rest, uniformly, in 2 minutes. How far does it travel? (16) Two particles fall from rest from the same point, at a short interval of time r ■ find how far they will be apart when the first par- ticle has fallen through a height A. Take e.g. A := 900 ft., r = ^-^ second. 75. The general problem of rectilinear motion requires the integration of the differential equation de=J' (5) where j is a function of s, i, and v, in connection with the equation ds . J. = ^- (^) 42 KINEMATICS. [76. As these two equations involve four quantities t, s, v,j, a third relation between them, say f{t,s,v,j) = o, (9) is always necessary in order to express three of these four quanti- ties in terms of the fourth. Next to the case of uniformly acceler- ated motion where the relation (9) is simply 7 = const., the most important cases are those when 7 is given as a function of t, of s, or of V, or of both s and v. 76. Whenever in nature we observe a motion not to remain uniform, we try to account for the change in the character of the motion by imagining a special cause for such change. In recti- linear motion, the only change that can occur in the motion is a change in the velocity, i.e. an acceleration (or retardation). It is often convenient to have a special name for this supposed cause producing acceleration or retardation ; we call it force (attraction, repulsion, pressure, tension, friction, resistance of a medium, elasticity, cohesion, etc.), and assume it to be proportional to the acceleration. A fuller discussion of the nature of force and its relation to mass will be found in Arts. 255—272. The present remark is only intended to make more intelligible the physical meaning and applications of the problems to be discussed in the following articles. 77. Acceleration inversely proportional to the square of the dis- tance, /. e. /= /i/j^ where /x is a constant (viz. the acceleration at the distance J = i) and i' is the distance of the moving point from a fixed point in the line of motion. The differential equation (5) becomes in this case ^ = ?' (^°) the first integration is readily performed by multiplying both members by ds\dt so as to make the left-hand member the exact derivative of \{ds\dff or \i?. Thus we find Cds u. h^'=f^J^=-i + C, (II) 78.] LINEAR KINEMATICS. 43 where the constant of integration, C, must be determined from the so-called initial conditions of the problem. For instance, if V = ■z'j, when s = s^, we have iv^^ = — fijs^ + C; hence, eliminat- ing C between this relation and ( 1 1 ), C^^-V) = -m(^-^J- (12) To perform the second integration we solve this equation for V and substitute dsjdt for v : ds , , 2/A 2^ ^'.+^- s or putting v^ + 21x1 s^ = 2fi/fJ,', ds \2n Is — fi' Here the variables s and t can be separated, and we find ^.S^J^^'"^'- (^4) 2M To integrate put s — fi' = ^^. The result will be different according to the signs of /it, /a', and v, which must be determined from the nature of the particular problem. It is easily seen that the methods of integration used in this problem apply whenever^ is given as a function of i' alone. 78. It is an empirical fact that the acceleration of bodies fall- ing in vacuo on the earth's surface is constant only for distances from the surface that are very small in comparison with the radius of the earth. For larger distances the acceleration is found inversely proportional to the square of the distance from the earth's center. By a bold generalization Newton assumed this law to hold generally between any two particles of matter, and this assump- tion has been verified by all subsequent observations. It can therefore be regarded as a general law of nature that any particle 44 KINEMATICS. [79- of matter produces in every other such particle, each particle be- ing regarded as concentrated at a point, an acceleration inversely- proportional to the square of the distance between these points. This is known as Newton's law of universal gravitation, the acceleration being regarded zs, caused by a force of attraction inherent in each particle of matter. It is shown in the theory of attraction that the attraction of a spherical mass, such as the earth, on any particle outside the sphere is the same as if the mass of the sphere were concentrated at its center. The acceleration produced by the earth on any particle outside it is therefore inversely pro- portional to the square of the distance of the particle from the center of the earth. 79. Let us now apply the general equa- tions of Art. "jy to the particular case of a body falling from a great height towards the center of the earth, the resistance of the air being neglected. Let be the center of the earth (Fig. 19), P^ a point on its surface, P^ the initial position of the moving point at the time t=o, P its position at the time t ; let OP^ = R, OP^ = i-„, OP = s ; and let g- be the acceleration at P^, j the acceleration at P, both in absolute value. Then, according to Newton's law, j : g = R" : .y^. This relation serves to determine the value of the con- stant /i in (10) ; for since the acceleration is to have the value g when s ^= R we have \dt'),^^- R'- ^' the minus sign being taken because the acceleration is directed toward the origin 0. We have therefore f^=-gR', 8o.J LINEAR KINEMATICS. 45 so that (lo) becomes in our case df~ s" (15) the minus sign indicating that the acceleration tends to diminish the distances counted from as origin. The integration can now be performed as in Art. jy. Multiply- ing by dsjdt and integrating, we find Jz^^ = gR^js + C. If the initial velocity be zero, we have z^ = o- for j = .y^j ; hence and Here again the minus sign before the radical is selected since the velocity v is directed in the sense opposite to that of the dis- tance s. Substituting dsjdt for v and separating the variables t and s we have R \ 2g y s^ — s ence, integrating as indicated at the end of Art. yy : I '=R ^,^(l/<.„-.) + .o-n-^l'^'). the constant of integration being zero since .«■ = .J,, for t = o. The last term can be slightly simplified by observing that -1 |/ 1 tl^ =i r-r^s-'^i sm ' y I — ti = cos ^u, whence finally : 80, Exercises. ( I ) Find the velocity with which the body arrives at the surface of the earth if it be dropped from a height equal to the earth's radius, and determine the time of falling through this height. Take I? = 4000 miles, ^=32. 46 KINEMATICS. [8i. (2) Interpret equation (17) geometrically. (3) Show that formula (16) reduces to z; = V^gk (Art. 73) with s = Ji ii s^ — J- = A is small in comparison with R. (4) Show that when s^ is large in comparison with R while s differs but slightly from R, the formulae (16) and (17) reduce approximately to »= — -j/^ _j t= ___«-_ l/j- 2\/2g R Hence iind the final velocity and time of fall of a body falling to the earth's surface (fl) from an infinite distance; (/') from the moon (^„=6o^). (5) Derive the expressions for v and t corresponding to (16) and (17) when the initial velocity is z'„. (6) Find the time of fall and final velocity of a meteor if (a) s^ = 2R, v^ = I mile per second, (i) s^ = t,R, v^ = 4 miles per sec- ond, (<:) j-„ = ^R, Vg= 10 miles per second. (7) A particle is projected vertically upwards from the earth's sur- face with an initial velocity z/^. How far and how long will it rise ? (8) If, in (7), the initial velocity be z'„ = 'y/gR, how high and how long will the particle rise ? How long will it take the particle to rise and fall back to the earth's surface ? (9) A body is projected vertically upwards. Find the least initial velocity that would prevent it from returning to the earth, taking g = 32, R = 4000 miles. 81. Acceleration directly proportional to the distance, i. e.j = ks, where /c is a constant and i' is the distance of the moving point from a fixed point in the line of motion. The equation of motion dh ~df = KS (18) can be integrated by the method used in Art. "j"]. The result of the second integration will again be different according to the sign of K. We shall study here only a special case, reserving the general discussion of this law of acceleration until later (see Arts. 125 sq.). 82.] LINEAR KINEMATICS. 47 82. It is shown in the theory of attraction that the attraction of a, spherical mass such as the earth on any point within the mass produces an acceleration directed to the center of the sphere and propor- tional to the distance from this center. Thus, if we imagine a particle moving along a diameter of the earth, say in a straight narrow tube passing through the center, we should have a case of^ the motion represented by equation (i8). To determine the value of k, for our problem we notice that at the earth's surface, that is, at the distance OP^ = R from the center (Fig. 2o), the acceleration must be g. If, there- fore, /denote the numerical value of the acceleration at any distance OP^ s (< i?), we havey : g — s : R, or j = gs j R. But the ac- d^s <^ celeration tends to diminish the distance s, hence —m= — %\S. dr R Denoting the positive constant gjR by /x^, the equation of motion d^'s dF = — u^s, where w = ^ •?. (19) Integrating as in Arts. 77 and 79, we find ^v' = - i^V + C. If the particle starts from rest at the surface, we have v — o when s = R; hence O = — J/t^-^^ + C ; and subtracting this from the preceding equation, we find v=- fiVR'' - s\ (20) where the minus sign of the square root is selected because s and V have opposite sense. 48 KINEMATICS. [83- Writing dsjdt for v and separating the variables, we have I ds dt= — whence I s /= -cos-^-5+ C. H K Ass = R when ^ = o, we have o = - cos"' I + C , or C = o. Solving for s, we find s = R cosfif. (21) Differentiating, we obtain ?' in terms of i : z) = — fiR sin/xi. (22) 83. The motion represented by equations (21) and (22) belongs to the important class of simple harmonic motions (see Arts. 121 sq.). The particle reaches the center when s = o, i. e., when fit = 7r/2, or at the time t = 7r/2/x. At this time the velocit)^ has its maximum'value. After passing through the center the point moves on to the other end, P^, of the diameter, reaches this point when s = — R, i. e. when fxt = tt, or at the time / = 7r//M. As the velocity then vanishes, the moving point begins the same motion in the opposite sense. The time of performing one complete oscillation (back and forth) is called the period of the simple harmonic motion ; it is evidently „ TT 27r r=4 =" 84. Exercises. (i) Equation (19) is a differential equation whose general integral is known to be of the form J = Cj &vap.t + Cj Qosp-t ; determine the constants Cj, C, and deduce equations (21) and (22). ( 2 ) Find the velocity at the center and the period, taking ^=32 and J? = 4000 miles. 87.] LINEAR KINEMATICS. 49 (3) If the acceleration, instead of being directed toward the center, is directed away from it, the equation of motion would be d^sjdt'^ ii^s. Investigate this motion, which can be imagined as produced by a force of repulsion emanating from the center. (4) A point whose acceleration is proportional to its distance from a fixed point O starts at the distance s^ from O with a velocity v^ directed away from O ; how far will it go before returning ? 4. ROTATION ; ANGULAR VELOCITY ; ANGULAR ACCELERATION. 85. A motion of rotation about a fixed axis can be treated in precisely the same way in which we have treated rectilinear motion in the preceding sections. It is only to be remembered that rotations are measured by angles (see Arts. 8— 10), while translations are measured by lengths. 86. The rotation of a rigid body (see Art. 5) about a fixed axis is said to be uniform if equal angles are described in equal times ; in other words, if the angle of rotation is proportional to the time in which it is described. In this case of uniform rotation, the quotient obtained by dividing the angle of rotation, 6, by the cor- responding time, t, is called the angular velocity. Denoting it by 0) we have to ^Ojt; and the equation of motion is e = ^, respectively : FOR TRANSLATION : FOR ROTATION : ^ = ■^o +/^' « = "o + '^A •f = V + 2/^'. ^ = «o^ + i«^'. (7) 92.] LINEAR KINEMATICS. 51 90. Let a point P, whose perpendicular distance from the axis of rotation is OP=r, rotate about the axis with the angular velocity =Ls. Theii, proceeding just as in the case of rectilinear motion (Art. 63) we rnay define the speed, or magnitude of the velocity, at the point P (or at the time i) as ,. Aj- ds V = hm -— = — . *i^o A? dt 95-] PLANE KINEMATICS. 53 It is however found convenient in curvilinear motion to incor- porate in the definition of velocity the idea of the varying direction of the motion by assigning to the velocity at P as its direction that of the tangent to the curve at the point P. The velocity at the point P is therefore represented by laying off on the tangent to the curve at P, in the sense of the motion, a segment PT propor- tional to dsjdt. 94. Velocity is thus defined as having both magnitude and direction and is represented by a rectilinear segment PT. It is assumed, moreover, that these segments representing velocities can be combined and resolved according to the parallelogram law (Arts. 38—45), so that velocity is a vector quantity. The results of this assumption are in complete agreement with observed facts ; a direct experimental verification would generally be difficult. Thus if a point be subjected to two or more simultaneous velocities, the velocity of the resulting motion will be represented by the vector found by geometrically adding the component velocities. A velocity may be resolved into any number of com- ponent velocities whose geometric sum is equal to the given velocity. W'e proceed to consider the most important cases of the reso- lution of a velocity in plane motion. 95. If the curve P^^P (Fig. 22) in which the point moves be re- y Py 1 /^ 1 -^...^ s^-^' y Po 'a 3C X / f Fif. 22. 54 KINEMATICS. [96. ferrea to rectangular cartesian co-ordinates x, y, it is convenient to resolve the velocity v along the axes Ox, Oy into v^, v . De- noting by a the angle at which the tangent to the curve at the point P is inclined to the axis Ox we have V ^ V cosa, V = V sina. As the point 7-* moves along the curve, its projection P on the axis Ox moves along this axis ; and since OP^ = x, the velocity of the rectilinear motion of the point P^ is dxjdt (Art. 63). Similarly, the projection P of /'on the axis (9/ moves along this axis Oy with the velocity dyjdt. It is easily seen that the com- ponents I' , V o{ V are equal respectively to these velocities of P^ and P ; that is : dx dy For we have ,. l^s l^s Ax V = V cosa — cosa lim --7- = cosa lim . — -: - I Ax dx = cosct . lim ~r— = —,' cosa At dt and similarly for 7'_ . As the components v^, v^ are rectangular we have In the language of infinitesimals these results are expressed by saying that the infinitesimal element ds of the path has the com- ponents dx = ds cosct, dy ^ ds sina, and that division by dt gives dx ds dy ds . — - = — - cosa = V cosa = v. —-= —- sma = v sma = v . dt dt " dt dt V 96. If the equation of the path be given in polar co-ordinates, it may be convenient to resolve the velocity v along the radius vector O/'and at right angles to it (Fig. 23). 97-] PLANE KINEMATICS. 55 Let r, 6 be the polar co-ordinates, i/r the angle between v and r ; then v^ = v cosyjr, v^ = v sinyjr. The element ds of the curve has in the same directions the components dr = ds cosi|r, rdO = ds sinyjr. Hence, dividing by dt, we find df ' dt U) 97. As the point P moves along the curve its radius vector OP sweeps out the polar area 5' of the curve, i. e., the area bounded by any two radii vectores and the arc of curve between their ends. If A5 be the increment of this area in the time A^, the limit of the ratio ^S/At, as A/" approaches zero, is called the sectorial velocity dSjdt of the point P (about the origin 0) : dS ,. AS —r- = lim -r — dt ^,=0 ^t It follows from the well-known expression for the element of polar area that in polar co-ordinates and in rectangular cartesian co-ordinates dS ,( dy dx ~dt =K"^-^^)- (') 56 KINEMATICS. [98. 98. In the case of relative motion we have to distinguish be- tween the absolute velocity v q{ 2. point, its relative velocity v^, and the velocity of the body of reference v^. To fix the ideas, imagine a man walking on the deck of a steamboat. His velocity v^ relative to the boat is his velocity of walking ; the velocity of the boat (say with respect to the water regarded as fixed), or more exactly speaking, the velocity of that point of the boat at which the man happens to be at the time, is the velocity j', of the body of reference ; and the velocity with which the man is moving with respect to the water, is his abso- lute velocity. Representing these three velocities by means of their vectors, we evidently find that the absolute velocity v is the geometric sum of the relative velocity v^ and the velocity v^ of the body of reference, just as in the case of displacements of translation (Art. 47). And conversely, the relative velocity is found by geometrically subtracting from the absolute velocity the velocity of the body of reference. It is often convenient to state the last proposition in a some- what different form. Imagine that we give the velocity — v both to the man and to the boat ; then the boat is brought to rest, and the resulting velocity of the man is what was before his relative velocity. Hence the relative velocity is found as the resultant of the absolute velocity and the velocity of the body of reference reversed. 99. Exercises. (i) The components of the velocity of a point are 5 and 3 ft. /sec. and enclose an angle of 135" ; find the resultant in magnitude and direction. (2) Find the components of a velocity of 10 ft. /sec, along two lines inclined to it at 30° and 90°, («) on opposite sides, (<5) on the same side. (3) A man jumps from a car at an angle of 60°, with a velocity of 9 ft. /sec. (relatively to the car). If the car is running 10 miles/hour with what velocity and in what direction does the man strike the ground ? 99-] PLANE KINEMATICS. 57 (4) Two men, A and B, walking at the rate of 3 and 4 miles/hour, respectively, cross each other at a rectangular street corner. Find the relative velocity of A with respect to B in magnitude and direc- tion. (5) How must a man throw a stone from a train running 15 miles an hour to make it move 10 ft. per second at right angles to the track? (6) The velocity of light being 300,000 km. /sec, the velocity of the earth in its orbit 30 km. /sec, determine approximately the con- stant of the aberration of the fixed stars. ( 7 ) A man on a wheel, riding along the railroad track at the rate of 9 miles an hour, observes that a train meeting him takes 3 seconds to pass him, while a train of equal length takes 5 seconds to overtake him. If the trains have the same speed, what is it? (8) A swimmer starting from a point A on one bank of a river wishes to reach a certain point B on the opposite bank. The velocity Z', of the current and the angle !?( < \tz') made hj AB with the current being given, determine the least relative velocity j/j of the swimmer in magnitude and direction. (9) A wheel of radius a rolls on a straight track with constant velocity (of its center) v^. Find the velocity w of a point P on the rim. What point of the rim has the velocity »„? (10) Show that the tangent to the cycloid described by P, Ex. (9), passes through the highest point of the wheel. (11) A straight line in a plane turns with constant angular velocity ti> about one of its points O, while a point P, starting from O, moves along the line with constant velocity v^. Determine the absolute path of P and its absolute velocity v. (12) Show how to construct the tangent and normal to the spiral of Archimedes, r = aO, where d = wt. (13) Show that the tangent to the ellipse bisects the angle between the radii vectores r,, r,, drawn from any point P on the ellipse to the foci. (14) Construct the tangent to any conic, a directrix and the corre- sponding focus being given. (15) Prove the relations (3), Art. 96, by the method of limits. 58 KINEMATICS. [loo. 2. APPLICATIONS. 100. The motion of the piston of a steam engine furnishes inter- esting illustrations of the application of graphical methods in kine- matics. In Fig. 24, let OQ = a he the crank arm, PQ = I = ma the connecting rod, P^P^ = s the "stroke," so that l=ma=^\ms. As P^P^ = A^A^ = 2a, we may regard A^A^ as representing the stroke. The position of the piston head P at the time when the crank pin is at Q will then be found as the intersection N oi A^A^ Fig. 24. with a circle of radius / described about P; in other words, N represents the position of the piston corresponding to the angle AfiQ = ^ in the forward stroke and to the angle AfiQ' = ztt — 6 in the return stroke. The crank may generally be assumed to turn uniformly, making n revolutions per second. The linear velocity of the crank pin Q is therefore n = 2Tra • n = irns. For the piston head P, or for the point JV, we must distinguish between its mean, or average, velocity V, and its variable instan- taneous velocity v at any particular moment. For each revolu- tion of the crank the piston head completes a double stroke so that its mean speed is V— 2ns. Hence we have for the mean speed V of the piston : V 2ns irns' or V= — u. IT 101. The instantaneous velocity v of the piston can be found graphically by constructing, as in Art. 23, the instantaneous I02.J PLANE KINEMATICS. 59 center C (Fig. 25) for the motion of the connecting rod. As the instantaneous motion of the rod FQ is a rotation about C, the c / R ^fV^ r ~ :::::r:^ R^ .A2 \ |a, / p Fig. 25. linear velocities of P and Q must be as their distances from C, i. e. v_CP^ u~~CQ' Through draw a perpendicular to 0P\ if PQ (produced if necessary) meet this perpendicular in R, the similar triangles CPQ and ORQ give : whence V _CP _0R V =- • OR, a i. e. V is proportional to OR. If the scale of velocities be selected so that the constant velocity n is represented by the length a of the crank, the instantaneous piston speed v is represented in length by OR. 102. The variation of the piston speed v in the course of the motion can best be exhibited graphically. Thus a polar curve of piston speed is obtained by laying off on OQ 2i length OR' = OR, for a number of positions of OQ, and joining the points R' by a continuous curve. 60 KINEMATICS. [103. Another convenient method consists in erecting perpendiculars to OP at the various positions of P and laying off, on these per- pendiculars, PR" = OR = V. 103. To derive an analytical expression for the piston speed v, let (^ be the angle OPQ which determines the position of the connecting rod. The triangle ROQ gives by. Art. lOi : V OR sin(6l + <^) - = -7=:-^ = — ^ — -. = smp + cosP tanm. u OQ C0S9 If, as is usually the case, the connecting rod is much longer than the crank arm, (j> will be a small angle, and we may substitute sin^ for tan<^. But from the triangle OPQ we have sm4> OQ a i sin^^7^^ l^ln' Hence V =u\ sin^ + COS0 — sin0 \=u\ sin0 + — sin2^ )• \ m ) \ 2m I 104. The motion of the piston head being rectilinear, we find its acceleration j by differentiating with respect to t the expression for V found in the preceding article : ■ ^^ ( ■ r, ^ ■ n\du ( . I ^ \ ^^ ; = -TT = I smc/ + sm26' — p + ?M cos^ -\ cos2y 1 ~r, dt \ 27)1 ) dt \ m ) dt' or, sin"e dO j aV = to = uj a, j = \ sin6 H sin2^ ) -rr + ( cos^ -\ cos2^ ) ~, ^ \ 2m J dt \ m j a where diildt = o if the crank motion can be regarded as uniform. 105. If the connecting rod were of infinite length so as to make PQ (in Fig. 24) parallel to A^A^, the position of the piston cor- responding to the position Q of the crank pin would be repre- sented by the projection M oi Q ox\. A^A^ ; that is NM would be zero. This length NM is therefore called the deviation due to the obliquity of the connecting rod. lo8.] PLANE KINEMATICS. 6 1 If m = Ija is so large that the connecting rod can be regarded as infinite, the cartesian curve of piston speed (Art. 102) be- comes a circle. And as AW= o the expression for the accelera- tion (Art. 104) reduces to dvfdt = (u'j a) cos0, representing a simple harmonic motion (see Art. 121). 106. The slide valve of a steam engine is generally worked by an eccentric whose radius is set on the shaft at such an angle as to shut off the steam when the crank makes a certain angle 6 with the direction of motion of the piston. It follows that the fraction of stroke completed before cut-off takes place is affected by the obliquity of the connecting rod. The rates of cut-off are there- fore different in the forward and backward strokes. In the for- ward stroke, the effect of the obliquity is to put the piston in advance of the position it would have if the connecting rod were of infinite length ; in the return stroke, t. e. when G is greater than TT, the piston lags behind. 107. An analytical expression for the deviation due to obliquity is readily obtained from Fig. 24. We have MN= PN— PM= l{ I — cos^) = tns sin'i^ = lms{2 sin|<^)^ or approximately, since is small, MN= lins sin^(^. Also, as in Art. 103, sinc^/sin^ = ijm; hence The greatest value of MN is thus seen to be sj^m ; for instance, if the connecting rod be four times the length of the crank, the deviation due to obliquity cannot exceed jJg of the stroke. 108. Exercises.* ( I ) Construct a polar diagram exhibiting the position of the piston * Some of these problems are taken with slight modification from Cotterill's Ap- plied mechanics, 1884, p. 112. 62 KINEMATICS. [109. for all angles by laying off on the crank arm OQ a length OJV' = ON and joining the points JV' by a continuous curve. (2) Construct the curves of piston speed indicated in Art. 102. (3) Show that for a connecting rod of infinite length the two loops of the curve of Ex. (i) reduce to two equal circles. (4) The driving wheels of a locomotive are 6 feet in diameter ; find the number of revolutions per minute and the angular velocity when running at 45 miles per hour. If the stroke be 2 feet, find the mean speed of the piston. (5) The pitch of a screw is 24 ft., and the number of revolutions 70 per minute. Find the speed in knots, a knot being a sea mile per hour = 6090 ft. /hour. If the stroke is 4 ft., find the speed of piston in feet per minute. (6) The stroke of a piston is 4 ft., and the connecting rod is 9 ft. long. Find the position of the crank, when the piston has completed the first quarter of the forward and backward strokes respectively. Also find the position of the piston when the crank is upright. (7) The valve gear is so arranged in the last question as to cut off the steam when the crank is 45° from the dead-points both in the for- ward and backward strokes. Find the point at which steam will be cut off in the two strokes. Also when the obliquity of the connecting rod is neglected. (8) If, in Fig. 24, the crank a = i ft., the connecting rod /= 4 ft., the number of revolutions = 90 per minute, what is the velocity of the cross-head J' when the crank stands at tf ^ 30° ? (9) With m ^ 4, u = const., in what position of the piston in the stroke is its velocity greatest? (10) Determine the velocity and acceleration of a point J? on the connecting rod QP if QJ? = n ■ QP, assuming the rotation of the crank as uniform. 3. ACCELERATION IN CURVILINEAR MOTION. 109. Let the velocity of a moving point be represented by the vector V = PT 2X the time t, and by the vector v' = P T' at the time t + ^t (Fig. 26). Then, drawing from any point O, OVand OV respectively equal and parallel to PT Rnd P'T', the vector VV represents the geometric difference between v' and v ; in no.] PLANE KINEMATICS. 63 Other words, VV is the velocity which, geometrically added to V, produces v'. As At approaches zero, the vector VV ap- proaches zero. But in general the quotient W'/At will approach a definite finite limit and the direction of VV will at the same time approach a definite limiting direction. A rectilinear segment of length . ,. VV laid off in this Hmiting direction, is called the acceleration of the point P at the time t. It can be regarded as the geo- metric derivative with re- spect to the time of the vector representing the velocity. The segments representing accelerations are assumed to follow the parallelogram law of composition and resolution (Arts. 38-45), just like the segments representing velocities (Art. 94) and trans- lations. Acceleration is thus defined as a vector. It will be noticed that the sense of the acceleration is towards that side of the tangent of the curve on which the center of cur- vature is situated. 110. Suppose a point P to move along a curve P^P^P^ . . . with variable velocity v (Fig. 27). From any fixed origin draw a vector 0V^ = v^, equal and parallel to the velocity %\ of P^, and repeat this construction for every position of the moving point P. The ends V,, V^, V^, , . . of all these radii vectores drawn from will form a continuous curve V^V^V^. . . which is called the hodo- graph of the motion of F. If we imagine a point V describing this curve Fj K-, V^ . . . at the same time that P describes the curve P^P^P^ . . ., it is evident that Fig, 26. 64 KINEMATICS. [III. the velocity of V, i. e., lim {VV ji^t), laid off on the tangent of the curve V^V^V^..., represents the acceleration of the point P both in magnitude and direction ; i. e., the velocity on the hodo- graph is the acceleration of the original motion. Fig. 27. 111. Acceleration having been defined as a vector, the rules for vector composition and resolution may be applied to accelera- tion just as they were before applied to displacements and to velocities. Thus, a point subjected to two or more simultaneous accelerations will have a resulting acceleration found by geo- metrically adding the component accelerations ; and conversely, the acceleration of a point may be resolved in various ways. 112, Let the vector / which represents the acceleration of the point P at the time t, make an angle i^ with the vector represent- ing the velocity v at the same time (see Fig. 26). Resolving the vector j along the tangent and normal at P, we obtain the tan- gential acceleration j^ =J cos t/t and the normal acceleration 7;=ysint. To find expressions for these components, let us put arc PP' = As, PT=OV=v, PT = 0V' = v' =^v + Av, and T^VOV =Aa. As Av approaches zero, the direction of VV approaches that of the acceleration / which makes the angle -v/r with the velocity V. By resolving the vector VV along Fand at right angles to it, it appears that 1 1 3-] PLANE KINEMATICS. 6$ cos-f = lim y^,, sin-f = hm yp^ • We find therefore VV 7, =j cosyjr = cosi/c lim — r — VV Av Av dv = cosi/r lim and = cosi|r lim --r — -;— = lim -r— = ■ , ^ Az' A^ At dt W A =^ sini|r = sm-f hm —r— . , ,. VV vAoi ,. Aa ^a = smiir hm — T T-— = hm z/ 7— = z' -r-- ^ vAa At At dt As da da ds 1 = __ = _. J, dt ds dt p where p = dsjda is the radius of curvature at P, we have finally : dv whence 113, When rectangular cartesian co-ordinates are used we may resolve the acceleration^;' into components/^ =y cos <^,j =j sin ^, along the axes Ox, Oy ; ^ being the angle at which the accelera- tion j is inclined to the axis Ox. We obtain an expression fory"^ by projecting the triangle VV (Fig. 26) on the axis Ox and de- noting the projections of the velocities OV, OV by v^, vj. This gives W cosd) ■= V ' —V = Av , whence, dividing by At and passing to the limit : j^ = dvjdt. PART I — 5 66 KINEMATICS. [114. Similarly we find j^ =dvjdt. Hence, by the formulae (i) of Art. 95 : J. dv x_ dt dh df^ Jy dv d'^y 'dt^dt^' (4) These so-called equations of motion offer the advantage that the curvihnear motion is replaced by two rectilinear motions, thus avoiding the use of vectors. By composition, we have of course r- ^y^'-AShm- , sin(£-g) 2Z'/ sin(£ — g) cose J-S — ;; — ) and -ft 9 — ^ • g cosO g cos'9 (7) What elevation gives the greatest range on a horizontal plane? (8) Show that on a plane rising at an angle 6 to the horizon, to obtain the greatest range, the direction of the initial velocity should bisect the angle between the plane and the vertical. (9) A stone is dropped from a balloon which, at a height of 625 ft., is carried along by a horizontal air-current at the rate of 15 miles an hour, (a) Where, (3) when, and (f) with what velocity will it reach the ground ? (10) What must be the initial velocity z'^ of a projectile if, with an elevation of 30°, it is to strike an object 100 ft. above the horizontal plane of the starting point at a horizontal distance from the latter of 1200 ft. ? (11) What must be the elevation e to strike an object 100 ft. above the horizontal plane of the starting point and 5000 ft. distant, if the initial velocity be 1200 ft. per second? (12) Show that to strike an object situated in the horizontal plane of the starting point at a distance x from the latter, the elevation must be £ or 90° — £, where £ = -j- sin~' {_gx/zi^). (13) The initial velocity v^ being given in magnitude and direction, show how to construct the path graphically. (14) The solution of Ex. (11) shows that a point that can be reached with a given initial velocity can in general be reached by two different elevations. Find the locus of the points that can be reached I20.] PLANE KINEMATICS. 73 by only one elevation, and show that it is the envelope of all the para- bolas that can be described with the same initial velocity (in one vertical plane). (15) If it be known that the path of a point is a parabola and that the acceleration is parallel to its axis, show that the acceleration is constant. (16) Prove that a projectile whose elevation is 60'' rises three times as high as when its elevation is 30°, the magnitude of the initial velocity being the same in each case. (17) Construct the hodograph for parabolic motion, taking the focus as pole and drawing the radii vectores at right angles to the velocities. (18) A stone slides down a roof sloping 30° to the horizon, through a distance of 12 ft. If the lower edge of the roof be 50 ft. above the ground, (a) when, (^b) where, ( sm{^cos{t -\- e.,), and being in the same line they can be added algebraically, giving the resultant displacement So KINEMATICS. [i3i- x= X^ + x^ = a^^ cos(&)^ + e,) + a^ cos(a)/ + e^) = («j cose^ + «2 cose^) cosco/ — (a^ sine^ + a^ sine^) sincot. Putting (comp. Art. 125) «j cose^ + a^ cose^ = a cose, «^ sine^ + a^ sine^ = « sine, we have X = a cose coso)^ — rt sine sinta^ = a cos(tB^ + e), where a' = (a^ cosej 4- a^ cose J' + (a^ sine^ + a^ sine^)^ = «j^ + is:^^ + 2(a;j«2 cos(e2 — ej and (2, sine, + a, sine, tane = -^ ^-^ — ^ ^. «j cosej + «2 cose^ 131. A geometrical illustration of the preceding proposition is ob- tained by considering the uniform circular motions corresponding to the two simple harmonic motions (Fig. 36). Fig. 36. Drawing the radii OP^ = a^, OP^ = a^ so as to include an angle equal to the difference of phase e^ — s^ and completing the parallelo- gram OP^PP.^, it appears from the figure that the diagonal OP of this parallelogram represents the resulting amplitude a. As P^P is equal and parallel to OP^, we have for the projections on any axis Ox the relation OP^^ + OP^^ = OP^, or x^ + x^ = x. If now the axis Ox be drawn so as to make the angle xOP^ equal to the epoch-angle ej, and hence xC/'j = e^, the angle .a; OP represents the epoch e of the resulting motion. We thus have a simple geometrical construction for the elements a, £ of the resulting motion from the elements a^, e^ and a^, e^ of the I33-J PLANE KINEMATICS. 8i component motions. As the period is the same for the two com- ponent motions, the points P^ and F^ describe their respective circles with equal angular velocity so that the parallelogram OP^PP^ does not change its form in the course of the motion. 132. The construction given in the preceding article can be de- scribed briefly by saying that two simple harmonic motions of equal period in the same line are compounded by geometrically adding their amplitudes, it being understood that the phase-angles determine the directions in which the amplitudes are to be drawn. Analytically, this appears of course directly from the formula of Art. 130. It follows at once that not only two, but any number of simple har- monic motions, of equal period in the same dne, can be compounded by geometric addition of their amplitudes into a single simple harmonic mo- tion in the same line and of the same period. Conversely, any given simple harmonic motion can be resolved into two or more components in the same line and of the same period. 133. The kinematical meaning of this composition of simple har- monic motions of equal period . _. in the same line will perhaps be best understood from the mechanism sketched in Fig. 37. A cord runs from the fixed point A over the movable pulleys B, D and the fixed pulleys C, E, and ends in F. Each of the J[ If 'I'F movable pulleys receives a verti- cal simple harmonic motion from the T bars EG and DH, just as in Fig. 34 (Art. 127). The mechanism is set in mo- tion by imparting uniform rota- tions to the wheels G, H. If the free end F of the cord be just kept tight, its vertical dis- placement will be twice the sum of the vertical displacements of B and D, and as these points have PART I — 6 82 KINEMATICS. [i34- simple harmonic motions, the motion of ./<'will be twice the resultant simple harmonic motion. The idea of this mechanism is due to Lord Kelvin. 134. Exercises. (i) Find the resultant of three simple harmonic motions in the same line, and all of period 7"= lo seconds, the amplitudes being 5, 3, and 4 cm., and the phase differences 30" and 60"^, respectively, be- tween the first and second, and the first and third motions. (2) If in the proposition of Art. 130 the amplitudes are equal, a^ = a^^ a, while the phase-angles differ by e., — fj = 5, show that the resulting motion has the amplitude 2a cos|-(5 and the phase-angle \^ : (a) directly, (i5) from the formula of Art. 130, ((t) by the geo- metric method of Art. 131. (3) Find the resultant of two simple harmonic motions in the same line and of equal period when the amplitudes are equal and the phases differ: {a) by an even multiple of n-, ((5) by an odd multiple of tt. (4) Resolve jc= 10 cos(7r/+45°) into two components in the same line with a phase difference of 30°, one of the components hav- ing the epoch o. (5) Trace the curves representing the component motions as well as the resultant motion in Ex. (i), taking the time as abscissa and the displacement as ordinate. (6) Show that the resultant oi n simple harmonic motions of equal period T in the same line, viz. : 'f^^+^ij' -^2 = «2 cos ^ ^ / -I- £ A ••• /27r \ x, = a„cosl — /-Fe J, is the isochronous simple harmonic motion X ^ a cos where (?' + ■)• / « \ ( n ^ E^iSine, ' = ( X;«i cose, j -I- [ X;a, sinc^ ) , tane = ■\ ZCl. COS&, 136.] PLANE KINEMATICS. 83 135, The composition of two or more simple harmonic motions in the same line can readily be effected, even when the components differ in period. But the resultant motion is in general not simply liarmonic. Thus, with two components x^ = a^ cos((Bj/' + 6;), x^ = «2 cos(o)/ + ej, putting 0)/ + e^ = o)^t + (co^ — (o^t -{■ e^ = ft)/ + Cj + S, say, where S = (ftjj — oi^t + e^ — e^ is the difference of phase at the time t, we have for the resulting motion x= x^Jr x^= a^ cos((B/ + ej + a^ cos(ft)/ +6^ + 8); = (a^ + «2 cosS) cos(2 ^ =*= '^^r The formula can be interpreted geometrically by Fig. 36, as in Art. 131. But as in the present case the angle 8, and con- sequently the quantities a and e in the expression for x, vary with the time, the parallelogram OP^PP^ while having constant sides has variable angles and changes its form in the course of the motion. A mechanism similar to that of Fig. 37 (Art. 133) can be used to effect mechanically the composition of simple harmonic motions in the same line whether the periods be equal or not. This is the principle of the tide -predicting machine devised by Lord Kelvin. * 136. To show the connection of the present subject with the theory of wave motion, imagine a flexible cord AB of which one * See Thomson and Tait, Natural philosophy. Vol. I., Parti., new edition, 1879, p. 43 sq. and p. 479 sq., and J. D. Evlrett, Vibratory motion and sound, 1882. 84 KINEMATICS. [137. end B is fixed, while the other A is given a sudden jerk or trans- verse motion fi-om A to Cand back through A to D, etc. (Fig. 38). The displacement given to A will, so to speak, run along the cord, travelling from A to B and producing a wave, while any particular point of the cord has approximately a rectilinear motion at right angles to AB. The figure exhibits the successive stages of the Fig. 38. motion up to the time when a complete wave A ' K has been produced. The distance A'K= X is called the length of the wave. Let T be the time in which the motion spreads from A' to K, that is, the time of a complete vibration of the point A, from A to C, back to D, and back again to A ; then is called the velocity of propagation of the wave. 137. Suppose now that the vibration of yi is a simple harmonic motion, say y ^ a sinca/. As the time of vibration of ^ is T we must have <» = 2^1 T, and hence, by (17), A, 138.] PLANE KINEMATICS. 85 If we assume that the vibrations of the successive points of the cord differ from the motion of A only in phase, the displacements of all points of the cord at any time t can be represented by y = a six\{wt — e), where e varies from o to 27r as we pass from A' to K. If we further assume that the phase-angle e of any point of the cord is proportional to the distance x of the point from A' we have e = kx, or since e = 27r for x =X: 277 Substituting the values of a> and e we find 7 = «sin -^(Fi- — x) (18) The assumptions here made can be regarded as roughly sug- gested by the experiment of Art. 136 or .similar observations. The motion represented by the final equation (18) may be called simple harmonic wave motion. 138. To understand the full meaning of the equation (18) it may be observed that, as (in accordance with the assumptions of Art. 137) the quantities a, \, V axe. regarded as constant, the dis- placement J/ is a function of the two variables t and x. If t be given a particular value t^, equation (18) represents the displacements of all points of the cord at the time ty The sub- stitution for xofx+ n\, where n is any positive or negative integer, changes the angle {itt/Xj (Vi — x) by 27r« and hence leaves y unchanged. This means that the displacements of all points whose distances from A differ by whole wave-lengths are the same ; in other words, the state of motion at any instant is given by a series of equal waves. If, on the other hand, we assign a particular value ;r, to x and let t vary, the equation represents the rectilinear vibration of the point whose abscissa is x^ Substituting for ( the value t + nT = i + nXj V, the angle (2'n- /X)(Vt — x) is again changed by 27rn, 86 KINEMATICS. [139. so that y remains unchanged. This shows the periodicity of the motion of any point. 139. It may be well to state once more, and as briefly as pos- sible, the fundamental assumptions that underlie the important formula (18). The idea of simple harmonic wave motion implies that the dis- placement y should be a periodic function of x and t such as to fulfil the following conditions : y must assume the same value (a) when X is changed into n\, {b) when t is changed into t + T, (r) when both changes are made simultaneously ; the constants X and T being connected by the relation X = VT. The condition [c] requires y to be of the form y =/{ Vt — x); for Vt — X remains unchanged when x is replaced by ;r -f- VT and at the same time / by ^ -|- T. A particular case of such a function is r = « smcf^ Vt — x\ As y should remain unchanged when t is replaced \yy t -\- T, we must have c = 27r/ VT = 2irl\. Thus the function 27r jc = « sin — ( F? — jr) fulfils the three conditions («), [b), (c). Putting 27r;r/\ = — e we have y = a sin (^^+«)- (19) The importance of this particular solution of our problem lies in the fact that, according to Fourier's theorem* any single-valued periodic function of period T can be expanded, between definite limits of the variable, in a series of the form : *For a discussion of Fourier's theorem and its applications the student is referred to Thomson and Tait, Natural philosophy, I., i, London, Macmillan, 1879 ; G. M. MlNCHIN, Uniplanar Kinematics, Oxford, Clarendon Press, 1882, pp. 13 sq. ; W. E. Byerly, An elementary treatise on Fourier's series and spherical, cylindrical, and ellipsoidal harmonics, with applications to problems in mathematical physics, Boston, Ginn, 1893 ; and H. Weber, Die partiellen Differentialgleichungen der mathemati- schen Physik, nach Riemann's Vorlesungen, I., Braunschweig, Vieweg, 1900, pp. 32-80. 141.] PLANE KINEMATICS. 8/ /(O = «o + «i sin (-^■f + eA + a^ sin [-^ ■ 2t + eA + «3 sin f y • 32- + Cjj + . . . . (2o) As applied to the theory of wave motion this means that any- wave motion, however complex, can be regarded as made up of a series of superposed simple harmonic wave motions of periods T, ^T, ^T, . ■ ., or since T^\/V, of wave-lengths X, JX, ^X, ••■. For, if the point A (Fig. 38) be subjected simultaneously to more than one simple harmonic motion, the displacements resulting from each can be added algebraically, thus forming a compound wave which can readily be traced by first tracing the component waves and then adding their ordinates. The motion due to the superposition of two or more simple harmonic waves may be called compound harmonic wave motion. 140. Exercises. (i) Trace the wave produced by the superposition of two simple harmonic wave motions in the same line of equal amplitudes, the periods being as 2 : i, (a) when they do not differ in phase, {b) when their epochs differ by 7/16 of the period. (2) In the problem of Art. 135, determine the maximum and mini- mum of the resulting amplitude a and show that the number of maxima per second is equal to the difference of the number of vibrations per second. 141. We proceed to the composition of simple harmonic motions not in the same line. We shall, however, assume that all the component motions lie in the same plane. It is evident that the projection of a simple harmonic motion on any line is again a simple harmonic motion of the same period and phase and with an amplitude equal to the projection of the original amplitude. Hence, to compound any number of simple harmonic motions along hnes lying in the same plane, we may project all these motions on any two rectangular axes Ox, Oy taken in this plane, 88 KINEMATICS. [142. and compound, by Art. 130 or 135, the components lying on the same axis. It then only remains to compound the two motions, one along Ox, the other along Oy, into a single motion. 142. Just as in Arts. 130, 135, we must distinguish two cases: («) When the given motions have all the same period, and (F) when they have not. In the former case, by Art. 130, the two components along Ox and Oj will have equal periods, i. e. they will be of the form X = a cosdit, y = b cos(a>i + B). (21) The path of the resulting motion is obtained by eliminating / between these equations. We have y -7- = cos(ot cosB — sintB^ sinS = — cosS — -»|i ^sinS. Writing this equation in the form (i-^^°^^) =('-^)^'"'^' x^ 2xy y^ ^, ^.^ ^ ,1- / • II cos^^ /sinS\=' we see that it represents an ellipse (since — ^ -tj 2^2 = ( " a a'P ~\ab is positive) whose center is at the origin. The resultant motion is therefore called elliptic harmonic motion. We have thus the general result that any number of simple har- monic motions of the same period and in the same plane, whatever may be tlieir directions, amplitudes, and phases, compotind into a single elliptic harmonic motion. 143. A few particular cases may be noticed. The equation (22) will represent a (double) straight line, and hence the elliptic vibration will degenerate into a simple harmonic vibration, whenever sin^5 = o. I44-] PLANE KINEMATICS. 89 i. e. when 5 = nit, where « is a positive or negative integer. In this case cos5 is + i or — i, and (22) reduces to X y .^ ^ = o, II = 2nnt, a X V and to — h 4 = o, if 5= (2/«-f- i) TT. a b Thus two rectangular vibrations of the same period compound into a simple harmonic vibration when they differ in phase by an integral multiple of tt, that is when one lags behind the other by half a wave- length. Again, the ellipse (22) reduces to a circle only when cos5 = o, i. e. 5 = (2»2 -f i)7r/2, and in addition a=b, the co-ordinates being assumed orthogonal. Thus two rectangular vibrations of equal period and amplitude com- pound into a circular vibration if they differ in phase by 51/2, /. e. if one lags behind the other by a quarter of a wave-length. This circular harmonic motion is evidently nothing but uniform motion in a circle ; and we have seen in Art. 121 that, conversely, uniform circular motion can be resolved into two rectangular simple harmonic vibrations of equal period and amplitude, but differing in phase by 7r/2. The results of Arts. 141-143 can also be established by purely geometrical methods of an elementary character.* 144. It remains to consider the case when the given simple harmonic motions do not all have the same period. It follows from Art. 135 that in this case, if we again project the given motions on two rectangular axes Ox, Oy, the resulting motions along Ox, Oy are in general not simply harmonic. The elimination of t between the expressions for x and y may present difficulties. But, of course, the curve can always be traced by points, graphically. We shall here consider only the case when the motions along Ox and Oy are simply harmonic. * See, for instance, J. G. Macgregor, An elementary treatise on kinematics and dynamics, London, Macmillan, 1887, pp. 115 sq. go KINEMATICS. [i45- 145. 7/" two simple harinonic motions along the rectangular direc- tions Ox, Oy, viz. : X = a^ cos((B/ + ej, y = a.^ cos(tB/ + e^), of different amplitudes, phases, and periods are to be compounded, the resulting motion will be confined within a rectangle whose sides are 7,a^, 2a^, since these are the maximum values of 2X and 2y. The path of the moving point will be a closed curve only when the quotient o'j/'i'i = T^j T^ is a rational number, say = mjn, where m. is prime to n. The x co-ordinate of the curve will have m maxima, the y co-ordinate n, and the whole curve will be traversed after m vibrations along Ox and n along Oy. The formation of the resulting curve will best be understood from the following example. 146. Let a^ = a^^ a, e^ = o, e-^ = d, and let the ratio of the periods be T^j T^ =2/1. The equations of the component simple harmonic vibrations are X =^ a cosuit, J' = a cos(2(u/ + 5). Here it is easy to eliminate t. We have y = a C052uit cos5 — a &{n2(iif sind (x' \ X \ x' 2 -r — I I cos(5 — 2a — -V I J sin5. a' J a \ or Hence the equation of the path is : ay = (^2x' — «') cos5 — zx y^ a^ — x' sin(5. If there be no difference of phase between the components, i. e. if 5 = o, this reduces to the equation of a parabola : o^ = \a,i,y-{- a). For 5 ^ 7r/2, the equation also assumes a simple form : a^y^ = 4x^(a^ — x^). 147. It is instructive to trace the resulting curves for a given ratio of periods and for a series of successive differences of phase (Zissa/ous' s Curves). I47-] PLANE KINEMATICS. 91 Thus in Fig. 39, the curve for TJT^ = ^, and for a phase differ- ence 5 = o is the heavily drawn curve, while the dotted curve repre- sents the path for the same ratio of the periods when the phase differ- ence is one-twelfth of the smaller period. The equations of the components are for the heavy curve 27r X = 6 cos — t, 3 y- 2-K S COS — t, 4 and for the dotted curve (2-K 2-k\ 2-K cos — f. 4 In tracing these curves, imagine the simple harmonic motions re- placed by the corresponding uniform circular motions (Fig. 39). Fig. 39. With the amplitudes 6, 5, as radii, describe the semi-circles ADB, A EC, so that BC 'vs, the rectangle within which the curves are con- fined ; the intersection of the diagonals of this rectangle is the origin O, AB is parallel to the axis oi x, AC to the axis of ^. Next divide the circles on AB, AC into parts corresponding to equal intervals of time. In the present case, the periods for AB, AC being as 3 to 4, the circle on AB must be divided into 3^ equal parts, that on A C into 4«. In the figure, n is taken as 4, the circles being divided into 12 and 16 equal parts, respectively. 92 KINEMATICS. [148- The first point of the heavily drawn curve corresponds to i = o, that is a; = 6, J' = 5 ; this gives the upper right-hand corner of the rectangle. The next point is the intersection of the vertical line through D and the horizontal line through E, the arcs £D =1/12 of the circle over AB, and CE ^ 1/16 of that over AC being de- scribed in the same time, so that the co-ordinates of the correspond- ing point are /^'^ 3 \ f I \ x= 6 cos I — 1 = 6 cosl 2-K • — |> ■>'=5cos(^^-^) = Scos(^2..^). Similarly the next point X = 6 cos I 27r ■ — 1, y = t; cos ( 2n ■ —; \ is found from the next two points of division on the circles, etc. To construct the dotted curve, it is only necessary to begin on the circle over AB with D as first point of division. 148. Exercises. (i) With the data of Art. 147 construct the curves for phase differ- ences of 2/12, 3/12, • •• 1 1 / 1 2 of the smaller period. (2) Construct the curves (Art. 146) X ^ a cosulf, y =: a cos(2(ii/ -f 5) for d = o, 7tJ4, 7r/2, 371/4, TT, 51/4, 37r/2, tk/4, 2:1. (3) Trace the path of a point subjected to two circular vibrations of the same amplitude, but differing in period : (a) when the sense is the same ; (1^) when it is opposite. 149. The mathematical pendulum is a point compelled to move in a vertical circle under the acceleration of gravity. Let be the center (Fig. 40), A the lowest, and B the highest point of the circle. The radius OA = / of the circle is called the length of the pendulum. Any position P of the moving point is determined by the angle AOP= 6 counted from the vertical radius OA in the positive (counterclockwise) sense of rotation. I50.J PLANE KINEMATICS. 93 If /'j be the initial position of the moving point at the time t=0, and 2^ A OP^ = ^5, then the arc PJr' = s described in the time t is s = 1(6^ — 6) ; hence v — dsjdt = — Idd jdt, and dvjdt = — ld^6 jdt^, the negative sign indicating that 6 diminishes as s and t increase. Resolving the acceleration of gravity, g, into its normal and tangential components g cos 6, g sin 6, and considering that the former is without effect owing to the condition that the point is M/^ R ^\n / 0- , \ j Y Q \/r7 Nv yiK \ -^ ^ / J- ii \ y Fig. 40. constrained to move in a circle, we obtain the equation of motion in the form dvjdt = g sin^, or d'e /~+gsme^O. (23) 150. The first integration is readily performed by multiplying the equation by dd jdt which makes the left-hand member an exact derivative, ".[Kl)"---]^ dt\ hence integrating, we obtain or considering that » = — IdOjdt, ^■!/-glcosd= CI. 94 KINEMATICS. [151. To determine the constant C, the initial velocity v^ at the time t=0 must be given. We then have \v^ — gl cosO^ = CI ; hence 2^ = i^o^ - £^^ cos0„ + g/ cos6' = ^( -i - / cos^i^ + / cos6i j. The right-hand member can readily be interpreted geometri- cally; v^jig is the height by falling through which the point would acquire the initial velocity v^ (see Art. 73); / cosO — / cos^^, = OQ — 0Q^= Q^Q, \{ Q, 0„ are the projections of P, P^ on the vertical AB. If we draw a horizontal line MN at the height v^jig above P^ and if this line intersect the vertical AB at R, we have for the velocity v the expression 1 If the initial velocity be zero, the equation would be At the points M, N where the horizontal line MN intersects the circle the velocity becomes zero. The point can therefore never rise above these points. Now, according to the value of the initial velocity v^, the line MN may intersect the circle in two real points M, N, or touch it at B, or not meet it at all. In the first case the point P performs oscillations, passing from its initial position P^ through A up to M, then falling back to A and rising to N, etc. In the third case P makes complete revolutions. 151. The second integration of the equation of motion cannot be effected in finite terms, without introducing elliptic functions. But for the case of most practical importance, viz. for very small values of d, it is easy to obtain an approximate solution. In this case can be substituted for sin0, and the equation becomes : 152.] PLANE KINEMATICS. 95 or, putting^// =^2; This is a well known differential equation (compare Art. 82, eq. (19), and Art. 125), whose general integral is 6 =C^ cosfit + C^ sin/xz". The constants C^, C^ can be determined from the initial condi- tions for which we shall now take 6 = 6^ and v = o when ^ = o ; this gives C^ = 0^, C^ = o ; hence T 6 6=6. cosut, t = - cos~^ -?r. The last equation gives with 6 = — 6^ the time ^j of one swing or deai, that is, half the period : ?=- = ,rJ-. (26) The time of a small oscillation or swing is thus seen to be inde- pendent of the arc through which the pendulum swings ; in other words, for all small arcs the times of s\ving of the same pendulum are very nearly the same ; such oscillations are therefore called isochronous. 152, The formula (26) shows that for a pendulum of given length /j the time of one swing t^ varies for different places owing to the variation of g. As l^ and t^ can be measured very accu- rately, the pendulum can be used to determine g, the acceleration of gravity at any place ; (26) gives : s = -^- (27) Now let /j be the length of a pendulum which beats seconds, i. e. makes just one swing per second ; by (26) and (27) we find for the length /^ of such a seconds pendulum : g L h=-^ = J^- (28) 96 KINEMATICS. [i53- The length /„ of the seconds pendulum is therefore found by measuring the length /^ and the time of swing t^ of any pendulum. This length /^ is very nearly a meter ; it varies slightly with g ; thus, for points at the sea level it varies from /„ = 99.103 cm. at the equator to /^ = 99.610 at the poles. If ^j be the value of ^ at sea level, i. e. at the distance R from the center of the earth, g^ the value of g at an elevation h above sea level in the same latitude, it is known that ^>o {R + Kf Hence, if g^^ be known, pendulum experiments might serve to find the altitude of a place above sea level ; but the observations would have to be of very great accuracy. 153. Let n be the number of swings made by a pendulum of length / in any time T'so that t^ = Tjn. Then, by (26), T fJ — = 7rJ- (29) n \g If T and one of the three quantities n, I, g in this equation be regarded as constant, the small variations of the two others can be found approximately by differentiation. For instance, if the daily number of oscillations of a pendulum of constant length be observed at two different places, T and / keep the same values while n and g vary by small amounts, say Ak and Ag-. Now the differentiation of (29) gives T irV~i dg ^ dn= °,, nr 2 gi or, dividing by (29) : dn dg We have therefore approximately, for small variations A«, Lg : Ln Ag -F = *^- (30) 1 55- J PLANE KINEMATICS. 97 154. Exercises. (i) Find the number of swings made in a second and in a day by a pendulum i meter long, at a placa where ^= 980.5. (2) Find the length of the seconds pendulum at a place where g= 32-17. (3) Find the value of ^ at a place where a pendulum of length 3.249 ft. is found to make 86522 swings in 24 hours. (4) A chandelier suspended from the ceiling is seen to make 20 swings a minute ; find its distance from the ceiling. (5) A pendulum of kngth i meter is carried from the equator where g= 978.1 to another latitude; if it gains 100 swings a day find the value of g there. (6) Investigate whether the approximate formula (30) is sufficiently accurate for Ex. (5). (7) If the length of a pendulum be increased by a small amount Al, show that the daily number of swings, n, will be diminished by An so that approximately An _ _^Al (8) A clock beating seconds is gaining 5 minutes a day ; how much should the pendulum bob be screwed up or down ? (9) A clock beating seconds at a place where g= 32.20 is carried to a place where ^= 32.15 ; how much will it gain or lose per day if the length of the pendulum be not changed ? (10) A pendulum of length 100.18 cm. is found to beat 3585 times per hour ; find the elevation of the place if in the same latitude g= 981.02 at sea level. (11) Show that for small oscillations the motion of the pendulum bob is nearly a simple harmonic motion, and deduce from this fact the relation (26), Art. 151. 155. When the oscillations of a pendulum are not so small that the angle can be substituted for its sine as was done in Art. 151, an expression for the time t^ of one swing can be obtained as follows. We have by (24), Art. 1 50 : \ir - \v^ = glicosd - cos0„). (24) PART I — 7 98 KINEMATICS. [155. Let the time be counted from the instant when the moving point has its highest position {N in Fig. 40), so that v^ = o. Substituting v— — IdOjdt and applying the formula cos^ = I — 2 sin^l^ we find : i/(^^y=2^(sin^H-sin^i0), whence g Vsm%e^ - sin2i6l Integrating from = o to ^ = ^u and multiplying by 2 we find for the time t-^ of one swing : ^ _ \T r"" de As 6 cannot become greater than 0^ we may put sin ^6 = sin J^u sin<^, thus introducing a new variable (^ for which the limits are o and 7r/2. Differentiating the equation of substitution, we have J cosj^ dd = sinj^j cos The integral in this expression is called the complete elliptic integral of the first species, and is usually denoted by K. Its value can be found from tables of elliptic integrals or by expand- ing the argument into an infinite series by the binomial theorem hence 157.] PLANE KINEMATICS. 99 (since k sinc^ is less than i), and then performing the integration. We have (I - «2 sin»-4 = I + Jk2 sin^c^ + ^«* sin^ + • - • ; 2 • 4 '■=-#-(ip^(^y'*— ]■ (34) If 77 be the height of the initial point N(6 = 6^ above the low- est point A of the circle, we have by (3 2) «_ sin 2^0- 2 2/' so that (34) can be written in the form 156. Exercises. (i) Show that /, = 7r J-(i + ^+ -?-4.-ii- + ... jifthe ^ ^ ' yg\ 16 1024 16384 j angle 28^ of the swing is 120°. ( 2 ) Show that as second approximation to the time of a small swing we have ^, = tt |/^(i +-h^o)- (3) Find the time of oscillation of a pendulum whose length is i meter at a place where g= 980.8, to four decimal places, the ampli- tude 0^ of the swing being 6°. (4) Denoting by /„ the first approximation, ■r:V^7/g; to the time /j of one swing, the quotient (/j — ''o)/^o ^^ called the correction for amplitude. Show that its value is 0.0005 for 6^ = 5°. ( 5 ) A pendulum hanging at rest is given an initial velocity »,. Find to what height k^ it will rise. (6) Discuss the pendulum problem in the particular case when MN (Fig. 40) touches the circle at B, that is when the initial velocity is due to falling from the highest point of the circle. 157. Central Motion. The motion of a point P is called central if the direction of the acceleration constantly passes through a lOO KINEMATICS. [158. fixed point 0. The most important case is that when, moreover, the magnitude of the acceleration is a function of the distance OP = r alone, say j=f{r). The fixed point O is in this case usually regarded as the seat of an attractive or repulsive force producing the acceleration, and is therefore called the center of force. Harmonic motion as discussed in Arts. 1 21-148 is a special case of central motion, viz. the case in which the acceleration j is directly proportional to the distance from the fixed center 0, i. e. f{r)=i.r. Another very important particular case is that of Newton's law, i. e.f{r)^= fJi^ji'^; this will be discussed below, Arts. 170—173. 158. Any central motion is fully determined if in addition to the form of the function /(r) we know the " initial conditions," say the initial distance OP^ = r^ (Fig- 41) and the initial velocity 5^^ of the point at the time t = o. As v^ must be given both in magnitude and direction, the angle t/tj between r^ and v^ must be known. It is evident, geomet- rically, that the motion is confined to the plane determined by and v^ since the acceleration always lies in this plane. Hence, any cen- tral motion, whatever may be the law of acceleration, is a plane motion. 159. Another fundamental property is that in any central motion, whatever the law of acceleration, the sectorial velocity is constant. This is most readily proved by taking the center as origin for polar co-ordinates r, 6. As by the definition of central motion (Art. 1 5 7) the acceleration / is directed along the radius vector Fig. 41. i6o.] PLANE KINEMATICS. lOI OP = r drawn from the center to the moving point P, the com- ponent j\ of the acceleration, at right angles to the radius vector, is always zero. We have therefore by the last of the equations (6) of Art. 114: ^'- rdt\ dt)-°' whence ^^ = c, (35) where c is the constant of integration. By Art. 97 this equation means that the sectorial velocity is constant and equal to ^c. 160. Let 6" be the sector PfiP described by the radius vector r in the time t, so that dS = ^r'dd is the elementary sector described in the element of time dt. Then (35) can be written dS_ ^ ~di~^^' whence integrating, since S = o for ^ = o : 5 = let. This shows that tke sector is proportional to the time in which it is described, which is merely another way of stating that the sectorial velocity is constant. It can be shown conversely, by reversing the steps of the above argument, that if in a plane motion the areas swept out by the radius vector drawn from a fixed point of the plane are propor- tional to the time, the acceleration must constantly pass through that point. It is well known that Kepler had found by a careful examina- tion of the observations available to him that the orbits described by the planets are plane curves, and the sector described by the radius vector drawn from the sun to any planet is proportional to the time in which it is described. This constitutes Kepler's first law of planetary motion. I02 KINEMATICS. [i6i. He concluded from it that the acceleration must constantly pass through the sun. 161. To express the value of the constant of integration c in terms of the given initial conditions (Art. 158), i. e. by means of ''o' ^'o' "^0' observe that at any time t ^dd rdO ds dt ds dt ^ hence at the time / = o we have c = v^r^ sin-f „. Denoting by p^ and / the perpendiculars let fall from on v^ and V we have r^ snv^^ = p^, r sirn//- = p ; hence i. e. the velocity at any time is inversely proportional to its distance from the center. 162. Let us now assume that the acceleration 7' of a central mo- tion is a given function, /(r), of the radius vector OP = r drawn from the center to the moving point P. With as origin, let X, y be the rectangular cartesian co-ordinates of the moving point P, and ;', its polar co-ordinates, at any time t. Then cos^ = x\r, smO = y/r are the direction cosines of OP = r, and, therefore, those of the acceleration j, provided the sense of j be away from the center, i. e. provided the force causing the accelera- tion be repulsive. In the case of attraction, the direction cosines ofy are of course — xjr, — yjr. Thus the equations of motion are in the case of attraction : d'^x ^, , X , d^y , 1/ ■^^ = ^=-A'')^' A-^=-/(0^ (36) For repulsion, it would only be necessary to change the sign of To integrate the equations (36) we cannot, in general, treat each equation by itself ; for, as r = i/jr^ -f- y', each equation con- 164.] PLANE KINEMATICS. 103 tains three variables x, y, t. We must therefore try to combine the equations so as to form integrable combinations. 163. Let us first multiply the equations (36) by y, x and sub- tract ; the right-hand member of the resulting equation is zero while the left-hand member is an exact derivative : d'^y d^x _ d ( dy dx\ ^dt^ ~^'di~^ = Jt yJt ~^~dt) = °- Integrating we find dy dx or, introducing polar co-ordinates : ^^ = -. (35) which is the equation (35), Art. 159; comp. Art. 97. 164. Next multiply the equations (36) by dxjdt, dyjdt and add. The left-hand member of the resulting equation is dj^d^ dyd^_dVJdA; (dy\-]^_±.y.. dt dt^ ^ dt df ~ dt \^ \dt) ^ ^ \dt) J ~ dt^^ > ' the right-hand member becomes The resulting equation d{\iF) = -f(f) dr can be integrated and gives \i^-\v^=-{^f{ryr; {37] i. e. it determines the velocity t/ as a function of r. 104 KINEMATICS. [165. 165. The two methods of combining the differential equations of motion (36) used in Arts. 163 and 164 are known, respectively, as the principle of areas and the principle of kifietic energy and work. The former name explains itself (see Arts. 159, 160). The latter is due to the fact (to be more fully explained in kinetics) that if equation (37) be multiplied by the mass of the moving body, the left-hand member will represent the increase in kinetic energy while the right-hand member is the work of the central force. Each of these methods of preparing the equations of motion for integration consists merely in combining the equations so as to obtain an exact derivative in the left-hand member of the resulting equation. If by this combination the right-hand member happens to vanish or to become likewise an exact derivative, an integration can at once be performed. This is the case in our problem. 166. The two equations (35) and (37), each of which was found by a first integration, are called first integrals of the equations of motion. By combining them and integrating again, the equation of the path is found. We have, by (4), Art. 96, for any curvilinear motion -=(S)^Kfy=(")T(sy-'']^ eliminating ddjdthy means of (35) we find for any central motion : Substituting this expression of 5^ in (37) we have the differential equation of the path in which the variables are separable. Shorter methods may occasionally suggest themselves in particular cases ; see, for instance, Art. 171. 167. To solve the converse problem, viz. to find the law of acceleration when the path is known, we have only to substitute the expression (38) of zi^ in the equation preceding (37); this gives : 169.J PLANE KINEMATICS. lOS ~~2de\\ddr) ^\r) \jr dr hence / d^ I d I I d i\ ^~ \de^r' deV^rdOr) 168. Kepler in his second law had established the empirical fact that the orbits of the planets are ellipses, with the sun at one of tJie foci. From this Newton concluded that the law of acceleration must be that of the inverse square of the distance from the sun. Our equation (39) enables us to draw this conclusion. The polar equation of an ellipse referred to focus and major axis is 1 -\- e cosO' where / = ii^/a = a( i — ^); a, b being the semi-axes, / the semi- latus rectum, and e the eccentricity. Hence l = i, + |cos^, -^,i=_Jcos0, so that (39) becomes 169. The third law of Kepler, found by him likewise as an empirical fact, asserts that the squares of the periodic times of dif- ferent planets are as the cubes of the tnajor axes of their orbits. Io6 KINEMATICS. [i/?- From this fact Newton drew the conclusion that in the law of acceleration, the constant /i has the same value for all the planets. Our formulae show this as follows. Let 7" be the periodic time of any planet, i. e. the time of describing an ellipse whose semi- axes are a, b. Then, since the sector described in the time T is the area irab of the whole ellipse, we have by Art. i6o ■7V ab = \cT. Substituting in (40) the value of c found from this equation we have 47rVi5^ I 47rW I fy) — ij-i ' p — j-i ' p ■ Hence 47rV is constant by Kepler's third law. 170. Planetary motion in its simplest form is that particular case of central motion in which the acceleration is inversely pro- portional to the square of the distance from the center so that where /x is a constant, viz., the acceleration at the distance r = i from 0. The equations of modon (36) are in this case, with as origin, (Px X d'^y y "^ = "'*?' ~df=~^^' (41) Combining these by the principle of energy (Arts. 164, 165), we find 172.] PLANE KINEMATICS. 107 ^g-^ (»■{ dx dy\ y.d(x'+f\ dt ~ r^ydt^^ dt)~ r^dt\ 2 ) yi. dr d\ hence integrating ~ r^dt~'^dtr' h^"-W = --z- (42) 171. To find the equation of the path, or orMt, write the equa- tions (41) in the form and ehminate r^ by means of (35) ; d^x IX ^de d^y fi . „de -j^ = cos^ —f, — y-, = smc' —f ■ df c df df' c dt Each of these equations can be integrated by itself: %-.. — 't.^e. |-=, = ^^(co,.-,), (43) where v^, v^ are the components of the velocity when = o. Multiplying by j, x and subtracting we find, by Art. 163 : (ft-.v\x + v^y + c = ^ (j; cos^ + 7 sin0) = -^/^^"y (4^) 172. The geometrical meaning of this equation is that the radius vector r = y/x^ -)- y^ drawn from the fixed point to the moving point P is proportional to the distance of P from the fixed straight line \'^-vAx+v^y + c = 0. (45) It represents, therefore, a conic section having for a focus and the line (45) for the corresponding directrix. 108 KINEMATICS. [i73- The character of the conic depends on the absolute value of the ratio of the radius vector to the distance from the directrix ; according as this ratio, is < I, = I, or > I, the conic will be an ellipse, a parabola, or a hyperbola. This criterion can be simplified. Multiplying by fijc and squaring, we have c ^ or since v^ + v^ = v^ and c = r^v^ sm-^^ = r^v^ : ^ with the initial radius vector r^. Determine the semi-axes a, b of the ellipse in magnitude and position. (7) The rectangular components of the acceleration of a point are both constant ; the initial velocity v^ is parallel to one of these com- ponents ; find the path. 5. VELOCITIES IN THE RIGID BODY. 175. A rigid body is said to have plane motion when all its points move in parallel planes. Its motion is then fully determined by the motion of any plane sec- tion of the body in its plane. It has been shown in Arts. 12-18 that the contin- uous motion of an invariable plane figure in its plane can be regarded as the limit of a series of instantaneous ro- _.. , _ Fig. 42. tations about the successive instantaneous centers, i. e. about the points of the fixed centrode. no KINEMATICS. [176. If at any instant the center of rotation C and the angular veloc- iiy s) about it be known, the linear velocity of any point P of the plane figure at the distance CP = r from C can be found, beitzg V = wr, at right angles to r. The components v^, v of this velocity along rectangular axes through C are evidently (Fig. 42) v^ = wr-y-^^ = -u>y, v^=o>ry-j = eox. (I) These results can also be obtained by differentiating, with respect to the time t, the relations X = r cosO, y = r sin^ between the cartesian and polar co-ordinates of the point P and observing that dS jdt is the angular velocity ta about the instan- taneous center C. For we thus find : dx dt „de r svs\0 -7- = — cay, dy ^dO v^^ = —f- = r cos6 -J- = (OX. " dt dt 176. In studying the motion of an invariable plane figure in its plane it is generally con- venient to use two sets of rectangular co-ordinate axes ; one set Ox, Oy, fixed in the plane, called the space axes or fixed axes, the other 0'^, O'v, fixed in the figure and moving with it, called the moving axes (Fig. 43). At any instant t, let the moving origin 0' have the co-ordinates x' ,y' and let the moving axis C'f make an angle Q with the I77.J PLANE KINEMATICS. Ill fixed axis Ox ; let x, y be the co-ordinates of any point P of the moving figure with respect to the fixed axes ; ^, 7; the co-ordi- nates of the same point P with respect to the rr\oving axes. Then we have the obvious relations x= x' 4- ^ cos6 — rj sin0, y = y' -t- | sinO -f- rj cos6, in which ^, 77 are constant while x, j, x:' ,y' , 6 are functions of the time. Differentiating we find for the component velocities of P par- allel to the fixed axes Ox, Oy : (2) Now, ddjdt is the angular velocity (o about the point 0' while dx' Idt, dy' jdt are the velocities of 0' parallel to the fixed axes, say v^, V ,. Considering moreover that | sin^ -\- rj cos6 =y — y' ^ ^ cos6 — r] sin6 = x— x', we have ^x =■^.'-(1 -/)'•'> ^v = ^/ + (^ - ^')^- (3) The velocity of P with respect to the fixed axes Ox, Oy consists, therefore, of two parts, a velocity of translation equal to that of 0' and a velocity of rotation about equal to that of P about 0' . 177. The instantaneous center being the point whose velocity- is zero at the given instant, we find its co-ordinates x,,, y^ from the equations o = z'^ - Oo - y'y, o = ^j,' + (•*'o - ^'y^ whence -o=-'-5^. ^o=y+^"- (4) When the angular velocity o), the co-ordinates x' , y* , and the velocity (^j, vj) of any point 0' are known as functions of the time, the elimination of t between the equations (4) gives the equation of the fixed centrode. 112 KINEMATICS. [178. The co-ordinates f„, tj^ of the instantaneous center referred to the moving axes are found in a similar way from the equations (2): ^0 = -(^x' sin^ — •ZV cos^, 77„ = -(v^ cos0 + v^y sin^), (5) to " ft) from which the body centrode can be found by eliminating t. 178. Exercises. ( 1 ) When a straight line moves in a plane show that the velocities of all its points have equal projections on the line. Hence show how to construct the velocity v of the end P of the connecting rod (Fig. 25, p. 59) when the velocity m of the other end Q is known. (2) Two points A, A' of a plane figure move on two fixed circles described with radii a, a' about O, O' ; show that the angular velocities w, w' of OA, O'A' about O, O' are inversely proportional to OM, O'M, J/being the point of intersection of OO' with AA' . (3) Given the magnitudes v, v' of the velocities of two points A, A' of an invariable plane figure and the angle {v, v') formed by their directions ; find the instantaneous center C and the angular velocity which are the components of the vector a along any three rec- tangular axes meeting at any point of /. It is easily proved that, if X, y, z are the co-ordinates of any point P of the body with respect to these axes, the components of the linear velocity v of P{x, y, z) are dx dy dt " '■^' y dt ' dz V ^ —r = 0) y — as X. ' dt '-^ « For, the component (o^ produces at /^ a velocity whose com- ponents, by Art. 175, are o, — (oj:, us^y ; similarly, to gives the components cc z, o, — (o,x; and o)^ gives — m^y, mx, o. By com- bining the velocities having the same direction the above equa- tions are obtained. 6. APPLICATIONS. 186, Kinematics of Machinery. A large majority of the cases of motion that are of importance in mechanical engineering can be reduced to plane motion. At first glance the application of theoretical kinematics to machines might seem to lead to rather complicated problems owing to the fact that a machine is never formed by a single rigid body, but always consists of an assemblage of several bodies some of which may even be not rigid (belting, springs, water, steam). The problem is, however, very much simplified by a characteristic of all machines, properly so called, that was first pointed out and insisted upon by recent writers on applied kine- matics, in particular by Reuleaux. This characteristic is the complete constraint of the motions of the parts of a machine. Il8 KINEMATICS. [i87- Thus Professor Kennedy defines a machine as " a combination of resistant bodies. wh.ose_ relative motions are completely con- strained, and by means of which the natural energies at our dis- posal may be transformed into any special form of work." With the latter clause of this definition we are not at present concerned ; it will be considered in kinetics. To explain • the former in detail would lead us too far into the domain of applied mechanics. A brief indication of the fundamental ideas must be sufficient. 187. By considering machines of various types it appears that the bodies, or elements, composing a machine always occur in pairs. Thus a single rigid bar will form a lever only when taken in connection with a support, or fulcrum ; a shaft to be used in a machine must rest in AP-^^^^ bearings ; a screw must turn in a nut. To take a more complex illustra- p,g' 47' tion, consider the mech- anism formed by the crank and connecting rod of a steam-engine (Fig. 47). It may be regarded as composed of four pairs, three so-called turning pairs at 0, A, B, and a sliding pair at B ; and these are connected by two rigid bars, called links, OA, AB, and a fixed link OB of variable length. 188. A sliding pair is formed by two bodies so connected that one is constrained to have a motion of translation relatively to the other. A pin moving in a groove or slot, a sleeve sliding along a shaft, are familiar examples. A turning pair constrains one body to rotate about a fixed axis in another, as in the case of a shaft turning in its bearings. A twisting pair makes one body have a screw motion about an axis fixed in the other. These three pairs are the only so-called lower pairs. They are characterized as such by the fact that their elements have surface contact. I9I-] PLANE KINEMATICS. 1 19 189. All other pairs are called higher pairs. The contact in such pairs is usually line contact. Higher pairs are of far less frequent occurrence in ordinary- machines than lower pairs. The only very common example of higher pairing is found in toothed wheel gearing. In any pair, whether higher or lower, the relative motion of either element with respect to the other is motioji with one degree of freedom (Art. 31). 190. For the purposes of kinematics a machine may be regarded as consisting of a number of bodies [links) connected by pairs in such a way that when one of the links is fixed all other links are constrained in their motion. In most cases this constraint is such as to leave but one degree of freedom to every link. A system of links of this kind forming, so to speak, a skeleton of the machine is called a kinematic chain (Reuleaux). When one link of such a chain is fixed, the chain becomes a mechanism. As a typical example we may take the •' slider crank " in Fig. 47. If the pairs are all turning pairs with parallel axes, the chain is called a linkage (Sylvester). A typical ex- ample is the four bar linkage in Fig. 48. A linkage with one link fixed has been called a linkwork (Sylvester). The four bar linkwork in Fig. 48 is also called a "lever crank" (Kennedy). 191. The Four Bar Linkage 1234 (Fig. 49). Whatever may be its motion, each link considered separately moves as an invariable plane figure and has therefore at any moment an instantaneous center C and an angular velocity w about this center. The center C^^ of i 2 and the center Cj, of 2 3 must always lie on a line passing through 2 since the velocity of 2 is perpendicular to both C;, 2 and C,3 2. Similarly, 3 must lie on the line joining the centers C,, and C^^; and so on. I20 KINEMATICS. [i9i- The quadrilateral i 2 3 4 is therefore, and always remains, inscribed in the quadrangle C^^C^^C^^C^^. This can be shown to hold even for the complete quadrilateral and quadrangle. The complete quadrilateral, or four-side, 1234 has- six vertices, viz. the six intersections i, 2, 3, 4, 5, 6 of its four sides , the complete quadrangle, or four-point, C,,C^,C.^^C\^ has six sides, viz. the six lines C,^C^^, C^^C^^, C^,C^^, C^C^^, CjjCj^, QaQi joining its four vertices ; and these six sides of the quad- rangle pass through the six vertices of the quadrilateral, respectively. Fig. 49. It remains to prove that Cj^Q passes through 5 and that C.^^C^^ passes through 6. Now the velocity of 2 can be expressed by u,^ ■ C^^2 and also by ">■, • Qa 2 ; hence C,, 2/C^^2 = o>Jw^ ; similarly C^^ 3/ Q 3 = wju,^. We have therefore, by the proposition of Menelaus,* for the intersec- tion of 2 3 with CjjCj^ : * If the sides of a triangle ABC be cut by any transversal, in the points A^, B' C BA' CB' AC ^ , '^''WC-bU-C''B = -'- See for ins solid geometry, Boston, Ginn, 1899, p. 240. BA^ CB' AC ^ , WC ' Wa ' CB = ~ '■ ^'^^ ^°'^ instance Beman and Smith, New plane and 192.] PLANE KINEMATICS. 121 sc„ 5C„ <«, The same value is obtained by determining the intersection of i 4 with Cj2 C^ ; the two intersections must therefore coincide. The proof for the point 6 is analogous. A corresponding proposition holds of course for four bar linkages ■with crossed bars i 2 4 3, or i 3 2 4. 192. Lever-crank. The linkage considered in the preceding article becomes a mechanism, or linkwork, as soon as one of its four links is fixed. It occurs in machines under a variety of forms some of which are referred to below. Let the link 3 4 be fixed ; then the center Cjj (Fig. 49) disappears ; Q falls into 4, Q into 3, and C,j becomes the inter- \ 5 section 5 of 4 i and 3 2. If I 2 were fixed instead of 3 4, 34 would have its center at 5. Similarly, if either 4 i or 2 3 be fixed, the center of the other is 6. Hence whichever of the four links be fixed, the cen- ters of all the links lie at some of the six vertices of the complete quad- rilateral 1234. If 34 be the fixed link (Fig. 50), the ratio of the angular velocities (u of 4 I and (Uj of 3 2 can be found. For if m denote the angular velocity of i 2 about 5, we have Fie. so. hence 4 I . <«! = 5 I • sin^, dyldt = aw cos^ ; hence V = au>, as is otherwise apparent. 194. If in the parallelogram 1234 the point 4 alone be fixed, we have a linkage called the pantograph. _, .„ It can serve to trace a curve similar to rig. 52, a given curve. Indeed, any line through 4 (Fig. 52) cuts the opposite links i 2, 2 3 (produced if necessary) I95-] PLANE KINEMATICS. 123 in points A, A' whose paths are homothetic (similar and similarly- situated) curves. For the points 4, A, A' remain always in line and the ratio 4A/4A' remains constant. Hence if a pencil be attached to A' and A be made to trace a given curve, A' will trace a similar curve. Instead of fixing 4, the point A' might be fixed ; then 4 and A will describe similar curves. This property is utilized in Watt's parallel motion (see Art. 199). The parallelogram linkage furnishes also a simple instrument for describing ellipses. Let the sides of the parallelogram be 23 = 41 ^ a, i2 = 34 = (5; and let a point A' on 2 ^ produced, at the dis- tance i from 2, be fixed (Fig. 53). Then, if r be made to describe a Fig. 53. Straight line passing through A', 4 will describe an ellipse. For, taking A' as origin and ^' i as axis of x, we have for the co-ordinates of 4 : X ^ (a + 26) cosjB, y = a sincp, whence x' /__ 195. In the parallelogram 1234, let the link i 2 be turned so as to coincide in direction with 4 3, and then give the links 4 i and 3 2 rotations of opposite sense. We thus obtain a linkage with equal, but intersecting, opposite sides, a so-called anti-parallelogram (Fig. 54). If 3 4 be fixed, the instantaneous center of i 2 is the intersection 5 of 4 i and 2 3. To obtain the centrodes in this case, notice that as the triangles 152 and 534 are equal, the triangle 5 4 2 is isosceles ; hence 51 = 5 3, and 45 — 35 = 41^ a. The difference of the radii vectores of 124 KINEMATICS. [196. 5 drawn from 4 and 3 being thus constant, it follows that the fixed centrode is a hyperbola whose foci are 4, 3, and whose real axis = a. As 4 3 = I 2 = i^, the equation of this hyperbola is SI? y iiy b-" - d' for 4 3 as axis of x and the midpoint of 4 3 as origin. It is easy to see that the fixed centrode becomes an ellipse when b <^a. As the triangles 152 and 354 are equal the body centrode is an Fig. 54. equal hyperbola or ellipse. The two centrodes lie symmetrically with respect to their common tangent at 5. For a given anti-parallelogram the centrodes are hyperbolas when one of the larger links is fixed ; they are ellipses when one of the shorter links is fixed. 196. If in the an ti -parallelogram only one point, say 4, be fixed, it can be used as an inversor, /. e. as an instrument for describing the inverse of a given curve. Let r = OP be the radius vector drawn from an arbitrary fixed origin, or pole, (9 to a given curve ; on OP lay off a length OP' =r' = K jr, where « is a constant ; then P' is said to describe the inverse of the given curve. Fig. 55. I97-] PLANE KINEMATICS. 125 The theory of inversors is based on the following geometrical prop- osition : If three lines CA = a, CA' = a, CO=l) (Tig. 55) turn about C so that O, A, A' are always in line, the product OA ■ OA' remains constant, viz. OA ■ OA' = ^^ — a^. For if the circle of radius a described about C intersect the line OC in B and JS' , we have OA- OA' =iOB- OB' = {b — a){l> + a). This proposition shows that in the anti-parallelogram 1234 (Fig. 56), with the vertex 4 fixed, the line joining the vertices 4 and z in- Flg. 56. tersects the circle described about 3 with radius 3 2 in a point 2' such that 2 and 2' describe inverse curves with respect to 4 as pole. For we have 4 2' • 4 2 = 4 3^ — 2 3^ = i5^ — a^ Moreover, any parallel to 4 2 will intersect the links 4 i, 4 3, 2 i in points O, A, A' dividing the three lines in the same ratio ; hence 4 2'( = i3) ^(9^' OA 42' /. e. OA ■ OA' = 4 2' ■ 4 2 = ^' — a', so that if O be fixed, A and A' will describe inverse curves for O as pole. This is the principle on which Hart's inversor is based. 197. Peaucellier's cell is another inversor (Fig. 57). It consists of the Hnked rhombus A B A' B' whose side we denote by a, and the two equal hnks OB, OB' of length b. If O be fixed, A and A' evi- dently describe inverse curves for O as pole. The figure shows that with ■^^ A O B ^ x> i^A' A B = 4> we have OA = b cos/ — a cos^, OA' = b cos/ + a cos^, whence OA ■ OA' = b"" cos'/ — a' cos%S 126 KINEMATICS. [198. and as, moreover, b sin;^ = a sin^, we find by adding to the preceding equation the relation o ^ b'' sin^;^ — a' sin'V' : OA ■ OA' = b' — a\ The practical application of inversors is based on the property that they enable us to transform circular motion into rectilinear motion (see Art. 199). The inverse of a circle r=: 2c cos^ passing through the pole is a straight line ; for we have for the radius vector r' of the inverse curve \b 7 \ \ A' OL, c r/ \ A' when the angular velocity w of the link OB is given, we notice that the instan- taneous center C of the link BA' lies at the intersection oi OB with the line drawn through A' parallel to 00'. Let 0/ be the angular I99-] PLANE KINEMATICS. 127 velocity of £A' about C. Then z; = w' • CA' ; also since the point B describes a circle about O, wb ^ ut' • CB ; hence If £A' intersect 00' in E, we have from similar triangles CA' : CB = OE: OB; hence V =^ (u ■ OE. The variable length OE depends on the angles EOB=d and BEO =

a -G a V C > < I) 1 i b FlE. 65. hence [2a/3 -f 2(b — /3)a] . J' = 2a/S . - + 2{b — /3)a ■ -, _ ^a' /3 + ba? - a?^ 233-] CENTROIDS. 145 If a, (8 are nearly equal and very small in comparison with a, b, we have approximately - _ 1 ^^^ba ''=^ a+b-a or still more roughly ^- ^a + b' 232. The area of a homogeneous circular sector (Fig. 6i, p. 139), of radius r and angle AOB = 2a can be resolved into triangular elements POP' = ^r^dO, the bisecting radius OC being taken as polar axis. The centroid of such an element lies, by Art. 227, at the distance f r from the center 0. Regarding the mass, p' ■ ^I'^dO, of each element as concentrated at its centroid, the sector is re- placed by a homogeneous circular arc of radius |r and density ^p'r'dO. By Art. 223, the centroid of such an arc, which is the required centroid of the sector, lies on the bisecting radius OC at the distance -^r-sina/a from the center 0. Hence _ sina X = ir ^ a 233. In general, for areas bounded by curves we must resort to integration, using the general formulae of Art. 214. If the area 5 be plane, we have in rectangular co-ordinates = I I p'dxdy, M-x=\ I p'xdxdy, M.y=\ I p'ydxdy; and if the mass be homogeneous, i. e. p' = const, since then the first integration can at once be effected : ■5'= (-yz-J^iK-^. Sx=\ x{y^ -y^dx, S-y = l\ (7/ - y^^)dx, or similar expressions for/ as independent variable. PART II — 10 146 INTRODUCTION TO DYNAMICS. [234. In polar co-ordinates, the element of area is rdrdd, and we have X =^ r cos^, y = r sin^; hence S=^ j^rdrdO, S-x =//^ cosddrde, S-y =ffr^ sinOdrde,- or, performing the first integration. S-x=^\ r^ cosddO, S-y = ^\ r It will be noticed that these last formulje express also that the infinitesimal sector ^r^d6 is taken as element, the centroid of this element having the co-ordinates fr cos^, f r sin^. 234. As a somewhat more complicated example let us consider a circular disk of radius a, in which the density varies directly as the distance from the center. Let a circle described upon a radius as diameter be cut out of this disk ; it is required to find the centroid of the remainder. Let O be the center of the disk of radius a, C that of the disk of radius ^a ; (7, the centroid of the latter, G the required centroid ; and put (9(7, = S,, OG =x. Then if M^ be the mass of the smaller disk, M^ that of the larger, we must have {M^ — M^ • *■ = M:^y The equation of the smaller circle is r = a cosi9. Taking as ele- ment of the mass of the smaller disk the mass contained between two arcs of radii r and r -f- dr, we have for this element : dM^ = p' ■ 20rdr, or since p' := kr, r = a cos^, dM^ = 2ka.'^0 cos'Od (cose). Hence 7j/^ = |^a' f ed (cos'0) = ^ka'(o cos'O — Ccoa^ OdoX = \kd C'''\os'OdO = ^ka' ■ I = *ka\ 236.] CENTROIDS. 147 The centroid of the element dAf^ lies, according to Art. 223, at the distance r sin6'/(? from O. We have therefore sin^ f _ no M.x. = — 2^3' I d sini? zo^ede • r = 2ka* T" sin'(? cos'ddO = yV^*- The mass of the larger disk is kr • 2iir ■ dr =^ 2nk I r'dr = ^Ttka^. Substituting these values in the equation of moments we find : ■«= ,, ,, = —, r« = O.1616 fl. M,-M^ 5(3^-2) 235. Proceeding to the determination of the centroids of curved surface-areas, we begin with the special case of the homogeneous area of a surface of revolution, bounded by two planes at right angles to the axis. The centroid evidently lies on the axis of revolution, which we take as axis of x; it is therefore sufficient to take moments with respect to the jr-plane. As element we take the strip contained between two planes, parallel to the j'j- plane, at the distances x and x -\- dx from it ; if the equation of the meridian section be 77 =/"(x), where 77 is the distance of any point of the surface from the axis of revolution, we have for this element : dS = 27nqds = 27r7y Vdx'^ + dy^ = 27rf{x) V i +/'V;r. Hence, if the bounding planes have the distances x^, x^ from the js-p\a.ne : 5 = 27r I /(x) VT+f'dx, S-x=27rj xf{x) Vx +f''dx. 236. When the area is bounded by two planes perpendicular to the axis of revolution and any two meridian planes, inclined to 148 INTRODUCTION TO DYNAMICS. [23/- each other at an angle cf>, we may take one of these meridian planes as xj-p\ane and find, with dS = 7y^0 ds = r, Vdx' + dfd4> =/{x) V7+f^dxd : 5 = r f f{x) Vi +/''dxd4> = 4> f /(x) Vx +f''dx, Sx= ff xf(x) Vi +/''dxd =

Vi +/'^dxd4> = sin^ r yVi+f'dx, S-s= P ( P sm y'i+/''dxd = (I - cos<^) r y^vi +f'''dx, where/" and/' are known functions of ,r. Instead of x, tj might be taken as independent variable. 237. In the case of spherical suifaces, although the preceding formulas can of course be used, it is often more convenient to make use of the geometrical property of the sphere that any spherical area is equal to the area of its projection on a cylinder circumscribed about the sphere. Thus the area on the sphere contained between two parallel planes is equal to the area cut out by the same two planes from the circumscribed cylinder whose axis is perpendicular to the planes. The centroid of such a spherical area is therefore on the radius at right angles to the bounding planes midway between these planes. 238. The Second Proposition of Pappus and Guldinus (compare Art. 224). A plane area 5" (Fig. 6&) rotating about any axis situated in its plane generates a solid of revolution whose volijme is V='w^(^y^ — y^dx, if the axis of revolution is taken as axis 239-] CENTROIDS. 149 Fig. 66. of X and j'j, y^ are the two ordinates of the curve bounding the area. On the other hand, if y be the distance of the centroid G of the plane area from the axis, we have by Art. 233. Combining these two results, we find V = 27rj7 • S, i. e., the volume of a solid of revolution is obtained by multiplying the generating area into the path described by its centroid. The proposition evidently holds even for a partial revolution. 239. To find the centroid of a portion of any curved surface F{x, y, z) = o, we have only to substitute dM ^ p'dS in the general formulae of Art. 2 1 4, and then express dS by the ordi- nary methods of analytic geometry. Denoting by /, m, 7t the direction cosines of the normal to the surface at the point {x, y, z), and put*:^ng for shortness dFjdx^^F,, dFldy = F^, dFlds = F^, wc .. _■ dydz dzdx c m n dS = I m n I Hence, substituting dS = dxdy VF' + F' + F' F in the formulae of Art. 214, we find M ^rr,'dxdy^~^i^i±^ F where the integration is to be extended over the projection of the portion of surface under consideration on the plane xy. The equation of the curve bounding this projection must be given ; 15° INTRODUCTION TO DYNAMICS. [240. it determines the limits of integration. It is obvious how the formula has to be modified when the projection of the area on either of the other co-ordinate planes is given. The expressions for M-x, M-y, M-z differ from the above expression for M only in containing the additional factor x, y, b, respectively, under the integral sign. 240. If the equation of the surface be given in the form z =/(x, y), as is frequently the case, we have F{x, y, z)=z -f{x, y) ; hence with the usual Gaussian notation dzjdx = dfjdx = p, dsjdy = Sfjdy = g, which eives P.= -P, F„=-<1< ^.= 1. y J mil c^ while M-l:, M-y, M-z contain the additional factor x, y, s, re- spectively, under the integral sign. In the case of a homogeneous spherical surface x' ■i-y'^ + z^ = a^, we have/ = dzjdx = — xjz, q = dzjdy = —yjz ; hence zVl + p^ + q^ = a, so that S-z= a j I dxdy = a- S^, where ^ is the area of the surface and 5. the area of its projec- tion on the plane xy. The formula shows that the distance 5 of the centroid of any spherical area 5 from a plane passing through the center is equal to the radius a multiplied by the ratio of the projection 5, of the area on the plane to the area itself 241. Exercises. ( 1 ) The sides of a right-angled triangle are a and b. Find the dis- tances of the centroid of the triangular area from the vertices. (2) From a square ABCD one corner EAF\& cut off so that AE = fa, AF^ \a, a being the side of the square. Find the centroid of the remaining area. 24I.J CENTROIDS. 151 (3) An isosceles right-angled triangle of sides a being cutout of the area of its circumscribed circle, iind the centroid of the remaining area. (4) If one-fourth be cut away from a triangle by a parallel to the base, show that in the remaining trapezoid the centroid divides the median in the ratio 4:5. (5) Prove that the centroid of any plane quadrilateral ABCD cova- cides with that of the triangle A CF, if the point F be constructed by laying off BF^ DE on the diagonal BD, E being the intersection of the diagonals. (6) Find the centroid of the cross-section of a retaining wall in the form of a trapezoid with two perpendicular sides, the lower base being a, the upper b, the height h. ( 7 ) Find the centroid of the cross-section of an angle-iron, or L, the sides being a, b, the thickness of each flange S (Fig. 67). (8) Find the centroid of the cross-section of a bar formed by placing four angle-irons with their edges together, two of the irons having the -6— -&-^ b Fig. 67. 6 Fig. 68. l._ dimensions a, b, a, /J, while the other two have the dimension a differ- ent, say a' (Fig. 68). (9) Find the centroid of the cross-section of a U-iron, the length of the flanges being a = 12 in., that of the web 2i^ ^ 8 in., and the thickness 5 = r in. Deduce the general formula for x, and an approxi- mate formula for a small S, and compare the numerical results. (10) In the cross-section of an unsymmetrical double T (Fig. 69), the flanges are 2b = 16 in., 2b' = 10 in. ; the web is a = 10 in. ; and the thickness of each of the two channel-irons forming the bar is 5 = i in. throughout ; find the centroid. 152 INTRODUCTION TO DYNAMICS. [241. (ii) In a T-iron (Fig. 70) the width of the flange is b, its thick- ness a ; the depth of the web is a, its thickness /S. Find the distance of the centroid from the outer side of the flange ; give an approximate expression and investigate it for a = 1^, a^ ^ ^ \a. (12) Find the centroid of a circular sector (comp. Art. 232) by- integration, using the formula of Art. 233, and deduce from the result k 6^ M< --6' ^ 1 1 1 1 1 - ( t \ 1 1 K h — ^i* h 4 1 1 — -p f 1 1 1 — 1 — Fig. 69. Fig. 70. that the centroid of a homogeneous semicircular area of radius r lies at the distance 5; =(4/371)^ from the center. (13) The centroid of the area of a homogeneous circular segment of radius r subtending at the center an angle 2a is at the distance sin°a V a_ sma cosa 6 rs — ch if c is the chord, h its distance from the center, and j the arc. (14) A painter's palette is formed by cutting a small circle of radius b out of a circular disk of radius a, the distance between the centers being c. It is required to find the distance of the centroid of the remainder from the center of the larger circle. (Routh. ) (15) The arch constructed of brick over a door is in the form of a quadrant of a circular ring. The door is 5 ft. wide ; i \ lengths of brick are used (say 12 in.). Find the centroid of the arch. Find the co-ordinates of the centroid for the following plane areas : (16) Area bounded by the parabola/ = i^ax, the axis of .x, and the ordinate J'. (17) Area bounded by the curve 7 = sinjc from x=oXo x=tz and the axis of x. (18) Quadrant of an ellipse. 244.] CENTROIDS. 153 (19) Elliptic segment bounded by the chord joining the ends of the major and minor axes. (20) Show, by Art. 222, that the centroid of the surface of a right circular cone lies at a distance from the base equal to one third of the height. (21) Find the centroid of the portion of the surface of a right cir- cular cone cut out by two planes through the axis inclined at an angle (Fig. 72), the element of volume is an infinitesimal rectangular parallelepiped having the concur- rent edges dr, rdO, r sm6dc}> ; hence dv = 1^ sinddrdOd^). As x= r cos6, y = r sin0 cos<^, z = r sin^ sin^, the centroid is determined by the equations : X R |-^ % \ ^^^e \ 1 \ — A _ ^::r4. Q Fig. 72. M=fffp,'^ smedrd6d4>, M-x=fffpr^ sin6l cosOdrdedcf), M-y =fj'fpr^ sin'e cos4>drded4>, M-z =fffpr' sin^^l sm^ = F^J>^ (Art. 297), where p^, p^ are the (perpendicular or obhque) distances of the resultant from F and F^, respectively. This resultant F^ + F.^ can now be com- bined with ^3 to form a resultant F^-^ F^^ F^, whose distances from F^ + F^ and F^ in the plane determined by these two forces are as F^ is to F^ + F^. This process can be continued until all forces have been combined ; the final resultant is F, + F^+... + F^. Any number of parallel forces are, therefore, in general equiva- lent to a single resultant equal to their algebraic sum (see Art. 303). 301. To find the position of this resultant analytically, let the points of application of the forces F^, F^, ■ ■ ■ F^ be (x , j/ , .? ), {^v Jv ^2)' • • • (•*■»' A' ^n)- The point of application of the result- ant F^ + F^ of F^ and F^ may be taken so as to divide the dis- tance of the points of application of F^ and F^ in the ratio F : F ■ hence, denoting its co-ordinates by x' ^ y' , z' , we have Fix' — x) = F^x^ — x'), or {F^ + F,)x' = F,x, + F^„ and similarly for y' and z' . 303.] PARALLEL FORCES. 1 89 The force ^j + F^ combines with F^ to form a resultant F^'\- F^-\- F^, whose point of application (^x", y, z") is given by {F^ + F, + Fy = F,x, + F^, + F^^, £md similar expressions ior y" , z". Proceeding in this way, we find for the point of application (x, y, z) of the resultant of all the given forces (F^+F, + ... + FJx=F,x, + F^,+ ... + Fx„, with corresponding equations for y and z. We may write these equations in the form : IFx _ IFy IFz ^=Yf' ^ = Tf' (0 '^~ sh- unless S/^= o (see Art. 303). As these expressions for x, y, z are independent of the direction of the parallel forces it follows that the same point {x, y, z) would be found if the forces were all turned in any way about their points of application, provided they remain parallel. The point (x, y, z) is for this reason called the center of the system of parallel forces. It is nothing but the centroid of the points of application if these points are regarded as possessing masses equal to the magnitudes of the forces. 302. As the origin of co-ordinates in the last article is arbitrary, the equations (i) evidently express the proposition that in any system of parallel forces whose algebraic sum is different from zero the sutn of their moments about a?iy point is equal to the moment of their resultant about the same point. In particular, the sum of the m.oments about any point on the resultant is zero. This proposition may be regarded as a generalization of the principle of the lever rekvred to in Art. 297. It furnishes the convenient method of " taking moments " for the purpose of de- termining the position of the resultant. 303. Couple of Forces. The construction given in Arts. 294 and 298 for the resultant of two parallel forces fails only when the two igo STATICS. [304. given forces are equal and of opposite sense. In this case, the lines pP' and qQ' in Fig. 83, and the lines I and III of the funicular polygon (Fig. 84), become parallel, so that their intersection r lies at infinity. The magnitude of the resultant is of course zero. The combination of two equal and opposite parallel forces [F, — F) acting on a rigid body is called a couple. A couple is, therefore, properly speaking, not equivalent to a single force, although it may be said to be equivalent to a force of magnitude zero at an infinite distance. The theory of couples will be con- sidered in detail in Arts. 319—329. A system of n parallel forces F^, F^, ■■ ■ F^^ has been shown in Arts. 300 and 301 to reduce to a single resultant unless 2/^= o. In this exceptional case the system may either reduce to a couple or it may be in equilibrium. To distinguish which of these cases arises it suffices to combine all the forces of the same sense ; we obtain thus, since 2/^= o, two equal and opposite forces. If these do not lie in the same line, the system reduces to a couple ; otherwise it is in equilibrium. 304. Conditions of Equilibrium. It follows from the preceding article that for the equilibrium of a system of parallel forces the condition 2i^ =0, or R= o, though always necessary, is not sufficient. Now, if the resultant R of the n parallel forces F^, F^, ■ ■ ■ F is zero, the resultant R' of the n — i forces F^, F^, ■ ■ ■ F^_^ cannot be zero, and its point of application is found (by Art. 301) from X = {F^x, + F^x^^...+^ F^_,x^_,)l{F^ + yr + . . . + ir_j and similar expressions for y and z. The whole system of parallel forces is therefore equivalent to the two parallel forces R' and F . Two such forces can be in equilibrium only when they he in the same straight hne ; i. e. F^ must lie in the same line with R' and must therefore pass through the point (J, y, z), which is a point oiR'. The additional condition of equilibrium is, therefore, 'x — X y — y z — z cosa cos/3 C0S7 3o6.] PARALLEL FORCES. 191 where a, /3, 7 are the angles made by the direction of the forces with the axes. 305. For practical application it is usually best to replace the last condition by taking moments about a convenient point. Thus, the analytical conditions of equilibrium can be written in the form 1F= o, I^Fp = o. Graphically, to the former corresponds the closing of the force- polygon, to the latter the closing of the funicular polygon. 306. "Weight ; Center of Gravity. The most important special case of parallel forces is that of the force of gravity which acts at any given place near the earth's surface in approximately parallel lines on every particle of matter. If g- be the acceleration of gravity, the force of gravity on a particle of mass /« is w = rng, and is called the weight of the particle or of the mass m. For a system of particles of masses m^, m^, ■ ■ ■ m^ we have w^ = m^g, Wj = m^g, • • • w^ = m^g. If the particles are rigidly connected, the resultant W of these parallel forces, W= Wj + Wj + \-w^=(m^ + m^+ \- m^g = Mg, where M is the mass of the system, is called the weight of the system. The center of the parallel forces of gravity of a system of rigidly connected particles has, by Art. 301, the co-ordinates _ "^mgx _ l^ingy _ "^mgz Zmg Z4mg 2,mg or since the constant g cancels, _ S;«x _ '^my _ "^ins z-m z^ni z^i'i 192 STATICS. [307- This point is called the center of gravity of the system, and is evidently identical with the center of mass, or centroid (see Art. 212). For continuous masses the same formulae hold, except that the summations become integrations. The weight IV of a physical body of mass M is therefore a ver- tical force passing through the centroid of its mass. 307, Exercises. (i) A straight rod (lever) of 6 feet length has suspended from its ends masses of 7 and 18 pounds, respectively. Find the point (/«/- cruni) on which it balances in a horizontal position : (a) if its own weight be neglected ; {b~) if it is homogeneous and weighs i-| pounds per running foot. ( 2 ) A straight beam rests in a horizontal position on two supports A, B. The distance between the supports (the span) is 24 ft. The beam carries a weight of 11 tons at a distance of 8 ft. from A, and a weight of 6 tons at 16 ft. from ^. Find the pressures on the sup- ports (or the reactions of the supports) : (a) when the proper weight of the beam is neglected ; {¥) when the beam weighs \ ton per run- ning foot; (<:) when the first third of the beam (from A) weighs \ ton, the second i ton, the third \ ton per running foot. (3) A homogeneous circular plate weighing W pounds rests in a horizontal position on three equidistant supports near its edge, (a) What is the least weight P that will upset it when placed on the plate ? {F) If there be four equidistant supports near the edge, what is the least weight that will upset the plate ? (4) Construct the resultant of two parallel forces of opposite sense by the graphical method of Art. 298. ( 5 ) Solve exercises ( i ) and ( 2 ) by the graphical method. -Q Q Q 8 4 22580 6 10 23560 Fig. 86. .a 7 4 22560 13200 (6) Find the reactions of the supports of a bridge truss of 50 ft. span, produced by a, freight locomotive whose weight is distributed 3o8.] PARALLEL FORCES. 193 over the three pairs of driving wheels and the front truck, as indicated in Fig. 86 : (a) when it stands in the middle of the span ; (/') when its front truck stands over one support. ( 7 ) Explain how the centroid of a plane area can be found graphi- cally by dividing the area into narrow parallel strips. (8) A homogeneous rectangular plate is pivoted on a horizontal axis through its center so as to turn freely in a vertical plane. If weights IJ\, JF^, Jr,, IV^ be suspended from its vertices, what is its position of equilibrium ? (9) The ends of a straight lever of length / are acted upon by two forces F^, F^ in the same plane with it, but inclined to the lever at angles i/j, a^. Determine the position of the fulcrum. (10) To resolve a force /F graphically along two lines a, b parallel to W {e. g. to find the pressures on the two supports produced by a weightless beam carrying a single load W) let W= 1 ^ represent the force polygon. Take any point A on a as pole, draw A 1 and take this line as the first side I of the funicular polygon. The third side III must then be the parallel i ^ to ^ 3 ; and the second side II is found by joining A to the intersection £ of III with b. As A is the pole, this line AB meets i 3 at a point 2 such that the required compo- nents are i 2 along a and 2 3 along b. Draw the figure. (11) The safety-valve of a steam-boiler consists of a horizontal arm, 32 in. long, hinged at one end A, and weighing 3 lbs. ; the vertical stem of the valve is attached at 4 in. from A and weighs, with the valve disk, i}^ lb. ; a ball weighing 20 lbs., suspended from the arm, 28 in. from A, presses the disk on the mouth of the valve. If the diameter of the mouth be 2 in. , what is the blowing-off pressure ? {JThe Locomotive, Vol. 16, pp. 113-120.) (12) If in a safety-valve the arm is 66 in. long and weighs 18 lbs., while the valve-stem and disk, 3 in. from the hinge, weigh 7 lbs. and the diameter of the disk is 4 in., at what distance from the hinge must a 50-lb. weight be suspended to make the valve blow at a pres- sure of 100 lbs. / sq. in. ? (7/5.) 308. FunicTilar Polygons and Catenaries, T\\^ funicular polygon in its original meaning represents the form of equilibrium assumed by a string or cord suspended from two fixed points and acted upon by any forces in the vertical plane. The " cord " is sup- PART II — 13 194 STATICS. [309- posed to be perfectly flexible, inextensible, inelastic and without weight. When the number of forces is made infinite, the polygon becomes a continuous curve called a catenaiy. The present discussion is confined to the case when the forces are all vertical so that they can be regarded as weights. 309. Let A, B (Fig. Zj) be the fixed points, and let there be five weights, W^, W^, W^, W^, W^, suspended from the cord which will form a polygon whose sides are denoted by I, II, III, IV, V, VI. If the cord be cut on both sides of the first vertex. III, and the corresponding tensions T^, T^ be introduced, this vertex must be 1 -^:^ / Fig. 87. in equilibrium under the action of the three forces W^, T^, T^, Hence drawing a line i 2 to represent the weight W^ and draw- ing through its ends i, 2 parallels to I and II, respectively, we have the force polygon of the first vertex. Its sides i and 2 represent in magnitude, direction, and sense the tensions 7j, T^ ; in other words, the weight W^ has thus been resolved into its com- ponents along the adjacent sides. The same can be done at every vertex of the polygon I II ■ • • VI, and all the tensions can thus be found. But as the tension T in II occurs again (with sense reversed) in the force polygon for the next vertex, and so on, the successive force polygons can be 3II-] PARALLEL FORCES. 19s fitted together, every triangle having one side in common with the next one. Thus the complete force polygon of the whole cord is formed, as shown on the right, in Fig. 87. Its vertical line represents the successive weights W^= i 2, W^= 2 ;^, JVs = 3 4. f^4 = 45/ ^5=56, while the lines radiating from the pole represent on the same scale the tensions or stresses in I, II, III, IV, V, VI. 310. The polygon I II • • • VI is called the funicular polygon. It will be noticed that if we have given the fixed points A, B, the magnitudes of the weights, their horizontal distances, say from A, and the directions of the first and last sides (whatever may be the number of the forces), the remaining sides of the funicular poly- gon can be found by laying off on a vertical line the weights ll\ = 12, ^Fj = 2 3, etc., in succession, drawing through i a parallel to the first side, through the end of the last weight (6 in Fig. 87) a parallel to the last side, and joining the intersection of these parallels to the points 2, 3, etc. The sides of the funicular poly- gon must be parallel to the lines radiating from ; at the same time these lines represent the tensions in these sides. 311. For the analytical investiga- tion, let P^ be that vertex of a funicu- lar polygon of any number of sides at which the ith. and {i -\- i)th sides intersect ; let a., a^_^^ be the angles at which these sides are inclined to the horizon, and W^ the weight sus- pended from the vertex P^ (Fig. 88). Cutting the cord on both sides of P^, and introducing the ten- sions 7^ and 7^.^,, the conditions of equilibrium of the point P^ are found by resolving the three forces W^, T^, T^^^ horizontally and vertically (Art. 280) : r.+j cosa.+i = r. cosa., (i) ^i+i sma,+i = T. sina. + W^. (2) Fig. 88. 196 STATICS. [312. The former of these equations shows that, whatever the weights W and the lengths and inclinations of the sides, tke horizontal components of the tensions T are all equal. Denoting this con- stant value by H, we have T^ cosfltj = T^ cosotj = ...= T^ cosa. = . . • = Zf. (3) Substituting the values of Z and T.^j as obtained from these re- lations, in (2), this equation becomes W. tana^+j = tana. -|- -^% (4) which shows that as soon as all the weights' and the inclination and tension of any one side are given, the inclinations and ten- sions of all the other sides can be found. 312. Let us now assume that the weights W are all equal. Then the values of tana._,.j given by (4) form an arithmetical pro- gression. If, in addition, we assume that the sides of the polygon are such as to have equal horizontal projections, i. e., if we assume the weights to be equally spaced horizontally, the vertices of the polygon will lie on a parabola whose axis is vertical. To find its equation, let us suppose, for the sake of simplicity, that one side of the polygon, say the kCa, is horizontal so that o!j. = o. Taking this side as axis of x, its middle point as origin, the co-ordinates of the vertex P^. are \a, o, if a be the length of the horizontal side and hence also that of the horizontal projec- tion of every side. Putting W\H=T, we have tana^ = o, tana^j = t, tana^+j = 2T,-.-; hence the co-ordinates of P,^.,., are x = ^a, y = ar ; those of /'j+2 are x = ^a, y = ar -\- 2aT = ^ar ; those of P^_^.^ are X = ^a, y = aT -\- 2aT -f ^ar = 6aT, etc.; those of the nth vertex after F^ are 2n -\- I n{n -\- i) X = a, y = -ar. 2 ' -^ 2 Eliminating n, we find the equation , 2a ( ar\ 314-] PARALLEL FORCES. 197 which represents a parabola whose axis is the axis of y, and whose vertex lies at the distance \ar=\aWIH below the origin 0. 313, Let the number of sides be increased indefinitely, the length a and the weight W approaching the limit o, but so that the quotient a/fF remains finite, say lim {ajW)= ijw. Then lim {ajr) = Hjw, lim (ax) = o ; so that the equation of the para- bola becomes , 2H w y> where w is evidently the weight of the cord, or chain, per unit of horizontal length. The parabola is, therefore, the form of equilibrium of a cord suspended from two points when the weight of the cord is uniformly distributed over its horizontal projection. This is, for instance, the case approximately in a suspension bridge with uniformly loaded roadbed, the proper weight of the chains being neglected. 314. This result can easily be derived independently of Art. 312, by considering the equilibrium of any portion OP of the chain beginning at the lowest point (Fig. 89). The forces act- ing on this portion are the horizontal tension H at 0, the tension T along the tangent at P, and the proper weight W of the chain. As this weight is assumed to be uniformly distributed over the horizontal projection OP' = x oi OP, the weight is IV= wx, and bisects OP' . For equilibrium the three forces W, H, T must pass through a point ; hence T must pass through the middle point Q of OP'. 198 STATICS. [315. Resolving the forces in the horizontal and vertical directions, we find, as conditions of equilibrium, -H+ T-y-^o, -wx-\- T-^ = 0; ds ds whence, eliminating ds, dy w Integrating and considering that x ^ o when j = O, we find the equation of the parabola as above. This relation can be directly read off from the figure ; for, tak- ing moments about P, we have o = — Hy + wx ■ \x, whence y = {wl2H)x^. 315. The force polygon of H, T, W is evidently a triangle similar to the triangle QPP' . Hence, if the height of a suspension bridge be h, its span 2 /, its total weight 2 W, we have for the horizontal tension H and the tension T at the point of support : H T W \l -//^^ + j/2 h' 316. The form of equilibrium assumed by a homogeneous cord is a common catenary. To find its equation, we again consider the equilibrium of a portion OP — s (Fig. 90) of the cord, beginning at the lowest point 0. The weight of this portion is now W^ ws, and if a be the angle made by the tangent at P with a horizontal line, we have the con- ditions of equilibrium T cos a = T ~- = B, Tsin a= T^ = ws. ds ds Dividing and putting H/w = c, we have the differential equation 3i6.] PARALLEL FORCES. 1 99 of the curve in the form dx c dy s Substituting this value of dxjdy in the relation ds'' = dx^ + dy^, we obtain ' ^ ^ sds I +-„ or dy = ±^ m s- Vs' + c' which gives by integration y + C= j/s' + c^, the minus sign being rejected since y increases with 5. The constant C can be made to disappear by taking the origin Fig. 90. 0' on the vertical through at the distance O'O = c below the lowest point 0. We have, therefore, f = s^ + ^. By means of this relation, i' can be eliminated from the original differential equation, and the result, ^^-^ =dx, Vf-c" can be integrated : c log {y + Vf — ^) = x-\- C. As y ^ c when x-= o, we find C ^ c log c ; hence j+ i/y — c^ = 200 STATICS. [317- Taking reciprocals and rationalizing the denominator we find hence, adding and subtracting, y = ^c\e'' + e " )= c cosh (x/c), s=\c\e''—e '')= c smh [xjc). 317. The first equations of Art. 316, T cosa = H = wc, T^sina = ws, give for the total tension T at any point P T" = zi/{c' + 5") = (ivyf. Thus, while the horizontal component is constant, the vertical component at any point P is equal to the weight of the portion of the cord from the lowest point to the point P, and the total ten- sion is equal to the weight of a portion of the cord equal to the ordinate of the point P. Let Q be the foot of the ordinate of P (Fig. 90), N the inter- section of the normal with the axis 0' x, and draw QR perpen- dicular to the tangent. Then PR = y sina = s, since T sinct = ws and T=wy; also QR = y cosa= c. Dividing, we have tanot = sjc ; hence, differentiating, I da I , ds c = — ; and p cos^a ds c ' da cos^a ' The figure shows that the radius of curvature p is equal to the length of the normal PN. The relation p cos^at = c shows further that at the vertex (a = o) the radius of curvature is p^ = c. It follows that for a cord or chain suspended from two points B, C in the same horizontal line c (and consequently H~) is large when p^ is large, i. e. when the curve is flat at the vertex ; in other words, when B and C are far apart. 319- J COUPLES. 20 r 318. Exercises. ( 1 ) A weightless cord ABCDEF is suspended from the fixed points A, F, and carries weights at the intermediate points B, C, D, E. Taking A as origin, the axis of x horizontal, the axis of y vertically upwards, the co-ordinates of the points B, C, D, E, E axe (2, — i), (4, — i-S)> (7, — i-S). (8-5. — I). (lO) 2)- If the weight at ^ be one pound, what are the weights at C, £>, Ef What are the tensions of the segments of the cord ? What are the reactions of the fixed points A, El (2) The total weight of a suspension bridge is iW =^ 60 tons ; the span is 2/= 250 ft.; the height is /^ = 25 ft. Find the tension of the chain at the ends and in the middle, both graphically and analytically. (3) A uniform wire of length 2s is stretched between two points in the same horizontal line whose distance 1 x'ys, very nearly equal to 2s. Find an approximate expression for the parameter c of the catenary and hence for the tension of the wire. (4) Find the tension of a telegraph wire No. 8 (diameter =; 0.165 in., weight 378.1 lbs. per mile) stretched between poles 100 ft. apart, if the length of the wire between two poles is 100 ft. 4 in. Determine also the sagging of the wire {y — c'). (5) A chain, 64 ft. long, and weighing i lb. /ft. is suspended between two points in the same horizontal line, 60 ft. apart ; find the tensions at the lowest point and at the ends, and the depth of the ver- tex below the points of suspension. (6) If the load of a suspension bridge is proportional to the dis- tance X from the middle point, show that the cable forms a cubical parabola. (7) Show that when xlc is so small that its third and higher powers can be neglected the arc of the catenary in Art. 316 can be regarded as an arc of a parabola. IV. Theory of Couples. 319. The combination of two equal forces of opposite sense F, F, acting on a rigid body along parallel lines, is called a couple of forces, or simply a couple (Art. 303). The perpendicular distance AB = p (Fig. 91) of the forces of the couple is called the arm, and the product Fp of the force F 202 STATICS. [320. B -F -F; Fig. 91. into the arm p is called the moment of the couple. The moment, or the couple itself, is also called a torque. If we imagine the couple {F, p) to act upon an invariable plane figure in its plane, and if the middle point of its arm be a fixed point of this figure, the couple will evidently tend to turn the figure about this middle point. (It is to be observed that it is not true, in general, that a couple acting on a rigid body- produces rotation about an axis at right angles to its plane.) A couple of the type {F, p) or {F', p') (see Fig. 91) will tend to rotate coun- ter-clockwise, while a couple of the type {F", p") tends to turn clockwise. Couples in the same plane, or in par- allel planes; are therefore distinguished as to their sense; and this sense is expressed by the algebraic sign attributed to the moment. Thus, the moment of the couple {F, p) in Fig. 9 1 , is -I- Fp, that of the couple {F", p") is - F"p". 320. The effect of a couple is not changed by translation, i. e. by moving its plane parallel to itself without rotating it. Let .^i?=/(Fig. 92) be the arm of the couple {F,p^ in its original position, and A' B' the same arm in a new position parallel to the original one in the same plane, or in any parallel plane. By introduc- ing at each end of the new arm A' B' two opposite forces F, — F, each equal and parallel to the origi- nal forces F, the given system is not changed (Art. 273). But the two equal and parallel forces F -sX A and B' form a resultant 2F at the middle point of the diagonal AB' of the parallelogram Fig. 92. 322.] COUPLES. 203 ABB' A'. Similarly, the two forces — FaX B and A' are together equivalent to a resultant — 2F at the same point 0. These tvvo resultants, being equal and opposite and acting in the same line, are together equivalent to zero. Hence the whole system reduces to the force F 2X A' and the force — F at B' , which form, there- fore, a couple equivalent to the original couple at AB. 321. T/u effect of a couple is not changed by rotation in its plane. Let AB (Fig. 93) be the arm of the couple in the original position, C its middle point, and let the arm be turned about C into the position A'B' . Applying again at A' , B' equal and opposite forces each equal to F, the forces — F at A' and F at A will form a resultant acting along CD, while F at B' and -Fat B give an equal and opposite resultant along CE. These two resultants destroy each other and leave nothing but the couple formed \iy F at A' and — F at B' , which is therefore equivalent to the original couple. Any other displacement of the couple in its plane, or to a parallel plane, can be effected by a translation combined with a rotation in its plane about the middle point of its arm. The effect of a couple is therefore not changed by any displaceinent in its plane or to a parallel plane. 322. The effect of a couple is not changed if its force F and its arm p be changed simultaneously in any way, provided their product Fp remain the same. Let AB =phe the original arm (Fig. 94), F the original force of the couple ; and let A'B' = p' be the new arm. The intro- duction of two equal and opposite forces F' at A'^ and also at 204 STATICS. [323- -F F A' -F' Fig. 94. B' , will not change the given system F, — F. Now, selecting for F' a magnitude such that F'p' = Fp, the force i^ at ^ and the force —F' at A' combine (Art. 297) to form a parallel re- sultant through C, the middle point of the arm, since for this point F- \p + (- F') ■ \p' = O. Similarly, — FdXB and F' at B' give a resultant of the same magnitude, in the same line through C, but of opposite sense. These two resultants thus destroying each other, there remains only the couple formed by F' at A' and — F' at B' , for which Fp = F'p'. 323. It results from the last three articles that the only essen- tial characteristics of a couple are : {a) the numerical value of the moment ; {b) the sense, or direction of rotation ; and {c) what has been called the "aspect" of its plane, i. e. the direction of any normal to this plane. It is to be noticed that the plane of the two forces forming the couple is not an essential characteristic of the couple ; just as the point of application of a force is not an essential characteristic of the force (see Art. 283) ; provided, of course, that the couple (or force) is act- ing on a rigid body. Now the three charac- teristics enumerated above can all be indicated by a vector which can therefore serve as the geometrical representative of the cou- ple. Thus, the couple formed by the forces F, — i^(Fig. 95), whose perpendicular distance is /, is represented by the vector AB = Fp laid off on any normal to the plane of the couple. The sense is indicated by drawing the vector toward Fig. 95. 325 ] COUPLES. 205 that side of the plane from which the couple is seen to rotate counter-clockwise. We shall call this geometrical representative AB of the couple simply the vector of the couple. It is sometimes called its moment, or its axis, or its axial inoment. 324. As was pointed out in Art. 303, a couple can be regarded as the limit of a force whose magnitude approaches zero while its line of action is removed to infinity. Similarly, in kinematics a rotation whose angle approaches zero while its axis is removed to infinity, reduces to a translation, and an angular velocity whose magnitude tends to zero while its axis is removed indefinitely becomes in the limit a velocity of translation. Just as, in kinematics (see Art. 184), two equal and opposite angular velocities about parallel axes produce a velocity of translation, so in statics two equal and opposite forces along parallel lines form a new kind of quantity called a couple. It should, however, be noticed that while rotations, angular veloc- ities, and forces are represented by rotors, i. e. by vectors confined to definite lines, translations, velocities of translation, and couples have for their geometrical representatives vectors not confined to particular lines. Just as in the case of couples of angular velocities, the vector repre- senting a couple of forces has for its magnitude and sense those of the moment of the couple, and for its direction that perpendicular to the plane of the couple. It is due to this analogy between the two fundamental conceptions that a certain dualism exists between the theories of statics and kine- matics, so that a large portion of the theory of kinematics of a rigid body might be made directly available for statics by simply substituting for angular velocity and velocity of translation the corresponding ideas of force and couple. 325. It is easily seen how, by means of Arts. 320—322, any number of couples acting on a rigid body can be reduced to a single resultant couple. It can also be proved without much difficulty that the vector of the resultant couple is the geometric sum of the vectors of the given couples ; in other words, vectors representing couples acting on the same rigid body are combined by the parallelogram law (Art. 40). 2o6 STATICS. [326. , p P . A -F < p' > < p f \ > In the particular case when the couples all lie in parallel planes, or in the same plane, their vectors may be taken in the same line, and can, therefore, be added algebraically. Hence, t/ie resultayit of any number of couples is a single couple whose vector is the geometric sum of the vectors of the given couples. Conversely, a couple can be resolved into components by resolv- ing its vector into components. 326. To combine a single force P with a couple {F, p) lying in the same plane it is only neces- sary to place the couple in its plane in such a position (Fig. 96) that one of its forces, say — F, shall lie in the same line and in opposite sense with the single force P, and to transform the couple {F, p) into a couple {P, p'), by Art. 322, so that Fp = Pp' . The original force P and the force — P of the transformed couple destroying each other at A, there remains only the other force P, at A' , of the transformed couple, that is, a force parallel and equal to the original single force P, at the distance , F P' = -pP from it. Hence, a couple and a single force in the same plane are together equivalent to a single force eqiLal and parallel to, and of the same sense with, the given force , but at a distance from it zvhich is found by dividing the moment of the couple by the single force. 327. Conversely, a single fai^ce P applied at a point A of a rigid body can always be replaced by an equal and parallel force P of the same sense, applied at any other point A' of the same body, in com- bination with the couple formed by P at A and — P at A' . Fig. 96. 329.] COUPLES. 207 This follows at once by applying at A' two equal and opposite forces each equal and parallel to P. 328. The proposition of Art. 326 applies even when the force lies in a plane parallel to that of the couple, since the couple can be transferred to any parallel plane without changing its effect. If the single force intersects the plane of the couple, it can be resolved into two components, one lying in the plane of the couple, while the other is at right angles to this plane. On the former component the couple has, according to Art. 326, the effect of transferring it to a parallel line. We thus obtain two non-intersecting, or skew, forces at right angles to each other. Let P be the given force, and let it make the angle a. with the plane of the given couple, whose force is F and whose arm is /. Then P sina is the component at right angles to the plane of the couple, while P cosa combines with the couple whose moment is Fp into a force P cosa in the plane of the couple ; this force P cosa is parallel to the projection of P on the plane, and has the distance FpjP cosa from this projection. Hence, in the most general case, the combination of a single force and a couple can be replaced by the coinbination of two single forces crossing each other (without meeting) at right angles ; it can be reduced to a single force only when the force is parallel to the plane of the couple. 329. Exercises. ( 1 ) Show that the moment of a couple can be represented by the area of the parallelogram formed by the two forces of the couple, or by twice the area of the triangle formed by joining any point on the line of one of the forces to the ends of the other force. (2) Show that the sum of the moments of two forces forming a couple is the same for any point in the plane of the couple (comp. Art. 289). (3) Show, by means of Arts. 320-322, how to combine any number of couples situated in the same plane, or in parallel planes. (4) Find the resultant of two couples situated in non-parallel planes, without using the vectors of the couples. 208 STATICS. [330. (5) Prove that the vector of the resultant of two couples in different planes is the geometric sum of the vectors of the given couples. V. Plane Statics. I. THE CONDITIONS OF EQUILIBRIUM. 330. Suppose a rigid body to be acted upon by any number of forces, all of which are situated in the same plane. To reduce such a plane system of forces to its simplest form the proposition of Art. 327 may be used. This proposition enables us to transfer all the forces to a common origin, by introducing, in addition to each force, a certain couple in the same plane. The concurrent forces can then be combined into their resultant by geometric addi- tion, i. e., by forming their force polygon (Art. 277) ; and the couples lying all in the same plane combine by algebraic addition of their moments into a resultant couple (Art. 325). Thus, let F {¥ig. 97) be one of the forces of the given plane system, P its point of application. Selecting any point O in the plane as origin, apply at O two equal and opposite forces F, — F, each equal and paral- lel to the given force F; and let p be the perpendicular distance of the origin from the line of action of the given force F. The force F a.t P is equivalent to the force F at together with the couple formed by F at P and — F at ; the moment of this couple is Fp, and its vector is perpendicular to the plane of the system. Proceeding in the same way with every force of the given sys- tem, all forces are transferred to the common origin 0. The whole system is therefore equivalent to the resultant R passing through 0, together with the resulting couple //= 1,Fp. 331, The given system of forces is said to be in equilibrium if the two following conditions of equilibrium are fulfilled : Ji = o, H=o. Fig. 97. 333-] PLANE STATICS. 209 It will be noticed that the moment Fp of the couple introduced by transferring the force F to the point is the moment of the force F with respect to this point 0. Hence, a plane system of forces is in equilibrium if {a) its result- ant is zero, and ib) the algebraic sum of the moments of all its forces is zero with respect to any point in its plane. 332. It is evident that the magnitude and direction of the result- ant R do not depend on the selection of the origin 0. But the position of this resultant and the magnitude of the resulting couple H will in general differ with the point selected as origin. Indeed, the origin can be so taken as to make the couple H van- ish (unless the resultant R be zero) ; that is, the whole system can be reduced to a single resultant. To do this (see Art. 326), it is only necessary, after determin- ing R and H for some point 0, to transfer RXa z. parallel line at such a distance r from its original position as to make the moment Rr of the couple introduced by the transfer equal and opposite to the moment "^Fp ; i. e., we must take (Art. 326) _ ff ^~ ~R' The line along which this single resultant acts is called the central axis of the given system of forces. 333. For a purely analytical reduction of a plane system of forces the system is referred to rectangular axes Ox, Oy, arbi- trarily assumed in the plane (Fig. 98). Every force F is resolved at its point of application P (x, y) into two components X, V, parallel to the axes, so that X= F cosflt, F= F sina, a being the angle made by F with the axis Ox. At the origin two equal and opposite forces X, — X are applied along Ox, and two equal and opposite forces Y, — F along Oy. Thus, X ■a.\. P is equivalent to ^ at together with the couple formed by X ^\. P and — X sX. O ; the moment of this couple is evidently PART II — 14 2IO STATICS. [334- — yX. Similarly, F at P is replaced by F at together with a couple whose moment is ^K The force i^ at P vs, therefore equivalent to the two forces X, Y at together with a couple whose moment is xY-jX. Proceeding in the same way with every given force, we obtain a number of forces X along Ox whose algebraic sum we call I^X, and a number of forces Y along Of which give 2K These two rectangu- . .Y -X — ^ X — *- -Y Fig. 98. lar forces form the resultant R = V^xf+JTVy whose direction is given by tana = 2F 25" where a is the angle between Ox and R. In addition to this, we obtain a number of couples xY — j/X whose algebraic sum forms the resulting couple H='2{xY-jyX). The whole system is thus found equivalent to a resultant force R together with a resultant couple N in the same plane with R. The conditions of equilibriutn R = o, II = o (Art. 331) can there- fore be expressed analytically by the three equations '2X= o, 2 Y= o, ^{xY-yX) = o. 334. If R be not zero, R and H can be combined into a single resultant R' equal and parallel to R at the distance — HI R from it (see Art. 332). The equation of the line of this single result- ant R' , i. e., the central axis of the system of forces, is found by considering that it makes the angle a with the axis of x and that 336.] PLANE STATICS. 211 its distance from the origin is HjR = 1{xY-yX)lV(^Xf-\- {^Y)\ Hence its equation is If i? = o, the system is equivalent to the couple H=^xY-yX). If H itself be also zero, the system is in equilibrium. 335. The following examples will illustrate the application of the conditions of equilibrium. To establish these conditions in any particular problem it will generally be found best to resolve the forces along two rectangular directions and equate the sums of the components to zero ; and then to "take moments," i. e., equate to zero the sum of the moments of all the forces with re- spect to some point conveniently selected as origin. The principle that three forces acting on a rigid body can be in equilibrium only if they are concurrent or parallel and lie in the same plane is often used to advantage although its proof can only be derived from the general conditions of equilibrium of a rigid body. 336. A homogeneous straight rod AB ^ 2/ (Fig. 99) of weight W rests with one end A on a smooth horizontal plane AH, and with the point £{AE ^= e) on a cylindrical support, the axis of the cylinder being at right angles to the vertical plane containing the rod. V^ „ Determine what horizontal \^ ^^^^^ ^ force F must be applied at a . l 3r''T; given point F of the rod ^.^-""^"''^ {AF=f~> «) to keep the rod — ~a~^ "^^ ^' ,; ,;.,;-■;; ^jv^ in equilibrium when inclined to ^ the horizon at an angle o. The rod exerts a certain unknown pressure on each of the supports at A and E, in the direction of the normals to the surfaces of contact, provided there be no friction, as is here assumed. The supports may therefore be imagined removed if forces A, E, equal and opposite to 212 STATICS. [337- these pressures, be introduced ; these forces A, E are called the reactions of the supports. The rod itself is here regarded as a straight line ; its weight W\% applied at its middle point C. Taking A as origin and AH as axis of x, the resolution of the forces gives E'ivnd = o, (i) sy= A — W+ E cose = o. Taking moments about A, we find E-e— W- Icoad — F- fsmO = o. Eliminating F{x sin(^ W P' while from the triangle AB C sm^o sinO 2/ _ \BC'' hence BC= 2hP/ TV, i.e. if we take h to represent W, P will be represented by \BC. For the total pressure A we have A'' -\- A^= IV' + P' — 2J^-Pcos,p, 214 STATICS. [339- /. e. A is the third side of the triangle having Wa.nd P for the other two sides and (p for the included angle. The magnitude of A is therefore represented by the median from A in the triangle ABC on the same scale on which Wis represented by h. But this median gives also the direction of A ; for we have A^ IV—Fcosp _ A —^BCcos

e; (J>) when/<^. 2l6 STATICS. [340- (8) A smooth weightless rod AB = I rests at Con a smooth hori- zontal cylinder whose axis is at right angles to the vertical plane through the rod ; its lower end A leans against a smooth vertical wall whose distance from C is C£> = a ; from its upper end £ a weight W is suspended. Determine the distance AC= x for equilibrium, and the reactions at A and C. (9) A homogeneous rod of weight Wis hinged at its lower end A, while its upper end J3 leans against a smooth vertical wall. The rod is inclined at an angle to the vertical, and carries three weights, each equal to w, at three points dividing the rod into four equal parts. Determine the pressure on the wall and the reaction of the hinge. ( 10) A homogeneous rod A£ = 2/ of weight fF rests with one end A on the inside of a fixed hemispherical bowl of diameter 2a and leans at C on the horizontal rim of the bowl, so that the other end B is out- side. Determine the inclination to the horizon d in the position of equilibrium. (11) A homogeneous rod AB = /, of weight TV, is hinged at its upper end A so as to turn in a vertical plane ; a cord attached to its lower end B runs over a fixed pulley C (regarded as a point, on the same level with ^ ) and carries a weight B. li AC = c and 2^ CAB = 9 be given find F and the reaction at A for equilibrium. (12) A homogeneous rod AB ^ 2 1 is free to turn in a vertical plane about a horizontal axis C dividing its length into segments AC= a, CB = b. It is in equilibrium in a horizontal position under the action of its own weight W and two forces F aX A and F' at B, inclined downward and outward at angles a, a' to the horizon. Find the ratio 711 = ajb and the pressure on the axis at C. (13) A homogeneous rod AB ^ 2/, of weight W, rests at ^ on a smooth floor and leans at B against a smooth wall inclined to the horizon at an angle a. A cord attached at B runs over a pulley C at the top of the wall and carries a weight F. Find F and the reactions at A and B for equilibrium. (14) What horizontal force P must be applied to the axle of a wheel of weight W to just pull it over an obstacle B rising one wth of the radius a above the horizontal road ? What is the pressure F' on the obstacle ? (15) Determine the force F which, applied at the center of a wheel and inclined at a given angle a to the horizon, will just start the 342] PLANE STATICS. 217 wheel over a fixed cylindrical log, whose diameter is one tiih of that of the wheel. For what value of a is ^ least ? 2. STABILITY. 341. The equilibrium of the forces acting on a rigid body may subsist while the body is in motion. Thus, if the motion consist in a mere translation with constant velocity, the equilibrium will not be disturbed during the motion if the forces remain equal and parallel to themselves. If, however, the body be subjected to a rotation, this will in general not be the case. The present considerations are restricted to the case of plane motion ; the forces are supposed to lie in the plane of the motion and to remain equal and parallel to themselves and applied at the same points of the body. 342. Let A^A^ (Figs. 103 and 104) be a rigid rod having two equal and opposite forces F^, F^ applied at its extremities in the direction of the line A^A^. Let this rod be turned through an A 2 A, A, r P F F, 2^^ 1^- ''. F2 Fig. 104. Fig. 103. angle ^ about an axis at right angles to A^A^. In the new posi- tion the forces F^, F^ instead of being in equilibrium, form a couple whose moment is ± F^- A^A^ sin0. If in the original position of the rod the forces tend to increase the distance A^A^ (Fig- I03)> the couple in the new position will tend to bring the rod back to its position of equilibrium. In this case the original position of the rod is said to be a position of stable equilibrium. The effect of the earth's magnetism on the needle of a compass offers a familiar example. If, however, in the original position the forces tend to diminish the distance A^A^ (Fig. 104), the couple arising after displacement 2l8 STATICS. [343- tends to increase the displacement and thus to remove the rod still farther from equilibrium. The original position in this case is said to be one of unstable equilibrium. The weight of a rod balanced in a vertical position and the reaction of the support may be taken as an illustration. Finally, a third case would arise if the forces F^, F^, being still equal and opposite, were applied at one and the same point of the rod. The forces would then remain in equilibrium after a dis- placement of the rod; such equilibrium is called neutral or astatic. Generally, the equilibrium of a rigid body is said to be stable if, after a sufficiently small displacement of the body, the forces tend to bring the body back to its position of equilibrium ; un- stable if, after a sufficiently small displacement, the forces tend to remove the body still farther from its position of equilibrium ; neu- tral if, after a small displacement, the body remains in equilibrium. These three types of equilibrium are also well illustrated by a homogeneous sphere placed within a hollow sphere (stable equi- librium), or balanced on top of another sphere (unstable equilib- rium), or placed on a horizontal plane (neutral equilibrium). - 343. The different cases of equilibrium can be distinguished by the algebraic sign of the product A^A.^ F^ = A^A^ F^, which is negative for stable equilibrium, since A^A^ and F^ have opposite sense (Fig. 103), positive for unstable equilibrium (Fig. 104), and indeterminate (since A^A.-^ = o) for neutral equilibrium. It is to be noticed that these considerations will hold whether the rotation of angle <^ take place in the positive or negative sense. But they hold only within certain limits for the angle of rotation. Thus, in the example illustrated by Figs. 103 and 104, if (^ were taken as large as vr, a new position of equilibrium would be reached ; but it would be unstable for Fig. 103, stable for Fig. 104. 344. Exercises. ( I ) Explain the nature of the equilibrium of a body of weight W supported at a single point according to the position of that point above the centroid G, below G, and at G (^common balance). 345-J PLANE STATICS. 219 ( 2 ) A body of weight W^is placed on a horizontal plane. Show that the equilibrium is stable if fF' meets the horizontal plane at a point A within the area of contact and that it is unstable HA lies on the con- tour of this area. If the actual area of contact have re-entrant angles, or consist of several detached portions, the area bounded by a thread drawn tightly around the actual area, or areas, of contact must be substituted. (3) An oblique cylinder rests with its circular base on a horizontal plane in unstable equilibrium. If the length of its axis be twice the diameter of its base, what is the inclination of the axis to the horizon ? (4) Show how to determine graphically the stability of a retaining wall against toppling over the front edge of the base, the pressure of the earth behind the wall being given in magnitude, direction and position. ( 5 ) A cube, a hemisphere, a right circular cone of height h and base radius a, and a right pyramid of height h and square base of side a are placed on a plane inclined to the horizon at an angle 0, the cube and the pyramid so that one side of the base is horizontal. If the lowest line or point be fixed, for what inclination will they topple over ? (6) A solid is formed by gluing the base of a homogeneous hemi- sphere of radius a and density f>^ on to the base of a homogeneous right cone whose base has the radius a and whose density is p^. The solid is placed on a horizontal table, with the axis of the cone vertical. What must be the height h of the cone to make the equilibrium neutral ? (7) In Ex. (6), substitute a cylinder for the cone. 3. JOINTED FRAMES. 345. The equations of equilibrium are derived on the suppo- sitions that all the forces of the given system act on one and the same rigid body and that this body is perfectly free to move. Hence, in applying these equations to determine the equilibrium of an engineering structure, a machine, etc., each rigid body must be considered separately, and the reactions required to make the body free mu.st be introduced. It will be shown in a subsequent section how the principle of virtual work makes it possible to dis- pense with some of these precautions. 220 STATICS. [346. When two rigid rods are connected by a frictionless pin -joint whose axis is perpendicular to the plane of the rods, the action of either rod on the other at the joint is represented by a single force whose direction is in general unknown. Sometimes con- siderations of symmetry will enable us to determine this direction. If a rigid rod, in equilibrium, be hinged at both ends and not acted upon by any other forces, the reactions of the hinges must of course be along the rod, and must be equal and opposite. 346. Two rods AC, £C (Fig. 105) in a vertical plane, hinged together at C, rest with the ends A, B on a horizontal plane, a?id carry a weight W suspended frojn the joint C. If the proper weight of the rods be neglected, determine the normal pressures A , B and the horizontal thrusts A , B at A, B. X' X ' Resolving the weight W along CA, CB Fig. 105. into W/^, Wg and considering the rod A C alone, it appears that the total reaction at A is along A C and = W^^ ; hence resolving lF^ in the horizontal and vertical directions, A^ and A are found ; similarly for BC. If (/, /9 be the angles at A and B in the triangle ABC, we find cos,9 cosa cosacos;9 _ sin g cos;? cos« sin/? ^ ^ sin(a + /?)'^^' ^^ sin(a-i-/?) ' ^ sin(a + /?)^- As the horizontal thrusts at A and B are equal it makes no differ- ence whether the rods be hinged to the supports at A and B, or whether the thrust is taken up by lateral supports, or by a string connecting the ends A, B of the rods. 347. T7V0 equal homogeneous rods AC, BC (Fig. 106) are hinged at A, B, C so as to form a triangle whose height h is vertical and whose base AB ^ 2b is horizontal. The weight of each rod being W, find the reactions at the joints. Owing to the symmetry of the figure, the reactions at C must be equal and opposite and horizontal. The rod y4C is subject to three forces only, viz. the horizontal reaction C, the weight W, and the 348.] PLANE STATICS. 221 reaction A ; the latter must therefore pass through the intersection D of C and W. If the direction of fF intersect AB at E and the scale of forces be taken so as to have fF represented by DE = h, DEA will be the force polygon ; hence EA represents C and AD represents A on the same scale on which Whs, represented by h. Analytically, the reactions are found by re- solving the forces horizontally and vertically and taking moments about A : A^ =C, A^= W, whence Ch= W-\b; C=mW, A= VA' + A' = Vm' + i • W^, where m = bl2h. 348. Two equal homogeneous rods AC, BC, each of -weight W, are hinged at C ; their ends A, B rest on a smooth horizontal plane ; a third rod DE is hinged to them, connecting their middle points (Fig. 107). The plane AB being smooth, the reaction at A is vertical ; the reaction at Cis horizontal owing to the symmetry ; that at Z> is likewise horizontal if the weight of the rod DE be neglected, for then this rod is subject only to the reac- tions at its ends. Resolving horizontally and ver- tically and taking moments about D, we find in this case C= D = IVcota, Fig. 107. where A= W, a=-4BAC. If, however, the weight w of the rod DE cannot be neglected, we have at Z> a horizontal reaction D^ and a vertical reaction D^. The equilibrium of DE requires that 2D^ = w. Hence resolving and taking moments as before, we find A= IV+iw, C= D^=( W+ \w) cota, D^ = Iw. 222 STATICS. [349- 349. Exercises. (i) Solve the problem of Art. 347 by replacing each weight W by J?Fat each end of each rod, then resolving the load WsX C along the rods, etc. (2) Apply to the problem of Art. 348 the principle that three forces in equilibrium must pass through a point. (3) Two homogeneous rods AC, BC oi equal weight, but unequal length are hinged together at C while their other ends are attached to fixed hinges A, B in the same vertical line. Show that the line of action of the reaction at C bisects AB. (4) Two homogeneous rods AC= BC ^ 2/, in a vertical plane, each of weight li^, are connected by a pin-joint at C while at A and B there are fixed pin-joints, the point A lying a ft. above B. Find the reactions of A and B if the inclinations a, ^ oi AC, BC to the horizon are given. 350. A triangular frame formed of rigid rods is rigid as a whole, even when the connections are pin-joints. A quadrangular frame with pin -joints becomes rigid only by the insertion of a diagonal. The iron and steel trusses used for roofs and bridges generally consist of a system of triangles, or quadrangles with diagonals, so that the whole truss can be regarded as one rigid body, at least in first approximation. Any one rod, or member, of the frame-work is thus acted upon by two equal and opposite forces, i. c., by a stress, in the direction of its length, the external forces, including the proper weight, being regarded as applied at the joints only. If the stress be a tension, i. e., if the forces tend to stretch or elongate the member, the latter is called a tie ; a member subject to compression or crushing is called a strut. Strictly speaking, owing to elasticity, the members of the frame sHghtly change their length and the supports yield. The points of application of the forces change therefore their position under any loading. These changes are neglected in the first approxi- mation, the supports being assumed fixed and the frame rigid. 351. For the purpose of dimensioning the members, it is neces- 35I-] PLANE STATICS. 223 sary to know the stress in every member. The following exam- ple illustrates a simple method for finding these stresses when the external forces are given. Let the frame-work represented in Fig. 108 be cut in two along any line a/3 ; the portion on either side of this line must be in equilibrium under the action of its external forces and the Fig. 108. stresses in the members intersected by a/3. Thus, in the figure, the forces A, JV^, F^, F^, F^ form a system in equilibrium ; hence, the sum of the moments of these forces with respect to any point must vanish. To determine F^, take moments about the mtersection of F.^ and F^ ; thus F^ and F^ are eliminated from the equation of moments, and F^ is found. Similarly F^ is obtained by taking moments about the intersection of F^ and F^. The arms of the moments are best taken from an accurately drawn diagram of the frame-work. If only two members be intersected by a/S, the origin for the moments is taken first on one, then on the other, of the two members intersected. 224 STATICS. [352- By beginning at one of the supports and taking sections through the successive panels, it will in the more simple cases be possible to draw the line a/3 so as to intersect not more than three members whose stresses are unknown. Thus the stresses in all the members can be determined. (i) Find the stresses in the braced beam AB (Fig. 109), carrying a weight of 6 tons at each joint of the up- per chord. The horizontal width of the panels is 1 2 ft. , the middle vertical is 9 ft. (2) In Fig. no, the dimensions are in feet, the loads in tons. After the first panel the sections cannot be so taken as to intersect not more than three unknown stresses. But the girder can be regarded as obtained by the superposition of two girders (each carrying half the 352. Exercises. A • ■ ' B Fig. 109. >^ ■^ '// load), in one of which the diagonals B' E, CD are wanting, while in the other D' C, EB are wanting. Each of these can readily be computed. 4. GRAPHICAL METHODS. 353. The graphical method explained m Art. 299 for deter- mining the resultant of a system of parallel forces can be ex- tended without difficulty to the general case of a plane system of forces. The only difference will appear in the form of the force polygon, which lor parallel forces collapses into a straight line, while in the general case it is an ordinary (unclosedj polygon whose closing line represents the resultant In magnitude and 355-] PLANE STATICS. 225 direction. In other words, when the forces are not parallel, they must be added geometrically, and not algebraically. The construction of the funicular polygon and its properties are the same as for parallel forces. If the force polygon does not close, the given system is equiva- lent to a single resultant represented in magnitude, direction, and sense by the closing line ; its position is obtained from the funicular polygon whose initial and final lines must intersect on the resultant. If, however, the force polygon closes, the system may be equivalent to a couple, or it may be in equilibrium. The distinc- tion between these two cases is indicated by the funicular polygon. If the initial and final lines of this polygon coincide, the system is in equilibrium ; if they are merely parallel, these lines are the directions of the forces of the couple to which the whole system reduces. The magnitude and sense of the forces of the resulting couple are obtained from the force polygon. 354. Thus it follows from the graphical as well as from the analytical method that a plane system may be equivalerit to a single force, or to a couple, or to zero. In the first case, the force polygon does not close, and the initial and final sides of the funicular polygon intersect at a finite distance. In the second case, the force polygon closes, and the initial and final sides of the funicular polygon are parallel. In the third case, the force poly- gon closes, and the initial and final sides of the funicular polygon coincide. The graphical conditions of equilibrium of a plane system are therefore, two: (i) the force polygon must close; (2) the funic- ular polygon must have its initial and final sides coincident. 355. To every vertex of the force polygon corresponds a side of the funicular polygon, and vice versa. The force polygon is said to close if the last vertex coincides with the first ; similarly, the funicular polygon might be said to close when its last side coincides with the first. With this convention, we may say that PART II — 15. 226 STATICS. [356. the conditions of equilibrium of a plane system require the closing of both the force polygon and the funicular polygon. 356. One of the most important applications of the graphical meth- ods is found in the determination of the stresses in the frame-works used for bridges, roofs, cranes, etc. The following example will illustrate the method. Fig. Ill represents the skeleton frame of a roof truss subjected to the "loads" W^, IV^, W^z.nA the reactions of the supports A, B. The members of the frame in connection with the lines of action of these forces (imagined as drawn from in- finity up to the points of appli- cation) divide the whole plane into a number of compartments marked in the figure by the let- ters a, h, c, d, ■ ^ . ■ The ex- ternal forces as well as the members of the frame (or the stresses acting along them) can thus be designated by the two letters of the two portions of the plane separated by the force or stress. For instance, the re- action A is denoted by at), and the stresses in the two members concurring at^ are l>c and ca. The figure just described may be called the frame diagram ; and we proceed now to con- struct its stress diagram .^ Laying off on a verticalline i^(? := W^,eg= 'f^j, ^'= ?^, and bisect- ing bj at a, we have the polygon of the external forces which gives the reactions A = ab, B =ja. * The student is advised to draw the stress diagram himself step by step as indicated in the text. 3 59- J PLANE STATICS. 227 Next, beginning at the vertex A the stresses in the two members intersecting at A are found by resolving the reaction A along the direc- tions of these members ; and this is done in the stress diagram by drawing parallels to these directions through the points a and b. The intersection is denoted by c. 357. It will be noticed that the three lines meeting at A have cor- responding to them, in the stress diagram, the three sides ab, be, ca of a triangle. The force A =■ ab vs, represented by ab ; the stress in the member be (/. e. in the member separating the compartments b, c in the frame diagram) is represented in magnitude, direction, and sense by the side be in the stress diagram ; and the stress in the member ea is given by the side ca of the triangle abe. To obtain the sense of each stress correctly, the triangle abc in the stress diagram must be traversed in the sense of the known force A = ab ; this shows that the member be is compressed, the stress at A acting towards A, while ea is subject to tension. It will be found in general that the lines of the stress diagram corre- sponding to all the lines meeting at any one vertex of the frame diagram form a closed polygon. The reason is obvious : the forces at the vertex must be in equilibrium. 358. To continue the construction of the stress diagram, we pass to another vertex of the frame diagram, selecting one at which not more than two stresses are unkno^vn. Thus at the vertex acd the stress in ac is known, being represented by ac in the stress diagram. Hence draw- ing through a a parallel to da, through c a parallel to cd, we find the point d of the stress diagram. The vertex dcbef can now be attacked ; dc, eb, be are already drawn, and it only remains to draw ^parallel to if/ and (^parallel to df. The rest explains itself Considerations of symmetry are frequently helpful in affording checks. 359. Exercises. (i) Check the computed stresses of Exercises (i) and (2), Art. 352, by constructing the stress diagrams. (2) Find the stresses in the frame (Fig. 112) if AA' =^ i^o ft., BB' ^= 12 ft., the distance between these lines is 3 ft., and BA = BC. 228 STATICS. [360. (3) Determine the stresses in the frame, Fig. 113, if the load con- sists of seven weights, each of 2 tons, apphed at the joints of the upper chord. Owing to the symmetry of the figure, it is sufficient to construct the stress diagram for half the frame. Beginning at ^ a difficulty arises at B and E since three members with unknown stresses pass through each of these points. The difficulty can be overcome in various ways ; for instance, by observing that, owing to symmetry, the stresses v!\ EG and EH must be equal ; or by taking a vertical section near C and moments about C, whereby the stress in BB' can be found. Fig. 113. 360. Shearing^ Force and Bending Moment. Consider a hori- zontal beam fixed at one end A (Fig. 1 14), and acted upon at the other end B hy ■&. vertical force F. If the beam be cut at any point C and the equilibrium of the portion AC he considered, the action on AC of the portion removed must be replaced by its equivalent. Now the force F at B is equivalent, by Art. 327, to an equal and parallel force F a.t C in connection with a couple whose moment is F- EC The force F 2XCis called the shear- ing force of the cross-section C, and the moment/^- .5 C the bending moment at C. Both are of great importance in engineering, as their combined effect p, ,1^ B 36I.J PLANE STATICS. 229 represents what must be overcome by the resistance of the material of the beam, i. e. by the internal forces holding together its fibers. These definitions are readily generalized. Let any beam or girder, supported in any manner, and acted upon by any number of vertical forces, be divided by a vertical cross-section into two portions A and B. For the portion A the shearing force at the cross-section is the sum of all the external forces acting on B ; and the bending moment is the sum of the moments of all these forces with respect to some point in the cross-section. Fig. 1 1 5. 361. According to its definition the bending moment of a beam at any cross-section is found by adding the moments, with respect to the cross-section, of all the external forces on one side of the section. Graphically, the bending moment is readily derived from the funicular polygon. Thus in Fig. 115, for the cross-section a^, the resultant of the forces on the left is R' ^ A — W^ — W^ = 03 230 STATICS. [362. in the force polygon. Its position is found by determining the intersection of the two sides A' B' and III of the funicular polygon met by the section a/3. For the funicular polygon resolves A along A'B' and I, W^ along I and II, W^ along II and III. The com- ponents falling into the same line being equal and opposite (as appears from the force polygon), the forces A, W^, W^ are to- gether equivalent to the components along A' B' and III ; their resultant R' must therefore pass through the intersection S of these lines. Now if p be the horizontal distance of the point 5' from a/3, the bending moment at a^ is i?' •/ = o 3 -p. If a^ intersect A' B' at P, III at Q, the triangles SPQ and (9 o 3 are similar, so that their altitudes / and //are as the corresponding sides PQ and o 3 ; hence PQ ^ 03 and the value of the bending moment is H ■ PQ. As H is con- stant, we find that the betiding moment is proportional to the vertical height, or ordinate, of the fnnicidar polygon. 5. FRICTION. 362, The reaction between two surfaces in contact has so far been regarded as directed along the common normal of the sur- faces. This is true when the surfaces are perfectly smooth. The surfaces of physical bodies are roiigJi, i. e. they pi'esent small elevations and depressions ; when two such surfaces are " in contact" the projections of one will more or less enter into de- pressions of the other ; the greater the normal pressure between the surfaces, the more will this be the case. Hence when a tan- gential force acting on one of the bodies tends to slide its surface over that of the other body, a resistance will be developed whose magnitude must depend on the roughness of the surfaces and on the normal pressure between them. This resistance is called the force of sliding friction, or simply the friction. The study of friction belongs properly to applied mechanics, and will here only be touched upon very briefly. 364.] PLANE STATICS. 23 1 363. Imagine a body resting with a plane surface on a hori- zontal plane. Let a small horizontal force P be applied at its centroid (which is supposed to be situated so low that the body is not overturned), and let the force P be gradually increased until motion ensues. At any instant before motion sets in, the friction is equal to the value of P at that instant. The value of P at the moment when motion just begins is equal and opposite to ihs. frictional resistance F between the surfaces at this moment, and this resistance is called the limiting static friction. Careful experiments with dry solids in contact have shown this force to be subject to the following laws : ( 1 ) The viagnitude of the limiting friction F bears a constant ratio to the nonjial pressure N between the surfaces in contact ; that is F=i.N, where /.i is a constant depending on the condition and nature of the surfaces in contact. This constant which must be determined experimentally for different substances and surface conditions is called the coefficient of static friction. It is in general a proper fraction ; for perfectly smooth surfaces /x = o. (2) For a given normal pressure the limiting static friction, and hence the coefficient of static friction, is independent of the area of contact, provided the pressure be not so great as to produce cutting or crushing. 364. The frictional resistance between two surfaces in relative motion is called kinetic friction. It is subject, in addition to the two laws just mentioned, to the third law : (3) For moderate velocities, kinetic friction is nearly independent of the velocities of the bodies in contact. The coefficient of static friction is somewhat greater than that of kinetic friction. A slight jarring will often reduce the coeffi- cient from its static to its kinetic value. It must not be forgotten that these so-called laws of friction are experimental laws, and therefore true only approximately and 232 STATICS. [365- within the limits of the experiments from which they were deduced. When the relative velocity of the surfaces in contact is high, or when, as is usually the case in machinery, a lubricating material is introduced between the two surfaces, the frictional resistance is found to depend on a number of other circumstances, such as the temperature, the form of the surfaces, the velocity, the nature of the lubricator, etc. Indeed, when the supply of the lubricant is sufficient, the two solid surfaces are kept by it out of actual contact ; the coefficient of friction in this case varies with the pressure, area of contact, velocity, and temperature. 365. Consider again a body resting on a horizontal plane (Fig. 1 16) and acted upon by a horizon- tal force P just large enough to equal the limiting friction F. The normal reaction N of the plane is equal and opposite to the weight W. The body is thus in equilib- rium under the action of the two pairs of equal and opposite forces ; but motion will ensue as soon as P is increased. If P be decreased, F will decrease at the same rate, so that the equilibrium remains undisturbed. The force of friction F can be combined with the normal reac- tion N to form a resultant, R,v- — N FL fi ^ .P ^^msm^^ ' W R = i//r2 + 7V^2 = -^//52 -f w\ which represents the total 7'eaction of the horizontal plane. If ^ be the angle between N and R when F has its limiting value F= ixN (Art. 363), we have, since tan^ = FJN, tan^ = /u. The angle (^ thus furnishes a graphical representation for the coefficient of friction /i ; it is called the angle of friction. 367-] PLANE STATICS. 233 366. If the plane be not horizontal, but inclined to the horizon at an angle 6, the weight W of the body (regarded as a particle) resting on the plane can be re- solved into a component W sm6 along the plane, and a compo- nent W cosd perpendicular to it (Fig. 1 1 7). Hence, if no other forces act on the body it will be in equilibrium, provided the com- ponent W smO be not greater than the limiting friction F= fiJVcosO. The hmiting condition of equilibrium is, therefore, ix W cosO = W sin^, or /a = tan^ ; in other words, if the angle 6 be gradually increased, the body -will not slide down the plane until ^ > ^. This furnishes an experi- mental method of determining the angle of friction <^, which on this account is sometimes called the angle of repose. 367. A particle P (Fig. 118) will be in equilibrium on any rough surface, if the total reaction of the surface, i. e. the resultant R of the normal reaction N and the friction F, is equal and opposite to the resultant R' of all the other forces acting on the particle. The limiting value of the angle between N and R is <^, so that the particle can be in equilib- rium only if the resultant R ' makes with the normal an angle ^^. Hence, if about the normal PN as axis, and with P as vertex, a cone be described whose vertical angle is 2(^, the condition of equilib- Fig. 118. rium is thati?' must lie within this cone. The cone is called the cone of friction. 234 STATICS. [368. 368. The idea of the angle of friction suggests a graphical method for problems on equilibrium with friction. The case of a rod resting on two inclined planes, Art. 339, Fig. 102, may serve as an example. If the intersection E, of the normal reactions A and B lies on the vertical through D, the rod will be in equilibrium whether there be friction at A and B or not. When this condition is not fulfilled, the rod may still be in equilibrium if there be sufficient friction between the ends of the rod and the supporting planes. Let Ai ^ tan^ be the coefficient of friction on the plane CA, ix' = tanp' that on CB ; then the total reactions at A and B will, by Art. 365, make angles not greater than

- -^/^ y c , ^w ' w Fig. 119. the weight JV, in a given position of the rod, can be found by deter- mining the intersection of the lines of these total reactions ; the limit- ing position of ^is the vertical through this intersection. Thus, to prevent the rod from sliding up the plane CA and down the plane CB, the friction angles p, . The other position D" is found by applying the friction angles in the positive sense. Equilibrium will therefore sub- sist if the weight be placed anywhere between D' and Z>". 370.] PLANE STATICS. 235 The construction is somewhat simpHfied when (p = y', since then the intersections of the total reactions he on the circle described about ABC(Y\g. 119). 369. As another example consider the ordinary jack intended to raise an eccentric load ff 'acting vertically downwards through A (Fig. 120) by a force P passing vertically upwards through the pitch line B of the rack. Near C and D the rack is pressed against the casing. The directions of the total reactions C, D at these points are found by applying the friction angle to the normals. The four forces IV, P, C, D can be in equilibrium only if the resultant of ]V and D is equal and opposite to the resultant oi P and C; hence, if ^ be the intersection of lVa.nd D, F that of P and C, each of these resultants must act along EF. If the load W be known, the other forces can now be found by constructing the force polygon. Draw \2 = Wvs\ position (J. e. through^); draw 2 3 parallel to C; 41 parallel to D ; and through the intersec- tion 4 of 4 I with EF draw the vertical 3 4 to the intersection 3 with 23- 370. Exercises. ( 1 ) Determine the tractive force required to haul a train of 160 tons with constant velocity up a grade of 2 per cent, if the coefficient of friction is 1/200. (2) A weight ^is to be hauled along a horizontal plane, the coeffi- cient of friction being ix = tan tp. Determine the required tractive force P if it is to act at an inclination a to the horizon, and show that this force is least when a = ^. (3) A particle of weight ^ is in equilibrium on a rough plane incHned to the horizon at an angle e, under the action of a force P parallel to the plane along its greatest slope. Determine P : (a) 236 STATICS. [371. when d^ by siny; ; the moment of the friction couple is then = p. Wr, i. e. the frictional force /i IVcun be regarded as acting at the lever-arm r (Fig. 121). 372. The effect of journal friction can also be illustrated as follows (Fig. 123). The shaft and journal may be regarded as rotating uniformly about their common horizontal axis under the action of a driving force whose moment about O would have to be equal and opposite to that of the resistance, or load, if there were no journal friction. For, in this case, the reaction of the bearing to the weight JVof the shaft would act ver- tically upwards through the axis of the shaft, so that its moment would be zero. The existence of friction between journal and bearing requires an increase of the driving force, which may be regarded as a small tan- gential force F appHed at any point £ such that its moment !"■ OB equals the moment about O of the frictional resistance. Let C be the intersection of the direc- tion of this force P with the vertical through O and A, which is the line of action of the weight W of the shaft. The resultant of P and /Fpasses through C, and intersects the circumference of the journal at a point D near A ; the to- tal reaction of the bearing is equal and opposite to this resultant. As this total reaction must make an angle p with the normal at D, we have for the perpen- dicular OE dropped from O on CD: OE = p ^ r sinp, where r is the radius of the journal. A circle described about O, with p as radius, has the total reaction of the bearing as a tangent. This circle is called the friction circle. As is generally very small in the case of journal friction, p. = X3.n

becomes a right angle. 386. It is evident that since H^ = H cos4>, the product RHcos^ is a constant quantity for a given system of forces. It has been called the invariant of the system. Fig. 129. 387.] SOLID STATICS. 24; If the elements of reduction for the central axis (R, H^ be given, those for any parallel line / at the distance r^ from the central axis are determined by the equations Rr. H'^H^-VR'r^, tan,^=^ ra Fig. 130. To sum up the results of the preced- ing articles, it has been shown that miy H^- system of forces acting on a rigid body can be reduced, in an infinite number of ways, to a residtant R in combination with a couple H. For all these reduc- tions the magnitude, direction, and sense of the resultant R are the same, but the vector H of the couple changes accord- ing to the position assumed for the line of R. There is one, and only one, position of R, called the cen- tral axis of the system, for which the vector H is parallel to R and has at the same time its least value, //j, ; this value H^ is equal to the projection of any other vector //^ on the direction of the resultant R. 387. While, in general, a system of forces cannot be reduced to a single resultant, it can always be reduced to two non-intersecting forces. This easily follows by considering the system reduced to its resultant R and resulting couple H for any origin (Fig. 131). Let i^, —i^ be the forces, / the arm of the couple H, and place this couple so that one of the forces, say —F, intersects R at O. Then, combining R and —F into their resultant F' , the given system of forces is evidently equiv- alent to the two non-intersecting forces F, F' (compare Art. 328). Fig. 131. 248 STATICS. [388. 388. The two forces F, F' determine a tetrahedron OABC; and it can be shown that the volume of this tetrahedron is constant and equal to one sixth of the invai'iant of the system (Art. 386). The proof readily appears from Fig. 131. The volume of the tetrahedron OABC is evidently one half of the volume of the quadrangular pyramid whose vertex is C and whose base is the parallelogram OB AD. The area of this parallelogram is Fp = H; and the altitude of the pyramid is = 7? cos, p cos^. To proceed systematically, we may tabulate the components of the forces, and the co-ordinates of their points of application, and then Fig. 1 34. 254 STATICS. [395- form the component couples, as shown below. The components of the unknown reactions A, B of the hinges are called A^, A , A , B , B , B . g A w F A B Components. Co-ordinates. Couples. X V z Jtr y z yZ—zV zX—jcZ xY—yX — IV sine o i: o lVcos9 o o a — a b sin(^ h costf) P sinijt) J>cos ^y< -^z + ^z- The two ^•-components cannot be found separately, since they act in the same straight line. A. Fig. 135. 40I.J SOLID STATICS. 257 Hence, a rigid body having a fixed axis is in equilibrium if the sum of the moments of all the forces vanishes for the fixed axis. 400. If, in the preceding article, the axis be not absolutely fixed, but only fixed in direction so that the body can rotate about the axis and also slide along it, we have evidently hence, by the third equation of equilibrium, SZ=o, as an additional condition of equilibrium. The body has in this case two degrees of freedom. 401. Eigid Body witli a Fixed Plane. A body constrained to slide on a fixed plane (that is, to move so that the paths of all its points lie in parallel planes) has three degrees of freedom. At every point of contact between the body and the plane, the latter exerts a reaction. As all these reactions are parallel, they can be combined into a single resultant N. Taking the fixed plane as the. plane xy, N will be parallel to the axis of s ; hence, if a, b, o be the co-ordinates of its point of application, the six equa- tions of equilibrium are 2X=o, SF=o, ^Z+N=o, ^{yZ - sY) + bN = o, lisX - xZ) - aN = o, '2(xV-yX) = o. The third, fourth, and fifth equations determine the reaction Al and the co-ordinates a, b of its point of application. The three other equations are the actual conditions of equilibrium ; they agree, of course, with the three conditions of equilibrium of a plane system as found in Art. 333. If there be not more than three points of contact (or supports) between the body and the 'fixed plane, the reactions of these points can be found separately. Let A^, A.^, A^ be the three points of contact ; N^, N^, N^ the required reactions ; a^, b^, a^, b^, a^, b^ PART II — 17 I I I «1 «2 «3 K K ^3 258 STATICS. [402. the co-ordinates of A^, A^, A^ ; then N must be resolved into three parallel forces passing through these points, and the conditions are iv; + TV, + 7V3 = N, These three equations always determine JV^, N^, Ny For if the determinant of the coefficients of N^, N^, N^ vanished, = 0, the three points A^, A^, A^ would He in a straight line, and hence the body would not be properly constrained. The reactions become indeterminate whenever there are more than three points of contact. VII. The Principle of Virtual Work. 402. "Work has been defined in Art. 266 as the product of a constant force into the displacement of its point of application in the direction of the force. Thus the expansive force F of the steam in the cylinder of a steam-engine, in pushing the piston through a distance s, is said to do work, and this work is measured by the product Fs, if F is constant. Similarly the force of gravity, i. e. the attractive force of the earth's mass, does work on a falling body. The resistance to be overcome by the engine, in the former case, and the resistance of the air in the latter, are also forces acting on the body during its displacement. But as the sense of the displacement is opposite to that of these forces, their work is negative ; work is done against these forces. Thus the muscular force of a man who raises a weight does work against gravity ; if the weight he holds is so heavy as to pull him down, gravity does work against his force ; if he merely tugs at a 404.] PRINCIPLE OF WORK. 259 weight without being able to Hft it, the work is zero, because the displacement is zero. 403. In general, the point of application of a force F may be acted upon by a number of different forces, so that the displace- ment J of this point will not necessarily take place in the direction of F. In this general case the work of a cojistant force is defined as the product of the force into the projection of the displace- ment of its point of application on the direc- tion of the force. In Fig. 136, for instance, the particle P while acted upon by the force F (and any number of other forces) is displaced from P to P'\ hence '\{ PP' = s, and ^P'PQ = j>, the work of the force F is W=F-scos4>. (i) It is obvious that this work might also be defined as the product of the displacement into the projection of the force on the displacement ; for we have Fscoscj) = FPQ=^sPR. The work of a force is evidently positive or negative accord- ing as the angle (p is less or greater than |7r, provided we select for always that angle between F and .y which is not greater than TT. 404. The above definition of work assumes that the force F remains constant, both in magnitude and direction, while the dis- placement .y takes place, and that this displacement is rectilinear. If either, or both, of these conditions be not fulfilled, the definition can be applied for small displacements, approximately. In the language of infinitesimals, the infinitesimal work done by a (finite) force /^during an infinitesimal displacement ds is denoted hy dW so that dW = Fds coscf), (2) and the total work done by any variable force F, while its point 26o STATICS. [405- of application is displaced along any straight or curvilinear path PQ, is obtained by integrating from Pto Q: W = I Fcos(f>ds. (3) 405. Since work can always be regarded as the product of a force into a length, its dimensions are found by multiplying those of force, MLT-2 (Art. 258), by L ; hence, the dimensions of work are W = ML2T-2. The unit of work is the work of a unit force (poundal, dyne) through a unit distance (foot, centimeter). The unit of work in the F.P.S. system is called the foot-poundal ; in the C.G.S. system, the erg. Thus, the erg is the amount of work done by a force of one dyne acting through a distance of one centimeter. These are the absolute units. In the gravitation system where the pound, or the kilogram, is taken as unit of force, the British unit of work is the foot-pound, while in the metric system it is customary to use the kilogram- meter as unit. 406. The numerical relations between these units are obtained as follows. Let x be the number of ergs in the foot-poundal, then (comp. Art. 259), gm. cm.^ lb. ft.^ \cm.) hence ^=^-( £r ) = 453-59 X 30.4797^=4.2139 X io«; i. e. I foot-poundal = 4.2139 x 10' ergs, and i erg = 2.3721 x lO'^ = 0.000 002 372 I foot-poundal. Again, let x be the number of kilogram-meters in i foot- pound, then xV'g. m. = I ft. lb., hence 4o8.] PRINCIPLE OF WORK. 26 1 lb. ft. ^=i|^-~ = 0.45359 X 0.3048 = 0.13825, i. e. I foot-pound = 0.13825 kilogram-meters. Finally, i foot-pound = g foot-poundals (Art. 262) ; hence i foot-pound = 1.356 X 10'' ergs, and i erg = 7.3737 X iO~^ foot- pounds, if ^= 981. 407. Exercises. (i) A joule being defined as 10' ergs, show that i foot-pound = 1.356 joules, and that i joule is about 3/4 foot-pound. (2) Show that a kilogram-meter is nearly 10* ergs. (3) What is the work done against gravity in raising 300 lbs. through a height of 25 ft.: (a) in foot-pounds, {b) in ergs? (4) Find the work done against friction in moving a car weighing 3 tons through a distance of 50 yards on a level road, the coefficient of friction being 0.02. (5) A mass of 12 lbs. slides down a smooth plane inclined at an angle of 30° to the horizon, through a distance of 25 ft.; what is the work done by gravity ? 408. It follows from the definition of work that, if any number of forces F^, F^, , , . F^ act on a particle P, the sum of their works for any displacement PP' = ds is equal to the zvork of their result- ant R for the same displacement. For, the resultant R being the closing line of the polygon constructed by adding the forces F^, •^2' • • • ^n geometrically, the projection of R on any direction, such as PP , is equal to the sum of the projections of the forces F on the same line (Art. 285); that is, if a^, a^, ... a^ be the angles made by F-^, F^, , , , F^ with PP' , and a the angle between R and PP' , we have F^ cosa^ -f F^ CQsa^ + ■ • ■ -\- F^ cosa^ = R cosa ; multiplying this equation by ds, we obtain the above proposition F, cosa.ds -4- F. cosa.ds 4- ■ ■ ■ 4- F cosa ds = R cosads, which expresses the so-called principle of work for a single particle. 262 STATICS. [409. 409. When the particle is in equilibrium, so that the forces do not actually change the motion, we may derive from this proposition a convenient expression for the conditions of equi- librium by considering displacements that might be given to the particle. Such displacements are called virtual, and the corre- sponding work of any of the forces is called virtual work. It is customary to denote a virtual displacement by hs, the letter S being used to distinguish from the actual displacement ds ; this distinction becomes of importance in kinetics. 410. The resultant being zero in the case of equilibrium, the sum of the virtual works of all forces acting on the particle must be zero for any virtiial displacement, i. e. F^ cosa^s -f- F^ cosa^Ss + ■ ■■ + F^ cosa^Ss = o. (4) As the resultant must vanish if its three projections vanish for any three axes not lying in the same plane, the necessary and suffi- cient conditions of equilibrium of a single particle are that the sum of the virtual works of all forces must be zero for any three virtual displacements not all in the same plane. This is the principle of virtual work for a single particle. If the particle be referred to a rectangular system of co-ordi- nates, its displacement Zs can be resolved into three component displacements hx, Zy, hz, parallel to the axes. The forces acting on the particle being replaced by their components X, Y, Z, the sum of their virtual works, for the displacement hs is SX • &v 4- 2F- Sj/ -f- 2Z- S^. Hence the analytical expression for the principle of virtual work : ^X-hx+^Y-lyJr^Z-hz = o. (s) As the displacements hx, hy, Bz are independent of each other, and perfectly arbitraiy, this single equation is equivalent to the three equations SX=o, SF=o, SZ=o, which are the ordinary conditions of equilibrium of a single particle. 41 3-] PRINCIPLE OF WORK. 263 411. The principle of virtual work is particularly useful in elimi- nating the unknown reactions arising from constraints. Suppose the particle be constrained to a smooth surface or curve. After introducing the normal reaction of the surface or curve the particle can be regarded as free ; and the equation of virtual work can be used to express the conditions of equilibrium. This equation will, in general, contain the unknown reaction. But as this reaction has the direction of the normal, it will be eliminated if the virtual displacement be selected along a tangent. Hence, in the case of constraint to a surface, tlie two conditio7is of equilibrium independejit of the reaction are foimd by forming the equation of virtual work for virtual displacements along any two tangents to the surface ; and in the case of constraint to a curve the one such condition is found from a virtual displacement along the tangent. If it be required to find the normal pressure on the surface or curve, which is of course equal and opposite to the reaction, it can be found from a virtual displacement along the normal. 412. If the equation (4) which expresses the principle of virtual work be divided by the element of time it, during which the dis- placement hs would take place, the factor Ss/Bt^ v represents a virtual velocity, and the equation becomes F^ cosa^ -v ■\- F^ cosa^ v -\- ■ ■ ■ -\- F^ cosa^ ■ z' = o. On account of this form, the proposition is often called the prin- ciple of virtual velocities. The product of a force into the virtual velocity of its point of application in the direction of the force, F cosa • v, is sometimes called the virtual moment of the force. 413. The principle of virtual work can readily be extended to the case of a rigid body acted upon by any number of forces. The forces acting on a rigid body can always be reduced to a resultant R and a resulting couple H [Art. 380). This reduction is based on the supposition (Art. 283) that the point of applica- 264 STATICS, [414. Fig. 137. tion of a force can be displaced arbitrarily along the line of the force. It can be shown that such a displacement of the point of application Pof a force F (Fig. 137), from P \.o Q along the line of the force, does not affect the work done by the force in any infinitesimal dis- placement of the body. Let PP' — & be the displacement of P, QQ' =&' that of Q; let / and q be the projections of P' and Q! on the line of the force F ; then, since the body is rigid, P' Q' = PQ ; and consequently Qq will differ from Pp only by an infinitesimal of an order higher than the order of the displacement PP' = &. Hence, F-Pp = F-Qq. It may here be noted that, in general, the principle of virtual work must be understood to mean that the sum of the works of the forces differs from the work of their resultant by an infinitesimal of an order higher than that of the virtual displacement. It does not mean that the difference is absolutely zero. 414. Owing to the proposition proved in the preceding article, the sum of the works of all the forces acting on a rigid body is equal to the sum of the works of the resultant R and the result- ing couple H for any infinitesimal displacement of the body, and the work of the forces is not changed by such a displacement. It follows that the necessary and sufficient conditions of equi- librium of a rigid body (Art. 381), viz. 7? = 0, ^7=0, can be expressed by saying that the sum of the virtual works of all the forces must be zero for any infinitesimal displaccmeiit of the body. For when the forces are in equilibrium, this condition is evi- dently fulfilled. To prove that there must be equilibrium when- 415]- PRINCIPLE OF WORK. 265 ever this condition is fulfilled, it is only necessary to show that both R and H must vanish if the sum of their works is zero for any infinitesimal displacement. To see this, consider first a displacement of translation, Bs, parallel to R. The work of R will be RSs while the works of the two forces constituting H are equal and opposite, so that the work of H is zero. As the sum of the works of R and H must vanish by hypothesis, it follows that R = O. Next consider a displacement of rotation B6 about an axis parallel to the vector H. Taking this axis so as to intersect R and bisect the arm p of the couple H, the work of R will be zero while that of each of the forces F of the couple /fwill be ^pB0 ■ F; hence the whole work of H is FpBd = HBO. As the sum of the works of R and H must vanish by hypothesis, it follows that The two conditions R — o, 11= o are, therefore, both fulfilled. 415. The following examples may serve to illustrate the appli- cation of the principle of virtual work. To find the force just necessary to move a cylinder of radius r and weight W up a plane inclined at an angle a to the horizon by means of a crow-bar of length I set at an a^igle /S to the horizon (Fig. 138). Let s be the distance from the fulcrum A of the crow-bar to the point of contact B of the cylinder with the plane. If the crow-bar be turned about A through an angle o/S, the work of the force F acting at the end of the bar is F . /5/3. The corresponding displacement of the center C of the '^' circle, which is the point of application of the force W, is parallel to the inclined plane, and may be regarded as the differential bs of the distance AB = s. The work of W is, therefore, IVSs sina. This gives the equation of work 266 STATICS. [416. F. I 8§= W. is sine ; hence J, W . Ss / 8S The relation between j and /3 can be found by projecting ABCDA on the vertical line ; this gives r cosa + J sina = r cos/3 + s sin/3, whence cos/3 — cosa s = r - sina — sin/3 Differentiating the former equation, we find sina ds= — rsin,5 5/3 + j cos/3 5/3 + sin/3 8s, Ss J cos/3 — rsin/9 cos^/9 — cosa cos/3 — sina sin/3 + sin'/3 or ' 5/3 sina — sin/3 (sina — sin/3)' I 5j- I _ cos(a — /9) == I r 5/3 (sina — sin/3)' I + cos(a + /3) Hence, finally, -^ = r 7 sina _^ 416. A weightless rod of length AB = / rests at C on a horizontal cylinder whose axis is at right angles to the vertical plane through the rod ; its lower end A leans against a ver- B tical wall, and from its tipper end B a y weight W is suspended. Determine the "^ reactions at A and C, and the distance AC "y\/ = X for equilibrium, if the distance CD = a of the point of support from the vertical wall is given (Fig. 139). (a) Let A glide vertically upwards, C remaining in contact. At A as well as at C the forces are perpendicular to the displacements ; hence, putting EB =y, we have W 8y=o. Also, i_x 4i8.] PRINCIPLE OF WORK. 267 whence (/ — x)x' — /(jc' -^ a') = o, or x" = a'/. {i) Give the rod a vertical displacement to a parallel position : a X 3 [T — W8y+ C-Sy^o; .-. C=-lV=Jl. W. X '^ \ a (f) Give the rod a displacement in its own direction : A -Ss + Ccos k^Ss — W}Lf- — f-. Ss = o. ^_ v^-^\ ^^ _^ ^ •=x-)'- 417. /« a parallelogram formed by four rods with hinges at the ver- tices, elastic strings are stretched along the diagonals. Determine the ratio of the tensions in these strings. Let m, m' be the lengths of the diagonals, T, T' the tensions, and 5m, Sm' the changes of length of the diagonals when the parallelogram is sHghtly deformed ; then by the principle of virtual work Tdm + T'Sm' = o. (i) From geometry we have, if a, b are the sides of the parallelogram, w' + ;«" = 2 a' + 2IP, hence, differentiating, m dm + m' dm' = o. (2) From (i) and (2) we find T/T' = m/m'. (3) 418. As any body or system of bodies can be regarded as made up of particles and as the work of such a system is the algebraic sum of the works of the forces acting on each particle it follows from Art. 410 that it is a necessary condition of equilibrium of any 268 STATICS. [419- system that the sum of the virtual works of all the forces must be zero for any virtual displacement. Without entering into an elaborate discussion of the limitations of this very general principle it will be well to show its meaning and usefulness by means of some simple illustrations of its application. 419 The systems that we shall consider are mostly mechanisms and simple machines. The object of a mechanism in machinery is to change the direc- tion and magnitude of an available force /' so as to make it more useful in overcoming a resistance Q. The ratio QIP of the re- sistance Q to the given force P is called the mechanical advantage of the mechanism. It is the object of every machine to do work in a certain pre- scribed way, i. e. to exert force, or overcome a resistance, through a certain distance. The various forces of nature, such as the muscular force of man and other animals, the force of gravity, the pressure of the wind, electricity, the expansive force of steam or gas, etc., are called upon for this purpose. In most cases it would not do to apply these forces directly ; they must be con- trolled, guided, and transformed in various ways to become use- ful, and this is done by interposing the machine between the given driving force, sometimes called the power, and the force against which the final work is to be done, usually called the resistance, load, or weight. We shall in general denote the driving force by P, the resistance by Q. The term power is objectionable in this connection, being here used to denote a force, while in kinetics it is used for the rate of doing work (see Art. 496). Under the action of the driving force P its own point of applica- tion as well as that of the resistance Q is displaced. The cor- responding work of the force P may be called the available or total work , that done against the force Q is called the useful work. The ratio of the useful work to the total work is called the efficiency of the machine. 421.] PRINCIPLE OF WORK. 269 In all machines this efficiency is a proper fraction, owing to the fact that the work done by P must balance not only the useful work, but also the so-called wasteful work due to friction, stiffness of ropes, slipping of belts, lack of rigidity, etc. 420. In the straight lever AB, with fulcrum at C(Fig. 140), let CA =■ a, CB = b. If the lever be turned through a small angle -r- Qt 4^ ^ ^3^ ^^V W? Fie. 140. AC A' -= W, the work of the force P ^t Ai?. = P ■ alQ, that of the resistance Q at B is = — Q • b 80. According to the principle of virtual work (Art. 418) the sum of the works must be zero for equilibrium ; hence : PaBe-Qne=o,or^=^, P b the well-known law of the lever (Art. 297). The mechanical ad- vantage of the lever is therefore the ratio of its arms. 421. In writing down the sum of the works in this case it is not necessary to consider among the forces acting on the lever the unknown reaction of the fulcrum at C because, for the displace- ment selected, the work of this force is evidently zero (the point of application of the force remaining fixed). The main advan- tage of the principle of virtual work lies in this fact that if the displacement be selected so as to be compatible with the coitdi- tions and constraints of the system, the unknown reactions and internal forces of the system will, for the most part, do no work and need not be considered. The most important exception to 270 STATICS. [422. this rule is formed by the force of friction ; if not so small as to be neghgible, its work must be included in the equation of virtual work. Thus, if the displacement be selected so as to be compatible with the conditions, the equation of work will in general not contain the unknown reactions ; in this form it can therefore not serve to determine these reactions. But by selecting the displace- ment differently, the principle of virtual work can be used to find these unknown forces. Thus, in the case of the lever (Fig. 140), if instead of turning the rod AB about C, it be given a vertical translation Ss, say downward, the equation of virtual work will be PBs+ QSs- CSs= o, where C is the reaction of the support at C which is thus found to be C=P+Q. 422. Let one face BC of the wedge (Fig. 141) slide along a fixed plane ; the vertical force F applied at right angles to AB can Fig. 141. then serve to overcome a horizontal resistance Q acting on the face AC. If the wedge be given a displacement compatible with 423-] PRINCIPLE OF WORK. 271 the condition that BC remains in its plane, the equation of virtual work (neglecting friction) is Pip -QZq = o, where hp, hq are the displacements of the points of application of/* and Q in the directions of these forces. It is easily seen that hq= 2hp tana, where a is half the angle of the wedge. Substi- tuting and dropping the factor S/, we find for the mechanical advantage of the wedge -p = icota. The normal pressure on each of the faces AC, BC does no work since the displacement is at right angles to the force. By resolv- ing P at right angles to the faces, this pressure is found = ^Psina. If the friction on the faces is taken into account the equation of virtual work is PSp — 2QSp tana — 2 x i/J-P sina —^ = o, '' cosa ' whence ■p = i (cota - fi). 423. In the clamp or binding screw (Fig. 142), the driving force P is applied tangentially to the rim of the screw head A ; the resistance Q is along the axis of the screw. If the piece B be fixed, the screw in turning exerts a pressure on the piece C. If the screw be turned through an angle S0 and the radius of the head be a, the work of Pis P- a SO. If s be the pitch of the screw, that is, the displacement along the axis corresponding to one revolution of the head, we have for the displacement Sq of the piece C corresponding to the rotation B6 : s Sq sS6 2'n-a a 80' ' ' ^ 27r 272 STATICS. Hence, by the principle of virtual work, [424. whence P.aBd= Q' Q 2'ira 27r I 1 1 t 1 1 + -. A /] i! B m Fig. 142. This shows the advantage of a large head and a small pitch. 424 In the mechanism formed by the crank and connecting rod of a steam-engine (Fig. 143), known as slider-crank, the driving force P acts at P in the direction through the center of Fig. 143. the crank circle while the resistance Q is applied at Q tangentially to the crank circle. For the only displacement compatible with the conditions of the mechanism we must again have p.hp=Q.lq, where Bp is the distance through which P advances along PO and Bq is the infinitesimal circular arc a SB described by Q. To ex- 424.J PRINCIPLE OF WORK. 273 press S^ in terms of B6 and the given lengths OQ = a, PQ = /, we regard it as the diiTerential of the distance OP = / for which we have the relation P =a' -^ f —2a p zosQ ; differentiating this equation we find 0= 2pdp — 2a COS0 dp -\- 2a p sinO dd, whence Substituting this value in the equation of virtual work we find : Q Bp p sinO P hq p — a COS0 If through we draw the radius perpendicular to OP and intersecting PQ produced at R, and also drop from Q the perpen- dicular SQ on OP, we have p — a cos6 = SP, a sin^ = SQ ; hence P ~ 'a{p - a cos"^ ~ aSP ' or smc& SQ I SP = OR I p: P ~ a p ~ a This result follows at once by the principle of virtual velocities (Art. 412) from Art. 10 1, where it was proved that the velocities of P and Q are as OR is to a. The effect of friction at P can be taken into account by observ- ing that the normal pressure on the guides at Pis P tan<^ (see Art. 292, Ex. (15)), where hp, whence Q / ^P or with the above value of Spjhq : ..OR = (i -/i taiK^)—-. 42S-] KINETICS. 275 PART III: KINETICS. CHAPTER V. e:inetics op the particle. I. Impulses ; Impact of Homogeneous Spheres. 425. Momentum and Impulse. A particle of mass m, moving with the velocity v, is said to have the momentum mv (see Art. 252). As long as this momentum remains constant, the particle will move in a straight line with constant velocity v (Newton's first law of motion, Art. 268). Any change occurring in the momentum is ascribed to the action of a force F on the particle. If the motion is rectilinear, the rate at which the momentum, changes with the time t is taken as the measure of the force (Newton's second law of motion. Arts. 255, 268, 271): dt ^ ' Integrating from the time /q when the velocity is v^ to the time t when the velocity is v, we have i Fdt = mv — mVQ. (2) If, in particular, the momentum varies uniformly with the time, the force F'vs, constant, and equation (2) reduces to F{t — Iq) = mv — ihVq. (2') The left-hand member of (2') for a constant force, of (2) for a variable force, is called the impulse of the force F during the time t — t^ (Art. 256). Hence, in rectilinear motion, the impulse of the force dicring any tivie is equal to the change of momentum during that time. 2/6 KINETICS OF THE PARTICLE. [426. 426. It appears, from equations (2) and (2'), that a very large force may produce a finite change of momentum in a very short interval of time, but that it would require an infinite force to produce an instantaneous change of momentum of finite amount. The impact of one billiard ball on another, the blow of a hammer, the stroke of the ram of a pile driver, the shock imparted by a falling body, by a projectile, by a railway train in motion, by the explosion of the powder in a gun, are familiar instances of large forces acting for only a very short time and yet producing a very appreciable change of velocity. The time of action, t — t^, of such a force is the very brief period during which the colliding bodies are in contact. The force, F, is a pressure or an elastic stress exerted by either body on the other during this time. Forces of this kind are called impulsive, or instantatieous, forces. 427. In the case of such impulsive forces, it is generally diffi- cult or impossible by direct observation or experiment to deter- mine separately the very brief time of action, t — 4, as well as the magnitude F of the impulsive force. Moreover, what is of most practical importance and interest in such cases of impact is, generally, not the force itself, but the change of momentum produced, z. e., the impulse of the impulsive force. In the present section, which is devoted to the study of the simplest cases of impact, we shall therefore deal rather with impulses and momenta than with forces. 428. It should be observed that some authors use the name impulsive, or instantaneous, force for what has here been called the impulse of the impulsive force. They define an impulsive force as the limiting value of the integral I Fdt when F increases indefinitely, while at the same time the difference of the limits, t — 4, is indefinitely diminished ; in other words, as the impulse of an infinite force producing a finite change of momentum in an infinitesimal time. According to this definition, an impulsive or instantaneous force is a magnitude of a character different from that of an ordinary force, and 431 •] IMPACT. 277 is measured by a different unit. Its dimensions are MLT"', and not MLT~^- Its unit is the same as the unit of momentum. Indeed, it is not a force, but an impulse. 429. The momentum mv of a particle, being a mere multiple of its velocity, can be represented by a vector (more correctly a vector confined to a line, or rotor), i. ^., by a rectilinear segment drawn through the particle and representing by its length the magnitude of the momentum, by its direction and sense the direction and sense of the velocity (Art. 267). Hence the mo- menta of rigidly connected particles are compounded and resolved according to the rules that hold for forces in statics, the trans- ferability of a momentum along its line being assumed (comp. Art. 283). Thus, in particular, by Arts. 294, 297, for two rigidly con- nected particles m, m! whose velocities are equal and parallel (Fig. 144), the momenta mv, m'v have a resultant momentum {m + m')v which is parallel to the common velocity v of the particles and divides their distance in the inverse ratio of the masses m, m' . It follows that this resultant momentum passes through the centroid G of the masses w, wi' (see Art. 219). ' pig. 144. 430. This result is readily generalized. By reasoning as in Art. 306, it can be proved that for a rigid body or mass M, hav- ing a velocity v of translation (so that all its particles have equal and parallel velocities w), the resultant of the momenta of all its particles is equal to Mv, is parallel to v, and passes through the centroid G of the mass M of the body. This momentum Mv, applied at G, is called the momentum of the rigid body, or the mo7nentjLm of its centroid. 431. The term momentum of the centroid is also used in a more general sense. It has been shown in Art. 212 that, for 278 KINETICS OF THE PARTICLE. [432. any system of particles, whether rigidly connected or not, whether at rest or in motion, there exists at every instant a cer- tain point G, the centroid, whose co-ordinates with respect to a fixed set of rectangular axes are x- y. 2m 2w2 If the particles are in motion, this point G will, in general, change its position in the course of time. Differentiating the equations with respect to the time, and putting 2w« = M, we find : M^=^m^A ilf^=2;«^, M^^^vi^. (3) dt dt dt dt dt dt ' These equations determine the rectangular components dx/dt, dy/dt, dz/dt of the velocity v of the centroid G. The product Mv is called the momentum of the centroid of the system of particles. If each of the particles moves with constant velocity in a straight line, it follows from the equations (3) that the centroid has a constant momentum ; i. e., it will also move with constant velocity in a straight line. This proposition can be regarded as a generalization of Newton's first law of motion. In particular, if two particles in, m' move with the velocities V, v' in the same straight line, the velocity of their centroid is - mv + m'v' m + nt' (4) 432. Direct Impact. We proceed to consider the particular case of two homogeneous spheres of masses m, m' , whose centers C, C move with velocities n, 71' in the same straight line. The spheres are supposed not to rotate, but to have a motion of pure translation ; then their Fig. 145. momenta are mu, m'u' , and can be represented by two vectors drawn from the centers C, 434-] IMPACT. 279 C along the line CC (Fig. 145). To fix the ideas, we assume the velocities it, 11' to have the same sense and « > ?/, so that in will finally impinge upon m' . The case when the velocities are of opposite sense will not require special investigation, as only the sign of «' would have to be changed. It is otir object to determine the velocities v, v' of m, m' im- mediately after impact, when the velocities jt, u' immediately before impact are given. The results here derived for homogeneous spheres hold generally, whatever the shape of the impinging bodies, provided that they do not rotate, and that the common normal at the point of contact passes through both centroids. 433. If the spheres were perfectly rigid, the problem would be indeterminate, for there is no way of deciding how the velocities would' be affected by the collision. Natural bodies are not perfectly rigid. The effect of the impact will, in general, consist in a compression of the portions of the bodies brought into contact. Moreover, all natural bodies possess a certain degree of elasticity ; the compression will therefore be followed by an extension, each sphere tending to regain its shape at least partially. The compression acts as a retarding force on the impinging sphere m, and as an accelerating force on -in' . It will last until the velocities 11, u' have become equal, say = iv. During the subsequent period of extension, or restitution, the elastic stress still further diminishes the velocity of m, and increases that of m\ until they become, say, v, v' . 434. The stress varies, of course, during the whole time t of compression and restitution. But, according to Newton's third law, the pressure F exerted at any instant by m on m' must be equal and opposite to the pressure F' exerted by m' on m at the same instant. Since F= mdu/dt, F' — m'du' /dt, and F= — F' at any instant during the time t, we have 28o KINETICS OF THE PARTICLE. [435- 1 Fdt= — I F'dt, or ;« I du = — m' \ du', whence mv — mu = — {m'v' — m'u'), or mv + m'v' = mu + m' u'; (S) that is, ^/^^ total inoinentum after impact is equal to that before impact. 435. This proposition will evidently hold for any number of spheres whose centers move in the same line, and can then be expressed in the form "^mv = '^m.u. It can be regarded as a special case of the so-called principle of the conservation of the motion of the centroid to be proved hereafter for any system not acted upon by external forces. On the other hand, the proposition can be looked upon as a further generalization of Newton's first law of motion. While the latter asserts that the momentum of a particle remains unchanged as long as no external forces act upon it, our law of impact asserts the same thing for the momentum of a system. 436. If the spheres were perfectly non-elastic, there would be only compression and no subsequent extension. As at the end of the period of compression, the velocities ii, u' have both become equal, viz., = w (Art. 433), the spheres after impact would have the common velocity m.u + m-'n' w = m + 437. If the spheres were perfectly elastic, i. e., if the elastic stress following the compression, or the so-C3.\\eA force of restitu- tion, were just equal to the preceding stress of compression, the spheres would completely regain their original shape. In this case, the elastic stress causes the impinging sphere m to lose during the period of restitution an amount of momentum 440.] IMPACT. 281 m(w — ti) equal to that lost during the period of compression. Hence, the final velocity of m. after impact would be V = W — {tl — IV) = 2 W — 21. Similarly, we have for the other sphere ?«' v' = w -\- (w — u') = 2 w — u'. As w is known from Art. 436, the velocities after impact can be determined for perfectly elastic spheres by means of these formulae. 438. In general, physical bodies are imperfectly elastic, the force of restitution being less than that of the original compres- sion ; that is, we have w — V ^= e{ii — w), v' — w = e{w — u'), where ^ is a proper fraction whose limiting values are O for perfectly inelastic bodies and i for perfectly elastic bodies. This fraction e, whose value for different materials must be determined experimentally, is called the coefficient of restitittion (or, less properly, the coefficient of elasticity). 439. To eliminate iv we have only to add the last two equa- tions ; this gives v' — V = e{ii — ti') ; (6) that is, the ratio of the relative velocity after impact to the rela- tive velocity before impact is constant and equal to the coefficient of restitution. This proposition, in connection with the proposition of Art. 434, expressed by formula (5), is sufficient to solve all problems of so-called direct impact, i. e., when the centers of the spheres move in the same line. 440. As the coefficient e is frequently difficult to determine, the limiting cases ^ = o, ^ = i are important as giving approxi- mate solutions for certain classes of substances. 282 KINETICS OF THE PARTICLE, [44' • Thus, for nearly inelastic bodies (such as clay, lead, etc.) we may put e = o, whence, by (6), v' = v, i. e., the velocities of the spheres become equal after impact; and the value of the common velocity is found from (S) as mil + m' u' V = m + m which agrees with the result found in Art. 436. For perfectly elastic bodies m^, m^ will descend while m^ ascends. The force producing the acceleration of the system formed by the two par- ticles is evidently the difference of the weights of the particles, viz., Wi—lVi = {}ny—7?i.^g, while the whole mass moved (neglecting the mass of the cord and of the pulley) is m^ + m^. Hence, we have for the acceleration J of the system, J = , _ g- _, „ (4) nil -\- VI2 This acceleration is constant, and the relations between space, time, and velocity are found just as for a single particle falling freely, except that the acceleration of gravity g is replaced by the frac- tion (otj — m^/(ni-^ -\- OTj) of g. It follows that if the masses m^, OTj be selected nearly equal, the accelera- tion will be small, and the motion can be observed more conveniently than that of a freely falling body. The tension T of the cord is, of course, the same at every point of the cord if, as is here assumed, the weight of the cord, the inertia of the pulley, and the axle-friction of the pulley be neglected. To deter- mine this tension, we have only to consider either particle separately. If the cord be cut just above OTj, and the tension T be introduced, the particle wzj will move like a free particle under the action of the resultant force W-^— T= in^g — T. Hence, as the sense of the accelera- tion/ of nix agrees with that of ^, m^g - T W, W, Fig. 149. / = (s) Similarly, we have for the acceleration of m-i, whose sense is opposite to that of ^, ™ — 2 Eliminating/ between any two of the equations (4), (5), (6), le tension ^_ 2 OT1OT2 OTl + Wj we find (7) 465.J RECTILINEAR MOTION. 299 465. If the two particles of Art. 464 move on inclined planes inter- secting in the horizontal axis of the pulley (Fig. 150), it is only necessary to resolve the weights niig and m^g into two components, one parallel, the other perpendicular, to the inclined plane. If the planes be smooth. Fig. 150. the system formed by the two particles is made free by introducing the normal reactions of the planes which counterbalance the perpendicular components of the weights. The resultant force is therefore the differ- ence of the parallel components, and the acceleration is . _ nil sin 61 — m.2 sin Q^ (8) where 6^, Bi are the angles of inclination of the planes to the horizon. The tension 7" of the connecting cord is again determined by equating this value of/ to the one obtained by considering either of the two particles separately. Thus, m^ taken by itself, becomes free if we intro- duce not only the normal reaction of the plane, but also the tension of the string. This gives m-^g sin 61 — T ; = (9) Similarly, we have for m^ / = T — m^g sin $2 (10) Hence, T^-^^^gisme. + sme^)- (") With ^1 = ^2 = V^ the formulae (8) to (11) reduce, of course, to the formulae (4) to (7). 300 KINETICS OF THE PARTICLE. [466. 466. Exercises. (i) A stone weighing 200 lbs. is raised vertically by means of a chain running over a fixed pulley. Determine the tension of the chain : (a) when the motion is uniform ; (l>) when the motion is uniformly accelerated upwards at the rate of 8 ft./sec.^ ; (c) when the acceleration is 32 ft./sec.- downwards. Neglect the weight of the chain and the axle- friction of the pulley. (2) A railroad car weighing 4 tons is pushed by four men over a smooth horizontal track. If each man exerts a constant pressure of 100 lbs. : (3) what is the velocity acquired by the car at the end of 5 sec? {i) what is the distance passed over in these 5 sec? (3) (a) Determine the constant force required to give a train of 100 tons a velocity of 30 miles an hour in 6 min. after starting from rest. (F) How far does the train go in this time? (<:) If the same velocity is to be acquired at the end of the first mile, what must be the tractive force of the engine? (4) If in Atwood's machine (Fig. 149) the two masses are each 2 lbs., and an additional mass of i oz. be placed on one of these masses, how long will it take this mass to descend 6 ft.? (g= 32.2.) (5) A cord passing over a pulley, fixed 40 ft. above the ground, carries a mass of 50 lbs. at one end and a mass of 51 lbs. at the other. If initially both masses are 20 ft. above the ground, where will they be after 10 sec? What is the tension of the cord? When does the heavier mass reach the ground? (g= 32.) (6) Solve the problem of Art. 465 when the inclined planes are rough, the coefficients of friction being /ij, fx^. (7) A mass of 9 lbs. rests on a smooth horizontal table, and has a cord attached which runs over a smooth pulley on the edge of the table. If a mass of 2 lbs. be suspended from the cord, find the acceleration and the tension of the cord. (8) A sleigh weighing 450 lbs. is drawn over a horizontal road, the coefficient of friction being -^-j-. Find the pull exerted by the horses when the motion is uniform. (9) When the U.S.S. Raritan was launched she was observed to pass in II sec. over an incline of 3° 40', 54 ft. long. Find the coeffi- cient of friction. (10) A coaster, after coming down a hill, runs up another hill, of 466.] RECTILINEAR MOTION. 301 slope I : 10, and comes to rest in 12 sec. How far up did it go if the coefficient of friction is ^^P (i i) A train of 120 tons is running 35 miles an hour. Find what con- stant force is required to bring it to rest : (a) in 4 min. ; (1^) in half a mile. (12) If it takes i min. to coast down a hill on a uniformly sloping road of i mile length, and the coefficient of friction be 0.02, what is the height of the hill? (13) If a man is able to lift 150 lbs. on the ground, how much can he lift in an elevator (a) ascending, (i) descending, with an acceleration of 8 ft./sec.2? (14) A train of 180 tons reaches its regular velocity of 45 miles per hour 3 min. after starting. What is the accelerating force if regarded as constant ? (15) The barrel of a rifle is 30 in. long; the ball, weighing i oz., leaves the barrel with a velocity of 1200 ft./sec. What is the average pressure of the powder gases ? (16) In a coal-pit shaft, a cage (a) ascends, (i) descends, with an acceleration of 10 ft./sec.^ Find the tension of the rope when the cage is empty and when it carries two men, of 165 lbs. each. What is the pressure exerted by each man on the bottom of the cage ? (17) A train of 60 tons runs one mile with constant speed; if the resistances be 8 lbs. per ton, find the work done by the engine : (a) on a level track ; (i) on an average grade of i % . (<:) On a i % grade, what is the ratio of the work done against gravity to that done against the resistances? (d) If, while the train is running 30 miles per hour, the steam is shut off, how far on the level will the train go before its speed is reduced to 10 miles per hour? (18) A train of 100 tons (excluding the engine) starting from rest acquires a velocity of 30 miles an hour on a level road at the end of the first mile. Determine the average tractive force of the engine, that is, the pull on the drawbar between engine and train : (a) if the frictional resistances be neglected ; (i) if these resistances be estimated at 8 lbs. per ton. (c) What tractive force is required to haul the same train at a constant speed over a level road ? (d) up a grade of i in 160 ? (19) Determine the work expended in raising from the ground the materials for a brick wall 30 ft. high, 40 ft. long, and 2 ft. thick, the weight of a cubic foot of brickwork being 1 1 2 lbs. 302 KINETICS OF THE PARTICLE. [467. (20) A chain 350 ft. long and weighing 10 lbs. per foot is hanging down a shaft, with a weight of 520 lbs. attached at the end ; what work is done by a capstan in winding it up ? (21) An elevator weighing 800 lbs. is raised by a hoisting engine; if the steam is shut off when the upward speed is 6 ft./sec, how much higher will the elevator rise ? (22) To what height could a train be lifted vertically by the work spent in giving it a speed of 30 miles per hour? (23) The interior diameter of a chimney is 10 ft. ; the exterior diame- ter is 28 ft. at the base, 14 ft. at the top ; the height is 300 ft. Find the work done in raising the building material from the ground to its place. 467. Variable Force. The two fundamental propositions of Arts. 458, 459 are true for rectilinear motion, even when the resultant force R is variable. For in this more general case we find by integrating the equation of motion (i) Art. 457, in each of its two forms : ^t mv — viVq= I Rdt, (12) v,^=CRds; (13) -^ niv — A mz and according to the definitions of the impulse (Art. 256) and of the work (Art. 266) of a variable force, these equations mean that the increase of tnoinentinn in any time is equal to the impulse of the force during this time, and that tlie increase in kinetic energy in any distance is equal to the work done by the resultant force in this distance (comp. Arts. 425, 442). If R is constant, the equations (12), (13) evidently reduce to (2), (3) respectively. It should also be observed that the equations (12) and (13) are equivalent, respectively, to _ d(inv) R = ; — dt and R = di^mv^). ds 469.] RECTILINEAR MOTION. 303 that is, they merely express the two fundamental definitions of force as the tune-rate of change of viomentitni and the space- rate of chajige of kinetic energy (comp. Art. 442). 468. The equation (13) expresses the principle of kinetic energy and work for the case of the rectiUnear motion of a particle (comp. Art. 459). The integral in the right-hand member, I Rds, which represents the work of the force R in the distance s — s^, can be evaluated analytically only when R is a given function of the distance s in the interval s — j-q. Graphical or mechanical methods are therefore often used to find its value approximately. We can always put I Rds = R -{s — j-q), where R is the space-average of the force R for the distance s— s^; in other words, the average, or mean, force R is that constant force which acting through the distance s — s^ does the same work as the variable force R. Similar considerations apply to the relation (12) and lead to the time-average of the force R. 469. Pressure of a Piston. The work of a variable force is well illustrated by the expansion of gas or steam in a cylinder with a movable piston (Fig. 151). Let f be the radius of the cylinder, / the pressure (in pounds) at any instant of the gas per square inch of surface ; then the total pressure of the gas on the inside of the piston is F= nt^p pounds, and if P^ be the pressure on the "q " outside (say the atmospheric pressure), the resultant force acting on the piston is R =z P— P^, friction being neglected. The force R is variable, since the pressure / varies with the volume v occu- pied by the gas. This volume being in the present case proportional to the distance s of the piston from the 2r 304 KINETICS OF THE PARTICLE. [470. fixed base of the cylinder, the force J? is a. function of j. The variation of £ can therefore be represented graphically by a curve having s for abscissa and ^ for ordinate (Fig. 151); and f/i^ area of this curve, I. e., the area contained between the curve, the axis of s, and two ordinates whose abscissas are s^ and s, being given by the integral JRds, represents the work done on the piston when pushed through the distance s — s^^. 470. In the case of a perfect gas, Boyle's law gives the relation pv = k, where k is constant if the temperature remains constant. Hence, s where K and /"„ are constants. This equation represents an equilateral hyperbola, whose asymptotes are the axis of R and a line parallel to the axis of s. For steam, the law connecting pressure and volume is more complicated, but the curve may be taken as very near hyperbolic. 471. The Steam-engine Indicator is an apparatus for measuring the pressure of the steam in the cylinder and at the same time recording it automatically on a drum revolving as the piston moves. Thus, if the indicator be put in connection with the interior of the cylinder, the curve traced by the indicator has for its abscissas the distances .r of the piston from one end, and for its ordinates the corresponding pressures P of the steam on the inside of the piston. At the beginning of the stroke, steam is admitted and acts with nearly constant pressure on the piston j the line AB (Fig. 152) traced by the indicator will therefore be nearly parallel to the axis of s. As soon as the steam is shut off by the slide- valve, the steam, being now confined within the cylinder, begins to expand nearly accord- ing to the law pv = const., or I's = const. ; the curve traced by the indicator is there- fore approximately an equilateral hyperbola BC, having the axes as asymptotes. When the slide-valve connects the cylinder with the con- denser, a partial vacuum is estabhshed behind the piston, and the pressure curve is approximately a line CD, parallel to the axis of P. P A B ^ "^N C ^^^ J s D Fig. 152. 472.] RECTILINEAR MOTION. 305 The area ABCDO evidently represents approximately the work of the pressure on the inside of the piston in a double (forward and backward) stroke. In reality, various circumstances produce deviations from the regular shape ABCDO, and the actual trace, obtained by means of an indicator for a double (forward and backward) stroke, usually called the indicator diagram, forms a loop somewhat like that indicated by the dotted curve in Fig. 152. The area of this loop, which represents the work in question, can readily be found approximately by dividing it up into narrow rectangular strips, or with the aid of a planimeter. The resultant or effective piston pressure is of course the difference between the pressures on the two sides of the piston. A diagram should therefore be obtained for each side of the piston ; from these two diagrams the curve of effective piston pressure is then derived by constructing the curve whose ordinates are the differences of the corre- sponding pressures on the two sides. By dividing the area contained between this curve and the axes by the length of the stroke, the average, or mean, piston pressure is finally found (see Art. 468). For details the student is referred to special works on the steam engine. 472. Attractive and Repulsive Forces. According to Newton's law of universal gravitation any particle of matter, of mass m, is attracted by any other such particle, of mass m\ with a force R, directed along the line joining the particles and in magnitude directly proportional to the masses m, m! and inversely proportional to the sqimre of the distance s of the particles. If we regard the attracting mass m! as fixed, say at the point O, and take this point as origin of the distance s = OP of m from m' (Fig. 153), the force R with which m' attracts m is directed toward O, i. e., in the negative sense of s\ we have therefore Fig. 153. R=-K—^, (14) where « is a positive constant, called the constant of gravitation. R"! ?o 771' 306 KINETICS OF THE PARTICLE. [473. 473. The constant k evidently represents the force with which two particles, each of mass i, attract each other when at the distance i. It is a physical constant to be determined by experiment, and its numeri- cal value depends on the units of measurement adopted. What can be directly observed is of course not the force itself, but the acceleration it produces. Dividing the force R, as given by formula (14), by the mass m of the particle on which it acts, we find for the accekration J produced by the attraction of the mass m' in the mass m at the distance r from m' : m' r- It is shown in the theory of attraction that the attraction of a homo- geneous sphere at an external point is the same as if the mass of the sphere were concentrated at its center. Thus, if ;«' be the mass of the earth (here assumed as a homogeneous sphere), the acceleration j it produces in any mass m situated at a point P above its surface, at the distance 0P= r from the center O, is = km' /r^. Now for points /'near the earth's surface this acceleration/ is known from experiments ; it is the acceleration of gravity, usually denoted by g. As the radius of the earth, ?-=6.37X 10* centimeters, and its mean density p = 5^, are also known, the value of the constant k can be found from the formula 1 whence « = !■ * TTpr With ^= 980 we find in C. G. S. units I -ij = 0.000 000 067. 1.50 X 10' This, then, is approximately the force in dynes with which two masses of I gram each would attract each other if concentrated at two points i centimeter apart. 474. The force between two electric charges as well as that between two magnetic poles follows Newton's law (Art. 472); that is, the force is directly proportional to the charges, or pole-strengths, and inversely proportional to the distance. But the constant k has a very different value. It is customary in electricity and magnetism to select the units 476-] RECTILINEAR MOTION. 307 for electric charges and magnetic pole-strengths so that the constant K= I. In astronomy and in the general theory of attraction the unit of mass is also often selected so as to make k = i . 475. Let us now return to the problem (Art. 472) of the motion of a particle ni when attracted according to Newton's law by a fixed mass ;«' (Fig. 153). This motion will be recti- linear if the initial velocity Vq of m is directed along the line OP, and of course also when Wg = o. The equation of motion is in this case Upon division by m it reduces to the equation discussed in Kinematics, Arts. 77-80. It is there shown how the integration can be carried through ; the only thing to be added here is the interpretation of the equation of kinetic energy and work (13), which, with the above value of R, reduces to 1 21 2 - C'ds (tn' ni'\ , „n \niv' — \mv^= — Kimn I —^ = ian\ ■ (15) 476. The quantity ni'/s, or, in absolute measure, icm'/s, is called the potential at P, due to the attracting mass vi' ; so that Kin' / s^ is the value of the same potential at P^. The right- hand member of equation (15), which represents the work done upon the particle in by the attraction of the mass ;«' in the dis- tance /'Pq = s — s^, is therefore ni times the difference of the potentials at P and Pg, due to m'. In other words, the differ- ence of potential between two positions, P and Pq, is the work that would have to be done by the attraction of m' to briitg a unit mass from Pg to P. It also follows from equations (14) and (15) that the force of attraction exerted by a mass m' {at O) on a unit mass at any point, P, is the space-derivative of the potential at P : m ds\ s J 308 KINETICS OF THE PARTICLE. [477- 477. The negative of the potential multiplied by the mass m, i. £., the quantity — Kmm! /s, is called the potential energy of the moving particle m. Denoting this by V, and the kinetic energy by T, the last equation becomes or r+ F= 7;+F;,= const. ; (i6) i. c, the sum of the kinetic and potential energies remains constant during the motion. This is the principle of the conservation of energy for this particular problem. 478. The physical idea to which the term potential energy is due may perhaps require some explanation. The region surrounding an attracting mass m' is called the field of the force of attraction R of m'. Wherever in this field a particle m be placed (say, with zero velocity), it will become subject to the attraction R of m' and move toward ;«' with increasing velocity, thus acquiring kinetic energy ; at the same time the force R does an amount of work on m which is exactly equivalent to the kinetic energy gained by m. It follows that, the farther away from m' the particle m is placed, initially, the greater will be the amount of work that m! can do upon it. It is this " potentiality" for doing work, due to the distance of m from m\ which is denoted as energy of positio7i, or potential energy. The equation (i6), or the equation (15) which differs from (16) merely in notation, shows that what the particle m in moving toward m' gains in kinetic energy it loses in potential energy so that the sum of kinetic and potential energy always remains constant. 479. By properly generalizing the idea of potential and potential energy the principle of the conservation of energy (Art. 477) can be extended to the more general case of any force R that is a function of the distance i' alone. For, if R = F{s^, the principle of kinetic energy and work (Arts. 467, 468) gives for rectilinear motion \ mv^ — \ mv^ = SlF{s)ds; (17) 48o.] RECTILINEAR MOTION. 309 hence, putting I F(s)ds = mf(s) and defining/(i-) as \ki^ poten- tial, and — mf{s) as the potential energy, due to the force R = F(s), we have \ mv^ - \ mv^ = nif{s) - mf{s^, or, with the notation of Art. 477, r + F= To +Fo = const. (18) It appears from these definitions that (just as in the particular case of Art. 476) the force exerted on unit mass at any point is the space-derivative of the potential at that point : in ds-^ ' 480. Free Oscillations. Among the forces of the form R = F{s), next to the Newtonian forces (Art. 472) which are inversely pro- portional to the square of the distance s, the most important on account of their applications are forces directly proportional to the distance s from a iixed point O. With the origin at O, the equation of rectilinear motion under such a force is m ■ ' dt"^ = — mK^s, (19) if the force be attractive, i. e., directed toward O. The less im- portant case of repulsion for which the minus sign would have to be replaced by plus, will not be considered here. It has been shown in Kinematics (Arts. 81-84, 120-128, see especially Art. 125) that the rectilinear motion defined by this equation (19) is a simple harmonic oscillation or vibration, about the point O as center. This point O, at which the force R = — mK^s is zero, is therefore a position of equilibrium for the particle. The potential energy V due to the force R = — mk^s is, by Art. 479. ^ ^ V=— \ Rds = niK^ ( sds = \ mk^s'^ + C. 3IO KINETICS OF THE PARTICLE. [481- Hence the principle of the conservation of energy gives iP- + /t^j^ = const. If the initial velocity be zero for s = s^, we have 481. As in the applications the moving particle m is generally subject to the constant force of gravity, it is important to notice that the intro- duction of a constant force ^ along the line of motion does not essentially change the character of the motion. For, the equation of motion d^s ■! , 7-. ••[ F m — = — niKS + 7' = — niK' s dr \ n reduces, with s — Fjmi?- = x, d'^x 9 to m — - = — niKX, which is of the same form as (17). The only change in the results is that the center of the oscillations, i. e., the position of equilibrium of the particle m, is not the point O, but a point at the distance c^F/iuk- from O. 482. Forces proportional to a distance, or length, are directly observed in the stretching of so-called elastic materials. Thus, a homogeneous straight steel wire when suspended vertically from one end and weighted at the other end is found to stretch ; and careful measurements have shown that the extension, or change of length, is directly proportional to the weight applied (the weight of the wire itself being assumed, for the sake of simplicity, as very small in comparison with the load applied) . Conversely, the tension, or elastic stress, of the wire is proportional to the extension produced. Moreover, when the weight is removed the wire is found to contract to its original length. This physical law, known as Hooke's law of elastic stress, holds only within certain limits. If the weight exceeds a certain limiting value, the extension is no longer proportional to the weight, and after removing the weight, the wire does not regain its original length, but is found to 484-] RECTILINEAR MOTION. 31I have acquired a permanent set, or lengthening ; it is said in this case that the elastic limits have been exceeded. Materials for which Hooke's law holds exactly within certain limits of tension and extension are called perfectly elastic. Strictly speaking, such materials probably do not exist ; but many materials follow Hooke's law very closely within proper limits. Thus, elastic strings, such as rubber bands, and spiral steel springs show these phenomena very clearly on account of the large extensions allowable within the elastic limits. 483. The elastic constant mi^. Let an elastic string whose natural length is / assume the length l-\- x when the tension is F, so that, accord- ing to Hooke's law, /^= — mi^x. To determine the factor of proportionality mi^ for a given string, we may observe the length /^ assumed by the string under a known tension, e.g., the tension — m-^g produced by suspending a given mass m^ from the string (the weight of the string itself being neglected). AA'e then have — mig = — mk' (Ji — I), whence mii .2 _ m^K m. J^-Oi: J and F=:-f^x. 484. Let the same string be placed on a smooth horizontal table, one end being fixed at a point O (Fig. 154), while a particle of mass m is attached to the other end. I Stretch the string to a | length OP,>=l+Xo (within -^^^^^^^^^^^^ii^^SB^-T: the limits of elasticity) and let go ; the particle m will move under the action of the tension F alone, its weight being balanced by the reaction of the table. The equation of motion is m — ^, = !^ X, dt'' k-l Fig. 154. 312 KINETICS OF THE PARTICLE. [485- the distance 0P= x being counted from the point Q at the distance 0(2 = /from the fixed point O. Putting again (Art. 483) and integrating, we find X = (Ti cos Kt-\- C2 sin Kt, whence = — == — kc^ sin k/+ k^2 cos Kt. (it As X = Xn and w = o for /= o, we have c^ = ^o; '^2 = 0; hence x=:X(, cos k/, » = — K.To sin k/. It should be noticed that these equations hold only as long as the string is actually stretched, i. e., as long as jc > o. The subsequent motion is, however, easily determined from the velocity for x = o. 485. It was assumed, in the preceding article, that the particle m is let go from its initial position P^ with zero velocity. This can be brought about by pulling the particle from Q to P^ with a gradually increasing force which at any point P is just equal and opposite to the corresponding elastic tension, or stress, 7^= — mi^x. The work thus done against the tension, i. e., in stretching or straining the string, is stored in the particle 711 as potential energy, or strain energy, V. To find its amount, observe that, as the particle m is pulled through the short distance A.v, the work of the force is = Wk'.vA.v ; this being the potential energy A V gained in the distance Ax, we have A V= tiD^xt^x ; hence 'Xf, . Thus, in the initial position P^ the particle m possesses this potential energy, but no kinetic energy. During its motion from Po to Q, the particle gains kinetic energy and loses potential energy. At any inter- mediate point P, for which QP=x, the kinetic energy is T=\mv', while the potential energy is V= 1 vik^xK By the principle of the con- servation of energy (Art. 479), the sum of these two quantities, the so-called total energy, E, remains constant as long as no other forces besides the elastic stress act on the particle : \ mv^ + \ ;«kV = const. 486.] RECTILINEAR MOTION. 313 The value of the constant is = ^ nii^x^, since this is the total energy at /(, ; hence, (Comp. Art. 480.) This relation also follows from the values of x and V given in Art. 484, upon eliminating /. When the particle arrives at the position of equilibrium Q, the po- tential, or strain, energy has been consumed, having been converted completely into kinetic energy. 486. Exercises. (i) In a steam engine, let /=35 lbs. per square inch be the mean piston pressure during one stroke, -f = 15 in. the length of the stroke, and d= 1.5 ft. the diameter of the cylinder, {a) What is the work in one stroke ? (b) To what height could a mass of 500 lbs. be raised by this work ? (2) The work done by an ideal gas in expanding from a volume i\ to J" 02 pdv, where pv = const, if the change goes on at "1 constant temperature. Find the final pressure and the work done when 12 cu. ft. of air expand at constant temperature to 60 cu. ft., the initial pressure being 30 lbs. per square inch. (3) Show that, in F. P. S. units, the constant of gravitation is about 1/9-3 X lol (4) Knowing that on the surface of the earth the attraction per unit of mass is^= 32, find what it would be on the sun if the density of the sun be \ of that of the earth, and its diameter 108 times that of the earth. (5) Describe in words the motion of the particle in the problem of Art. 484 ; determine the time of one complete (back and forth) oscilla- tion, and the work done by the tension in a quarter oscillation. (6) In the problem of Art. 484, let the string be a rubber band whose natural length of i ft. is increased 3 in. when a weight of 4 oz. is suspended from it; determine the motion of a i-oz. particle attached to one end, the band being initially stretched to a length of i^ ft. ; find (a) the greatest tension of the band, {b") the greatest velocity of the particle, {c^ the period, {d) the work done by the tension in a quarter oscillation. 314 KINETICS OF THE PARTICLE. [486. (7) Discuss the effect of friction, of coefficient /n, in the problem of Art. 484. (8) The length (9(2 = / of an elastic string is increased to C(2i = 4 = / + ,? if a mass m is suspended from its lower end, the upper end O being fixed (Fig. 155). The mass m is pulled down to the distance Q^P^ = Xf, from the position of equilibrium (2i and then released. Prove the following results : With Q^ as origin the equation of motion of m is d'^x Ig —rn = — K^x, where k =\/- , ar ^ e whence x = Xq cos k/, z' = — io:„ sin Kt. If Xa e, the tension vanishes for ;<; = — e, i. e., IT' at Q; the velocity at this point is Vi = — k^x^ — e', and J ii„ the particle rises to the height h = {Xff — e-) /2 e above Q. ^'> The total time of one up and down motion is Fig. 155 2V^/g-[i TT + cos-'(^/xo) + V(.r„/^)^- i]. (9) How is the motion of Ex. (8) modified if the elastic string be replaced by a spiral spring suspended vertically from one end ? Assume the resistance of the spring to compression equal to its resistance to extension. (10) The particle in Ex. (8) is let fall from a height h above Q; determine the greatest extension of the string. (11) An elastic string whose natural length is / is suspended from a fixed point. A mass m-^ attached to its lower end stretches it to a length /, ; another mass m^ stretches it to a length /.. If both these masses be attached and then the mass m.i be cut off, what will be the motion of m{i (12) If a straight smooth hole be bored through the earth, connecting any two points A, B on the surface, in what time would a particle slide from A to B 7 The attraction in the interior is directly proportional to the distance from the center of the earth. (13) A straight rod of length / ft. and cross-section A sq. ft. is loaded at one end so as to swim upright when k ft. of its length are immersed in water. The weight mg of the rod is then balanced by its buoyancy, i. e., a force equal and opposite to the weight, 62.5 Ah lbs., of the water 48;.] RECTILINEAR MOTION. 315 displaced by the immersed part. If hi,{> h) ft. were immersed, the result- ant force acting at the centroid vertically upward, would be = 62.5 Ah^ — mg. Show that the centroid will perform simple harmonic oscillations provided the rod remains always partly immersed. (14) If the rod in Ex. (13) be dropped upright into the water, its lower end being initially h^ ft. above the water, find, by the principle of the conservation of energy, the depth x-^ of immersion, provided the upper end does not pass below the surface. Determine the relation between / and h necessary for the latter condition. Take /=3 ft., h = i ft., hi = I ft. 487. Resistance of a Medium. It is known from observation that the velocity v of a rigid body moving in a liquid or gas is continually diminished, the medium apparently exerting on the body a retarding force which is called the resistance of the medium. This force F is found to be roughly proportional to the density p of the medium, the greatest cross-section A of the body (at right angles to the velocity v), and generally, at least for large velocities, to the square of the velocity v : F= kpAv\ where /^ is a coefificient depending on the shape and physical condition of the surface of the body. This expression for the resistance F can be made plausible by the following consideration. As the body moves through the medium, say with constant velocity v, it imparts this velocity to the particles of the medium it meets. The portion of the medium so affected in the unit of time can be regarded as a cylinder of cross-section A and length v, and hence of mass pAv. To increase the velocity of this mass from o to w in the unit of time requires, by equation (2') of Art. 458, a force P^^UL^pAv-^. I The retarding force of the medium must be equal and opposite to this force multiplied by a coefificient k to take into account various disturbing influences. 3l6 KINETICS OF THE PARTICLE. [488. For small velocities, however, the resistance can be assumed proportional to the velocity, F=kv, the coefficient k to be determined by experiment. 488. The above consideration is only a very rough approximation. Thus the particles of the medium are not simply given the velocity v in the direction of motion ; they are partly pushed aside and move in curves backwards, causing often whirls or eddies alongside and behind the body. If the medium is a gas, it is compressed in front, and rare- fied behind the body ; indeed, when the velocity is great (greater than that of sound in the gas), a vacuum will be formed behind the body. Moreover, a layer of the medium adheres to and moves with the body, thus increasing the cross-section. It is therefore often found necessary to assume a more general expression for the resistance ; and this is, in ballistics, generally written in the form F = KpAiPfiv) . The careful experiments that have been made to determine the resistance offered by the air to the motion of projectiles have shown that for velocities up to about 250 meters per second, as well as for veloci- ties above 420 m./sec, f{v) can be regarded as constant, i. e., the re- sistance is proportional to the square of the velocity. But for veloci- ties between 250 and 420 m./sec, /. e., in the vicinity of the velocity of sound in air (330-340 m./sec), the law of resistance is more com- plicated. 489. Falling Body in Resisting Medium. Assuming the resist- ance proportional to the square of the velocity, the equation of motion for a body falling (without rotating) in a medium of constant density is d^s dv , ™ , > ;« — ^ =ni—- = mg — mkv^, (20) where /& is a positive constant. To simplify the resulting for- mulae, put 2 g then the separation of the variables v and t gives 49°-] RECTILINEAR MOTION. 317 whence ^ = J_ log ^1±-^ . (21) 2 it. g - fX.V the constant of integration being zero if the initial velocity is zero. Solving for v, we have S. . ^ ~ ^~ = -tanhMA (22) /A «•''' + ^-'" At Writing ^j/^^ for •?/ and integrating again, we find, since s = o for t = o, ^ = S ^°s ^ (^^ + ^"'") = S ^°g ^0^^ '*^- (23) The relation between ^y and s can be obtained by eliminating t between the expressions for v and s, or more conveniently by eliminating t from the original differential equation by means of the relation j j j > av _ dv as __ dv dt ds dt ds This gives ds = -f^-Arv g^ — iX^V^ whence, with v = o for s = o, ^log ^ 2 /U.^ ^ — /.i^z/^ 490. Exercises. (i) Show that, as t increases, the motion considered in Art. 489 approaches more and more a state of uniform motion without ever reaching it. (2) Determine the motion of a body projected vertically upward in the air with given initial velocity Wq, the resistance of the air being pro- portional to the square of the velocity. (3) In Ex. (2) find the whole time of ascent and the height reached by the particle. (4) Show that, owing to the resistance of the air, a body projected vertically upward returns to the starting point with a velocity less than the initial velocity of projection. ' = tvi^°^ ^ 1 ,.%x (24) 3l8 KINETICS OF THE PARTICLE. [491. (5) A ball, 6 in. in diameter, falls from a height of 300 ft.; find how much its final velocity is diminished by the resistance of the air, if k = 0.00090. (6) Determine the rectilinear motion of a body in a medium whose resistance is proportional to the velocity, when no other forces act on it. (7) A body falls from rest in a medium whose resistance is propor- tional to the velocity ; find v and j in terms of /, v in terms of s. 491. Damped Oscillations. Let a particle of mass in be at- tracted by a fixed center O, with a force proportional to the distance from 0, and move in a medium whose resistance is proportional to the velocity. If the initial velocity be directed through (or be zero), the motion will be rectilinear, and the equation of motion is m — - = — mK'^s — mkv, at'' or, putting k = 2X, dh , , ds — ; -f- 2 X — dt^ dt __. + 2X- + /.-^ = 0. (25) This is a homogeneous linear differential equation of the second order with constant coefficients, which can be integrated by a well-known process. The roots of the auxiliary equation, - X ± VX2 - «2_ are real or imaginary according as X > «:, or X < /c. The limit- ing cases X = /c, X = o, k = o, also deserve special mention. (a) If X > /c, the roots are real and different, and as X is positive, both roots are negative ; denoting them hy — a and — i>, so that a and d are positive constants, and i> a, the gen- eral solution is _„, , _,,, As the force has a finite value at the center O, we can take j- = o, V = Vq for / = o as initial conditions. This gives s = ^^ (e""' - e''"), V = -^^^ (be-"' - ae""). b — a b — a 491.] RECTILINEAR MOTION. 319 The velocity reduces to zero at the time h = 7 log - — a a As a and b are positive and b > a, s has always the sign of I'fl, t. e., the particle remains always on the same side of O ; it reaches its elongation at the time /'j, for which v vanishes, and then approaches the point O asymptotically. Hence, in this case, the damping effect of the medium is sufficiently great to prevent actual oscillations. Such motions are sometimes called aperiodic. (^) If X = «:, the roots are real and equal, viz., = — X, and the general solution is With s = o, 7' = v^ for / = O, we find s = v^fe~^\ V = ■z'o( I — X/)^~". The velocity vanishes for t-^=i/\, and then only. The nature of the motion is essentially the same as in the previous case. (c) If \ ; ds , , dE dE ds dE for we have — - = — — r = ^^ -—' dt ds dt ds 495-] RECTILINEAR MOTION. 321 494. Forced Oscillations. In the case of free simple harmonic oscillations, while the force regarded as a function of the dis- tance J is directly proportional to s, the same force regarded as a function of the time is of the form R = — mK?s^ cos Kt, since s = Sq cos Kt. Conversely, a particle acted upon by a single force R = mk cos fxt, ox R = ink sin fit, directed toward a fixed center O, will, if the initial velocity passes through O, have a simple harmonic motion. Suppose that such a force in the line of motion be super- imposed in the case of Art. 491 so that the equation of motion becomes w— T = — mic^s — 2 m\v + mk cos ^Lt, or --4-|-2\-i-|- A = ^sin/iA (26) dt^ dt ^ ^ ' The particle is then said to be subject to forced oscillations. For a particle suspended from a spiral spring this could be realized by subjecting the point of suspension to a vertical simple harmonic motion of amplitude k and period 2 tt/jx. The non-homogeneous linear differential equation (26) with constant coefficients can be integrated by well-known methods. 495. Exercises. (i) With ju,= 2, »o=4, sketch the curves representing j as a function of / in the five cases of Art. 491 ; take (a) X = 3, {b) A = 2, {c) \ = \, {e)\ = 2. (2) Compare the cases (c) and {d) of Art. 491 ; show that the os- cillations in a resisting medium are isochronous, but of greater period than in vacuo. The ratio of the amplitude at any time to the initial amplitude is called the damping ratio ; show that the logarithm of this ratio, the so-called logarithmic decrement, is proportional to the time. (3) Derive the equation of motion in the case of free oscillations from the principle of the conservation of energy. 322 KINETICS OF THE PARTICLE. [496. d-s (4) Integrate and discuss the equation — -\-\^s = a sin /x?' ; show that the amplitude of the forced oscillation becomes very large if the periods of the free and forced oscillations are nearly equal. Discuss the limiting case when ju, = k. (5) Integrate (26), assuming a particular integral of the form where p is the radius of curvature of the path. Hence, if the resultant force R which has the direction of / be resolved into a tangential force Ri = R cos i/f, and a normal force i?„ = R sin i/r, the above equation of motion will be replaced by the following two equations : Fig. 156. dv m- ■ R„ (0 501. The formulae (i) show how the force R affects the veloc- ity of the particle and the curvature of the path. The change of the magnitude of the velocity is due to the tangential force Ri alone. If this component be zero, i. e., if the resultant force R be constantly normal to the path, the velocity v will remain of constant magnitude. The curvature of the path, i/p, and hence the direction of v, depends on the normal component i?„. If this component be zero, the curvature is zero ; i. e., the path is rectilinear. 502. Instead of resolving the resultant force R along the tan- gent and normal, it is often more convenient to resolve it into three components, R cos 0. = X,R cos (3=Y,R cos 7 = Z, parallel to three fixed rectangular axes of co-ordinates Ox, Oy, Oz, to which the whole motion is then referred. If x, y, z be the co-ordinates of the particle m at the time t, the equations of motion assume the form (comp. Art. 113) d'^x T^ d'^y -r^ d^z -7 (2) Thus, the curvilinear motion is replaced by three rectilinear motions. 504.] FREE CURVILINEAR MOTION. 327 503. If the components A', V, Z were given as functions of the time / alone, each of the three equations (2) could be inte- grated separately. In general, however, these components will be functions of the co-ordinates, and perhaps also of the veloc- ity and of the time. No general rules can be given for integrat- ing the equations in this case. By combining the equations (2) in such a way as to produce exact derivatives in the resulting equation, it is sometimes possible to effect an integration. Two methods of this kind have been indicated for the case of two dimensions in a particular example in Kinematics, Arts. 163-165. We now proceed to study thtse. pi'inciples of integration from a more general point of view, and to point out the physical mean- ing of the expressions involved. 504. Tlie Principle of Kinetic Energy and Work. Let us com- bine the equations of motion (2) by multiplying them by dx/dt, dyjdt, dzjdt respectively, and then adding. The left-hand member of the resulting equation will be the derivative with respect to t of m a-m-m. r^ 1 tllni^ 2 inv'' We find, therefore, dj^^mv'^) _ j^dx Y^-\- Z'^ dt dt dt dt Let Vf^ be the velocity at any point P^ of the path, v the velocity at any other point P, and let arc P^P = s — s^, the distances s^, s being counted along the path. Then, integrating from /"„ to P, we find : \ nnl^ - \ mv^ = £{Xdx + Ydy + Zdz). (3) The left-hand member represents the increase in the kinetic energy of the particle ; the right-hand member represents the work done by the resultant force R, since its work is equal to the sum of the works of its components X, V, Z (comp. Arts. 404, 408). Equation (3) states, therefore, that tke amount by 328 KINETICS OF THE PARTICLE. [505- which the kinetic energy increases, as the particle passes from P^ to P, is equal to the work done by the resultant force R on the particle (comp. Arts. 467, 468). 505. The principle of kinetic energy and work can also be deduced from the former of the two equations (i). Multiplying this equation by w = ds/dt, we have d{}ymv^) T3 ds T3 , ds ^-^=/e.- = 7?cost^^; hence, integrating as in Art. 504 : i mv^ — A mv^ = I R cos -^ds. (4) 506. The evaluation of the work integral in the right-hand member of the equations (3) or (4) requires in general a knowl- edge of the path, since the integration is to be extended along this path. As in many problems the path is not known before- hand, but is one of the things to be determined, it is very im- portant to notice that the work integral f\Xdx + Ydy + Zdz) has a value independent of the path connecting the initial and final positions Pq. P^ wJicnever the expression Xdx -\- Ydy -\- Zdz is the exact differential of a one-valued function U of x, y, z; for in this case C{Xdx + Ydy + Zdz) = C dU= U- U^, where C^ is the value of U{x, y, z) at P^, U that at P- 507. Now the expression Xdx -\- Ydy -\- Zdz is certainly an exact differential whenever there exists a function U of the co-ordinates x, y, z alone {i. e., not involving the velocities or the time explicitly), such that dx &y bz ' ^^' for then Xdx + Ydy + Zdz = —dx \ ^dy + ^dz = dU. dx dy dz S08.] FREE CURVILINEAR MOTION. 329 The function U is called the force-function, and forces for which a force-function exists are called conservative forces. The conditions (5) for the existence of a force-function U can be put into a different analytical form which is frequently useful. Differentiating the second of the equations (5) with respect to z, the third with respect to y, we find dzdy dz dydz dy ' whence dY/dz=dZ/dy. If we proceed in a similar way with the other equations (5), it appears that they can be replaced by the following conditions : dY dZ dZ dX dX dV .,. — = — , — = — . — = (o) dz dy dx dz dy dx The equations (5) or the equivalent equations (6) are there- fore sufficient conditions for the existence of a force-function U. 508. The dynamical meaning of the existence of a force- function U lies mainly in the fact that, if a one-valued force- function exists, the work done by the forces as the particle passes from its initial to its final position depends only on these positions, and not on the intervening path. The equation (3) then gives \mv^^\mv,^ = U -U,; (7) the work done on the particle as it passes from P^, to P is = U- U,. It follows that the work of conservative forces is zero if the particle returns finally to its original position, that is, if it de- scribes a closed path, provided that the force-function U is one- valued, an assumption which will here always be made. In the case of central forces depending only on the distance, for which a force-function can always be shown to exist, the force-function, divided by the mass of the particle, is usually called the potential (see Arts. 476, 479). The negative of the force-function, say v =— U 330 KINETICS OF THE PARTICLE. [509. is called the potential energy. If this quantity be introduced, and the kinetic energy be denoted by T (as in Art. 477), the equation (7) assumes the form r+F=ro+Fo, (8) which expresses the principle of the conservation of energy for a particle : the total energy, i. e., tlie sum of the kinetic and potential energies, remains constant throughout the motion whenever there exists a force-function. In other words, whatever is gained in kinetic is lost in potential energy, and vice versa. 509. As the force-function f/ is a function of the co-ordinates X, y, z alone, an equation of the form U = c, where c is a constant, represents a surface which is the locus of all points of space at which the force-function has the same value c. By giving to c different values, a family of surfaces is obtained, and these surfaces are called level, or equipotential, surfaces. The values of the derivatives of U at any point P (x, y, z) are proportional to the direction-cosines of the normal to the equipotential surface U = c at P. But, by (s), they are also proportional to the direction-cosines of the resultant force R at this point. It follows that tJie resultant force R at any point P is always normal to the equipotential surface passing through P. If the equation of the equipotential surfaces be given, the resultant force R at any point {x, y, s) is readily found, both in magnitude and direction, from its components (5): 510. The force-function U determines, as has been shown, a family of equipotential surfaces f/= const. Starting from a point P on one of these surfaces, say f/=c (Fig. 157), let us draw through P the direction of the resultant force, which is 51 1-] FREE CURVILINEAR MOTION. 331 normal to the surface U=c. Let this direction intersect at P' a near surface, U=c'. At P' draw the normal to U=c', and let it intersect a near surface, U= c", at P". Proceeding in this way, we obtain a series of points P, P', P", P'", ■ ■ ., which in the limit will form a continuous curve whose direction at any point coin- cides with the direction of the re- sultant force at that point. Such a line is called a line of force. The lines of force evidently form the orthogonal system to the family of equipotential surfaces. The differential equations of the lines of force are therefore : dx _ dy ds dx dy dz 511. Exercises. (i) Show that a force-function exists when the resultant force is con- stant in magnitude and direction. (2) Find the force-function in the case of a free particle moving under the action of the constant force of gravity alone (projectile in vacuo) ; determine the equipotential surfaces and the potential energy. (3) Show the existence of a force-function when the direction of the resultant force is constantly perpendicular to a fixed plane, say the jy-plane, and its magnitude is a given function /(z) of the distance z from the plane. (4) Find the force-function, the equipotential surfaces, and the kinetic energy when the force is a function /(r) of the perpendicular distance r from a fixed line, and is directed towards this line at right angles to it. (s) Show the existence of a force-function for a central force, i. e., a force passing through a fixed point {x^jyo, ^o), if the force is a function of the distance r from this point. What are the level surfaces? (6) Show that a force-function exists when a particle moves under the action of any number of such central forces as in Ex. (5). 332 KINETICS OF THE PARTICLE. [512. 512. The Principle of Angular Momentum or of Areas. We con- fine ourselves to the case of plane motion, so that we have only two equations of motion : Combining these by multiplying the former by y, the latter by X, and subtracting the former from the latter, we find mx —^ — my—— = xV — yX. dfi dfi The right-hand member is evidently the moment (about the origin) of the resultant force R, whose components are X and Y. The left-hand member is an exact derivative, viz., the derivative with respect to the time of mxdy/di — mydx/dt, as is easily verified by differentiating this quantity. The equation can therefore be written —-[mx — — my — \ = xY—yX. (q) dt\ dt dtj ^^' Integrating from /"g to t, we find mx-^ — my — = i (xY—yX^dt. (\o\ dt dt Jh ' This equation expresses the principle of angular momentum or of areas for plane motion. 513. The name is due to the following interpretation of the left-hand member of equation (10). Divided by m, this left- hand member is, by Art. 97, twice the sectorial velocity. Equa- tion (9), after division by m, can therefore be expressed in words as follows : The time-rate of change of twice the sec- torial velocity about any point is equal to the moment of the acceleration about that point. This kinematical interpretation accounts for the name principle of areas. 5I5-] FREE CURVILINEAR MOTION. 333 Fig. 158. 514. The dynamical meaning of equation (lo) appears by con- sidering that mdx/dt, mdy/dt are the components of the momentum mv of the moving particle (Fig. 158). The prod- uct mvp of the momentum and its perpendicular distance from the origin is called the moment of momentum, or the angular momentum, of the particle about the origin. As the moment of mv is equal to the algebraic sum of the moments of its components, we have ay dx nivp = mx^^ — my — ^ dt ^ dt The angular momentum is evidently nothing but twice the sectorial velocity multiplied by the mass, just as linear momen- tum is linear velocity times mass. The dynamical meaning of equation (9) can therefore be expressed as follows : tlie time-rate of change of angular momen- tum about any fixed point is equal to the moment of the resultant force abotit the same point. 515. The most important case in which the integration in (10) can be performed is the case when xY-yX=o, which evidently means that the direction of the resultant force R passes through the origin. If this condition be fulfilled, equation (10) reduces to the form dy dx mx-=^ — my — = c, dt dt (11) where ^ is a constant of integration to be determined from the initial position and velocity. 334 KINETICS OF THE PARTICLE. [516. Kinematically, this equation means that the sectorial velocity remains constant. It can be put into the form dS _ c _ , dt 2 7)1 whence, by integration, we find S-S^^c\t-Q. (12) Hence, if the acceleration passes consta7itly throiigli a fixed point, the sector S — S^^ described aboiU this point in any time t — t^ is proportional to this time. This is the principle of the conservation of area for plane motion. Dynamically, equation (11) means that if the resultatit force passes constantly througli a fixed point, the angular momentum about this point remains constant. This proposition is called the principle of the conservation of angtdar momentum. If ^'q be the initial velocity, /q the perpendicular to v^ from the fixed point, equation (11) can also be written in the form ^/' = ^'oA- (13) 516. Exercise. The equation (9) can be written d{nnp)/dt= xY — yX. Show that the two terms of d{mvp)/di = ??ipdv/di-\- mvdp/dt xeiiresent the moments of the tangential and normal components of the resultant force R, respectively. 517. The Principle of d'Alembert. Let us consider a particle of mass m moving under the action of any forces F^, F^, ■ ■ ■ F„, whose resultant is R. The total acceleration j of the particle has the components d\r/dt^, d^y/dt'^, d'^z/dt^ parallel to the rectangular axes Ox, Oy, Oz. If the forces F-^, F^, ■ ■ ■ F„ be imagined removed, a force equal to mj would be required to give the particle the same acceleration j that it had under the action of the forces F^, F^, ■ ■ ■ F^. This fictitious force, mj, whose com- ponents are md'^x/df', md'^y/dt'^, md'^z/dt'^, is called the effective 519.] FREE CURVILINEAR MOTION. 335 force. For the sake of distinction, the forces F-^, F^, • • • Fn, which actually produce the motion, are called the inifj'essed forces. The ordinary equations of motion of a particle, d X d V ci "^ m^,=X, m^,= Y,m^, = Z, (14) where X, V, Z are the components of the resultant R of the impressed forces, express merely the equality between the effective force mf and the resultant impressed force R. It fol- lows that if the reversed effective force —inj, or its components, — md'^x/dt'^, — md'^y/dt'^, — md'^s/dt'^, be combined with the impressed forces Fp F^, ■ • ■ F„, we have a system in equilibrium. This is the fundamental idea of d'Alembert's principle. 518. The reversed effective force, — mj, is sometimes called the force of inertia of the particle. To understand the idea underlying this ex- pression, imagine the impressed forces to be removed, and then push the particle, say with the hand, so as to give it the same motion that it had under the action of the impressed forces. The pressure of the hand on the particle must at every instant be equal to the resultant R, or to the effective force mj, while the equal and opposite pressure of the particle on the hand represents the force of inertia. It must, how- ever, be clearly understood that this force of inertia, or inertia-resist- ance, is a force exerted on the hand and not on the particle. 519. Owing to the fact that, by combining with the impressed forces the reversed effective force, we obtain at any given in- stant a system in equilibrium, it becomes possible to apply to kinetical problems the statical conditions of equilibrium. Since in the case of a single particle the forces are all con- current, the conditions of equilibrium are obtained by equating to zero the sum of the components of the forces along each axis. This gives and these are the ordinary dynamical equations of motion (see (14), Art. 517). 336 KINETICS OF THE PARTICLE. [520. 520. The conditions of equilibrium of a system of forces can also be expressed by means of the principle of virtual work (Art. 410). Thus, let S.r, Bf, Bz be the components of any virtual displacement Bs of the particle ; then the principle of virtual work applied to our system of forces gives the single condition which is of course equivalent to the three equations (14) on account of the arbitrariness of the displacement Bs. The equation (15), which may also be written in the form expresses d'Alembert's principle for a single particle : /or any virtual displacement the stem of the virtu-al works of the im- pressed forces is equal to that of the effective force. 521, The advantage of using the equations of motion in the form given to them by d'Alembert arises mainly from the application of the principle of virtual work which thus becomes possible ; this will be seen more clearly later on, in the treat- ment of constrained motion. For the present it may suffice to notice that, if the actual displacement ds of the particle in its path be selected as the virtual displacement Bs, equation (16) becomes ^(^«'^+ ^^' + 5^^)= Xdx+ Ydy + Zdz. (17) This is the equation of kinetic energy and work (Art. 504) ; for the left-hand member is the exact differential d{\ mi^) of the kinetic energy, while the right-hand member represents the ele- ment of work of the impressed forces. In the particular, but very common, case of conservative im- pressed forces, the right-hand member is likewise an exact 523-] FREE CURVILINEAR MOTION. 337 differential, dU ; hence, in this case a first integration can at once be performed, and we find, as in Art. 508, \ mv^ - 1 mv} = ^{Xdx + Ydy + Zdz) =U-U^. (18) 522. There is an essential distinction between the principle of d'Alembert on the one hand, and the principles of kinetic energy and of areas on the other. D'Alembert's principle merely gives a con- venient form and interpretation to the dynamical equations of motion, through the application of the principle of virtual work ; but it does not show how to integrate these equations. The principle of kinetic energy and work and the principle of areas are really methods for integrating the equations of motion under certain conditions. The fact that these particular methods of combining the differential equations so frequently furnish the solution of physical prob- lems, is the best proof of the adequacy of the fundamental definitions and assumptions of mechanics. It has led the physicist to ascribe real existence to the quantities whose exact differentials are introduced by the combination, viz., to force and work, to kinetic and potential energy, and to regard the conservation of energy as a law of nature. While this view may often be useful, it must not be forgotten that the question of the objective reahty of these abstractions is beyond the ken of exact science. 2. CENTRAL FORCES. 523. We proceed to apply the general principles developed in the preceding articles to the motion of a particle under the action of a central force, i. e., a force whose direction always passes through a fixed point called the center of force (see Art. 1 5 7). We shall here consider only central forces whose mag- nitude is a function of the distance from the center alone. Thus, let O be the center of force, P the position of the moving par- ticle at any time t, m the mass of the particle, and OP = r its distance from the center ; then the general expression for such a central force F is ^^ p^^^ ^ ^^^^^^ where the function F{r) represents the law of force, and the function /(r) the law of the acceleration produced by this force in the particle m. 338 KINETICS OF THE PARTICLE. [524. 524. The most important special case is that of a force pro- portional to some power of the distance r, say F{r) = ix.r'\ where /u. and n are constants. The constant /li, which represents the value of the force at unit distance from the center, is often called the intensity of the force, or of the center. In the case of Newton's law of universal gravitation (Arts. 472, 473) we have « = — 2, yn = Kmm', where /c is a constant, viz., the acceleration produced by a unit of mass acting on a unit of mass at unit distance, while ;« is the mass of the attracted par- ticle, and m' that of the attracting center; that is, Newton's law is expressed by the formula r- mm' 525. From the physical point of view, attractions following Newton's law, and indeed, central forces generally, are usually regarded as due to the presence of mass (or an electric charge, etc.), not only in the mov- ing particle, but also at the center of force ; and the action between these two masses is then a mutual action, being of the nature of a stress, i. e., consisting of two equal and opposite forces. It follows that what we have called the center of force is not a fixed point. It will, however, be shown later (Arts. 567, 568) that a simple modi- fication allows us to apply to this case the results deduced on the assump- tion that the center is fixed. Again, the attracting or repeUing masses will here be regarded as concentrated at points. It is shown in the theory of attraction that a homogeneous sphere, according to Newton's law, attracts a particle outside of its mass as if the whole mass of the sphere were concentrated at the center of the sphere. The attraction of any mass on a particle can, of course, always be reduced to a single force ; but as the particle moves, the direction of this force will not in general pass through a fixed point ; such a force is, therefore, not central. 526. If a particle P of mass m be acted upon by a single central force c- ^//.a t = mjyr), 5270 FREE CaRVILINEAR MOTION. 339 its acceleration y = F/m=f{r) will pass through the center of force and be a function of r alone. The problem reduces, therefore, at once to the kinematical problem of central motion (Art. 157). Although the leading ideas of the solution of this problem have been indicated in Kinematics (Arts. 157-174), the importance of the subject of central forces demands a restate- ment in this place of some of the results in the language of kinetics, and a more complete exposition of some special cases. 527. A particle of mass ni acted upon by a single central force F = mf{r) will describe a curvilinear path whenever the initial velocity is different from zero and does not pass through the center of force. For the case of rectilinear motion under a central force see Arts. 472-486. All central motions, whatever may be the law of the force, have two properties in common : {a) the path of the particle, here often called the orbit, is 2. plane curve (Art. 158); {U) the sectorial velocity is constant (Arts. 1 59-161). Taking the plane of motion as x;/-plane and the center of force O as origin (Fig. 159), the direction cosines of the force F are =F xjr, T yjr, the upper sign corresponding to an at- tractive force, the lower to a repulsion. Hence, the dynamical equations of motion are Fig. 159. dt^ r at-' r If mf{r) be substituted for F, the factor m disappears, and the equations become purely kinematical. To avoid the use of the double sign, we shall give the equa- tions in the form corresponding to the more important case of attraction ; for a repulsive force it will only be necessary to change throughout the sign of F or /(r). Thus the funda- mental equations of motion are (comp. Art. 162): 340 KINETICS OF THE PARTICLE. [528. If polar co-ordinates r, 6 (Fig. 159), with the center of force as pole, be used, the equations of motion are, since the total acceleration is along the radius vector (see Art. 114): ^'- dt-^ \dt)~ ^^'^^' ^'-rdtK dt (3) 528. Two principal problems present themselves : {a) the problem of finding the orbit for a given law of force, and (b) the converse problem of determining the law of force, i. e., the function /(r), when the orbit is given. The solution of the former problem is effected by obtaining first integrals of the equations of motion from the principle of areas and from the principle of kinetic energy, and by combining these integrals so as to effect a second integration. Formula for the solution of the latter problem will be found incidentally. 529. As shown in Kinematics (Arts. i59-i6i)the second of the equations (3) gives the first integral ,de v — = r dt ' (4) where c is twice the sectorial velocity ; and with the notation indicated in Fig. 160 it readily follows that C=/t;=^„t;(,= .yrsin -«|r = 7'oroSini/r(,; (5) i. c, the velocity is in- versely proportional to Its perpendicular distance from the center, or, as it is sometimes expressed, the moment of the veloc- ity about the center of force is constant. The relation (4) can also be obtained from the equations (2) by applying the principle of areas (Art. 512, see Art. 163). Fig 160. 532.J FREE CURVILINEAR MOTION. 341 530. Another first integral of the equations of motion is obtained by combining the equations (i) according to the prin- ciple of kinetic energy and work (Art. 504, comp. Art. 164). This gives ^(-|- mv^) = — Fdr, or d{\ v^) = —f{r)dr, (6) whence v^ — v^ — z I f{r)dr; (7) i. e., the velocity at any distance r depends only on this distance (besides the initial radius vector and velocity) and is indepen- dent of the path described, being the same as if the particle had been projected with the initial velocity along the straight line joining the initial position to the center. 531. To perform the second integration we have only to substitute in (7) for v its value in terms of r and / or r and 6. Now the general expression for the velocity in any curvilinear motion is (Art. 96) ,2_f^^Y-U.2^^^V-^«'^V ''=Kjt)'^'\-dt)-K'dt) Lvij-^: From these expressions one of the variables 6 and t can be eliminated by substituting for dO/dt its value c/r"^ from (4) ; this del (8) It is often convenient to replace the radius vector r by its reciprocal u=i/r; we then have n^\dtJ sT-"^: (9) 532. The formulas (4) and (7), together with the expression (8) or (9), contain the complete solution of the two principal problems mentioned in Art. 528. Thus, if the law of force be given, the form of the function f{r) is known, and v can be found from (7) in function of r or ti; substituting this value of V in either (8) or (9), we have a differential equation of the first 342 KINETICS OF THE PARTICLE. [533. order between ;' and t, or between r and 0. The integration of the latter equation gives the integral equation of the orbit. On the other hand, if the equation of the path be given, the expressions (8) or (9) furnish the value of v'-, which, substituted in (6), determines the law of force f{i'). When the equation of the orbit is known, i. e., when ris known as a function of 6, or vice versa, the time t of the motion can be found from (4), which gives '='-J'rV6l. 533. If the second expression for ii"^ in (9) be introduced in (6), we find, as shown in Art. 167, /(o=^'-^(g+4 (10) This will generally be found the most convenient form for find- ing the law of force when the polar equation of the orbit is given. Again, when _/(;-) is given, the integration of this differ- ential equation of the second order is often more convenient for finding the equation of the orbit than the method indicated in Art. 532. It may be noted that the important relation (10) can be de- rived directly from the equations of motion (3), by eliminating t by means of (4) and introducing n for i/r. We have dr _ dr dO _ c dr _ du Jt' d~e~dt~ 7^de~ ~ '^de' d'^j' _ _ dhi dd _~ _ 2 2 ^'^" . ^ ~ '^ de^~dt~ ^ " ^ ' if these values be substituted in the first of the equations (3), the relation (10) will result. 534. When the equation of the orbit can be expressed con- veniently in terms of the radius vector r and the perpendicular / from the center to the velocity, as is, for instance, the case for the conic sections, it is of advantage to combine the aqua- S37-] FREE CURVILINEAR MOTION. 343 tion of kinetic energy (6) directly with the equation resulting from the principle of areas, /i' = ^ (Art. 529). This gives ■^^^"^ dr ~2drf~ p^dr '^"^ 535. Exercises. (i) Find the la.w of force when the equation of the orbit is r"^q"/{i +ecoinQ), e being positive, and investigate the particular cases n = I, « = 2, n = — i, « = — 2. (2) Find the law of the central force directed to the origin under whose action a particle will describe the following curves : (a) the spiral of Archimedes r-=aQ; {b) the hyperbolic spiral fir = a ; (c) the logarithmic or equiangular spiral r = ae"^ ; {d) the curve r=a cos nO. (3) A particle moves in a circle under the action of a central force directed towards a point on the circumference. Find the law of force. (4) A particle is acted upon by a force perpendicular to a given plane and inversely proportional to the cube of tlie distance from the plane. Determine its motion. (5) A particle moves in a semi-ellipse under the action of a force perpendicular to the axis joining the ends of the semi-elhpse. Deter- mine the law of force and the velocity at the ends. 536. Force Proportional to the Distance : f{r) = K-r. The equa- tions of motion (2) are in this case the upper sign holding for attraction, the lower for repulsion. Their solution is very simple, because each equation can be integrated sepa- rately. We find, in the case of attraction, X = ai cos Kt -f a^ sin k/, y = 61 cos Kt -\- ^2 sin «/, and in the case of repulsion, X = a^e'^' -\- (Ja^""', y = b^e"' -\- b2e~'' ; Uy, a^, by, b-i, being the constants of integration. 537. To find the equation of the orbit, it is only necessary to elimi- nate t in each case. 344 KINETICS OF THE PARTICLE. [538- In the case of attraction, this elimination can be performed by solv- ing for cos Kt, sin «/, squaring and adding. The result is {a^y — b^xf + {a^y — b^J' = {a-J)^ — aj>^^, and this represents an ellipse, since ia^ + a^)(b^ + b^ - {a A + a.^b^j' = {a^b^ - ajiif is always positive. The center of the ellipse is at the origin, and the lines a-^y — b^x, a^y = b.^ are a pair of conjugate diameters. 538. In the case of repulsion, solve for e"' and ^"''', and multiply. The resulting equation, {a-^y — b-^x){b^ — a^y) = {a^b^ — aj>^'^, represents a hyperbola whose asymptotes are the lines aiy = b-iX, a^y = b^. 539. It is worthy of notice that the more general problem of the motion of a particle attracted by any number of fixed centers, with forces directly proportional to the distances from these centers, can be reduced to the problem of Art. 536. Let X, y, z be the co-ordinates of the particle, r^ its distance from the center Oi ; x„ y^, Zi the co-ordinates of Oi ; and — KjVj the acceleration produced by Cj. Then the :v:-component of the resultant acceleration is = - 2k,V< . ^^^' = - S/c,^ {x - X,) = - x%K,' + %K^x, ; and similar expressions obtain for the y and 2 components. Hence, the equations of motion are As the right-hand members are linear in x, y, z, there is one, and only one, point at which the resultant acceleration is zero. Denoting its co-ordinates by x, y, z, we have wWchT °^'!"'' ''^"''^°''' *°"^ '^^' '^^^ P°i"t of -ero acceleration, of ore fTh!™" ""'t ''' '^"'" ''"''"' '^ '"^^ «"'™d of the center It "Tidllt TrdToint!^"^""^^ -'^-^^ ~ --• - ^'^- 542.] FREE CURVILINEAR MOTION. 345 540. By introducing the co-ordinates of the mean center, we can now reduce the equations of motion to the simple form where f? = 2kA Finally, taking the mean center as origin, we have It thus appears that the motion of the particle is the same as if there were only a single center of force, viz., the mean center (x, y, z), attracting with a force proportional to the distance from this center. The plane of the orbit is, of course, determined by the mean center and the initial velocity. 541. It is easy to see that most of the considerations of Art. 539 apply even when some or all of the centers repel the particle with forces proportional to the distance. It may, however, happen in this case that the mean center lies at infinity, in which case, of course, it can not be taken as origin. Simple geometrical considerations can also be used to solve such problems. Thus, in the case of two attractive centers O^, 0^ (Fig. 161) of -equal intensity k^, the forces can evidently be represented by the dis- tances POi = ri, PO2 = r^ of the par- ticle P from the centers. Their resultant is therefore = 2 PO, if O "' denotes the point midway between O^ pig. 16I. and O^; and this resultant always passes through this fixed point O, and is proportional to the distance PO from this point. 542. Exercises. (i) Determine the constants of integration in Art. 535, \i x^, y^, are the co-ordinates of the particle at the time i=o and v-^, v^ the com- ponents of its velocity v^ at the same time. The equation of the orbit will assume the form 346 KINETICS OF THE PARTICLE. [542- for attraction, and K\xoy —yoxf — (viy — v^)' = — (x„z'2 —yo^if for repulsion. (2) Show that the semi-diameter conjugate to the initial radius vector has the length Vo/k, where v„- = v^- + vi- As any point of the orbit can be regarded as initial point, it follows that the velocity at any point is proportional to the parallel diameter of the orbit. (3) Find what the initial velocity must be to make the orbit a circle in the case of attraction, and an equilateral hyperbola in the case of repulsion. (4) The initial radius vector r^ and the initial velocity v^ being given geometrically, show how to construct the axes of the orbit described under the action of a central force (of given intensity k^) proportional to the distance from the origin. (5) A particle describes an ellipse under the action of a central force proportional to the distance ; show that the eccentric angle is proportional to the time, and find the corresponding relation for a hyperbolic orbit. (6) A particle of mass m describes a conic under the action of a central force F=^ mK-r. Show that the sectorial velocity is ^c = ^ Kab, a and b being the semi-axes of the conic. (7) In Ex. (6) show that the time of revolution is T= 2 tt/k, if the conic is an eUipse. (8) A particle describes a conic under the action of a force whose direction passes through the center of the conic. Show that the force is proportional to the distance from the center. (9) A particle is acted upon by two central forces of the same intensity (k^, each proportional to the distance from a fixed center. Determine the orbit : {a) when both forces are attractive ; {b) when both are repulsive; {c) when one is an attraction, the other a repulsion. (10) A particle of mass m is attracted by two centers Oj, O2 of equal mass m' and repelled by a third center Oj, whose mass is »?"= 2 m'. If the forces are all directly proportional to the respective distances, determine and construct the orbit. 543-] FREE CURVILINEAR MOTION. 347 (11) When a particle moves in an ellipse under a force directed towards the center, find the time of moving from the end of the major axis to a point whose polar angle is 6. (12) Prove that if, in the problem of Art. 541, the intensities of C] and O2 are kj, k^, the resultant attraction F passes through the centroid G of two masses kj, k^, placed at Oi, O2, and that F= {ki + k,)PG. (13) In Art. 536, in the case of attraction, the component motions are evidently simple harmonic oscillations. Show that the equation of the path can be put in the form (comp. Art. 142) -!-^^sin8+-^' = cos2S. a^ ab r (14) Show that the total energy of a particle of mass m describing an ellipse of semi-axes a, b under a force w;/cV directed to the center is = \ mK\a' + b"). 543. Force Inversely Proportional to the Square of the Distance : f{r)= /ti/r^ (Newton's law). It has been shown in Kinematics (Arts. 157-169) how this law of acceleration can be deduced from Kepler's laws of planetary motion. From Kepler's first law Newton concluded that the acceleration of a planet (regarded as a point of mass m) is con- stantly directed towards the sun ; from the second he found that this acceleration is inversely proportional to the square of the distance. The motion of a planet can therefore be explained on the hypothesis of an attractive force, issuing from the sun. The value of /a, which represents the acceleration at unit dis- tance or the so-called intensity of the force, was found to be (Art. 169; or below, Art. 556) ,3 ;u, = 4 7r' a" 7^' and as, according to Kepler's third law, the quantity cfi /T'^ has the same value for all the planets, Newton inferred that the 348 KINETICS OF THE PARTICLE. [544. intensity of the attracting force is the same for all planets ; in other words, that it is one and the same central force that keeps the different planets in their orbits. 544. It was further shown by Newton and H alley that the motions of the comets are due to the same attractive force. The orbits of the comets are generally ellipses of great eccen- tricity, with the sun at one of the foci. As a comet is within range of observation only while in that portion of its path which lies nearest to the sun, a portion of a parabola, with the same focus and vertex, can be substituted for this portion of the elliptic orbit, as a first approximation. It is also found from observation that the motions of the moons or satellites around the planets follow very nearly Kep- ler's laws. A planet can therefore be regarded as attracting each of its satellites with a force proportional to the mass of the satellite and inversely proportional to the square of the distance. 545. All these facts led Newton to suspect that the force of terrestrial gravitation, as observed in the case of falling bodies on the earth's surface, might be the same as the force that keeps the moon in its orbit around the earth. This inference could easily be tested, since the acceleration g of falling bodies as well as the moon's distance and time of revolution were known. Let m be the mass of the moon, a the major semi-axis of its orbit, T the time of revolution, r the distance between the centers of earth and moon ; then the earth's attraction on the moon is (Art. 543) TV- or, since the eccentricity of the moon's orbit is so small that the orbit can be regarded as nearly circular. 547-] FREE CURVILINEAR MOTION. 349 On the other hand, the attraction exerted by the earth on a mass m on its surface, i. e., at the distance R= 3963 miles from the center, is F' = mg. Now, if these forces are actually in the inverse ratio of the squares of the distances, we must have F' F a" or, since the distance of the moon is nearly = 60 if. F' = --6o^F. Substituting the above values of F and F' , we find P=: A 1 2_ 6o'i?_ With i?= 3963 miles, T= 2f 7*43'", this gives a value which agrees sufficiently with the observed value of g, consider- ing the rough degree of approximation used. 546. In this way Newton was finally led to his law of universal gravitation, which asserts that every particle of mass m attracts every other particle of mass m' with a force F=K-~, (12) r^ where r is the distance of the particles and k a constant, viz., the acceleration produced by a unit of mass in a unit of mass at unit distance (see Arts. 472, 473). The best test of this hypothesis as an actual law of physical nature is found in the close agreement of the results of theo- retical astronomy based on this law with the observed celestial phenomena. 547. Taking Newton's law as a basis, let us now turn to the converse problem of determining the motion of a particle acted upon by a single cetttral force for which f{r)= /Lt/r^ (problem of planetary motion). 350 KINETICS OF THE PARTICLE. [548. It has been shown in Kinematics (Arts. 170-173) that if tlie force be attractive, the particle will describe a conic section with one of the foci at the center of force, the conic being an ellipse, parabola, or hyperbola, according as V, 2<2M (13) If the force be repulsive, the same reasoning will apply, except that /u. is then a negative quantity. The orbit is, therefore, in this case always hyperbolic ; the branch of the hyperbola that forms the orbit must evidently turn its convex side towards the focus at which the center of force is situated, since the force always lies on the concave side of the path. 548. To exhibit fully the determination of the constants and the dependence of the nature of the orbit on the initial conditions, a solution somewhat different from that given in Kinematics will here be given for the problem of planetary motion in its simplest form. With f(f) = ;u,/r^, the equation of kinetic energy and work, (7), Art. 530, gives C'dr r ro or, if the constant of integration be denoted briefly by h and u = i/r be introduced, tJ' — 2 fxti -\- h, where h=^v^ ^- (14) Substituting this expression of v^ in the equation (9), Art. 531, we find the differential equation of the orbit in the form ^J+«.^ = i(2;uz^ + /^), (15) To integrate, we introduce a new variable u' by putting the resulting equation. S49-] FREE CURVILINEAR MOTION. du' 351 = I - u'^, or d6 = ±- Vi - «'" has the general integral — a = =F COS"' u', or u' = cos {0 — a), where a is the constant of integration. The orbit has, therefore, the equation (16) -:=^+V?^-^(^-«)' which agrees with the equation (47) given in Kinematics, Art. 173, excepting the different notation used for the constants. 549. The equation (16) represents a conic section referred to its focus as origin. The general focal equation of a conic is - = - + -cos{0-a), r i I (17) where / is the semi-latus rectum, or parameter, e the eccentricity, and a the angle made with the polar axis by the line joining the focus to the nearest vertex. In a planetary orbit (Fig. 162), the sun S being at one of the foci, the nearest vertex A is called ik^e. perihelion, the other vertex A' the aphelion, and the angle Q — a made by any radius vector SP= r with the peri- helion distance SA is called the true anomaly. Comparing equations (17) and (16), we find, for the determina- tion of the constants : hence, /=£?, . = ^1+^, (18) II. ' IX. or, solving for c and h, c=^lil, h = IX — - — (19) Fig. 162. 352 KINETICS OF THE PARTICLE. [550. 550. The expression for the eccentricity ^ in (18) determines the nature of the conic ; the orbit is an ellipse, parabola, or hyperbola, according as ^= i ; hence, by (18), according as the constant h of the equation of kinetic energy is negative, zero, or positive. Owing to the value of h given in (14), this criterion agrees with the form (13), Art. 547. It should be obseiTcd that it follows from (13) that the nature of the conic is independent of the direction of the initial velocity. 551. The criterion (13) can be given the following interpretation. Consider a particle attracted by a fixed center according to Newton's law. If it move in a straight Hne passing through the center, the principle of kinetic energy gives for its velocity, at the distance r, o 2 C^ dr 2 a , 2 2U. J't, r- r ^0 hence, if it start from rest at an infinite distance from the center, it would acquire the velocity V2 /x/r at the distance r. The criterion (13) is therefore equivalent to saying that the orbit is an ellipse, a parabola, or a hyperbola, according as the velocity at any point is less than, equal to, or greater than the velocity luhich the particle tvould have acquired at that point by falling towards the center from infinity (comp. Art. 475). 552. For a central conic, whose axes are 2 a, 2 b, we have / = b^/a, e = Va^ =F b''/a (the upper sign relating to the ellipse, the lower to the hyperbola), so that the equations (19) reduce to the following : c=^bJf^, h = Tl^. (20) ^a a The latter relation, with the value of h from (14), gives for the major or focal semi-axis a : „ ±i = ^-^; (21) a ro fx. while the former, with the value of c as given in (5), Art. 529, deter- mines the minor or transverse axis b : b = c\- = r^Vo sin i//„-v / - • (22) 55S-} FREE CURVILINEAR MOTION. 353 553. The magnitudes of the axes having thus been found, their directions can be determined by a simple construction which furnishes the second focus. In the ellipse, the focal radii have a constant sum = 2 a, and lie on the same side of the tangent, making equal angles with it. In the hyperbola, they have a con- stant difference = 2 a, and lie on opposite sides of the tan- gent. Hence, determining the point C" (Fig. 163), which is symmetrical to the center of force O with respect to the initial velocity, and drawing the line Pf,0" , we have only to lay off on this line from Po a length PfiO' = ±{2 a — r^ ; then O' is the second focus, which for an elliptic orbit must be taken with O on the same side of the tangent PaT, and for a hyperbolic orbit on the opposite side. 554. For z. parabola, since ^= i, we find, from (19), Fig. 163. , / of the eUipse as a new variable, and to express /, r, and in terms of (p. In astronomy, the polar angle 6 is known as the frue anomaly, and the eccentric angle ^ as the eccentric anomaly. 5S9-] FREE CURVILINEAR MOTION. 355 558. The relation of the eccentric angle (j> to the polar co-ordinates r, 6 will appear from Fig. 165, in which F is the position of the planet Fig. 165. at the time /, P' the corresponding point on the circumscribed circle, 2C AOP= 6 the true anomaly, and ^ ACP' = (ft the eccentric anomaly. The focal equation of the ellipse I + e cos ( I + e cos ( gives r+i?rcos 0=a — ae' ; and the figure shows that rcos 0=a cos <^ — ae; hence r = a(i — f cos <^), or a — r = flf cos <^. (28) Equating this value of-r to that given by the polar equation of the ellipse, we have cos — e 1 — e cos = : or cos = - I + I? cos 0' I — e cos A more symmetrical form can be given to this relation by computing I — cos 6=2 sin^ i 6 = (i + e) I + cos 6=2 cos^ i = (i — e) whence, by division, tai cos (^ I — e cos (t> I -|- cos 4> \ — e cos <^ ' = \ Y^3^ tan \ <^. (29) 559. To find / in terms of r, we have only to substitute in (24) for ip- its value from (8), Art. 531, and to integrate the resulting differential equation 3S6 KINETICS OF THE PARTICLE. [560. ~dij r'~~~a' As, by (20), Art. 552, c'^ = ixb''-/a = iji.a{i —e'), this equation becomes a rdr or dt=X- The integration is easily performed by introducing the eccentric angle ^ as variable by means of (28) ; this gives di = \- ■ a(i — e cos d>) d. If the time be counted from the perihelion passage of the planet, we have /= o when r= a — ae, i. e., when <^ = o ; hence, putting ^ /jl/o? = n, as in Art. 556, we find . nf= (j> — e sm — e sin A), 2 TT and this agrees with (30) since, by (25), 2 7r/7'= n. S62.] FREE CURVILINEAR MOTION. 357 561. Kepler's equation (30) gives the time as a function of 1^ ; by means of (28), it establishes the relation between / and r ; by means of (29), it connects t with Q. It is, however, a transcendental equation and cannot be solved for <^ in a finite form. For orbits with a small eccentricity e, an approximate solution can be obtained by writing the equation in the form <^= nt-\- e sin <^, and substituting under the sine for <^ its approximate value nt: (j) = 7i/+ e sin nt (31) This amounts to neglecting terms containing powers of r ^'^y Mm yl —y '"w= - ^ ■ r dh Mm z' — z _ " r" ' r while the equations of motion of the sun are M^i- mM = K — r- X ~x' d/^ 7^ r M'^'y'-. mM ^ If .L -y df ^ r M^'i-. mM = K z -2' df- t'- r (==) 566. By adding the corresponding equations of the two sets, we find — {mx + Mx'\ = o, — - (my + My^ =0, — (mz-\- Mz') = o. If it be remembered that the centroid of the two masses m, M has the co-ordinates - _ mx-\- Mx' - _ my + My' -^ _mz-\- Mz' ~ m + M ' " 7n-\-M ' ~ i?i + M ' it appears that these equations can be written in the form d^x dj dH ^=°' ^=°' ;^=°^ in words : the acceleration of the common centroid of planet and sun is zero ; i. e., this centroid moves with constant velocity in a straight line. 567. The integration of the equations (i) would give the absolute path of the planet. But the constants could not be determined, because the absolute initial position and velocity of the planet are, of course, not S69-] FREE CURVILINEAR MOTION. 361 known. The same holds for the absolute path of the sun. All we can do is to determine the relative motion, and we proceed to find the motion of the planet relative to the sun. Taking the sun's center as new origin for parallel axes, we have for the co-ordinates i, t), t, of the planet in this new system, i = x — x',r) =y—y\ t,=z — z\ Now, dividing the equations (i) by m, the equations (2) by M, and sub- tracting the equations of set (2) from the corresponding equations of set (i), we find for the relative acceleration of the planet r (if r^ M-\-m Y] r (3) ^^_ M+m ^^ df '^ r^ r The form of these equations shows that the relative motion of the planet with respect to the sun is the same as if the sun were fixed and contained the mass M-\- m. Thus the problem is reduced to that of a fixed center, the only modification being that the mass of the center M should be increased by that of the attracted particle m. 568. This result can also be obtained by the following simple con- sideration. The relative motion of the planet with respect to the sun would obviously not be altered if geometrically equal accelerations were appHed to both. Let us, therefore, subject each body to an additional acceleration equal and opposite to the actual acceleration of the sun (whose components are obtained by dividing the equations (2) by M^. Then the sun will be reduced to equilibrium, while the resulting accel- eration of the planet, which is its relative acceleration with respect to the sun, will evidently be the sum of the acceleration exerted on it by the sun and the acceleration exerted on the sun by the planet. This is just the result expressed by the equations (3). 569. It can here only be mentioned in passing that, while the problem of two bodies thus leads to equations that can easily be integrated, the problem of three bodies is one of exceed- 362 KINETICS OF THE PARTICLE. [57°- ing difficulty, and has been solved only in a few very special cases. Much less has it been possible to integrate the 3 n equations of the problem of n bodies. 570. According to the equations (3), the first and second laws of Kepler can be said to hold for the relative motion of a planet about the sun (or of a satellite about its primary). The third law of Kepler requires some modification, since the intensity of the center jx should not be kM, but k {M + m). We have, by (26), Art. 556, IX = k{M + m) = ^TT^ — ; in other words, the quotient a^/T"^ is not independent of the mass in of the planet. Thus, if ;«j, m^ be the masses of two planets, a-^, a^ the major semi-axes of their orbits, and 7"j, T,^ their periodic times, we have a^^/T^^ ^ M + m, ^ i + mJM This quotient is approximately equal to i if iW is very large in comparison with both ?«j and m^ ; hence, for the orbits of the planets about the sun, Kepler's third law is very nearly true. 571. Exercises. ( 1 ) Two particles of masses ;//i, m-i, attract each other with a force which is any function of the distance r between them, say F=mim.2f{f). Show that their common centroid moves uniformly in a straight line, and find the equations of this line. (2) In Ex. (i), write out the equations for the relative motion of either particle with respect to the common centroid. IV. Constrained Motion. I. INTRODUCTION. 572. The motion of a free particle is fully determined if all the forces acting upon the particle, as well as the so-called initial conditions, are given. The motion of a particle may, however, S74-] CONSTRAINED MOTION. 363 depend not only on given forces, but on other conditions not directly expressed in terms of forces. The motion is then said to be constrained (comp. Art. 396). To mention some examples : a heavy particle sliding down a smooth inclined plane is subject not only to the force of gravity, but also to the condition that it cannot pass through the plane ; a railway train running on the rails, a piece of machinery slid- ing in a groove or between guides, can, for many purposes, be regarded as a particle constrained to a curve ; the bob of a pendulum, a stone attached to a cord and swung around by the hand, may be regarded as constrained to a surface. 573. Sometimes these constraining conditions can be easily replaced by forces. Thus, in the first illustration above, the condition that the particle cannot pass through the inclined plane can be expressed by introducing the reaction of the plane, i. e., a force acting on the particle at right angles to the plane, so as to prevent it from passing through the plane. Similarly, in the case of the stone attached to the cord, we may imagine the cord cut and its tension introduced so as to replace the condition by a force. Whenever the constraints to which a particle is subjected can thus be expressed by means of forces, these forces can be combined with the other impressed forces, and then, of course, the equations of motion for a free particle can be applied. Thus, let X' , V, Z' be the components of the resultant of all the constraints ; X, Y, Z those of the resultant of all the other impressed forces. Then the equations of motion are : ^%-X-,X<, n.'i^=y+Y', -f^=^+^'- (0 It must, however, be noticed that the reactions representing the constraints, such as the tension of the string in the example referred to, are generally not given beforehand. 574. On the other hand, the constraints are often expressed more conveniently by conditional equations. Thus, if the motion 364 KINETICS OF THE PARTICLE. [575- of a particle be restricted to a surface, the equation of this sur- ^^'^''^y {x,y,.) = o, (2) may be given as a constraining condition to be fulfilled by the co-ordinates of the moving particle. As a particle has but three degrees of freedom, it can be subjected to only one or two conditions of the form (2). One such condition confines it to a surface ; two to the curve of intersec- tion of the two surfaces represented by the two conditional equations ; three conditions would evidently prevent it entirely from moving. 575. The curve or surface to which a particle is constrained may vary its position and even its shape in the course of time. In this case the conditional equations, referred to fixed axes, will contain not only the co-ordinates, but also the time. That is, they will be of the more general form ^{x,y, z, t)=0. (3) 576. To constrain a particle completely to a surface, we may imagine it confined between two very near impenetrable sur- faces. The complete constraint to a curve may be imagined as due to a narrow tube having the shape of the curve, or by regarding the particle as a bead sliding along a wire. In many cases, however, the constraint is not complete, but only partial, or one-sided. Thus, the rails compelling the train to move in a definite curve do not prevent its being lifted verti- cally out of this curve, nor does the cord that confines the motion of the stone to a sphere prevent it from moving towards the inside of the spherical surface. While complete constraints are generally expressed by equa- tions, one-sided constraints should be properly expressed by inequalities. Thus, in the case of the stone, the condition is really that its distance r from the hand is not greater than the length / of the cord, i. e., ^ ,. 577-] CONSTRAINED MOTION. 365 but as soon as r becomes less than /, the constraining action ceases, and the stone becomes free. It is, therefore, in general sufficient to consider conditional equations ; but the nature of the constraint, whether complete or partial, must be taken into account to determine when and where the constraint ceases to exist. We now proceed to consider briefly the motion of a particle constrained to a fixed curve and that of a particle constrained to a fixed surface. 2. MOTION ON A FIXED CURVE. 577. The condition that a particle should move on a given fixed curve can always be replaced by introducing a single additional force F' called the constraining force, or the con- straint. Consider, for instance, a particle of mass m, subject to the force of gravity F= mg alone ; in general it will describe a parabola whose equation can be found if the initial conditions are known. To compel the particle to describe some other curve, say a vertical circle, a con- straining force F' (Fig. 166) must be introduced such that the result- ant F of F and F' shall produce the acceleration required for mo- tion in the circle. Thus, for in- stance, for uniform motion in the circle the resulting acceleration must be directed towards the center and must be = u?a, if a is the radius and 0) the constant angular velocity. We have, therefore, in this case R = mu?a along the radius, F= mg vertically downward ; and hence, denoting by & the angle made by the radius CP with the vertical (Fig. 166), r^=F' + R''+2.FRcoi& = m^ ig'' + <^'a- + 2 gw^a cos 9) . 366 KINETICS OF THE PARTICLE. [578. The constraint F', which is thus seen to vary with the angle Q, can be resolved into a tangential component F^ and a normal component F,l. As in our problem the velocity is to remain of constant magnitude, the tangential constraint must just counterbalance the tangential component of gravity F, = nig sin d. The normal constraint F^ not only counter- balances the normal component of gravity F^ = mg cos 6, but also fur- nishes the centripetal iorce R=miii''a required for motioninthe circle ; i.e., F^ = F + Fcos6 = m (u?a +gcosd). 578. In the general case of a particle of mass m acted upon by any given forces and constrained to any fixed curve, it is convenient to resolve both the resultant F of the given forces and the constraint F' along the tangent and the normal plane. The equations of motion (see Art. 500) can then be written in the form , m—= resultant of F^ and FJ, P where v is the velocity and p the radius of curvature of the path at the time t. It should be noticed that the components F„ and FJ, though both situated in the normal plane, do not in general have the same direction. But in the important special case of plane motion, i. e., when the path is a plane curve and the resultant F of the given forces lies in this plane, F„ and F^ are both directed along the radius of curvature so that the right-hand member of the second equation becomes the sum or difference of F^ and /^„'. 579. The normal component F^ of the constraining force is generally denoted by the letter N and is called the resistance or reaction of the curve; a force —N, equal and opposite to this reaction, represents the pressure exerted by the particle on the curve. The. tangential component F^ of the constraint will exist only 58i.] CONSTRAINED MOTION. 367 when the constraining curve is " rough," i. e., offers frictional re- sistance ; we have then, denoting the coefficient of friction by /i, We shall therefore write the equations of motion as follows : m-^=Ft-it.N, (i) at m~ = resultant of F„ and N. (2) P 580. The normal component, miP'/p, of the effective force is sometimes called the centripetal force ; it is directed along the principal normal of the path towards the center of curva- ture. A force equal and opposite to this centripetal force, i. e. , = — mv^/p, is called centrifugal force. It should be noticed that this is a force exerted not 07t the moving particle, but by it. It appears from equation (2) that the normal reaction N is the resultant of the centripetal force mv^jp and the reversed normal component of the given forces, — F^. Changing all the signs, we can express the same thing by saying that the presstire on the curve, — N, is the resultant of the centrifugal force, — miP'lp, and of the normal component F^ of the given forces. If, in particular, this normal component F^ is zero, the pres- sure on the curve is equal to the centrifugal force. This case is of frequent occurrence. Thus if a small stone attached to a cord be whirled around rapidly, the action of gravity on the stone can often be neglected in comparison with the centripetal force due to rotation; hence the centrifugal force measures approximately the tension of the cord, and may cause it to break. Again, when a railway train runs in a curve, the centrifu- gal force produces the horizontal pressure on the rails, which tends to displace and deform the rails. 581. It may happen that at a certain time t the pressure — N vanishes. If the constraint be complete (Art. 576), this would merely indicate that the pressure in passing through zero 368 KINETICS OF THE PARTICLE. [582. inverts its sense. If, however, the constraint be one-sided, the consequence will be that the particle at this time leaves the constraining curve ; for at the next moment the pressure will be exerted in a direction in which the particle is free to move. Now A^ vanishes when its components — F^ and mv'^/p become equal and opposite. The conditions under which the particle would leave the curve are, therefore, that the resultant F of the given forces should lie in the osculating plane of the path, and that F„ = mv'^/p. 582. To obtain the equations of motion expressed in rect- angular cartesian co-ordinates, let X, Y, Z be the components of the resultant F of the given forces, and N^, Ny, N^ those of the normal reaction N of the curve. If there be friction, the f ric- tional resistance ^iN, being directed along the tangent to the path opposite to the sense of the motion, has the direction cosines — dx/ds, — dyjds, — da/ds, so that the components of the force of friction are — p.Ndx/ds, — fjiNdy/ds, — fiNdz/ds. The general equations of motion are, therefore, -S=^+^.-.^| (3) dt^ ds If the acceleration of the particle be zero, the left-hand mem- bers are all = o, and the equations reduce to the conditions of equilibrium of a particle on a fixed curve. In addition to the equations (3) we have of course the equa- tions of the curve, say ^{x, y, z) = o, ^{x, y, z) =0, (4) and the relations N"^ = N^ + A^,J + N^, (^5 ) the latter expressing that iVis normal to the path. 584.] CONSTRAINED MOTION. 369 583. Multiplying the equations (3) by dx, dy, ds, and adding, we find the equation of kinetic energy 41 mi!^) = Xdv + Ydy + Zdz - ^iNds. (7) This relation might have been obtained directly from the con- sideration that for a displacement ds along the fixed curve the normal reaction N does no work, while the work of friction is — iiNds. If there is no friction along the curve (^u, = o), it follows from (7) that the velocity is independent of the reaction of the curve. 584. Exercises. (i) Show that when the given forces are zero and there is no friction, the particle moves uniformly on the curve, and the pressure on the curve is proportional to the curvature of the path. (2) A particle of mass m moves down a straight line inclined to the horizon at an angle Q, under the action of gravity alone. {a) If there be no friction, we have by Art. 5 79, since p = 00 (see Fig. 167) : dv ■ n dt ^ o = mg cos 6 — N. The first of these equations is the same as that derived in Kinematics for motion down an inclined plane (Arts. 116, 117). The second equation gives the normal reaction mg Fig. 167. of the line N'=mg cos 6; hence, the pressure on the line, —JV, is constant. {i) If the line be rough, the second equation remains the same, while the first must be replaced by the following : m — = mg sin — fxJV= mg(sin 6 — fj. cos $). dt As the acceleration is constant whether there be friction or not, the motion is uniformly accelerated, unless sin ^ — /x cos ^ = o, i. e., fj. = tan 6. Find V and s ; show that in the exceptional case fi. = tan &, the motion 370 KINETICS OF -rtlE PARTICLE. [584. is uniform unless the initial velocity be zero ; show that, for motion up the line, the first equation becomes i3'z//(// = — ^(sin 6 + /i cos ^), the motion being uniformly retarded until t= z'o/g'Csin ^ + /a cos B'), when the particle either begins to move down the line or remains at rest. (3) A cord of length / (ft.) carries at one end a mass of m lbs., while the other end is fixed at a point O on a smooth horizontal table. A blow, at right angles to the cord, imparts to m an initial velocity 710 ft. per second. Show that, as m describes the circle of radius / about O on the table, the velocity remains constant and the tension of the cord is = mvi Igl lbs. (4) In Ex. (3), let OT = 2 lbs. ; /= 3 ft. ; find the tension in pounds : {a) when the mass makes one revolution per second ; {F) when it makes 8 revolutions per second, (jc) If the cord cannot stand a tension of more than 300 lbs., what is the greatest allowable number of revolutions ? (5) A locomotive weighing 32 tons moves in a curve of 800 ft. radius with a velocity of 30 miles an hour ; find the horizontal pressure on the rails. (6) To prevent the lateral pressure on the rails in a curve, the track is inclined inwards. Determine the required elevation e (in inches) of the outer above the inner rail for a given velocity v and radius R if the gauge ii. e., the distance between the rails) is 4 ft. 8 in. Show that with R = 2000 ft. and z' = 45 miles per hour, e = 3.8 in. (7) A plummet is suspended from the roof of a railroad car ; how much will it be deflected from the vertical when the train is running 45 miles an hour in a curve of 300 yards radius? (8) A body on the surface of the earth partakes of the earth's daily rotation on its axis. The constraint holding it in its circular path is due to the attractive force of the earth. Taking the earth's equatorial radius as 3963 miles, show that the centripetal acceleration of a particle at the equator is about \ ft. per second, or about -^^^ of the actually observed acceleration g= 32.09 of a body falling in vacuo. (9) If the earth were at rest, what would be the acceleration of a body falling in vacuo at the equator? (10) Show that if the velocity of the earth's rotation were over 17 times as large as it actually is, the force of gravity would not be sufficient to detain a body near the surface at the equator (comp. Ex. (14), Art. 562). 585.] CONSTRAINED MOTION. 371 (ii) Show that in latitude (^ the acceleration of a falhng body, if the earth were at rest, would be ^1 = ^ +y cos-' = 45", g= 980.6 cm. ; hence ^1 = 982.3. (12) A chandelier weighing 80 lbs. is suspended from the ceiling of a hall by means of a chain 12 ft. long whose weight is neglected. By how much is the tension of the chain increased if it be set swinging so that the velocity at the lowest point is 6 ft. per second ? (13) A cord of 2 ft. length passes at its middle point through a hole in a smooth horizontal table. It carries at its lower end a mass of 2 lbs., at its other end a mass of i lb. The latter is set to revolve in a circle about the hole so as to stretch the cord and just prevent the mass of 2 lbs. from descending, (a) How many revolutions must it make ? (i) If only one fourth of the cord lie on the table while three fourths hang down, how many revolutions must be made ? (14) Show that, when a particle moves with constant velocity in a vertical circle, the constraining force I^' (Art. 577) is always directed towards a fixed point on the vertical diameter. (15) A bicycHst is rounding a curve of 50 ft. radius at the rate of 10 miles an hour; find his inclination to the vertical. (16) Owing to the earth's rotation on its axis the direction of a plumb-hne does not pass through the center of the earth, even when the earth, as here assumed, is regarded as a homogeneous sphere. Express the angle 8 of the deviation in latitude , and determine in what latitude S is greatest. (17) A cord, 4 ft. long, whose breaking strength is 24 lbs., is swung in a circle with a 3-lb. mass attached at its end. Find the greatest number of revolutions allowable : (a) when the circle is hori- zontal ; (6) when it is vertical. (18) At what speed would a locomotive whose centroid is 6 ft. above the track be upset in a curve of 400 ft. radius, the gauge being 4 ft. 8^ in., if the rails are on a level ? 585. A particle of mass m subject to gravity alone is corv- strained to move in a vertical circle of radius I. If there be no friction on the curve and the constraint be produced by a 372 KINETICS OF THE PARTICLE. [586. weightless rod or cord joining the particle to the center of the circle, we have the problem of the simple mathematical pendulum. Equation (i), Art. 579, is readily seen to reduce in this case (see Fig. 168) to the form /^+^sin^ = o. (8) A first integration gives, as shown in Kinematics (Arts. 149, ISO), o 1 7/2 = _^(/ cos 61 + ^ - / cos ex (9) 2^ where v^ is the velocity which the particle has at the time t = o when its radius makes the angle AOPf^ = 0Q with the vertical. Multiplying by m, we have, for the kinetic energy of the particle, i mv^ = mg {I cos 6 + h), (10) where h = v^/2 g — I cos ^g is a constant. If the horizontal line MN, drawn at the height v^/ig above the initial point Pg, inter- sect the vertical diameter AB a.t R, it appears from the figure that h = RO. Fig. 168. mg 586. Taking R as origin and the axis of z vertically down- wards, we have RQ = z = I cos Q ^ h ; hence the force-function U has the simple expression U = mgz ; and the velocity v= ^2gz is seen to become zero when the particle reaches the horizontal line MN. For the further treatment of the problem, three cases must be distinguished according as this line of zero-velocity MN SSg-] CONSTRAINED MOTION. 373 intersects the circle, touches it, or does not meet it at all ; i. e., according as 2 /^|/, or ^|2/COS2 10O. (II) 587. Equation (2), Art. 579, serves to determine the reaction N of the circle, or the pressure —N on the circle. We have m— = — mgcos 6 -\- N, whence JV = m(— + £-cos Substituting for 7^ its value from (10), we find N= mg-f 2 j+ s cos dj. (12) The pressure on the curve has therefore its greatest value when ^ = o, i. e., at the lowest point A. It becomes zero for / cos 0^ = — ^h, which is easily constructed. 588. If the constraint be complete as for a bead sliding along a circular wire, or a small ball moving within a tube, the pres- sure merely changes sign at the point 6 = 6-^. But if the con- straint be one-sided, the particle may at this point leave the circle. The one-sided constraint may be such that 0P<1, as when the particle runs in a groove cut on the inside of a ring, or when it is joined to the center by a cord ; in this case the particle may leave the circle at some point of its upper half. Again, the one-sided constraint may be such that OP^l, as when the particle runs in a groove cut on the rim of a disk ; in this case the particle can of course only move on the upper half of the circle. 589. Exercises. (i) For h = l, equation (10) can be integrated in finite terms. Show that in this limiting case the particle approaches the highest point B of the circle asymptotically, reaching it only in an infinite time. 374 KINETICS OF THE PARTICLE. [589. (2) Derive the equations of motion for the problem of the simple pendulum (Art. 585) from the general equations of Arts. 582, 583. (3) For ^0 = 60°, ''=1 ft-, z'o = 9 ft. per second, show that the par- ticle will leave the circle very nearly at the point 61= 120°, if the con- straint be such that OP^ I (Art. 588). (4) For z'o= 10 ft. per second, everything else being as in Ex. (3), show that the particle will leave the circle at the point Q^ = i34i°, nearly. (5) A particle, subject to gravity and constrained to the inside of a vertical circle {OP^l), makes complete revolutions. Show that it cannot leave the circle at any point, if ^h> I; and that it will leave the circle at the point for which cos Q = — ^ hjl, if \h<,l. (6) In the experiment of swinging in a vertical circle a glass contain- ing water, and suspended by means of a cord, if the cord be 2 ft. long, what must be the velocity at the lowest point if the experiment is to succeed ? (7) A particle subject to gravity moves on the outside of a vertical circle; determine where it will leave the circle : {a) \{ MN (Fig. 168) intersects the circle ; (li) if MN touches the circle ; (c) if MN does not meet the circle. (8) A particle subject to gravity is compelled to move on any verti- cal curve z =f(x) without friction. Show that the velocity at any point is v = ^2 gz (comp. Art. 586) if the horizontal axis of x be taken at a height above the initial point equal to the " height due to the initial velocity,'' i.e., Vo/^g. (9) A particle slides on the outside of a smooth vertical circle, start- ing from rest at the highest point of the circle. Find where it will meet the horizontal plane through the lowest point of the circle. (10) A heavy particle is constrained (without friction) to a common cycloid in a vertical plane, the axis of the cycloid being vertical and the concavity turned upward. Counting the arc j- of the cycloid from the vertex, the equation of motion is d'^s/di'^ = — gsj/^ a. If z' = o when s = Jo, this gives J = Jq cos vV/4 a • t. Show that the motion is isochro- nous, i. e., that the time t-^ of reaching the lowest point is independent of ^0- (11) The involute of a cycloid being an equal cycloid, with its vertex at the cusp, its cusp on the axis, of the original cycloid, the particle in Ex. (10) can be constrained to the cycloid by means of a cord of S9I-] CONSTRAINED MOTION. 375 length 2 a, attached to the cusp of the involute, and wrapping itself on a cylinder erected on the involute as base. Show that, if the particle starts from rest at the cusp of the original cycloid, the tension of the cord is twice the normal component of the weight of the particle. 3. MOTION ON A FIXED SURFACE. 590. Just as for motion on a curve (Art. 582), we find the general equations of motion The normal reaction N=-^Nj+W+^'' (2) being at right angles to the constraining surface 4>{x,y,z) = o, (3) the following conditions must be satisfied : ^ = ^ = ^. (4) 4>z -Pp 4>z where (f>^, ^y, (f)^, denote, as usual, the partial derivatives of (j)(x, y, 3) with regard to x, y, z, respectively. If the acceleration of the particle be zero, the equations (i) reduce to the conditions of equilibrium of a particle on a surface. 591. A particle of mass m, subject to gravity, is constrained to remain on the surface of a sphere of radius r. If the constraint is produced by a weightless rod or cord joining the particle to the center of the sphere, the rod or cord describes a cone, and the apparatus is called a conical or spherical pendulum. 376 KINETICS OF THE PARTICLE. [592 Taking the center O of the sphere as origin (Fig. 169), and the axis of z vertically down- wards, we have for the equation of the sphere x^+y^^z^-r^^O, (S) whence (^^/x=^y/y=<^^/z. The direction cosines of iVare —x/r, —y/r, —z/r. Hence, the equa- tions of motion, as there is no friction : (t^Z z •.-^ = mg-N-^ (6) As the resistance N does no work, the principle of kinetic / '^ X \^-~-. y/\ / \r "fp j \z 1 y/ z ■=r^^;;-\ ^ — -/ p Fig. 169. d^x ^^x dr r ^mv'^= mgz + C, energy gives or, dividing by J w, v'^= 2{gz + h). (7) To determine the constant of integration h, we have v = Vq when s = Zq-, hence V+2^(.J-^o)- (8) 592. That particular case of the problem of the conical pendulum in which the particle moves in a horizontal circle can be treated directly in an elementary manner. It finds its application in the theory of the governor of a steam engine. Let OQ (Fig. 170) represent a vertical shaft turning about its axis with angular velocity w ; 0P= I a rigid arm hinged to the shaft at O so as to turn freely in the vertical plane POQ. This arm, whose weight is neglected, carries at /* a heavy mass m. The forces acting on m are its weight mg and the tension N of the arm PO ; the resultant R of these forces lies in the vertical plane. If the point P is to describe a horizontal circle about the axis OQ, the resultant R must be perpen- dicular to OQ, and the angular velocity u> and the angle QOP= must be such that R = m V^// to make the bob fly out ; for smaller velocities the pendulum would hang down vertically. 693. Exercises. (i) A conical pendulum makes 60 revolutions per minute, (a) What is the height h of the cone ? {b) If the bob weighs i lb., and the length of the arm is i ft., what is the tension of the arm ? {c) What is the angle 6 ? Answer the same questions when the number of revo- lutions per minute is 180. (2) A straight rod AB is fixed at an angle a to a vertical axis AC, the angle a opening upward. A bead of mass m slides without friction along A£. When AB turns about A C with angular velocity co, what is the position of equilibrium of m ? (3) If the bead in Ex. (2) be prevented from shding along AB by a projecting ring, find the pressure on this ring : (a) when a opens upward ; {b) when a opens downward. (4) If in Ex. (2) the straight rod AB be replaced by an arc of a circle with its center on A C, we have a kind of centrifugal governor. In practice, however, the mass m is generally not made to shde along a material circular arc, but is constrained to a circle by being suspended from a point C on the axis by means of a rigid arm. For the sake of symmetry, two equal arras and masses are generally used. 378 KINETICS OF THE PARTICLE. [593- (5) Determine the curve that should be substituted for the circular arc in Ex. (4) if the mass m is to be in equilibrium, for a given angular velocity m, at every point of the arc. (6) From the equations (5) and (6), Art. 591, derive the approximate path of the bob of a conical pendulum when the angle B remains very small. 596.] GENERAL PRINCIPLES. 379 CHAPTER VI. KINETICS OF THE RIGID BODY. I. General Principles. 594. In kinetics the term rigid body means 'any system or aggregate of mass-particles whose mutual distances remain invariable. A rigid body may therefore consist of a finite number of rigidly connected particles or of a continuous mass of one, two, or three dimensions. Its motion depends not only on the forces acting on the body, but also on the way in which the mass is distributed throughout the body. In the present section the rigid body is assumed to be free unless the contrary be stated explicitly. For the sake of sim- plicity the body is conceived as a rigidly connected system of a fiiiite number of particles. 595. Let us consider any one particle m of the body ; at any time t, let j be its acceleration and F the resultant of all the forces acting on the particle. Then the motion of this particle (see Art. 500) is determined by the equation mj = F. ( I ) It should be noticed that among the forces acting on the particle are included not only those external forces acting on the rigid body that happen to be applied at in, but also the so-called internal forces which would replace the rigid con- nection of the particle ni with the rest of the body. 596. If, at the time t, x, y, z are the co-ordinates of the par- ticle m with respect to a fixed set of rectangular axes, then the components of its velocity v may be denoted by x,y,z; those 38o KINETICS OF THE RIGID BODY. [597- of its acceleration y by i^, j, z* And if the components of F along the same axes are X, Y, Z, the equation (i) can be replaced by the following three : — mx +X=0, —tnj/+Y=o, —in3 + Z=0. (2) Such a set of three equations can be written down for each particle ; hence, if the body consist of n particles, there would be in all 3 n equations. 597. For th-e solution of particular problems these 3 n equa- tions are of little use, not only because their number would in general be very great, but mainly because the forces X, V, Z include the unknown reactions between the particles. It is, however, possible to deduce certain general propositions from these equations. The 3 n equations express the equilibrium of the system formed by all the forces, both internal and external, acting on the particles, and the reversed effective forces. To apply the principle of virtual work to this system, let us multiply the three equations (2) by the components 8x, hy, hs of some virtual displacement of the particle m ; let the same thing be done for every other particle of the body ; and let all the resulting equations be added : S(— mx + X)^x + S(— my -\- F)Sy + '2{—mz + Z)8z = o. (3) 598. It is important to notice that the internal reactions between the particles which make the body rigid occur in pairs of equal and opposite forces, and form, therefore, a system which is in equilibrium by itself. This may be regarded as an assumption which should be included in the definition of the rigid body. Hence, while these internal forces enter into * This is the so-called fluxional notation, according to which derivatives with respect to the time are denoted by dots ; thus x stands for dx / dt, x for d'^x/dt'^, etc. Derivatives were calledy?«;r« by Newton ; thus the component of the accelera- tion of a point in any direction is the time-flux of its velocity in that direction ; the component of its effective force in any direction is the time-flux of its momentum. 6oo.] GENERAL PRINCIPLES. 38 1 the equations (2), they do not appear in equation (3), since the equal and opposite forces cancel in the summation. Thus, equation (3) expresses that the external forces acting on the rigid body and the reversed effective forces form a system in equilib- rium ; and this is d'Alembert's Principle for the rigid body. It must, however, not be forgotten that the displacements hx, hy, hz should be so selected as to be compatible with the nature of the rigid body ; i. e., with the conditions that the distances between the particles should not be disturbed. 599. The number of conditions expressing the invariability of the distances between n particles is 3 « — 6. For if there were but 3 particles, the number of independent conditions would evidently be 3 ; for every additional particle, 3 additional conditions are required. Hence, the total number of condi- tions is 3 + 3 (« — 3) = 3 « - 6. It follows that if a rigid body be subject to no other con- straining conditions, the number of its equations of motion must be 3 « — (3 « — 6) = 6. Hence, a free rigid body has six independent equations of motion (comp. Art. 31). 600. The six equations of motion of the rigid body can be obtained as follows. Imagine the equations (2), viz. : mx = X, my = Y, mi = Z, written down for every particle, and add the corresponding equations. This gives the first 3 of the 6 equations of motion : 1.mx = 2X, 2;«j/ = 2 F, 'Lmz = '^Z. (4) As the internal forces cancel in the summation, the right-hand members of these equations represent the components R„ Ry, R^ of the resultant R. of all the external forces acting on the body. The left-hand members can be written in the form d{^7nx)/dt, d{'^my)/dt, d{'Lmz)/dt: these are the time derivatives or fluxes of the sums of the linear momenta of all the particles parallel 382 KINETICS OF THE RIGID BODY. [6oi. to the axes. The equations (4) can therefore be written in the form d d d — ^mx = R„ —1,mj/=Ry, —Imz^R,. (5) dt dt dt The axes of co-ordinates are arbitrary. Hence, if we agree to call linear momentum of the body in any direction the algebraic sum of the linear momenta of all the particles in that direction, the equations (5) express the proposition that the rate at which the linear mom.entuni of a rigid body in any direction changes with the tim.e is equal to the sum of the components of all -the external forces in that direction. 601. Let us now combine the second and third of the equa- tions (2) by multiplying the former by s, the latter by y, and subtracting the former from the latter. If this be done for each particle, and the resulting equations be added, we find ^m{yz — zy) — ^{yZ— zV). Similarly, we can proceed with the third and first, and with the first and second of the equations (2). The result is : tmi^yz — zy) = 1.{yZ — zV), 'Zm{zx — xz) = 'Z{zX — xZ), 1,7n{xy — yx) = 2 {xY — yX\ (6) Here again the internal forces disappear in the summation, so that the right-hand members are the components H„ Hy, H„ of the vector H of the resultant couple, found by reducing all the external forces for the origin of co-ordinates. The left-hand members are the components of the resultant couple of the effective forces for the same origin. We can also say that the right-hand members are the sums of the moments of the external forces about the co-ordinate axes (Art. 391), while the left-hand members represent the moments of the effective forces about the same axes. The latter quantities are exact derivatives, as shown in Art. 512. The equations (6) can therefore be written in the form — ^m{yz — zy)=H„ — %m{zx—xz) = Hy,--1,m{xy—yx)=H^. (7) Cl'C do iAfl' 6o2.] GENERAL PRINCIPLES. 383 As explained in Art. 514, the quantity m{yz — zy) is called the angular momentum (or the moment of momentum) of the particle m about the axis of x. We may now agree to call the quantity S;« {y'z — zy) the angular momentum of the body about the axis of x, just as ^mx is the linear momentum of the body along this axis; and similarly for the other axes. The meaning of the equations (7) can then be stated as follows : The rate at which the angular momentum of a rigid body about any axis changes with the time is equal to the sum of the moments of all the external forces about this line. The equations (4) and (6), or (5) and (7), are the six equations of motion of the rigid body. The three equations (4) or ( 5 ) may be called the equations of linear momentum, while (6) or (7) are the equations of angular momentum. 602. The equations (4) and (6) can also be derived from the equa- tion (3), which expresses d'Alembert's principle, by selecdng for Ix, Sy, S2 convenient displacements. Thus, the rigidity of the body will evidently not be disturbed if we give to all its points equal and parallel infinitesimal displacements, since this merely amounts to subjecting the whole body to an infinitesimal translation. Equation (3) can in this case be written 8a-S(- wi: + X) + lyt{- my + F) + S:2(- mz + Z) = o, and is therefore equivalent to the three equations (4), since Sx, By, Sz are independent and arbitrary. Again, let the body be subjected to an infinitesimal rotation of angle SO about any line /. As shown in Art. 185, the linear velocities of any point (x, y, 2) of a rigid body, due to a rotation of angular velocity u) = 8d/St about any line / through the origin, are, if : X = (u^ — u)j_)', y = o)jX — (1)^, z = lo-^y — WjX. Hence, putting u)^8/=8^j, ^Bt=S6^, we have for the displacements of the point (.r, y, 2), due to a rotation of angle 8^, &x = z8ey-y86,, By = xBe,-zW,, 8z=yW,-x&e,. 384 KINETICS OF THE RIGID BODY. [603. If these values be introduced in d'Alembert's equation (3) and the terms in 8^^, 8^,, 8^^ be collected, it assumes the form 8e^2[ -m{yz — zy) +yZ-zY] +Sd^'%[— m{zx — xz) + zX — xZ^ + W^^\_—m{xy —yx) + xY—yX'\ = o ; as 8^^, 8^y, W^ are independent and arbitrary, their coefficients must vanish separately, and this gives the equations (6). 603. The equations of linear momentum, (4) or (s), admit of a further simplification, owing to the fundamental property of the centroid. By Art. 212, the co-ordinates x,y, z of the cen- troid satisfy the relations Mx = '2mx, My = 'l.my, Ms = Xms, where M = %nt is the whole mass of the body. Differentiating- these equations, we find Mx = '^mx, My = 'S.my, Ms = "Emz, and Mx = 'S.mx, My = "Imy, Ms = 2;«^, where Ic, y, i are the components of the velocity v, and x, y, z those of the acceleration y, of the centroid. The equations (4) or (5) can therefore be reduced to the form Mi = ^Mi =R„ Mj = ^M'y = R^, Ms = ^Ms = ^- (8) at at at whence Mj ^ —Mi = R\ (9) dt i. e., if the whole mass of the body be regarded as concentrated at the centroid, the effective force of the centroid, or the time- rate of change of its momentum, is equal to the resultant of all the external forces. It follows that the centroid of a rigid body moves as if it contained the whole mass, and all the external forces were applied at this point parallel to their original directions. 604. If, in particular, the resultant R vanish (while there may be a couple H acting on the body), we have by (8) and 6°S-] GENERAL PRINCIPLES. 385 (9) 7=0; hence v = const. ; /. e., if the resultant force be zero, the centroid moves uniformly in a straight line. This proposition, which can also be expressed by saying that, if 7? = O, the momentum Mv of the centroid remains constant, or, using the form (5) of the equations of motion, that the linear momentum of the body in any direction is constant, is known as the principle of the conservation of linear momentum, or the principle of the conservation of the motion of the centroid. 605. Let us next consider the equations of angular momen- tum, (6) or (7). To introduce the properties of the centroid, let us put .r — ,r = f , jj/ — 7 = 77, 2 — :? = if, so that ^, t), f are the - co-ordinates of the point (.*•, y, z) with respect to parallel axes through the centroid. The substitution oi z = x + ^, y =y -f rj, s=s + ^ and their derivatives in the expression ys — zy gives y'z — zy =yz—zy+y^ — zr] + 7]Z— ^J' + V^— ^V- To form 'S.in(yz — zy) we must multiply by m and sum through- out the body ; in this summation, y, i, y, z are constant and, by the property of the centroid, 'Emr] = o, 2wzf=0, tm-q^o, 'Znit= o. Hence we find tm{yz — sy) = 2 m{r)^— ^v) + M(yz — zy). The second term in the right-hand member is the angular momentum of the centroid about the axis of x (the whole mass M of the body being regarded as concentrated at this point), while the first term is the angular momentum of the body (in its motion relatively to the centroid) about a parallel to the axis of ;ir, drawn through the centroid. Similar relations hold for the angular momenta about the axes of y and z ; and as these axes are arbitrary, we conclude that the angular momentiLm of a rigid body about any line is equal to its angular momentum about a parallel through the centroid plus tlie angular momentum of the centroid about the former line. 2C 386 KINETICS OF THE RIGID BODY. [606. 606. Differentiating tiie above expression, we find at at Tiie first of tlie equations (7) can therefore be written at Now, if at any time t the centroid were taken as origin, so that j/ = o, i=o, this equation would reduce to the form which is entirely independent of the co-ordinates of the cen- troid. On the other hand, wherever the origin is taken, if the centroid were a fixed point, the same equation would be obtained. Similar considerations apply of course to the other two equa- tions (7). It follows that the motion of a rigid body relative to the centroid is the same as if the centroid were fixed. 607. If, in particular, the resultant couple H be zero for any particular origin (which will be the case not only when all external forces are zero, but whenever the directions of all the forces pass through the point 0), the equations (7) can be inte- grated and give 'l.m{ j/s — zj) = C^, '2m(zx — xz) = C^, '2m{xy — yx)= C^, (10) where C-^, C^, Cg are constants of integration. Hence, if the external forces pass through a fixed point, the angular momentnm of the body about any line through this point is constant ; if there are no external forces, the angular montentttm is constant for any line whatever. This is the principle of the conservation of angular momentum. 608. Another interpretation can be given to these equations. As shown in Arts. 97, 513, the quantities yz — sy, zx — xb, 609.] GENERAL PRINCIPLES. 387 xj> — yx can be regarded as sectorial velocities. Thus, if the radius vector, drawn from the origin to the particle m, be pro- jected on the j/.s'-plane, ji — sy is twice the sectorial velocity of this radius vector in the /^^-plane, ;} {yds — zdy) being the ele- mentary sector described in the element of time dt. Let us denote by dS^ the sum of all these elementary sectors for the various particles, each multiplied by the mass of the particle ; and similarly by dSy, dS^ the corresponding sums of the projections on the other co-ordinate planes. Then the equations (10) can be written in the form 2 5^= Cj, 2Sy= C2, 2 5^ = Cg. (11) Hence the proposition of Art. 607 might be called the principle of the conservation of sectorial velocities ; it is more commonly called the principle of the conservation of areas. The equations (11) can be integrated again and give, if the sectors be measured from the positions of the radii vectores at the time t=o, 609. If the radii vectores be projected on any plane through the origin whose normal has the direction cosines «, /3, 7, the sum of the elementary sectors described in this plane, each multiplied by the mass, will be ^5 = l(Ci« -h Q/S 4- C^'^)dt; hence 5 = J (Ci« -I- C^^ -1- C37)/. On the other hand, by (10), the angular momentum of the body about the normal of this plane has the expression Cj" -f Cg/S -t- C37, as it must be equal to the sum of the projec- tions on this normal of the angular momenta about the axes of co-ordinates, which can be regarded as vectors laid off on these axes. Now it is easy to see that this angular momentum C^a- -|- C^^ -t- C37, and hence the quantity 5 at a given time t, is greatest 388 KINETICS OF THE RIGID BODY. [6io. for the diagonal of the parallelepiped, whose edges are equal to Cj, C^, C^ along the axes, i. e., for the normal to the plane C^x + C^y + C^z = o. (12) For, the direction cosines of this normal are cc' = CJ D, ^'=C^/D, r^'=CJD, where D=-s/Q + Q+Q; and the quantity CyO. + C^^ + C37 can be written in the form ^(§« +§/8 + §7) = D{a'a + /3'/3 + 7V), where the quantity in parenthesis is the cosine of the angle between the directions («', 0, 7') and (a, /3, 7), and is therefore greatest when these directions coincide. The plane (12) about whose normal the angular momentum is greatest, and by projection on which the area S is made greatest, is called Laplace's invariable plane. As its equation is independent of t, it remains fixed. The normal of this plane is sometimes called the invariable line or direction. 610. Let us now return to the general case of the motion of a rigid body acted upon by any forces whatever. The propositions of Arts. 603 and 606 together establish the so-called principle of the independence of the motions of translation and rotation. In studying the motion of a rigid body it is pos- sible, according to this principle, to consider separately the motion of translation of the centroid, and the rotation of the body about the centroid. By Art. 603, the motion of the centroid is the same as that of a particle of mass M acted upon by all the external forces transferred parallel to themselves to the centroid. As the motion of a particle has been discussed in Chapter V., nothing further need be said about this part of the problem. By Art. 606, the motion of the body about the centroid is the same as if the centroid were fixed. The problem of the motion of a rigid body with a fixed point is therefore of great impor- tance ; it will, however, not be discussed in its generality in this 6i2.] GENERAL PRINCIPLES. 389 elementary work. The more simple special case of a rigid body with a fixed axis is treated in Section III. The solution of both these problems depends on the equations (6) or (7). 611. In d'Alembert's equation (3) it is of course allowable to substitute for the virtual displacements Bx, Bj/, Bz the actual dis- placements Ax, A.J/, Az of the particles in any small motion of a free rigid body, since these actual displacements are certainly compatible with the condition of rigidity. The equation can then be written '2m(xAx + yAy + zAz) = I.{XAx + YAy + ZAa). (13) After dividing by the time At of the small displacement {Ax, Ay, Az) and passing to the limit for At = o, the left-hand member of this equation becomes the exact time-derivative of the kinetic energy T=^\mv^=^lm{x^+f + k'^) (14) of the body. The right-hand member of (13) represents approxi- mately the elementary work done by the external forces during the small displacement. Hence integrating, say from ^ = o to t=t, we find the relation r- T^=l.\mv^ -^\mv^ = ^^{Xdx + Xdy + Zdz), (15) where the right-hand member represents the work W done by the external forces on the body during the time t. This equa- tion expresses the principle of kinetic energy and work, for a free rigid body : in any motion of tlie body, the increase of the kinetic energy is equal to the work done by the external forces. The relation (15) is often written in the equivalent differential form : dT=d^\ mv^ = Xdx -|- Ydy + Zdz = dW. 612. By introducing the co-ordinates of the centroid, i. e., by putting x = x-\-^, y=y + r], z = z + ^, as in Art. 605, the 390 KINETICS OF THE RIGID BODY. [613. expression for the kinetic energy assumes the form (since 2w/f = o, S;«T/ = o, S;«^= o): r = 2 1 m(x^ +j^ + s^)+-2 1 ;«(f + V^+^) where v is the velocity of the centroid and ti the relative velocity of any particle m with respect to the centroid. Thus, it appears that the kinetic energy of a free rigid body consists of tivo parts, one of which is the kinetic energy of tJie cen- troid (the whole mass being regarded as concentrated at this point), while the other may be called the relative kinetic energy zvith respect to the cetitroid. 613. By the same substitution the right-hand member of equation (13), i. e., the elementary work 2(A'A.r + FAj/ + ^As-), resolves itself into the two parts (AJ2X+ A7S F+ AiSZ) + 2(XA^ + FAt; + ZA^. The first parenthesis contains the work that would be done by all the external forces if they were applied at the centroid ; it is therefore equal to the kinetic energy of the centroid, that is, to t^(^\ M-iF). The equation of kinetic energy (13) reduces, there- fore, to the following : A(2 i mi?) = 2(XA| -f- FAt; -|- ZAf) ; (16) in other words, the principle of kinetic ejiergy holds for the rela- tive motion with respect to the centroid. 614. Impulses. The equations determining the effect of a system of impulses on a rigid body are readily obtained from the general equations of motion (4) and (6). We shall denote the impulse of a force /^by F. It will be remembered that the impulse F oiz. force F\^ its time integral; i. e., F= C Fdt. 615.] GENERAL PRINCIPLES. 391 We confine ourselves to the case when i' — t \s very small and F very large, in which case the action of the impulsive force F{Arts. 426, 427) is measured by its impulse F. If all the forces acting on a rigid body are of this nature, and the impulses of X, Y, Z during the short interval t' — t be denoted by X, V, Z, the integration of the equations (4) from / = / to t=t' gives S;«(x'-x)=SA', tm{y -y)=tY, lm(z'-z)=SZ, (17) where x, y, z denote the velocities of the particle m at the time t just before the impulse, and x\ y' , z' those at the time t^ just after the action of the impulse. Similarly the equations (6) give 2;«[j(i' - b) - z(y' -»] = l(yZ- z Y), tm[z{x' -x)- x{z' - i)] = t(zX - xZ),- ( 1 8) Xni\x{y'-y)-y{x'-x)'\ = 1{xY-yX\ 615. In determining the effect on a rigid body of a system of such impulses, any ordinary forces acting on the body at the same time are neglected because the changes of velocity pro- duced by them during the very short time t' ~ t are small in comparison with the changes of velocity x' — x,y — y, z' — z pro- duced by the impulses. If the impulse F oi an impulsive force F be defined as the limit of the integral I Fdt when t' — t approaches zero and ./^approaches infinity, it is strictly true that the effect of ordinary forces can be neglected when impulsive forces act on the body. If the rigid body be originally at rest, it will be convenient to denote by x, y, i the components of the velocity of the par- ticle m just after the action of the impulses. We may also denote by R the resultant of all the impulses, by H the result- ant impulsive couple for the reduction to the origin of co- ordinates, and mark the components of R and H by subscripts, 392 KINETICS OF THE RIGID BODY. [6i6- as in the case of forces. With these notations the effect of a system of impulses on a body at rest is given by the equations ^mx=R^, 1^my=Ry, 1-mk = R^, (19) ^m{y'z-zy)=H„ 1.m(z'x-xz)= Hy, 1.m{xy - yx) = H,. (20) In the equations (19) we have, of course, l.mx = Mx, Imy = My, 1m3 = Mz, where x, y, 'z are the components of the velocity of the centroid, and M is the mass of the body ; i. e., the momen- tum of the centroid is equal to the resultant impulse. The mean- ing of the equations (20) can be stated by saying that the angular momentum of the body about any axis is equal to the moment of all the impulses about the same axis. II. Moments of Inertia and Principal Axes. I. INTRODUCTION. 616. As will be shown in Section III., the rotation of a rigid body about any axis depends not only on the forces acting on the body, but also on the way in which the mass is distributed throughout the body. This distribution of mass is character- ized by the position of the centroid and by that of certain lines in the body called principal axes. It has been shown in Art. 212 that the centroid of a mass is found by determining the moments, or more precisely, the moments of the first order, '^jnx, ^iny, '^mz, of the mass with respect to the co-ordinate planes, i. e., the sums of all mass- particles m each multiplied by its distance from the co-ordinate plane. The principal axes of a mass or body can be found by deter- mining the moments of the second order, 'S^mx^, "Lmy"^, l.mz'^, 1,myz, "Lmzx, Itnxy of the mass with respect to the same planes. We proceed, therefore, to study the theory of such moments. 619-] MOMENTS OF INERTIA. 393 617. If in a rigid body the mass m of each particle be multi- plied by the square of its distance r from a given point, plane, or line, the sum .r, , , „ extended over the whole body, is called the quadratic moment, or, more commonly, the moment of inertia of the body for that point, plane, or line. If the body is not composed of discrete particles, but forms a continuous mass of one, two, or three dimensions, this mass can be resolved into elements of mass dm, and the sum Swzr^ becomes a single, double, or triple integral {r'^dm. Expressions of the form Imr^r^, or ^r-^r^din, where r^, r^ are the distances of m or of dm from two planes (usually at right angles), are called ■moments of deviation or products of inertia. 618. The determination of the moment of inertia of a con- tinuous mass is a mere problem of integration ; the methods are similar to those for finding the moments of mass of the first order required for determining centroids (Arts. 219-250), the only difference being that each element of mass must be multi- plied by the square, instead of the first power, of the distance. A moment of inertia is not a directed quantity ; it is not a vector, but a scalar ; indeed, it is a positive quantity, provided the masses are all positive, as we shall here assume. If the mass is homogeneous, the density appears merely as a constant factor; regarding the density in this case as =1, it is customary to speak of moments of inertia of volumes, areas, and lines. The moment of inertia of any number of bodies or masses for any given point, plane, or line is obviously the sum of the moments of inertia of the separate bodies or masses for the same point, plane, or line. 619. The moment of inertia "^mt^ of any body whose mass is J/= 2;« can always be expressed in the form 394 KINETICS OF THE RIGID BODY. [620. where ru is a length called the radius of inertia, arm of inertia, or radius of gyration. This length r^ is evidently a kind of average value of the distances r, its value being intermediate between the greatest ;-' and least r" of these distances r. For we have '2mr''^>1inr'^>^mr"'^, or, since 1.mr''^ = Mr''^, 1mr'^=Mi'^, ^mr"^^ = Mr"\ r'>r,>r". 620. As an example, let us determine the moment of inertia of a komogeiteotis rectilinear scgmetit (straight rod or wire of constant cross-section and density) for its middle point (or what amounts to the same thing, for a line or plane through this point at right angles to the segment). Let / be the length of the rod (Fig. 171), O its middle point, p" its density {i. e., the mass of unit length), x the distance OP il ■, X |dx Fig. 171. of any element dm=p"dx from the middle point. Observing that the moment of inertia for O of the whole rod AB is the sum of the moments of inertia of the halves AO and OB, and that the moments of inertia of these halves are equal, we have, for the moment of inertia /of AB, Jo and for the radius of inertia r^, since the whose mass is M=p"/, ' M ^2 621. Exercises. Determine the radius of inertia in the following cases. When nothing is said to the contrary, the masses are supposed to be homogeneous. (i) Segment of straight line of length /, for a perpendicular through one end. (2) Rectangular area of length / and width h : (a) for the side k ; {!>) for the side /; {c) for a line through the centroid parallel to the side h ; {d) for a line through the centroid parallel to the side /. 622.] MOMENTS OF INERTIA. 395 (3) Triangular area of base b and height /;, for a line through the vertex parallel to the base. (4) Square of side a, for a diagonal. (5) Regular hexagon of side a, for a diagonal. (6) Right cylinder or prism of height h, for the plane bisecting the height at right angles. (7) Segment of straight line of length /, for one end, when the density is proportional to the «th power of the distance from this end. Deduce from this : {a) the result of Ex. (i) ; (J>) that of Ex. (3) ; (c) the radius of inertia of a homogeneous pyramid or cone (right or oblique) of height h, for a plane through the vertex parallel to the base. (8) Circular area (plate, disk, lamina) of radius a, for any diameter. (9) Circular line (wire) of radius a, for a diameter. (10) Solid sphere, for a diametral plane. (11) SoUd elUpsoid, for the three principal planes. (12) Area of ring bounded by concentric circles of radii «,, a,^, for a diameter. (13) Area of the cross-section of a ±-iron : {a) for its line of sym- metry ; (U) for its base. (Dimensions as in Fig. 65, Art. 231.) (14) A rectangular door of width b and height h has a thickness S to a distance a from the edges, while the rectangular panel (whose dimen- sions are b — 2 a, h — 2 a) has half this thickness. Find the moment of inertia for a line through the centroid parallel to the side b. 622. The moment of inertia of any mass M for a point can easily be found if the moments of inertia of the same mass are known for any line passing through the point, and for the plane through the point perpendicular to the line. Let (Fig. 172) be the point, / the line, it the plane ; r, q, p the perpendicular distances of any particle of mass m from O, /, ir, respectively. Then we have, evidently, r^ = ^^ +/2. Hence, mul- tiplying by m, and summing over the whole mass M, ^mr^ = l.mq'^ + Sw/^ ; 396 KINETICS OF THE RIGID BODY. [623. or, putting 1nir'^ = Mr^, 1.7110^ = Mq^, '^mf = Mp^, where r^, q^,p^ are the radii of inertia for 0, I, tt, 623. The moment of inertia of any mass M for a line is equal to the sum of the moments of inertia of the same mass for any two rectangular planes passing through the hne. Thus, in particular, the moment of inertia for the axis of ;«• in a rectangular system of co-ordinates is equal to the sum of the moments of inertia for the ^■^-plane and xj-plzne. This fol- lows at once by considering that the square of the distance of any point from the hne is equal to the sum of the squares of the distances of the same point from the two planes. Thus, if q be the distance of any point {x, y, z) from the axis of x, we have q"^ =y^ + z^ ; whence limq"^ = "^my^ + 'S.mz'^. 624. It follows, from the last article, that the moment of inertia I^ of a plane area, for any line perpendicular to its plane, is t — t j^t if Ty, /j are the moments of inertia of the area for any two rectangular lines in the plane through the foot of the perpen- dicular line. 625. The problem ot finding the moment of inertia of a given mass for a line I' , when it is known for a parallel line I, is of great importance. Let S;«^2 ]3g ^-jjg moment of inertia of the given mass for the hne / (Fig. 173), 'l.mq''^ that for a parallel line /' at the distance d from /. The distances q, q' of any particle m from /, /' form with d a triangle which gives the relation Fig. 1 73. ^'2 = 5-2 + ^^ - 2 qd cos {q, d ). 627.J MOMENTS OF INERTIA. 397 Multiplying by m, and summing over the whole mass M, we l.tnq''^ = '2mq'^ + Md'^ — 2 d^mq cos {q, d). Now the figure shows that the product q cos {q, d) in the last term is the distance / of the particle ;« from a plane through / at right angles to the plane determined by / and /'. We have, therefore, ^^^,i ^ ^^^^2 + ^^2 _ 3 ^S;«/, (2) where the last term contains the moment of the first order 'S.mp = Mp of the given mass M for the plane just mentioned. If, in particular, this plane contains the centroid G of the mass M, we have %mp = o, so that the formula reduces to tmq'^ = tmq^ + Md^ (3) Introducing the radii of inertia q^', q^, this can be written 626. Similar considerations hold for the moments of inertia t.ntp'^, 1.mp'^ with respect to two parallel planes tt, tt' at the distance d from each other. We have, in this case, p' =p — d\ hence, ^^^^a = Jw/ + Md"^ - 2 dtmp, ^4) and if the plane tt contain the centroid G, lmp'^=l.mp'^ + Md'^. (5) 627. Of special importance is the case in which one of the lines (or planes), say / (tt), contains the centroid. The formulae (3), (3'), and (5) hold in this case; and if we agree to designate any line (plane) passing through the centroid as a centroidal line (plane), our proposition can be expressed as follows : The moment of inertia for aity liite {plane) is fonnd from the moment of inertia for the parallel centroidal line {plane) by adding to the latter the product Md'^ of the whole mass into the square of the distance of the lines {planes). It will be noticed that of all parallel lines (planes) the centroidal line (plane) has the least moment of inertia. 398 KINETICS OF THE RIGID BODY. [628. 628. Exercises. Determine the radius of inertia of the following homogeneous masses : (i) Rectangular plate of length / and width h, for a centroidal line perpendicular to its plane. (2) Area of equilateral triangle of side a : (a) for a centroidal line parallel to the base ; (d) for an altitude ; (<:) for a centroidal Hne per- pendicular to its plane. (3) Circular disk of radius a : (a) for a tangent ; (i) for a line through the center perpendicular to the plane of the disk ; (c) for a perpendicular to its plane through a point in the circumference. (4) Solid sphere, for a diameter. (5) Area of ring bounded by concentric circles of radii «,, ^2, for a line through the center perpendicular to the plane of the ring. (6) Right circular cylinder, of radius a and height h : (a) for its axis ; (i) for a generating line ; (c) for a centroidal line in the middle cross-section. (7) By Ex. (3) (6), the moment of inertia of the area of a circle of radius a, for its axis (i. e., the perpendicular to its plane, passing through the center), is /= J ira'^. Differentiating with respect to a, we find : ,, a/ , , — = 2 ira" = 2 7r« • a j ^a hence, approximately for small Aa : A7= 2 Tra^Aa = 2 waAa • a^. This is the moment of inertia of the thin ring, of thickness Aa, for its axis. (Comp. Ex. (5).) If the constant surface density (Art. 214) of the circle be p\ we have I=\(l-Ka'^; hence A/= 2 Trap'Aa • a^, where p'A« is the linear density p" of the ring. (8) Apply the method of Ex. (7) to derive from Ex. (4) the moment of inertia of a thin spherical shell, of radius a and thickness Aa, for a diameter. (9) Area of ellipse : (a) for the major axis ; {b) for the minor axis ; {f) for the perpendicular to its plane through the center. (10) Solid ellipsoid, for each of the three axes. 629.] MOMENTS OF INERTIA. 399 (11) Area of the cross-section of a T-iron, for a centroidal line par- allel to the flange. (Comp. Art. 621, Ex. (13), and Art. 231.) (12) Area of the cross-section of a symmetrical double T-iron, width of flanges 2 b, thickness of flanges 8, height of web a, thickness of web 2 S (Fig. 69, Art. 241, with b' = b) ; for the two axes of symmetry, and for a centroidal line perpendicular to its plane. (13) Wire bent into an equilateral triangle of side a, for a centroidal line at right angles to the plane of the triangle. (14) Paraboloid of revolution, bounded by the plane through the focus at right angles to the axis, for the axis. (15) Anchor-ring, produced by the revolution of a circle of radius a about a line in its plane at the distance b (> a) from the center, for the axis of revolution. (16) Prove that the moment of inertia of the solid generated by the revolution of any plane area about a line /, situated in its plane, but not intersecting it, is I = M {j, q^ -\- IP-') , where b is the distance of the centroid of the area from /, and q the radius of inertia of the area for the centroidal line parallel to /. (17) Show that the moment of inertia of a homogeneous triangular plate for the centroidal line parallel to the base is equal to that of one eighth of the whole mass concentrated at the vertex, or to that of one half the whole mass concentrated at the base. (18) Find the radius of inertia of a parallelogram whose sides a, b include an angle B, for the centroidal line perpendicular to its plane. (19) Find the radius of inertia of the area of a trapezoid (parallel sides a, b, height h) for the centroidal line parallel to the parallel sides. (20) Prove that the radius of inertia q of any homogeneous right prism or cylinder, for the centroidal line perpendicular to the axis {i. e., the line joining the centroids of the bases), can be found from the for- mula if = q^ -{- q^j where q^ is the radius of inertia of the axis, q^ that of the middle cross- section, for the same centroidal line. 2. ELLIPSOIDS OF INERTIA. 629. The moments of inertia of a given mass for the different lines of space are not independent of each other. Several ex- amples of this have already been given. It has been shown, in 400 KINETICS OF THE RIGID BODY. [630. particular (Art. 625), that if the moment of inertia be known for any line, it can be found for any parallel line. It follows that if the moments be known for all lines through any given point, the moments for all hnes of space can be found. We now pro- ceed to study the relations between the moments of inertia for all the Hnes passing through any given point 0. 630. It will here be convenient to refer the given mass Mtoa rectangular system of co-ordinates with the origin at the point O. Let X, y, z be the co-ordinates of any particle m of the mass ; and let us denote hy A, B, C the moments of inertia of M for the axes of x, y, s; by A', B' , C those for the planes yz, zx, xy ; by D, E, F the products of inertia (Art. 617) for the co-ordinate planes ; i. e., let us put A = '^m{y'^ + z^). A' = ^mx^, D=1,myz, B = ^m{z^ + x^), B' = ^my'^, E = l^mzx, (6) C ='S^m{x'^ + y"^), C = '2,mz'^, F='Lmxy. 631. These nine quantities are not independent of each other. We have evidently A = B' + C', B=C'+A', C=A'+B'; hence, solving for A', B', C , A' = \{B+C-A), B' = i(C+A-B), C' = ^(A + B-C). The moment of inertia for the origin is 2;«r2 = ^m(x^ +/ + 2'^)= A' + B' + C = 1{A + B + C). (7) 632. The moment of inertia / for any line through O can be expressed by means of the six quantities A, B, C, D, E, F; and the moment of inertia T' for any plane through O can be ex- pressed by means of A' , B', C , D, E, F. Let TT (Fig. 174) be any plane passing through O; /its normal ; 633-] MOMENTS OF INERTIA. 401 l/ia,^,y) Fig. 174. a, /3, 7 the direction cosines of /; and, as before (Art. 622), /, g, r the distances of any point {x, y, z) of the given mass from ir, I, and O, respectively. Then, projecting the closed polygon formed by r, x, y, z on the line /, we have p = ax -\- ^y ■\- 'iz ; hence, squaring, multiplying by m, and summing over the whole mass, we find t^l.mx'- + ^'^my'^ + 'f'Lmz'^ + 2 ^ /z > -4. and hence q\> qi> q^, q must be ^ q^/q^ and ^ q^lqx- As long as q is less than the middle semi-axis (f Jq^ of the ellipsoid, the axis of the cone coincides with the axis of x ; but when q > q^ Iq^, the axis of z is the axis of the cone. For q = q^ Iq^, i. e., q = q-i, the cone (15') degenerates into the pair of planes {q^— q2)x'^ — {q^— q^)z'^ = 0. These are the planes of the central circular (or cyclic) sections of the ellipsoid ; they divide the eUipsoid into four wedges, of which one pair contains all the equimomental cones whose axes coincide with the greatest axis of the ellipsoid, while the other pair contains all those whose axes he along the least axis of the eUipsoid. 643-] MOMENTS OF INERTIA. 407 642. There is another ellipsoid closely connected with the theory of principal axes ; it is obtained from the momental ellipsoid by the process of reciprocation. About any point O (Fig. 175) taken as center let us describe a sphere of radius c, and construct for every point P its polar plane it with regard to the sphere. If P describe any surface, the plane tt will envelop another surface which is called the polar reciprocal of the former surface with regard to the sphere. Let Q be the intersection of OP with -k, and put OP=p, OQ = q; then it appears from the figure that pq = A (16) 643. It is easy to see that the polar reciprocal of the momental ellipsoid (n') with respect to the sphere of radius e is the elhpsoid , , , ?1 ?2 ?3 To prove this it is only necessary to show that the relation (16) is fulfilled for p as radius vector of (11'), and q as perpendicular to the tangent plane of (17). Now this tangent plane has the equation i?l 92 ?3 hence we have for the direction cosines a, fi, y, and for the length q, of the perpendicular to the tangent plane These relations give qia = (x/qi)q, qS = {yh2) ^'^d this is what we wished to prove. 408 KINETICS OF THE RIGID BODY. [644- 644. The surface (17) has variously been called the ellipsoid of gyra- tion, the ellipsoid of inertia, the reciprocal ellipsoid. We shall adopt the last name. The semi-axes of this ellipsoid are equal to the princi- pal radii of inertia at the point O. The directions of its axes coincide with those of the momental ellipsoid ; but the greatest axis of the former coincides with the least of the latter, and vice versa. By comparing the equations (12') and (18) it will be seen that q is the radius of inertia of the line (a, /3, y) on which it lies. Thus, while the radius vector OP = p of the momental ellipsoid is inversely propor- tional to the radius of inertia, i. e., p = ^ Jq, the reciprocal ellipsoid gives the radius of inertia q for a line I as the segment cut off on this line by the perpendicular tangent plane. 645. We are now prepared to determine the moment of inertia for any line in space. Let us construct at the centroid G of the given mass or body both the momental eHipsoid and its polar reciprocal. The former is usually called the central ellipsoid of the body; the latter we may call the fundamental ellipsoid of the body. As soon as this fundamental ellipsoid ?i f2 qi is known, the moment of inertia of the body for any line whatever can readily be found. For, by Art. 644, the radius of inertia q for any line /o passing through the centroid is equal to the segment OQ cut off on the line /« by the perpendicular tangent plane of the fundamental elhpsoid ; and for any line / not passing through the centroid, the square of the radius of inertia can be determined by first finding the square of the radius of inertia for the parallel centroidal line 4 and then, by Art. 627, adding to it the square of the distance d of the centroid from the line /. 646. In the problem of determining the ellipsoids of inertia for a given body at any point, considerations of symmetry are of great assistance, just as in the problem of finding the centroid (comp. Art. 244). Suppose a given mass to have a plane of symmetry; then taking this plane as the jcz-plane, and a perpendicular to it as the axis of 647-] MOMENTS OF INERTIA. 409 X, there must be, for every particle of mass m, whose co-ordinates are x,y, z, another particle of equal mass m, whose co-ordinates are —x,y, z. It follows that the two products of inertia Imzx and 'S,mxy both vanish, whatever the position of the other two co-ordinate planes. Hence, any perpendicular to the plane of symmetry is a principal axis at its point of intersection with this plane. Let the mass have two planes of symmetry at right angles to each other ; then taking one as yz-p\a.ne, the other as z:e-plane, and hence their intersection as axis of x, it is evident that all three products of inertia vanish, 'S,myz = o, ^mzx = o, 'Xfnxy = o, wherever the origin be taken on the intersection of the two planes. Hence, for any point on this intersection, the principal axes are the line of intersection of the two planes of symmetry, and the two per- pendiculars to it, drawn in each plane. If there be three planes of symmetry, their point of intersection is the centroid, and their hnes of intersection are the principal axes at the centroid. 647. Exercises. Determine the principal axes and radii at the centroid, the central and fundamental eUipsoids, and show how to find the moment of inertia for any line, in the following Exercises (i), (2), (3). (i) Rectangular parallelepiped, the edges being 2 a, 2 b, 2 c. Find also the moments of inertia for the edges and diagonals, and specialize for the cube. (2) Ellipsoid of semi-axes a, b, c. Determine also the radius of inertia for a parallel / to the shortest axis passing through the extremity of the longest axis. (3) Right circular cone of height h and radius of base a. Find first the principal moments at the vertex ; then transfer to the centroid. (4) Determine the momental ellipsoid and the principal axes at a vertex of a cube whose edge is a. (5) Determine the radius of inertia of a thin wire bent into a circle, for a line through the center inclined at an angle « to the plane of the circle. 4IO KINETICS OF THE RIGID BODY. [648- (6) A peg-top is composed of a cone of height JI and radius a, and a hemispherical cap of the same radius. The pointed end, to a distance /i from the vertex of the cone, is made of a material three times as heavy as the rest. Find the moment of inertia for the axis of rotation ; specialize for h ^= a ^ \ H. (7) Show that the principal axes at any point P, situated on one of the principal axes of a body, are parallel to the centroidal principal axes, and find their moments of inertia. (8) For a given body of mass il/ find the points {spherical points of inertia) at which the momenta! ellipsoid reduces to a sphere. (9) Determine a homogeneous ellipsoid having the same mass as a given body, and such that its moment of inertia for every line shall be the same as that of the given body. (10) For a given body M, whose centroidal principal radii are ^1, q^, ^3, determine three homogeneous straight rods intersecting at right angles, of such lengths 2 a, 2 b, 2 c, and such hnear density p", that they have the same mass and the same moment of inertia (for any line) as the given body. 3. DISTRIBUTION OF PRINCIPAL AXES IN SPACE. 648. It has been shown in the preceding articles how the principal axes can be determined at any particular point. The distribution of the principal axes throughout space and their position at the different points is brought out very graphically by means of the theory of con- focal quadrics. It can be shown that the directions of the principal axes at any point are those of the principal diameters of the tangent cone drawn from this point as vertex to the fundamental ellipsoid ; or, what amounts to the same thing, they are the directions of the normals of the three quadric surfaces passing through the point and confocal to the fundamental ellipsoid. In order to explain and prove these propositions it will be necessary to give a short sketch of the theory of confocal conies and quadrics. 649. Two conic sections are said to be confocal when they have the same foci. The directions of the axes of all conies having the same two points S, S' as foci must evidently coincide, and the equation of such conies can be written in the form 652.] MOMENTS OF INERTIA. 41I where X is an arbitrary parameter. For, whatever value m^y be assigned in this equation to A, the distance of the center O from either focus will always be Va" + X — {b- + X) = Va' — b' ; it is therefore constant. 650. The individual curves of the whole system of confocal conies represented by (19) are obtained by giving to X any particular value between — 00 and + co ; thus we may speak of the conic X of the system. For X = o we have the so-called fundamental conic x'/a' +f/l^ = i ; this is an ellipse. To fix the ideas let us assume a> b. For all values of X> — ^-, i.e., as long as — i^-c,vre have OS3 > OSy ; i. e., Si, S^' he between S3, Ss on the axis of x. The same holds for hyperboloids. 655. Two quadric surfaces are said to be confocal when their princi- pal sections are confocal conies. Now this will be the case for two quadric surfaces whose semi-axes are a^, b^, c-^, and a^^, b^, C2, if the directions of their axes coincide and if (72— /^^— «2_/, 2 i2_.2_z2_^2 -2 ,2_^2 ^2 «! Oi — U2 O2 , C^i ti — O2 — 1 2 , «! — Ci ^ U2 — C2 . Writing these conditions in the form 657-] MOMENTS OF INERTIA. 413 a2 — a-c = l>2 — l>i = ci — Ci, say = X, we find a^ = a^ + \, bi — b} + X, (Tj^ = (Ti^ + X. Hence the equation ct:^ 1,2 72 (Z^ + X ^2 + X f^ + X where X is a variable parameter, represents a system of confocal quadric surfaces. 656. As long as X is algebraically greater than — c^, the equation (20) represents ellipsoids. For X = — i:^ the surface collapses into the interior area of the ellipse in the jy-plane whose vertices are the foci 8-2, S2 and S^, S3'. For as X approaches the hmit — r, the three semi- axes of (20) approach the hmits -Va- — r, V(5^ — r, o, respectively. This hmiting ellipse is called the focal ellipse. Its foci are the points ^i, Si, since d^ — r — (b^ — c^) = a'- — <5l When X is algebraically < — <^, but > — a^, the equation (20) repre- sents hyperboloids ; for values of X < — a^ it is not satisfied by any real points. As long as —b'^ Ist fall from the origin O on these tangent planes tto, ttj^, are given by the relations (the proof being the same as in Art. 643) g,^=aV + b'li' + c-'-y\ (21) g,' = {a' + A) «^ + {b^- + A) /3^ + {c^ + A) /, (22) where «, fi, y are the direction cosines of the common normal of the planes TTo, TT^. Subtracting (21) from (22), we find, since a^ + ^'^ + ■/= i , ?a'-?o'=^; (23) i. e., t/te parameter A of any one of the confocal surfaces (20) is equal to the difference of the squares of the perpendiculars let fall from the common center on any tangent plane to the surface A, and on the parallel tangent plane to the fundamental ellipsoid A = o. 659. Let us now apply these results to the question of the distri- bution of the principal axes throughout space. We take the centroid G of the given body as origin, and select as fundamental ellipsoid of our confocal system the polar reciprocal of the central ellipsoid, i.e., the elUpsoid (17) formed for the centroid, for which the name " fundamental ellipsoid of the body " was introduced in Art. 645. Its equation is x^ y 2^ ?i qi qi if ?i> li, ?3 are the principal radii of inertia of the body. The radius of inertia q^ for any centroidal line /„ can be constructed (Art. 644) by laying a tangent plane to this ellipsoid perpendicular to the hne /„; if this hne meets the tangent plane at Q^ (Fig. 176), then qo = GQa. Analytically, if a, /3, y be the direction cosines of 4 qo is given by formula (21) or (12'). 662.] MOMENTS OF INERTIA. 415 660. To find the radius of inertia q for a ]ine /, parallel to 4, and passing through any point P, we lay through P a plane 77;^, perpendicular to /, and a parallel plane ito, tangent to the fundamental ellipsoid ; let Q)^, (?o be the intersections of these planes with the centroidal line /q. Then, putting GQ^ = q„, GQ>, = g^, GP==r, PQ>^ = d, we have, by q^ = qi-^d^. The figure gives the relation d^ = r'^ — q^, which, in combination with (23), reduces the expression for the radius of inertia for the line / to the simple form f = r''-\. (24) 661. The value of r^ — \, and hence the value of q, remains the same for the perpendiculars to all planes through P, tangent to the same quadric surface A : these per- pendiculars form, therefore, an equimomental cone at P. By varying A we thus obtain all the equimomental cones at P. The principal diame- ters of all these cones coin- cide in direction, since they coincide with the directions of the principal axes of the momental ellipsoid at P (see Art. 640) ;■ but they also coin- cide with the principal diam- Flg 176. eters of the cones enveloped by the tangent planes ir^. It thus appears that the prificipal axes at the point P coincide in direction with the prin- cipal diameters of the tangent cone from P as vertex to the fundamental ellipsoid x'/q,' + f/qi + zyqi = I . 662. Instead of the fundamental ellipsoid, we might have used any quadric surface \ confocal to it. In particular, we may select the con- focal surfaces Xi, A2, X3 that pass through P. For each of these the cone of the tangent planes collapses into a plane, viz., the tangent plane to the surface at P, while the cone of the perpendiculars reduces to a single line, viz., the normal to the surface at P. Thus we find that the prin- 4l6 KINETICS OF THE RIGID BODY. [663. cipal axes at any point P coincide in direction with the normals to the three qtiadric surfaces, con/ocal to the fundamental ellipsoid and passing through P. For the magnitudes of the principal radii q^, g^, q^ at P, we evidently have q\=r^-X„ ?/ = r^ - X„ q!=^r--K 663. Exercise. (i) The principal radii q^, q^, q^ of a body being given, find the equa- tion of the momental ellipsoid at any point P, referred to axes through this point P parallel to the principal axes of the body ; determine the directions of the principal axes at P, and show that these directions coincide with the normals of the three surfaces passing through P and confocal to the fundamental ellipsoid of the body. III. Rigid Body with a Fixed Axis. 664. A rigid body with a fixed axis has but one degree of freedom. Its motion is fully determined by the motion of any one of its points (not situated on the axis), and any such point must move in a circle about the axis. Any particular position of the body is, therefore, determined by a single variable, or co-ordinate, such as the angle of rotation. Just as the equi- librium of such a body depends on a single condition (see Art. 399), so its motion is given by a single equation. This equation is obtained at once by " taking moments about the fixed axis." For, according to the proposition of angular momentum (Art. 601), the time-rate of change of angular mo- mentum about any axis is equal to the moment of the external forces about this axis. Hence, denoting this moment by H and taking the fixed axis as axis of z, we have as equation of motion the last of the equations (7), Art. 601, viz., — '2m(xj — yx) = H. ( i ) dt 665. The angular fnomentum, '2,m{xy — yx), about the fixed axis can be reduced to a more simple form. For rotation of 666.] BODY WITH FIXED AXIS. 417 angular velocity m about the ^-axis we have (Art. 175) i-= — ooj, y = (iix, so that "^mixy — yx) = &)S;«(x^+_y2^ = (b . '^mr^ = /w, where r is the distance of the particle m from the axis and /= Inir'^ the moment of inertia of the body for this axis. This expression for the angular momentum can be derived without reference to any co-ordinate system. For evidently mar is the linear momentum of the particle m, mwr'^ is its moment, i. e., the angular momentum of the particle, about the axis ; and 'Lmwr^ = mXinr^ = la is the angular momentum of the body about the axis. It thus appears that, just as in translation the linear momen- tum of a body is the product of its mass into its linear velocity, so in the case of rotation the angular viomenUmi of the body about the axis of rotation is the product of its moment of inertia (for this axis) into the angular velocity. As regards the right-hand member of equation (i), the reactions of the axis need not be taken into account in form- ing the moment H\ for as these reactions meet the axis, their moments about this axis are zero. 666. Substituting /o) for '^m{xy—yx) in equation (i), and observing that the moment of inertia / about a fixed axis re- mains constant, we find the equation of motion in the form i. e., for rotation about a fixed axis, the product of the moment of inertia for this axis into the angular acceleration equals the mo- ment of the external forces about this axis ; just as, in the case of rectilinear translation, the product of the mass of the body into the linear acceleration equals the resultant force R along the line of motion: dii „ m — = K. dt 4l8 KINETICS OF THE RIGID BODY. [667. And just as the latter equation may serve to determine experi- mentally the mass of a body by observing the acceleration pro- duced in it by a given force R, e. g., the force of gravity (as in the gravitation system, Art. 262), so the former equation, (2), may serve to determine experimentally the moment of inertia of a body about a line /, by observing the angular acceleration pro- duced in the body when rotating about / under given forces. 667. For the kinetic energy of a body rotating with angular velocity <« about any axis, we have T='^\ mv^ = '% \ maP'r'' = ^ loP', (3) an expression which is again similar in form to that for the kinetic energy of a body in translation, viz., T=\mv^- 668. When the axis is fixed so that /is constant, the equation of motion (2), multipHed by a and integrated, say from ^ = to t=t, gives the relation \I<^~\I<^,^ = £Ho.dt, (4) which expresses the principle of kinetic energy and work, and might have been derived from the general formula (15), Art. 611. For, taking the axis of rotation as axis of z, we have in the present case dx= —wydt, dy^coxdt, ds = o, so that ^{Xdx + Ydy + Zdz) = t{x Y—yX) wdt = Hwdt ; this, with the value (3) for r, reduces equation (15) of Art. 611 to the above equa- tion (4). Conversely, the equation of motion (2) can be derived from the principle of kinetic energy and work, dT=dW {Art. 61 1) or d^ 70)2 = -EiXdx +Ydy + Zdz) = Hadt. If, in particular, H is constant, and we put a = dO/dt, (4) re- duces to l^(„2_„^2) = ^(^_^^)^ where ^ — ^g is the angle through which the body turns in the time t in which the angular velocity increases from tUg to w. 670.] BODY WITH FIXED AXIS. 419 669. A rigid body with a fixed horizontal axis is called a compound pendulum if the only external force acting is the weight of the body. The plane through axis and centroid will make, with the vertical plane (downwards) through the axis, an angle Q, which we raay take as angle of rotation, so that m= d9 /dt {¥ig. 177). The weights of the par- ticles, being all parallel and proportional to their masses, have a single resultant Mg pass- ing through the centroid G. Hence, if h be the perpendicular distance OG of the centroid from the axis, the moment of the external forces is H= — Mgh sin Q ; and if the radius of inertia of the body for the centroidal axis par- allel to the axis of rotation be q, the moment of inertia for the latter axis is I=M{q^ + h^). With these values the equation of motion (2) assumes the simple form ;sin6'. (5) Fig. 177. dP±^__gh_ dfl q^ + B " As shown in Art. 585, the equation of the simple pendulum of length / is ,25/ ^ — - = — 2^ sin 6. dfl I The two equations differ only in the constant factor of sin 6, and it appears that the motion of a compound pendulum is the saine as that of a simple pendnliim whose length is l=h^%. h (6) 670. The problem of the compound pendulum has thus been reduced to that of the simple pendulum. The length / is called the length of the equivalent simple pendulmn. The foot O (Fig. 177) of the perpendicular let fall from the centroid on the axis is called the center of suspension. If on the line OG a length OC=l be laid off, the point C is called the center of 420 KINETICS OF THE RIGID BODY. [671. oscillation. It appears, from (6), that G lies between O and C. The relation (6) can be written in the form h{l — h) = g^, or OG ■ GC= const. As this relation is not altered by interchanging and C, it follows that the centers of oscillation and suspension are inter- changeable ; i.e., the period of a compound pendulum remains the same if it be made to swing about a parallel axis through the center of oscillation. 671. Exercises. (i) A pendulum, formed of a cylindrical rod of radius a and length L, swings about a diameter of one of the bases. Find the time of a small oscillation. (2) A cube, whose edge is a, swings as a pendulum about an edge. Find the length of the equivalent simple pendulum. (3) A circular disk of radius r revolves uniformly about its axis, making 100 rev./min. What is its kinetic energy ? (4) A fly-wheel of radius r, in which a mass, equal to that of the disk in Ex. (3), is distributed uniformly along the rim, has the same angular velocity as the disk. Neglecting the mass of the nave and spokes, determine its kinetic energy, and compare it with that of the disk. (5) A fly-wheel of 12 ft. diameter, whose rim weighs 10 tons, makes 50 rev./min. Find its kinetic energy in foot-tons. (6) A fly-wheel of 10 ft. diameter, weighing 5 tons, is making 40 revolutions when thrown out of gear. In what time does it come to rest if the diameter of the axle is 6 in. and the coefficient of friction iu. = 0.05? (7) A fly-wheel of radius r and mass m is making N rev./min. when the steam is shut off. If the radius of the shaft be r', and the coefficient of friction ^, find after how many revolutions the wheel will come to rest owing to the axle friction. (8) A homogeneous straight rod of length / is hinged at one end so as to turn freely in a vertical plane. If it be dropped from a horizontal position, with what angular velocity does it pass through the vertical position? (Equate the kinetic energy to the work of gravity.) 671.] BODY WITH FIXED AXIS. 42 1 (9) A homogeneous plate whose shape is that of the segment of a parabola bounded by the curve and its latus rectum swings about the latus rectum which is horizontal. Find the length of the equivalent simple pendulum. (10) A fly-wheel of 10 ft. radius makes 45 rev./min. Its rim (re- garded as a circular line) weighs 8000 lbs., while each of the 10 spokes (regarded as a straight line) weighs 200 lbs. Find the kinetic energy stored in the wheel. (11) Find the work that would have to be done by the engine to increase the number of revolutions to 60 per minute for the fly-wheel in Ex. (10). (12) When q is given while /and h vary, the equation (6) represents a hyperbola whose asymptotes are the axis of / and the bisector of the angle between the (positive) axes of h and /. Show that 4,;^ = 2 ^ for h=q ;■ also that /, and hence the period of oscillation, can be made very large by taking h either very large or very small. The latter case occurs for a ship whose metacenter (which plays the part of the point of suspen- sion) lies very near its centroid. (13) Explain how to determine experimentally the moment of inertia of a body for any line / by observing its small oscillations about / as axis. (14) Find the horse-power required to keep a wheel weighing k tons rotating with N rev./min., the radius of the axle being / ft. and the coefficient of friction /x. (15) If q be the radius of inertia of the wheel in Ex. (14) (including the shaft and other attachments), for its axis, determine : (a) after how many revolutions, {p) in what time, the wheel will come to rest if left to itself. (16) A homogeneous circular disk, i ft. in diameter and weighing 25 lbs., is making 240 rev./min. when left to itself. Determine the constant tangential force applied to its rim that would bring it to rest in I min. (17) A cord is wrapped around the horizontal axle of a heavy wheel ; the free end of the cord passes over a fixed vertical pulley and carries a mass of 75 lbs. It is found that when the mass has descended 10 ft. the wheel is making 67 rev./min. What is the moment of inertia of the wheel ? 422 KINETICS OF THE RIGID BODY. [672. (18) A homogeneous circular hoop of radius a is suspended by a point in its rim. Find the length of the equivalent simple pendulum when the hoop swings : (a) in its own plane, (b) at right angles to this plane ; {c) determine the ratio of the periods in the two cases. 672. Small Oscillations Due to Torsion. Let a homogeneous straight bar be suspended horizontally by means of a stout wire attached to its middle point. If the bar be turned, in a horizontal plane, out of its position of equilibrium, the torsion produced in the wire tends to bring the bar back to its original position ; thus, vibrations about this position are set up. The torsional stress of the wire acts on the bar as a couple in the horizontal plane, and within certain hmits, even for oscillations that are not very small, the moment of this couple can be taken proportional to the angle through which the bar is turned, say = f]id. The motion of the bar about the vertical axis of the wire is therefore given by the equation 4>-.^> (7) where / is the moment of inertia of the bar about the vertical centroidal axis, and /w, the moment of the torsional couple that would turn the bar through I radian. Comparing with the equation for the small oscillations of a pendulum, it appears that the length / of the equivalent simple pendulum is and hence the time of a complete oscillation T=2^yj^. (8) 673. This formula can be used to determine experimentally the moment of inertia of the bar. To eliminate the unit torsion moment /x, the time of oscillation is generally observed first for the bar alone, then for the bar loaded, /. c, in connection with pieces whose moment of inertia is known or easily determined. If the moment of inertia of the added pieces be /', the time of oscillation of the loaded bar T', we have 675-] BODY WITH FIXED AXIS. 423 Dividing this equation by (8), we find : T" I ' whence, J=I' -• (q) 674. Magnetic Needle. The moment Af of a. magnetic needle is defined as the moment of the couple acting on the needle when placed in a uniform field of unit strength, at right angles to the lines of force. The forces of this couple can be regarded as appHed at the poles and as equal to the pole-strengths. If the strength of the field is not i, but H, the moment of the couple acting on the needle is MN ; and if the needle is placed at an angle 6 with the lines of force, the moment of the so-called restoring, or directing, couple that tends to place the axis of the needle parallel to the Hnes of force is easily seen to be = MH%m e. The equation of motion for the oscillations of a magnetic needle, pivoted so as to turn freely in a horizontal plane, when the needle is placed in any position in the earth's magnetic field, is therefore I — = -Afffsme, (10) dt' ^ ^ where If is the strength of the earth's horizontal field. 675. When the oscillations are small, sin 5 can be replaced by 6, and we find for the length of the equivalent simple pendulum MB and for the time of one complete oscillation T^2Tr\-^- (11) This formula is used for determining MH by observing T; as the quotient M/H can be found from deflection observations, the formula serves ultimately for the determination of the earth's horizontal in- tensity H. 424 KINETICS OF THE RIGID BODY. [676- 676. The dynamical meaning of the radius of inertia of a body for any line / appears by considering that, according to equation (2), if the body be set revolving about the line / as a fixed axis, under the action of any forces, the motion, t. e., the variation of the angular velocity, depends only on the moment H of the forces and the moment of inertia /. The motion remains, therefore, the same when the body is replaced by any other body having the same moment of inertia for /, provided the moment H of the forces remains the same. Thus, in study- ing the motion of a rigid body about a fixed axis /, if the mass of the body be Ma.nd its moment of inertia 1= Mq^, the body might be replaced by a ring, or a cylindrical shell, of mass M and radius q about the axis /, or by a single particle of mass M placed at the distance q from the axis. Thus, the radius of inertia q receives its dynamical interpretation as that distance from the axis at which the mass M of the body must be concen- trated to produce the same motion of rotation as the actual motion. It should be carefully observed, however, that, in replacing the body by another of equal moment of inertia, the reactions of the axis, and hence the pressure on the axis, are in general changed, as will be explained later (Art. 699). 677. Reduced Mass. In substituting for a rotating body another of equal moment of inertia for the axis of rotation, it is not even necessary to keep the mass the same (as was done in Art. 6^6). Thus, a body of mass M and moment of inertia I=Mq'^ for the fixed axis of rotation / may be replaced by a mass M^ distributed uniformly along a circle of radius r, about / as axis, provided that M^ and r be selected so that M^-t^ = Mq^ = /. This equivalent mass M^ is called the mass reduced to the dis- tance r from the axis. As regards the -external forces, since only their moment H about the fixed axis is essential, they can be replaced by a single force F, perpendicular to the axis /, at such a distance / from it 679-] BODY WITH FIXED AXIS. 425 that its moment Fp is equal to H. By thus putting H= Fp the external forces are said to be reduced to the distance p from the axis. 678. If both the mass M of the rotating body and the exter- nal forces acting on it be reduced to the same distance r from the axis /, the equation of motion (2), Ida/dt= H, assumes the simple form ^^y „ Mr-r-^F; and as rdcuj dt is the linear (tangential) acceleration of a point at the distance r from the axis, the equation is exactly the same as that for rectilinear translation : ntdv/dt = F. The following exercises will show in what way the idea of reduced mass can be used to advantage. 679. Exercises. (i) Reduce the mass of a homogeneous circular plate to its circum- ference, for rotation about the axis of the plate. (2) Reduce the mass of a homogeneous straight rod of length 2 a to its middle point, for rotation about the perpendicular through one end. (3) Reduce the mass of a homogeneous solid sphere of radius a : (a) to the surface, for rotation about a diameter ; {d) to the center, for rotation about a tangent. (4) Two masses »?i, ;«2 hang from a weightless cord slung over a fixed pulley of mass m ; show how to determine the effect of the inertia of tlie pulley on the motion. Denoting by / the acceleration of the cord, by T^, T^ its tensions above m^ and m^, we have as equations of motion of my and m^ sepa- and for the motion of the pulley : m ■ -7, = T-iT — T^r, at where q is the radius of inertia, r the radius, of the pulley. If m^ be the mass of the pulley reduced to its circumference, the latter equation becomes da, . _, _ m,r-- = mj = 71 — 7^. at 426 KINETICS OF THE RIGID BODY. [679- Substituting for T^, T^ their values from the first two equations, and solving for j, we find : _ „ „, / = r ^g, where mr = \in if the pulley can be regarded as a homogeneous disk, and m^ = m if its mass be regarded as concentrated in the rim. For the tensions we find : 7; = , / ■ »hg, ^2 = , ' , ' • m,g. mi + m.2 + m^ mi + m.2+ m^ Compare the results of Art. 464, where the mass of the pulley was neglected. (5) In Ex. (4), show how to take account of axle- friction. The total pressure on the axle being T1 + T2 + mg, the frictional force is F = ii.{Ti-\- T2 + mg) ; it acts at an arm equal to the radius p of the axle. To reduce this force to the circumference of the pulley we have to find F, from ^_ .r=F.p = ^p{Ti + T, + mg). We then have (see Ex. (4)) : mJ=Ti-T,-F,. = T,-T,~p.P{Ti + T, + mg); r whence, substituting for 71 and Z'2 and solving fory : J = rn^ IX (m^ + OTg + m)p/T tn, — m, — u.(}n. + m, + m)a r m-i + OT2 + OTr — /"•('■''i — ''^2) pI '^ (6) The apparatus called "wheel and axle " can be regarded as con- sisting of two equal masses m suspended over two rigidly connected co- axial pulleys, with different radii ;- and p<.r ; find the acceleration j of the mass hanging from the larger pulley. Neglecting axle friction, we have 7] = mg — mj, T-i = mg-{- m • {fi/r)j, since the accelerations are evidently as r : p. Hence, m,r^ --^^T^r- T^p, at or smct rdio/dt= J, ;«(i--n/r) m(i + p-/r^)+ m,. In particular, for r = 2 p, we findy= 2 mg/(^ m + 4 m^. (7) A wheel, weighing 40 lbs., radius of inertia for its axis i ft., has a cord wrapped around its axle which is 6 in. in diameter ; this cord passes over a pulley, and has a weight of 12 lbs. suspended from its end. Find in what time the weight will descend 15 ft. 68i.] BODY WITH FIXED AXIS. 427 680. Many pieces of machinery are bodies turning about fixed axes. Thus in most machines we find shafts, or axles, kept in nearly uniform rotation by a driving force supplied by a prime mover (steam engine, water wheel, turbine, elec- tric motor, etc.); to the shaft we find rigidly attached, and turning with it, cranks, eccentrics, wheels, pulleys, etc., which, at a certain distance from the axis of the shaft, have to overcome a resistance and thus do useful work, such as lifting a hammer, or a rack, when the work is done against gravity, or shaping material, as in a planer, when the work is done against the cohesion of the material to be planed, and so forth. The resistance is generally directed at right angles to the axis of the shaft, this being evidently the most effective way of doing work in this case. 681. The work done in one revolution by a uniformly revolv- ing shaft against a constant perpendicular resistance R at the distance r from the axis of the shaft (Fig. 178) is = 2 irr • R, where r is called the leverage of the force R. J) [nx ] -U-5 The work done as the shaft turns through any angle 9 is Or-R; and as Rr is the ^'e- '^s- moment of R about the axis, we may say that the work done against the resistance is eqital to the moment of the resistance multiplied by the angle of rotation. The rate of work, or power, is then equal to the moment of the resistance miiltiplied by the angular velocity. This presupposes that the resistance is of constant magni- tude, always at right angles to and at the same distance from the axis, and that the rotation is uniform. Otherwise we have to fall back on the general expressions of Arts. 428 KINETICS OF THE RIGID BODY. [682. 682. Steam Engine. In the steam engine the pressure P of the steam on the piston (Fig. 179) is first resolved at the cross-head F into a component P — P sec 4> along the connecting rod PQ and a com- ponent /'" = /'tan <^at right angles to the guides of the cross-head. Fig. 179. Only the former of these components P' acts on the crank ; it can be resolved at the crank pin Q into a normal component along the crank QO : P„' = P' cos (e + ct>) = Psec cos {e + (t>) = P (cos ^ — sin ^ tan ), and a tangential component, at right angles to the crank, P,' = P' sin (6 + cl>) = P (sin 61 -H cos 61 tan ). As PJ passes through the center of rotation O it has no turning effect. The rotation of the crank is therefore due entirely to the force P,' ; its moment about O is H = PI a = a (sin Q + zo%B tan <^) P, where a= 0(2 is the length of the crank. Even if P were constant (which it is not, see Arts. 469-471), this moment H would vary with the angles Q and <^. The angle <^ can be eliminated, its variation depending on that of the angle 0. For, with OQ^a, PQ = /, the triangle OPQ gives sin is generally small, the length of the connecting rod being usually at least 3 or 4 times that of the crank, we can substitute sin (f> for tan <^ so that we obtain for the turning tnoment the simple approximate expression H = a (sin Q + — sin 2 ff) P. 2 m 683. Fly- Wheel. As the turning moment H varies in the course of the stroke, it follows that even if the resistance were constant, the angular velocity u of the crank, or the hnear velocity u of its end Q, will not remain constant. In order to diminish this irregularity in the rotation of the crank as much as possible, a heavy fly-wheel is fixed on the crank shaft. The function of the fly-wheel is not to create energy, but to store and distribute it. During that part of the stroke during which the turning moment ^is greater than its mean value for one stroke, energy is being accumulated in the mass of the fly-wheel ; and when H is less than its mean value, part of this energy is consumed in doing useful work and thus making up for the lack of turning moment. 684. It must here suffice to discuss a simple ideal case. Assuming the connecting rod of infinite length so that = 0, and the driving force P' = P constant, its tangential component (Fig. 180) is P,' =Psin6, and the turning moment is If = Pa sin d. The work done by the driving force in a double (forth and back) stroke is evidently W= 2- P- 2a = 4Pa. Fig. 180. If, for the sake of simplicity, we assume the resistance P to be overcome by the crank as constant in magnitude and always tangential to the crank circle, the work of this resistance in a double stroke is = (2 • 2 ira. This must be equal to the work of the driving force, so that ^ ^ 4 Pa = 2 TT Qa, whence Q = -P=o.6:iT P. 430 KINETICS OF THE RIGID BODY. [68s. cY ^ --^ \a / 1 \ / i\ 40'y\ /\^o° 1 \ ii / \ 1 / \ ^ - ^. 685. It is easy to determine the angle 6^ for which the effective driving force F,' = Psin 6^ is just equal to the resist- ance Q=2 F/n ; we find approximately 61 = 40°, 140°, 220°, 320°. As long as /'/> Q, i.e., from Cj to C^ -A^l — ! ""iy f Ci^" 1 — \.^^ and from C3 to C4 (Fig. 181), the rotation is accelerated ; from Cj to Q and from Q to Ci it is retarded. The velocity is there- fore greatest at Q and C4, least at C3 and Q. Now, in the interval C1C2 the work ?J^ ^'^^ '®'' of the driving force is JV12 = P ■ chord C1C2 = P • 2 aco?,di = TvQa cos 6j, since Q= 2 PJtt, while the work of the resistance Q is W^ = (2 ■ arc CiQ = (2 • a(7r - 2 ^O ; hence, the energy stored in the fly-wheel in this interval C1C2 is W^^ - W^' = ttQJcos 61-1 + 2^^= .21 TrQa =lQa, approximately. 686. On the other hand, if M], wj be the angular velocities of the crank, v^, v^ the linear velocities of the crank-pin, at Cj, C2, respectively, and if the mass of the fly-wheel reduced to the crank-pin be denoted by ntajVit find for the same energy stored in the fly-wheel in the period CjQ the other expression : Pi-Pi = ^ fn^a\wi — w/) = i m^{vi — vf). Now, the difference Vs — v^ between the greatest and least velocity, divided by their mean ^(?'i-)- z',) = J", measures the xtXz.'wt fluctuation in velocity ; the reciprocal of this quotient, _ V _ Z'l -|- I'a Z/j— Z'l 2(i72 — Wi)' can be regarded as a measure of the uniformity of the rotation ; it is called the degree of uniformity. Introducing this quantity k, we have T^-T^^-m^i?. 689.] BODY WITH FIXED AXIS. 43 1 Equating the two expressions found for the kinetic energy stored in the fly-wheel, we find for the reduced mass of the wheel, If the resistance Q be expressed in pounds, the mass m^ in pounds will be ^ The coefficient k is selected differently according to the nature of the engine and the object for which it is used. For slow-running engines it is between 10 and 203 for very fast engines it may reach 100 or more. 687. Exercises. (i) Show that »z„= 112,000 kVP/v'N, if hP is the horse-power, A^ the number of revolutions per minute. (2) Find m^ when |-P = 100, IV= 25, w = 56 ft. per second, k = 50. 688. Reactions of the Fixed Axis. A rigid body that turns about a fixed axis exerts an action on the fixed axis that tends partly to shift it bodily and partly to turn it out of its position. The axis must, therefore, be kept fixed by certain forces, called the reactions of the axis. As a straight line is determined by two of its points, to fix the axis it suffices to fix two of its points, say A and B. The reactions of the fixed axis can, therefore, be regarded as two (in general unknown and variable) forces, A at A and B at B. Like any system of forces acting on a rigid body, these two forces can be replaced by a single force 7? together with a couple H. Thus, if we apply at A two equal and opposite forces, each equal and parallel to B, the resultant of A and B at A will form a single force R, while B at B with — B at A forms a couple H. 689. In certain particular cases the reactions of the fixed axis may be zero. It is obvious that for a revolving piece of ma- chinery it is desirable to avoid as far as possible any action on 432 KINETICS OF THE RIGID BODY. [690. the bearings that keep the axis of the piece in a fixed position. It is, therefore, of importance to know under what conditions the reactions of the axis vanish. Before discussing the general case of the determination of the reactions for a body turning about a fixed axis (for which see Arts. 697-705), it will be well to treat some simple cases commonly occurring in machines by the most direct methods. 690. Consider a revolving shaft, with wheels, pulleys, cranks, etc., mounted on it. Every particle in rigidly attached to the revolving body describes a circle about the axis of the shaft. Jf the rotation be uniform, the acceleration of the particle m, at the distance r from the axis, is the centripetal acceleration mV required for uniform circular motion (see Art. 120); it is directed along r towards the axis. The effective force of the particle is therefore mu?r ; and by d'Alembert's principle (Art. 598) these effective forces, reversed in sense, that is, the centrifugal forces niufr, directed along r away from the axis, must be in equilibrium with the external forces. 691. If the mass of the revolving body be distributed symmetrically about the axis of rotation, the centrifugal forces mu?r are in equilibrium by themselves. For, this symmetry means that for every particle of mass m at the distance r from the axis there exists one of equal mass, at the same distance on the opposite side of the axis ; and the cen- trifugal forces of these particles, being equal and opposite and acting along the same line, are in equilibrium. Now, by Art. 690, if the centrifugal forces are in equilibrium by themselves, so must be the external forces. And if there be no external forces except the reactions of the fixed axis, these reactions will be zero, so that there is no tendency to either shift or turn the axis. This axis is then called a permanent axis of rotation. Among the external forces the most important is generally the weight of the revolving body. But if the speed of rotation is very high, the centrifugal forces may be so large that the weight is insignificant in comparison with them. Unless this is the case, the weight will produce a pressure on the bearings which is easily determined. 694-] BODY WITH FIXED AXIS. 433 692. The centrifugal forces may, however, be in equilibrium, and hence the reactions may be zero, even when the symmetry of mass distribution is not as perfect as assumed in Art. 691. Thus, let two particles of unequal ^^ masses Wj, m, be rigidly attached to the shaft, on opposite sides, on a line meeting the axis at right angles (Fig. 182). If their distances ^1, ^2 from the axis be so chosen that Q^ m^r^ = m^ri, their centrifugal forces, mix; hence, ir = — toj/ — o) V, y=^a)X—%nizx + uP'^myz = — Ew + Da?, "Zmzx = — wZmyz — aPSmsx = —Dd> — Ea>^, ^mixy —yx) = aiXmx^ — co^^mxy + wLniy"^ + w^'Lmxy = C(o, where C=^m{x'^ +y'^), D = '^myz, E = ^mzx are the notations introduced in Art. 630. With these values the equations of motion assume the form : - Mxw"^ -Myw = tX+ A^ + B„ - Myufl + Mxi, = ^Y+Ay + B^, o=2Z+A, + B,, (12) Dc? -Ew= ^(yZ- z Y) - aAy - bB^, - Ea? -Dw= ^{zX- xZ) + aA, + bB^, Ca}=^xY-yX). 700.] BODY WITH FIXED AXIS. 437 699. The last equation is identical with equation (2), Art. 666. The components of the reactions along the axis of rotation occur only in the third equation, and can therefore not be found separately. The longitudinal pressure on the axis is The remaining four equations are sufficient to determine A„ Ay, B„ By. The total stress to which the axis is subject, instead of being represented by the two forces, at A and B, can be reduced for the origin (9 to a force and a couple (comp. Art. 688). The equations (12) give for the components of the force -A^-B^ = 1,X+ Mx(^ + My a, -Ay-By = l.Y+M]>afi-Mxm, (13) -A,-B, = 1.Z. This force consists of the resultant of the external forces, i? = V(2X)2 + (2F)2+(EZ)2, and two forces in the xj/-p]a.ne which form the reversed effective force of the centroid ; for Mxm^ and Myofi give as resultant the centrifugal force MaP'^x^ +j^ = Mco^', directed from the origin towards the projection of the centroid on the xy-plane, while Myo), —Mx(o form the tangential resultant Mcor, perpendicular to the plane through axis and centroid. The couple has a component in the ys-p\zne, and one in the ^■;ir-plane, viz. : aAy + bBy = ^{yZ-zY)-Da>^ + Ew, -aA^-bB^ = l.{zX-xZ)-\-Eay^+Dm, ^^"^^ while the component in the xy-p\a.ne is zero. The resultant couple lies, therefore, in a plane passing through the axis of rotation. 700. In the particular case ivken no forces X, V, Z are acting on the body, the last of the equations (12), or equation (2), shows 438 KINETICS OF THE RIGID BODY. [701. that the angular velocity (o remains constant. The stress on the axis of rotation will, however, exist ; and the axis will in general tend to change both its direction, owing to the couple (14), and its position, owing to the force (13). If the axis be not fixed as a whole, but only one of its points, the origin, be fixed, the force (13) is taken up by the fixed point, while the couple (14) will change the direction of the axis. Now this couple vanishes if, in addition to the absence of external forces, the conditions D = Iviyz = 0, E= ^viax = (15) are fulfilled. In this case the body would continue to rotate about the axis of z even if this axis were not fixed, provided that the origin is a fixed point. A line having this property is called a permanent axis of rotation. As the meaning of the conditions (15) is that the axis of ^r is a principal axis of inertia at the origin (see Art. 639), we have the proposition that if a rigid body with a fixed point, not acted upon by any forces, begin to rotate about one of the principal axes at this point, it will continue to rotate uniformly about the same axis. In other words, the principal axes at any point are always, and are the only, permanent axes of rotation. This can be regarded as the dynamical definition of principal axes. 701. It appears from the equations (13) that the position of the axis of rotation will remain the same if, in addition to the absence of external forces, the conditions ^ = 0,7 = (16) be fulfilled ; for in this case the components of the force (13) all vanish. If, moreover, the axis of rotation be a principal axis, the rotation will continue to take place about the same line even when the body has no fixed point. The conditions (16) mean that the centroid lies on the axis of z ; and it is known (Art. 639) that a centroidal principal axis is 703-] BODY WITH FIXED AXIS. 439 a principal axis at every one of its points. The axis of z must therefore be a principal axis of the body, i. e., a principal axis at the centroid. We have thus the proposition : If a fr-ee rigid body, not acted upon by any forces, begin to rotate about one of its centroidal principal axes, it will continue to rotate iinifornily about the same line. 702. The determination of the reactions is much simplified in the case when both the mass of the body and the external forces are distributed symmetrically with respect to the centroidal plane perpendicular to the fixed axis. For it is then sufficient to sup- port the axis at a single point, viz., the point O (Fig. i86) where the axis meets the plane of symmetry ; this point O is called the center of suspension. The reaction of the axis is therefore a single force P, passing through 0, in the plane of symmetry ; — /" is the pressure on the axis. The whole problem becomes plane ; we take the fixed axis as axis of z, the plane of symmetry as ory-plane. Owing to the symmetry, the impressed forces will reduce for any point in the xy-p\a.ne, say for the centroid G of the body, to a single resultant R and a couple H in this plane. It will generally be convenient to resolve the forces along OG and at right angles to it ; thus the reaction P is replaced by its components /-"„ Pg, and the resultant R of the im- pressed forces by P^, Re- 703. The centroid G describes a circle about with radius OG = h\ its accelerations are therefore hO"^ along GO and JiQ at right angles to GO, being the angle of rotation ^'s- '86. counted from the arbitrary axis Ox in the plane of symmetry. The centroid moves as if all the forces were applied at this 440 KINETICS OF THE RIGID BODY. [704- point (Art. 603). Hence resolving along OG and at right angles to it, and taking moments about O, we find the three equations of motion : -mhe'^ = Rr + Pr, (17) m/ie = R, + P„ (18) m{g'^ + /?)9 = H, (19) where q is the radius of inertia for the centroidal axis parallel to the fixed axis. The equation (19), which is the same as equation (2), Art. 666, determines 6 and its derivatives ; substituting their values in (17) and (18), the reactions /"„ P9 a-re found. 704. If the fixed axis be assumed horizontal and gravity as the only impressed force, we have the case of the compound pendulum (Art. 669). Taking the axis of x vertically down- ward, we have R^ = mg cos 6, Rg= — mg sin d, H= — nigh sin 6. Equation (19) becomes and gives by integration if &> = co^ for 6 =6^: 0^ = 0,2 = 0,^2 + _lglL_ (cos e - cos e^). q'' + /r Substituting in (17) and (18), we find for the components of the pressure on the axis : — Pr=mgcosd-[- mhvP' = 7ng{ w? ^-^ cos ^n + ^ .7"^ f cos ^ ), V g q^+k^ q^+h^ J -Pe=- mg. -j^-— sin 6. q^ + ¥ The latter component is independent of the initial conditions, the former is not. The total pressure on the axis, ^ P^ + P^, varies in general in the course of the motion both in magnitude and in its direction in the body. In the particular case when 705 •] BODY WITH FIXED AXIS. 441 a>^= 2 gh COS O^/^q^ + h"^), i. e., when the initial kinetic energy \ m {^ + k^) &)q2 = mgh cos ^q (which means that w is zero when OG is horizontal), we have ~Pr= tng ^ -^ cos o, ■Pg= -mg sin I ^2 + /,2 — -' " -^ ^2 + /,2' and the inclination of the total pressure to OG is given by -P. d' tan o{ P to the rod is given by j, tan <^ = -^ = — jV tan 6. Discuss in a similar way the horizontal and vertical components ofP. 442 KINETICS OF THE RIGID BODY. [706. (2) A rapidly revolving axle whose axis is Az (Fig. 188) has the straight homogeneous rod yiC attached to it at a constant angle, by means of the link BC, at right angles to Az. The mass of the axle and of the VmkBC are neglected ; also the action of gravity on the whole revolving mass. If AB = b, BC = c, determine the pressure on the axis. As of the six equations of motion (12) (Art. 698) only the last, which does not involve the reactions, requires integration, so that the determination of the reactions, after u> has been found from the sixth equation, is a mere algebraical process, we may select the zjc-plane so as to pass through the centroid. Taking A as origin, and A and B as fixed points, we have in the equations (12) of Art. 698: a = o, b = b, X = ^ c, }• = o, z = \ b ; Pig. 188. and the sixth equation gives, as there are no impressed forces, w = const. It is easy to see that D = o, E = \ mbc. Hence the equations reduce to - i mc,>? = A, + B^, o = A,j + B^, = A, + B,_, o = — bBy, — \ mbcu? = bB^, C ■ io = o, whence A^ = — ^ otmV, B^ = — ^ ?nw'c, A,j = 0, ^j = o. The two parallel forces — A^, — B.j. are equivalent to a single result- ant = 1 m(ji-c, which is the centrifugal force of the centroid ; but it should be noticed that this resultant is not applied at the centroid G, but at the distance Zi = ^b from the .r;'-plane ; in other words, it passes through the centroid of the triangle ABC. (3) Solve (2) when the axis Az is vertical (downward) and the weight of the rod ^C is taken into account. (4) A homogeneous plate of the shape of a right-angled triangle of sides b, c, revolves about the side b, under no impressed forces ; deter- mine the pressure on the axis. (5) Solve (4) when the axis b is vertical and the weight cT the plate is taken into account. 706. Impulses. Suppose a rigid body with a fixed axis is acted upon, when at rest, by a single impulse F, in a plane perpen- 707.] BODY WITH FIXED AXIS. 443 dicular to the axis and at the- distance p from the axis. It is required to determine the initial motion of the body just after impact. As the impulsive reactions of the fixed axis have no moment about this axis, the initial angular momentum of the body about the fixed axis must be equal to the moment of the impulse F about the same axis ; i. e., to Fp. If co is the initial angular velocity, the angular momentum of the body about the fixed axis is, by Art. 665, = ml, where / is the moment of inertia of the body for the fixed axis. Hence we have o) = -^. (20) 707. Let the impulse F be produced by a particle of mass ni im- pinging on the body with the fixed axis /; the mass of this body we shall here denote by m'. The idea of reduced mass (Art. 677) can often be used to advantage to determine F. Let u be the projection of the velocity of the particle m on the plane perpendicular to the axis (the component parallel to / is evidently in- effective as regards rotation about /) ; and let the particle m strike the body at the point P where the direction of u meets the plane through the axis / perpendicular to u. Then we have only to reduce the mass m} of the body to the point P by putting /= m\ •/, where / is the moment of inerda of the body for the fixed axis /, / the distance of /"from /, i. e., the shortest distance of ic and /, and m\, the mass of the body reduced to P. The impact can now be treated like that of two homogeneous spheres (Arts. 432 sq.), except that the mass m' of the body impinged upon is replaced by m\. The velocity v of m and the velocity v' of the point P are therefore given by the equations mv -f- OTpW' = niu -\- m\u', v' — V = e{it — u'), where e is the coefficient of restitution and a' is the linear velocity of the point /"just before impact. 444 KINETICS OF THE RIGID BODY. [708. 708. In the case of inelastic impact we liave ^ = o, and hence v' = v ; the former of the two equations gives, therefore, if the body m' is ini- tially at rest, „„( m -\- m J, v' mup mup , V mup mup , s. whence co=— =-t- — r- = — ir' — V K^"^) p p^{ni-\-m\) mf + I The momentum imparted to the body by the impact of the particle is F=m'v' = ^ = ^^ --mu. (22) m-\- m\ 1 + mp' This value of ./^reduces the formula (20) to (21). The result (22) can be expressed by saying that, in the case of inelastic impact where the particle after impact moves along with the body, we may take for F the whole momentum of the impinging mass if at the same time we increase the moment of inertia / of the body by that of the particle, viz., mp^. The same result can be derived without using the idea of reduced mass. As the particle after impact moves along with the body, with the velocity v' of the point P, the momentum given up by it to the body is F = m (u — v') = mtt — miap ^ mu — mFp"^ / 1, by (20); hence F= muJ/{I-\- mp''), which agrees vdth (22). 709. To determine the impulsive stress produced on the axis by a single impulse F, we have to apply the general equations (17), (18) of Art. 614, or (19), (20) of Art. 615. But we can also proceed directly as follows : Take the fixed axis as the axis of z and the ^.r-plane through the centroid (?(Fig. 189) and let X, o, o be the co-ordinates of G, and jTj, j/j, ^•j those of the point of application P of the impulse F. The components of F may be denoted by Jf, Y' ^'' those of the reactions of the axis by A„ Ay, A„ B^, By, B, (comp. Art. 697). F\z- 189. 7IO.] BODY WITH FIXED AXIS. 445 As the initial motion after impact is a rotation about the axis of s, we have x= —eoy, y = a>x, s = o, so that the momentum of a particle of mass m has the components —maj, mcox, o. Reducing these momenta to the origin O, we find a resultant momentum whose components are —aljiij/^o, o^Zmx = Muix , o; and a resulting -couple whose vector has the components — wlinzx=—E(io, —a)1myz=—Dco, ai'2tii{x'^-\-y'^)=C(o, where C, D, E have the same meaning as in Art. 698. The six equations of motion just after the impact are there- fore, if the body was originally at rest, o = X + A, + B^, Mxa>= Y+Ay + B,j, o = Z+A, + B,, (23) — Eco =y^Z— z^ Y— a Ay — bBy, - Da, = s-^X—x^Z +aA^ + Bb^, Ca3=x-yY—y^X. It is easily verified that this is the form assumed by the equa- tions (19), (20) of Art. 615 in the present case. 710. The last of these equations is nothing but the equation (20) of Art. 706. The components A-, B, along the axis cannot be determined separately ; the other components of the reactions can be found from the first, second, fourth, and fifth equations. The impulsive stress to which the axis is subjected by the impulse, or the so-called percussion of the axis, instead of being represented by two impulses, A,B as above, can also be regarded as composed of an impulsive pressure at O whose components are -A,-B, = X, -Ay-By = V,-Mxm, -A,-B,=Z, and an impulsive couple whose vector has the components aAj, + bBy=yiZ— s^Y+E(o, —aA:, — b^B^ — z.^X~x-^Z-\-Du), o. The last component being zero, the resulting couple lies in a plane passing through the axis of z. 446 KINETICS OF THE RIGID BODY. [711. If there were any number of impulses acting on the body- simultaneously, the effect on the axis could be determined in the same way, except that the quantities X, Y, Z, jjZ— jjF, .CjA' — .fjZ', must be replaced by the corresponding sums. 711. It follows from the preceding article that the conditions under which a single impulse acting on a rigid body with a fixed axis will produce no stress on the axis are X=o, Y=Mx(D, Z = o, -z^Mx+E=o, D = o. (24) If these conditions are fulfilled, the resulting motion will be the same even when the axis is free. The first and third equations show that tlie impulse must be perpendicular to the plane passing through axis and ccntroid. The meaning of the fourth and fifth conditions becomes appar- ent if the ,t:y-plane be taken so as to pass through the point of application P of the impulse. The new origin O' is the foot of the perpendicular let fall from P on the fixed axis. To trans- form the conditions (24) to the new system it is only necessary to substitute z -\- s^ for .;■; the first three conditions are not affected, and the last two become — z-^]\Ix -f- 1.ni::x -{■ ^j 27;/x = o, ^myz -f- z^my = o, or, since '^mx = Mx, Imy = o, E' = o,I)'=o, where E', £>' are the products of inertia at O'. It thus appears that the axis of z must be a principal axis at the foot of the perpe7idicnlar let fall on this axis from the point of application of the impulse. 712. It should be noticed that a line taken at random in a body is not necessarily a principal axis at any one of its points. But if a line is a principal axis at a point O', then it is always possible to determine an impulse that will produce no stress on this line so that the body will begin to rotate about it as axis even though it be not fixed. As shown in the last article, the 7I3-] BODY WITH FIXED AXIS. 447 impulse must be =Mxu>, and must be directed at right angles to the plane through axis and centroid. The point where it meets this plane is called the center of percussion for the line. Its dis- tance -t-j from the axis is found from the equation of motion, viz., the last of the equations (23), which, owing to the conditions (24), reduces to C= Mxx^. If ^' be the radius of inertia of the body for a parallel centroidal axis, we have C — M{q''^ -\-x^); hence x^=x^'i^- (25) X Hence, if a given line / be a principal axis for one of its points O', there exists a center of percussion ; it lies on the intersection of the plane (/, G) with the plane through O' perpendicular to /, at the distance Xj, given by (25), from the line /. An impulse Mxcj through the center of percussion at right angles to the plane through axis and centroid, while producing no percussion on the axis, sets the body rotating with angular velocity w if it was originally at rest; on the other hand, if the body was originally in rotation about the axis, such an impulse can bring the body to rest without affecting the axis. 713. Exercises. (i) A homogeneous straight rod of length / and mass m is suspended vertically from a horizontal axis through its end. At what point and in what direction must it be struck to produce no shock on the axis? (2) Show that the center of percussion of a compound pendulum coincides with the center of oscillation (Art. 670). (3) A homogeneous straight rod of length / lies on a smooth hori- zontal table ; how should it be struck to begin turning about a vertical axis through one end ? (4) A homogeneous circular disk, of radius a and mass to, while turning about its axis with angular velocity w, is suddenly stopped by a peg (which projects radially from its circumference) striking a fixed obstacle. Determine the momentum of the impact and the stress on the bearings which are at distances a, b from the disk on opposite sides. 448 KINETICS OF THE RIGID BODY. [713 (5) "A pendulum is constructed of a sphere (radius a, mass M) attached to the end of a thin rod (length b, mass m). Where should it be struck at each oscillation that there may be no impulsive pressure to wear out the point of support ? " (Routh.) (6) Show that if a body with a fixed centroidal axis is struck a blow, the axis is always subject to an impulsive stress. (7) A trip-hammer consists essentially of a heavy mass carried by an arm which can turn about a fixed axis c\ A cam projecting from a revolving shaft strikes the arm of the hammer at each revolution, raising it a certain distance so as to let the head of the hammer drop on the anvil. Let m be the mass of this shaft (with its fly-wheel), / its moment of inertia for its axis c, p the distance of the point of impact P from the axis c of the shaft ; then m^ = ///^ is the mass reduced to P. Let ;;/ be the mass of the hammer and arm, /' their moment of inertia for the axis c\ and /' the distance of the point of impact P from c', so that ;«'j, = /y/'^ is the mass of the hammer reduced to P. Regarding the impact as inelastic, we then have for the velocity of the point P just after impact (see Art. 708), v = f-^ "' where u = up, w being the angular velocity of the shaft c. The lost kinetic energy is (by (10), Art. 444), £ = ^—r • h ni-oU^- OTp + m\ To make this as small as possible, mj,/m\ should be large ; /. e., the reduced mass of the fly-wheel should be large in comparison with that of the hammer. To reduce as much as possible the percussion of the axis c' of the hammer, the point of impact P should be near the center of percussion (Art. 712) of the hammer ; that is, p' should satisfy the relation r = m'xp', where x is the distance of the centroid of the hammer (with its arm) from the axis c' ; this gives , /' fn'x (8) In the trip-hammer, Ex. (7), let mi be the mass of the hammer- head regarded as concentrated at the distance a from the axis c' ; m^ the mass of the arm regarded as a straight rod of length l{>a). Find 7I3-] BODY WITH FIXED AXIS. 449 the center of percussion P, and show that with m-^/m^=:. 8, (3!//=o.75, we find /'//= 0.74. (9) In the problem of Arts. 707, 708, determine F for perfectly elastic impact (f = i). (10) The ballistic pendulum consists essentially of a heavy wooden block (or a case filled with sand) suspended from a horizontal axis. A bullet is shot into the block at rest, at right angles to the vertical plane through the axis of the pendulum. From the observed rise of the block due to the impact of the bullet, the velocity of the bullet is computed. The bullet enters a certain distance into the wood ; the time of this motion is neglected, /. e., the impact is supposed to take place at the point A where the bullet stops. Let a be the distance of this point from the axis, a the angle it makes with the vertical ; then the moment of the momentum mu of the bullet is mu ■ a cos « ; equating this to the angular momentum of pendulum and bullet (since after the impact the bullet moves with the pendulum), we have the equation of impact, which gives the initial angular velocity of the pendulum. If /«' be the mass of the pendulum, q its radius of inertia for the axis, h the distance of its centroid from the axis, the equation of motion of the pen- dulum (with the bullet) can be written down and integrated. This gives (m'(f + ma^^ w^ = 2 m'gh (i — cos ^i) -f 2 mga [cos « — cos (^1 — «)] = 2 {m} + ni)gh (i — cos ^i), nearly. The angle B^ through which the pendulum swings from the vertical, can be measured by a tape attached to the block, say at the distance b from the axis, and drawn out of a reel as the pendulum moves ; if the length of tape drawn out is c, we evidently have <: = 2 ;? sin i ^j. Combining the results, we find finally , m'\ he /—, m ) ab cos « where / is the length of the equivalent simple pendulum. (11) A homogeneous cube whose edges are a rests on a horizontal plane ; one of its base edges being a fixed axis. Determine the impulse Foi the least blow that will upset the cube about the fixed axis. Observe that the kinetic energy due to the blow must be sufficient to raise the centroid to the position vertically above the fixed axis. This work. 450 KINETICS OF THE RIGID BODY. [714. which must be done before the upsetting takes place, is sometimes called the dynamic stability. (12) In Ex. (11), if the edge of the cube is i ft, determine with what velocity a mass equal to one third of that of the cube would have to strike it (at the middle of the edge opposite the axis, at right angles to the diagonal plane) to upset the cube. (See Art. 708.) IV. Plane Motion. 714. As explained in Art. 610, the general problem of the motion of a rigid body resolves itself into two problems which may be treated separately : the motion of the centroid regarded as a particle containing the whole mass of the body and having all the forces acting on the body applied to it parallel to their actual directions, and the motion of the body about the centroid regarded as a fixed point. The former problem is solved by integrating the three equations (4) of Art. 600, which can also be written in the form (5), Art. 600, or (8) Art. 603 ; the latter by integrating the three equations (6) of Art. 601, which can also be written in the form (7), Art. 601. 715. If the centroid moves in a plane curve and the rotation about the centroid always takes place about axes perpendicular to the plane of this curve, the motion of the body is called //aw^ motion. It suffices to study the motion of the cross-section of the body in this plane. The motion of the centroid Gijc, y) is given by the two equations of linear momentum : M'i=^X,M'y = %Y, ■ (i) j^M,= R^,^^My=R,. ' (X') The motion about the centroid is given by a single equation of the form v / •■ --n v/ -^t v\ / \ z.m{xy — yx) = 2.{xY —yX), (2) or —'2m(xy—yx) = //, (2') the equation of angular momentum, obtained by "taking mo- ments " about any point of the plane. As the motion about the 7i6.] PLANE MOTION. 451 centroid takes place as if the centroid were fixed (Art. 606), it will often be convenient to take moments about the centroid. The equation (2') then assumes the form 'S-« (2") where / is the moment of inertia of the body about the centroidal axis perpendicular to the plane, H the sum of the moments of the external forces about the same axis, &> the angular velocity about the axis. Analogous considerations hold for the equations of the change of motion due to impulses (Arts. 614, 615). 716. As an instructive example of plane motion consider the motion of a homogeneous circular cylinder down tin inclined plane, the axis of the cylinder remaining horizontal. Let m be the mass, a the radius of Fig, 190. the cylinder. In Fig. 190, G(, is the initial position of the centroid (at the time t=6) when the linear velocity v of the centroid and the angu- lar velocity u about it are both zero ; G is the position of the centroid at any time t. The inclination of the plane to the horizon is a. If that point of the circumference which is initially in contact with the plane (at O) has at the time / the position O' , then 0'GC=6 is the angle through which the cylinder has turned about its axis in the 452 KINETICS OF THE RIGID BODY. [717. time t. This angle 6 and the distance OC = G^G = x through which the centroid has moved are the variables that determine the position of the cylinder at any time. 717. The motion of a cylinder on a fixed plane is called pure rolling if at every instant the line of contact is the instantaneous axis. Rolling is to be clearly distinguished not only from sliding, when the cylinder has a motion of translation parallel to the plane so that the instantaneous axis lies at infinity, but also from spinning, when the line of contact turns about one of its points in the plane. In the most general case the motion of the cylinder on the plane would be a combination of these three kinds of motion. It follows from the definition that in pure rolhng the length of the circular arc O' C = aO must equal the distance (9C= G(,G = x, through which the centroid has moved. The kinematical condition for pure rolling is therefore : x = ad; (3) , dx dO / s hence — = « — , or z' = am, (4) dt di , d'-x d'-Q dm ■ dv dm , ^ and — T = « — ^ = 1 — , orys — = a — , (i;) df dt- dt ^ dt dt ^^' where v and/ are the (Unear) velocity and acceleration of the centroid, o) and dui/dt the angular velocity and acceleration about the centroid, t. (?., about the axis of the cylinder. 718. Let the only impressed force be the weight 7ng of the cylinder. The reaction of the plane at the point of contact C can be resolved into its normal component N, at right angles to the plane, and the frictional resistance F, along the greatest slope of the plane. If the plane were perfectly smooth so that F= o, the cyhnder would necessarily slide down the plane, provided its initial angular velocity be zero. For, as the only forces acting, 77ig and IV, both pass through the centroid, no angular momentum about the axis can be generated. In this case the principle of kinetic energy and work gives, if the velocity at the time / for this limiting case be denoted by Vi, ^ nivi = mgh, or »j = ^ 2 gh, (6) where ^^ = ^ sin « is the vertical height through which the centroid G has descended. 720.] PLANE MOTION. 453 719. Pure rolling occurs only if the frictional force F is of sufficient amount to prevent sliding (or slipping, as it is called). The moment Fa of the friction produces angular momentum, giving the cylinder an angular velocity to (clockwise in Fig. 190) about its axis. Mechanical means might be substituted for friction to produce the same effect, e. g., gearing, or a flexible band wrapped around the cyhnder and stretched up the plane. In the case of pure rolling the principle of kinetic energy and work gives (rolling friction being neglected) ^ tnii- + ^ Iiii- = mgh, (7) the work of N as well as that of F being zero because the point of ap- plication C is instantaneously at rest. The left-hand member which represents the kinetic energy of the cylinder can be simphfied by introducing the mass reduced to the point of contact C, i. e., to the distance a from the centroid (Art. 677): ?ii'=Ila^; as in the case of pure rolling (Art. 717) a(D=v, (7) reduces to ^(m + m') if- = mgh. (8) Denoting, as in Art. 718, the velocity of frictionless sHding by v^, we find: V = ^ Wi- (9) ■yj i-\-m' Jm For the homogeneous cylinder, m'=\m (see Ex. (i). Art. 679), so that V = V-| Vi. The same reasoning applies when the cylinder is replaced by any other sohd of revolution whose mass is distributed symmetrically both with respect to the axis of revolution and with respect to the centroidal plane perpendicular to this axis ; the formute (6) to (9) hold without change in this more general case. But the quotient m'/m will not always be a pure number ; it will depend on the dimensions of the body and on the distance a of the centroid from the incHned plane. 720. Exercises. (i) A homogeneous cylinder of i ft. diameter rolls down a plane inclined at 30° to the horizon, over a distance of 20 ft. ; find the final linear and angular velocity. 454 KINETICS OF THE RIGID BODY. [721. (2) A sphere, a circular disk or cylinder, and a hoop or thin cylin- drical shell, all three homogeneous, roll down the same inclined plane through the same distance ; show that the velocities acquired are z/^=Vf 2^1 = 0.845 z'j, Z'.= V| 2^1 = 0.8162/1, v^=^^v^ = o.^o^Vl. (3) Two equal circular disks, each of radius R and mass m, are rigidly connected by a short cylindrical shaft, of radius a ( I + '-^ ) /x. (15) \ m j The motion will be a combination of rolling and shding. The distance x= GoG= OC{¥\g. 190) through which the centroid moves in any 456 KINETICS OF THE RIGID BODY. [725- time t is not equal to, but greater than, the arc O'C = a6 through which the body turns about the centroid in the same time ; hence the relations (3). (4). (5) between x and 6», v and w, dv/dt and du>/dt do not hold. The equations of motion still have the form (10) ; but as actual slid- ing takes place at the point of contact C, we now have F= ixN. (16) Combining this relation with the first two of the equations (10), we find -^=^(sina-/x,cosa). (17) df Hence the motion of the centroid is again uniformly accelerated and, owing to (15), the acceleration is positive. The third of the equations (10) gives du> _ fjimga cos a . , „^ 'Jt~ 7 ' ^' '' hence the rotation about the centroid is also uniformly accelerated. 725. Exercises. (i) For the three types of bodies mentioned in Ex. (2), Art. 720, the centroid having in each case the distance a from the inclined plane, determine the linear and angular accelerations : {a) in the case of pure rolling ; (^) in the case of rolling combined with sliding ; {c) determine the condition for pure rolling if the angle of friction is <^. (2) A homogeneous circular disk, 2 ft. in diameter, rolls down a plane sloping 1:5. Starting from rest, how far will it go in 10 sec, {a) if ;u, = o.i? (p) if /x = 0.05? (3) Discuss the motion of a homogeneous cylinder with horizontal axis 2ip an inclined plane, if initially v = Vf,, w = o : (a) when there is no friction ; (ff) when there is friction, but not sufficient to produce rolling. (4) In Ex. (3), if the friction is sufficient to produce rolling, the cylinder can be made to roll up the plane under the action of gravity alone by giving it an initial angular velocity wo about its axis. This can, e. g., be done by a blow directed up the plane and striking the cylinder above the centroid (z. e., at a distance from the plane > a). For such a blow is equivalent to an equal and parallel blow through the centroid, giving the centroid an initial velocity v^, together with an impulsive couple tending to make the body roll up the plane with 726.] PLANE MOTION. 457 initial angular velocity wg. It should be noticed that the frictional resistance J^, in the case of rolling up the plane, is directed up the plane, just as in the case of roUing down the plane. For, what produces, or in the present case destroys, angular momentum is, properly speaking, not the force F, but the component mg sin « of gravity which is directed down the plane. Write down the equations of motion and solve them. (5) The effect of rolling friction, hitherto neglected, can be expressed by a couple of moment \\! mga cos «, where fx! is the coefficient of rolling friction (see Arts. 378, 379). Its sense agrees with that of the moment of .F in downward motion, but is opposite to it in upward motion. Thus, for rolling down the plane we have m—- = mg sma — F, /—- = Fa + fi' mga cos a ; dt dt the condition for pure rolling (5) gives I* = (sm « — fi. — ^^cos oAing. m + m m 726. A homogeneous circular cylinder, of radius a and mass m, placed on a rough horizontal plane, is acted upon by gravity and a horizontal force X, passing through the centroid at right angles to the axis of the cylinder. The equations of motion, m— = X-F, o = N- mg, I— = Fa, (19) dt ' ^ dt ' ^ ^' give in the case of pure rolling, owing to (5), ^ dv , I du> , , t^dv / s X = OT— + -— = (;« + »?')— , (20) dt a dt dt where for the cylinder m' = \m. The same solution holds, with the proper value of m' , for the more general solid of revolution mentioned at the end of Art. 719. The expression for X shows that the centroid of the body moves like a particle, not of mass m, but of mass m + m' , under the force X alone. This result is used in studying the motion of a carriage or train to take into account the " rotary inertia" of the wheels by adding to the total mass the sum '2,m' of the masses of the wheels reduced to their circumference. Thus for a train running down an incline of angle «, 458 KINETICS OF THE RIGID BODY. [72/- under the action of gravity alone, if M be the mass of the train (includ- ing the wheels), m' the sum of the reduced masses of the wheels, the acceleration will be ^ -^sin a. 727. Exercise. (i) Determine the height at which the buffers and coupling chain of a railroad car should be placed to prevent "pitching," J/ being the mass of the car, m that of the wheels, m' = mq^la^ their mass reduced to the point of contact with the rails, h the height of the centroid of the car above the axles. 728. A homogeneous sphere, of radius a and mass m, is placed on a rough horizontal plane, the sphere having initially a velocity of trans- lation Wo parallel to the plane and an angular velocity wo about the hori- zontal diameter perpendicular to w,). Gravity and the reaction of the plane are the only forces acting on the sphere. As these cannot change the directions of either Wo or the axis of wo, the motion must be plane. The following discussion applies without change to the case of a homogeneous cylinder and of a hoop ; it is only necessary to give the proper value to the reduced mass m' . Two cases may be distinguished according as the initial angular velocity coo gives to the point of contact A a linear velocity auo of the same sense as Wq or of the opposite sense. 729. If a^i>a agrees in sense with Vs^ (Fig. 191), the frictional resist- ance F=^\t,mg is opposite to »o as well as to am,,, and tends to diminish , both »o and (dq. Hence the equa- ^,_1_^ tions of motion are, if the sense of /^ 1 \^ z'o be taken as positive, j ''g[— l^vo »i-^ = — f^'^g, o = N—mg, V *'"f / J N. y dt a Fig. 191. where m' = 1/ a^ is the mass re- duced to the point of contact. The first and last of these equations show that the motion of the centroid 73°-] PLANE MOTION. 459 as well as the rotation about the centroid is uniformly retarded. Inte- grating these equations, we find m g V = Vo — ixgi, (0 = Wo — /i — ; - /. Hence v would vanish at the time /i = Va/ (t-g, lo at the time 4 = m' a o, so that the initial motion is rolling combined with slid- ing, the instantaneous axis lying (parallel to the axis of co) at the dis- tance Wo/<"o above the centroid. At the time t, the velocity of A is v + aa — IJig(T. +'—\t; it vanishes at the time , _ Wo + awo ix.g{i + m/m'Y after which the motion becomes pure rolling. As V(, and aw^ are both positive, this time 4 must lie between the times ti and t.^. If /j > t^, so that w vanishes first and has become negative at the time t^ when pure rolling sets in, the sphere will, after the time /,, roll forward (i. e., in the sense of z'o). If, however, (^ < /,, V first reduces to zero, and is, therefore, negative when pure rolling begins ; the sphere will, therefore, after the time 4, roll backward. If ti = 4, the sphere comes to a stop at the time t^ = ti = t,,. The values of t^ and /■> show that we have /j = /, according as Wo = ^«a)o. < tn 730. Exercises. (i) Discuss that case of the problem of Art. 728, when the initial angular velocity wq gives to the lowest point of the sphere a velocity aa-^ opposite in sense to v^ ; distinguish the three cases Wj = | awf, \, taking the sense of Vi, as positive so that a is (apart from position) the fixed centrode for c < a, and vice versa. (5) f=2a{x + \a). (6) The fixed centrode is a circle passing through 0\ O"; the body centrode is a circle of twice the radius of the fixed centrode. The path of any point in the fixed plane is a Pascal hmagon ; the points of the body centrode describe cardioids. (8) Two equal parabolas ; the motion is the same as that of Ex. (5). (10) With O as pole and OB as polar axis the equation of the Jzxed centrode is r^ cos^ 9 — 2 ar cos^ 6 -\- a^ = P. With O as origin and OB as ;e-axis the equation in cartesian co-ordinates is {x'^a'- l^)^x^ +/ = 2 ax\ or (;c2 + a2-/2)V + /) = 4«V- This last equation represents, however, not only the centrode of AB as B moves on the positive x-axis, but also the centrode of AB when B 461 462 ANSWERS. moves on the negative x-axis. With l-\-a = s, l—a=.d, the cartesian equation can also be written : The equation of the body centi-ode, with A as pole and AB as polar axis, AC = r' , 'ifBAC= 6', is found by observing that r=r' + a, I sm 6' = OB s\n6 = r cos 6 sinQ ; substituting the resulting values of r and 6 in the equation of the fixed centrode, we find {a- - P cos^ ey - 2 aPr' sin^ ff + 1-{P - a"- cos Q') = o, which breaks up into the two equations ,_, /+acos^' ,_, l—acoiff a + /cos^' a — /cos 6 These relations can be read off directly from the figure if perpendicu- lars be dropped froQT O on AB and from ^ on AC. In cartesian co- ordinates each of these equations is of the fourth degree. For \\\& path of any point P \. After eliminating (j> by means of the relation //a = sin 6/sm <^, it re- mains to eUminate ; the result is somewhat complicated. For the path of the midpoint of AB the equations reduce to .T = a cos ^ -1- I / cos (^, y = a sin ^ — -^/sin <^, whence x = ^ a^ — 4/ + ^sl P — 4/, which is of the fourth degree. Page 27. (3) V3. (6) 2acos\a. (4) 9.3 miles; N. 10° E. (7) (b) 160° 48'.;. (8) (V^-i)«, iV2(V3-i)a. fio) Inclination to vertical: {a) ()\° ; (b) 22%° ; {c) 36!°; {d) 68|°. Page 33. (5) («) 5-9; (b) 40.6; (^) 73.3; (,/) 35.2; (o<^/^^^^-, ^=-7^-0 ^0 V 2 gR _ Vj (k'R -s)- \/so{k'R - So) + ^R(cos-^-\^- ■ COS" •l4 {b) If v„ = ^2gR^J^, t=:^yl^^(so^-s^). (c) If vo>V2gRyj~, ^Rlog ■V7+~i?R + Vs V^L VJ^+~i^R + VZ + Vso(so + ^R) - Vs(s + k'R) 464 ANSWERS. (6) {a) v=s ™les per second, t=26 min. {b) v=i miles per second, /= 27 min. ic) »= 12 miles per second, t= 19 min. (7) If Vti<^2 gR, the height above the earth's surface to which the particle rises is h = v,^R/{2 gR -v^^) ; the time of rising to this height is R 2gR — V 2gR ■ _i ■Vi) z'o + ^===,sm V2 gR — v^ V2 gR_ If Vo = \/2gR, the time of rising to the distance s from the center is /=_^(.i-^t), sRy/2g and the particle does not return. If Wo > V2 gR, VR+VR+7n /= RV2g where Vj(x + k') - V^ (i? + /c^) + k' log - »0^ — 2 ^^ (8) h = R, /= (i +^7r) Vj?/^ = 34 min. 50 sec, approximately. The time of falling back is the same as the time of rising ; comp. Ex. (i). (9) About 7 miles per second. Page 48. (i) Differentiating s = C^ sin ix.t + C^ cos fi-t, we find the velocity dsjdt = v zs, a. function of the time V ^ /xCi cos /a/ — /A C2 sin jjif. As s = R when /= o (in the problem of Art. 82), the former equation S'^^^ R=C, cos o, .-. C2 = R, while the latter gives, with w = o for /= o, = fxCi, ." . Ci = o. With these values of Q and C^ the two equations reduce to s — R cos fjif, v = — fji.R sin /*/, of which the latter can also be derived from the former by differentiation. (2) w = S miles per second ; T= 2 ir 'VR/g = i hr. 25 min. (3) s=lR (ei^' + e-i^') = R cosh ixf. (4) 4 1^. ANSWERS. 46s Page 51. (i) u) = TT radians ; ^=15.7 ft./sec. (2) (a) 3; (d) 28^. (3) -0.157 rad./sec'. (5) (a) 402 ; (i) 25560. (4) 5- (7) 31 ft./sec. (8) (a) 0.022 rad. /sec'*; (d) 15.7 ft./sec. ; (c) 7.8 ft./sec. Page 56. (3) z;= 21 ft./sec, at 22° to the train. (5) v= 24.2 ft./sec, at 24!° to the track. (6) About 20". (7) 36 miles per hour. (8) z^j = z/j sin ^. (9) Resolve v into Vo parallel to the track, and v^ along the tangent to the wheel; show that v bisects the angle between these components; it follows that Vo = v^, and hence z' = 2 z/o cos OCF, where O is the center of the wheel and C its point of contact. (11) r=vat, d = t; hence, eliminating /, r={vo/ui)0, a spiral of Archimedes. (12) At the pole O erect a perpendicular OF = a to the radius vector OP= r; then F'F is the normal. Proof by Ex. (10). (13) For the elhpse, rj + rj = 2 ff, whence dr^ldt= — dr^ldt. Notice that dr^jdt, dr2/dt are the projections (not the components) of the velocity v (with which the curve is described) on the radii vectores r,, r^. This is seen by observing that v can be resolved into components in two ways : (a) into dr^/dt along r, and a component _L r^ ; {b) into dr^/dt along r^ and a component _L ro. Hence the perpendiculars erected at the ends of dr^^/ dt and dr^ldt (laid off from F in the proper sense) must meet at the end of v. (14) The projections of the velocity on the radius vector and on the focal axis are in the constant ratio e of the focal radius vector to the dis- tance to the directrix. It follows that the tangent meets the directrix at the same point as does the perpendicular to the radius vector through the focus. Page 62. (4) 7V= 210, u>= 22, V— 14 ft./sec. (5) 16.5 knots; 9^ it./sec. (6) 55°, 66°, 2| in. (8) sf ft./sec (7) 0.174, 0.119, 0.146 of the stroke. (9) At 0.447 of the stroke. 466 ANSWERS. dx d6f n smzO \ (iy , ^ dd „ ^ ' dt dt\ zVot^ — sin^e/ «^ "' ,7^'.v (dB df~ ""Kdt itr cos 2 6 + sin 4 ( cos Q + n {m- — sin^ 6) d'y , , (dff^ . . \i m = l/a is large, we have approximately Page 67. (2) Follows from the last of the equations (6), Art. 114. (3) By (2), Art. 1 1 2, /„ = ^ = p • (^^ (4) By Art. 112, /„=/ sin i// = z'Yp ; hence, »-=/ - psmxp. (6) Since j is directed towards A, we have, with A as origin, /g — o, I.e., r' —r = const. dt (7) — =m=const. J r=const. ; hence, by (6), Art. 114,/=/,= —rwl (9) Differentiate twice with respect to t the equations of the cycloid d\x x = a(6 — sin 0), y = a{i —cos6). At the lowest point: — ^ = o, d'y fde\^ u ^- u . . , d:'x d'e d^'-y fdd\ J = \lt) ' at the highest pomt: ^=2«^, -^ = _a^-j. Page 69. (i) {a) 2360 ft. above the point ; {b) after 3 min. 2 sec. ; {c) 2670 ft. behind the train ; {d) 25 miles per hour. (4) 45°- (6) Construct a vertical circle having the given point as its highest point and touching, (a) the straight line, {b) the circle. Page 72. (9) (a) 137^ ft. from the vertical of the starting point ; {b) 6:^ sec. ; {c) 201 ft./sec, at 6\° to the vertical. (10) 227 ft./sec. (11) 4° 21' or 86° 48'. ANSWERS. 467 (13) Let OV—Va be the given initial velocity. On the vertical through O lay off OD = H=v^/2g; then the horizontal through Z> is the directrix. Double the angle DOV, making ^ VOF= ^ DOV, and lay off 0F= OD = H; then i^ is the focus. (14) With vil2 g = H, the locus is jc'^ = — 4 Hi^y — H), a parabola. (17) A horizontal line. (18) {a) 1.5 sec; {b) 25.1 ft. from the building; ( is obtained; determine the distances CA, CA'. (5) Vji = r cot , where — ;: is the velocity of A. (6) Vji= 28.3, v,^= 22.4 ft./sec. Page 140. (3) Based on the proposition that the bisector of an angle of a triangle divides the opposite side into segments proportional to the adjacent sides. (5) The center of the incircle of the triangle formed by the mid- points of the sides. (6) About 1000 miles below the earth's surface. (8) x=y={2/^)r. (9) Taking OA as axis of x, X = ^ (u sin a + cos « — i), v = ^ (sin a — a cos a). a 'a'' ' , _ 3 V2— log(i +V2) 3 _ 2 a/2 — I , (10) x= ~= £^^^ — ^ ■ -, y = — - . 4 a. V2+l0g(l+V2) 4 V2+l0g(l+V2) (11) x = i7a, y—\a. (12) x =y = \a. _ _ ^x - — - (13) X = o, y= \-iy, where s = <:{e°~e '). / ^ ~ sin ^ - I — cos - 17/1 (14) x=r- —^, y = r , z = J krO. Page 150. (i) \ V«^ + b\ i Va^ + 4 b\ i V4 «' + b'. (2) With AE, AF as axes, x = ^ti^ a, y = Jj% ANSWERS. 469 (3) With the sides of the triangle as axes, x=y=--^. -a; distance from center = -^-, a. 6(7r- i) (5) Resolve the area into elements parallel to BD. (6) With the lower base and the perpendicular side as axes, .V = \{a^-Vab^ b-^/{a ^b),y = \{a + 2 b)h/{a + b). (7) Compare Art. 231. I {a + a'YlS + 4 a'(b - jS)a (8) x = 2 {a + a')P + 2 (b — I3)a _. la' + bS-S' (9) ^=^___^= 4.90 m.; fi . . ,. ^ I a' + bS first approximation, x = 7 = = 4.93 in. ; 2 a -\- b — , . . - I a- second approximation, x = 7 = 4.50 in. ', ^_ ia(a + 2b')-(2 a — b + b')8 ('°) ^ = -, a + b + b'-2B ^ = 4-5 in- _ i a''l3 + ba''-a''fi i g-p + ba' i a'/3 ^'^'^' "^^1 aji + ba-aji' 2 a^ + ba- aji' 2 ap + bci' °-^^ '^' 0.33 a, 0.25 a. (14) x = -bh/{a^-b'). (15) For a segment of a ring of angle 2 a and radii r,, r^, the dis- , , . , , , . _ 2 sin « r,^ + Ti^a + ^2^ tance of the centroid from the center is x = - ■ ■ 3 « r^ + r^ Hence, x = (740 + 73 V2) = 3.65 ft. ; /. e., the centroid lies about 147 TT I in. above the lower arc. {j.(,)x = \x, y = \y. (17) x = ^, J'=|- 2 (18) x = — (7 = 0.4053(7, y= — I 2 (21) Take the vertex as origin, the axis of the cone as axis of x, and one of the bounding planes as plane of xy. Then, if a be the radius of the base, h the height, and 2 a the angle at the vertex of the cone, the formulae of Art. 236 give Tj = (a/A)x = tan a, • x, S =^ ah sec a • 470 ANSWERS. (22) About 2600 miles from the center. (23) At I ^ from the Hd. Page 158. / N - ■, / \ - 1 z, 3 >^i jM^VvW (3) Let F, be the volume of the supplementary pyramid, Fj that of the whole pyramid, V that of the frustum ; x\, x,, x the distances of their centroids from the lower base ; hi, h,,, h their heights. Then the equation of moments is ( F, — V^x = V^^ — VyXi. By geometry, V,/ V^ = ri/>\^ ; hence (;'/— r{^)x = r.^^ — r^x^. Also Ag/'^i^'V'']; K—h^=h, 'Xi=\h^,, x^ = h-\-\hx; whence finally x is found as in (2). ,^3(i5^. (5) x = lh. (7) J^ = AjV Z: (6) y = ^_y\. (8) j<: = |a, ;- = f ^, (9) («) x = -y=\a; (d) S = 4S-^+i^? 90 TT ,,, - 977^+16' _ „ /x - 2(1517—8) (^) •^■=T^«'^=^^' (^) "= ,5(3—4) ' 6(9^^-16) (10) Take as element a hemispherical shell of radius r and thickness dr;x = — — i-^2— a. 2 (« 4- 4) (11) Let /"i, /j, /'a be the vertices of the triangle, P^P^ a median, G the centroid ; then -J— = -, _L^ ^ -¥ - Xi . whence, S=Xi+f(x4-a-i)=i(jfi+jC2+.s;3). -^ i-* 4 3 ■'1-' 4 ^^ — -^1 For the tetrahedron P^P^P^P^ let P5 be the centroid of the base opposite /^i so that jCj = i-(,a;2 + *3 + *"4)- Then, proceeding as above, we find X = \{x-y + X2 + x^ + x^. (12) About 0.2 mile. (13) x = \{II+h). (14) Compare Art. 225, Ex. (5), Art. 224, and Ex. (11), Art. 238. V=^w{p + q + 7-)A, where A is the area of the triangle ; S=ir\_a{q + r) + b(r+p) + c{p + q)'\. ANSWERS. 471 (15) Taking the axis of the cup as axis of y, let (x^, y^) be the centroid of the mass m of cup and handle, (o, y^) that of the water whose mass m' = ky.,. Then the co-ordinates x, y of the centroid of the cup, handle, and water satisfy the equation {m 4- ^yi) x^ — kx^xy — 2 mx-^x + ni-^x^ = o. (16) Taking the axis of z parallel to the axis of the cylinder, and the origin in the line of intersection of the bases, we have V= ( \zdxdy, or if (^ be the angle of inchnation of the bases, V= tan <^ I iydxdy = tan <^ • J" I i dxdy. (17) Apply (16) twice. Page 161. (2) 34yi miles per hour. (3) 32,000 Ib.-ft. Page 165. (i) 6.4 X 10' poundals, 8.9 x 10' dynes. (2) 4.5 lb. (3) 0.14. (5) 60. Page 174. (3) 120°. (4) 218 lb. at 36° 35' to the force of 100 lbs. (5) 10.35,14.64. (7) Q = ^{-F + ^4^'-3n- (8) 569, inclination to horizon = 99° 26'. (9) Twice the focal distance. (10) 124° 14'. (11) 9°°- (12) 18.48. (13) J? = 6 and acts along 5. (14) The resultant acts along the diameter through A, and is in magnitude equal to the perimeter. (15) (1+V2)/'. fie) (a) Ws\ne, IVcosO; {b) Wta.ne, W/cosO; {c) IFsmd/cosa, IV cos (0 + a) /cos «. ( 1 9) Produce i? C to the intersection J? with the circumscribed circle ; then DA is equal and parallel to the resultant of OA, OB; DAO'Cis a parallelogram ; hence, Z>A = CO'. (20) Resolve the components /"i, F^ along the bisectors of $. 472 ANSWERS. Page 179. (2) T=IV- a/c, P=W ■ b/c. (3) ^C must bisect the angle BCW. (4) R- = A' + B- -V C^ ^- 2 BC cos « + 2 C^ cos /3 + 2 AB cos y. (5) See Arts. 286-288. (7) The sura of their moments must vanish for two points in the plane not in line with their point of intersection. (10) P=%W; T=%W. (i i) T= JV; P= 0.89 JF along the bisector oi'^BCW. (12) P= Wsm{a-\-li) sin /3 becomes a maximum for /3 = -J- (tt— «), i. e., when the sail bisects the angle between boat and wind. (13) w sin^ f.F ^^" " • ^ '^^ sin(« + y8) sin(« + ;8) (14) Tens ion in^.g and CD= W- 1 /.d, tension in.5C= W- {c—l)/2 d, where d=^r--\{c-iy. (is) /'„a.. = 0.2 6i?. (18) i^tana; 13.4, 28.9, 50, 86.6, 137.4, 186.6, 283.6, s7iS' «>• (19) 848, 282 ; 1000, 600. (20) 0.640 W. (21) (a) V2/>F; (3) 2 fFcosl(|7r±^). (23) 2 Jf^cosi(i7r + o!), etc. (a) a = 30°, ^=120°, 7 = 30°; (U) impossible. (25) r=o.s6fF; A = C=o.^2W, ^=0.67 /F. (26) T=A = C= IV3 /F, ^ = W. (27) 8 = 7r-3;3, 30°<;8<6o°. Page 192. (i) Take moments about the fulcrum, (a) 4.32 ; (J)) 3.94. (2) ia) ^ = 8|, B=-i\; {c) A=i^\, B=i^l (3) (a) P=W- ib) P=(i+V2)JV. (6) (a) 19.4 tons and 21. i tons; (3) 30.5, 9.9. (8) Let a be the angle subtended at the center by the side 12, and the angle at which the diagonal 13 is inclined to the horizon; then tan 6 = y-f -j CSC « + cot a. (9) X = FJ sin ('.^/{Fi sin «! + /^^ sin a^) . (11) 49 lbs. per square inch. (12) 63.1 in. ANSWERS. 473 Page 201. (i) C=i, r>=i\, E=6l, AB = ^.s, BC=^.i, CD = ^.o, DE = 4.2, EF= 8.9 ; reaction 3X A = 4.5, at E= 8.9. (2) JI=7S, T=8i. (t,) c=\ : H=wc. (4) 1185 lbs. per square inch; 3.5 ft. (5) H= 47.4 lbs., r= 57.2 lbs., j^ - ^ = 9.8 ft. Page 215. (i) r=7.68, ^ = 9.76, ^=12.80 lbs. (2) T= 2 mW, A = V4 ni^ — 2 m + 1 JV, where ;« = c/L (3) The three forces IV, T, A, must pass through a point ; cos 4> = 2V-j(i — m^, where m = l/l>; T= IVsec (j>, A= lVta.n 4>- (4) r= i fFcos 6l/sin {e-). (5) A^ = —B = "^ W A =W. (6) B = J^W,A^ = ~J^-^/¥^^'W,A,^=i^^-^YV. (7) {a) Equilibrium impossible; {b) E= E = ^-^^ JV, A = W. (8) X = am, A = ■\/ n? — i W, C = m W, where ;« = (JJa)'^. (9) ^ = 1.(3 w + ^F) tan e, ^ = (3 o' + fF)Vitarf6l+i. (10) cos ^ = i {m + Vot^ + 32). where m = //a. <^ being the angle at which .5C is inchned to the horizon. (12) m = (JV+ 2 E' sin a'')/(}V+ 2 i^sin a), C, = Ecos a-E' cos a', C^ = W+ i^sin « + E' sin a'. (13) A = \W, B = iWcosa,E=^Wsma. (IS) P= (n — i) cos a + 2 V« sin 1 f^sin K. 474 ANSWERS. Page 219. (3) 60^ (5) cot ^ = I for hemisphere, hj \ a for cone, ^/2 a for pyramid. (6) h = V3 p^lp^a. (7) h=^p-^/2pia. Page 222. (3) Let h be the altitude of AB C, through C, and a, b the segments into which this altitude divides AB ; then C^ = _ cosacosl3 j^^^ ^ ^ h lV,C,= ^-tz^JV. a + d (4) ^x sin (a + y8) 1 sin a cos ;S 2 sin (a + ;8) 2 ^+a ^F=2?F-^„, Page 235. (2) p^ s'"-^ (i) 4 tons. (2) ?= iiii^^^ — W. cos (<^ — «) (3) («) gm(i^^^^^sin(g+.^)^^ ^^ ^^ cos cos . F^2lV sin 61 ; (f) if P act up the plane, P^ sm( + e) ly . ^^ p ^^^ down the plane, F^ sm (.^ - 6) ^ cos (^ (5) 2 2 61 lbs.; 5 61 lbs. cos (6) (a) P=^JIL(I^^IV; (i) P^^Ml±^w. cos(« + <^) cos(a — <^) (7) e = ^-2<^. (8) g = tan-^^^'+/-'^'. (9) ^ = (i-;«cos2e)«^ C=OTcos6ifF, ja = ;« cos 6 sin I — m cos^ 6 , where wz cos 6 ?F, sin 2 6 = —tan 2 d>, m m — IJc , s y, sin (^ — V' = 7T-V. (6) If the original velocities are of the same sense, » = 42-I, &' = 46|- ; if not, v = — 19I, v' = io-|. ANSWERS. 475 (7) e = o gives (a) 44^^, (i) -2^; e=i gives (a) » = 38i v' = 49I, (d) z; = -s5l, »' = 3Sf. (8) v = -eu. (10) 8|ft. (11) (a) 0.31 ft.; (i) 9| sec. J (^) 66^^ ft. (13) (^^Y^^- (14) (a) 44 ft./sec. ; (^) 38I ft./sec. (is) For ^ = o: (a) v= '^ u ; {b) limz; = o; (c) limw==«'. 711 -\-nr ' \ / m — m , 2 m ■■ u, v' = m + m m + t?i limz'' = oj (c) \\mv = 2u'—u, \\mv' = u'. Interpret these results. For.= i: (^a) v="^^^,u, v' = -^^u; (^)Iim^; = . fn + m' m + m' ' Page 288. (i) The momenta are as 20 : i ; the kinetic energy is the same. (2) 3125 lbs. (3) gibs. (4) 6250 lbs. (5) About 450 lbs. ; about -^ of the available energy is wasted. (6) 4.9 ft.-lbs. (7) 363 ft. -tons ; 9 miles per hour. (10) 3 tons. (11) 13^ and 2 ft.-tons. (12) 144,000 lbs. (13) About 3300 lbs. (14) About 60 lbs. per sq. in. (15) About 400 lbs. per sq. in. (16) 449 Ib.-ft. (17) 26 lbs. Page 292. (i) 57 Ib.-ft. (2) 16 ft./sec. (3) ^-=8.8, ;8=59''; ^''=14-4, /?'= i7F- (4) v=(>\ ft. per second ; F— 160,000 lbs. (5) 5:1- (6) 17:1- (7) The impinging sphere is brought to rest ; »'=Vz?+«^, tan j8' = «'/«. Page 300. (i) (b) 250 lbs. (2) ia) 8 ft./sec. ; (b) 20 ft. (3) ia) 764 lbs. ; (K) i^mile; (c) 1 146 lbs. (4) 4.9 sec. (5) 35.8 and 4.2 ft. above the ground ; 50^ lbs. ; 11. 2 sec. (6) j = {nil sin B^ — m^ sin ^3 — /.iiOTj cos ^1 — /ij^Zj cos d^g/{m-i + m^, T= (sin $1 + sin 0^ — i"i cos di + /u,2 cos 6^ m^m^glim-^ + m^. (7)y=5.8ft./sec.^; r= i^l lbs. (9) 0.036. (10) 288 ft. 476 ANSWERS. (ii) (a) About 1600 lbs; (i) about 3750 lbs. (12) 589 ft. (Find first sin 6 by successive approximation.) (13) (a) 120; {&) 200 lbs. (14) 4125 lbs. (15) 562^ lbs. (16) (a) 2625,3058; (/;) 1375,1602; (a) 217; (6) 113. (17) (a) 1267.2 ft.-tons; (i) 4435.2 ft.-tons ; (^) 5 : 2 ; (d) about i^ mile. (18) (a) 1 146; (i) 1946; (c) 800; (d) 2050 lbs. (19) 2016 ft.-tons. (21) About 7 in. (23) 513,274 ft.-tons. (20) 39 7 J ft.-tons. (22) 30 ft. Page 313. (i) (a) 11,133 ft.-lbs.; (^) 22ift- (2) 6 lbs./in.2 ; 83,400 ft.-lbs. (4) 864 ft./sec.^ (5) Time = - ( tt -f 2 — ), work = | micx^, where k = •\/ — — — rr- (6) (a) ^ lb. ; (3) 11. 3 ft./sec. ; (c) 0.6 sec. (7) If ^0 < 2 yu,;«i (/i — /) Wi, the particle comes to rest between Pq and Q (Fig. 155). If Xo> 2 /x;«i (/j — /)/oti, but j^o (k^.»o— 2 jig)^ 2 l/fj.?ng, the particle comes to rest between (2 and Q' ; etc. (9) If .To < e, nothing is changed ; if ^Cq > e, the particle performs simple harmonic oscillations about (?], just as in the case x,^ < e. (10) The length /is increased to / + e-\-^e{e + 2/1). (11) Take x^ = l.^ — I vci Ex. (8). (12) 42I min. (14) X, = h (i 4-Vi + 2 hi/h). Page 317. (2) The equation of motions — = —ms;—mkv'- gives, with F UZ'o cos ix/— P'sin u/ jO' , /uWn . ?> = - r « ^ ^, .y = 4-, log^smu^-fcosui' ju, ^z^o sm(U./ + ^cos/i/ K \g / zy"' r + z^V 2/j ^g + kv' (3) Timeofascent= — = tan"M \/- • ^o ), height=— logf iH — v^ ■Vkg \^g J 2 A \ g (4) ANSWERS. 477 ^0 V^ + kv^^ (5) In vacuo z'l = 1 7 ft. per second ; in the air z/j = 122 ft. /sec. (6) J- = ^ (i - e-"'), V = v^e-"' = »„ - ks. k (7) » = |(i-^-«), ^ = f ^+|(^-"-i) = -^ + |log '^'^."■-'^» Page 321. (2) The logarithmic decrement is log if~" = — Xt. (4) If /u. =?^ K, J = (Tjcos k/ + (-jsin k/H — ;; 2 sin /A/" j if fJ. = K, at . K- — IA. s ^Ci cos Ki + 1:2 sm k/ -| sm «/. 2 K (5) The term due to the forced oscillation is , cos /x( t- /■„) ; hence, this oscillation lags behind the force by the phase difference /x4 ; the amplitude is less than for undamped oscillations. The free oscilla- tions (if any) will rapidly die out so that the motion soon approaches a state of motion given by the above term. Page 324. (i) I watt = 0.00134 H.P., I H.P. = 746 watts (with ^=981). (2) I metric H.P. = 736 watts = 0.986 British H.P. (3) 2o|. (s) (a) 64; (i) 224; (c) 384. (7) 35,200 gals. (4) 41^. (6) 188 gals. (8) About i hr. (9) (a) 177,000 ft.-tons; (d) 51 hr. ; (c) about 74 days (of 8 hr.). (10) 253% lbs. (12) 12.8. (14) 21J. (11) 15.6 lbs. (13) 4- Page 331. (i) I7=cz+C. (2) V=mg(z-Zo). (4) /7=-f/(ryr = -F(r). (5) As r = (x - XoY + (y— yoY + (z - ZoY, rdr = {x - x„) dx + (^y — y,^ dy + (z — z^ dz ; hence the direction cosines oi R=fif) are X - Xt, ^ J. 5x gj^ Hence X.= ± fif) ~, etc., and putting r dr or f^ f(r) dr = Fif), we find U= ± F{r). 4/8 ANSWERS. Page 343. (i) Put r= i/u, and find d-u/iW in terms of ti alone : <^^=:^u-\-{n-\){i-e') ^-'^»«-2"+i - (« - 2) ^-"«-''+i. Substituting in (lo), Art. 533, we find fir) = 4 [(« - i) (i - Or"»"'"^' - (« - 2) '^"""'']- n=\ gives an ellipse if i, all referred to focus and focal axis ; « = 2 gives conies referred to their axes ; « = — i gives Pascalian limagons (cardioids for 1? = ± i) ; « = — 2 gives a lemniscate if d'=± i. (^) C'^) <^V^)^ (^0 ^; W ^^^; C'^) ^(^+^> (3) 8«V/^- (4) Ellipse, parabola, or hyperbola according as ;u, = v-iyi, where y^ is the initial distance of the particle from the plane, v^ the component of its initial velocity at right angles to the plane. Page 346. (2) The equation of the orbit given in Ex. (i) is satisfied not only by (xo, jVo), but also by iyiJK, v-i/k) ; i. e., the orbit passes not only through the initial point /J,, but also through the point Q, which is the extremity of the radius vector OQ=:Va/K parallel to Vt^; OP^, and OQ are the conjugate semi-diameters whose equations are x^y =;)',><') v^y^v.^. (4) The problem reduces to that of constructing the axes of a conic from a pair of conjugate diameters. (5) The curve being referred to its axes, the equations of motion are X =^ a cos Kt, y = b sin Kt for the elhpse, and x=:\a{e'''+e~'^') = a cosh Kt, y = ^ b(e''- — e^"') = b sinh Kt for the hyperbola. (6) From the equations of Ex. (5) it follows that for the ellipse tan ^ = (^/a!) tan k/; \\eT\ct d6/dt= Kab/r- ; then apply (4), Art. 529. (8) Use the equations of the conic in terms of the eccentric angle 22 ft./sec. (7) To count the angles from the highest point of the circle, put ■77 — 6 = (j> ; then, putting h — 1= h', where h' is the height to which the velocity at the highest point is due, we have AT ( ± 2/+/«'\ iV^= — 3;«_f/cos <^ — -J. The particle remains on the curve as long as cos ^ >|^(/+ A')//. Distinguish the cases /?' > o, h' = o, h' <,o. (9) i.46i7«- (10) ti = TrVa/g. Page 377. (i) {a) 9.8 in.; {b) 1.23 lbs.; {c) 35!°; i.i in., 11. i lbs., 844°. (2) Distance from axis r= (g/w^ cot 6. (3) P— mg cos & (ruy'/g cot ^ — i). (5) Taking ^C as axis of *■, its intersection with the required curve as axis of j^, the equation of the curve is 7^= (2gfr Mx^, C = /s + Mx^. (8) The centroid may be such a point ; if the central elHpsoid be an oblate spheroid, the two points on the axis of revolution at the distance ± V(/i — /2)/^from the centroid are such points. (9) The ellipsoid must have the same central elhpsoid as the given body; its equation is x'^/A'+f/B' + z'/C' = c,/M, where M is the mass and A', B', C are the moments of inertia for the principal planes of the body at the centroid. (10) p" = MIN, where M 'z^ = I — f (?2^ + 1^ - 1^)' etc. r ANSWERS. 483 Page 420. (I) ^.Ji^ii^l (7) 6gL 2600 gfxr (^) i^"'""- (8) j (i^ = |. jU, - COS a, (o„ = 2 ;u, - COS Of, |«a)o|, /^ has the same sense as in the case of Art. 729 ; the equations are therefore the same ; but (Uq is a negative quantity. If Wi,<|aa)(i|, the sense oi F is reversed. In both cases the sphere rolls forward, i. e., in the sense of v,^. (3) 4=9| sec; z' = — 10.5 ft. /sec. ; number of revolutions per second = |. (4) la. (5) ^V INDEX. [The Numbers refer to the Pages.] ABERRATION' of light, 57. Absolute motion, 25. units, 163-166. Acceleration (in rectilinear motion), 36-49. (in curvilinear motion), 62-68. , angular, 50. in cartesian co-ordinates, 65-66. directly proportional to the distance, 46-49. of gravity, 37, 39. inversely proportional to square of distance, 42-46. , normal, 64-65, 67. in polar co-ordinates, 66-67. in simple harmonic motion, 76, 78. , tangential, 64-65. in uniform circular motion, 74. as vector, 63. Activity, 322. Advantage, mechanical, 26S. d'Alembert, principle of, 334-337, 381. Amplitude, 74. , correction for, 99. Anchor ring, moment of inertia of, 399. Angle of friction, 232. of incidence and reflection, 291. of repose, 233. Angle-iron, centroid of cross-section, 151. Angular acceleration, 50. momentum, 333. , conservation of, 386. , equations of, 383. about fixed axis, 416-417. in plane motion, 450. of rigid body, 383. Angular velocity, 49, 50, as rotor, 113. resolved along axes, 117. velocities, composition of, 114-117. , parallelogram of, 116-117. Anomaly, eccentric, 354. Anomaly, mean, 356. , true, 351, 354. Anti-parallelogram, 123-124. Aperiodic motion, 319. Aphelion, 351. Apparent solar day, 29. Appell, P., 179. Archimedean spiral, 57, 343. Arcs of curves, centroids, 135, 138-141. Areal acceleration, 51. density, 135. velo'city, 51. Areas, centroids of, 141-153. , conservation of, 387. , principle of, 104. Arm of a couple, 201. of inertia, 394. Astatic equilibrium, 218. Attraction and repulsion, 102, 305-309, 343-344, 350- Atwood's machine, 297-300, 425-426. Available work, 268. Average angular velocity, 50. force, 303. piston pressure, 305. velocity, 34. Axis of rotation, 5. Axle-friction, 239. in Atwood's machine, 426. Balance, common, 218. , running, 434. , standing, 433. Ballistic pendulum, 449. Beat, 95. Belt-friction, 240-242. Belt on pulleys, 51, 52. Bending moment, 228-230. Binding screw, 271-272. Body centrode, II, 112. falling in vacuo, 39-41. 485 486 INDEX. Body falling toward the earth, 44-46. through the earth, 47-49. projected upward, 40-41, 46. of reference, 25. See also Rigid body. Bole, 160. Boyle's law, 292, 304. Breaking strength of cord, 270-271. Buffers, height of, 458. Bullet, 34, 40-41, 70-73. Buoyancy, 314. Byerly, W. E., 86. Cardioid, centroid of arc, 141. Catenary, centroid of arc, 141. , common, 198-201. Catenaries, 193-201. Center of displacement, 8, 9. — force, 100, 337. gravity, 136, 191-192. inertia, 136. mass, 131, 134, 135-136. oscillation, 419-420. parallel forces, 189. percussion, 447. rotation, 6, 11. suspension, 419-420, 439. Centimeter, 7. Central axis, 20-21, 209, 210-21 1, 250- 251- ellipsoid, 408. forces, 351-362. motion, 99-100. Centrifugal couple, 434. force, 367, 370-371, 432. Centripetal force, 367. Centrode, lo-ll, III-II2. Centroid, 131, 134, 135-136, 192. of arcs of curves, 135, 138-141. areas, 141-153. I93- any solid, 155-156. r'gid body, motion of, 384. solid of revolution, 155, 158, volumes, 153-159. Centroidal line, 397. principal axes, 438-439. C. G. S. system, 6, 32, 36, 132, 160, 163, 260, 322. Chain, kinematic, 119. raised by capstan, 302-303. Circle, centroid of area, 141. Circle, motion in vertical, 92-99, 365-366, 371-375- , motion under central forces, 343, 357. Circular arc, centroid of, 139-140, 141. disk, centroid of, 146-147. . on horizontal plane, 459-460. , impact, 447. as pendulum, 420. , reduced mass, 425. rotating, 421, 459-460. line or wire, moment of inertia, 395, 398, 409- orbit, 357. ring, moment of inertia, 395, 398. sections of ellipsoid, 406. sector, centroid of, 145, 152. segment, centroid of, 152. Cissoid, centroid of area, 153. Clamp, 271-272. Coaster, 300-301. Coefficient of friction, 231. restitution, 281, 283. rolling friction, 243. Complete constraint, 11 7-1 19, 364. Component, 4, 23, 56-58, 170. Composition, 23. of angular velocities, 114-117. concurrent forces, 176. couples, 205-206, 207-208. force and couple, 206-207. forces acting on the same par- ticle, 170-175. forces in the same plane, 208, 209-211. intersecting rotors, 11 6-1 17. parallel forces, 183-184, 185- 190. parallel rotors, 114-116. simple harmonic motions, 79- 92. velocities, 53. Compound harmonic motion, 78-92. wave motion, 87. pendulum, 419-422, 439-442, 447, 448-450. Compression, 222. Conchoidal motion, 14-15, 16. Concurrent forces, 175-183. Conditions of equilibrium for forces acting on the same particle, 172-174. ■ for concurrent forces, 1 76. INDEX. 487 Conditions of equilibrium for parallel forces, 190-igi, 192-193. for forces in a plane, 20S- 209, 210-211, 211-217. for forces acting on any rigid body, 244, 250. Condition for pure rolling, 452. Cone, centroid of, 139, 153. , equimomental, 405. of friction, 233. , moment of inertia of, 395. , principal axes of, 409. Confocal conies, 410-412. quadrics, 412-414. Conical pendulum, 375-378. Conic sections as orbits, 343-347, 348-358. Conic, tangent to, 57. Connecting rod, motion of, 14, 16, 58-62, 112. and crank, forces acting on, iSi, 272-274, 428-429. Conservation of angular momentum, 334, 386. areas, 334, 387. energy, 286, 308, 308-309, 320, Z^Z' 330- • linear momentum, 385. motion of centroid, 280, 385. Conservative forces, 329. Constant, elastic, 311. of gravitation, 305-306, 313. Constrained motion, 362-378. Constraining force, 365-366. Constraints, 18-19, 11 7-1 19, 255-258, 263, 297, 363-366. Constraint, complete, 11 7-1 19. Cord, equilibrium of, 193—201. running over pulley, 297-300, 425- 426. Correction for amplitude in pendulum, 99. Cotterill, J. H., 61. Couple of forces, 189-190, 201-208. , centrifugal, 434. represented by vector, 204-205. of rolling friction, 343. Crane, 179-180. Crank and connecting rod, 272, 274, 428- 429. Cube, impact on, 449-450. as pendulum, 420. , principal axes of, 409. Curvilinear motion, 325-378. Cyclic sections of ellipsoid, 406. Cycloid, II, 57. Cycloid, centroid of arc, 141. , centroid of solids generated by revo- lution of, 158. , motion on, 374-375. Cylinder, centroid of, 153, 159. , moment of inertia of, 395, 398, 399. , moving down inclined plane, 451- 456. , moving up inclined plane, 456-457. on horizontal plane, 457-45S. Cylindrical rod as pendulum, 420. Dai.ey, W. E., 435. Damped oscillations, 318-320. Damping ratio, 321. Decrement, logarithmic, 321. Degree of uniformity, 430. Degrees of freedom, 18-19, "9. 255-258, 381- Density, 131-132, 135. Derived units, 31, 130. Deviation due to obliquity of connecting rod, 60. Dimensions, 31-32, 37, 160, 163, 260, 322. Direct impact of spheres, 278-293. Directing couple, 423. Displacement, 3, 4, 8, 17, 19. in simple harmonic motion, 74. Distribution of principal axes in space, 410-416. Door, moment of inertia, 395. on hinges, 253-255. Driving force, 268, 427. Dynamic stability, 449. Dynamical meaning of principal axes, 438. radius of inertia, 424. Dynamics, i, 129-169. Dyne, 163, 165-166. Earth and moon, 141, 348-349, 358. Eccentric anomaly, 354. Effective force, 334-335. Efficiency of machines, 26S, 322-325. Elastic constant, 311. stress or tension, 310. strings and springs, 310-315. Elasticity, perfect and imperfect, in im- pact, 280-281. 488 INDEX. Elasticity, perfect, 311. Elevation of outer rail, 370. of projectile, 70. Elevator, 301. Ellipse, centroid of area, 141, 152, 153. , focal, 413. , moment of inertia, 398. as orbit, 343-347. 348-358- , tangent to, 57. Ellipsograph, 123. Ellipsoid, central, 408. , centroid of octant, 158. , equivalent, 410. , fundamental, 408. of gyration or inertia, 408. , moment of inertia, 395, 398. , momental, 402. ■ , principal axes, 409. , reciprocal, 408. Ellipsoids of inertia, 399-410. Elliptic co-ordinates, 412, 413-414. harmonic motion, 88. ■ — — integral, 98-99. motion, 12-14, 16, 112. Energy, kinetic, 161-167. , potential, 308, 308-309, 330. • , total, 312, 323. Epicycloid, Epitrochoid, II. Epoch, epoch-angle, 75. Equation of the center, 357. Equation of motion of a particle, 293. about a fixed axis, 417. Equations of linear and angular momen- tum, 383, 450. Equations of motion of a rigid body, 381-383. Equiangular spiral, 343. Equilibrium, 170, 172. of cord or chain, 193-201. of forces in a plane, 208-209, 2lo- 211, 211-217. on inclined plane, 175, 233, 235-236. of rigid body, 244, 250. , stable, unstable, neutral (astatic), 217-219. Equimomental cone, 405. Equipotential surfaces, 330. Equivalence of couples, 202-204, 207-208. Equivalent ellipsoid, 410. simple pendulum, 419, 420-423. Erg, 260. Everett, J. D., 6, 32, 83, 163. Expanding gas, 313. Falling body, 39-41, 295-296. in resisting medium, 316-318. Field of force, 308. First integrals of equations of motion, 104. Fixed axis, body with, 416-450. , reactions of, 431-442. Fixed centrode, 10, III. curve, motion on, 365-375. ■ surface, motion on, 375-378. Flux, fluxional notation, 380. Fly-wheel, 51, 52, 420, 421, 429-431. Focal ellipse and hyperbola, 413. Foot, 7. Foot-pound, foot-poundal, 260. Force, 161-169, 275, 284. , attractive and repulsive, 305-309, 343-344, 35°- , average or mean, 303. , central, 337-362. , centrifugal and centripetal, 367. , conservative, 329. • , constant, 294-302. , effective, 334-335- , impressed, 335. of inertia, 160, 335. , intensity of, 338. , internal, 379. inversely proportional to square of distance, 305-309, 347-362. , law of, 168-169. , normal, 326. proportional to distance, 309, 343- 347- reduced to distance from axis, 425. of restitution, 280. , tangential, 326. , variable, 302-325. Forced oscillations, 321-322. Force-function, 329-331. Force polygon, 172, 185, 225-228. Four bar linkage, 119-121. Fourier's theorem, 86. F. P. S. system, 6,32, 36, 132,161,260,322. Free curvilinear motion, 325-363. Free oscillations, 309-315. Freedom, degrees of, 18-19, 119, 255- 258, 381. Frequency, 75. Friction, 236-243, 300, 314. INDEX. 489 Friction angle, 232, circle, 238. cone, 233. in pure rolling, 453, 455-457- , rolling, 242-243. Frustum of cone or pyramid, centroid of, 158. Fulcrum, 192. Fundamental units, 31, 130. ellipsoid, 408. Funicular polygon, 186-188, 193-201, 225-228, 229-230. Galileo's laws of falling bodies, 40. Gas engine, 325. , expanding, 313. Gases, kinetic theory of, 291-293. Geometric addition, 23. derivative, 63. difference, 25. subtraction, 24. sum, 23. Geometry of motion, I, 3-28. Governor, 376—378. Grain, gram, 130. Graphical methods in statics, 171, 172- 173. 183-184, 185-188, 191, I93-I95> 201, 224-228, 234, 236. time-table, 33. Gravitation, constant of, 305-306, 313. , terrestrial, 348-349. units, 163-166. , universal, 43-44, 349. Gravity, acceleration of, 37, 39. and centrifugal force, 370-371. Guldinus, first proposition of, 140. , second proposition of, 148-149. Gyration, ellipsoid of, 408. , radius of, 394. Halley, 348. Hammer and nail, 285-286, 288, Harmonic motion, 74. Hart's inversor, 125. Head or height due to a velocity, 40. Helix, centroid of arc, 141. Hemisphere, centroid of volume, 158. Heterogeneous mass, 131. Hexagon, moment of inertia, 395. Higher pairs, 1 18. Hodograph, 63-64, 67, 73, 358. Homogeneous mass, 131. Hoolce's law of elastic stress, 310. Hoop, equivalent simple pendulum, 422. on inclined plane, 454-457. Horse-power, 322, 324-325, 421. Hyperbola, as orbit, 343-347, 348-358. , focal, 413. Hyperbolic spiral, 343. Hypocycloid, Hypotrochoid, II. Impact, direct, 275-290, 291-293. , oblique, 290-291. of water in pipe, 289. Impressed force, 334. Impulse, 161-162, 275, 390-392. acting on body with fixed axis, 442- 450. Impulsive force, 276. reactions, 444-450. Inclined plane, 68-70, 175, 296, 299-300, 369-370,451-457. Independence of translation and rotation, 388. Indicator, 304-305. diagram, 305. Inertia, 160. , ellipsoids of, 399-410. , force of, 335. , law of, 168. , moment of, 393. , product of, 393. of pulley, 425-426. , radius of, 394. , spherical points of, 410. Instantaneous axis, 15, 21. center, 10. force, 276. Intensity of force, 338. Internal forces, 379. Invariable plane, line, direction, 388. Invariant of forces acting on rigid body, 246, 248. Inverse of a curve with respect to a circle, 124-125. Inversors, 124-127. Isochronous motions, 75, 374. Jack, 235. Jointed frames, 219-228. Joule, 261. Journal friction, 236-239. 490 INDEX. Kelvin, 19, 82, 83. Kennedy, 118, 119. Kepler's equation, 356-357. laws, 101-106, 347-348, 362. Kilogram (force), 165. Kilogram-meter, 260. Kinematical conditions for pure rolling, 452- Kinematic chain, 119. Kinematics, i, 29-128. of machinery, 1 17-128. and statics, 205. Kinetic energy, 166-167. , change from impact, 283-289. of centroid, 390. relative to centroidal axes, 390. of rotation, 418. and work, principle of, 104, 295. 303. 327-331. 389- friction, 231. theory of gases, 291-293. Kinetics, i, 275-460. of the particle, 275-378. of the rigid body, 379-460. Knot, 62. Laplace's invariable plane, 388. Law of universal gravitation, 43-44, 349. Laws of force, inertia, stress, 168-169. of friction, 231-232. of motion, Newton's, 167-169. Length of equivalent simple pendulum, 419. of wave, 84. Level surfaces, 330. Lever, 192, 193, 269-270. . crank, 119, 121. Limiting static friction, 231. Line of quickest descent, 69-70. of force, 331. Linear density, 135, 398. kinematics, 30-52. momentum, conservation of, 385. , equations of, 383. of rigid body, 3S2. in plane motion, 450. motion, 4, 6-7. velocity, 51. Link, linkage, linkwork, 118-119. Lissajous's curves, 90-92. Load, 268. Logarithmic decrement, 321. Logarithmic spiral, 343. Lorenz, P. H., 435. Lost kinetic energy in impact, 288. work, 286, 323. Lower pairs, 118. Macgregok, J. G., 89. Mach, E., 169. Machine, 118. Magnetic needle, 423. Mass, 129-133. moment, 133. , reduced, 424, 425-426. and weight, 130. Material lines and surfaces, 131. particle, 131, 159, 293-294. Mathematical pendulum, 92-99. Mean angular velocity, 50. anomaly, 356. force, 303. motion, 354. piston pressure, 305. solar day, 29-30. sun, 29. time, 30. velocity, 34. Mechanical advantage, 268. Mechanics, I. Mechanism, 119, 268. for compound harmonic motion, 81. for simple harmonic motion, 77, 78. Menelaus, proposition of, 120. Metacenter, 421. Minchin, G. M., 86. Modulus of a machine, 324. Moment, bending, 228-230. of a couple, 202. of first order, 392. of a force about an axis, 250. of a force about a point, 1 77-178. of inertia, 393, 392-416. of inertia, determined tally, 421, 422-423. of mass, 133. of momentum, 333. polygon, i86-i88. of second order, 392 . Momental ellipsoid, 402. Momentum, 159-161, 167, 275, 277. — , angular, 333. of centroid, 277-278. expenmen- INDEX. 491 Momentum, conservation of angular, 386. , conservation of linear, 385. , equations of linear and angular, 383- of rigid body, 277, 382, 3S3. INIoon and earth, 141, 348-349, 358. Motion, 3. on a fixed curve, 365-375. on a fixed surface, 375-378. , mean, 354. Neutral equilibrium, 218-219. Newton, 106, 347-348. Newton's laws of motion, 167-169. law of universal gravitation, 43-44, 349- Newtonian forces, 305-309. Normal acceleration, 64-65. force, 326. Oblique impact, 290-291. Octant of ellipsoid, centroid of volume, 158. One-sided constraint, 364. Orbit, 339, 348. Oscillations, damped, 318— 320. , due to torsion, 422-423. , forced, 321-322. , free, 309-315. Pairs, lower and higher, 118. Pantograph, 122-123. Pappus, first proposition of, 140, 141. , second proposition of, 148-149. Parabola as catenary, 197. , centroid of arc, 141. , centroid of area, 152. as orbit, 348-358. Parabolic segment as pendulum, 421. Paraboloid, centroid of volume, 158. , moment of inertia, 399. Parallel forces, 183-201. motion, 127-128. Parallelopiped, centroid of volume, 153. , principal axes, 409. Parallelogram, centroid of area, 141. (mechanism), 122. , moment of inertia, 399. of angular velocities, 116-117. of forces, 171. law, 23, 116. Parallelogram law for couples, 205-206. Particle, 131, 159, 293-294. Particles, centroid of, 137-138, 140. Peaucellier's cell, 125-127. Peg-top, moment of inertia, 410. Pendulum, 92-99. , compound, 419-422. , simple mathematical, 372-375. , spherical or conical, 375-378. Percussion of axis, 445. , center of, 447. Perfect elasticity, 311. Perfectly smooth, 236. Pericycloid, peritrochoid, II. Perihelion, 351. Period, periodic time, 74, 106, 354-358. Permanent axis of rotation, 432, 438. set, 310. Perry, J., 289, 325. Phase, phase-angle, 75. Pile-driver, 288-289. Pin-friction, 239. Piston-pressure, 303-305. Piston-rod motion, 58-62. Pivot-friction, 239-240. Plane area, centroid of, 145-146. kinematics, 52-128. motion, 4, 7-16,. 450-460. statics, 208-243. Planetary motion, 106-109, 347-362. Plumb-line, 371. Point of application, 172. Polar co-ordinates, 156. reciprocal of momental ellipsoid, 407-408. Potential, 307, 329. energy, 308, 308-309, 330. Pound (force), 165. (standard), 130. Poundal, 163, 165-166. Power, 268, 322, 427. Pressure, 179. on curve, 366-367. of piston, 303-305. on rails, 370. Principal axes, 402, 404-405, 409-410. , distribution in space, 410— 416. , dynamical meaning, 438. moments of inertia, 402. radii of inertia, 405. 492 INDEX. Principle of angular momentum or of areas, 104, 332-334. of conservation of angular momen- tum or of areas, 334, 386, 387. . ■ of energy, 286, 308, 308- 309, 320, 323, 330. of linear momentum or of motion of centroid, 385. of d'Alembert, 334-337- of independence of translation and rotation, 388. of kinetic energy and work, 104, 29s. 303. 327-331. 389, 418. of the lever, 185, 189. of virtual velocities, 263. work, 258-274, 262, 264- 265, 336, 380. of work, 261. Prism, centroid of volume, 153. , moment of inertia, 395. Problem of two bodies, 359-362. of three bodies, 361-362. Product of inertia, 393, 400. Projectile motion, 70-73, 358. Propagation, velocity of, 84. Pyramid, centroid of volume, 139, 153. , moment of inertia, 395. QuADRIC surfaces, confocal, 412-414. Quadrant of circle, centroid of area, 141. Quadrilateral, centroid of area, 142-143. Quantity of motion, 159-161. Quickest descent, line of, 69-70. Radian, 7. Radius of gyration, 394. of inertia, 394. ■ , dynamical meaning of, 424. Railroad train, 33, 34, 40-41, 69, 235, 288, 293, 300-302, 370-371, 457-458. Raindrop falling, 27. Range of projectile, 72. Reaction, 179, 192, 211, 255, 263, 366. , total, 232. Reactions in compound pendulum, 439- 442. of fixed axis, 431-442. , impulsive, 444-450. Reciprocal ellipsoid, 408. Recoil, 289-290, 293. Rectangle, moment of inertia, 394, 398. Rectangular door, moment of inertia, 395. Rectilinear motion of particle, 293-325. segment, centroid, 138-139. ■ , moment of inertia, 394, 395. Reduced mass, 424, 425-426. Reduction of forces in plane, 208, 209— 211. in space, 243-244, 244— 249. Relative motion, 25-26. of planet with respect to sun, 361-362. velocity, 56, 56-58. Repulsion and attraction, 343-344, 350. Resistance, 179, 268, 366, 427. of a medium, 3 1 5-3 1 8. Resolution of angular velocity along the axes, 117. of a force into parallel components, 193- of velocity, 53-55, 56-58. Restitution, 279-281. Restoring couple, 423. Resultant, 4, 23, 170, 172, 183, 186-190, 245. Resulting couple, 245. Reuleaux, 117, 119. Revolving shaft, work of, 427. Rifle-ball, 301. Rigid body, 4, 175-176, 379. , conditions of equilibrium, 250. , kinetics of, 379-460. supported at three points, 252— 253. 257-258. with fixed axis, 256-257, 416— 450. with fixed plane, 257-258. with fixed point, 255-256. Ring, circular, moment of inertia, 395, 398- Rod, equilibrium of, 211-217, 265-267. , oscillating vertically in water, 314— 315- , reduced mass, 425. , swinging about one end, 441-442, 447- Rods, equilibrium of jointed, 219-228, 267. Rolling, 452. cones, l8. friction, 242-243, 457. INDEX. 493 Rotation, 4, 5, 49-52, 113. Rotor, 113, 167, 205. Rough, 236. Running balance, 434. Safety-valve, 193. Sagging of telegraph wire, 201. Sanborn, F. B., 325. Screw, binding, 271-272. motion, 18-22. Second, 29-30. Seconds pendulum, 95-97. Sector of circle, centroid, 145, 152. of sphere, centroid, 154-155. Sectorial acceleration, 51. velocity, 51, 67, 100-103, 324- Segment of circle, centroid, 152. of sphere, centroid, 158. , rectilinear, moment of inertia, 394, 395; Semi-circle, centroid of arc, 141. Sense, 3, 22, 33, 202. Set, permanent, 3 10. Shearing force, 228. Sidereal day, 29-30. Similar curves, 122-123. Simple harmonic motion, 74-78, 347. wave motion, 85-87. mathematical pendulum, 372-375. Sine-curve, centroid of area, 152. Skew forces, 207, 247-248. Sleigh, 300. Slide valve, 61, 62. Slider crank, 119, 272-274. Sliding, 452. friction, 236-242. pair, 118. Small oscillations due to torsion, 422-423. Solid statics, 243-274. of revolution, centroid, 155, 158. Specific density, specific gravity, 132-133. Speed, 33. Sphere, centroid of volume, 154-155. on horizontal plane, 458-460. on inclined plane, 454-457. , moment of inertia, 395, 398. , reduced mass, 425. Spheres, impact of, 278-293. Spherical motion, 1 6-1 8. pendulum, 375-378. points of inertia, 410. Spherical sector, centroid, 154-155- segment, centroid, 158. shell, moment of inertia, 398. surface, centroid of area, 148. Spinning, 452. Spiral of Archimedes, 57, 343. , equiangular, hyperbolic, logarithmic, 343- spring, 314. Square, moment of inertia, 395. StabiUty, 217-219. , dynamic, 449. Stable equilibrium, 2 1 7-2 1 9. Standard mass, 130. Standards, 7. Standing balance, 433. Static friction, 231. Statics, I, 170-274. and kinematics compared, 205. Steam engine, 313, 324-325. indicator, 304-305. hammer, 2S9. Strain, strain energy, 312. Stress, 195, 222, 338, 359. diagram, 172, 185. during impact, 279. , elastic, 310. , law of, 169. Stresses in a frame, 222-228. Stroke, 58. Strut, 222. Surface area, centroid, 149-150. density, 135, 398. of revolution, centroid, 147-148. Suspension bridge, 197, 201. Swing, 95. Sylvester, 119. Symmetry, 136, 153-154, 408-409. T-IRON, centroid of cross-section, 144-145, 151-152. , moment of inertia of cross-section, 395. 399- Tacking against the wind, 180-181. Tangential acceleration, 64-65. force, 326. Tension, 179, 195, 222, 297. , elastic, 310. Terrestrial gravitation, 348-349. Thomson and Tait, 19, 83, 86. Three bodies, problem of, 361-362. 494 INDEX. Tie, 222. Time, 29-30. of flight, 72. . in planetary motion, 354-358. Toggle-joint press, 181. Top, moment of inertia, 410. Torque, 202. Torsion, oscillations due to, 422-423. Total energy, 312, 323. reaction, 232. work, 268, 323. Translation, 4, 9, 22-28, 112. Trapezoid, centroid of area, 143-144, 151. , moment of inertia, 399. Triangle, centroid of area, 13S-139, 141- 142, 150-151. of forces, 171. , moment of inertia, 395, 398, 399. Triangular frame, centroid, 141. plate swinging, 442. Trip-hammer, 448. Trochoid, 1 1. True anomaly, 351, 354. solar day, 29. Truncated cylinder, centroid, 159. Turning pair, 118. Twist, 21. Twisting pair, 118. Two bodies, problem of, 359-362. U-IKON, centroid of cross-section, 151. Uniform circular motion, 67-68, 73-74. mass distribution, 131. rectilinear motion, 30, 30—34, rotation, 49. Uniformly accelerated motion, 36, 37-41. rotation, 50. Unit of acceleration, 36-37. of angle, 7. of density, 132. of force, 163-166. of length, 7. of mass, 130. of momentum, 160-161. of power, 322, 324. of velocity, 31. of work, 260-261. Units, 6. of force and work, 286-287. , fundamental and derived, 130. Universal gravitation, 43-44, 349. Unstable equilibrium, 218-219. Useful work, 268, 286, 323. Valve-gear motion, 62. Variable force, 302-325. rectilinear motion, 34, 34-49. Varignon's theorem, 177-179, 184-185. Vector, 23, 23-28, 53, 204-205. Velocity, 34-35- of light, 34, 57. in plane motion, 51-52, 52-62. of propagation of wave, 84. , sectorial or areal, 51, 55, 334. of uniform motion, 30. Velocities in the rigid body, log-117. Virtual moment, 263. velocity, 263. work, 262. Volume density, 135. Wasted work, 286, 323. Water in a pipe, impact of, 289. Water-wheel, 325. Watt, 324. Watt's parallel motion, 127-128. Wave length, 84. motion, 83-87. Weber, H., 86. Wedge, 270-271. Weight, 164, 179, 191-192, 268. Wheel, pulled over obstacle, 216-217. , rolling, 16, 57, 68. , rotating, 51, 52. and axle, 251-252, 421, 426. Wire, stretched between two points, 201. Work, 166-167, 258-260, 284. against gravity, 297, 301. , available or total, 26S, 323. , lost or wasted, 323. of piston pressure, 303-305. of revolving shaft, 427. , useful, 268, 323. THE ELEMENTS OF PHYSICS BY EDWARD L. NICHOLS, B.S., Ph.D., Professor of Physics in Cornell University, AND WILLIAM S. FRANKLIN, M.S., Professor of Physics and Electrical Engineering at Lehigh University WITH NUMEROUS ILLUSTRATIONS IN THREE VOLUMES Vol. I. Mechanics and Heat, ^1.90 net. Vol. II. Electricity and Magnetism, $1.90 net. Vol. III. Light and Sound, $1.50 net. It has been written with a view to providing a text-book which shall correspond with the increasing strength of the mathematical teaching in our university classes. In most of the existing text-books it appears to have been assumed that the student possesses so scanty a mathematical knowledge that he cannot understand the natural language of physics, i.e., the language of the calculus. Some authors, on the other hand, have assumed a degree of mathematical training such that their work is un- readable for nearly all undergraduates. The present writers having had occasion to teach large classes, the members of which were acquainted with the elementary principle's of the calculus, have sorely felt the need of a text-book adapted to their students. The present work is an attempt on their part to supply this want. It is believed that in very many institutions a similar condition of affairs exists, and that there is a demand for a work of a grade intermediate between that of the existing elementary texts and the advanced manuals of physics. No attempt has been made in this work to produce a complete manual or com- pendium of experimental physics. The book is planned to be used in connection with illustrated lectures, in the course of which the phenomena are demonstrated and described. The authors have accordingly confined themselves to a statement of principles, leaving the lecturer to bring to notice the phenomena based upon them. In stating these principles, free use has been made of the calculus, but no demand has been made upon the student beyond that supplied by the ordinary elementary college courses on this subject. Certain parts of physics contain real and unavoidable difficulties. These have not been slurred over, nor have those portions of the subject which contain them been omitted. It has been thought more serviceable to the student and to the teacher who may have occasion to use the book to face such difficulties frankly, reducing the statements involving them to the simplest form which is compatible with accuracy. In a word, the Elements of Physics is a book which has been written for use in such institutions as give their undergraduates a reasonably good mathematical train- ing. It is intended for teachers who desire to treat their subject as an exact science, and who are prepared to supplement the brief subject-matter of the text by demon- stration, illustration, and discussion drawn from the fund of their own knowledge. THE MACMILLAN COMPANY 66 FIFTH AVENUE, NEW YORK CHICAGO BOSTON SAN FRANCISCO ATLANTA A Laboratory Manual OF Physics and Applied Electricity ARRANGED AND EDITED BY EDWARD L. NICHOLS Professor of Physics in Cornell University IN TWO VOLUMES Vol. I. JUNIOR COURSE IN GENERAL PHYSICS BY ERNEST MERRITT and FREDERICK J. ROGERS Cloth. $3.00. Vol. II. SENIOR COURSES AND OUTLINE OF ADVANCED WORK BY GEORGE S. MOLER, FREDERICK BEDELL, HOMER J. HOTCHKISS, CHARLES P. MATTHEWS, and THE EDITOR Cloth, pp. 444. $3.25. The first volume, intended for beginners, affords explicit directions adapted to a modern laboratory, together with demonstrations and elementary statements of prin- ciples. It is assumed that the student possesses some knowledge of analytical geom- etry, and of the calculus. In the second volume more is left to the individual effort and to the maturer intelligence of the practicant. A large proportion of the students for whom primarily this Manual is intended are preparing to become engineers, and especial attention has been devoted to the needs of that class of readers. In Vol. II., especially, a considerable amount of work in applied electricity, in photometry, and in heat has been introduced. COMMENTS "The work as a whole cannot be too highly commended. Its brief outlines of the various experiments are very satisfactory, its descriptions of apparatus are excel- lent ; its numerous suggestions are calculated to develop the thinking and reasoning powers of the student. The diagrams are carefully prepared, and its frequent cita- tions of original sources of information are of the greatest value." — Street Railway Journal. " The work is clearly and concisely written, the fact that it is edited by Pro- fessor Nichols being a sufficient guarantee of merit." — Electrical Engineering. " It will be a great aid to students. The notes of experiments and problems re- veal much original work, and the book will be sure to commend itself to instructors." — San F'raiuisco Chronicle, THE MACMILLAN COMPANY 66 FIFTH AVENXJE, NEW YORK CHICAGO BOSTON SAN FRANCISCO ATLANTA ¥ii.<«mi)^i»A'*.<>-im'>i'-.