A Course of Mathematics for Engineers and Scientists. Volume 3: Theoretical Mechanics

A COURSE OF

MATHEMATICS FOR ENGINEERS AND SCIENTISTS Volume 3 Theoretical Mechanics

A COURSE OF

MATHEMATICS FOR ENGINEERS AND SCIENTISTS Volume 3 Theoretical Mechanics BRIAN H. CHIRGWIN AND CHARLES PLUMPTON DEPARTMENT OF MATHEMATICS QUEEN MARY COLLEGE MILE END ROAD, LONDON E.1

A Pergamon Press Book THE MACMILLAN COMPANY NEW YORK 1963

This book is distributed by

THE MACMILLAN COMPANY NEW YORK pursuant to a special arrangement with

PERGAMON PRESS LIMITED Oxford, England

Copyright © 1963 Pergarnon Press Ltd.

Library of Congress Card No. 60-13894

Made in England

CONTENTS

CHAPTER

I. THE FOUNDATIONS OF MECHANICS

1

Introduction. Fundamental concepts. Newton's Laws of Motion. CHAPTER

II. PLANE STATICS AND VIRTUAL WORK

10

The equilibrium of a particle and a rigid body. Equivalent sets of forces. The work done by a force. The Principle of Virtual Work. Potential energy. Statically indeterminate systems. CHAPTER

III. CONTINUOUSLY DISTRIBUTED FORCES

45

A uniform, flexible, inelastic string hanging under gravity. Variable loading —the suspension bridge. The equilibrium of heavy elastic strings. Strings in contact with surfaces. Shearing forces and bending moments. The bending of thin elastic beams. Pure flexure. The properties of the Heaviside Unit function and its derivative. Macaulay's Method. Clapeyron's theorem of three moments. Struts. CHAPTER

IV. KINEMATICS

105

Introduction. Velocity and acceleration. Special representations. The Hodograph. Relative motion. Angular velocity. Instantaneous Centre : Centrodes. CHAPTER V. PARTICLE DYNAMICS I: MOTION WITH ONE DEGREE OF FREEDOM

139

Introduction; conservative forces. Motion on a straight line. Simple harmonic motion and damping. Motion in a circle. Motion on other curves. CHAPTER VI. PARTICLE DYNAMICS II: MOTION WITH TWO DEGREES OF FREEDOM

Introduction. The motion of a projectile in a resisting medium. Motion under a central force — general theorems. The differential equations of central orbits. The direct law of force. The inverse square law. Newton's theorem of the revolving orbit.

169

vi CHAPTER

CONTENTS

VII. SYSTEMS OF PARTICLES

206

The motion of two interacting particles. The encounter of two interacting particles. The motion of connected particles. The general theory of systems of particles. The motion of a rigid lamina. Applications of the principles of conservation of linear momentum, angular momentum and energy to the motion of rigid bodies. Moments and Products of Inertia of a plane system. CHAPTER VIII. THE UNIPLANAR MOTION OF A RIGID BODY

249

Introduction. The motion of a rigid body about a fixed axis. Systems without a fixed axis of rotation. Conservative systems of force. Rolling and sliding motion. The use of the Instantaneous Centre: Initial Motions. CHAPTER IX. STABILITY

292

The concept of stability. Stability of equilibrium. Oscillations about stable equilibrium. Motion on a rotating curve. Perturbation of an orbit. Systems with more than one degree of freedom. CHAPTER X. IMPULSIVE MOTION AND VARIABLE MASS

331

Introduction. The impulsive motion of a system of particles. Collisions between particles. Collisions between bodies: energy changes. The motion of a body with changing mass — Rockets. ANSWERS TO THE EXERCISES

365

INDEX

371

PREFACE WE continue our series for students in British and Commonwealth universities in this volume which deals mainly with Theoretical Mechanics. We introduce at an early stage the concepts of virtual work, generalised coordinates and the derivation of generalised forces from the potential energy function. The chapter on continuously distributed forces includes the use of Heaviside's unit function in a discussion of the bending of beams. Although the treatment is mainly two-dimensional, vector representations are used wherever possible so that the results can be later extended to three dimensions. The subject of kinematics is considered in a separate chapter. The problems of particle dynamics are classified into problems of one or two degrees of freedom. In particular the equation ±2 = f (x) is discussed and illustrated in detail and central orbits are considered via their differential equations. The encounter of two interacting particles is used to illustrate the chapter on systems of particles which includes the general theorems of the motion of a rigid body. In particular we emphasise the technique of taking moments about a moving point. Stability is treated as a matter of dynamics and a discussion of a variety of problems, including some simple cases of the stability of steady motion, is given. We conclude with a chapter on impulsive motion and bodies of variable mass. Our thanks are due to the University of Oxford, to the Syndics of the Cambridge University Press, to the Senate of the University of London and to the N.U.J.M.B. for permission to use questions set in their various examinations.

viii

PREFACE

We also acknowledge our indebtedness to our colleagues whom we have consulted so frequently, especially Professor V. C. A. Ferraro, Dr. A. Mary Tropper, and Mr. E. J. Le Fevre.

Queen Mary College, London, E. I. 30. June 1962

B. H. Chirgwin C. Plumpton

CHAPTER I

THE FOUNDATIONS OF MECHANICS 1:1 Introduction In studying the processes which take place in our physical surroundings we cannot give a complete description since this would be far too complicated. We select features in which we are interested and ignore those which we believe, or hope, do not affect the processes we are studying. The surveyor, for example, is only concerned with the distances and heights of a region, and not with the chemical composition of the rocks or soil. The geologist, on the other hand, is concerned with both these aspects. The science of mechanics, likewise, ignores many features of the bodies whose motions and interactions are the object of study. The colour, the temperature, and the chemical composition of the bodies, for example, are irrelevant for mechanics. Not only are such factors ignored, but approximations and simplifying assumptions are made so that we must say that theoretical mechanics is not concerned with real bodies as we find them in the physical world, but with abstractions which are limiting cases of bodies actually found. Nevertheless, these abstractions, simplifications and approximations (if they are made with wisdom) do give us an understanding of and insight into the processes of the world around us. The following lists the chief types of body with which we are concerned in theoretical mechanics.

(i) A particle. This is a geometrical point with the property of mass added.

(ii) A rigid body. This is a body which can be conceived as made up of a large number of connected particles. The defining characteristic of rigidity is that the distance between any two of the constituent particles remains the same in all circumstances. Nevertheless, for many

2

A COURSE OF MATHEMATICS

purposes a rigid body can also be considered as made up of a continuous distribution of mass to which the processes of infinitesimal calculus can be applied. A rigid lamina is a rigid body whose linear dimension in one direction is zero. It is part of a geometrical plane and has a mass associated with every element of its area. A rigid (heavy) rod is a rigid body with only one linear dimension (its length) whose cross-section is a geometrical point. The rod forms part of a geometrical straight line and has a mass associated with every element of its length. A light rod is geometrically similar but has no mass. Both of these are called wires when the 'lines' of which they are parts are curved instead of straight. (iii) An elastic body. An elastic body in theoretical mechanics usually takes the form of a string or spring and is a body whose length can be altered by the action of a longitudinal force, the change of length being related to the force by Hooke's Law T = 2x1a. Here T is the force which produces an extension x beyond the natural (or unstretched) length a, and 2 is a constant, the modulus of elasticity, which is independent of T, x, and a. A spring can be extended or compressed; unless an elastic string is extended it must be slack. Strings are usually 'perfectly flexible', i.e., they offer no resistance to bending without change of length and take up the form of any curve with which they are in contact. Their cross-sections are usually geometrical points, and strings usually have no mass. (iv) An inextensible or inelastic string. Such a string is perfectly flexible but always remains the same length and has no mass. A `heavy' string or chain is similar but has the additional property of mass associated with every element. From time to time we may postulate bodies with properties different from those above, (e.g. beams in Chap. III) but this will be indicated in the context. It is clear that no real physical bodies have exactly these properties, although many bodies approximate more or less closely to one or other of these idealised bodies.

1:2 Fundamental concepts In order to discuss the motions and interactions of the bodies listed above we need systems of units ; those in general use are described below.

§ 1:2

THE FOUNDATIONS OF MECHANICS

3

The different sets (or systems) of units used in mechanics are identified by three basic units of length, mass and time, in terms of which other units are defined. (i) The Systeme Internationale or MKS system uses the metre, kilogram and second as the respective basic units. (ii) The CGS system, most widely used in basic physics, is based on the centimetre, gram and second. (iii) The FPS system, which is not used outside the British Commonwealth and the U.S.A., is based on the foot, pound and second.

Time intervals Everyone has a subjective awareness of the passage of time but this sense is unsuitable as a basis for measurement. To obtain a suitable basis we define a unit of time and give a means of comparing the time interval between two events with this unit, and so ascribe a measure to the interval. The unit of time is the second. Until recently this was defined to be the fraction 1/86 164.09 of a sidereal day. This took the rotation of the earth relative to the stars as a measure of the time interval between two events. Modern apparatus for measuring time intervals has now been developed with a high accuracy using the oscillations of a crystal. Such accuracy has been achieved that these methods are now preferred to astronomical methods and the Mean Solar Second, defined by the International Astronomical Union, is the accepted unit. This unit is the appropriate fraction of the time of revolution of the earth relative to the stars on a given date near the beginning of this century. We cannot discuss here either the detailed methods of measuring time intervals or the nature of time ; we assume that any time interval can be measured with the desired degree of accuracy. We also make the fundamental assumption that all observers, moving or stationary, agree on the length of the time interval between two events. In fact, we assume that there exists a universal, or Newtonian, time-scale which can be used to time all events. (This assumption is not made in relativistic mechanics.)

Position Distances and lengths are measured in terms of internationally agreed units of length, the primary unit being the metre. The two units in general use are the metre, and its decimal multiples and submultiples

4

A COURSE OF MATHEMATICS

km, cm, mm, etc., in the metric systems, and the yard with its multiples and submultiples, mile, ft, in., etc., in the British systems. The metre is now (1961) defined as 1 657 763.73 times the wave-length of a specified orange line in the spectrum of krypton-86. The yard is defined to be 0.9144m, i.e. 1 ft = 30.48 cm and 1 in. = 25.4mm, all exactly. This makes the British length units non-decimal members of the metric family. We assume in mechanics that any observer can measure the distance between any two points (though in practice this may often be a difficult undertaking). A frame of reference is used to specify the position of a point by giving its coordinates which are distances or angles. Just as we have assumed the existence of a fundamental, universal time-scale, so we also assume the existence of a fundamental, so-called Newtonian, frame of reference. This is strictly a frame whose origin is at rest relative to the centre of mass of the solar system and whose axes are in directions fixed relative to the distant, fixed stars. It can be proved that any frame of reference moving with uniform (unaccelerated) velocity relative to this fixed frame serves equally well. In practice, a frame of reference at rest on the surface of the earth, or any frame in uniform motion relative to this terrestrial frame, is usually treated as a Newtonian frame. (We do not pursue this discrepancy in this volume but leave it for investigation in more advanced work.) A Newtonian frame is called an 'inertial' frame by some authors. Mass It is a matter of common experience that large bodies usually require a greater effort to set them in motion than do small bodies. For example, it requires more effort to set a liner in motion, even slowly, than it does to set a dinghy in motion. Further, once a large body is in motion it is more difficult to stop than is a small body. This reluctance to start moving or, when in motion, to stop is a property which is possessed to a greater or less extent by all bodies. This property is called inertia and the mass of a body is a measure of its inertia. The basic unit of mass is the kilogram (kg) which is the mass of a metal prototype kept at Sevres. The pound is now defined to be a mass of, exactly, 0.45359237kg . This makes the British unit a non-decimal member of the metric family. Other masses are defined as multiples or submultiples of the kilogram or the pound.

§ 1: 2

THE FOUNDATIONS OF MECHANICS

5

Before we can assign a mass to a given body we must have an unambiguous method of comparison with the unit mass. The commonest method of comparison is by means of a physical balance which utilises the fact that a body with inertia is also attracted by the earth; we assume that bodies which are equally attracted at a given place have the same mass. Another method, impossible to use in practice, but of `theoretical' significance, involves taking the unknown mass and a unit mass a long way from all other masses, setting them at a stated distance apart and measuring the acceleration of each. We define the masses of the bodies to be inversely proportional to the accelerations they undergo ; since one of them is a standard mass, this definition fixes the value of the unknown mass. Forces From our daily experience of 'pushes' and 'pulls' we have a crude idea of the effect of a force. Two important features of forces in physical experience are (i) that they change or tend to change the state of motion of a body, and (ii) that the action of a force extends the spring in a spring-balance or alters the shape of some other type of elastic body. Two agencies which produce identical effects, of either kind, at a given place exert equal forces. [It is a matter of observation that forces which are equal by test (i) are also equal by test (ii).] In principle then, we can, when armed with a spring balance, construct agencies to exert forces of any magnitude, once we have defined a force of unit magnitude. (Two unit forces acting together in the same direction produce the same effect as a single force of magnitude 2; three equal forces which together produce the same effect as a unit force each have magnitude and so on.) The definitions of all the units of force now in use in science and technology depend upon the fact that when an unbalanced force acts upon a body it produces an acceleration. There are several systems of units which have been in use in different countries and in different circumstances, but international agreements are gradually reducing the number in use and standardising the definitions. In the MKS system the unit of force is the Newton (N) which is that force which imparts an acceleration of 1m/s2 to a mass of lkg. In the CGS system the unit is the dyne (dyn) which is the force giving an acceleration of I cm/s2 to a mass of 1 g. In the FPS system the unit force is the poundal (pdl) which is the force giving an acceleration of

6

A COURSE OF MATHEMATICS

1 ft/s2 to a mass of 11b. These are all so-called absolute units for they contain no reference to local circumstances. The attraction of the earth was used to define gravitational units. These are not of invariable magnitude and are unsuitable for use as units. They have been superseded by technical units, defined in the following manner. An acceleration of gm =980.665 cm/s2 is called the standard acceleration. That force which imparts the standard acceleration to a unit of mass is a technical unit of force, and there is one such technical unit corresponding to each of the commoner units of mass. Thus, 1 kilogram-force (kgf) imparts an acceleration gm to a mass of 1 kg; 1 gram-force (gf) imparts an acceleration gm to a mass of 1 g, 1 pound-force (lbf) imparts an acceleration gm to a mass of lib, 1 tonforce (tonf) to a mass of 1 ton. Consequently we have the following equivalences 1 kgf = 9.80665 N (MKS system) 1 gf = 980.665 dyn (CGS system) 980.665 1 lbf pdl (exactly) (FPS) system) 30.48 = 32.174 pdl (approx.). To distinguish technical units from the former gravitational units the latter are designated kilogram-weight (kg-wt), gram-weight (g-wt), or pound-weight (lb-wt), in the respective systems, but, because the attraction of the earth varies with geographical position, these gravitational units are now dropping out of use. The value 9.806 65 m/s2 is chosen for the standard acceleration because it is within 1% of the value of the acceleration due to gravity at any point of the earth's surface. There is another unit of mass which is used in certain circumstances, the slug: it is such that 1 lbf imparts to 1 slug an acceleration of 1 ft/s2. There is no accepted name for the metric counterpart to a slug.

1:3 Newton's Laws of Motion The preceding sections have given an account of how the crude ideas obtained from our own physical experience are refined and given precise measures. We may, for the moment, call all these concepts and definitions the elements from which mechanics is built. We now introduce two laws which are abstractions from experience and are based on ex-

§ 1:3

THE FOUNDATIONS OF MECHANICS

7

perimental observations. From a mathematical point of view we regard these laws as axioms; we then use the methods of mathematics to work out the consequences of applying these axioms to the elements of mechanics in a wide variety of circumstances. In what follows we assume that the reader is familiar with vectors and with the use of calculus, and also that he has already acquired a knowledge of elementary mechanics. [See "Theoretical Mechanics for Sixth Forms."] Newton stated three laws of motion. The First Law merely states that a force changes or tends to change the state of motion of a body, i.e., states the effect of inertia. This law gives no method of measuring force or mass nor any numerical relation. We have already outlined (theoretical) procedures for giving numerical values to mass and force. We therefore adopt one Law of Motion, which is equivalent to Newton's Second Law. When a particle is subject to a force P the momentum of the particle is

changed according to the law P=k

dp

dt

where k is a constant of proportionality. The momentum of a particle is the vector p=

my.

(1.2)

If the mass of the particle is constant, the law of motion becomes P = kmf,

(1.3)

where f is the acceleration vector. The value of the constant k is determined by the system of units in use. If one of the standard systems, MKS, CGS or FPS is in use, IP = 1 when m = 1 and If! = 1. Therefore, eqn. (1.1) becomes dp dt

(1.4a)

P = mf.

(1.4b)

P and eqn. (1.3) becomes

When technical units are being used eqn. (1.4b) shows that unit force produces an acceleration of magnitude gn when m = 1. (Here gn stands for the appropriate numerical value, i.e., gn =9.80665, exactly, in the

8

A COURSE OF MATHEMATICS

MKS system, gn =32.174, approx. in the FPS system.) Equations (1.1) and (1.3) become respectively

p

1 d

gn dtp

p

mf g

(1.5)

If gravitational units are being used, the factor gn has the value of the acceleration due to gravity prevailing at the place where the gravitational unit is defined. Newton's Third Law of motion is usually briefly stated : "Action and reaction are equal and opposite". This is a statement about forces rather than about the motion of a particle. We shall call it the Law of Action and Reaction and express it as follows :

When one particle A exerts a force (the action) on another particle B, then B exerts a force (the reaction) on A equal in magnitude and opposite in direction (to the action). Both forces act along the line joining A and B. We assume that this law is true whether or not the particles actually touch, i.e. we contemplate the possibility of action at a distance. The actions and reactions of this law arise as tensions in strings, internal forces in rods, rigid bodies, etc., and forces exerted at hinges on bodies attached there, and so on. The forces which appear in the Law of Motion arise from gravitational, electric and magnetic fields, and from friction, as well as from the action of one body on another. There are two consequences of the Law of Motion which we examine now. Momentum, rate of change of momentum and acceleration are all vector quantities which are compounded by the parallelogram law of addition. We assume that, when a particle is subject to the action of several forces, each force produces its own corresponding acceleration, the resulting rate of change of momentum being the vector sum of these separate vectors. But this resultant rate of change of momentum would be produced by a single force R which must therefore be the vector sum of the original forces P1 , P2 , ..., i.e.,

R

+ P, + • .

This addition of forces by the parallelogram law, and eqns. (1.1) and (1.3) to (1.5) justify the use of vectors to denote forces. The other consequence of the Law of Motion concerns the special case in which f = 0, in which the particle is stationary or is moving with uniform velocity. In this case

R = + P2 + • • • = 0

§1:3

THE FOUNDATIONS OF MECHANICS

9

and the particle is said to be in equilibrium under the action of the forces, or the forces are said to be in equilibrium. We devote the next chapter to the consideration of conditions for equilibrium in a variety of circumstances, and to their formulation in different ways. Throughout this volume we confine our attention to Plane Mechanics. By this we mean situations where, effectively, the forces acting and any motions taking place are confined to one plane. Nevertheless, this does not prevent consideration of the motions of cylinders, spheres and other bodies where the circumstances are suitably restricted so that every particle of the system moves parallel to one plane. Such "plane motions" illustrate many of the general principles of mechanics without geometrical complications. When the principles are understood in the simpler cases they can be more readily appreciated and generalised in more advanced work.

CHAPTER II

PLANE STATICS AND VIRTUAL WORK 2:1 The equilibrium of a particle and a rigid body We assume that the reader is already familiar* with the methods and results of elementary statics, and we content ourselves here with stating the more important results, although problems with friction are treated in some detail. The necessary and sufficient condition for the equilibrium of a set of forces, of which Piis a typical member, acting on a particle is that R= P, = 0 and is expressed by resolution or by means of the polygon of forces. Resolution. If the forces are resolved into components parallel to the rectangular axes Oxy and Pi has components (Xi , Yi), then the necessary and sufficient conditions for equilibrium are X = XX„ = 0, Y -= EY, = O. The triangle or polygon of forces. A set of concurrent forces, when represented in magnitude and direction (but not in position) by the sides of a polygon which is closed, are in equilibrium. The fact that the polygon is closed is a necessary and sufficient condition for equilibrium. This forms the basis of graphical solutions of statical problems. When there are only three forces the closed polygon is a triangle and the magnitudes of the forces are given by Lami's Theorem, viz., each force has a magnitude proportional to the sine of the angle between the other two. The Law of Motion, as stated in Chap. I, refers to a particle. Later we develop the Law to apply to rigid bodies in motion but here we * The results quoted here are discussed in detail in Theoretical Mechanics for Sixth Forms by C. Plumpton and W. A. Tomkys.

g 2 •• 1

PLANE STATICS AND VIRTUAL WORK

11

consider the special case of the equilibrium of a lamina acted on by a number of forces in its own plane, the various forces acting at different points of the lamina. Each force not only tends to move the whole lamina in one direction, i.e., tends to cause translation, but also tends to rotate the lamina. Therefore, in equilibrium there must be no tendency either for rotation or for translation. The Principle of the Lever, established by Archimedes, provides the basis for considering the turning effect of a force. The (scalar) moment of a force F about a point 0 at a perpendicular distance p from the line of action of F is defined to be

N(0) = pF

(2.1)

Archimedes' result showed that the lever was in equilibrium when the moments about the fulcrum of the 'load' and the 'effort' were equal and opposite. If we attach a sign to N (0), a positive value denoting a turning effect in the anti-clockwise sense, the Principle of the Lever implies that the sum of the moments of the applied forces about the fulcrum is zero. The turning moment N (0) is the scalar magnitude of the vector product r x F, where r is the position vector w.r. to the origin 0 of any point on the line of action of F. Since r and F both lie in the plane of the lamina, their vector product is a vector perpendicular to this plane. In Vol. II the definition of the moment of F about the point 0 was given as the vector product r x F; the scalar moment N(0) is, therefore, strictly the turning moment of F about an axis through 0 perpendicular to the plane of the lamina. There is a formal theorem we mention (without proof) here. If R ( = P1+ P2) is the vector sum of the forces P1 and P2 acting at B, the moment of R about A is the sum of the separate moments of P1 and P2 about A . By induction, this result is also true for any number of forces acting at a point. This is Varignon's Theorem. If forces P1 , P2 , P3 , of a set act at the respective points with position vectors r1, r2 , r3, ..., we define equilibrium of this set of forces to be determined by the conditions

+ P, ± P3 + • • • = 0 ,

(2.2)

N (A) = N1(A) N 2(A) N ,(A) + • • • = 0.

(2.3)

R

Here N1(A) , N2(A) , . . . are the moments of the respective forces about an arbitrary point A. We adopt the conditions (2.2) and (2.3) to define

12

A COURSE FO MATHEMAITCS

equilibrium. The reasons for choosing these conditions are that eqn. (2.2) shows that the forces do not tend to cause translation and that condition (2.3), which is a generalisation of the Principle of the Lever, shows that the forces do not tend to cause rotation about any point. We prove later that a stationary lamina which is acted upon by forces satisfying conditions (2.2) and (2.3) does not move. The reader is probably familiar with the following alternative expressions of conditions (2.2) and (2.3). (i) If the forces are resolved into components parallel to the axes Ox,Oy of a rectangular frame of reference so that Pi has components (Xi , Yi), and A is an arbitrary point, then

X = E Xi =0, Y = EYi = 0, N(A) = EN,(A) = 0, (2.4) are the conditions for equilibrium. (ii) The three equations (2.4) can be replaced by three moment equations, obtained by taking moments about three non-collinear points

A, B, C. N(A) = 0,

N (B) = 0,

N (C) = 0.

(2.5)

(iii) If the set contains only three forces, then these three forces must either be concurrent, and satisfy Lami's Theorem, or they must be parallel. Friction. When a particle is in contact with a surface, or when two surfaces are in contact at a point, the action of the surface on the par tide or the action between the surfaces is a force which, in general, is inclined to the normal to the surface(s). This force is resolved into a component, 11, normal (perpendicular) to the tangent plane of the surface(s) at the point of contact, and a tangential component, F, which lies in the tangent plane. If the component F is zero, the surface is smooth. The laws of friction which we explain now are simplifications; the actions between real bodies in contact conform fairly closely but not exactly to these laws. We consider two situations. (a) Slipping does not take place. When a body resting on a rough surface is subject to a small force parallel to the surface it does not slip, but if the force is gradually increased, the body slips when the force reaches a certain critical value. We therefore have the following two laws for the action between

§ 2:1

PLANE STATICS AND VIRTUAL WORK

13

two rough surfaces (or between a particle and a rough surface) at a point. (i) The two components of this action, the normal component, R, and the tangential component, F, are just sufficient to maintain equilibrium provided that the ratio FIR is less than a certain critical value,u. (ii) The direction of F is always opposite to that in which relative motion tends to take place. The quantity ,u is the coefficient of friction and is assumed to be independent of both F and R and to be a joint characteristic of the two surfaces in contact. If F = R the equilibrium is called 'limiting equilibrium' or we say that 'slipping is about to take place'. If the two components, F, R, of the action between the two bodies are compounded into a single force through the point of contact, the resultant force, 5, is inclined to the normal there. The laws of friction imply that the line of action of this resultant lies within the 'cone of friction' ; this is a cone, with its vertex at the point of contact and its axis along the normal, whose semi-angle is 2, where tan 2 = u. The angle 2 is the 'angle of friction'. (b) Slipping takes place In this case the tangential component of the action between the surfaces is of magnitude F = ,a' R, where is a constant, independent of R, and is a joint characteristic of the two surfaces. The direction of F is opposite to that of relative motion. The coefficient it' is called the coefficient of sliding (dynamical) friction and is usually assumed to be the same as ft, the coefficient of static friction. In fact, it' is usually slightly less than ,u, but to adopt this smaller value greatly complicates calculations without altering the essential character of any motion which takes place. When these 'laws of friction' are used in solving problems the normal reactions and frictional forces should be inserted separately. Limiting friction, i.e., the relation F ,aR, should not be used unless there is a specific reason for so doing. In general, the conclusions drawn from statical problems involving friction are in the form of inequalities. Usually it is assumed that equilibrium is possible ; the forces acting are calculated on this basis and the assumption that equilibrium is possible is then examined in the light of inequalities of the form

F
14

A COURSE OF MATHEMATICS

Examples. (i) A uniform rod AB of length 2a rests in equilibrium with the end A resting against a rough vertical wall, coefficient of friction u = tan 2, and the end B on a rough horizontal floor, coefficient of friction y = tank'. The plane of the rod is perpendicular to the plane of the wall and the rod makes an angle 0 with the horizontal. Show that 1 — kty' tan 0 2y' • For equilibrium the resultant reaction at A (Fig. 1) cannot lie outside the angle made by the lines AO, AL, each of which makes an angle 2 with the normal at A. Similarly the resultant reaction at B cannot lie outside the angle LB M. But the only forces acting on the rod AB are (1) the reaction at A, (2) the reaction at B and (3) its weight which acts vertically downwards through its mid-point G. Clearly these forces are not parallel and hence, for equilibrium, they must be concurrent. This is only _ N possible if the vertical through G cuts the area 0_ I — bounded by the quadrilateral LMN 0. Thus, .172virT for equilibrium, we must have L\ -- 'NA A 0 cos 2 < a cos 0 . But \

A' I \ \

I

I/ B

AO sin (In — 0 —

AB —

sin an — + 2') •

These two relations lead to the inequality 1 — /tit' tan0

FIG. I. which must be satisfied if equilibrium is possible. If the vertical through G passes through 0, equilibrium is limiting and the forces of friction and the normal reactions can be calculated. Otherwise these forces are indeterminate since three independent equations are insufficient to determine four unknown forces. (ii) A uniform square lamina ABCD rests in a vertical plane with one edge on a rough horizontal floor. A force P acts at the corner C in an upward direction making an acute angle a with CD. The magnitude of P is gradually increased. Discuss the conditions which determine whether sliding or turning about A first takes place. (The following discussion also refers to a solid, which has a uniform crosssection ABCD, under the action of a force P in the plane of its mid-section.) The lowest edge AB (Fig. 2) is in contact with the floor at a number of points, and at each point of contact the action of the floor may be resolved into a normal component and a tangential (frictional) component. The normal components have a resultant which acts at some point M; the frictional components are together equivalent to a horizontal force which passes through M. Therefore we can represent the action of the rough floor by a single inclined force S acting through M.

§ 2:1

PLANE STATICS AND VIRTUAL WORK

15

If the lamina is not on the point of turning about A, M must lie between A and B and equilibrium can only be broken by sliding. If the lamina is on the point of turning about A, M coincides with A and the frictional force prevents slipping there. There is a transitional case when the lamina is on the point both of turning and of slipping.

FIG. 2. Since three forces act on the lamina, the action S must pass through N, where the weight W and the force P meet. Hence,

LM = a(2 tana) tau.O. Lami's Theorem gives sin 0 '

cos (a -- 0) i.e.,

tan 0 —

P cos a W — P sin a •

(This latter relation can equally well be obtained by resolution of forces.) (a) Sliding takes place if tan0 > ,a, provided that LM < a. .•. tan 0 < and

1 2

tan a '

P cos a = W — Psina) , giving Pa = cos a

W

it sin a

sin a -H-1cos a •

16

A COURSE OF MATHEMATICS

(b) If the lamina is on the point of turning about A, ,u .•. tan 61 — so that

1 2+ tan a

tan@ and LM = a.

1 2 + -Lana '

P cos a W— P sin a

Pb = 2 (sin a + cos a)

i.e., 1

Pa < Pb . Therefore P attains the value Pa first and equi2 tan a ' librium is broken by sliding. 1 If > Pc,> Pb. Therefore P attains the value Pb first and equi2 tang , 1 librium is broken by turning. In the transitional case = the fric2 + tana tional force at A when turning commences is the maximum (limiting) value possible. Exercises 2:1 If y <

1. A heavy circular cylinder of radius a rests with its curved surface in contact with a rough horizontal plane. A uniform bar AB, of length 4a, rests in equilibrium, in a plane perpendicular to the axis of the cylinder, with the end A on the horizontal plane and touching the cylinder at P, where AP = 3a. By considering the equilibrium of the cylinder, find the direction of the reaction between cylinder and bar. Show that the coefficient of friction at A must be at least 8/21. 2. Two uniform circular cylinders of equal radii rest in contact on an inclined plane of inclination a . The weight of the lower cylinder is W and the weight of the upper cylinder is nW (n > 1). If the coefficient of friction between any two surfaces in contact is y , show that equilibrium is not possible if y is less than (n 1)/(n — 1). Show further that, if y exceeds this value, equilibrium is possible for all values 1)1. of a less than tan-1{2 ny/(y + 1) (n 3. Two equal uniform ladders, AB, BC, each of length 2/ and weight W, are smoothly hinged together at B, while the ends A and C rest on a rough horizontal plane. The coefficient of friction is y at A and C, and the angle ABC is 2 a. Prove that a man of weight w can ascend a distance x given by

x—

2/ W(2y — tana) w (tan a — y)

provided that x < 2/ and y < tan a < 2y . 4. A uniform rod AB of weight W lies on rough horizontal ground, the angle of friction between the rod and the ground being A, where la> A> ;, a. A force of constant magnitude P acts at B inclined at an angle B to AB produced and in the vertical plane through AB. If B is gradually increased from zero, prove that (i) if 0 < P < z W the rod will remain in equilibrium; (ii) if -} W < P < W sin

§2:2

PLANE STATICS AND VIRTUAL WORK

17

equilibrium will be broken by the rod turning about A when sin0 = W/2 P; (iii) if W sin g, < P < i W sea, equilibrium will be broken by the rod sliding when P cos (0 — = W sink. 5. A framework consists of six equal uniform rods, each of weight W, freely jointed at their ends to form a regular hexagon ABCDEF, which hangs freely in equilibrium from the point A, being kept in shape by three light rods AC, AD, and AE. Find the reaction at the joint at B and show that the tensions in the rods AC and AD are WO and 2 W respectively. 6. Two smooth planes intersect in a horizontal line L. They lie on opposite sides of the vertical plane through L, making angles 0, 99 (< 7r/2) with this plane. Two smooth horizontal uniform circular cylinders, of equal weight and length but differing in radius, are in contact all along a generator and rest in equilibrium each with a single generator in contact with one inclined plane. Show that the plane through the axes makes an angle y with the vertical where 2 cot e = tan 0

tan q).

7. Four uniform rods AB, BC, CD, DA, each of weight W and length 1, are smoothly jointed at A, B, C and D, and rest in a vertical plane with A vertically above C and BC, DC in contact with two smooth horizontal pegs, which are at the same horizontal level and distant 2b apart. A weight 4n, W is attached to C, and equilibrium is maintained by a light rod joining A to C. If the angle between CB and CD is 20, show that the tension in the rod is b 2W {(n + 1) -7cosec30 — 11

2:2 Equivalent sets of forces Here we assume that the reader is familiar with the concept of a couple, which is a pair of forces equal in magnitude but acting in opposite directions, each along one of a pair of parallel lines. The couple is characterised by its torque, or moment, which is the product of the magnitude of either of the forces and the shortest distance between the lines. A positive torque tends to produce rotation in the anti-clockwise sense. The equilibrium of a rigid lamina, subject to the action of two couples with equal and opposite moments, is unaffected if either of the couples is replaced by a couple of equal moment but constituted from different forces with different lines of action. We therefore say that two couples are equivalent when their moments are equal. We now generalise this idea to the equivalence of two sets of forces. A set of forces consists of forces Pi , (Xi , Yi) , acting at the respective points ri (xi , yi), (i = 1, 2 , n). We introduce two additional ,

18

A COURSE OF MATHEMATICS

forces Pi and — P1each acting at the origin. This augmented set is equivalent to the old set because the two additional forces are in equilibrium. But the force Pi at ri and the force — Pi at 0 together constitute a couple whose torque is ri X P, = (xi i

yi j) X (Xi i

— yi X,„) k = .AT,„ (0) k,

Yip =

where k is a unit vector perpendicular to the plane (see Vol. II § 4 :6 b). We can carry out this procedure with each of the forces PI , P2 , P3 , . . . in succession. The final set of forces is therefore a number of forces P1, P2 , P3. . . all acting at 0 together with a couple. The resultant force and the couple are given by R = EPi ,

G(0) = EN i (0)

(2.6)

or, expressed in components, R is (X , Y) where

X = X 1, Y =

G(0) = N ,(0) .

Y

(2.7)

The origin 0 is an arbitrary point and we could equally well use the point 0' (xo , yo) whose position is ro (= 00') = xo I yo j. A reduction similar to that leading to (2.6) gives the same force at 0' but a different couple, i.e.,

X = 'X1, Y = E Yi Now

i.e.,

N ,(0')

,

(0') = N 1(0') .

(2.8)

(x,— x0) Y — (y, — yo) Xi .

.•. (0') = EN 2(0') = N ,(0) — (xo Y, — yo X,), 0(0') = 0(0) — x0 Y yo X

(2.9)

We may express this in terms of vectors by saying that the given set of forces PE (i = 1, 2, ..., n) reduces to R at 0' and a couple whose torque is k G (0') where R = EPi , kG (0') = kG (0) — ro x R. (2.10) If we look for a point 0' such that G' (0) vanishes, then xoY — yo X = 0(0). This equation implies that 0', (xo , yo), lies on a straight line, at every point of which the given set of forces reduces to a force R only, with no couple. This force is the resultant of the set. When both X = 0 and Y = 0, G(0') = (0) and the system reduces to a couple which (as with any couple) has the same moment about

§2:2

19

PLANE STATICS AND VIRTUAL WORK

any point in the plane. Of course, if X = 0 , Y = 0 , and G(0) = 0, then the set of forces is in equilibrium and the system cannot reduce to a force or couple. Thus we have shown that, in general, an arbitrary set of forces is equivalent either to a single force R, or to a couple, or is in equilibrium. (This result is not true for a three dimensional set of forces.) Later we show that two sets of forces which are equivalent, or reduce to the same R and G(0), give the same accelerations to a lamina. If two sets of forces are equivalent and all the forces in one set are reversed, then the first set is in equilibrium with the reversed set. Questions concerning the equivalence of sets of forces may be solved by the device of introducing 'null pairs'. The geometrical properties of triangles and vectors can be used in certain types of problem, and the theorem in vectors [see Vol. II § 4:4 (a) eqn. (4.6) and § 4:4 (c) example (iv)] that WA -}-,LOB = (2 ± ,a) OG, where G is a point in AB such that 2A0 !LOB, is frequently useful in finding equivalent sets of forces. Examples. (i) A system of coplanar forces acting on a rigid body is equivalent to forces X, Y along rectangular axes Ox, Oy together with a couple of moment G. Prove that the system is equivalent to a single force in the line

Xy — Yx G

0.

The moments of a system of coplanar forces about the points (2, 1), ( 3, 4) and (1, — 3) are 11, — 15 and 15 units respectively. Find the magnitude and the equation of the line of action of the force acting at the point (3, 2) which, together with the given system, constitute a couple. Show that the moment of the couple is 13 units. —

The moment of the given system (X, Y) and 0 about the point (x, y) is k0' = kG — (ix + jy) X (iX --F j Y) = (0 — x Y + yX)k. But at any point on the line of action of the resultant G' = 0. Hence the equation follows. Suppose the system is equivalent to (X, Y) at 0 and the couple G. Then by eqn. (2.9) (or the expression given above) for the point (2, 1), for the point (— 3, 4), for the point (1,

—

3),

X — 2 Y + G = 11, 4X + 3 Y +

= —15,

—3X — Y + G = 15.

Hence X = — 2, Y = — 4, 0 = 5.

20

A COURSE OF MATHEMATICS

We add a force (X', Y') at (3, 2). The new system is equivalent to (X' — 2, Y' — 4) at 0 and a couple of torque G', where Y'j),

G'k = 5k+ (3i + 2j) x (X'i i.e.,

G' = 5 + 3 Y' — 2X'.

Since the complete system is equivalent to a couple

X' — 2 = 0, Y' — 4 = 0 and the resultant torque is G' = 5 + 12 — 4 = 13. The equation of the line of action of the force is x —3y—2 X' Y" 2x — y — 4. (ii) Prove that any system of forces in a plane is equivalent to three suitably chosen forces X, Y, Z acting along the sides BC, CA, AB of a given triangle in

FIG. 3 (ii).

the plane. Prove that if the system of forces is equivalent to a couple G, then

X Y Z BC CA AB

0 2 '

where 4 is the area of the triangle ABC. D, E, F are the mid-points of the sides BC, CA, AB of the triangle. Forces 2EF, ,u FD, vDE act along EF, FD, DE respectively. Find the equivalent system of forces acting along the sides of the triangle ABC. Suppose that an arbitrary member F of the given set of forces cuts BA, BC in P, Q respectively [see Fig. 3(i)]. Then we can choose A., it so that F

(2+

)P

(1)

§2:2

PLANE STATICS AND VIRTUAL WORK

21

---,-

But we can replace (2 ratio 2.:,a so that

,a)PQ by forces APB and yPC provided we choose the ABQ = ,uQC.

(2)

Eqns. (1) and (2) determine the scalars A, y. By a similar process we replace

p PC by forces IAC and mBC where / m = ,a , 1AP = m,PB. The last two equations determine the scalars 1, m. Thus we can replace F by three —> ---> —> forces APB,1AC,mBC which are three 'suitably chosen' forces acting along the sides of the triangle. We can repeat the process for each force of the given set and the required result follows by addition. If the set is equivalent to a couple, the moment of the set about any point is G. By taking moments about A, B, C respectively we get

Xh A =a, Y hB =- G, Zito = where h A , h B ,Itcare the altitudes of the triangle through A, B, C respectively. But BC • h A 2.4, CA • hB =24, AB • hc, =- 24.

X Y Z BC CA AB --> The force AEF is equivalent to j-

--->

24•

(Fig. 3 (ii)] and an equal parallel force

' at A is equivalent to ]ACA + 12AB, in position as well -1-i1CB at A. But 12CB as magnitude, since all three forces pass through A. —> —>\ .•. AEF = 1A0B — BC CA). Similarly

yFE = ly(BC — CA + AB), vDE_ v (C A — AB+ B

Hence the equivalent set of forces is v — A)BC i(v + A — y)CA

y — v)AB.

Exercises 2:2 1. Forces of equal magnitude F act at opposite ends of the diameter x = 0 of the light circular disc of boundary x2 + y2 = a2 and in its plane. The force at (0, — a) is parallel to the negative x-axis, and the force at (0, a) makes an angle 0(0 < 0 < n/2) with the x-axis. Find the perpendicular distance of the line of action of the resultant of these forces from the centre of the disc and show that the disc cannot be kept in equilibrium by a third force whose line of action cuts the disc if 0 is less than 2 sin-4{-1 (V5 — 1)1.

22

A COURSE OF MATHEMATICS

2. The algebraic sums of the moments of a system of coplanar forces about three non-collinear points (a1, b1 ), (a„ b2), (a3, b3) are Gl , G2 , G3 respectively. Prove that the tangent of the angle which the direction of the resultant force makes with the x-axis is

c1(1)2 b3) G2(b3 — bi) + G3 (bi b2) G, (a2— a3) + 02 (a, — a,) 03 (a, — a2) —

and find the algebraic sum of the moments of the system about the origin. 3. Two forces, whose magnitudes P and Q are given, act in a plane at fixed points A and B respectively. If the angle between the forces is given, show that, whatever their directions, the resultant B of P and Q passes through a fixed point. If A and B have coordinates (a, 0) and (0, b) referred to fixed rectangular axes in the plane, and if P is in the positive direction of the xlaxis when Q is in the positive direction of the y-axis, show that the coordinates of the fixed point are

x = P(aP bQ)1B2, y = Q(aP bQ)IR2 • 4. The points P, Q, R on the sides BC, CA, AB respectively of a triangle ABC divide each of these sides in the ratio 1+ k:1—k. Show that forces represented by AP, BQ and CR are equivalent to a couple and express its moment in terms of the area of the triangle. 5. Forces 2BC,aCA,vBA act along the sides of a triangle ABC in directions indicated by the order of the letters. The resultant cuts BC, CA in P, Q respectively. Show that the resultant is v-1 (2

v)

v) /

2:3 The work done by a force Work and energy are important quantities which occur throughout mechanics and play a fundamental part in other sciences. Compared with many of the concepts of mechanics the concept of work has only a vague connection with our physical experience. The porter carrying a heavy load along a horizontal platform is doing "hard work" in the everyday sense but is doing none in the sense defined below; the popular and strict senses agree in the case of a joiner planing a piece of wood. The formal definition of work is as follows: A constant force F does work when its point of application moves through a displacement x. The amount of work is defined to be the scalar product F • x. The scalar product may be either positive or negative. The units in which work is measured are given in the following table. If the force varies with the position of its point of application, the definition above takes the form of a line-integral, f F • d s. In this

§2:3

23

PLANE STATICS AND .\ IRTUAL WORK

integral ds is an element of arc, where the force has the value F, of the curve followed by the point of application. An important related result concerns the work done by a couple when a rigid lamina on which it acts changes its position. In Fig. 4 the couple is constituted by the forces (— X, — Y) at A and (X, Y) at B.

System

Unit of force

Unit of displacement

FPS

newton (N) dyne (dyn) kilogramforce (kgf) poundal (pdl)

foot (ft)

British technical

pound-force (lbf)

foot (ft)

MKS CGS Metric technical

Unit of work

metre (m) centimetre (cm) metre (m)

joule (J) erg (erg) kilogram-force metre (kgf m) foot poundal (ft pdl) foot pound-force (ft lbf)

In order to specify the position of the lamina, we need the coordinates of A relative to the axes Ox, Oy and the angle 0 between Ox and a line A drawn in the lamina. We choose A to be one axis of a pair A , An drawn in the lamina. The position of B is given either by its coordinates (4:, 97) relative to the axes A , An or by its coordinates (xB, yB) referred to Ox, Oy. Since B is a point of a rigid lamina, $, remain unaltered by a change of position of the lamina. Now

x, = x A

cos 0 — n sin 0 , y, = yA

sin 0 + n cos 0 .

In a change of position of the lamina we can alter xA , yA and 0 by arbitrary small amounts; the corresponding changes in xB,y, are

ox, =- 6x, 6y, = oy A

(—

sin 0 —

cos 0) 60 + 0(602) = ox A— (y, — yA) 60 ± 0(602),

(2.10)

cos° —ri sine) 60 + O(602) (x, — x A) 60 + 0(602). = 6yA

The work done by the two forces at A and B is

6W = — X6x A— Y 6yA X6x1 Y 6y, [(x, — x A) Y — (Y B — Y A) Xi 6 0 + 0 (6 02), i.e.,

6W = N60 + 0(002).

(2.11)

24

A COURSE OF MATHEMATICS

Hence in a finite displacement of the lamina the work done is W =-iNdO,

(2.12)

where N is the moment of the couple (about A , or about any other point). This result is clearly independent of the precise forces constituting the couple : if N is independent of B , then W

N (P — oc),

where f3, oc are the final and initial values of 0. We give now examples of certain sets of forces which do no work in a displacement. (i) When a wheel or another rigid body rolls on a stationary surface without slipping the force acting on the wheel does no work in the course of the rolling because the point of contact, at which this force acts, is instantaneously stationary. If the wheel rolls on a moving surface, then the action of the wheel on the surface does work, but the reaction of the surface on the wheel does an equal and opposite amount of work. (ii) If a particle or a body slides on a smooth stationary surface, the force exerted by the surface on the body does no work because the displacement of the point of application is always perpendicular to the force, which acts along the normal. (iii) The action of a particle A of a rigid body on another particle B of the body acts along the line AB and is equal and opposite to the reaction of B upon A. Because the distance AB remains fixed, in any motion these two forces do no net work. (iv) A smooth hinge is one which exerts no couple on a body attached at the hinge ; it can, of course, exert a force on the body. When the body turns about a fixed hinge no work is done because the force acts at a fixed point. Most of the examples cited above involve 'internal' forces acting in some mechanical system, and do no work in the displacements specified. The reader, however, should not conclude that 'internal' forces do no work in displacements of non-rigid bodies. For example, if a wheel is not quite rigid, then the forces exerted by the wheel do some work when the wheel rolls. If the distance between two interacting particles of a body alters, then their action and reaction do different amounts of work in a displacement.

§2:4

PLANE STATICS AND VIRTUAL WORK

25

Exercises 2:3 1. One end of a light elastic string, of natural length a and modulus W, is attached to a fixed point 0 and the other end to a particle of weight W, which rests on a rough horizontal table, 3a/2 vertically below 0, the string being vertical. A horizontal force P is applied to the particle so that it moves infinitely slowly along a straight line on the table. Show that when the string makes an angle B with the vertical . P = W cos° + ltan 0 — sin 0 — Show that when the particle is about to lift off the table the angle the string makes with the vertical is 60°, and that the total work done by the applied horizontal force is then Wa[5 2/42 log(2 + Y3) — V3}], where it is the coefficient of friction between the particle and the table. 2. The force of attraction on a body of mass m when at a distance x above the surface of the earth is mga2 /(a x)2 , where g is the acceleration due to gravity at the surface and a is the radius of the earth. Show that the work which must be done against gravity in transporting the body from the earth's surface to a height h is mgah/(a h).

2:4 The Principle of Virtual Work The concept of work forms an essential part of an alternative method of investigating conditions of equilibrium or the equivalence of sets of forces. The adjective 'virtual' is used because the work is done, or would be done, by the forces in displacements which the system on which they act is imagined to undergo. (a) The equilibrium of a particle A particle is subject to the action of a number of forces which add up to a force R. We can imagine this particle given a small displacement 6s; if it actually made this displacement, the forces would do an amount of work R • S s; this work is the virtual work of the forces. If the particle is in equilibrium, R = 0 and the virtual work is zero. If the virtual work is zero for all possible virtual displacements then R must be zero. (The scalar product R • 6s cannot vanish for all possible displacements unless R = 0.) Hence a necessary and sufficient condition for the equilibrium of the particle is that the virtual work of the applied forces is zero, for all possible displacements. Expressed in terms of components referred to rectangular axes, if the forces have resultant sums (X, Y) and the displacement has components (6 x, 6 y) , then the virtual work is

6W = Xox+ Yoy.

(2.13)

26

A COURSE OF MATHEMATICS

If X = 0, Y = 0, then 6W = 0; if OW = 0 for arbitrary values of 6x, (Sy, we deduce that X -= 0, Y=O. (b) The equilibrium of a rigid lamina Suppose that a number of forces P1, (Xi , Yi), act at points (xi , yi) respectively of a rigid lamina. We specify the displacement of the lamina by means of the displacement of a point A, which we may suppose originally coincident with 0, together with the angle 60 through which

y

0

X FIG. 4.

the lamina is rotated. (We consider the motion of a lamina in its own plane in Chap. IV.) Using Fig. 4 and writing 6 xA= 6 xo, 0Y A = ayo , eqns. (2.10) give

Ox,= 6x0 — yi 6 0 + 0(602), oyi = 6y0 x i o 0 ± 0(602), since xA =yA =0. The virtual work of the set of forces Pi in this `virtual' displacement is

6W = E(Xi oxi + Yi 6yi)

= xo i.e.,

+ ay°EYi + 60 E(xiY, —

OW = X6x0

6y0 + G(0)60 + 0(602),

(2.14)

where we have used the notation of § 2: 2. We assume that the internal

forces of the lamina do no work so that (2.14) gives the total virtual work.

§ 2:4

PLANE STATICS AND VIRTUAL WORK

27

Now, if X = 0, Y = 0, G(0) = 0, then 6 W = 0 correct to the first order for all displacements of the lamina in its own plane. Also, if OW = 0 correct to the first order for all possible displacements, we deduce that X = 0, Y = 0, (0) = 0. (Suppose that X 0; since the displacement is arbitrary we choose one for which 6 xo 0 ,6 y 0= 0 , 6 0 = 0 : then O W = X 6 x0 0. This is impossible, for we are given that 6 W = 0; hence X = 0 . Similarly, Y = 0 and G(0) = 0.) Hence we recover the necessary and sufficient conditions (2.4) for the equilibrium of a rigid lamina. We therefore adopt the Principle of Virtual Work or Principle of Virtual Displacements in the following form. 11 a set of forces acts upon a mechanical system, a necessary and suf-

ficient condition for equilibrium is that the virtual work of the forces is zero, correct to the first order, when the system is given an arbitrary virtual displacement consistent with the constraints of the system. The phrase 'consistent with the constraints' needs some explanation. We have proved the Principle for a particle and for a rigid body. Most idealised mechanical systems consist of particles and rigid bodies which are connected by hinges, strings, etc. Conditions may also be imposed on the system such as one body remaining in contact with a given surface, a particle sliding on a smooth wire, and so on. When a system is given a virtual displacement 'consistent with the contraints', the displacement is such that these conditions (constraints) are not violated and that any forces acting on the system because of these constraints (e.g. the reactions between surfaces, the tensions in inextensible strings, the interactions between particles of a rigid body) do no work in the displacement. If work is done in the displacement (perhaps a surface may be rough) it is usual to remove the constraint and add appropriate (initially unknown) forces to the given set acting on the system. The following examples illustrate the use of the Principle of Virtual Work. Using the Principle frequently avoids the necessity for the introduction of unknown internal actions and reactions unless a particular reaction or 'force of constraint' is required. Such a force can often be determined by violating the corresponding constraint in the virtual displacement. (The force of constraint becomes one of the set of applied forces when the constraint is ignored.) The Principle of Virtual Work can be used to compare two sets of forces. If the two sets do the same work in an arbitrary virtual displace-

28

A COURSE OF MATHEMATICS

ment, they are equivalent sets of forces, in the sense of § 2:2. The Principle is capable of greater generalisation than the methods of resolution and is used extensively in more advanced mechanics to discuss very general systems and obtain results of a wide applicability. It must, however, be emphasised that we have proved the Principle only for the equilibrium of a particle and of a rigid body, or for systems formed from such bodies hinged together or rigidly connected. We assume its truth for a much broader class of mechanical systems including flexible A and elastic bodies, continuously distributed forces, gravitation, action at a distance, etc. In effect, we take it as an axiom determining the equilibrium of a set of forces. Examples. (i) Four smoothly jointed, weightless rods, each of length a, form a rhombus A D ABCD. It rests in a vertical plane with vertically above C, the lower rods BC, CD supported on fixed smooth pegs which are on the same level and distant b apart. A light rod BD maintains an angle 2 0 at A when a load W hangs from A. Show by the method of virtual work, or otherwise, that when BD acts a as strut, the thrust it exerts on the framework is C

FIG. 5.

4a cos()

(b cosec2 0 — 4a sin e) .

Show also that if ABCD is a square, BD can only act as a tie and find the tension in BD. The thrust in BD (Fig. 5) is an internal force of the system, so that, in order to find this force, we remove the 'constraint' BD, replace it by the forces T acting at B, D, and use the principle of virtual work. We consider the system in a general position such that
T • 6(2asin0) = T • 2a cos 060 + 0 (602).

(2)

§ 2:4

PLANE STATICS AND VIRTUAL WORK

29

The detailed reasons for writing down these expressions are as follows. Equation (1) is the form which the scalar product F • 6s takes in this case: F is here the weight which acts vertically downwards and Os is the displacement which takes place vertically upwards. Hence the virtual work is F • Ss= — Wo(2a cos° —

cot0).

To explain eqn. (2) we note that we have removed the `constraint' that BD is a rigid rod, i.e., is of fixed length, and have inserted the forces T acting at B and D in place of the constraint. The forces T each act in the line BD, so that contributions to the scalar product F • Os will only arise from components of Ss which take place in the line BD. Whatever may be the displacements of B and D separately 'in space', for the two forces T the virtual work is XF • Ss = T 6(BD) = T 6(2a sin0). (If T were a tension, the virtual work would be — T 6(BD) because then the displacement 6(B D) is opposite in direction to the force T.) The equation of virtual work is — W(-2a sine + Zb cosec2 0)60

T 2a cos° 60 = O.

(In the condition of zero virtual work we use the equilibrium values of 0 and retain only 1st order terms.)

T = W(b cosec2 0 — 4a sin0)/(4a cos 0). If 0 = n/4, T —

b W(2b — a —W —1 a 2 1/22 1/2) 1/2 a

< 1, i.e., T < 0. This shows that BD is a But b < BD =a j/2 and hence y2a tie when 0 = Tc14. (ii) Four uniform rods AB, BC, CD, DE, each of length a and weight W are smoothly jointed together at their ends B, C, and D, and the ends A, E, are smoothly jointed to fixed points at a distance 2a apart in the same horizontal line. If AB, BC make angles 0, q) respectively with the horizontal when the system hangs in equilibrium, show, by the principle of virtual work, that 3 cot 0 = cot 9). Show also that if B and D are now connected by an inextensible string of length a, the tension in this string is W/y3. When there is no string joining B and D we make variations in 0 and q) which maintain the symmetry of the system, Fig. 6. Such variations leave the length AN unaltered but alter the lengths NH,NK,NC (and BD). Because of these latter variations the weights of the rods contribute to the total virtual work but the reactions at the joints contribute nothing. In the position shown

NH

a sine , N = a sin0 + la sing), a(cos0 cosq9)=a.

When a force F, (0, Y), moves its point of application through a displacement as, (6x, 0y), the work done is F • S s = Yby . Hence, for forces with only vertical components , only the alteration Sy in the vertical height of the point of application

30

A COURSE OF MATHEMATICS

need be calculated to find the virtual work of the force. We use this result frequently in virtual work problems. In particular for this problem, in a virtual displacement given by 60, 6q), the virtual work of the weights of A B, DE is 2W 6(N H) = Wa cos060 + 0(602). The virtual work of the weights of BC, CD is

2W 6(NG)= 2W (a cos060 --k za cosq)60+ 0(602). The increments 60, 6q) must satisfy the condition 6 (A N) = 0, i.e., sin060

sin0q) = O.

(1)

N

C Fm. 6. The condition for equilibrium is, therefore, 3 Wa cos 0 (5 0 i.e.,

3cos0

Wa cos To qr) = 0,

cosq) 6q) = 0,

(2)

3 cot 0 = cot 99 .

subject to eqn. (1).

If there is a tension T in a string joining BD, the virtual work of this force in the general position shown is

—T 6(B D) = T 6(2a cosT) = 2Ta sincp 699. Hence the condition for equilibrium, i.e., zero virtual work, is 3 Wa cos0 60

Wa coscp 69)

2Ta

Sep = O.

(3)

Since, in equilibrium, BD = a, cos q) = 1, and therefore cos 0 = 2, eqn. (1) gives

V3 2

13 2

q) = 0, i.e., 60 = —6(p.

§2:4

PLANE STATICS AND VIRTUAL WORK

31

Also (3) gives 3W W 2 2

2T•

V3 = 0; i.e., 2

T = W/1/3.

Note that the tension in this special case has been calculated after writing down the equation of virtual work for the system under consideration in the general case. We do not insert the values of 0 and 97 corresponding to the specified configuration until the equation of virtual work (3) and the first order variation (1) of the equation of constraint have been written down. (iii) A frame is constructed of 5 heavy uniform bars which are smoothly jointed to form a regular pentagon ABCDE. The frame is suspended from B and E so that BE is horizontal and A is uppermost. The frame is kept in shape by two light struts AC, AD. Find the tension in the bar CD. The tension T, in CD is an internal force so we must violate the condition that CD is fixed in length in order to find T. We preserve the remaining constraints, viz., AB, BC, CA etc. are all fixed in length and B and E remain on the same level (Fig. 7). We therefore consider the framework in which < DAC = 0,
A:

B, E: y2 =0.

yi---- a cos (0 + n/5);

C, D: y3 =a cos (0

g•r/5) — 2a cos (gt/5) cos 0 .

The length CD = 4a cos (x/5) sin 0 . The equation of virtual work leads to the equilibrium condition — 2 W 6(1 Yi) — 2W (zYa) — W Oy3 — T•SCD = 0. .•.

Wa sin (0 + n/5) — 2 W[— a sin (0 + 7r/(5) + 2a cos (n/5) sin 0]

—T • 4a cos (a/5) cos0 = 0. In the equilibrium position 0 = n/10. (3 sin

37-c

10 4 cos

4 cos sin 5 10 5

cos

W.

10

(iv) A circular opening of radius a in the side of a box is closed by a light rigid concentric circular disc of radius b (< a) fixed to the box by a light annular inextensible membrane. If the box is exhausted of air, show that the force required to hold the disc in its normal position flush with the side of the box is

np (a2 where p is the atmospheric pressure.

ab

b2) ,

32

A COURSE OF MATHEMATICS

Consider a virtual displacement which moves the circular disc a distance Sx as shown in Fig. 8. If S V is the volume of the conical frustrum, the equation of virtual work gives — FSx + p SV = 0 . (Although it appears from Fig. 8 that the 'inextensible' membrane is stretched,

FIG. 7. since S x is small this stretching is of the second order, 0 (6 x2) , and so does not affect the first order equation of virtual work.) a 27rr(a — r) dr But OV = 7rb2 ox + ox

a -- b

= o x {b2+ a (a + b) = 3 ir 6x(a2 Hence

ab

2 (a — b3) 3 (a —

f

b2) .

F = 1pn(a2 ab +b2).

(v) A plane framework consisting of seven equal uniform rods, each of weight W and length 2a, freely jointed together, is kept in the shape of two squares A BCD,

§ 2 :4

PLANE STATICS AND VIRTUAL WORK

33

CDEF by two light struts DB, EC which are parallel, and freely jointed at their ends. The framework hangs freely under gravity from A. Show that the thrusts in the struts DB, EC are W 1/10, W 1/(8/5) respectively. To apply the Principle of Virtual Work in such a way that we determine the thrusts T1and T 2, Fig. 9(i), we consider the framework in an arbitrary position specified by the angles 0 (the angle between AC and the vertical), y (the angle DAC), and y (the angle CFD), Fig. 9 (ii). In the equilibrium position the line APF is vertical, and the forces T1and T2 are such that y =- v = n/4. In the general position

DB = 4a siny,

EC --- 4a sinv.

The vertical distances below A of the mid-points of the rods are :

L:

a cos(0 + 99) ,

Q: 2a cos(0 + 99) + a cos(0 + 2v —

M:

a cos (99 — 0),

R:

N:

2a cos(y — 0) + a cos (y ± 0) ,

P:

a cos(y — 0) + 2a cos(y + 0) , S: 2a cos(0 y) + 2a cos(9) — 0)

a cos(99 — 0) + 2a cos(0 + 99)

+ 2a cos(0 ± 2y — 9),

+ a cos(0 + 2v — y). (The inclination of DE to the vertical is 0 + q +2(y — y) = 0 + 2y — 99.) In a virtual displacement given by 60, 693, dv the equation of virtual work gives — W {10a sin(e + 92) (60 + 99) + 7a sin(y — 0) (6 — S0) + --I- 4a sin(0 + 2v — 9)) (60 +

— (50} T 4a cosw 699 +

+ T24a cosv Sv = 0. Since 60, ov, Sy, are arbitrary, we equate their coefficients to zero separately. The coefficient of 60 gives 10 sin(0 y) — 7 sin(y — 0) + 4 sin(0 ± 2y — 99) -= 0. Since 99 =- v = nI4, 10

7 4 (sine + cos 0) — w(cos 0 — sin 9) + w(sin 0 + cos e) = 0 . .•. 21 sin 0 + 7 cos = 0, or tan° =

i.e., sine =

1

yio

, cos °

3

= 0.0

1

--y ,

, verifying that the line APF is vertical.

The coefficient of (599 gives 4/11cos y = W {10 sin (0 + 92) + 7 sin (99 — 0) — 4 sin (0 + 2v — 99)).

34

A COURSE OF MATHEMATICS

§2:4

35

PLANE STATICS AND VIRTUAL WORK

With the equilibrium values for 0, (fi, y, we obtain W 4T5 — {10 (sin0 + cos 0) + 7 (cos 0 — sin0) — 4 (sin0 1'2

n

i.e.,

W

4 T, = no {10( 1 + 3) + 7(3 + 1) 4(— 1 + 3)) —

cos0)}, 40 W 1/10

.•. Tl = W V10. From the coefficient of Sip we obtain 4 T2 siny, = 8 W sin(0

21,v — 99)

whence 772—

yio

w

\ 5)

Exercises 2:4 1. A uniform rod AB of weight W and length 2/ has one end A in contact with a smooth horizontal plane and it rests against a small smooth peg C at a distance n l from the plane (n < 2). A horizontal force P is applied at the end B in a vertical plane through AB. Find the work done by the forces acting on the rod when the angle 0 between AB and the horizontal is increased by a small amount 60. If the rod is in equilibrium when W = P1/3 and 0 is 30°, find the value of n. 2. A framework consists of four equal uniform rods, each of weight W, freely jointed at their ends to form a square ABCD, which hangs freely in equilibrium from the point A, being kept in shape by an endless light inextensible string passing through three light smooth rings, one fixed at C and the other two fixed at X and Y, the mid-points of AB and AD respectively. Show, by the principle of virtual work, that the tension in this string is 4(5 + 4 1/10) W/27. 3. Three equal uniform rods AB, BC, CD, each of weight W, are smoothly jointed to one another at B and C and to two fixed points A, D in the same horizontal line. A light inextensible string joining B to D maintains ABCD in the shape of a rhombus with < BCD = 60°. Prove that when a weight 4 W is suspended from B the tension in the string is 2 1/3 W. 4. Six rods are freely jointed at their ends to form a hexagon ABCDEF with AB = CD = DE = FA = 2a each of weight W, , and BC = EF = 2 b each of weight W2 . The joints B and F and the joints C and E are joined by light struts and the framework is freely suspended from A. It is in equilibrium in a vertical plane with the struts BF, CE horizontal. If AB, BC, CD make acute angles 0, q), y) respectively with the vertical, and CE is not greater than BF, find the thrusts in CE and BF. 5. Two uniform rods, AB, BC, of unequal lengths but equal weights W, smoothly hinged at B, rest in a vertical plane, the end A being smoothly pivoted at the point (a, 0), B and C being above A, and C sliding on the smooth vertical constraint x = 0. The system is kept from collapse by a light string of length 1 joining the point (0, 0) to the hinge B, and AB, CB make angles + a, — a respectively, with y = 0. Show that the tension in the string is given by T = 2 W(Ila) cot a.

36

A COURSE OF MATHEMATICS

6. Six equal uniform rods, each of weight W, are freely jointed at their ends to form a hexagon ABCDEF. The framework is maintained in the form of a regular hexagon by three light rods connecting B and D, B and F, and D and F respectively. If the framework rests in a vertical plane on two supports at A and B, so that AB is horizontal and below DE, prove that the tensions in the rods BF and DF are each of magnitude W and the thrust in BD is 2 W. 7. A rhombus ABCD consists of four equal uniform rods, of length a and weight W, freely jointed together at their ends. The corners A and C, B and D are joined by two light elastic strings under tension, of equal natural length and each having a modulus of elasticity equal to 2 W. The system is freely suspended from A so as to hang in a vertical plane. Show that in the position of equilibrium AC = 4 a/v5, provided the natural length of each string is less than 2 a/V5. 8. Four equal uniform rods each of length a are freely jointed together to form a framework ABCD which is suspended from the joint A. The framework is kept in the form of a square by a light string joining A and C. The weight of the framework is W, and a weight w is suspended from B and D by two equal light strings each of length b (> a/V2) . Show that the tension in the string AC is

1 {HT

w

wa V(2 b2 — a2)

} •

9. A framework consists of six equal uniform rods, each of weight W, freely jointed at their ends to form a regular hexagon ABCDEF, which hangs freely in equilibrium from the point A, being kept in shape by three light rods AC, AD, and AE. Show that the tensions in the rods A C and AD are W V3 and 2 W respectively. 10. Six equal rods, each of length 2a and weight W, are freely jointed together at their ends to form a hexagon ABCDEF, not necessarily regular. The joints B and F, and the joints C and E are joined by light struts, and the framework is freely suspended from A, the struts BF, CE being horizontal and inside the hexagon. If AB, BC, CD make acute angles 0, q), y) with the vertical, and CE is not greater than BF, prove that the thrusts in BF, CE are respectively W (5 tan

3 tan cp) ,

z W (tan 1p — 3 tan 9)) .

Find the reaction at the joint D. 11. Five equal uniform rods AB, BC, CD, DE, EA, each of weight W and length 2a, are smoothly jointed together at A, B, C, D, E and rest in equilibrium in a vertical plane with the joint A uppermost, and AB, AE equally inclined to the vertical and in contact with two smooth fixed pegs, on the same horizontal level and distant 2 c (< 2a) apart. If E and B are joined by a light rod of length 2a, show by the principle of virtual work that the compression in the rod is 2 W (1

5 c )/VS a

§2:5

PLANE STATICS AND VIRTUAL WORK

37

2:5 Potential energy If a particle, subject to gravity, is displaced around a closed path returning to its starting point, the net work done by the gravitational force, the weight, is zero. On the other hand, if a particle is dragged around a closed curve on a rough surface, the net work done by the frictional force is negative (or the net work done by the dragging agent is positive). If F is the force (e.g., gravitational or frictional) acting at a typical point of the path in the above cases, then 915 F • ds, the line-integral of F around a closed contour, is the net work done by F. We distinguish the two cases in general: (i) if fi F • ds = 0 for an arbitrary closed Contour, F is a conservative force : (ii) if F • d s 0 for an arbitrary closed contour, F is a non-conservative (usually a dissipative) force. With a conservative force the work done in one part of the circuit is 'given back' in another part. A lifting agent does an amount of work mg y against uniform gravity in lifting a particle of mass m to a height y above ground-level. By allowing the particle to return to ground level this work can be 'given back' to the agent. We say that the particle has potential energy of amount mgy in the raised position. An agency does work against the tension in a string (or spring) when it stretches the string from its natural length, a. When the length of the string is a + s the tension is 2s/a, and the work done by the agent is

(2s1a)ds in stretching the string to length a + x. This work is

`given back' by allowing the string to contract. The potential energy of the string is V in its stretched state, where x

V=

a

f A s ds

2 x2 2a

(2.15)

0

In the general case of a particle subject to a conservative force F, ( X, Y), the work done by an agent in displacing the particle from A to P (see Fig. 10) against F is independent of the path joining A and P; the work depends only on the positions of A and P. If we assume A to be the

38

A COURSE OF MATHEMATICS

`standard position' of the particle, the potential energy in the position P is the work done by the agent against F, viz., V (X, y)=

—

IF•ds =

f (X dx

—

Ydy).

(2.16)

A The potential energy at a neighbouring point 131, (x + ox, y), is P, 17 (X +

P,

y) = — f F•ds = — f (X dx Y dy). P,

I7 (X

± ox, y) — V (x, y) = — f

P,

F • ds = — f (X dx Y dy)

— X6x + 00x2)

.

y

x

Fm. 10. Hence, if we divide by S x and proceed to the limit, we find

aV Similarly,

ax aV

ay

X.

(2.17)

= — Y.

(2.18)

=

These relations are very important. The minus sign implies that the forces act in a direction which tends to decrease the potential energy.

§2:5

PLANE STATICS AND VIRTUAL WORK

39

Examples. (i) The potential energy of a particle at (x, y) under gravity is V =mgy. The force acting on the particle is

av

av

X= ax _ 0 Y =

ay

'

—

—mg.

This is the gravitational force, the weight, which tends to increase y the negative sign implying, of course, that the weight acts downwards. (ii) The potential energy of a stretched string is V = A x2 /2 a. The force tending Ax aV to increase the coordinate x is X = . This is the tension in the 0 x a string. — -

— —

These ideas can be generalised to apply to any mechanical system acted on by a set of forces. A general system requires a number of parameters such as lengths, angles, coordinates, etc., to specify its 'configuration' ; we denote these by q1, g2 , . . . ,. gn . (The worked example (v) of § 2: 4 uses three such parameters q1 =0, q2 = p , q3 = y .) If some agency causes a change in this configuration, it will do work against the set of forces acting on the system, the work being given by an integral

2' Qid g„. A system for which this work is zero when i=i the system returns to its initial configuration after an otherwise arbitrary displacement is subject to a conservative set of forces ; (these 'forces' may include couples). If some suitable configuration is chosen as the `standard' position, A, the potential energy in any other position, P, is defined to be P of the form —

.

IL

V (P) = —I i Qi dqi . i=i

(2.19)

A

This is the general definition of potential energy and can be expressed in words thus : "A set of forces acting on a system is conservative if the net work done by an agency in displacing the system from a configuration A to an arbitrary configuration P is independent of the 'path' followed by the system from A to P. The potential energy of the system is the work done by the agency against the set of forces in moving the system from a standard configuration A to the configuration P." This definition implies that the potential energy is arbitrary to the extent of an additive constant ; an alteration of the 'standard' position A adds a constant, independent of P, to the expression for V (P) given in eqn. (2.19).

40

A COURSE OF MATHEMATICS

If the parameters q1, q2, , qnare such that any one of them can be altered independently of the others without violating any of the constraints of the system, then an argument similar to that leading to (2.17) and (2.18), shows that (2.20) Q, = (i = 1, 2, ... , n).

av d q,

The 'generalised force' Q, is the force tending to increase the parameter q,„; Q is a force if qt is a distance ; Q, is a torque if q, is an angle. Some authors use slightly different conventions and names for the function V. Sometimes it is called the 'work function' and is defined so C

Fm. 11.

that X = a Via x , etc. Here we keep V to denote the potential energy satisfying (2.19) and (2.20); we use different notations for alternative conventions. The reader should refer, in connection with this section, to pure mathematical discussions of perfect differentials (see Vol. II § 5.4). The condition that the work done in a displacement around a closed circuit is zero ensures that d V, where

dV = — Qdq„,

(2.21)

i=

is a perfect differential. It is the differential of a function V (q1, q2, • • • >qn) such that (2.20) is satisfied. Example. A uniform rod of weight W and length 2a can turn freely about one extremity A, which is fixed. A light inextensible string attached to the other extremity passes through a small smooth ring fixed at a point C, distant 2a from A and at the same level as A, and carries at its other end a weight w. When the rod is at an inclination 0 to the horizontal and below it, determine the magnitude of the couple which tends to increase 0. In Fig. 11 the moment of the forces about A in the clockwise direction, i.e., in the sense of 0 increasing, is

G(A) = W a cos° w 2a cosie.

41

PLANE STATICS AND VIRTUAL WORK

§2:6

If we take the standard position to be the one in which AB coincides with AC, the potential energy in the position shown is V = — Wa sin° + w 4a sin-10. (The centre of mass, where the weight W acts, is at a vertical distance a sin() below its standard position, and the weight w is a distance BC above its standard position.) dV — Wa cos/9 + w 2a cos+0 = —G(A). • • • dO This shows that the derivative of V gives the negative value of the 'force' (torque) tending to increase O.

2:6 Statically indeterminate systems The reader should not imagine that the conditions of equilibrium already discussed can be used to find the forces acting in every statical system. The systems discussed and the problems solved so far have all been chosen so that the conditions of equilibrium and the geometrical R

Q

S

Fm. 12.

information provided are together sufficient to give unique answers to the problems concerned. We now give examples in which it is impossible to find all the forces uniquely and which correspond to situations of very common occurrence in practice. Consider a rigid rod AB pinned at each end and carrying a weight attached at a point of its length, Fig. 12. (This is an approximation to the situation of a beam or girder supporting the roof of a building.) It is impossible, by the methods of statics alone, to determine uniquely the forces acting on this system. The conditions of equilibrium furnish three linear equations between the four unknown forces P, Q, R, S. Therefore, there is no unique solution unless additional information is provided. In statical problems, for example, this may take the form of saying that the rod is supported at B on a smooth support. This gives the additional relation that S = 0

42

A COURSE OF MATHEMATICS

if the support has a horizontal surface, or that R = — S cota if the surface of the support makes an angle a with the horizontal. It may be a better approximation to physical arrangements to assume that the rod is not rigid. The rod therefore departs slightly from the strict linear shape, the deflection at a point depending on the stresses there and the elastic properties. We discuss this type of problem in Chap. III. Miscellaneous Exercises II 1. Three smooth circular cylinders, of equal radii a and equal weights W, are held together in contact along generators by a light elastic band of natural length 2na and modulus nn W. Two of the cylinders rest on a horizontal table with the third above them. Show that the reaction of the upper cylinder on either of the 1 lower cylinders is W and the reaction between the lower cylinders is (3n1/3, either zero or W (3n,

2

according as n < or >

1/ 3

18 1 ) 2. Two thin uniform rectangular boards ABB' A' and CBB'C' of weights W and W' respectively are freely hinged along the edge BB' and rest with the edges AA', CC' on a rough horizontal floor so that the angles BAC and BOA are acute and have tangents p and p', where Wp> W'p'. The middle points of the boards are joined by a straight light inelastic string whose tension is originally zero and is gradually increased. Find which board slips first, and show that the tension of the string when this happens is {

W(1 + P'it)

W' (1 ±_PP + 21/1, )1/(P

p'),

the coefficient of friction between either board and the floor being p. 3. Two equal rough uniform circular cylinders, of radius a and weight W, rest on a rough horizontal plane with their axes 2a V2 apart. A third cylinder, of radius a and weight 2 W, rests symmetrically on the first two. Show that for equilibrium the angle of friction for the cylinders in contact must not be less than n/8. Show also that the coefficient of friction between the cylinders and the ground must not be less than ztan (n/8). 4. Two uniform circular cylinders, of equal weight but unequal radii, are placed in contact with their axes horizontal on a rough plane inclined at an angle a to the horizontal, the lower cylinder having the smaller radius. The plane containing the axes of the cylinders makes an angle a + /3 with the horizontal, p is the coefficient of friction between the cylinders and piis the coefficient of friction bet ween each cylinder and the plane. If the cylinders are in equilibrium, show that: (i) the frictional force is the same at all contacts, (ii) for equilibrium to be maintained at the line of contact of the cylinders /1 „>- cot/3 , (iii) for the upper cylinder not to slip on the plane tan a a

1 + p, sec/3

PLANE STATICS AND VIRTUAL WORK

43

5. Three uniform rods OA, OB and 0 C are each of length 1 and of weight W; they are smoothly jointed together at 0 and are placed symmetrically over a smooth fixed sphere of radius a, the joint 0 being vertically above the centre of the sphere and the rods resting on its surface in equilibrium. Show that

1 sin3 a = 2a cos a, where a is the angle each rod makes with the downward vertical, and deduce that the rods will be at right angles if (and only if) 1 = 3a/ V2 . 6. A framework of smoothly jointed rods in the form of a pentagon ABCDE of equal sides of length a is kept just stiff by rods AC,AD of length b (a/2 < b< 2a) , and rests flat on a smooth horizontal plane. If the rods AB, BC, CD, DE, EA are subject to a uniform normal inward pressure p per unit length, apply the method of virtual work to show that the thrusts in AC and AD are

p r 4

2a2 — b'

ab (b2

1,0)2

( a2

•

7. Four uniform rods AB, BC, CD, DE, each of length 5a and weight W, are freely jointed at B, C, D. The joints B and D are connected by a light rod of length 8a; the ends A, E are attached to small smooth fixed hinges at the same horizontal level and distant 14a apart. If the system hangs in equilibrium in a vertical plane with BD below AE and C below BD, show that the stress in the rod BD is 11 W/24. 8. Four uniform rods, each of length a and weight w, are freely jointed to form a rhombus ABCD which is freely suspended from A. A smooth uniform disc, of radius r and weight w, rests between the rods CB and CD and in the same vertical plane. The joints B and D are connected by a light elastic string of natural length 3 a/4 and, in equilibrium, BD = a . Use the principle of virtual work to show that the modulus of elasticity of the string is 3 (2 r — 1/3a) w/a if r > 1/3a/2. 9. Three uniform rods AB, BC, CD each of weight W and length 2a are freely jointed at B and C. They rest in equilibrium with BC horizontal and uppermost, with AB, CD making equal angles of 0 = 30° with the vertical and resting over two smooth pegs E and F respectively at the same horizontal level. Prove by the principle of virtual work that EF = 13a/6. If 0 is decreased to sin-1(5/13) by means of a light elastic string of natural length 2a joining A and D, show that the modulus of the string is approximately 0.32 W. 10. Two equal uniform heavy rods AB, BC are freely pivoted at B, and stand in a vertical plane with the ends A and C resting on a rough horizontal board and the angle ABC equal to 2 a. The board is gradually tilted about a horizontal axis perpendicular to the plane of the rods. Show that, if motion is confined to the vertical plane of the rods, equilibrium can never be broken by slipping up the plane, but that slipping down the plane will occur first at the upper or lower of the two points of contact according as the coefficient of friction is greater than or less than tan a. 11. A polygon A,... An(not necessarily regular) is made up of 92 rods

A,A„

An ,A„, Anil„

44

A COURSE OF MATHEMATICS

freely jointed at their ends, and placed on a smooth horizontal table. Each rod experiences a force proportional to its length, applied at its mid-point, and directed along the outward normal. Show that in equilibrium the joints A1, An lie on a circle, and that the reaction at each joint is the same. If all the rods are of the same length, and the common force applied to each of them is F, calculate the reaction at the joints. A rectangle has two rigid sides each of length 2a, which are opposite sides of the rectangle, and two perfectly flexible sides, each of length s, freely jointed together. The sides are then subjected to a constant pressure p per unit length, directed along the outward normal. Show that in equilibrium the rigid sides are parallel and a distance 2a tan0 apart, where 0 is the solution of the equation

s cos0

2a0.

12. A framework consists of two uniform heavy rods AC, BC each of weight W and of length 2a, and a light rod AB also of length 2a, with AC horizontal and B below AC. The framework rests in equilibrium with A fixed to a vertical wall and is kept in position in a vertical plane by a horizontal rod BD of weight W and length a, where D is fixed to the wall vertically below A. All the rods are pin-jointed at A, B, C and D and a weight w is attached to C. A virtual displacement is made in the system such that the length of AB is increased, the others remaining constant in length. If 2 80 is the increase in the angle AC B and lq is the increase in BAD, prove that 60 + S qq = 0. Hence prove that the tension in AB is (7W + 4 w)12 V3.

CHAPTER III

CONTINUOUSLY DISTRIBUTED FORCES 3:1 A uniform, flexible, inelastic string hanging under gravity In this and the following three sections we consider the equilibrium of a perfectly flexible string or chain. Such a body, described briefly in Chap. 1, is regarded as a continuous line of particles in which the only action one element can exert on a neighbouring element is a force

x

(the tension) along the line joining the elements, i.e., along the tangent. There can be no couple or transverse force ; consequently the string offers no resistance to bending. We first discuss the equilibrium of a uniform inelastic string hanging under gravity. We use the notation of Fig. 13. The tension at the lowest point A is To and acts on the section AP as shown; the tension at P acts along the

46

A COURSE OF MATHEMATICS

tangent there on the section AP in the direction given by p. The arc length AP is denoted by s, so that (s, v) are the intrinsic coordinates of P. The weight of unit length of the string is denoted by w, and we consider the equilibrium of the section AP under the action of T, To and its weight ws. Resolving horizontally and vertically for these forces gives T cosy = To , (3.1)

T sin v = ws

(3.2)

.

Equation (3.1) implies that the horizontal component of the tension is constant and we therefore define the parameter c by the relation To =we.

(3.3)

Dividing eqn. (3.2) by eqn. (3.1) gives

s = c tang)

(3.4)

which is the intrinsic equation of the curve in which the string hangs. Then we obtain the parametric equations of this curve, i.e., x and y in terms of v, by the method of Vol. I § 4:7. Thus, dx = cosy ds

leads to

dx dx ds ds dyeds d = v cosy dv

c secy.

Integrating and noting that x = 0 when v = 0 gives

x = c log (secip + tan v) .

(3.5)

Similarly dy dye

dy ds ds — c sec v tan v = sin v ds dv dv

gives y = c secv;

(3.6)

the constant of integration vanishes if we take 0 to be at a distance c below A. Therefore eqns. (3.1), (3.3) and (3.6) give

T = wy

(3.7)

and eqns. (3.4) and (3.6) imply that y2 =_.

e2

s2

(3.8)

§3:1

CONTINUOUSLY DISTRIBUTED FORCES

47

To obtain the cartesian equation of the curve we write (3.5) in the form secy tany = ex/e. The reciprocal of this equation is see y — tany = e-xie. Then addition and subtraction lead to the equations y = c cosh (x/c),

(3.9)

s = c sinh (x/c).

(3.10)

The curve in which the string hangs is called a (uniform) catenary of parameter c, A is called the vertex of the catenary and the x-axis is sometimes called the directrix of the catenary. [This does not imply that the catenary is a conic.] Examples. (i) A uniform flexible chain, of length 2/ and weight W, hangs from two fixed points H, K on the same level, so that the lowest point of the chain is at a depth h below the line HK. Show that the parameter of the catenary in which the chain hangs is given by 2hc = 12 — h2, and deduce that the tension at either point H or K is W (11h + h11). If the span H K is 2a, show that 2ha -= (Z2 — h2) log (1 + h)/(1

h)}.

Equation (3.8) implies that, in Fig. 14,

yir =l2 +c2.

But

PH = c h and hence

2hc = 12 — h2.

Also H, K are on the same level and (3.7) gives W TH - TK 21 ?/R• But yn

c -I- h and hence (1) and (2) give TH = IV (11h + h11).

if the span HK = 2a, then (3.9) and (3.10) give yH = c

h = c cosh(a/c), sH = / = c sinh(alc). .•.1+ c+h=cealc,

i.e.,

a

on using (1) above.

c log / c c

h

12

h2 2h

log

(1 + h h)

48

A COURSE OF MATHEMATICS

(ii) A uniform heavy chain BC, of length 2/ and weight w per unit length, hangs symmetrically over two fixed smooth pegs H, K at the same level, and the lowest point A of the portion of the chain between the pegs is at a depth //18 below the level of the pegs. Find the lengths of the portions of the chain which hang vertically, and the distance between the pegs. Show further that the thrust on either peg is * V13 of the weight of the chain.

FIG. 14. Since the peg H is small and smooth, the tension in the chain immediately to the left (see Fig. 14) of the peg is equal to that immediately to the right. Since this tension supports the vertical portion HB, TR = wHB; but T H = wyH. .•.

yH = HB.

Similarly

yK =KC. Hence the free ends B, C both lie on the directrix (y = 0) of the catenary as shown in Fig. 14. By symmetry yH =yK . The arc AH = (1 — yH) and hence by (3.8)

= C2 + (1— YH)2.

(1)

But since the sag (the depth of A below KH) is 1/18, yH = c 08. Substitution in (1) gives 9c2 — 18c/ + 8/2 -= (3c — 4/) (3c — 2/) = 0. Hence c = Then

(We reject the root c = 11 since it implies HB + KC > 2/.)

HB = yH 13//18.

§ 3:1

CONTINUOUSLY DISTRIBUTED FORCES

49

Also sec V H = .•. xH = c log (sec Hence the span HK =.2xH =

4/

yH

c

13 12 •

tampH) = / log (1-: + 15i). log

(3\ 2

/

Since TH = wyH =13w1/18 and cos /pH = lt, the force exerted by the chain on either peg is the resultant of two forces, each of magnitude 13 w//18, inclined at an angle cos-1(A) to each other. These forces give the required thrust. (iii) A string AB, of length 1 and weight w/, will break if subjected to a tension in excess of 2w1. It is fixed at B and is in equilibrium under the action of a

TB

FIG. 15 (i).

FIG. 15 (ii).

horizontal force F at A . Show, if the string is just about to break, that F = w/ V3, and find the horizontal and vertical distances of B from A. The tangent to the catenary at A must be horizontal. If the tangent at A is inclined at an angle a to the horizontal, see Fig. 15(E), vertical and horizontal resolutions for the equilibrium of the particle of the chain at A give

T sina = 0, T cosa = F( 4- 0), and so a = 0. Since the chain is in equilibrium, the tension at B balances the force F and the weight w/, see Fig. 15(i), .• TB = (W1)2 +F2 . But the greatest tension in the string is at B. Hence T B G 2w1 and thus F < w11/3 . In the critical case when the string is about to break at B, F = w11/3 Then the horizontal resolute at A gives F = To =we whence c = 1V3 since A is the vertex of the catenary.

50

A COURSE OF MATHEMATICS

The equation sB = c tanyBgives tanyB= 1/1/3 so that yB= n/6. Then the horizontal and vertical distances of B from A, i.e., the horizontal and vertical projections of AB, are given respectively by

x B— x A = c log (secn/6 tanx/6) = /1/3 log3, yB -YA= c(secn/6 — 1) = (2 — I/3)1.

(iv) ABD is a rope, of length lla and of weight w per unit length, with one end tied to a small ring A free to slide on a smooth fixed horizontal rail. After passing through a smooth ring B fixed to the rail it hangs with BD vertical. The string is kept in equilibrium by a force 4a w applied horizontally at A. Show that at A and B tan Y = and that the distance AB = 8a log2. (Fig. 16.) A

D FIG. 16.

The equilibrium of A (horizontal resolution) gives 170 =4 wa, i.e., c = 4a. Also by symmetry yA = yB and hence the part AB of the string is of length 2c tanyB. But as in example (ii) above BD = yB = c secyB . Hence the total length of the string ABD is c sec yB + 2c tanyB. But this is given as 11a and so yBsatisfies the equation (1) secyB + 2 tanyB = 11/4. The 1.h. side of (1) is a strictly increasing function of yB as yBincreases from 0 to an and hence (1) has one root only in this range, given by tanyB = 1 :1as required. Then AB = 2c log.(secy B tantpB) = 8a log2. (v) A uniform chain of length 2/, and of weight 2w/ hangs in a vertical plane with one end 0 fixed and the other end A attached to a uniform rod AB of weight w 1. The rod is kept in equilibrium at an angle in to the horizontal in the plane of the chain, the points A and B being on the same side of the vertical through 0,

§3:1

CONTINUOUSLY DISTRIBUTED FORCES

51

by the action of a horizontal force at B. Show that c = land find the vertical and horizontal distances from A to 0. The equilibrium configuration is shown in Fig. 17 and the reader should verify that this gives the only possible equilibrium position, since in all other cases the forces acting on the rod cannot be in equilibrium.

The forces on the rod are parallel, respectively, to the sides of the triangle GKL in which GL = 1BK =1-GK. Using GKL as the triangle of forces gives

GL GK 2w1 • F F = wl. Also

we F GL — cosy A = TA LK TA

The geometry of the figure gives tantpA = 2. But 2/ — so — s A =c tanvo — c tanipA

52

A COURSE OF MATHEMATICS

Hence tally° = 4. Then the vertical and horizontal distances of 0 from A are respectively c sectpo — c secvA= 1(1/17 — 1/5), c log

(4 1/17 2 + V15 seeVA tanYA— / log sec yo

tamp()

(vi) A uniform string, of length 2/ and weight 2 w1, is slung symmetrically over two small smooth pegs A and B in the same horizontal line. A particle of weight

B

A

2w1I3 is fixed to the middle point of the string and in the position of equilibrium of the system the particle is at a depth 51/27 below AB. Prove that the distance 8/ (3 A B is — log -9 2 ). The arc AC is part of a catenary which is shown in Fig. 18 extended to its vertex V. The coordinate axes Ox, Oy are such that M lies on Ox [see ex. (ii)] and V lies on O y. Now AM = y A =c sec v A, and vertical resolution for the symmetrical forces on the particle at C gives 2 To sinve = 2w1/3. But

T = wyc= we sect')(. .•. tanve = //3c. The length AC of chain is c(tanvA — tantpc) so that

1= AM + are A C = c sec TA c(tanyA — tanzpo). 1 c

.•. secvA tarivA = —

1 3c

—

4/ 3c

§3:1

CONTINUOUSLY DISTRIBUTED FORCES

53

and 3c 4/

secyA — tanyA

.•. secyA —

1 ( 4/ 3c 2 3c + 4/ )• 5/ 27

Since c secyA — c sec/p, = yA ye

.•. 1 = sec2

13/ 27c

5/ 27c

3c \ 1( 4/ 2 3c + 4/ )

SeCIPC

— tanzy —

\2

13/ 27c

12

3c 8/ )

23 (c\ 2 88 36 / ) -1- 272

9(c\ 4 64 ( 1 )

3c 81

9 c2 •

0.

Hence 2 ) ( 3 c2 8/2 27

( 3 c2 8/2

44 ) 27 - 0

and so c 41/9. [The other positive root must be discarded since it would imply that the length of the string exceeds 2/.] Then ye =tan-'(a), y A =tan-1( 4,) and AB= 2c log

secyA tanyA secyc tanyc )

8/ _o 3 ) 9 1 g( 2

The tightly stretched wire If a uniform wire of length 2l is tightly stretched between two points A and B in the same horizontal line at a distance 2a apart, and I only slightly exceeds a so that 6 = (1 — a)/a is small, then the inclination, yA, of the wire at A or B must be small, Fig. 19. Since 6 is small we determine the magnitudes of other quantities in terms of 6. If c is the parameter of the catenary,

l

1 = c sinh (a/c) = c j

a + a3 6c c

as .•.

1

=

a /1 +

/—a a

6c2

=-

a5 120 c5

0

a4 + 0 (f' 120c4 66

a2 a2 (l {1 ± ± 0 c44 6 c2 20 c2

c7 a7

.

▪ 54

A COURSE OF MATHEMATICS

Hence, to a first approximation a2

ii.e.,

= 66

e2

c

II

1 16(1—

l

a)

J

More precisely

o= .• . c =

a2 (1 + 3 6 1 + 0(62 \ ) 10 6c2 1 )f .

11 j a j 36 38 1+ 20 11 + 10 + "52)i — j(6 6) I

a

1(66)

+0(624.

a ..... .... ...........

d

...... ...........

.........................................

Fie. 19. The sag

d at the lowest point is

d = c cosh (-a--) — c = c But a2 2c2

a4 2 c2 ▪ 24c4

a" \

a2

c6 )1 •

2 a2 a4 +0 ( 6 ) = 2ac2 {1 2462 \c/ + 12c2 36

= 36

10 +

0(62)1

4

° C4 ) 1

_6 1 {1 + + O(624 2

3 6 i1 + — + 0(62) . 5

d

36

146 6)

a

1+ 20

0(62)} 3 6 76 20 +

( 36)11 2

0 62

6 -5

)1,

i.e., to a first approximation

d=a

(

36

\

2

) = 1,I ) -3a / — a)} . /12

0(62) c

§3:I

CONTINUOUSLY DISTRIBUTED FORCES

55

The tension at a point of the wire whose horizontal distance from the mid-point is x, is T = wy = To cosh (x/c).

To < T < Tocosh (a/c). But cosh a/c = 1 -I-

a2

-1- 0

2 c2

a4

3 6 + 0 (62).

=1

To < TG

+ 3 6 + 0(62)1.

Hence, to a first approximation, we may regard T as a constant (= To) along the wire, and, to this approximation, To =

-:—

a3

wa 10 (5)

6 (/ — a)} •

w

It is of interest to note that when c is large the form of the catenary approximates to a parabola. For in this case y

c co h (

c

_ v

so that, neglecting terms of order

+

(

x4 ) 3

C

2 4_ 0( ;44 )} s 2c2

'

x2 = 2 c (y — c) which represents a parabola of latus rectum 2c with its axis upwards and vertex at (0, c). This corresponds to the case of uniform horizontal loading discussed in the example of § 3:2. Example. A telephone wire, of mass 100lb per mile, has a span of 80 yards and a tension of 1351bf. Find the sag. Measuring all lengths in feet, 100 135 5280 c• Also the sag 1202 2c 1202 • 100 2 • 135 • 5280

100 2 1 ft. 99 • -

56

A COURSE OF MATHEMATICS

Exercises 3 :1 1. A uniform flexible string hangs in equilibrium symmetrically over two small smooth pegs at the same level, distance 2a apart. Find the relation between the length 2/ of the string and the parameter c of the catenary in which the portion of the string between the pegs hangs. Hence show that equilibrium is not possible unless 1 o ae and find, when 1= a e , the acute angle which the tangent to the string at a peg makes with the horizontal. 2. A uniform flexible chain, of length 54a, hangs over two small smooth pegs, the lengths of the vertical portions being 20a and 13a respectively. Show that the parameter of the catenary between the pegs is 12a, and that the horizontal distance between the pegs is 12a log(9/2). 3. A chain, of weight per unit length w, has its ends attached to two points A and B distance 2a apart and at the same horizontal level, and hangs freely between them. The lowest point of the chain is at a distance k below AB. Prove that if k is small the length of the chain is approximately 2a +

4k2 , 3a

and find an approximate value for the tension at the ends. 4. A uniform flexible string PQR of length 2a has one end attached to a fixed point P above a rough plane which is inclined to the horizontal at an angle of 30°, The string hangs in equilibrium from P with the part QR, of length a, in contact with the plane. The coefficient of friction between the plane and the string is 1/1/3 and R is on the point of moving up the plane. Show that the parameter of the catenary in which the portion PQ hangs is 403. Show, also, that the difference between the heights of P and R is la (21/3 — 1). 5. The ends of a uniform heavy chain of length 2l can slide on a fixed rough horizontal rod. If the angle of friction between the rod and the chain is A, show that the least possible depth of the mid-point of the chain below the rod is gr A l tan (— — — and that the corresponding distance between the ends of the chain 4 2' is 2/ tanA log (cotlA). 6. A uniform heavy chain ACBD, of length 23a and weight w per unit length, has one end tied to a small ring A, free to slide on a smooth fixed horizontal rail, and after passing through a small smooth ring B fixed to the rail, hangs freely with BD vertical. The part ACB hangs in an arc of a catenary, and equilibrium is maintained by a horizontal force 12a w applied to A. Show that, at A and B, tan ip = 5/12, and find the length BD. Show also that AB = 24a log(3/2). 7. A uniform heavy flexible string AB, of length 2/, has its ends A, B attached to light rough rings which can slide on a fixed horizontal rod. The coefficient of friction between the rings and the rod is 1/V3. Show that when equilibrium is limiting at both ends the depth of the mid-pointM of AB below the rod is //1/3. If a particle of the same weight as the string is now attached to M and the string takes up a new symmetrical position of limiting equilibrium, show that M falls a distance 1(3 — 1/7)/1/3.

3:1

CONTINUOUSLY DISTRIBUTED FORCES

57

8. A uniform heavy flexible string of length 1 rests in a vertical plane with a length //2 in contact with a smooth plane inclined at an angle 7r/4 to the horizontal. The upper end of the string is attached to a point A above the plane. Show that the tension of the string at A is w/v(5/8), and find the horizontal and vertical distances of A from the lower end of the string. 9. A D-shaped wire frame is formed from a semi-circular wire of radius a and a diameter. The frame is fixed in a vertical plane so that the diameter is vertical. The ends of a uniform heavy flexible string of length 1(1 < a) are attached to two small smooth light rings, one of which is free to slide on the diameter and the other on the semi-circular part of the frame. If the vertical distance between the rings is b(b 1), find the horizontal distance between them and show that sinh

j tab

1

12

b2

2 lb 12 — b2 •

10. A uniform heavy flexible string hangs freely under gravity from two points A and B. A is below B, the lowest point of the string is below A, and the vertical distance between A and B is k times the length of the string (k < 1). If a and /3 are the angles which the string makes with the downward vertical at A and B respectively, prove that (1 — tanl a — (1 [- k) tan12 /3. 11. A uniform heavy chain AB is of length 8l and weight w per unit length. A smooth ring of weight wl fixed to the end A is free to slide on a fixed vertical wire, and the chain is slung over a smooth peg C so that a length 5/ of the chain hangs vertically. Find the parameter of the catenary AC, and show that the distance of the peg C from the vertical wire is 3l log (1'10 — 1) . 12. A heavy uniform chain of length 1 is attached at one end to a point A at height Is < 1 above a rough horizontal table. The point A is moved parallel to the table with constant velocity. Show that the length s of the chain which remains free of the table is the positive root of

s2+ 2 kilts = h2 + 2 phi, where p is the coefficient of friction between the chain and the table. 13. A perfectly flexible uniform heavy chain has one end fixed at a point A and the other end at a point B, which is higher than A. If the inclination of the chain at A to the horizontal is 30° and the lowest point of the chain divides its length in the ratio 1:3, show that the inclination of the chain at B to the horizontal is 60°, and that AB is inclined to the horizontal at an angle cot-1LI (3 + 1/3) log (3 + 2 {/3)}. 14. One end of a uniform heavy flexible chain of length 1 is attached to a light ring which is free to slide on a smooth vertical wire, and the other end of the chain passes over a smooth peg at a distance a from the wire, the whole system lying in a vertical plane. If 1> a e, prove that there are two positions of equilibrium, and show that in one of these positions the length of that part of the chain which hangs vertically lies between 2l and -P/2 a2)I1 , whilst in the other position it exceeds(/2 a2)/1.

58

A COURSE OF MATHEMATICS

3 : 2 Variable loading — the suspension bridge If an inextensible string hangs in equilibrium under gravity and carries a load or is of variable line density so that the mass per unit length is not uniform, the curve in which it hangs can be determined by methods similar to those used for the uniform string. Example. A light string hangs under gravity and is loaded so that the weight on each element is proportional to the horizontal projection of that element. Show that the string will hang in the form of a parabola. If 4a is the latus rectum of the parabola and w is the weight of the loading per unit of horizontal span, show that the tension T, at a height y above the vertex, is given by T = 2w(ay a2) 2 . Taking A in Fig. 13 as the lowest point of the wire and To as the (horizontal) tension there, resolutions for the equilibrium of the arc AP give

T cosy

T,,

T

wx,

(1) (2)

where w is now the load per unit horizontal projection. Hence tamp =

wx T,

dy dx •

Integration gives

w x2 Y — 2 To

'

the constant of integration vanishing if we choose A as origin so that y = 0 when x = 0. This is a parabola of latus rectum 2Tolw. Since the latus rectum of the parabola is given to be 4a we must have To =2 w a. Squaring equations (1) and (2) gives T2 =Tg

iox2

=

w2(4 a2 + x2) = w2(4 a2 + 4 ay) .

.•. T = 2w(ay

a2)+.

The suspension bridge The ends of a light chain are attached to two fixed points A0 , An„and to the intermediate points A1, A2 , ..., An are attached weights W1, W2 , ..., W.?, respectively so that the portion Ar_i Ar is in tension Tr and makes an angle Or with the horizontal (Fig. 20). Then, resolving vertically and horizontally for the equilibrium of the particle of the chain at A, gives Tr±i sin 0,+, — Tr sin Or = Wr , T,+, COS Or+ 1

=

TrCOS Or = H,

§3:2

CONTINUOUSLY DISTRIBUTED FORCES

59

where H is the (constant) horizontal component of the tension in the chain. (Each section is straight since the chain is light.) Division of the first of these equations by the second gives tan 0„4 — tan Or =

Hr

(r = 1, 2, ... , ) .

(3.11)

Further, if the horizontal and vertical distances of An+1from Ao are known (h, k say), then n+1 n +1 Ar _i ArCOS Or =h, Ar_,A,. sin Or =k. 2=1

(3.12)

r=1

Equations (3.11) and (3.12) suffice to determine the n +2 unknowns 0„„, H. In particular, if the weights are all equal, then the 01, 0,, tangents of the inclinations of the portions of the chain form an arithmetical progression. If, in addition, the horizontal separations of the loads are equal, the points Ao , , A n±1lie on a parabola and the problem corresponds to that of a suspension bridge, i.e., a uniformly load ed horizontal roadway suspended from a cable by a number of equispaced vertical ties. Exercises 3:2 1. A non-uniform flexible chain hangs in the form of a smooth curve from two Fm. 20. points at the same horizontal level, which are at a distance 1 apart, with its lowest point at 0; the curve is referred to horizontal and upward vertical axes 0 x, 0 y respectively. The tension T and mass per unit length m vary continuously along the chain, and have values T, and me respectively at 0. Show that cm0 dp p2) dx '

T = T c 1/( 1 + p2), m — 1/(1

where p = dy/dx and c = T 01(mo g). The ratio y = T/(mg) measures the stress. If this is constant along the chain, show that the curve has equation y = c log sec () .

60

A COURSE OF MATHEMATICS

hang from various points of a light string 2. A number of weights W1, W2, whose ends are fixed; if a and /1 are the inclinations to the horizontal of the extreme portions of the string, prove that the horizontal component of the pull on the points at which the string is attached is

+ W2 + tan a

tan/3

are attached at successive points of a light string at3. Weights W1, W2, tached at its ends to fixed points ; if the two sections of the string on either side of the weight Ws are inclined at as _ „ asto the horizontal, show that Ws =c (tan. as — tan as_1), where c is a constant. Deduce that if a chain suspended by its ends assumes the form of a circular arc, the density at any point varies inversely as the square of the depth of the point below the centre of the circle. 4. If a light string is loaded with equal particles W at equal horizontal intervals a and if yn is the vertical ordinate of the n-th weight, prove that y„, — 2 yn y n = Wa/H, where H is the horizontal tension. Hence show that, if A and B are constants,

?I'

Wa ne 2H + Alt • B

5. A suspension bridge consists of a heavy horizontal road, weight w per unit length, supported by a light chain which is attached to the road along its length by a large number of close vertical light ties. The chain passes over two pulleys at a height h above the lowest point of the chain, equilibrium being maintained by two weights hanging freely from the ends of the vertical portions of the chain. Prove that the form of the chain is a parabola. If the total span is 2R, prove that the weights must each have the value

wR(R2 4h2A/2 h.

3:3 The equilibrium of heavy elastic strings The investigation of the equilibrium of heavy elastic strings is based on the assumption that Hooke's law applies to every element of the string, each element being in equilibrium under the action of its neighbouring elements and its own weight, as illustrated in the following examples. Examples. (i) Find the length of a heavy elastic string AB under the action of its own weight and a weight W attached at B. The unstretched string has a length a and weight w per unit length, and the modulus of elasticity is A. Figure 21 (i) refers to the unstretched state and Fig. 21 (ii) to the stretched state of the string so that the element PA of length assat a distance so from A in the

§3:3

CONTINUOUSLY DISTRIBUTED FORCES

61

unstretched state becomes the element PQ of length S s distant s from A in the stretched state. The relation between stretched lengths and corresponding unstretched lengths of the string is obtained by applying Hooke's law to a typical element. The tension, T, in the element PQ is T

2(6s — 680) so

whence, in the limit as Ss -÷ 0, T— 2

(d d so

1).

(1)

—

A

A

so

Po Co

8s0

P

Ss

Bo no. 21 (i). B

w FIG. 21

(ii).

But T supports the weight W and the portion PB of the string. T = W w(a — 80). Equations (1) and (2) give ds W w — 1+ (a 80). dso A A

(2)

Integrating and choosing the constant of integration so that s = 0 when so = 0 gives Wso w 8 = so ± A + (2aso — 4). 22

62

A COURSE OF MATHEMATICS

Hence when se, = a, i.e., at B

,

s a +

?t, a2

Wa

and the extension of the string is la (2 W wa)/A (ii) An elastic string which is uniform when unstretched hangs under gravity between two fixed points. Find the intrinsic equation of the curve in which the string hangs. Using the notation of Fig. 13, and so , s to denote the unstretched and stretched lengths, respectively, of the arc AP, horizontal and vertical resolutions for the equilibrium of this section give

T cosy = To = wc, T sing = wso .

.T=w

c2),

so =c tan y

But T

ds clso

ds w = A f(.3(; d so

c2 ) +

integration gives

s =- 2w2. [so 1(1 + c2)

c2

so

r(s°

G2) 11

so .

Fence the intrinsic equation is , c2

s = c tan y

22 {sec y tan y + log (see y

thaw)} .

When is large this approximates to a catenary, but when is small to a parabola. Exercises 3 : 3 1. The string of example (i) above rests on a rough horizontal plane; the coefficient of friction is p. The string is pulled at one end with a gradually increasing force F. Show that a portion of the string remains unstretched provided F < pw a and that in this case the extension of the string is F2 /(2 pw A). 2. A heavy uniform elastic string, of natural length 2l has mass m per unit length when unstretched, and modulus of elasticity . Show that, when the string is stretched at a uniform tension T, the mass per unit length is mAl(T + 2). The string can just support a tension equal to the weight of an unstretched length L(> 1) of the string. It is suspended from two points at the same height and at a distance 2d apart. Obtain the equations

T cosy = To , T (T ))

d = m,y1, dx

X

3: 4

CONTINUOUSLY DISTRIBUTED FORCES

63

where To is the tension at the lowest point, y is the inclination of the string to the horizontal and x is the horizontal distance. Hence or otherwise show that

d < (L2

12)'

mgl 1.

+tarth-i

1

.

3:4 Strings in contact with surfaces Here we consider the equilibrium of a heavy inelastic string of weight w per unit length which lies in a vertical plane and rests in limiting equilibrium in contact with a rough surface, the coefficient of friction being u.

The results for light strings and/or smooth surfaces can be derived by putting w = 0 and/or kc = 0. We consider an element PP' (Fig. 22) of length Os which is about to slip in the direction PP'. If R is the normal reaction per unit length, the frictional force is pRos acting as shown. Resolution along and perpendicular to the tangent at P gives the equations

(T 6T) cos ov — T ?Dos siny (T oT)sin6v-:= wos cosy + Ros, or, correct to the first order in 6y, 6T, 6s, 671— itRos = w6s siny, Thy — R6s = w6s cosy.

64

A COURSE OF MATHEMATICS

Division by 68 and letting os —› 0 gives dT ds

(3.13)

,uR = w sin p,

(3.14)

T dy —R=wcosy. ds Elimination of R gives dT dy

,uT = w(siny

ds tj c cosy) thp

whence

T = Ceiv

e"1 W

V — du cosy) e-PW

ds dy

(3.15) dy

(3.16)

where C is a constant. Examples. (i) A light string rests in contact with a smooth surface. Then (3.13) gives T = constant along the whole length of the string. (ii) A light string rests in contact with a rough surface. Then (3.16) gives

T = To eitv, where To is the tension where y) = 0. For example, in a capstan, when a rope is wrapped 12 times round a circular cylinder and is about to slip round the cylinder, v = 3n and the ratio of the tensions at its ends is therefore e3 Ya. (iii) A heavy uniform string rests in contact with a smooth curve which lies in a vertical plane. Then (3.13) can be written dT ds

dy w sin v =w d ,

dT dT ds dy ds dy

s

w•

.•. T = wy + constant.

( 1)

(2

(iv) A uniform chain passes over a number of smooth surfaces and is at rest in a vertical plane with its ends hanging freely. Prove that the curved portions of the chain, not in contact with the surfaces, are catenaries which have a common directrix and that the free ends of the chain are on this directrix. This result follows at once from eqn. (1) of example (iii). Clearly this equation holds for a uniform chain hanging freely under gravity also, and so the tension throughout the chain is given by eqn. (2) in which, without loss of generality, we can put the constant of integration as zero. But the parts of the chain hanging under gravity must be the portions of catenaries and since the relation T = wy holds when Ox is the directrix of a catenary, these catenaries must have a corn-

§ 3:4

CONTINUOUSLY DISTRIBUTED FORCES

65

mon directrix. Further, at the free ends T = 0, i.e., y = 0, and hence the free ends of the chain lie on this directrix. (v) A uniform heavy string rests partly on the upper surface of a rough vertical circle of radius a and partly hangs vertically. Prove that, if one end is at the highest point of the circle, the greatest length that can hang freely is 2 y a + ( 42 — 1)aeiAnt2 /42 The equation connecting the tension T with the angle 0 (Fig. 23) when the string is about to slip in the direction AB is easily found to be dT d0

µT = wa(pcos0 — sine).

The solution of this equation (a linear differential equation with constant coefficients) for which T = 0 when 0 = 0 is

T —

wa {(O — 1) erne — (y2 — 1) cos° + 2 y sin0}. it2 +1

Hence, at 0 = 1-7r,

T—

wa

2 2 + 1 [ 14 +(µ2 — 1) et"T12)

which supports the stated length of the string. It should be noted that in all cases a chain partly in contact with a surface leaves that surface tangentially. This result follows at once from consideration of the equilibrium of an infinitesimal element at the point where the chain leaves the surface. [See example p. 49.

Exercises 3:4 1. A uniform chain of length 2/ hangs symmetrically over two fixed smooth circular cylinders of equal radii r. The axes of the cylinders are horizontal and parallel and at a distance 2a apart on the same level, the chain hangs in a plane perpendicular to the axes. Prove that

=

(a — r sin 0) (sec 0 + tan 0) r (2– — cos 0 ± 0), log (sec0 tan0)

where 0 is the angle the tangents at the highest points of the catenary make with the horizontal. 2. A light rough rope passes over a circular cylinder in a plane normal to its axis, and is about to slip. Shew that, if 0 is the angle subtended at the axis by two points A, B on the rope, T A = 6±1"TB, where the sign depends upon the direction of slip, it is the coefficient of friction between the rope and the cylinder, and TA, TB are the tensions of the rope at A and B.

66

A COURSE OF MATHEMATICS

A horizontal rough circular beam is fixed in position; a parallel circular beam of weight W is suspended by a rope which is fixed to the bottom of the upper beam and passes half-way round both beams twice; the distance between the beams is sufficiently large to allow all connecting sections of the rope to be regarded as vertical. Calculate the tension on the free end of the rope, (a) when the lower beam is just rising, (b) when it is just falling. 3. A heavy chain of length / rests partly in a straight line on a rough table and the remainder, after passing over the smooth edge of the table which is rounded off in the form of a cylinder of radius a, hangs freely. If the coefficient of friction is It, show that the least length on the table is 1

/

+ 1 (

na + a) . 2

4. A uniform string ABCD of weight w per unit length hangs in equilibrium over a rotating pulley of radius a in a vertical plane. The end A is attached to a fixed point. The portion BC, of length 1, is in contact with the pulley, and the portion CD, of length 1', hangs vertically. The pulley is rotating in the direction BC, and the coefficient of friction between the string and the pulley is ,a . Prove that the dependence of the normal reaction R per unit length between the string and the pulley upon angular distance 0 from C is given by

dR

= 2w cosh,

and that the couple opposing the motion of the pulley is 2wa2 /410,14 112 + 1

I

(

war cos 1 + ttsin -a1 + a

— 1).

5. One end of a light string is attached to a fixed point P on a rough vertical wall, and the other end supports a mass m. The string passes in a thin rough groove over a wheel in a vertical plane perpendicular to the wall. The wheel, which is of radius a and mass M, is supported only by frictional forces at the wall and on the part of the string, of length 260 < an, in contact with it. The coefficient of friction between string and wheel is le, and the wall is perfectly rough. Prove that the string will not move if 0 lies between the two roots 991, 972 in the range 0 < co < Pc of the equation M e2 /1 9,cos2 = 1 + . 2m Prove also that equilibrium is not possible unless es/tan- 'µ

M < 2m,

+ 1,2

§ 3: 5

67

CONTINUOUSLY DISTRIBUTED FORCES

6. A rough circular cylinder, of radius a, is fixed with its axis horizontal. A uniform string of length na hangs over the cylinder in a vertical plane perpendicular to its axis, a length a q) being in contact with the cylinder and the remainder hanging vertically. Prove that, if the equilibrium is limiting, eq, tan A sin( 99 +22) =

99 —

21 +

sin22,

where is the angle of friction between the string and cylinder. 7. A heavy inextensible string rests over a circular cylinder in a vertical plane at right angles to the horizontal axis of the cylinder. It occupies the quadrant a — g/2 < 8 s a of the circular cross-section, where 0 is measured from the topmost point of the cross-section, and slipping is about to occur in the sense of a increasing. Prove that eM n/2 tan(a — 271) = 1, where ,a = tan R, and is the angle of limiting friction between the cylinder and string.

3:5 Shearing forces and bending moments We turn now to the consideration of bodies which are neither perfectly rigid nor perfectly flexible, although we do not consider this question in general but confine our attention to nearly straight, thin beams which are deformed by the action of external forces. We first investigate the F

P

A

M

8

F Fm. 24.

forces involved and introduce a convenient representation of the internal stress system ; we leave consideration of the change of shape, or strain, resulting from these forces until § 3 :6. Consider a beam AB under the action of a set of external forces and couples which are all in the vertical plane containing AB. Each of these external forces may either be continuously distributed, e.g., the weight of a heavy beam, or they may be concentrated at certain points, e.g., at points where the beam is supported. We represent the internal forces in the beam as shown in Fig. 24. We imagine the beam AB divided into two parts at an arbitrary point P by a normal section, i.e., a section

68

A COURSE OF MATHEMATICS

whose plane is perpendicular to the tangent to A B at P. Then the forces exerted across the section at P by the r.h. portion on the l.h. portion must balance the external forces on the 1.h. portion. (Similarly, the forces exerted by the l.h. portion on the r.h. portion balance the external forces on the latter.) Now any plane set of forces is equivalent to a force at some point together with a couple (see § 2:2). Therefore, in general, we can represent the action of the r.h. portion on the l.h. portion by two component forces T, F and a couple M, at some point of the section. (See Fig. 24.) The component F is the shearing force, the component T is the tension, and the couple M is the bending moment. The sign convention used in this book gives F, T, M positive values where they act in the directions shown in Fig. 24. Other writers use different conventions but that adopted here leads to differential equations with positive signs throughout. The Law of Action and Reaction requires that the forces exerted across a section by the l.h. portion on the r.h. portion are as shown in Fig. 24 where they are positive. A stress is, strictly, a force per unit area; if the area of the section of the beam at P has a value oc then F/cc is the shear-stress and T/cc is the longitudinal stress.

Relations between the bending moment, shearing force and loading Suppose that the beam is straight and horizontal and that there is a continuous vertical load w per unit length. Measure x along the beam and consider the equilibrium of the element PQ of length 6 x where AP = x, (Fig. 25), and suppose that there are no concentrated external forces (loads, reactions at supports, clamping couples, etc.) acting on PQ. Then resolving horizontally and vertically and taking moments about P for PQ gives (T 611) — T = 0 ,

(F ± 6F) — F — wax = 0, (M 6M) — M — (F 6F) Ox — w(6x)2 = . Dividing by a x and letting 6 x

0 we find dT

= 0,

(3.17)

dF dx

w,

(3.18)

d /I/ dx

F.

(3.19)

dx

§3:5

CONTINUOUSLY DISTRIBUTED FORCES

69

Equation (3.17) implies that, if all the external forces act vertically, then the tension T vanishes at all points of the beam.

Concentrated loads If a concentrated load W acts at a point between P and Q, the set of forces acting is shown in Fig. 26. Here the increments d T, d F, d M x

w8x FIG. 26.

0 . In fact, resolution and taking moments may not tend to zero as a x leads to (F + ZIF)ox + 0(6x). ZIT = 0, dF = W wox, Hence, in the limit as 8 x 0 in such a way that the point of application of W remains inside 6x, F W, (3.20)

Z1M --- 0,

AT --- 0.

(3.21)

This shows that the shearing force is discontinuous and the bending moment and tension continuous at concentrated loads. Examples. (i) A uniform beam AB of length 3a and weight w per unit length is supported in a horizontal position by vertical forces at its points of trisection C, D and a load 5 wa is suspended from the midpoint G of the beam. Find the shear-

70

A COURSE OF MATHEMATICS

ing force and bending moment at a point P where AP = x and draw the shearing force and bending moment diagrams. The diagram, Fig. 27(i) shows typical positions, P1, .P„ P3, P4 , of P, one in each of the ranges of x given below. The conditions of equilibrium of that portion of the beam lying to the left of P lead to the results in the table. For example, C A

G

n

D P3

P2

P4 B

4wa 5wa 4wa FIG. 27 (i). suppose Plies in the second range a < x < 3a/2; vertical resolution and moments about P for the left hand portion give

—wx

F = 0, —wx(ix)+ 4wa(x — a) ± M = 0.

4wa

.•. F = w(x — 4a), IV

— 4wa(x — a).

[The values of F and M tabulated below can each be displayed by a single formula using the Heaviside unit function, see p. 81.]

F

Range

<x
3a/2 < x < 2a 2a < x < 3a

M

wx

1-wx2

w(x — 4a)

iwx2 — 4wa(x — a)

w(x + a)

-1-wx2 — 4wa(x — a) + 5wa(x — 3a/2)

w(x — 3a)

lwxz — 4wa(x — a) + 5wa(x — 3a/2) — 4wa(x — 2a)

The shearing force and bending moment diagrams are shown in Figs. 27(ii), 27(iii). (ii) A beam AB is supported in a horizontal position by vertical forces at its ends A and B. If the weight per unit length of the beam at a distance x from A is w sin(rx/2a), where w is a constant, and AB = 2a, sketch the graphs of F and M, and find the total weight of the beam and the greatest value of M. The total weight W of the beam is given by 2a

W

fw sin 0

( TEX ) 2a

dx —

4a w

71

CONTINUOUSLY DISTRIBUTED FORCES

§ 3 :5

3wa

a

2a

- 3wa FIG. 27

(ii).

3a

x

72

A COURSE OF MATHEMATICS

Equation (3.18) gives dF = w sin dx

x 2a

so that

F=K

2aw x

cos

( otx \ 2a ) '

where K is constant. But (3.20) gives F = — 1 W ( = — 2aw/n) at x = 0 and hence K = 0. 2aw nx cos( 2a ) Then (3.19) gives dM 2aw cos dx 2a whence 4a2 w 7r x M — 70 sin 2a the constant of integration vanishing since M = 0 when x = 0. Clearly the greatest (numerical) value of M occurs when x = a and is 4a2 w/n2. The graphs of F and M are shown in Fig. 28 (i), (ii).

M

a x 4a2 w 2-2

FIG. 28 (ii).

3:6 The bending of thin elastic beams A complete analysis of stress and strain in an elastic solid is outside the scope of this book. Here we consider the deflection of an initially straight beam which is bent uniformly. When a beam is bent uniformly every particle is displaced parallel to a fixed plane, the plane of bending, and any line drawn in the beam parallel to the length of the beam before

§3:6

CONTINUOUSLY DISTRIBUTED FORCES

73

deformation (a fibre of the beam) is bent into a plane curve parallel to the plane of bending ; also, particles which lie initially in a plane normal section are still in one plane in the displaced position and this plane is still a normal section, i.e., the fibres still intersect it at right angles. Suppose the beam is initially straight and horizontal and has a crosssection which is symmetrical about a vertical plane and that the plane of bending is vertical. Consider the deformation of a small portion ABCD of length 8s , Fig. 29 (i). Since the bending is uniform, the crossD

SS C

FIG. 29 (i). y

sections AD, BC remain plane but are displaced to the positions ad, be so that they meet in a line of which 0 is the trace in the vertical plane, Fig. 29 (ii). If the bending is concave downwards, so that the upper part of the portion is stretched and the lower part contracted, somewhere in between there will be a layer of fibres which are unaltered in length. This layer is called the neutral surface of the bent beam. Let MN ,mn, respectively, be the traces of the neutral surface in the unstrained and strained states. Then Om = R where R is the radius of curvature of the neutral fibres after bending so that, if the angle mOn is dye, then

6s = Roy. Initially all the fibres ab, mn, pq, dc have the same length. Let the fibres pq be at a distance nbelow mn, Fig. 29 (iii). Then initially arcpq = arc mn = os = R Sy; since the beam is thin, after bending arc pq (R — ri)6 vcorrect to the first order in 6y. Hence the fibres pq are compressed by the amount n 6v. We consider those fibres passing through the element 6S of the normal section ap' m' dmp [see Fig. 29 (iv)]. These fibres are compressed, nSy and the strain is the fractional decrease of length _ This strain

R 6v

R.

74

A COURSE OF MATHEMATICS

is produced by the stress acting across 8 S, the stress being the force acting along the fibres per unit area in the region (58. Now Young's Modulus is

E

_

stress strain • C

M

N

A

B

a FIG. 29 (iv). Hence the contribution from 6 S to the tension (§ 3 : 5) is negative being

6T —

6S.

Hence, by integration

T

J.

1

dS

B

d S,

where the integral is taken over the normal section a p' m' dmp . But this is a 'first-moment' integral (see Vol. I § 6:4). Hence

T=—

E _ R

§3:6

CONTINUOUSLY DISTRIBUTED FORCES

75

where S is the area ap'm'dmp and iis the distance of the centroid of this area from the neutral section mn. Two cases now arise. (1) If this thrust vanishes, then i = 0 and the neutral surface passes through the centroid of the section. This is the case of pure flexure. (2) If the total thrust (or tension) does not vanish, the neutral surface does not pass through the centroid of the cross-section and may even be outside the beam (in which case the fibres would all be contracted or all extended). In either case the forces acting across the normal section ap'm'dmp can be reduced to forces acting along md and mn together with a couple obtained by taking moments about the axis mm' , Fig. 29 (iv). The anti-clockwise moment about the neutral line mm' of the force exerted across 8S on the material lying on the r.h. side of the section En S through AD is R 97 . Hence, with the sign convention of § 3:5 the total bending moment is given by En2 EI u = f (3.22) R where I is the second moment of the cross-section about mm'. The product E I is called the flexural rigidity. The differential equations of bending It can be shown from the general theory of elasticity that the deformations in a beam produced by shearing forces are very small compared with the deformations due to bending moments. To obtain the downward deflection y at any point of a loaded beam, we take the moments of the external forces about that point and substitute in eqn. (3.22). Taking coordinate axes Oxy [Fig. 29 (i)] with Ox along the undisturbed neutral line of the beam and with Oy vertically downwards and assuming that the deflection and gradient of the beam are small 1 d2y d2y d2y/dx2 dx2 R {1 + (dy/dx)21312 d x2 El ±°CY' 2)1-1Hence d2 y EI = M. (3.23) d x2 Equations (3.18), (3.19) taken with (3.23) lead to the differential equations for F, M and y . These equations can be combined into d2 d2 y = w. (3.24) dx2 dx2

76

A COURSE OF MATHEMATICS

The values of the constants of integration and of any unknown reactions or couples occurring in the equations of deflection are found from the values of y, y, , y2 , y3at supports or at free ends. (Here, for boundary conditions, suffixes denote differentiations w.r. to x.) In particular : (1) If a point is supported, y is given (and, if at the same level as the origin, y = 0 there). (2) If a point is freely supported, usually only the condition for y is given but, if the point of support is at a free end (where M = 0), y, = 0 there also. (3) If one end is clamped, y and y„ are each given there. (4) At a free overhanging end (where M = 0 = F), y2 = 0 = y3. (5) At any point of support not at the ends of the beam, y and y, are each continuous.

3:7 Pure flexure We now consider some simple illustrative examples of pure flexure concerning beams initially straight under various types of loading and with prescribed supports and end conditions. Examples. (i) A light cantilever OA of length 1 is clamped horizontally at one end and a load W is suspended from the other end. (Fig. 30). To balance the load W the forces exerted by the clamp on the beam at 0 must be an upward force W and a counter-clockwise couple W/. For the equilibrium of the portion PA the bending moment at P (x , y) must be M = W (1 — x) .

.• . El

. • . EI

d2

de dy

dx

=

W(1— x).

— Wlx — W x2 + A.

.•. Ely =i-W1x2— 1,W x3 + Ax+ B. The constants of integration each vanish since y = 0 = yi at the clamped end (x = 0). Hence the (downward) deflection at any point is y = Wx2 (3/ — x)I6 E I and the deflection of the free end is W/3/3E1. (ii) A uniform cantilever, of length 1 and weight w per unit length is clamped horizontally at one end and is free at the other end.

§3:7

CONTINUOUSLY DISTRIBUTED FORCES

77

With the notation of Fig. 30 (with If = 0), the bending moment at P, obtained by considering the equilibrium of PA , is w (1 — x)2 . d2 y

. • E I dx2— iw(1 — x)2 . The solution of this equation which satisfies the end conditions y = 0 = y, at x = 0 is

y

w EI

(1 — x)4 + 13 x 24

6

14 24

and hence the deflection of the free end is w/4/8E/. (iii) A uniform beam OA, of length 1 and weight w per unit length, is clamped horizontally at 0 and freely supported at A at the same level, i.e., there is no ex0

y K / 0

A R2

FIG. 31. ternal couple applied at A. Find the reaction at A, the clamping couple, and the maximum deflection of the beam.

With the forces as marked in Fig. 31 the equation of vertical equilibrium gives R1 + R, = wl. Then by considering the equilibrium of PA we find the bending moment at P is

w (1 — x)2 — R2(1 — x) . c12 y

.• E I de = w(1 — x)2 — R2(1 — x). 2.

78

A COURSE OF MATHEMATICS

Integration gives

w 12 x2 EIY — 2 ( 2

lx3

x4

(le

+ 12 )

3

x3

R2 2

6);

the constants of integration must both vanish since y = 0 = y, at x = 0. The end condition y = 0 at x =1 gives R, = 3w1/8. Then taking moments about 0 gives the clamping couple K = E I y2 = w/2 /8. The deflection y is given by

EIy — w

(12 x2 16

5/ x3 x4 48 + 24

The maximum deflection occurs when dy/dx = 0, i.e., when 612 — 15/x + 8x2 = whence x = (15 — 133) 1/16. (iv) A light beam OH of length 1 is freely supported at each end and carries a distributed load of intensity Wx//2where x is the distance from O. Find the equation of deflection. Four successive integrations of the equation of loading, (3.24),

RI

Wx 12

d4 y d x4

give

W x5 Ely

120/2

A x3 6

+

B x2 2

Cx + D.

The four constants of integration A, B, C, D are determined from the end conditions y = 0 = y2 at x = 0 and x = 1. These conditions give B = 0 = D, A = — W/6, C =7 W/2 /360, and imply

E

Wx (3x4 36012

10x212

7/4) —

Wx 36012 (12

— 3x2) . x2) (712

Note that the shearing force at the end x = 0 is given by

dM d'y — (E/ ° — ( dx)x ° —0 dx

F

= A=—

W 6

Hence the reaction at the support 0 is W/6, i.e., of the weight + W of the beam. This is otherwise obvious by taking moments about 0. However, in more complicated examples the reactions etc. cannot be calculated by elementary methods and eqns. (3.18), (3.19), (3.20) must be used. (v) A uniform beam AB, of length 6a and weight 6 w a, rests on two supports C, D each at the same level. If AC = CD = DB, find the elevation of the midpoint 0 above the line of the supports.

§3:7

79

CONTINUOUSLY DISTRIBUTED FORCES

Taking axes through the mid-point 0 of the beam as shown in Fig. 32 the equation of loading is dly

EI

dx4

— w.

This equation applies for all points of the beam except at the points C, D. At these points the loading is infinite, because of the concentrated forces from the supports, and the shearing force has discontinuities at each of these points. Integration of the loading equation gives, when Ix I < a ,

El

d3 y — wx Cl . d x3

(1)

x A

By symmetry the shearing force F(= El d3 y/dx3) vanishes when x = 0. Hence = 0 and d3

EI —d— x3

wx.

d2

.•. El dx2 — lwx2

C,

(2)

when x I < a. But consideration of the equilibrium of DB gives M = 2wa2 at D, i.e., EIy2 = 2wa2 when x = a. Hence we put C2 = 3 wa2 /2 and

El

d2 y

dx2

lw x2 ±

a2 '

<

a.

Two successive integrations (putting y = y, = 0 at x = 0 to show that the new constants of integration both vanish) give

wx4 EIY

24

+

3wa2 x2 , 4

< a.

(3)

Hence the elevation of 0 above CD is 19wa4 /24E/. (A knowledge of y when loci> a is not required for the solution of this particular problem. However, equations for y and its first three derivatives in the range a < <3a are similar to eqns. (1), (2), (3) but have different constants of integration. These constants are obtained from the requirements that y, y„ y2,

80

A COURSE OF MATHEMATICS

(= IE I), are continuous at C, D and that F (= E I y3) is discontinuous at these points; the discontinuity is an increase in the shearing force equal to the force from the support, see eqn. (3.20). Exercises 3:7 1. A heavy uniform beam AB, 10 feet long, is supported in a horizontal position by vertical forces at A, and at a point C such that AC = a feet and a > 5. Draw the curves of shear and bending moment, and find the value of a if the bending moment vanishes at a point distant 4 feet from A. 2. A uniform beam AC, of length 21 and of weight 2 w1, rests on three supports A, B, C at the same level, where B is the mid-point of AC. Assuming that the vertical deflection at all points is small, show that the reaction at B is equal to fiveeighths of the weight of the beam. 3. A uniform beam A B, of length land weight w / , rests on smooth rigid supports at its ends A, B and at its middle point M. A and B are on the same level but M is at a small depth d below AB. Determine d in order that the reactions from the three supports shall be equal. 4. A light beam of length 1 rests on two supports at the same level, one at each end. The beam is subjected to a continuous loading, proportional to distance from one end. Determine the point of the beam at which the deflection is a maximum. 5. A uniform beam of length 2a ft, mass w lb per ft run and constant flexural rigidity El rests on three supports, one at each end and one in the middle. The ends are at the same level but the middle support is b ft below the others. If at any section of the beam distant x ft from one end w = k (a — x) where x < a and k is constant, show by integration that an end reaction is 3E./b/a3 +11 ka2 /40. 6. A uniform beam of weight W and length 41 rests symmetrically on two supports at the same level and at a distance 2/ apart. Show that the shearing force and bending moment vanish at the middle point, and that, with the usual notation, the elevation of the middle point above the level of the supports is W/3/96EI . 7. A uniform beam AC, of length 2/ and weight 2w1, rests on the three supports A, B, C at the same level, where B is the mid-point of AC. If the vertical deflection at all points is small, show that the radius of curvature of the beam at the mid-point of BC is twice the radius of curvature at B. 8. A uniform beam is clamped horizontally at one end and carries a load which is uniformly distributed between the centre and the free end. If this load has the same linear density as the beam, show that the ratio of the deflections of the free end and of the centre is 89:31. 9. A uniform beam, of length 2/ and constant flexural rigidity EI, is clamped horizontally at one end and pinned to the same level at the other. The load intensity at any point is proportional to the product of the distances of the point from the ends. Show that the reaction at the clamp is 13/20 of the total load. 10. A uniform rod AB, of length 2/ and weight 2 w/ , is supported at its ends A and B so that AB is horizontal. A particle, of weight 2nwl, is placed at a point P on the rod at a distance z from A. Draw the shearing stress diagram (i) when / > z >n + 1 ' (ii) whenn + 1 > z .

§ 3:8

CONTINUOUSLY DISTRIBUTED FORCES

81

Prove that, if 1> z > n + 1 ' the bending moment is greatest at P and that its magnitude at Pis Prove also that, if

(2n + 1) z (2/ — z).

z /

+ 1 > z, the bending moment is greatest at a point Q

distant 1 — nz from A and that its magnitude at Q is

(1 + nz)2.

11. A heavy uniform rod AB, of length 2a, is supported horizontally on two pegs one placed under the end B of the rod and the other under a point C at a distance c (< a) from A. Sketch the shearing stress and bending moment diagrams. Prove that the bending moment is greatest at C if c > a (2 — I/2). 12. A rod, of length 1 and weight w per unit length, is clamped at one end and supports a weight W attached to the other end. Show that if both ends are to be at the same level the clamp must be inclined to the horizontal at the small

12 (W wl

, where K is the coefficient of flexural rigidity. angle 0 given by 0 = 17 — + 3 13. A light uniform beam, clamped horizontally at one end and supported at the same level at the other end, carries a load W at its mid-point. Find the reactions at the ends, and show that if the support at the non-clamped end is lowered a small distance 6, the reaction upon it is reduced by an amount 3E I 61P , where 1 is the length of the beam.

3:8 The properties of the Heaviside Unit function and its derivative The discussion of the "functions" concerned in this section is not intended to be rigorons, but to indicate how the discontinuities can be handled by the same formal methods as continuous functions. The results can be justified by pure mathematical methods of a somewhat different and abstract character. The Heaviside Unit function is defined by

H(x — a) = 0 if

x < a,

=1 if

x > a.

It is a function with a simple discontinuity at x = a and has a graph shown in Fig. 33 (i). Suppose now that 1(x) is a continuous function for some range of x including the point x = a and that F (x) is an indefinite integral, i.e., F' (x) = / (x); then, as shown in Fig. 33 (ii), 1 (x) H (x — a) is a continuous function except for a simple discontinuity at x = a.

82

A COURSE OF MATHEMATICS

We consider ! 1(x) H (x — a) d x for various values of x, , x,. Using x, Fig. 33 (ii) we easily obtain the following results / (x) H (x — a) dx = 0 if x,, x, < a, = f f(x)dx = F (x,) — F (a) if x1 < a, x,

a,

/(x)dx = — F (xi) -1- F (a) if x, > a, x2 < a, rr, = f f (x) d x = F(x2) — F (x„) if x„, x2 >a.

y=f(x)H(x-a)

a

x

FIG. 33 (ii).

All these results can be expressed in the single formula x, f f (x) H (x — a) dx = (x2) — F (a)} H (x2— a) — x, — {F (xi) — F (a)} H (xi— a) .

(3.25)

This is equivalent to saying that f f (x) H (x — a) dx = {F (x) — F (a)} H (x — a) + constant gives an indefinite integral of (x) H (x — a) .

(3.26)

§ 3:8

CONTINUOUSLY DISTRIBUTED FORCES

83

The 6-function, 6 (x — a), is defined by the equations (x — a) = 0 if x + a, xz f f(x) (x — a) dx = f (a), (3.27) z, for all x1, x2 such that x1< a < x2 and an arbitrary integrable function f(x) . The 6-function, also called the Impulse function, has a graph

0

a

x

FIG. 34.

which is the limiting case of Fig. 34, in which the peak encloses unit area, having zero width and infinite height. If we choose f(x) = 1 in z, the definition, f 8 (x — a)dx = 1, showing that the area between the x, curve and the x-axis is unit area. If we put f(x) = F' (x) into eqn. (3.25), we can formally integrate by parts and obtain x, x, f F' (x) H (x — a) dx = [F (x) H (x — a)]x* — f F (x) Hi (x — a) dx xi = F (x2) H (x2— a) — F(x1) H (x, — a) x, — f F (x) (x — a) dx . x, Comparison with eqn. (3.25) leads to the result x, f F (x) H' (x — a) dx = F (a) (H (x 2 — a) — H(xi— a)} .

(3.28)

84

A COURSE OF MATHEMATICS

(Strictly, of course, the integral on the left hand side does not exist because H' (x — a) is not defined at x = a, and in any case cannot be finite there.) We investigate the values of the r.h. side of eqn. (3.28) for different values of x1, x2.

F (a) {H (x2— a) — H (x, — a)} = 0

if

x1 , x2 < a,

= F (a) (1 — 0) = F (a)

if

x1 < a, x, > a,

= F (a) (0 — 1) = — F (a)

if

x, < a, xi > a,

= F (a) (1 — 1)

if

=

0

X2 >

a.

Hence eqns. (3.28) and (3.25) are consistent if we ascribe a value to the 1.h. side of (3.28) given by

x, f F(x) H' (x — a) d x =F (a) if xi.
(x — a) =- (x — a).

(3.29)

Example. The graph of the function h,A(x) = (1 tanh A x) resembles that of the Heaviside Unit function 'with the corners rounded off'. The shape approximates more closely to that of the Unit function for larger values of A. Increasing A `sharpens the corners' so that, in the limit, 71(x) = lim h,. (x) = Fun (l

tanh A x) .

The graph of the function d A(x) = +2 sech2A x is symmetrical about the y-axis and has a single peak, of height IA , on the y-axis. The area between the curve and the x-axis is unity. The peak is higher and narrower the larger the value of A, and, in the limit, 6(x) = lim c 1A(x) = lim + 2 sech2 Ax . 24.0

1.40.0

We note that c/A(x) —

d 16 2. (x) dx

corresponding to the relation (3.29).

3 : 9 Macaulay's Method In § 3:7 the flexure of a beam was investigated by writing down and integrating a differential equation. If there was a concentrated load at a point x = a, solutions were found for x < a and for x > a. These

§ 3:9

CONTINUOUSLY DISTRIBUTED FORCES

85

solutions were then 'fitted together' at x = a with the help of eqns. (3.20), (3.21). (See also § 3 :10.) Use of the Heaviside and 6-functions enables us to handle these situations by the same formal methods as for continuous distributions and shearing forces. A concentrated load W at x = a corresponds formally to a distributed load w(x) = W6(x — a); the corresponding discontinuity in the shearing force is given by the term W H (x — a) in the expression for the shearing force. This agrees with eqns. (3.18) and (3.20) for d w (x) = — W H' (x — a) = W (x — a) d by eqn. (3.29). We give now a modified form of Macaulay's method by using the Heaviside and 6-functions and regarding each problem as one of inte-

Fiu. 35.

grating the 4th order differential equation (3.24), the loading equation, subject to the appropriate end conditions. Some of the examples given below could equally well be solved by writing down the equation of bending at once, as illustrated in the examples of § 3:7. Examples. (i) A light beam OL of length 1 is freely supported at its ends and a concentrated load W is applied at C where OC = c. (Fig. 35.) Find the equation of deflection. The 'distributed' load w (x) = IV (5 (x — c) leads to the differential equation, using (3.24) with El constant,

EI

d4 dy

w(x) = W (5(x — c) = W H' (x — c).

This has to be integrated subject to the following boundary conditions :

EIy3 = —R, at

x = 0;

EIy2 = 0

x

Ely,

=R2 at

x = /;

0;

=0 at

x = /;

is unknown at both

x = 0 and

x —1;

x = 0;

=0 at

Ely = 0

at

at

x

/.

(1 )

86

A COURSE OF MATHEMATICS

The first integration of (1) gives

EI

d3

d x3

— W H(x — c) ± A .

Substitution of the values of Ely3 at x = 0, x =1 give —Rl = WH(—c)-F A;

i.e.,

R2 = W H(1 — c) — R, = W — Ri ;

A = —R1;

i.e., R1 + R2 = W.

(2)

The second integration, after substituting for A , gives

EI

d2 y — — Ri x ± (x — c)W H(x — c) B, dx2

where we have used eqn. (3.26) with F (x) = x W . The boundary conditions on the values of Ely2 at x = 0, x = 1, give = —cWH(—c)+ B, 0 = — Ri

c) W (1 — c), i.e., Ri l

(/

EI

i.e., B = 0; W (1 — c).

d2 y — Ri x -I- (x — c)W If (x — c). d x2

(3) (4)

(This equation could be obtained by taking moments about P for OP first with P to the left of C, and then with P to the right of C.) Two more integrations give, after allowing for the fact that y = 0 at x = 0,

Ely — —

Rix' 6

+

(x — c)3 6

WH(x— c)-F Cx,

where C is an arbitrary constant. The condition that y = 0 at x =1 determines the value of C from R113 (1 — c)3W 0—— + C/. 6 6 .•. C = c(/ — c) (2/ — c)W/(6/). Hence

EIy——

(1 — c) Wx3 6l

c(/ — c) (2/ — c)Wx (x — c)3W H (x — c) 61 6

where we have used the value of R1given by eqn. (3). The reader will note that eqns. (2) and (3) are also obtainable by considering the equilibrium of the whole rod under the action of Ri , R2 and W. (ii) We modify example (i) by having the rod clamped horizontally at 0, and we take c = +1.

§3:9

CONTINUOUSLY DISTRIBUTED FORCES

87

In this case we integrate the equation d4 y EI dx4 = W H' (x — 41) subject to the boundary conditions:

x = 0;

= R2 at x 1;

E I y, = K

'at x = 0;

=0 at x=/;

Ely,- 0

at

x = 0;

E Iy = 0

at

x = 0;

E y, — R, at

=0 at

x = /.

The clamping couple K is unknown and R, and B2 are different from R, and R2 of example (i). As before we obtain

EI where

d'y = —R1 x d x2

-b (x +1) W H (x — +1) + 13,

lel R, = W. This time the boundary conditions give K — B; 0 = 1111 + +1W H(41) -N B,

i.e.,

—2R11-1- 2K + 1W = 0.

(I)

The next integration gives dy

EI dx —

R1 x2

(x — 1)2

2

2

WH(x-41)+ Kx + C

and the boundary condition on yl gives C = 0. Remembering that y = 0 where x = 0, the final integration leads to

Ely--

R,x3 6

(x — 41)3 6

WH(x 11)+

Kx2

The condition y = 0 where x = 1 gives 0—— i.e.,

81 13

13 W

K12

6

48

2

—8R11+ 24K + /W

O.

Solution of (1) and (2) gives

R, = 11 W/16,

K = 3 W//16.

Hence 96EIy = W{91x2— 11x3+ 16(x — +1)3H (x — +1)).

(2)

88

A COURSE OF MATHEMATICS

(iii) A light beam OA of length 3/ carries a load W at B where 0 B = 2/. Both ends are clamped horizontally at the same level. (See Fig. 36.) Find the equation of deflection and the maximum sag. In this case we solve

d4y

El

= W H'(x — 21)

dx4

subject to the boundary conditions

EIya

—R,

at

x=0;

= I1

at

x= 31;

EIy2 = M,

at

x = 0;

=M2

at

x = 31;

E Iy, = 0

at

x = 0;

=0

at

x = 31;

Ely = 0

at

x = 0;

=0

at

x = 31.

Hence the equation of bending is

E

d2 y — M,— R,x (x — 21)W H(x — 21) dx2

whence

E y = 1_111 ,x2 — ?1 x3

(x — 21)3 W H (x — 21) Ci x

+

C2

Mi A

-77;

L

5tL

r 9!

R

R

ui

M2

FIG. 36.

C,

0 = C2 since y -= 0 -= yi at x 9l2 2

111

M, 3/

R,

313

O. Also at x

W13 270 6 + 6 9P 2

3/, y = 0 = yi and hence

0'

W/2 2 = 0,

giving M1 =2 W//9, RI = 7 W/27. Thus the equation of deflection is

E Iy —

W / x2 9

7 Wx3 162

W (x — 2 /)3 H(x — 21). 6

The maximum sag occurs where y1 = 0, i.e., where 2 W/x 9

7 Wx2 1 W(x — 21)2 H(x — 21) = 0. 54 + 2

§ 3:9

CONTINUOUSLY DISTRIBUTED FORCES

89

For x < 21 this gives 121x — 7x2=- 0, with roots x = 0, 12 //7 ; for x > 2/ this gives 17x2 — 601x 5412 = 0 which has complex roots. Hence the maximum sag 16 W131147 EI occurs when x = 121/7. (iv) A cantilever OA, of constant flexural rigidity EI, uniform weight w per unit length and length 1, is clamped horizontally at 0 and is supported at the same level at B where 0 B = 1/. A load W is hung from A . Find the reaction at B and the clamping couple at 0. With the notation of Fig. 37

w x2 — R2(x — 11) H (x — 1).

E/d2 y/dx2 = M, —

Integrating twice and using the conditions y = 0 = y1 at x = 0 we find

Ely = z Mo x2 — R1 x3 When x

— fi R2 (x — “)3 H(x — “).

y = 0 (same level as 0).

R,

2 9 1231°

8P w 27 + 24

6

1614 81

= 0.

( 1)

When x =1, y2 -= 0 (M = 0). .•. M0 —

11.1)12 — R21 --- O.

(2)

Resolving vertically for the system

R, = W wl. Solution of (1), (2), (3) gives R, =

Lwi — W, R2 = Nwi

W,

mo

4w/2

- W/.

(v) A light beam ABC, of length 2/, flexural rigidity El and midpoint B, is supported at the same level at its ends A ,C C. The beam carries a uniform loading w per unit length on AB and a uniform loading 2w per unit length on BC. Find the equation of deflection. The distributed load in Fig. 38 can be expressed with the help of the Heaviside function as w wH(x — 1). Hence we have to integrate

EI

d4 — dx4

+ H (X - l)

90

A COURSE OF MATHEMATICS

subject to the conditions

E ly, = — R, at

x = 0;

= R2 at x=21;

Ely,- 0

x = 0;

=0 at

Ely,

at

x = 21;

is not given at either end;

EIy=0

x

at

0;

=0 at

x 21.

The first integration gives

EI

da y

dx3

— wx w(x — 1) H(x —1) + A.

,,,,,,,,,, ,,,,,,,, ,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,

Fie. 38. The boundary conditions on y3 at x = 0, x = 1, give

A = — R1, R2 = 2w1+ wl — R„ i.e., The second integration gives

El

c1.2x2 y = d

R2= 3w1.

R,

R1 x +

w 2x2

+

w 2

(1)

(x — 1)2H (x — 1),

no constant of integration being needed because y2 = 0 at x = 0; the boundary condition at x = 21 gives 0= — 2_11,1+ 2w12 +

R,=

i.e.,

wl.

.•. R2

(2)

Equation (1) and the result (2) could be obtained directly by considering the equilibrium of the whole beam. The next two integrations give

Ely =

5w1 x3

woe'

4 6

24

+

w (x —1)4 H (x —1) + Cx, 24

§ 3 : 10

91

CONTINUOUSLY DISTRIBUTED FORCES

where the condition y = 0 at x = 0 is satisfied. Since y = 0 at x = 21

i.e., Hence

Ely =

—

0—

w/4 ( 40 + 16 + 1) + 2C1, 24

C —

23w13 48

24wlx3

'4 w + 48wl3 x

x

2

—

1)'411(x

—

1).

3:10 Clapeyron's theorem of three moments Suppose we have a uniformly loaded uniform beam resting on three or more supports and let A, B, C be any three consecutive supports (Fig. 39) with AB = a, BC = b and A, C at small depths p, q below B. Let RA, {C '

Rc S RC

RB, Rebe the reactions at the supports and MA, MB, Ma the bending moments there. Since the beam is supported at ... A, B, C, ... there is a discontinuity in the shearing force at each of these supports, but the bending moment has no discontinuities. Between any pair of supports the differential equation (3.24) is EI

d4 y =w d x4

since the load is uniform. We integrate this equation for each section of the beam between the supports and 'fit the solutions together' at each support. —a<x<0 0 <x
wx a2 .

,

Because of the discontinuity in F(= E I d3y/d x3) at x= 0 we have the boundary condition at x= 0

—B B

= a2

°CI*

92

A COURSE OF MATHEMATICS

The next integration gives 2y El d d x2

1 w ± Ni x + flu , w

oc2x

p2•

(3.30)

At x = 0 the bending moment 111(= EId2y/d x2) is continuous and has the value MB • MB= NI P2• •. •

-

The third integration gives EI

dy = dx

2o1x2 + 111B x yi , w + cx2x2 + MB x + 2 •

The gradient of the beam is continuous at x = 0 having the value tan BB .•. El tan°, = yi = y2 . The final integration gives EIy=

wx4 24

W X4

24

ai x3 6 H

M x2 B +x El tan 0, , — a < x < 0, 2

111 X2 00 2 X3 + + X EItan0B , 6 2

0 < x < b.

(3.31)

Here we have assumed that y is continuous at x = 0 and vanishes there. From eqns. (3.30) we obtain the bending moments at A, C by putting x = — a, + b respectively into the appropriate expression. Hence + MB, M 0 =2wb2 + c 2b+MB.

MA = 2,va2

— M B Wa = MA a 2

Mc —M B Wb b 062 = 2

(3.32)

From eqns. (3.31) we find the deflections p, q at A, B respectively EIp =

EIq =

24 w b4 24

—

cqa3 6

111B a2 2

aEItan0B ,

a2 b3 6

11/,b2 2

tan°, .

• CONTINUOUSLY DISTRIBUTED FORCES

ti 3 : 10

93

Eliminating al , nc,, tan BR from these, equations by means of (3.32) gives

a E I (P

172 (a3 +,b3) 2-13 ( a ± b) — —

bq =24 (a3 b 3) ▪ (MA — MB) a

(Mc — -211-B) b

6

6 24

aM A + 2 (a

(a3

b) MB ±

aM A + 2 (a

b3)

6

a P b) MB ±b.M,-=6E1(—

+ I w (a3 H-b3). (3.33)

This is Clapeyron's equation of Three Moments. To obtain the reaction at B we use the relation RB = c xi — a, and egns. (3.32) and find Mc 1 1 MA ± (3.34) b) BB = a a ± b) b Similar results can be obtained for light beams with concentrated loads and for the combination of uniform loading and concentrated loads. (See exercises 3:10 No. 1). Examples. (i) A uniform beam rests on three supports at the same level. (Fig. 40.)

Me- 0

B

A Mgr 0

•

AMMAN'

•

RB

RA

C

NIIIRAII.W.1,44■

.11.11.11.1.11I

Rc

HF

b

Fie. 40. Since M = 0 at a free end, MA = 0 = Mc . Hence Clapeyron's theorem gives 2 (a + b) M B =

(a3 +b3 ) .

.•. MB = 8 w(a2 — ab

b2).

Then by eqn. (3.34) 1 1\ — RB 42V (a + b) M B (— + — a b

b) (a2 3ab 8a b

b2)

The reactions at A and C are obtained by the ordinary rules of statics (moments and/or vertical resolution) as RA

w(3a2 ab — 8a

w(3b2 ,

BC

ab — a2) • 8b

94

A COURSE OF MATHEMATICS

(ii) The problem of a uniformly loaded cantilever supporting a point load W at the end can be solved by taking A, B close together and C at the other end of the beam, Fig. 41. Then the theorem gives 1 6EId 21M B = But from elementary statics mB

HT/ + wi2 .

.•. EId = ;',W13

iw14 .

... 0,m,ff"""ff'0..."0..,.../~seawaresw.•••••• ■ •••■ •••••sosso ,. /syrey.sysye,o

d C

Flu. 41.

Fm. 42. Exercises 3:10 I. A light beam of constant flexural rigidity EI, rests on a number of supports at the same level with one concentrated load in each bay, Fig. 42. Show that /, a M A ± 2 (a + b)M B + bMo = W1 — (a — 11) 2 a — a 12 12) (2 b —12), + W2 — b (b (

(1 -27-8 MB

1 b )

—

Me

MA a

b

W„ (a — /1) a

W 2 (b — 12) b

Show also that if the beam is uniformly loaded and with several loads in each bay 1 a MA+ 2(a + b)M B + bMc = — w (as b3) + 4 +

W2 1z(b

— /2) ( 2 b

2,

W1/1(a — /1) (2a — li) a

12)

b

MB(

+

1\ b)

1

=

a

MA M C

a

W, (a — 1,)

a

b

+

w (a + b) 2

W2 (b — 1 2)

+

b

§ 3 : 10

CONTINUOUSLY DISTRIBUTED FORCES

95

2. A uniform beam AB, of length 1, flexural rigidity RI and weight w per unit length, is clamped horizontally at A and simply supported at B which is at the same level as A. A concentrated load W is hung from C where AC =--1. Show that the clamping couple at A is w12 /8 + 15 W//128 and find the reaction at B. 3. A light beam ABCD, of length 3a and flexural rigidity El, is simply supported at its ends A and D and carries loads each of weight W at its points of trisection B and C. Show that the deflection at B is 5 Wa316EI. 4. One end A of a uniform beam of length 21 is clamped horizontally; the beam is raised by a support at the middle so that the other end B, which is free, is at the same level as A. If w is the weight of the beam per unit length, show that the force exerted by the support is 12 w//5. 5. A uniform beam AB, of length l and weight W, is clamped horizontally at B and rests on a support, on the same level as B, at the point C where AC = *AB. Determine the reaction at the support and show that the deflection at A is 5 W/3/1944E/. 6. A uniform beam of length 2/ is clamped horizontally at one end and supported at the same level at its mid-point. It carries a uniformly distributed load of intensity 2w between the support and the clamped end and one of intensity w over the remainder. If, at the clamp, the bending moment is M and the vertical reaction R, find the bending moment at a distance x from the clamp for (a) x <1 and (b)

x > 1. Hence show that M = 0, R =

wl

and find the deflection at the free end. 2 7. A uniform straight beam of length 2a, weight 2 wa , and flexural rigidity El rests on three supports A, B, and C at the same horizontal level and such that AB=BC=a. If the central support is gradually raised, show that when its height above the original level is h, the reaction on it is equal to (5 wa4 24E/h)/4a3 or to 2 wa according as h is less or greater than wa4 /8E./. 8. A light uniform beam of length 6a is freely supported at the same level at four equidistant points A, B, C, D, two of which are the ends of the beam. It carries concentrated loads W at the mid-point E of the beam and at the midpoint of each end-span. Set out the equations for the bending moment, slope and deflection at any point in the part AE, and hence show that the reaction at each end support is 7 W/20. 9. A light uniform bar AB of length 1 has a weight suspended from B. The bar is supported horizontally by a peg above the bar at A and a second peg below the bar at distance a from A. Assuming that bending is slight, show that the greatest deflection of the part of the bar between the pegs occurs at a point C at distance a/V3 from A. Show further that the deflection at C will attain its greatest value if the second peg is placed at a point such that a = 1. 10. The cross-section of a beam AB of length 1 is a circle whose area is Tr a2 x3, x being measured from A. The end B is clamped horizontally, and the beam bends slightly under its own weight. Show that the displacement of A is 2eg/4

9Ea2 where E is Young's modulus and is the density.

96

A COURSE OF MATHEMATICS

11. A light straight rod of length a is smoothly pivoted at two points distant b (< a) apart, one pivot being at one end of the rod. The other end of the rod is then displaced a small distance c from the straight line through the pivots. The rod has a large uniform flexural rigidity El, and is liable to fracture if the bending moment exceeds M. Prove that the largest safe value of c is -ka (a — b)MI(EI).

3:11 Struts We now consider the deflections of beams which are subject to thrusts (struts) or tensions (ties) in addition to transverse loads. In this case the relation d2 y M EI d x2 is still true but I is the second moment of the area of the cross-section about the neutral axis which in this case does not pass through the centroid of a normal plane section. TA

BT x

P(x,y) FIG. 43.

y

Since nis the distance between the centroid and neutral line of any normal section

=10 + Ae where /0 is the second moment of the area of the section w.r. to a horizontal axis through its centroid. Usually Ti is small and we may therefore take I = Io , and use the same value of I for a strut as for flexure. Consider a long, thin, light strut, of length 1 and flexural rigidity El, whose ends are hinged and constrained to move along a straight line and are subjected to thrusts T, Fig. 43. At P (x, y) the bending moment is obtained by taking moments for the section AP about P. The bending moment is — T.

.•. EI

d2 y d x2

= —Ty,

d2 y = n2 y d x2

§ 3: 11

CONTINUOUSLY DISTRIBUTED FORCES

97

where n2 =TIEl. The solution of this equation is y = A cosnx

B sinnx.

The end conditions y = 0 when x = 0 and when x =1 give

A = 0, B sinnl = 0 so that either B = 0 or sin nl = 0. If B = 0 or n = 0, the solution is y = 0 and the strut remains straight. If sinnl = 0, n > 0, B can take any value. This occurs when n/ = rn, where r is an integer, i.e., T = r2 E In2112. In this case the strut takes up the shape of a sine-curve of arbitrary amplitude and hence will collapse. The first critical thrust for which the strut collapses is T = E In2112and in practical applications this is the only critical thrust of importance.

Strut clamped at both ends If the strut is clamped at its ends (with clamping couples M0), the equation of bending becomes d2 ±n2) y = EI d x2 with boundary conditions y = 0 = dy/dx at x = 0 and at x = 1. The solution is M0 y= (1 — cosnx)

El

where

31° (1 El

cosnl) = 0 ---

El

sinnl.

Hence either M0 = 0 or cosn/ = 1 and sin nl = 0, so that the (first) critical thrust is given by nl = 2n, i.e., T = 4n2 E '112. Examples. (i) A strut under axial thrust T is clamped horizontally at one end 0 and hinged at the other end A which is constrained to move horizontally. We must assume a clamping couple Mo at 0 and forces S and T at the ends of the beam as shown in Fig. 44. Then the bending moment at P is Mo — Sx — Ty. Hence d2 y

EI

where n2 =T/EI.

dx2

— M, — Sx — Ty.

(D2 + n2) y —

.M, — Sx El

(1

)

98

A COURSE OF MATHEMATICS

The general solution of (1) is

Mo — Sx T

y = C„ cosnx C2 sinnx

The end conditions at x = 0 are y = 0 = dy/dx; at the end x = 1 there is no clamping couple so that the bending moment is zero there. Hence, at x = 1, y = 0 = d 2 y/dx2. These give, respectively,

C1

M

— T a0

2 —

— 2

C, cosni + C2 sinn/ C, cosn/

M

—

oT

T

=0,

Si

— 0,

C, sinnl = 0.

Hence

mo = Si,

Cl Mon — C, S

tann/ = — tann/ = nl.

(2)

The only other possibility is Mo = 0 implying that S = 0 = Cl = C2 in which case the strut remains straight. If M, 0 the solution is S

y = T (1 — cosnx)

S

nT

(sinnx — nx),

(3)

where n satisfies (2).

T

The first positive root of (2) is approximately nl = 4.494 and hence the (first) critical thrust is T = 20.25 E///2approximately. (ii) A uniform horizontal strut OA, of length 1 and flexural rigidity El, carries a uniform load w per unit length and has both ends clamped horizontally. The ends are constrained to move in a horizontal straight line and are subject to axial thrust T. With the external forces as shown in Fig. 45 the equation of bending at P is

EI i.e.,

d2 y = T y d x2

lwx2,

(D2 + n2)y = (M 0 — lw/x lwx2)/E/,

(1)

Sg 3 •• 11

CONTINUOUSLY DISTRIBUTED FORCES

where, as before, n2 =T/EI. The solution of (1) is y = Cl cosnx

C2 sinnx

1

2T

(2M,

2w n'

The end conditions y = 0 = dy/dx at both x = 0 and x =

1110 —

Y

w n2

—

wl 2n

cot

w sin lnx sin +n(1 — x) nT sin÷,n1

w x (1 — x) 2T

From (4) y -- co if sin ln/ = 0 so that the critical loading is the same as for a strut clamped at both ends with no transverse loading. Exercises 3:11 1. A uniform thin lath, of length 1 and constant flexural rigidity El, is clamped vertically at its lower end and at its upper end carries a small light bracket of length a fixed perpendicularly to the lath. When a load W is hung from the bracket it deflects a small horizontal distance b and negligible vertical distance. State the bending moment at a point on the lath distant x vertically and y horizontally from the clamped end. Find b and the bending moment at the clamp in terms of the other quantities given. Evaluate W when b = a. 2. A light uniform pole, of length 1 and constant flexural rigidity EI, is fixed vertically in the ground at its lower end A, and its upper end B is acted upon by a force T which makes an angle a with the downward vertical. The consequent small horizontal deflection of B is a. Taking the origin at A, measuring x vertically up and y horizontally, state the bending moment at any point P (x, y) of the pole and show that (D2 n') y = n' a + n2 (/ — x) tan a where D d/dx and Eln2 = T cosa. Solve this differential equation and show that na = tana(tannl — n1). 3. A uniform light bar, of length land constant flexural rigidity El, has its lower and clamped at a small angle a to the vertical and carries a vertical load W at its upper, free end. Assuming that a is small enough for sina to be taken as approx-

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A COURSE OF MATHEMATICS

imately equal to a, show that the differential equation for the displacement y (from the undeflected position of the bar) of a point distant x from the clamp is approximately EIcPyldx2= W{o — y a(/ — x)} where y = 6 at the free end. Show also that 6 = al(tannl — nl)/(nl)

where

E I n2 = W.

4. A light thin uniform strut, of length l and flexural rigidity EI, is clamped at one end A and subjected to an axial thrust T at the other end B. The point B is at the same horizontal level as A and is prevented from moving perpendicularly to AB by a force F. Show that if any point P in the strut distant x from A has a small transverse deflection y the corresponding differential equation is EI cl2 y1dx2 = F(x —1)— Ty. Show also that if a deflection is possible then T must satisfy the equation tannl = nl where EIn2 = T. 5. A horizontal light strut A B, of length a + b and flexural rigidity El., is hinged at each end and subject to end thrusts T. A concentrated load W is placed at P where AP = a. Show that the deflection at P is W sinna sinnb n T sinn (a + b)

Wab T (a b)

where n2 =T1EI. 6. A light strut of length l has end thrusts T applied eccentrically at distance a from the neutral axis and acting in the same line. Assuming that the ends are constrained to move horizontally, prove that the equation of deflection is y = 2a seci n1 sinInx sin-in(/ — x), where n2 =T/EI. Miscellaneous Exercises III 1. A uniform heavy flexible string A B, of length 2/, has its endsA,B attached to light rough rings which can slide on a fixed horizontal rod. The coefficient of friction between the rings and the rod is 1/1/3. Show that when equilibrium is limiting at both ends the depth of the mid-point M of AB below the rod is //1/3 . If a particle of the same weight as the string is now attached to M and the string takes up a new symmetrical position of limiting equilibrium, show that M falls a distance 1(3 — 1/7)/1/ 3 . 2. A uniform chain AB is attached to a support at A and to a particle at B. The particle weighs is times the weight of the chain and it rests on a rough plane of inclination a, the angle of friction between the particle and the plane being (> a). AB is horizontal and in a vertical plane through a line of greatest slope of the inclined plane, and no part of the chain is in contact with the plane. If B is in limiting equilibrium prove that the inclination of the chain to the horizontal at either end is cot-1{ (2n + 1) tan (A. — a)} . 3. A uniform rod AB of length 2a can turn freely in a vertical plane about an end A which is fixed. A uniform heavy chain has one of its ends attached to B,

CONTINUOUSLY DISTRIBUTED FORCES

101

passes over a smooth peg C at the level of A, and hangs freely. The weight per unit length of the rod is twice that of the chain In equilibrium the rod is inclined to the vertical at 45°, and the tangent at B to the chain is horizontal. Prove that the tangent at C to the chain makes with the horizontal an angle y given by (1 + I/2) cosy = I/2 ; and find the length of the chain. 4. A chain consists of two uniform portions AC, CB of equal weight and of respective lengths a, b. The chain is suspended from A and B so that C is its lowest point. Prove that the tangents at A and B make the same angle with the the difference in level of A and B is horizontal and that, if this angle is (a — b)tan-ifl. Prove, also, that the horizontal distance apart of A and B is (a + b) cot (3 log (sec i3 + tan M . 5. A uniform heavy chain of length 2l is stretched between two supports at the same horizontal level. The lowest point of the chain is at depth h below the supports. Prove that the centre of gravity of the chain is at a depth h2

12 4h

2h1 (12— h2)2 8h2/ sinh-1/2 — h2

below the supports. 6. A loaded beam AB of length / is clamped horizontally at one end A. The weight of the beam and its load are uniformly distributed along AB but the moment of inertia I of the cross section of the beam is k(21 — x), where k is constant and x is the distance of the section from the end A. Find the deflection of the free end B. 7. A uniform rod, of weight W and flexural rigidity El, has its ends constrained by smooth horizontal guides which are in the same horizontal line and at a distance 2l apart. If a weight W' is supported at the middle point, prove that the bending W/ W'/ couples at the ends are of magnitude + and calculate the deflection at 6 4 the middle point. 8. A uniform beam, of weight W and length 6a, is supported symmetrically at its centre C and at two points B, D at the same level, each at a distance 2a from C. Express the bending moment at any point P in BC, distant x from the end A nearer to B, in terms of x, a, W and R, the reaction at B. By integration show that this reaction is N W and find the deflection of A below B. 9. A and B are two consecutive supports, at the same level, of a light uniform beam, AB being of length 1. There is a concentrated load W at a point P between A and B, where AP = a. If the slope of the beam at A is zero show that /2 (2MA + M B) = W a (1 — a) (21 — a) where MA and MB are the bending moments at A and B. A light uniform beam of length / is clamped horizontally at one end and freely supported at the same level at the other. It carries a concentrate d load W at a

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A COURSE OF MATHEMATICS

distance 4 1 from the clamped end. Find the bending moment and the vertical reaction at the clamp. 10. A uniform beam, of length 1 and constant flexural rigidity El, is clamped horizontally at one end and is subject to a uniformly distributed load of intensity w per unit length. The other end is free but is acted upon by a horizontal thrust T. If the end deflection is a show that, in equilibrium, the bending moment at any point (x, y) distant x from the clamp is

Ta w12/2 — wlx wx2/2 — Ty. State and solve the differential equation for y in terms of x and show that

w12

a T = wl — tann/2 — n

w n2 (seen/ — 1),

where E In' = T. 11. A light uniform tie rod of length 2/ has its ends kept horizontal and at the same level. It is subject to constant horizontal tensions T acting through the ends and carries a concentrated load W at its mid-point. Show that the differential equation for the deflection y at a distance x(
RI d2 y/dx2 = Ty + M — Wx/2 where M is the bending moment at an end. w nl Solve this, using hyperbolic functions, and show that M is 2— n tank — and the 2 deflection at the load is

W/ 2T

1

2

n1

n/ ) tanh 2 \ where

Fin' = T.

12. A uniform light strut, of length 1 and flexural rigidity E I, is subject to equal and opposite applied torques at its ends so that the small deflection at the middle of the strut is a. Show that the form of the strut is a parabola, and that the applied torques are of magnitude 8E1a112. If now the strut is subjected to compressive forces P acting through the ends, show that the deflection y at a distance x from one end satisfies the differential equation dz y 8a n2Y 1" where Kin' = P. Hence find the greatest bending moment in the strut. 13. Three small smooth pegs are fixed at intervals of length 2a in the same horizontal line. A uniform chain passes over the pegs and is in equilibrium in a vertical plane with its ends free. Prove that the least length of the chain is a (3 et + e-2), where t is the real root of the equation 30 2 (1 — t) = (1 t) and 0 < < 1. 14. The wireless aerial of an aircraft in horizontal flight with constant speed V is a light flexible cable carrying a mass m at the lower end. Assuming the air resistance on an element ds of the cable, of inclination 99 to the horizontal, to be perpendicular to the element and of amount k V 2 sin2 q ds, show that the tension

CONTINUOUSLY DISTRIBUTED FORCES

103

in the cable is constant throughout, and that its shape is that of an arc of a catenary with vertical directrix. If the drag on the mass m is negligible, show that if the length of the cable is 1, the mass is at a horizontal distance

V

mv k g2 g: 22v : (12 + 17

behind the point at which the cable leaves the aircraft. 15. The ends of a uniform chain AB, of length 2/ and weight W, are fixed to two points in a horizontal line and a load is W is hung from the middle point C of the chain. Prove that the tensions at A and C are in the ratio (2n + 1 + 22): (2n + 1 — 22), where 2l is the depth of C below AB. 16. A heavy uniform chain of weight w per unit length hangs in a catenary of parameter c. Show that the vertices of a light string can fall on the same curve, if it supports an odd number of weights at equal horizontal intervals nc such that the weight Wr at a horizontal distance me from the lowest point is given by n Wr = 4 we sinh2 z n coshrn, the horizontal tensions in the two cases being equal. 17. A uniform beam of length 21 is supported freely at its ends, which are at the same level, and sags a small distance a in the middle. The beam is now clamped horizontally at one end and the support at the other end is removed. Show that the deflection of the free end is (48 a/5) . 18. A uniform heavy bar ABC of length a + b is clamped horizontally at A, and rests on a support at B at the same level as A and at a distance b from the free end C. Show that, if C is below AB,

a3 < 6 (a + b) b2. 19. A uniform rod AB, of length 2a and weight 2wa, rests on two pegs at its ends A and B, the pegs being in the same horizontal plane. A uniform rod CD, of length a and weight wa, rests along AB with C and D at distances y and a + y respectively from A, y being less than a. Draw the shearing stress and bending moment diagrams for the rod AB and prove that the greatest bending moment 7w is 64 (7a2 + 4ay — 4y2). 20. The line-density of a thin straight rod AB, of length a, at a distance x from the end A is w (a + x), where w is constant. The rod is supported in a horizontal position by vertical forces applied to it at A and at a point distance 20a/21 from A. Find the magnitudes of the greatest shearing force and bending moment to which the rod is subjected. 21. A uniform heavy rod AB, of length 2a and weight 2 wa , is supported in a horizontal position by vertical forces applied to it at points C and D such that

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A COURSE OF MATHEMATICS

AC = DP = b < a. Find the range of values of b for which the bending moment at every point in the rod is less than wa2 /8. 22. A uniform straight rod of length 1 has a rigid arm of length a rigidly attached to it at each end, the arms being perpendicular to the rod and on the same side of it, and in the same plane. The rod is in compression under the action of equal and opposite forces P applied to the ends of the arms parallel to the axis of the rod when undeflected. Show that the displacement of the rod from the line joining its ends is a[sec(m//2) cos(ml/2 — mx) — 1] at a distance x from one end, where m2 =PIEI, and that the greatest value of the bending moment in the rod is equal to aP sec (m //2).

CHAPTER IV

KINEMATICS 4:1 Introduction The topic 'kinematics' is concerned with the description of the motion of particles and rigid bodies. Here we introduce some of the more useful methods of expressing algebraically and geometrically by vectors, components, etc., the motion of a point or of a rigid body. Later we use the methods and results of this chapter when we consider such motions and the forces which cause these motions. This chapter falls broadly into two sections in which we consider the motion of a particle and the uniplanar motion of a rigid body (or lamina). We use methods and express the results in forms which are connected with the corresponding results for three dimensional motion and so familiarise the reader with the ideas in the simpler context of two dimensional motion.

4:2 Velocity and acceleration A point has a velocity when its position is altering and an acceleration when its velocity is altering. In Vol. I § 5: 9 the velocity of a point moving in one dimension was obtained by differentiation of the position coordinate, the acceleration by differentiating the velocity. We now consider velocity and acceleration in a more general context making use of vectors (see Vol. II, Chap. IV). First, we introduce the operation of differentiating a vector w.r. to a scalar parameter. We use the letter t to denote the scalar because in most of what follows the vectors concerned may depend upon the time ; nevertheless the result of differentiation is quite general. To make it completely so we give the results for vectors in three dimensions although we shall apply them in this volume almost exclusively to situations where we require only two-dimensional vectors. (A two-dimensional

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A COURSE OF MATHEMATICS

vector can be regarded as a three-dimensional vector whose third component is zero.) A vector a (t) , which depends upon the parameter t, has components a, (t), a2 (t), a3(t) referred to a fixed frame of reference (i.e., the frame does not alter if t alters). An increment 6 t in the parameter t corresponds to an increment 6a in a, where 6a = a(t

St) — a(t) ,

(4.1)

and the components of 6a are 6a, = ai (t + 6t) — al (t), 6a2 = a2 (t ot) — a3(t)

5a3 = a3 (t

6t) — a2 (t), (4.2)

da to be the vector which has components dt . [If one (or more) of the components al , a2 , a3 can( dt dt dt) not be differentiated—it may be a discontinuous function of t —then da/dt does not exist for that value of t.] Because multiplication of eqn. (4.1) by the scalar 1/6 t implies multiplication of the component eqns. (4.2) by llot the definition of da/dt implies that We define the vector

da a (t = Ern dt 0

at) — a(t) 6t

(4.3)

Briefly then, the derivative of a vector is obtained by differentiating its components. It is important to remember that, in general, d a/d t has a different direction from the direction of a. The following results are easily obtained from applications of the definition of eqn. (4.3). (We leave the formal proofs for the reader, see Exercises 4:2 No. 1.) (1) If 2, a both depend on t and b = 2a, then db d2 da = a H- A . dt dt dt

(4.4)

(ii) If a, b depend upon t and u = a • b, then du dt (iii) If c = a x b, then de dt

da dt

db . dt

(4.5)

da xb+a x db dt dt

(4.6)

b + a•

§4:2

107

KINEMATICS

The order a before b must be retained in eqn. (4.6). Equation (4.5) has an important consequence which we develop here. If a is a vector whose magnitude, a, is constant but whose direction may depend upon t, then a2 = a2 =a • a. Hence 0

—

d (a2) dt

d dt (a • a)

da da dt . a +a •dt

2a •

da dt •

(4.7)

Therefore the vectors a and da/dt are perpendicular. If a is a unit vector in two dimensions, we can write i cos 0 da .•. dt

i0 sine

j sine. j0 cos@ = et,

(4.8)

where b = —1 sin j cos 0 = i cos (0 + 1,7) j sin (0 + la). Therefore b is another unit vector obtained by rotating a through one right angle in the positive sense. If k is a unit vector perpendicular to the plane of i, j and forms a r.h. set with them, we can also write eqn. (4.8) in the form d = 0 (k X a). (4.9) dt If the position of a point is given with reference to a standard, inertial or Newtonian (see § 1:2) frame of reference with origin 0 by a position vector r, with components (x, y) in two dimensions, when the point is moving the velocity of the point is defined to be the vector dr dt

v

(4.10)

We usually denote the components of v by (v1, v2) so that dx dy v, =, v = . dt 2 dt

(4.11)

Similarly the acceleration of a point is defined to be the vector f=

dv dt

d2 r . dt2

(4.12)

The cartesian components of the acceleration are obtained by differentiating the corresponding °artesian velocity components.

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A COURSE OF MATHEMATICS

Since the operation d/dt when applied to a vector is formally similar to the operation with scalars, we can, in many cases, also integrate w.r. to t. Consider a particle moving in a plane with a constant acceleration — gj . (If j is vertical and i is horizontal the discussion applies to the motion of an unresisted projectile.) In this case f—

dv d t = gi •

:.v = u

—

gtj,

(4.13)

where u is an arbitrary, constant vector (the velocity of projection of the particle). We can integrate again and obtain r = a + ut — igt2 j,

(4.14)

where a is another arbitrary, constant vector (the position vector of the point of projection). If we take the scalar product of eqn. (4.13) with i, we obtain V1 =v • i = u • i = tcj.= constant . This demonstrates one of the important characteristics of projectile motion viz., the horizontal component of velocity is constant. The component equations corresponding to eqn. (4.14) give the path of the projectile in terms of t, and elimination of t shows that the path is a parabola. Examples. (i) Drops of water are thrown tangentially off the rim of a wheel of radius a which is rotating in a vertical plane about its axis, which is fixed and horizontal, with angular velocity co > (gla) 1. Show that if a horizontal ceiling, at height It > a above the axis of the wheel, is not to be spattered with water w < fg h g( 42— a2)11/a A typical drop will leave the wheel at A (Fig. 46). Its velocity and position after time t are given by (4.13) and (4.14) if we put a = a(i cos°

j sine), u

aco(—i sine

j cos0).

Hence v = a (— i sin r = a(i cos 0

j cos 0) — gtj,

j sine) + a cot (— i sine

j cos 0) — 1-gt2 j.

The particle ceases to rise when the component v2 vanishes, i.e., when aco cos° — gt = 0 so that t = (aw/g) cos° .

§4:2

109

KINEMATICS

For fixed 0 the greatest height above 0 reached by the particle is y = a sin0 + a cot cos0 — i gt 2 =a sin° (a2 co2 /2g) cos2 0. As varies y has a maximum value ym when a co2 sin° = g, ym =

a2

2wp If the ceiling is not to be spattered, ym < h. a2 w2

+

2g

2co2 + 2g < h ' a2co4 — 2g h co2

g2 < .

This implies that the value of co2lies between the zeros of the quadratic expression a2 t2 — 2ght g2, so that gh — gy(h2 — a2) < a2w2 < gh gy(h2— a2) . The required result follows.

(ii) Show that simple harmonic motion is given by the projection of uniform circular motion. A point P describes a circle with uniform speed and Q is the foot of the perpendicular from P to a diameter. The coordinates of P can be written (a cos cot, a sin cot); if we choose the origin at the centre of the circle and the x-axis along the diameter through Q, then Q is the point (a cos cot, 0). The velocity of Q is (— a co sin cot, 0), and the acceleration of Q is (— a co2cos cot, 0) , i.e., f, = — a co2 cos cot = — co2 x, ,f, = 0. This shows that the acceleration of Q is directed towards 0 and is proportional to OQ; this is the definition of simple harmonic motion. (Most of the properties of S. H.M. can be obtained from this method.)

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(iii) A point P(x, y) moves round a circle of radius a and centre the origin 0. If, between the points Po (a, 0) and A (a/V2, a/I/2), P moves with constant speed a co, show that the acceleration of the point Q (x, 0) is then directed towards 0 and is proportional to x, and that the speed of Q when P is at A is a co/V2. Between A and the point B(— a/1/2, a/V2) P moves so that Q has constant speed a co/V2; show that the acceleration of P is perpendicular to the x-axis and inversely proportional to y3, whilst the speed of P is inversely proportional to y. If, after reaching B, P again moves with constant speed a co, show that the time taken by Q in traversing the diameter from (a, 0) to (—a, 0) is (4 + 7t)/2 co. As P moves from Po to A, Fig. 47, P is the point (a cos cot, a sin cot) and 9 is the point (a cos cot, 0). Hence the acceleration of Q has components

A = —a co2 coscot = —co2 x, f2 =O. When t = i/4w the velocity of Q has components v, = —a co/V2, v2 = 0. As P moves from A to B, P is at (a cos 0, a sin 0) where n/4 < 0 < 3n/4. The velocity of Q has components aco v, = a0 sine= v o = 0. ' •

—

aq/2 2 sin 0

The velocity of P has components (— a e sin 0, ab cose) =

cola , (a co cot 0)42} .

The acceleration of P has components, by differentiating the last expression,

(0

a co • a co 2 cosec2 0) = (0 , — V2 2n si' 0

But the y-coordinate of P is a sin 0; hence the acceleration of P is perpendicular to Ox and is proportional to 1/y3. The speed, V, of P is given by

V2 = ia2co2 (1 + cot2 0) = 1a2co' cosec20 , V —

a col/2 2 sin°

showing that V is proportional to 1/y. The time taken for P to travel from Po to A, and from B to P1 is in each 4 co 2a v2 2 case. The time taken for Q to travel from Al to B1 is — — = — 0) since Q has V2 aco constant speed. Hence the total time required for Q to traverse the diameter

§4:2

KINEMATICS

111

c PO P, is —

2 1 — = co (n + 4). (The motion of Q is similar to that of a particle 2 co co 2 on the end of an elastic string which becomes slack when Q lies between Al and B1 .) Exercises 4:2 1. Write out the formal proof of eqns. (4.4) to (4.6) by differentiating the components. 2. Find the acceleration of a point moving around a circle with speed a co by differentiating its position vector r = a (i cos cot + j sin cot). 3. A particle is projected in a vertical plane from a point 0 with velocity u at angle a to the horizontal. Show that the time at which the particle is moving at an angle B to the horizontal is u cos a g

(tana — tan 0).

Find the time from the moment of projection until the particle P is moving perpendicular to OP and show that for this to be possible tana 4 21/2. 4. A particle is projected with velocity V in different directions from a point 0 in a plane inclined at an angle a to the horizontal and moves under gravity. Find, in terms of V, a, 99 and g, the maximum ranges up and down the plane in a direction making an angle o with the line of greatest slope in the plane. If cartesian co-ordinates (x, y) are taken in the inclined plane with 0 as origin, the x-axis horizontal and the y-axis along the line of greatest slope through 0, show that the locus of the furthest points reached in different directions is the conic x2 + y2 cos2 a + 2 ky sin a = k2, where kg = V 2. 5. A ball is projected from a point on a smooth fixed plane inclined at an angle a to the horizontal. The initial direction of projection is up the plane, makes an angle a + /3 with the horizontal, and is in the vertical plane containing the line of greatest slope through the point of projection. Show that the condition that the ball will have just ceased bouncing when it returns to its point of projection is tana tan/3 = 1 — e, where e is the coefficient of restitution for the ball and the plane. 6. Two smooth vertical walls stand on a horizontal plane at a distance d apart. A ball is projected with velocity V and at elevation 0 from the foot of one wall, and after striking the walls 2n times in all first returns to the level of projection lth time). The coefficient of restitution beby striking the other wall (for the n tween the ball and the walls is e. Prove that V2On (1 — e) sin2 0 = gd(1 — 02+1). 7. A particle is projected from the foot of a smooth vertical wall with velocity

V, at an inclination a in a vertical plane perpendicular to the wall, so as to strike a similar parallel wall distant a from the first wall. Show that if the particle

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A COURSE OF MATHEMATICS

traverses the space between the walls at least n times before reaching the ground again then the coefficient of restitution e must satisfy the condition (A — 1) en — A en-1+ 1 < 0 where A = (V2/ga) sin2 a . 8. Air flows horizontally along a pipe of radius a. The velocity of the air at distance r from the axis of the pipe is k (a2 — r2) where k is constant. Particles of dust move with the horizontal velocity of the air and with constant downward velocity u. The projection of the path of a particle on to a vertical plane cutting the pipe at right angles is a vertical chord of the circular cross-section, subtending an angle 2 a at the centre. If a particle enters the pipe at the top of the chord show that after time t its horizontal velocity is k (2 aut sina — u2t2) and deduce that it travels horizontally a total distance 4 ka3 sin3a/3u . What is the least length of pipe such that all dust entering it will settle within the pipe?

4:3 Special representations Here we give some formulae for the components of the velocity and acceleration of a point in terms of different representations of its position. (a) Cartesian coordinates. We repeat this here for completeness and to illustrate the use of vector notation. The position vector is r = xi + yj and the velocity is, by differentiation, dr dx =_ i v= j = vli + v2 j (4.15) dt dt dt

dt

since di/dt = 0 = dj/dt, i and j being fixed vectors. The acceleration is, similarly,

f = dv = d2x dt

d2 y dt2

dt2

_=

dt

+

dv22 d dt

(4.16)

(b) Polar coordinates. (Fig. 48) Using a fundamental cartesian frame of reference we write the position vector r = ar, where a = i cos 0 by differentiation

j sin 0 is a unit vector in the radial direction. Then,

di = a + 1.6; • (4.17) dt Here we have used eqn. (4.8), and 1, = — i sin 0 j cos 0 is a unit vector in the transverse direction. By differentiating eqn. (4.17) again we obtain the acceleration di di) f = dv = i* a + (i• 6 + r e) ru dt dt dt v =--

dr 6

—

dt

a+r

§4:3 db Now dt

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KINEMATICS

6 a, by eqn. (4.8). Hence f =(r— re2) a ± (2 7,e

ro) L.

(4.18)

A very common form for the transverse component is 2r0

9'0 = r dt (r20 ) .

(4.19)

In working with polar coordinates it is most useful to give the velocity and acceleration by means of radial and transverse components. These

components are given respectively as the coefficients of a and b in eqns. (4.17) and (4.18). (c) Intrinsic coordinates. (Fig. 49.) This representation can be used when a particle moves along a curve or on a surface. One of the intrinsic coordinates is the arc length s along the curve from some fixed point to P, and the other is the angle p between the tangent at P and a fixed direction. The speed of the particle along the curve is v = ds/dt; we denote the direction of the tangent at P by the unit vector s" and the direction of the normal toward the centre of curvature by n. " . (In Fig. 49 the curvature is such that pincreases as s increases, i.e., d y/ds 1/e >0. The velocity vector is given by ds „ v = vs = s. (4.20) dt By differentiating this we obtain the acceleration

f=

dv dv -= s v dt dt dt

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A COURSE OF MATHEMATICS

But d; dt

dy, dt

v

ds dt ds

f=

dv v2 s + —n. dt e

(4.21)

ds 2 dv The component — is called the tangential component v dt ate ds) ds of the acceleration; the component v2Ip is called the normal component of the acceleration. (The reader can easily see that the same formula holds even if the curvature is in the opposite direction, i.e., when dye/ds < 0, provided that ñ is always directed towards the centre of curvature.) Examples. (i) A particle describes the equiangular spiral, r = a e°, in such a manner that its acceleration has no radial component. Prove that the magnitudes of the velocity and acceleration are each proportional to r. Since r = ae0, then i• = 8 ae0 = re. Hence 1 . ; -= 9412 + ro, and

—

412 =

We are given that the acceleration has no radial component so that r8 = 0, i.e., = 0 so that 8 = w = constant. .•. r = car, ro = cor. The magnitude of the velocity is v2 = 2:2 + (r 8)2 = 2 012.

.•.v = v2 (or , showing that v is proportional to r. Since the radial component of acceleration is zero the resultant acceleration, if any, must be in the transverse direction. Hence the acceleration is of magnitude f = 2 ie +

2rco • a) = 2r co2,

showing that f is also proportional to r. (ii) A point moves in a plane curve and at time t passes through the point of the curve at which the curvature is x with velocity v. If d2 v/dt2 = av, where a is a constant, show that x = b/v3, where b is a constant, and that 2) 2 = av2 + C— b21v2, where c is a constant.

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Let a, n be unit vectors in the tangential, normal directions respectively; then v = va, dv/dt = (dv/dt)

(xv2)

Hence d2 v dt2

d2 v dv dt2 a+dt "n

n.

d 2) — x2 ndt (xv v2a '

where we have used the relations da/dt = xv a, and dii/dt = — xvi. Then av = ava = (:ei —x2v3) i

{xv dv ct

(ne) } ii

We identify the two rectangular components in the directions of a and ri on each side of this equation. dv .•. av = v — x2v3, xv2 v (xv2) = 0. (1) ds ds In the latter form we have used the relationdt

v where 8 is the arc length ds ' between the point and an arbitrary origin on the curve. The second of eqns. (1) leads to dv dx 3xv v2 O. ds ds dv dx This is equivalent to the separable differential equation 3 + 0 with v x solution xv3 = b(= constant). We use this result in the first of eqns. (1) which we write in the form av = dv

b2 (/;'2)

X2V3 =-CFI/ (.“2)

7) •

This can be integrated to give z v2

= av2 — 2v2 b + 40,

where C is an arbitrary constant. This is equivalent to the required relation. Exercises 4:3 1. A particle describes a cardioid r = a (1 + cos 0) in such a manner that the radius vector from the origin rotates with uniform angular velocity w. Show that the acceleration consists of a component 2a co2 parallel to the initial line and a component (4r — 3a) co2 towards the origin. 2. If the position vector of a point P moving in a plane relative to an origin 0 --> in the plane is OP = r R, where R is a unit vector making an angle 0 with a given direction in the plane, obtain expressions for the radial and cross-radial resolutes of the acceleration of P relative to 0.

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If the velocity and acceleration vectors always make angles with the position vector equal to a, 2a respectively (measured in the same sense), show that = kr,, where k is a constant independent of a. 3. If R and S are unit vectors in a plane in directions making angles 0 and (0 + n/2) with a fixed line in the plane, obtain expressions for the radial and crossradial resolutes of acceleration of a point P moving in the plane, the position vector of P being r R, referred to a fixed point of the plane. If the acceleration of P is (k re — yr) R — ki. S, where p and k are constants, and if the velocity of P is ÷, k a S when r = a, show that P will again be moving at right angles to the radius vector when

r = 2ak(k2 +4 i)-'/2. 4. A particle moves with constant speed v along the cardioid r = a (1 + cos 0) . Show that the radial component of the acceleration is constant, and that both d 0/dt and the magnitude of the resultant acceleration are proportional to r-1/2.

4:4 The Hodograph The hodograph of the motion of a point P is defined by the following construction : at an arbitrary point of its motion P has a velocity v; drawn from a fixed origin 0', the vector 0' Q = v fixes the position of Q . As P follows its motion so Q traces out a curve ; the motion of Q is the hodograph of the motion of P. The important and self-evident property of the hodograph is that the velocity of Q is the same as the acceleration of P. If the hodograph is a motion whose characteristics are obvious or well-known, then the acceleration in the original motion can be easily deduced. Examples. (i) Suppose P traces out a circle of radius a with uniform speed ace, Fig. 50. The hodograph is the motion of Q where 0'Q = a w and the direction of 0' Q is perpendicular to OP. Hence Q traces out a circle of radius ace with uniform speed a co2 perpendicular to 0' Q. Hence the acceleration of P is a co2 directed along P 0 .

(ii) The velocity of a point P relative to an origin 0 is given by v=u

(k x r)/r,

where r = OP, and u and k are constant vectors such that u • k = 0 = r • k. Prove that the hodograph of P is a circle of radius k, and that the path of P is a conic of eccentricity u lk described about 0 as focus, under an acceleration proportional to 1/r2 and directed towards 0 . Since r • k = 0, P always lies in a plane perpendicular to k. Also since (v

v = u (k x r)/r, u)2 = (k x r)2/r2 = {k2,2 (k • r)2}/r2 = k2.

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Hence the hodograph is a circle of radius k whose centre is at C where 0' C = u, Fig. 51. But ---> 0' Q v 0' C+CQ=u+ (k x r)/r .

.• CQ = (k x r)/r. This implies that CQ is perpendicular to r (= 0 P) . Hence Q moves around the hodograph circle with speed Ice (k = 1 k > 0) at any point. The velocity of Q

P

Fro. 51. is the acceleration of P which therefore has magnitude k, 0 in the direction of — r. (We assume, without loss of generality, that 0 > 0.) Therefore the transverse acceleration of P is zero, i.e., 1 d (r2b) r dt which leads to r2 0 = h (= constant). Also, for the radial component, which is the only component of acceleration, — ro2

—k(5 -= —1c,h1r2.

This proves that P has an acceleration proportional to 1/r2 directed to ward 0.

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To find the path we resolve the velocity O'Q along and perpendicular to the direction OP to obtain the radial and transverse components of velocity. For the radial component we get, since O'Q = O'C CQ = u+ CQ, = u cos 0 . For the transverse component we get

ro = —ic sin° f k, since CQ has magnitude it and is in the transverse direction and O'C makes the angle -1-76 + 0 with the transverse direction. Division of these two equations gives r

dO

—

dr

k — usin0 u cos°

This is a separable differential equation which integrates to give

A

=1 — — sin0. k

This is the polar equation of a conic of eccentricity

whose major axis is at

right angles to the initial line which we chose in the direction of the vector'u.

4:5 Relative motion If two points A and P are both in motion, Fig. 52 the vector rule of addition for displacements gives r = rA +R.

(4.22)

The vector It is the position vector of the point P "relative to the origin A", and eqn. (4.22) simply gives, in vector form, the law of transformation for a change of origin (see Vol. I § 4:10). In terms of observers, R is the position vector an observer moving with A would ascribe to P. (It is assumed that the directions of the axes of the frames of reference used by A and 0 are the same. We shall consider change of direction of the axes later.) We differentiate eqn. (4.22) to give two equations, one relating velocities, (4.23) vA + V, and one relating accelerations, f = 4 + F.

(4.24)

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The velocity V and the acceleration F are vectors describing the motion of P relative to A. (The vectors v, f are sometimes called the 'actual' velocity, acceleration of P, a phrase meaning that 0 is taken as a stationary point.) Problems on relative motion reduce to a correct use and interpretation of eqns. (4.22), (4.23), (4.24). If the terms in eqns. (4.22), (4.23), (4.24) are transposed so that, e.g., V = v — VA , the relative velocity (or acceleration) is given as the sum A

no. 52.

Fie. 53.

of the actual velocity (or acceleration) and one equal and opposite to that of the point of reference. In other words we add to the system a velocity (or acceleration) which brings the point of reference to rest (or reduces its acceleration to zero) to obtain the velocities (or accelerations) relative to the 'point of reference' A . Examples. (i) A smooth wire ABC is bent at right angles at B and is fixed with AC vertical and A uppermost. A bead can slide from A to B and another bead can slide from B to C. If the beads leave A and B simultaneously, starting from rest, prove that their nearest subsequent approach is ac/1/(a2 c 2) , where a= BC and c = AB. Find the hodograph of the motion of the lower bead relative to the upper bead. If we work with accelerations, velocities or displacements relative to P, we can regard P as though it were stationary at A (Fig. 53). The 'actual' acceleration of Q is g sin 0 in the direction of BC, and that of Pis y cos 0 in the direction of A B. From eqn. (4.24) we see that the acceleration F of Q relative to P has components g sin 0 in the direction BC and g cos 0 in the direction BA (F = f fA) . These two components are together equivalent to an acceleration g in a direction BO, where 0 is the mid-point of AC. —

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Hence, 'viewed from 13' the motion of Q is one from rest at B towards 0 with an acceleration g. The nearest subsequent approach of Q to P is then the displacement AN = AB sin 0 = ca/V(a2 c 2) . Since the acceleration of Q relative to P is constant, the relative velocity is in a fixed direction. Therefore the hodograph of the relative motion is a straight line in the direction B 0. (ii) An aircraft of speed v m.p.h. has a range (out and back) of R miles in calm weather. There is a wind w m.p.h. from a° E. of N. Show that in order to fly

FIG. 54. (i) Outward journey.

FIG. 54 (ii) Inward journey.

in a direction whose true bearing is 99° E. of N., the apparent bearing must make an angle /1° with the true one, where v sin/3 = w sin(q9 — a). Prove also that the range in miles for a true bearing 99° is R(v2 — w2) v V[v2 — w2sine (99 — a)] The velocity of the aircraft relative to the air is of magnitude v in an appropriate direction. The actual velocity, V1 or V2 , is the vector sum of v, the velocity of the aircraft relative to the air, and the wind velocity w. The additions of these velocities for the outward and inward journeys are shown in Fig. 54 (i) and (ii) respectively: the directions of V1and V2 are opposite on the outward and return journeys. We apply the sine formula and obtain for the outward journey,

V1 sin(9) —

pi -a)

—

sin (3,

—

sin (99 — a) '

(1)

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and for the return journey,

V2 sin (99 — a + /3„)

sin

(2)

sin(q) — a)

These equations show that /31 = /32 = /3, and eqn. (1) shows that v sin

g = w sin (q) —

a) .

(3)

The time of flight in still air is 2 R/v, and the time of flight for a range X in the direction q9° E. of N. is

X

X +

Now V, + V2 —

2R

172

i.e.,

X—

2R V1V2 v(Vi + V2) •

V

sin(q9 — a)

{sin(q) — s — a) + sin(q) +

- a)}

= 2v cos/3 = 2 1/{v2 — w2 sin2 (9) — a)}, on using eqn. (3). Also

VI V2 =

v2 sin (q) —

p - a) sin (q) + — a) sin2(99 — a)

v2{sin2 (co — a) cost — cos2(9) — a) sin2 /3)} sin2 (99 — a) =v2 {1

v2 {sin2 (c7 — a) — sin2 /3} sin2(q) — a)

sines sin2(99 a) — v2 — w2.

Hence

X=

R (v2— w2) 2; {v2— w2 sin2 (99 — a)} • Exercises 4:5

1. Any number of particles are projected under gravity from the same point at the same instant. Prove that (i) at any subsequent instant the directions of their velocities are concurrent; (ii) the point of concurrence rises with constant acceleration. 2. A ship A is moving due E. with constant speed u, whilst a second ship P is moving due N. with constant speed 2u, and a third ship Q is moving N.-E. with constant speed 2 12u. When A is at a point 0, it is observed that P and Q cross the track of A simultaneously at distances a and 2a respectively ahead of A. Prove that, when OA = x, the line joining A to the mid-point of PQ is rotating in space about the vertical through A with angular velocity 12au/(9a2 + 16x2). 3. An aeroplane, flying, with constant speed u, in a horizontal straight line at a height It above a gun, passes over it, and a shell is fired, with velocity v, directly at the plane when its elevation from the gun is a. Show that the plane will be hit if gh cot2 a = 2u(v cos a — u).

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If v and a are fixed, show that the plane cannot be hit if its speed is less than cot a {v sin a — 1/(v2 sin2 — 2gh)}. 4. A ship X, whose speed is u, sights a ship Y which is at a distance c due E. of X and is steaming due N. at speed v (> u). If X steams in a direction E. of N., prove that the velocity of X relative to Y is in a direction 13 S. of E., where tan/3

V - u COO)

u sin 0

on the assumption that u and v remain fixed. Show that, if u, v and c are fixed, the distance between X and Y can never be less than — V(v2 — u2) whatever the value of 8. v 5. A flying target moves with uniform velocity v along the line y = h in the x, y plane, starting at time t = 0 from the point (0, It). The x-axis is horizontal and the y-axis vertical. A guided missile P starts from the origin when t = 0 and moves with constant speed 2v in such a way that its velocity is always directed towards the target. Show that, if at time t the coordinates of P are (x, y), then, denoting dx/dy by q,

q(h — y)

vt — x

and 1

q2-=- 4 v2 (dt/dy)2.

By differentiating the first of these equations with respect to y and eliminating dt/dy, derive a differential equation relating q and y. Deduce that {q + V(1 + q2))2 =hl(h — y). 6. A river of width 2a flows between parallel banks. Rectangular axes are taken so that the banks lie along the lines x = ± a. The water flows in the positive direction of the y-axis and at the point (x, y) its speed v is vo (a2— x2)/a2, where v, is a constant. A small boat, which travels at a constant speed V (> v0) relative to the water, starts at the point (— a, 0) and is so steered as to cross the river in the shortest possible time. Show that its motion relative to the water must be in the direction perpendicular to the banks and that the least time is 2a/ V. Show also that the gradient of the curve followed by the boat is v/ V and find the equation of the curve. Find how far the boat has drifted down the river when it reaches the opposite bank. If the boat were steered so that it moved directly across the river, along the line y = 0, show that the time taken to cross the river would be

0

f

a cos0 dO

(V2 --vii cos4 8)0 •

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123

4:6 Angular velocity (a) The angular velocity of a point. When a point P is in motion, the line joining P to a fixed origin 0 rotates about 0 (see Fig. 48 p. 113). If we take the direction of the initial line through 0 to be 'fixed in space', then d O/d t is defined to be the angular velocity of P about 0. It is necessary to specify the point 0 for, in general, the angular velocity of P about some other point will be different from that about 0. (b) The angular velocity of a lamina. The concept of angular velocity is of more importance in connection with the motion of a lamina in its own plane. For the purposes of this discussion the moving lamina is assumed to be of infinite extent; the motion we consider resembles the sliding of a sheet of glass on a table-top. If we wish to consider the motion of a special body such as a rod, or a wheel, this body is imagined to be part of the extensive lamina, and the rod, or wheel, is imagined to be drawn upon the sheet of glass. The discussion also applies to the motion of a three dimensional body when every particle of the body moves parallel to one plane ; in this case we may imagine the body as rigidly attached to the sheet of glass. If one point of the lamina is fixed in position, the only motion possible is one of rotation about an axis perpendicular to its plane passing through that point. The most general motion of the lamina can be specified by the motion of an arbitrary point A and the rotation of the lamina about A. The rotation is specified by the angular velocity of the lamina which is, by definition, the rate of increase of the angle between a line fixed in the lamina (i.e., drawn on the glass) and a line fixed in space (i.e., drawn on the table-top). Choosing different lines in the lamina or in space merely alters the angle by a constant and does not affect the rate of increase of that angle and so leads to a unique value for the angular velocity. Hence we refer to the angular velocity of the lamina without specifying an origin. If one point of the lamina is fixed at 0, any other point P (see Fig. 48) can only have a transverse velocity re (OP = r), since the lamina is rotating about 0. This is conveniently expressed by means of vectors if we introduce a unit vector k perpendicular to the plane of the lamina ; then (4.25) v = B (k x r) , where r i s the position vector OP. If we specify the motion of the lamina by giving the velocity VA of a point A, then the velocity of any other

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point P relative to A is given by (4.25) [the motion of P relative to A being in a circle with A as centre] and the actual velocity of P is v = VA + B (k x r),

(4.26)

where r = AP. The cartesian form of this result is obtained as follows. In Fig. 55 the axes A$, An are fixed in the lamina (drawn on the glass) and the axes Ox, Oy are fixed in space (drawn on the table-top). The velocity vA of A y

0 Fm. 55.

has components (vA1,v A2) referred to the axes Ox, Oy. A point P of the lamina, whose coordinates in the frame A are ($, n) has coordinates (x, y) where

x -= x A

e cos° —

sin0,

y=yA +esin°

cost),

(4.27)

in the frame Oxy. Since Pis a point fixed in the lamina, the coordinates $, i do not change during the motion, so that the components of the velocity of P are

v1 =dx/dt=(d x A/dt) -0($sin0+rl cos0)= v A1 -0(y— y A), (4.28) v2 d y/dt = (d yA/d t)

($ cos@ — n sin 0) = v A ± (x — x A) . (4.29)

The eqns. (4.28), (4.29) are the component equations of eqn. (4.26) and provide an independent proof of that result. A lamina which is moving so that 0 = 0 is undergoing a translation; this is a motion in which the velocities of all points of the lamina are parallel to one direction,

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It is sometimes convenient to use the angular velocity of one lamina

S relative to another lamina T. For linear velocities (accelerations) eqns. (4.23), (4.24) give the relative velocity as the sum of the actual velocity and a velocity which brings the point of reference to rest. To find the angular velocity of lamina S relative to lamina T we add to the angular velocity of S an angular velocity equal and opposite to that of lamina T. (This does not necessarily bring T to rest; it may leave it with a translatory motion). When working with angular velocities, the reader must be careful to see that they are obtained as rates of increase of angles measured from lines fixed in space. Failure to ensure this is the source of many errors as is pointed out in some of the examples which follow. Examples. (i) A cylinder of radius b rolls on the inside of a larger cylinder of radius a. If the outer cylinder is rotating about its own axis, which is fixed, with angular velocity S2 and the plane containing the axes of the two cylinders is turning about the fixed axis with angular velocity co, find the angular velocity of the small cylinder. (Assume that w, 12 are both positive.)

RB

FIG. 56. In Fig. 56 the lines AQ and 0R are vertical and the marked point C of the small cylinder was initially in contact with the marked point B of the large cylinder. Because the cylinders are rolling on each other, arc BP = arc CP.

.• b (0 + 99) = a (0 — 1p).

(1)

Since 0 B is fixed in the large cylinder and 0 R is fixed in space, the angular velocity of the large cylinder is = dy/dt = Since AC is fixed in the small cylinder and AQ is vertical the angular velocity of the small cylinder is — dT/dt, measured in the positive sense of rotation. Also the angular velocity of 0 A P is = c10/dt. Hence, on differentiating (1), we find

b(co

12) = a(co — ,Q).

b 99 = (a — b) co — a S2 .

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In particular, when the large cylinder is fixed, = 0 and = (a — b) co lb . (ii) A circle of radius a rotates in its own plane with uniform angular velocity co about a fixed point 0 of the circumference. A point moves round the circumference with uniform speed V relative to the circle and in the same sense as co. Prove that the locus of the point with respect to the plane in which motion takes place may be written

r-=- 2a sin (

V0 V+ 2aco

0 being the pole. Find the accelerations of the point along and perpendicular to the line joining it to 0 when it is at a distance r from 0.

In Fig. 57 the line OA is fixed in space. The angle COA = 0 + z —

co = 0 — 0. The velocity of P relative to the disc isV . Since 0C is fixed in the disc, V = a (2 p). Initially we take P to be at 0 so that 0 = = 0 at that point. By integration we find that

cot = 0

— c9

and

V t = 2a 0 • Elimination of t gives 2a co 0 = V(0 — q)) .

VO V + 2a co But

V0

0 P r = 2a sing) =2a sin (

V

2 a co )

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127

The acceleration of P relative to C is V' la along PC and the acceleration of C is a (02 along CO. We add the components of these parallel to OP to give the radial acceleration V2 + a2co2 V2 — — sin 92 -a co2sin 99 — — a 2a • a The corresponding components in the transverse direction are ( V 21a) cos q), — a w2 cos 99, giving a transverse acceleration ( V2 — a2 w2) V2 — a2 1/(4 — r2). cos g) — a 2 a2 (iii) A circle, of radius b and centre C, rolls with angular velocity w without slipping on a circle of radius a and fixed centre 0, itself rotating with angular velocity S2, measured in the same sense. Find the angular velocity of OC. If .S2 and co are constant and P is a point on the circumference of the circle of centre C, prove that the acceleration of P is w2 PR, where R divides OC in a constant ratio throughout the motion, and the velocity of Pis w PS in a direction perpendicular to PS, where S divides OC so that

CS2= CR • CO.

FIG. 58.

In Fig. 58, the velocity of B is ca2 in a direction perpendicular to OB. The velocity of C relative to B is btu perpendicular to OB. Hence, the velocity of C is a gl b w in this same direction. Since OBC is always a straight line, the angba))/(a b). ular velocity of OC is (aS2

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Because C is moving uniformly in a circle about 0, the acceleration of C is --> ( ap 4- bw 2CO. The acceleration of P relative to C is co2 PC. Hence the accelera-

t

ion, f, of P is f

aS2±bw )2 C 70+ co2 P —C >.

—( a + b

But CO = PO — PC. .•. f =A PO + aPC = (A+ µ) PR, where ( c tS2 +b coy ' a+ b

= co2— A, A OR = ,uCR.

.•. f = (02 PR, where R is a fixed point such that CR OR

CR A lc ' Le" CO

cc,f2 + bco\2 ccc.o+ bc o ) •

A A+µ

We use k as a unit vector perpendicular to the plane of the figure. The ye(a.52 ba)) , —›-->\ locity of Cis x OC). The velocity of P relative to Cisco k x CP. a +b Hence the velocity of P is v k x {( ap a ++ bbw ) 0C+a)CP}=kx(2,0P+iii CP), where aD+ bco a+b'

14'

cp

21.

Hence v = (21+ pe1)

x S = co(k x S ,

where A,0 S = µ,C S. CS • • C0

A, + pi

CR CS2 • • COCO

a 12 + b co a co + b co •

A, cu i.e.,

CS2 =CR • CO.

Exercises 4:6 1. A circular disc of radius r is pivoted about an axle through its centre 0 perpendicular to its plane. To a point A of its rim a rod AB of length 4r is freely jointed; and to B is freely jointed a piston rod BC which is constrained to move

§4:7

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129

in a straight line through 0. The disc rotates with constant angular velocity w. Initially OA BC (in order) is a straight line. Prove that after time t the angular velocity of AB is w cos cot 1(16 — sin2 co t) and that the velocity of B is then cos cot rco sincot {1 + y(16 — sin2 cot) Prove also that the accelerations of BC at the ends of its path are in the ratio 5:3. 2. A straight rod AB of length c has angular velocity co radians per second clockwise about the end B which is pinned to a horizontal table. At a certain instant the direction from B to A is northward. One second later, the table is continuously translated eastward with constant acceleration f . Given that the eastward component of A's velocity is never zero, show that w cannot lie between n/2 and 3x/2. Assuming that co is less than n/2 and that the eastward component of A's velocity is never zero, show that this component has alternate maxima and minima, provided f < c O. Show also that the time interval between any maximum and the succeeding minimum is 2 cos-'

f

and is less than n/co. 3. A point P is on the circumference of a circle of centre C and radius b which rolls without slipping upon a fixed circle of centre 0 and radius a. If the angular velocity w of 0 C is constant and 95, is the angle P C 0, show that the speed of P is 2 (a + b) w sin-192 and that the radius of curvature of the path of 0 is 4b

(a + b) sin (a + 2b)

s '

4. A point A moves in a counter clockwise direction round a circle of centre 0 and radius a with angular velocity cot , and a point B moves counter clockwise round a concentric circle of radius 2a with angular velocity co,. If 0 is the angle 0 AB when the relative velocity of A and B is parallel to the line AB, prove that tang 0 —

co — 4co: 3 co:

provided that co, > 2 co,. Find also the speed of B relative to A at that instant.

4:7 Instantaneous Centre: Centrodes There is one important feature of the motion of a lamina. With the ex ception of the case for which 0 = 0 eqns. (4.28), (4.29) show that the point with coordinates (4.30) { X A - (V A 20) Y A + (VA /0)}

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is instantaneously stationary. This means that, at any instant, the lamina is rotating about a point given by (4.30) which is called the instantaneous centre of rotation. The important feature is that, with the exception of translatory motion, any motion of a lamina in its own plane is a rotation at that instant about some point I which is stationary. At that instant the velocity of any other point P is given by v = w(k x

(4.31)

where w is the angular velocity of the lamina at the same instant, i.e., the velocity is perpendicular to IP and of magnitude wIP. Apart from using the coordinates (4.30) there are two chief ways of finding the position of I. When the directions of the velocities of two points P1and P2 are known, I lies at the intersection of lines through 131, P2perpendicular to their respective velocities. When the magnitude and direction of v are known for a point P and the angular velocity of the lamina is known also, the distance I P is vIco and the direction of I P is perpendicular to v. Both these methods are of use in applications. (Translation can be regarded as a motion in which I is at an infinite distance.) The above considerations apply to the motion of a lamina at one instant. As the lamina moves the position of I will alter from instant to instant both in the body and in space. The path followed by I in the body is called the body-centrode and the path followed by I in space is the space-centrode. A simple example is given by the motion of a wheel rolling along the ground. The point of the wheel in contact with the ground is stationary at that instant and is therefore the instantaneous centre. As the wheel rolls the point of contact traces out a line on the ground and traces out the circumference in the wheel; these are the two centrode curves for this motion. We now prove the fundamental theorem that the motion of a lamina in its own plane is given by the rolling of the body-centrode (taking the body with it) on the fixed space-centrode curve. (This is obvious for the motion of the rolling wheel.) In Fig. 55 let P represent the instantaneous centre I with coordinates (x, y) referred to Oxy and n ) referred to A$71. Hence, at this instant v1 =v2 = 0. •• • v AI = ($sine +

cose), vA2 =e(— cos°

n sin 0) . (4.32)

Now we consider the change of position of I. The space centrode will be represented by an equation connecting x, y which each vary with the

§4:7

131

KINEMATICS

time ; the body centrode will be represented by an equation between $,97which also vary with time. Now eqns. (4.27) relate x, y, e, n at any instant of the motion and (4.32) gives the velocity of A at any instant of the motion. Differentiating eqns. (4.27) we find =

e cos — 9) sin 0 — 0 (e sin 0 + 92 cost)) = cos° — 2.7 sin 0 , (4.33)

-= vA 2

e sin 0 ± n cos 0 + 0 ($ cos 0 —

7)

sin 0) = e sin 0 + n cos 0,

(4.34) where we have used eqn. (4.32) to reduce the expressions. (The reader should clearly understand why, in eqns. (4.28) and (4.29) we regard e,

Body centrode

7

Space centrode

as constants, but in eqns. (4.33) and (4.34) we have treated them as with which variables.) These eqns. (4.33), (4.34) relate the velocity I traces out the space-centrode to the velocity (e, 77') with which I traces out the body-centrode, (Fig. 59). We note that +

(4.35)

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A COURSE OF MATHEMATICS

showing that I moves along the two curves with the same speed. Also sin 0 —•

+ ii cos 0 = tan° + (ii/e)

cos 0 — 9) sing

.•. tan a =

1 — (-;//) tan 0

tan g + tang 1 — tan 0 tang = tan(0 + g),

(4.36)

where tang = , and tang = 77/. These identifications mean that a is the angle between the tangent at I to the space-centrode and Ox, and that p is the angle between the tangent at I to the body-centrode and Hence the curves touch at I having the same tangent. The results (4.35), (4.36) imply that in the motion of the lamina I is always at the point of contact of the two curves and moves along each with the same speed. Hence one curve rolls on the other. Because the particle of the body which is at I at any instant is stationary at that instant it does not follow that that particle has no acceleration. A particle which is instantaneously stationary may have a finite acceleration. In general there is one point of a moving lamina which has no acceleration; however, the 'centre of no acceleration' is seldom used whereas the instantaneous centre is of practical importance in the theory of machines. The examples which follow illustrate the use of the instantaneous centre of rotation in a variety of problems. Examples. (i) A rod AB of length 4a and mid-point C passes through a ring P and moves so that C always lies on the circumference of a circle, of radius a and centre Q, which passes through P. Prove that the instantaneous centre I lies on the circle and find the locus of I relative to the rod. When the rod makes an angle 0 with PQ, show that the velocities V A , vn of A, B respectively, satisfy the relation /J A M B =tan(0/2). Since QC is perpendicular to the velocity of C and the rod at P must move parallel to itself (Fig. 60), I is at the intersection of CQ and PI, where CPI is a right angle. Hence I is at the other end of the diameter through C. Since CI = 2a, being a diameter of the circle, for all positions of the rod the locus of I relative to the rod is a circle centre C and radius 2a. This circle is the body centrode, and the given circle is the space centrode. Now

VA vB

IA IB •

§4:7

133

KINEMATICS

Also CP = 2a cos0, AP = 2a(1 — cos0), BP = 2a(1 cos0), PI = 2a sin0. 4 a2(1 — cos 0)2 = 8a2 (1 — cos 0).

.•. A I2 =4a2 sin2 0

BI 2 =8a2 (1

Similarly

AI • • BI

ii tl— cosO 1H- cos° )

cos0). 2 sinzl 0 2 cost 1-0

tan40 = :A B •

(ii) A thin uniform rod AB moves in the plane x0y so that A slides on Ox and AB touches the parabola Icy = x2. Prove that the space-centrode of the motion of AB is an arc of the parabola

j-k2.

ky = 4x2

Find also the rate at which the instantaneous centre is describing this curve when A is at a distance a from 0 and is moving with speed u.

Flo. 60. Suppose the coordinates of B, Fig. 61, are given parametrically by

x = P, Y = P2 /k. The equation of the normal at B is

k2 x + 2 kpy —

— 2 IP = 0.

The equation of the tangent at B is 2px — ky — p2 =0, so that A is the point (Z p, 0). Since the direction of motion of B is along the tangent and the direction of motion of A is along the x-axis, I is the intersection of the normal and the line x = zp. Hence the coordinates of I are p2

x = 11), y= --k— Therefore the space centrode has the equation

Icy = 4x2

k

•

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A COITBSE OF MATHEMATICS

When x = a, p = 2a and u =4-j). Therefore

Y

2p:p

2 • 2a

k

k

2u— u2

8au

, and x= u.

u2 (1 +

64 a2 k2 /I

where v is the speed of I along the space centrode. (iii) Two rods AB, BC, each of length a, are freely jointed at B and placed on a horizontal table. The end A is freely pivoted to a fixed point on the table and the angle ABC is x/3. The system is then set in motion in such a way that the end B

FIG. 62.

C begins to move with velocity V and acceleration V 2/a, both in the direction of CB. Find the initial angular velocities of the rods and show that their initial angular accelerations are 8 V2 /(3 1/3a2) and 10 V2 /(3 V3 a2) In Fig. 62, AX is a fixed direction and the angles have deliberately been drawn so as to be different from those given in the enunciation of the question. We suppose that col and w2 are the angular velocities of the rods. The angular velocity of AB is col = 0, since 0 is the angle between AB and the fixed direction AX. However, the angular velocity of BC is not 4): the angle between BC and AX is 0 + w, hence w2 = 0+ (j). The velocity of B is awl perpendicular to AB; the velocity of C relative to B is awe perpendicular to BC. The actual velocity of C is the sum of these two velocities and is specified as of magnitude V and directed along CB. We take components perpendicular and parallel to BC, and find

awlcos 97 — awe =0, a co, sin q9 = V.

§4:7

135

KINEMATICS

In the given position 0 = 3 x, so that

co, —

2V a i/3 '

V a1/3

The acceleration of B has components a wl along BA and awl perpendicular to AB. The acceleration of C relative to B has components a co; along CB and ad), perpendicular to BC; the actual acceleration of C is the sum of these. We find the components along and perpendicular to the initial (instantaneous) position of BC. Component in the direction CB, acb, singe — act); cosq)

aco: = V' la,

and in the direction perpendicular to BC,

acb, cos q9 -I- act)! sin 95, — acb2 =0. In these equations we put 99 = 91 and the values of co, and co, obtained above and solve for thiand cb2 thereby obtaining the stated results. Exercises 4:7 1. A circular disc, of radius a and centre 0, turns in its own plane about its centre with angular speed to. A second circular disc, of radius b and centre C, coplanar with the first, rolls in contact with it and has an angular speed co,. Find the space and body centrodes for the motion of the second disc. 2. A lamina is moving in its own plane. At a certain instant the directions of motion of two points A and B meet at C, and the motion of C is parallel to AB. Prove that A and B have equal speeds, which are less than the speed of C. 3. Two rods AB, BC are rigidly joined together at B, so that angle ABC = a, and move in a plane so that AB passes through a fixed point D and BC passes through a fixed point E, where DE = 1. Show that the space centrode is a circle, passing through D and E, with diameter 1 cosec a, and that the body centrode is a circle with this distance as radius. 4. A rod AB moves so that AB is always a tangent to a fixed circle, and the end A lies on a fixed tangent to the same circle. If the motion takes place in the plane of the circle, show that the locus of the instantaneous centre in space is a parabola, having the centre of the circle as focus and the fixed tangent as directrix. Show, further, that the locus of the instantaneous centre relative to the rod is a parabola having A as focus, and find its directrix. 5. A body consisting of two thin straight rods 0 X, 0 Y which are joined at right angles moves in a plane so that OX touches a fixed circle of radius a and 0 Y passes through a fixed point A on the circle. Determine the position of the instantaneous centre of rotation of the body and prove that it lies (i) on a fixed circle of radius +a, (ii) on a circle fixed with respect to the body and of radius a. Find the velocity of the instantaneous centre in terms of the angular velocity of the body.

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A COURSE OF MATHEMATICS

Miscellaneous Exercises IV 1. A particle P describes the curve whose polar equation is r = a e0coto,. Find its radial and transverse accelerations in terms of r, 0 and 0. Show that its tangential and normal accelerations are r sec a and r 02cosec a respectively. The particle is made to describe this curve in such a manner that r 02is constant. Show that the acceleration makes an angle tan-1(2 tan a) with the tangent at P. 2. A straight line AB of length b turns in one plane with uniform angular velocity co about its end A, which is fixed. A circle of radius a, which is turning in the same plane with angular velocity 2 w in the same sense, rolls without sliding along the line towards A from initial contact at B. Show that the instantaneous centre of rotation of the circle lies in the line AB and find the time that elapses before the circle is in contact at A. If a lamina turns in the same plane of motion with angular velocity co about a given point P on the circle in the same sense as the angular velocities of the circle and the line A B, show that its instantaneous centre of rotation is the vertex R of the parallelogram APCR, where C is the point of contact of the circle at the instant considered. 3. A small ring is pivoted to a plane at a distance c from a fixed straight line. A rod AB passes through the ring and the end A is constrained to move along the fixed straight line. Show that the locus of the instantaneous centre in space is a parabola with latus rectum c and vertex at the ring. Show further that, if A moves with uniform velocity along the fixed straight line, the acceleration at every point of the rod is inclined to the rod at an angle tan-1(2 cot 0) , where 0 is the angle which the rod makes with the fixed straight line. 4. A circle of centre A and radius 2a turns in its own plane about a point 0 of its circumference with angular velocity w. A second circle of centre B and radius a rolls without slip on the outside of the first circle with angular velocity ,52 in the same sense. Find the angular velocity of AB. If co and S2 are constant prove that the velocity of B is proportional to and perpendicular to BX, where X divides OA in a constant ratio, and prove that the acceleration of B is directed to a point Yin OA such that

AY=

AX AX. AO

5. A circular cylinder of radius a and axis OQ touches two planes hinged together along the line AB, OQBA being a rectangle. One plane is fixed horizontally and the other inclined at an obtuse angle r — 0 to it. Contact between the cylinder and the inclined plane is perfectly rough, but the horizontal plane is smooth. If contact is maintained between the cylinder and the planes as 0 varies, show that the spin of the cylinder is 0(1 - i sec210). Taking A as origin, Ax in the horizontal plane perpendicular to AB and Ay vertically downwards, show that the space centrode is given by y (a' — x2) = 2ax2.

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KINEMATICS

6. A bar AB moves in the plane of rectangular axes Ox, Oy so that it always touches the parabola y2 =4ax and the end A of the bar moves along the x-axis. Find the coordinates of the instantaneous centre in terms of the angle 0 between AB and Oy and prove that the equation of the space centrode is ay2

4x(a — x)2

0.

Prove also that, referred to A as pole and AB as initial line, the equation of the body centrode is r = 2a sin 0 sec3 0 7. The boundary of a circular lamina of radius a touches a fixed line OA, and a fixed point P of the lamina at distance b (< a) from its centre C moves on the fixed line OB, which makes an acute angle a with OA. Show that the space-centrode of the motion is a circle of radius b cosec a whose centre is the intersection of OB with a line through C parallel to OA. Show also that the body-centrode is a circle of one-half the radius of the space-centrode. 8. Three unequal coplanar rods AB, BC, CD are hinged at B and C and move in their own plane about the fixed points A and D. Show that the angular velocities of AB andDC are in the ratio EB:AB, where E is the point on AB (produced if necessary) such that ED is parallel to BC. 9. Two circular discs, each of radius a, are in the same vertical plane and roll with angular velocity a), without slipping, on a horizontal floor so that the distance between their centres A and B is constant and equal to 2c (> 2a). A third circular disc, of radius b (> c — a), moves in the same plane so as to be in contact with each of the first two and so that its centre is higher than AB. Assuming that there is no slipping at any point of contact, find the angular velocity of the third disc and also its instantaneous centre. 10. 5, H are two points of a fixed plane at a distance c apart. 5', H' are two points of a moving lamina also at a distance c apart. The lamina moves in the plane so that SS' = HH' = k(> c) , while the joins SS' and HH' meet in a point P which lies between S and S' and between H and H'. Show that Pis the instantaneous centre of rotation and that the body centrode and space centrode are each an ellipse of major axis k and eccentricity c/k. 11. A cyclist rides along a straight road with uniform speed V. At a given instant, the cyclist is at a point 0 on the road and a dog is at A, a point distant a from the road. The dog runs with speed V towards the cyclist. Taking OA and the road as axes of x and y respectively, and the coordinates of the dog as x, y, show that d2 y

dy)2 0. dx f '

x dx2

and hence that the equation of the dog's path is

x

1 { x2 y — 2 2a

—

a log 7t—

a1

Show further that the distance between the cyclist and the dog tends to la. 12. An aeroplane P receives wireless signals sent out from a point 0 enabling it to steer on a bearing always at right angles to OP. If its air-speed is constant and

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A COURSE OF MATHEMATICS

equal to v, and it encounters a delaying wind of constant velocity u (u < v) which is a direct head wind initially when OP = r0, show that the path of the aeroplane is an ellipse of eccentricity u/v, and that the nearest approach to 0 occurs when —v u , when the wind is a directly following one. OP = ro u+ 13. If the radial and cross-radial resolutes of acceleration are constant multiples (A) of the cross-radial and radial resolutes of velocity respectively, and p is the perpendicular from 0 upon the tangent to the path of P, show that the p–r equation of the path is given by 4p2 (7L2 r2 + 22p, logr + v) = (A r2 + ,a)2 , where and and v are constants. 14. A point describes the curve whose polar equation is r2 = a + b cos20, a > b, so that the time t is given by 2ht = 2a0 b sin20, where a, b, and h are constants. Prove that the acceleration is radial and express it as a function of r only. 15. Show that in the two-dimensional motion of a lamina with an angular velocity co, there are, at any instant, points I and J such that the velocity v and acceleration f of any point P of the lamina are given by v = to X IP,

f=cux JP — co2 JP,

provided to 0 in the case of the point I, and provided to and io are not both zero in the case of the point J. 16. The motion of a particle in the (x, y)-plane is governed by the equations d2 x dy + 2a +bx = 0, dt2 dt d2 y dt2

2a

dx + by = 0, dt

where a and b are constants. The particle leaves the origin at t= 0 and returns at t=1. Show that, for given a, there is a least value of b for which this is possible and that in this case the total distance travelled by the particle is

(1—

21 J

3. b 2 sin2 0) dO,

yt

o

where 1 is the maximum distance of the particle from the origin.

CHAPTER V

PARTICLE DYNAMICS MOTION WITH ONE DEGREE OF FREEDOM 5:1 Introduction: Conservative forces This chapter is concerned with the motion of a particle which has one degree of freedom and includes motion along a straight line and motion in which the particle is constrained to move along some prescribed curve. Such a constraint is achieved by restricting a bead to slide on a fixed wire, by attaching a particle to an inextensible string, or by similar devices. Consideration of the motion of a particle with two degrees of freedom is deferred until Chap. VI. The problems discussed in this chapter usually involve the solution of a (second order) ordinary differential equation but, in some cases, it may be convenient to start the solution by introducing more than one variable and then reducing to one variable by elimination. When the particle is of mass m and the resultant force on the particle is F, the equation of motion is F = mf,

(5.1)

where f is the acceleration. The resolutes of eqn. (5.1) give rise to the differential equations associated with the motion under investigation. As in other branches of physics, the most convenient coordinate system for use in any one problem is essentially determined by the nature of that problem. In fact, correct choice of the coordinate system in which to write down the resolutes of eqn. (5.1) is the first step in most problems of dynamics.

Conservative forces The force F in eqn. (6.1) is the resultant of all the forces acting on the particle ; these forces arise from constraints, surfaces in contact with

140

A COURSE OF MATHEMATICS

the particle, elastic strings, friction, air-resistance, gravitation and other causes. In certain circumstances we can write down an integral of eqn. (5.1) called the energy equation. When the particle undergoes a displacement or, the force F does an amount work 6W = F • Or (cf. § 2 :3). If the particle undergoes a sequence of infinitesimal displacements from position A to position B, the work WAB = fAB dW, done by the force F, will, in general, depend upon the path followed by the particle. If the force F is such that WAB is independent of the path followed from A to B, the forces acting on the particle are said to be conservative. An equivalent statement of this condition is that F does no net work when the particle moves round a closed circuit returning to its starting point and, in the terminology of § 5:4 of Vol. II, dW is a perfect differential. In this case f AB d W = VA — VB, where V p is the potential energy of the particle at the point P. We take V I to be the work done against the force F when the particle undergoes a displacement from some standard position to the point P (by any arbitrary path). (See also § 2:5.) When the particle moves, the force F is doing work on the particle at a rate d W/dt = F • i; when F is conservative, F • r = —d V/dt. If we multiply eqn. (5.1) scalarly by r we get mi. • -= F d

dt

mi2) =

dV dt

since m is constant. But z mi2 =T, the kinetic energy of the particle. d g' dt

dV dt

.*. T + V = constant.

(5.2)

This is the energy equation which can always be written down when the forces are conservative. In general frictional forces, or forces which depend upon the magnitude or direction of the velocity of the particle, cannot be conservative. In these circumstances we cannot integrate eqn. (5.1) to give an energy equation.

§ 5 :2

PARTICLE DYNAMICS

I

141

5:2 Motion on a straight line If the particle P, of mass m, Fig. 63 moves along a straight line so that, at time t, OP = x, where 0 is a fixed point on the line, and is acted upon by a force Fin the direction x increasing, then the equation of motion (5.1) has only one component equation, which is

mf = F,

(5.3)

where the acceleration f and velocity v (both measured in the direction x increasing) are related by the equations

f=

d2 x dv dv =v -= dt2 dx dt

The various classes of problems which arise can be summarised as follows :

(1) F is a function of t only (including the case F = constant); (2) F is a function of v only; (3) F is a function of x only; (4) F is a function of any two or all three of the variables t, x and v. „P 0

FIG. 63.

Problems of class (4) are usually extremely difficult ; some of the solvable types are discussed in the remaining sections of this chapter. Problems of class (1) can be solved at once by direct integration(s) of the equations d2 x dv dt2 dt and have been discussed in Vol. I. Problems of class (2) were discussed in Vol. II § 1:7. Here for revision purposes we give an illustrative example. Example. A particle, moving in a straight line, is subject to a retardation of amount kvn per unit mass, where v is the speed at time t and k> O. Show that, if n< 1, the particle will come to rest at a distance u2-n/{k (2 — n)) from the point of projection at time t = u'-n/{k (1 — n)), where u is the initial speed.

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What happens when (a) n =1, (b) l< n < 2, (c) n =2, (d) n> 2? The equation of motion can be expressed in the forms dv dv v — = - ley" — = -kvn. , dt dx

(1)

If n< 1, separation and integration of each of eqns. (1), the constants of integration being determined by the initial conditions v = u, x = 0 at t = 0, gives kx -

u2-n (2 - n)

v2-1,4 kt(2 - n) ' 1 - n

(2)

1 - n

The particle comes to rest when v =0 giving the required results. (a) If n 1, the second of eqns. (1) becomes dv

at

kv

which has solution v = ue-kt. Another integration gives kx = u(1

(3)

e-kt).

(4)

Hence, as co, v-- 0 and x-->ulk; i.e., the particle never comes to rest in a finite time but the displacement tends asymptotically to the finite value u/k. (b) If 1 < n < 2, eqns. (2) hold but the second of these equations, written in the form

1

1

kt = (n -

(n - 1)0-1

shows that v-+ 0 as t -- oo. Hence the particle never comes to rest but the displacement tends asymptotically to u2-n/{k (2 - n)} . (c) If n =2, eqns. (1) become dv dx

dv

=

kv, dt =

k v2

with solutions ekx

kut, v=

u

1+ kut •

Hence x = k-1log(1 kut) and so x-> DO and v -- 0 as t-÷ co. [Here x as log t.] (d) If n > 2, eqns. (2) hold and can be expressed in the forms (n - 2)kx = Hence as t -> oo, v

1 1 1 1 , (n -1)kt = vn-2 un-2 vn-1

0 and x--> oo. [Here x-> oo as en-2)/(n-1).]

.

co

§ 5:2

PARTICLE DYNAMICS I

143

Problems of class (3) lead to differential equations of a type which also arises in problems concerning central orbits, motion on surfaces of revolution (e.g., the spherical pendulum) and the oscillations of tops and gyroscopes. For this class, writing f as v d v/dx and integrating, eqn. (5.3) gives (5.4) ±2 =i(x), where f (x) is a known function of x. This equation is equivalent to the energy equation (5.2), the acceleration being given by x = i f' (x). The character of the motion is determined entirely by the properties of the function f (x) and in particular by its zeros x = al , a2 , ..., an (in ascending order). We suppose first that f (x) and f' (x) are continuous, that ap , a„/ are apair of consecutive simple zeros, and that f (x) > 0 when ap <x < a„ 1; then f (x) < 0 for ap_1 < x < % and a„/ < x < ap+2 so that f' (%) > 0 and f' (a„/) < 0. Since in any real motion the velocity X must be given by a real number, x must have a value such that f (x) > 0, i.e., x may lie between ap and a„1. Interpreted in terms of the motion these facts are : (i) when x = ap , x = 0, x > 0 and so the particle is instantaneously at rest but not in equilibrium and moves away from this point in the direction of x increasing : (ii) when x = a„/ , x = 0, x < 0 and so the particle is instantaneously at rest but not in equilibrium and moves away from this point in the direction of x decreasing. In short, the motion is a 'to and fro' motion between the positions x = a and x = ap+1) called a libration motion. The time required for the motion from ap to ap+1(or the reverse) is a p+ i

T=

dx {(x — ap) (ap+, — x) g(x)}

(5.5)

ap

where the function g (x) is given by

1(x) = (x — ap) (ap +1— x) g(x) and is such that g (x) > 0 when a, < x < (Iv+ , . The integral in eqn. (5.5) is convergent at both limits (see Vol. II § 6:6), and hence the time between x = ap and x = ap+1is finite. The motion is periodic with period 2T .

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A COURSE OF MATHEMATICS

If a, is a repeated (double) zero of f(x) , then x = aq is a possible position of equilibrium. In this case, if x approaches aq from below, from x = b say, the time T1to reach aq is given by aq

dx

T, = f

(aq — x) f {k (x)} '

where f(x) = (a, — x)2 k (x). This integral diverges like lim log (a, — x) and hence the particle approaches asymptotically (but never reaches) x = a 5 . Such a motion is called a limitation motion. A similar result applies whenever x approaches a multiple zero of 1(x) from above or below. If f (x) > 0 for x > an , where an is the greatest zero of 1(x), then > 0 for x > an . If ± > 0 also, the particle recedes to infinity. Similarly, if f(x) > 0 and x < 0 for x < al , where al is the least zero of f (x), the particle recedes to infinity in the negative direction. Example. A particle of unit mass moves on a straight line Ox under the action of a force in the direction 0 x whose value at x is a single-valued function f (x). Derive an energy equation of the form 4.42 =9)(x) • Show that the motion is not periodic if q' (x) = Ile (a + x) (a — x)2, where k> 0 and a> 0. Prove also that in this case the particle moves from x = — a to x = 0 2 in time (i/ ) log (1 + i'2). ak The equation of motion of the particle is dv v — = f (x) dx so that integration gives

f

+, v2 = f (x) dx + constant. Since v = x this relation can be written ,

= w(x) and is the required energy equation. If 99 (x) =

k (a + x) (a — x)2, where k> 0, then V.C 2 = k (a

x) (a — x)2,

so that x — a. The value x = a is a repeated zero of q9 (x) and is a position of equilibrium.

§5:2

145

PARTICLE DYNAMICS I

Consider first motion in the range — a < x < a. The velocity is zero at x = — a and x = a, and at no other points. Since X = —14(a — x) (a + 3x), the acceleration is positive at x = — a and is zero at x = a, this latter point being a position of equilibrium. If the particle moves towards x = — a , the direction of the velocity is reversed when it gets there and the particle subsequently moves towards x = a. If x > 0, the time to reach x = a from x -= c(< a) is a T — 1

dx

I (a — x)y {k (a + x)}

This integral diverges at the limit x = a as does lim log (a — x) and so the particle x—>anever reaches x = a from below. Consider now motion in the range x a. In this range X > 0, since x = 0 only where x = a. Therefore, if the particle starts to move away from x = a (x increasing), it will continue to do so with ever increasing velocity. If the particle moves towards x= a (x decreasing), it does so with diminishing velocity. If X < 0, the time required to move from x = c1(c1 >a) to x = a given by a

c,

dx

dx

TZ

f ilk (a + x) (a — x)2)

f (x _ a) V {k (a ± x))

c,

This integral diverges at the limit x = a as doeslimlog x,a± (x — a) and so the particle never reaches x = a from above. Hence no possible motion is periodic. The time from x = — a to x = 0 is 0

1

T 3

dx

j/k f j

(a — x) V(a

x)

—a

The substitution a + x = z2 gives a T3

—

2 r dz Vk j 2a — z2

1 1' (2a k)

V(2a)

z yla

V(2a) — z jo

1

/(2 ak))

log(1 + V2).

0

Exercises 5: 2 1. A circular hole of radius a is cut in the top of a smooth horizontal table and is 1 covered by a uniform circular disc of radius a (1 + — and mass m1, whose centre n C is vertically above the centre of the hole. A small rough lump of lead of mass m2 is placed on the disc at C and can slide, but not roll, on the disc. The disc is acted on by a constant horizontal force P whose line of action passes through C. Prove that the lump of lead will not slide on the disc if

P < (m1 + m2)14, where it is the coefficient of friction between the lead and the disc.

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A COURSE OF MATHEMATICS

Prove also that, when P exceeds this value, the lump of lead will fall through the hole provided that

1

P>u j g*

+ m2}

2. A particle of mass m moves in a straight line between two centres of force A, B, distant 2a apart, whose attractions on the particle when it is at a point P are n ,uP A-2, m ktP B-2respectively. Find the velocity with which it must be projected from the point C, where AC = 9CB, in order that it may just come to rest at the middle point 0 of AB. Prove that the time taken by the particle to travel from C to D, where 50D = 30B, is I /_ 3 log

—

I \ a2 \ )•

3. A particle P, of unit mass, moves under the action of a repulsive force OOP. The particle is initially projected from a point A, with speed a co I/6, in the direction AOB, where OA = a and OB = 2a. Show that the time from A to B is 1

— to log (I + I/6). If there is a fixed elastic obstacle at B, the coefficient of restitution between the particle and the obstacle being e, show that the particle will not return to A if e < 2/3. 4. A particle of mass m moves on the straight line Ox, the displacement of the particle from the origin 0 being x at time t. The particle is acted on by a force mX(x) in the direction Ox, where X (x) is a given function of x, and x = 0, = u, at t = 0. Prove that x satisfies the differential equation

142 = 99 (X), where

99 (x) =

11'62 f X () d$.

Consider the particular case in which x2

X = —2n2 x (1 — —), u =- na, where n and a are positive constants. Determine the function q9(x), and give a sketch of the graph y = 99(x). Find the value of x for t> 0, and prove that a as t--->- Do. 5. A particle of unit mass moves in a straight line, and at time tits displacement from the origin is x. It is acted on by a time-dependent force F(1), a restoring force k2 x, and a frictional resistance 2cd.(c > 0). Write down the differential equation satisfied by x. Discuss the forms the complementary function takes for

§ 5 :2

147

PARTICLE DYNAMICS I

different values of c2/10. If k2 — c2 = n2 (n real and non-zero), derive the general solution of the equation in the form

x = (A cosnt

B sinnt) e-et

1

F (u) e-e (6- u) sinn (t

— n

u) du.

0

Show also that if F (t) = a e-et, with c constant, the position at time t of a particle initially at rest at the origin is a n

x = 2 (1 — cosnt) e-ct. 6. A particle moves with velocity v in a straight line against a resistance a v b v3 per unit mass. If initially the velocity is u, show that it will be u/2 after the part1 u (a b) icle has described a distance tan-1 , and find the time taken. 2a ± bu2 f(ab) 7. A particle slides on a plane inclined at an angle a to the horizontal. The coefficient of friction ,u is proportional to the distance r from a point 0 of the plane, i.e., it = k r . The particle is projected from 0 with velocity u up the line of greatest slope. Prove that the particle will remain at rest when it first comes to rest if u2

3g sin' al(k cosa).

8. The resistance to the motion of a vehicle of mass M gm, when moving in a straight line with velocity v cm per sec, is M k 0, and the engine propelling the vehicle works at constant power M k 03 ergs per sec. Show that C has the dimensions of a velocity; and, if the vehicle starts from rest, that v is always less than C and that the time taken to acquire a velocity 40 is 1

kC

f

1 63 f

2 1 6 log 7 tan-1 — — — r3

secs.

Find the distance traversed in this time. 9. A car of mass m starts with its engine exerting a constant pull mP . At any subsequent instant it can be switched over to work at a constant rate mH. Show that if a possible speed V (> HIP) is to be attained from rest against a resistance mk v2, where v is the speed, then the distance covered will be a maximum if the engine is switched over when the speed is HIP. Show also that this maximum distance is P3 H2 1 log

6k

(P3— k H2) (H — k V 3)2

10. A particle of mass m, moving in a straight line, is acted on by an attractive force towards a fixed origin in the line, given by m j a2 /x2 when x i a, and by m ,uxla when x < a. If the particle starts from rest at a distance 4a from the origin, prove that it will reach the origin with velocity 41/10 it a.

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A COURSE OF MATHEMATICS

Show also that the time taken to reach the origin is V( [V:

(4n + 3 V3) --V sin-1

V( 120)

11. A particle of mass m is projected vertically upwards, the resistance of the air being assumed of the form mc v2, where c is a constant and v is the velocity. During its motion, the particle has equal velocities at two points whose heights above the point of projection are x while moving upwards and y while moving downwards. Show that e 2c(a—x) e-2c(a—V) = 2 , where a is the greatest height.

5 : 3 Simple harmonic motion and damping In Vol. II § 2 : 7, we considered problems of simple harmonic motion and damping as illustrations of differential equations. The derivation of these equations is illustrated in examples (i)–(iii) below. (For a discussion of simple harmonic motion considered as the projection onto a diameter of motion with uniform speed around a circle see Theoretical Mechanics for Sixth Forms). Here we consider simple harmonic motion (S.H.M.) as the solution of the equation = — n2 x .

(5.6)

The solution can be written

x = A cosnt

B sinnt

x= a cos (nt

p),

or

(5.7) (5.7 a)

where A, B, a, f3 are arbitrary constants, showing that the motion is periodic with period 2n/n. In fact the (two or three dimensional) equation of motion ___ n2, (5.8) has solution r = A cosnt B sinnt, (5.9) where A and B are arbitrary constant vectors, since the resolutes of eqn. (5.8) are of the form (5.6). In eqn. (5.7a) the constant a, the amount by which x swings on either side of its mean position (zero), is called the amplitude of the motion.

§5:3

PARTICLE DYNAMICS I

149

Small oscillations Suppose the equation of motion of a particle moving on a straight line is = (x), (5.10) where 99 (a) = 0 , and the particle is slightly disturbed from the equilibrium position x = a. By the term 'slightly disturbed' we mean that initially the displacement x — a, from the equilibrium position, and the velocity, t are both small. We write x = a + e and use Taylor's Theorem in eqn. (5.10) and obtain ,

= 92(a + a) = 92(a) + 899'(a) + -.1-1 E2T" (a) + • • • e.w' (a) + 0 (e2).

(5.11)

When (a) < 0, a first approximation to the solution of (5.11) is a simple harmonic variation of 8 with period 2z/j/{ — (p' (a)} . In this case, the displacement, x — a = E , and the velocity, t = E remain small and 'small oscillations' occur near the equilibrium position. If (a) 0, the expression for a obtained from (5.11) by ignoring the initial value of O(E2 ) does not, in general, remain small. Thus, ,

if (p' (a) = 0, then

a = A + Bt,

if (p' (a) > 0, then e = A coshmt

B sinhmt;

neither of these expressions remains small for large values of t and arbitrary values of A, B. Hence, so' (a) < 0 is a sufficient condition that small oscillations occur. In this case the equilibrium at x = a is stable (see Chap. IX). Examples. (i) An endless light elastic string of unstretched length 2a passes round two small smooth pegs A, B, at a distance a apart in the same horizontal line. A particle P of mass m is attached to the string which, in equilibrium, forms an equilateral triangle. Show that, if P is displaced vertically through a (2 Y3 a )1/2 small distance and then released, it will oscillate with a period 27c

7g

In equilibrium the particle rests at C, Fig. 64, and vertical resolution gives 2T cos (x/6) = mg, whence the tension of the string T = mg/Y3. But, if 2 is the modulus of the string, T = 2(3a — 2a)/2a = +2. Hence

A = 2mg/J/3.

(1)

If P is displaced vertically downwards through a small distance x, so that CP = x and CPB = 0 , the tension of the string becomes 1-2(cosec 0 — 1) and the equation of motion of P resolved vertically downwards is mx = mg —

(cosec 0 — 1) cos 0 .

(2)

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A COURSE OF MATHEMATICS

After use of (1) this equation can be written g f (x), where f(x) — 1

4x a V3

(3)

2a 4x/V3 2x)2 a2} VRaV3

Expanding f(x) in ascending powers of x/a by the binomial theorem gives 7 gx x2 A f(x) — 2aV3 + Hence (3) can be written 7g x x2 --(7 2 2aV3 + 0 ( A •

•

P

B FIG. 65.

FIG. 64.

If x is small, this represents S.H.M. of the stated period. (ii) Two particles A, B, of masses km, m respectively, are connected by a light elastic string of natural length a and modulus of elasticity A. Initially they lie at rest with the string just taut and perpendicular to the edge of a smooth horizontal table, and B is close to the edge. If B is then pushed gently over the edge and released, show that A will still be on the table when the string becomes slack provided that A. is greater than 27E2k mg 1(k + 1)2. Suppose that at time t after the start of the motion the string has not become slack, that A remains on the table and has moved a distance x from its initial position 0 and that B has fallen a distance y, Fig. 65. Then the tension of the string is A (y — x)la and the equations of motion of the particles are A,

km4 = 2(y — x)/a,

(1)

B,

m = mg — A(y — x)la.

(2)

Hence —g

A (y — x) , ma

so that by subtraction d2 dt2 (Y x) — g —

.t — (k

2,(y — x) kma

1) (y — x) • kma

)-

(3)

§ 5:3

151

PARTICLE DYNAMICS I

The solution of this differential equation in y — x, for which y — x = 0, — X = 0 at t 0, is kmag (1 — cosnt) (4) y— X —

2(k + 1)

where n2 = A(k 1)/kma. The string becomes (just) slack when nt = 2gt. But substituting from (4) in the second of eqns. (3) and using the conditions x = 0, x = 0 at t = 0 we find

x—

gt2

kmag (1 — cosnt) A(k + 1)2

2(k + 1)

A (5)

T

But A is still on the table when the string becomes slack if x < a when t = 27t/n, i.e., if

27c2 g (k + 1) n2
0

which reduces to the given condition on substitution for n2. (iii) A mass M is suspended in a medium whose resistance is 2kM X (velocity) by a light spring of modulus .Ml and natural length 1 from a point A which moves up and down with a simple harmonic motion so that its height y above a fixed point 0 is y = a(1 — cospt), M being at rest and in equilibrium at time t = 0. If A < k2, find the complete expression for the downward displacement of M at time t, and show that, when t is large, the

l+x

average rate of dissipation of energy is approximately

kMa2A' p2 (2 — p2)2+ 410 p2 •

ME — FIG. 66.

Suppose that at time t after the start of the motion 3/ is at a distance 1 x below 0, Fig. 66. Then x= = 0 at t O. (1) At time t the length and tension of the string are respectively 1+ x y, y). The equation of (downward) motion of 1ff is

T = A M (x

.111=

AM(x+ y)± Mg, Ax g — Aa(1 — cospt).

+ 2kX

(2)

The general solution of (2) is x =e- k t (C1 ell (k' -Az) t g

a+

C2 e--1/(k.- Az )t)

Aa {(2L — p2) cospt 2kp sinpt} (A — p2 )2 4k2 p2

The constants C„ C2 can be chosen to satisfy the initial conditions (1). We do not find C1 , C2 here but note that the terms in which they arise [the complementary

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A COURSE OF MATHEMATICS

function of eqn. (2)] tend to zero as t --> oo and can be neglected for large values of t The average rate of dissipation of energy must be the average rate at which the external force P = T applied to A does work. At time t the rate at which P does work is

11 = AM (x y) and the average rate, (i.e., the mean value over a period 2n/p), is

f

2 nlp

27c ip

2 n/p

AM (x y)

ApM — 27r

0

X y dt . 0

Integration gives the required result (neglecting the terms involving Cl and C2) .

Exercises 5:3 1. A particle of mass m is attached to the mid-point of an unstretched elastic string of natural length a and modulus mg. The string, with the mass attached, is then stretched between two points in the same vertical line, distant 2a apart. Find the position of equilibrium of the particle. If the particle is now slightly displaced from its equilibrium position in either a vertical or a horizontal direction, show that in each case the ensuing motion is simple harmonic, but that the period of a horizontal oscillation is (15/7)1 times the period of a vertical oscillation. 2. A straight light elastic string A B, of natural length a and modulus of elasticity mg, lies unstretched on a rough horizontal table. A particle of mass m is attached to the string at A, the coefficient of friction between the particle and the table being p. The end B of the string is moved with uniform velocity u in the direction AB. Show that, at time t after the end A begins to move, the length of the string is — (0 silica

(p ± 1) a, where g = a to2, and find the greatest tension in the

string during the motion, assuming that u < pato. 3. A particle hangs from the lower end B of an inelastic string AB, and the upper end A moves in a vertical line with simple harmonic motion of amplitude b and period 271/p. Prove that the string remains taut if b < g/p2. Consider the same problem for an elastic string. The unstretched length of the string is a, and initially the upper end A is at rest, and the particle hangs at rest, the stretched length of the string being a + c. At the instant t = 0 the end A starts to move, its downward displacement at time t being bsinpt, where 0 < p < n and n = 1/(g/c). If x is the downward displacement of the particle at time t from its initial position, prove that, so long as the string remains taut,

n2 x = n2 b sinpt. Prove that the string remains taut throughout the motion if

pb n2 — p2

(n sinnt — p sinpt) c

§5:3

PARTICLE DYNAMICS I

153

for all values of t. Hence prove that the string remains taut if

b<

c. -

p)

4. A particle of mass m moves on the straight line Ox, its displacement from the origin 0 at time t being x. It is attracted to 0 by a force of magnitude mn2 x1, and the motion is resisted by a force 2 km! 1, where 0 < k < n. Prove that x satisfies the differential equation d2 x dx + 21c — d t2 at

n2 x = O.

Find the solution of this equation if x = 0, x = u (u > 0), at t = 0. Prove that the particle is again moving through the origin in the positive sense at the instants t = ra(r = 1, 2, 3, ...), where a = 27t/it and it = i/(n2 — k2) . Prove that the velocity at time t = ra is A'' u, where 2 = e-2 5. A light spring AB is constrained to be in 'a vertical line and is fixed at its lower end B. At its upper end A the spring carries a cup of mass m. In the equilibrium state the cup compresses the spring, which obeys Hooke's law, by the amount a. Show that if a particle of mass m is gently placed in the cup, the ensuing motion is governed by the equation 2 d2 x/dt2

gxla = 2g,

where x is the compression of the spring. If the system is released from rest with x = 6a, show that the particle will leave the cup at time (27c/3) 1/(2 a/g) after the start of the motion. 6. A particle, of mass m, is fixed to one end of an elastic string of natural length a and modulus 2, the other end of the string being fixed to a point 0 on a rough horizontal table. The only frictional force is that between the particle and the table and is equal to k A, where k is a constant and 1 > k> 1. The string is stretched on the table and the particle is released from rest at a distance 2a from 0. Prove that the particle will come to rest at a distance 2ka from 0 after a time n 1/ (a m/A). 7. An elastic string of natural length c and modulus A has its ends attached to two points at a distance a ( > c) apart on a smooth horizontal plane. A particle of mass m is attached to a point of trisection of the string and is displaced through a distance x < (a — c)/3 in the direction of the string and released. Show that it V( 2 m c will oscillate with period 27r . If the particle is released from rest when 3 the smaller portion of the string is of length c/3, and the two portions of the string are in the same straight line, find its velocity when it passes through the equilibrium position. 8. A mass m lies at rest on a horizontal table and is attached to one end of a light spring which, when stretched, exerts a tension of amount m co2 times its extension (where co is constant). If the other end of the spring is now moved with uniform velocity u along the table in a direction away from the mass, and the table offers a resistance to the motion of the mass of an amount mk times its

154

A COURSE OF MATHEMATICS

speed (where k is constant), obtain the differential equation for the extension x of the spring after time t. If k = 2 co show that x =

[2

(2 + cot) e-°it].

9. A particle of unit mass can describe simple harmonic oscillations about a point 0 as centre with a time period 2x/k }/2. If now a resistance of amount 2kx speed is introduced to oppose the motion, show that the quantity xekt varies in a simple harmonic manner with period 2n/k , x being the displacement from 0. Sketch the space-time curve when the particle is projected from 0 with speed V and show that the total distance traversed by the particle before ultimately coming to rest is V y2 e-70/k (1 — e-Th 10. A particle of mass m, is fixed to the middle point of an elastic string, of natural length 2a and modulus a mn2. The ends of the string are fixed to points A and B at a distance 4a apart on a smooth horizontal table. If the particle is pulled to the point B and released from rest at B, prove that it will return to B after a time 2 2 — [2 cos-1 — + Y2 cos-1 1/(.7 -)1 n 3 ) .

5 :4 Motion in a circle When a particle P moves in a circle, of centre 0 and radius a, then the acceleration of P has components a 02, a 0 along and perpendicular respectively to PO, (§ 4 :3). Here 0 is the angle between OP and a fixed direction OA. The forces acting on the particle are (1) the applied forces and (2) the forces exerted by the constraint, e.g., the reaction of a wire on a bead. The latter forces are usually unknown. The choice of method for the solution of any one problem is dependent on whether the forces are or are not conservative. For example, if a bead is threaded on a rough wire, a displacement of the bead once around this wire, i.e., a displacement which brings the bead back to its starting point, requires that a positive amount of work must be done against the frictional force so that the forces in this case are not conservative. On the other hand, a bead attached to the end of an inextensible string can be under action of conservative forces for in any displacement the tension of the string does no work. When the forces are conservative, we write down the energy equation (5.2) in the form v2 ± V = constant,

§ 5:4

PARTICLE DYNAMICS I

155

where V is the potential energy function. This usually takes the form

R0)0

h(0) = E,

(5.12)

where the constant E is determined from the initial values of 0, 0, or the known values at some other point of the motion. Differentiation of this equation w. r. to the time t and division by 0 gives

26/(0) ± 62/1 (0)

h' (0) = 0.

(5.13)

The vector equation of motion (5.1) when resolved in two perpendicular directions, combined with eqns. (5.12), (5.13) gives the means for solving these problems. We may summarise the steps by Energy equation -s- Acceleration Unknown forces. Integration of the energy equation w. r. to the time t gives the position of the particle at any instant. Examples (i) and (ii) below illustrate the method. If the forces are not conservative, then the equations of motion, with the frictional forces opposing the motion, must be used as illustrated in example (iii) below. Examples. (i) A light inextensible string of length a is attached at one end to a particle P, of mass m. To the other end of the string is attached a ring Q, of mass 3m/2, which is threaded on a fixed rough horizontal wire. The ring is at rest and the particle hangs in equilibrium under gravity. The particle is then projected horizontally and parallel to the wire with velocity 1/(2ga). If 0 is the inclination of the string to the downward vertical in the subsequent motion, show that, provided the ring does not slip on the wire, the tension in the string is 3mg cos0 . Deduce that the ring will not slip if the coefficient of friction between it and the wire exceeds 1 /173. We suppose that the ring does not slip and that the string remains taut, so that the forces acting on P are conservative. If at time t the string makes the angle 0 with the downward vertical and is in tension T, Fig. 67, then the energy equation is lma 2 e2 mga(1 — cos0) = Zm 2ga. .•. ae2 == 2g cos0.

( 1)

Since P is moving in a circle with centre Q, its equation of motion resolved along PQ is (2) mae2 =T — mg cost). Equations (1) and (2) give T= 3nig cos 0 . (3)

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A COURSE OF MATHEMATICS

< 0 < Zn during the motion and hence Since 62 > 0, eqn. (1) implies that (3) implies T > 0. Thus our assumption that the string remains taut is justified. Resolving horizontally and vertically for the equilibrium of Q gives

F = T sine = 3mg sin° cos°, B = 3mg/2 T cos 0 = 3mg(4 cos2 0). 2 sin° cos 0 sin 2 0 F •• B 1 ± 2 cos2 0 — 2 + cos2 0 d F\ 2(1 ± 2 cos28) (2 + cos20)2 • • d0 B )

(4)

) changes from posHence as 2 0 increases through the value cos-1( I), d0 B itiveto negative and so the maximum value of F B (occurring when 0 = 5x/12) R

mg FIG. 67. is 1/1/3. Therefore, if the coefficient of friction between Q and the wire exceeds 1/V3, the friction at Q never attains its limiting value and the assumption that no slipping takes place is justified. (ii) A particle P of mass m is free to move on the inner surface of a smooth fixed sphere, of centre 0 and internal radius a. The particle is projected horizontally from the lowest point A of the sphere with speed u. Show that if 2 g a < u2 < 5g a the particle will leave the sphere. If u2 = 4ga, show that the particle leaves the sphere after a time i/(a/g) log (1'6 + i'5) and that it strikes the sphere again after a further time (4/3) 1/(10 a/3 g) .

§ 5:4

PARTICLE DYNAMICS I

157

Since the forces acting on P are gravity and the reaction of the smooth sphere, we may use the energy equation. Clearly P moves in the vertical plane through the direction of projection. Assuming that P remains in contact with the sphere and that the (normal) reaction of the sphere on P is R, Fig. 68, the equations of energy and motion along the inward normal can be written lma2 62 = lmu2 — mga(1 — cost)),

(1)

ma02 =- R — mg cos0.

(2)

. R = m(u2— 2ga

3ga cosO)/a.

(3)

If u2 < 2ga, then (1) and (3) imply

R a m(2ga — u2)/2a

2ga cos0 2ga — u2,

and in this case P does not leave the sphere. Since u2 < 2ga the particle does not rise to the level of 0 and so oscillates through an angle less than 180°. Since dR/dO = — 3mg sin0, R is a strictly decreasing function of 0 for Zn e 0 a x. But when 0 = 17r,

R = m(u2— 2ga)/a;

when 0 7r,

R =- m(u2— 5ga)la.

Hence, if 2ga < u2 < 5ga, R vanishes for some value of 0 in the range < ac — 11 where cosfl (u2 — 2ga)/3ga. This cannot be so and hence P leaves the sphere at B where 0 = az — /3. If u2 = 4ga, then (1) can be written

a02= 2g(1 + cos0).

(4)

Hence the time T for motion from A to B is given by n-I3

T—f 0

d.t9

0

n-13

—

0

Yl

2g) f 1/(1

(a = I/ —) [log (see

cos 0)

2

0

1/(lf c sec — 8 d0 g 2 0

+ tan

P 2/10

= /(--) log (cosec 12- /3 + cot -I 88).

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A COURSE OF MATHEMATICS

2 But in this case cos/ = — so that cosec (3/2) = i/6, cot (0/2) = 1/5. 3

.• T =

/

() log (V6 + I/5).

Referred to horizontal and vertical axes, Ox, Oy through 0, B has coordinates (a sin/3, a cosP). Also the speed of the particle when at B is V = 1/(2ga/3). But the particle moves freely under gravity when it leaves B until it strikes the sphere again at C. Hence its coordinates at time t after leaving B and before striking the sphere are

x = a sin/3 — V t cos/3, y = a cos/3 + V t sinfl — +gt 2.

(5)

To find the time at which P strikes the sphere we substitute eqn. (5) into the equation of the vertical section of the sphere, viz., the circle x2 + y2 = a2, and obtain (a sinfl — Vt cos/3)2 + (a cos/3 Vt sinfl — i g )2 =a2 (6) which reduces to

gt4 — 4 Vt3 sin/3 = 0. 4V sin/3

(7) 4 1 /( 10a \ — 3 3g )

r

Hence the particle strikes the sphere after a further time g as required. Note. The simplification of eqns. (6) to (7) can be seen on general grounds. The circle x2 + y2 = a2 and the parabola (5) intersect in four points. But the circle is the circle of curvature of the parabola at B and hence three of these points of intersection coincide at B. Therefore the quartic equation (6) has three coincident roots (t = 0) and thus reduces to (7). (iii) A rough wire in the form of a circle, of radius a, coefficient of friction kt and centre 0, is fixed with its plane vertical. If a bead P, projected with speed V from the lowest point A, just reaches the level of the horizontal diameter, prove that (1 + 4/12) V2 = 2ga (3,u ei47( + 1 — 2µ2). When OP makes the angle 0 with OA (An extra force, the friction, should be added to Fig. 68.) the equation of motion resolved along the tangent and inward normal gives

mae = —mg sin0 —

( 1)

ma02=- R — mg cos0.

(2)

In this case cos 0 and 02 are each positive for the motion and hence IR! = R = mg cose

mab2

so that (1) can be written = —g (sing

y cos 0) — µa02.

(3)

§5:4

PARTICLE DYNAMICS I

159

Writing 19 = co (the angular velocity of OP) so that 13 = w d co/de = z d(w2)/de, eqn. (3) can be expressed in the form d(0) 2g 2 lc (02 = s( ing + ju cos B). (4) dB a This ordinary differential equation (in cu2 with constant coefficients) has solution cu2

C e-2P0

2 g {(1 — 2 a2) cos 0 — 3 sin 0} (1 ± 4,0) a

(5)

But a w = V when (9 = 0. Hence the constant 2g(1 — 20) V2 C —(6) a2 (1 + 4,u2) a • Also cu = 0 when 13 = In. Substitution in (5) and use of (6) gives the required result.

The simple pendulum If a particle P of mass m is attached to a fixed point 0 by a light inextensible string of length land moves freely in a vertical plane, then

the system is called a simple pendulum. When OP makes an angle with the downward vertical OA, Fig. 69, the equation of motion of P resolved along the tangent to its (circular) path is

10 = — g sin 0 .

(5.14)

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A COURSE OF MATHEMATICS

When 0 is small eqn. (5.14) can be written (5.15)

/6 =

approximately, the error being of 0 (Os) . Also the vertical displacement of P (above A) is 1 (1 — cos 0) = 0(02). Hence for small 0, P moves horizontally with S.H.M. of period 2z f(//g). If the equation of motion of a dynamical system S with one degree of freedom can be written exactly

L0 = —g sin 0 ,

(5.16)

a comparison of eqns. (5.14) and (5.16) shows that the system oscillates in exact synchronism with a simple pendulum of length L when the angular amplitudes, a, of the two oscillations are equal. The amplitude a need not be small. Here L is the length of the equivalent simple pendulum, ESP. On the other hand, if the equation of motion of the system reduces to (5.17) L'0 = — g 0

for small 0, then L' is the length of the equivalent simple pendulum for small oscillations. For example, in example (i) of p. 149 the length of the ESP for small oscillations is (2a )/3)/7. Exercises 5:4 1. A string of length 1 has its ends tied to fixed points A and B, A at a height a vertically above B. A smooth heavy ring P threaded on the string rotates about AB so that BP is horizontal. Prove that

l /1 2g w I II la (12 a2) where w is the angular velocity. 2. A railway curve is an arc of a circle of radius R. The track is banked at an angle a to the horizontal such that there is no sideways force on the rails when a vehicle travels along the track with speed V. Express R in terms of V and a. If a vehicle of weight W traverses the same curve at a speed 2 V show that the sideways force on the rails is 3 W sin a . 3. One end of a light inelastic string of length a is attached to a fixed point A, distant 13a/27 below a horizontal ceiling. The other end supports a particle which is projected from its lowest position with a horizontal velocity 1/(2. g a). Find the smallest value of 2 for which the particle will just reach the ceiling. 4. A particle B of mass m is attached to a fixed point A by a light string AB. The particle moves (as a conical pendulum) in a circle lying in a horizontal plane at a depth h below A. If w is the angular velocity, prove that w2 '

§5:5

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161

If the string is an elastic string of natural length a, and the tension needed to double the length of the string is A, prove that a 2 cannot attain or exceed the value A/ma, and that is must be greater than mga/2.. 5. A particle rests at the top of a smooth fixed hemisphere of radius a, which stands, with its plane face horizontal, on a horizontal plane. If the particle is slightly displaced from its equilibrium position, show that it leaves the surface of the sphere at a height 2a/3 above the plane with velocity 7/(2ga/3). Show further that the particle hits the plane at a distance (5 7/5 + 4 V23) a/27 from the centre of the plane face of the hemisphere with velocity V(2ga). 6. Two particles are connected by a light string which passes over a fixed smooth vertical disc of radius a. The system starts from rest with the lighter particle, of mass m, at one end of the horizontal diameter of the disc and the heavier particle, of mass M, hanging freely. Find the reaction between the lighter particle and the disc in terms of B, the angle which the radius drawn to the particle makes with the horizontal. Find also the tension in the string at this instant. In the subsequent motion it is found that the particle leaves the disc having risen a height a/2. Find the velocity of the particle when contact ceases, and the value of the ratio m/M. 7. A smooth thin tube in the form of a circle of radius a is rigidly fixed by spokes to a horizontal axis through its centre and normal to its plane. The ends of the axis rest in two smooth fixed vertical slots. The mass of the tube, spokes and axis together, is km. A small particle of mass m resting at the lowest point of the tube is given a velocity A V(ag) along the tube. Show that the axis will rise in the slots when the angle a, which the radius to the particle makes with the upward vertical, is given by k = 0. 3 cos2 a — (A2 — 2) cosa Hence show that when A = 4 the particle will describe complete revolutions without the axis rising, provided k > 11. 8. A bead, threaded on a rough fixed circular wire of radius a whose plane is horizontal, is projected along the wire with velocity V. Show that the bead comes to rest after covering a distance

2a

sinh-1 2 it

-172 ) ,

ga

where it is the coefficient of friction.

5:5 Motion on other curves Problems concerning the motion of a particle on a plane curve can be solved by the use of the energy equation (for unresisted motion only) together with one or both of the equations of motion as for motion on a circle (see § 5 : 4). In this case the tangential and inward normal components of acceleration, ;q = v dv/ds and v2/Q respectively (§ 4:3), are used.

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The cycloidal pendulum If a bead is threaded on a smooth wire bent into the form of the cycloid s = 4a sin g) with vertex downwards and tangent at the vertex horizontal, Fig. 70, then the equation of tangential motion of the particle when at P, (s, y)), is ms = — mg sin y . Replacing sin s by s/4 a we have gs 4a which represents S.H.M. of period 4n (al g) whatever the amplitude of the oscillations. Therefore the motion is strictly simple harmonic for oscillations of finite amplitude.

Example. The normal cross section of a cylindrical surface is a complete arch of a cycloid with vertex uppermost whose intrinsic equation is s = 4a sirup . The surface of the cylinder is smooth, and it is held fixed with its rectangular face in contact with a horizontal plane. A particle is projected from a point on the highest generator of the cylinder along the tangent to the normal cross section through the point of projection. Show that, if the speed of projection is i/(2 a g), the speed of the particle while still in contact with the surface is V{2ag(1 + 2 sin2 w)}, and that the particle will leave the surface when it has descended a vertical distance la. Fig. 71 shows the vertical section in which the particle moves. Taking Ox, Oy horizontally and vertically downwards through the vertex of the cycloid, v is measured as shown. Then dy dy ds =silly) • 4a cos v. dy,cis .•. y = 2a sin2 v; the constant of integration has been chosen so that y = 0 when v = 0.

(1)

§5:5

PARTICLE DYNAMICS I

163

During the time that the particle remains in contact with the surface, the speed v is given by the energy equation lm,v2 — mgy = 4m • 2ag , whence

sin2 y), v2 =2ag(1 on using (1). At P, the (normal) reaction R is given by

(2)

mg cosy — B = mv2 /g. Using (2) and e = ds/dy = 4a cosy we find R

mg(cos2 p — 3 sin2 y) 2 cosy

Hence R will vanish and the particle will leave the surface when y =- r/6. Thus the particle will have descended a vertical distance Ia when it leaves the surface.

y FIG. 71.

The uniplanar motion of a particle attached to a light string wound on a fixed plane curve. Suppose that a light inextensible string is completely wound on a fixed curve and that a particle of mass m is attached to the end P which is at the point A of the curve (Fig. '72). P is now projected with speed V at right angles to the tangent at A. Suppose at time t a length s = PC = arc A C of the string has been unwound and P has moved over the are AP of length 1. Let the tension in the string be T and the components of external force on m be R, S along and perpendicular to CP as shown. Then, since C is the centre of curvature of the path of P, the equations of tangential and normal motion of P are respectively dv mv d /

M. V2

S,

8

T

—

R,

(5.18)

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A COURSE OF MATHEMATICS

where v is the speed of P. But d/ = ad y, where y is the angle the line CP makes with the tangent at A, and so the first of eqns. (5.18) can be written my dv = S. (5.19) 8 d

Example. A string of length na is wrapped round the circumference of a circle of radius a on a smooth horizontal table. One end of the string is fastened to a point A of the circumference and to the other end is attached a particle of mass m. The particle is projected with velocity V perpendicular to the string from the opposite end of the diameter through A so as to unwind the string from the circle. Prove that the string is completely unwound after a time a7t2 /2 V and that the tension of the string while unwinding at time t from the beginning of the motion ism 1/(2aW. In this case, using the notation of Fig. 72 with S = R = 0, the tangential and normal resolutes of the equation of motion are my dv — 0, ay chp

m v2 ay

T,

( 1)

where we have used the relation s = ay. The first of eqns. (1) gives v = V. (This follows also from the energy equation.) But v = s yi = ay y. Hence ayy = V. lay2 = Vt.

§5:5

PARTICLE DYNAMICS I

165

Hence the string is completely unwound, 1,v = n, after a time ant/2 V . The second of eqns. (1) then gives T—

m V2

mV'

al (2 V t la)

(2 a t)'

, •

Exercises 5:5 1. A ring of mass m is threaded on a smooth wire in the shape of the curve y = a cos (x/c) which is fixed in a vertical plane with the y-axis directed vertically upwards. The ring is slightly disturbed from rest at the point (0, a) . Show that in the subsequent motion its speed is 2 (ga)4sin (x/2c). Show also that when the ring passes through the point (nc/3, a/2) the reaction between it and the wire is 3 4c2) 2mgc(a2 4c2)/(3a2 2. A particle of mass m slides on a smooth cycloid s = 4a sing which is held fixed with axis vertical and vertex downwards. If the particle is released from rest at a point whose distance from the vertex, measured along the arc, is c, prove that it passes through the vertex with speed c 1/(g/4 a) and that the thrust on the curve is then mg (1 + c2 /16a2) 3. A particle A of mass 2m can move in a smooth groove in the shape of a cycloid 8 = 4a sin , whose axis is vertical with vertex upwards. A light inextensible string (of length greater than 4a) is fastened to A and, stretched along the groove, passes over a small smooth pulley at the vertex carrying another particle B of mass m hanging vertically. If A is projected along the groove from the vertex with velocity V (g a), show that the reaction of the groove vanishes at the point given by s = 18a/5. 4. A smooth wire in the form of a parabola, of latus rectum 4a, is fixed with its axis vertical and vertex upwards. A small ring of mass m can slide on the wire and is attached to the focus of the parabola by means of an elastic string of natural length a and modulus 2 m g . If the ring is released from rest when at the end of the latus rectum and if 0 is the angle between the string and the upward vertical, show that .

6 = - {(g12a) cos° sin2 0} 5. A rough cycloidal tube is fixed with its axis vertical and vertex uppermost. A particle is projected within the tube from the vertex with a velocity V(4ag) sin a, where a is the radius of the generating circle of the cycloid and tan a is the coefficient of friction. Prove that the particle will reach the cusp with velocity V, where V2 = 4ag costa. 6. A particle is suspended by a light string from the circumference of a cylinder of radius a whose axis is horizontal, the string being tangential to the cylinder and its unwound length being a/3. The particle is projected horizontally in a plane perpendicular to the axis of the cylinder so as to pass under it. Show that the least velocity it can have so that the string may wind itself completely up is I/2ga (fl — sin/3) .

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A COURSE OF MATHEMATICS

7. A fine rough wire in the shape of the cycloid 8 = 4a sin y is fixed in a vertical plane with its axis vertical and vertex downwards. A heavy bead of mass m is threaded on the wire and held at rest at the point where y = In. If the coefficient of friction between the bead and the wire is tz, show that when the bead is released it will slide down the wire provided that y < 1. If u = 1, show that the reaction between the wire and the bead at the vertex of the cycloid is

V5

20 (8 + e-7q4) mg.

Miscellaneous Exercises V 1. A particle of unit mass is projected vertically upwards with speed a V in a medium that produces on it a retardation g(vIV)211+' when its speed is v, where g is the acceleration of gravity and n is a positive integer. Show that the particle returns to the starting point with speed 13 V, where /3 is determined by the relation a

r xdx

1 + x211+1

= 0,

-13 and that the total time taken is given by 0 VIg, where 0, d 0 — 1 xx2 n +1 If n = 0, show that (1 + a) (1 — e-11) = 0. 2. A particle of mass m falls from rest in a medium exerting a resistance mgv4 /c4 where v is the speed and c is a constant. Show that it acquires a speed c/2 after falling a distance {c2 log (5/3)} /4g in a time c {2 tan-1(4) + log 3} /4g . 3. The engine of a motor car of mass m exerts a constant driving force m 2 and when the engine is shut off the brakes can exert an equal retarding force ma. Show that, if the resistance to the motion is proportional to the square of the velocity, the maximum speed attained on a journney from rest to rest over a distance s is given by VItanh(28/ V2)}°- where V is the limiting velocity. 4. A particle, of mass m, moves in a horizontal straight line under a force equal to m n2 times the displacement from a point 0 in the line and directed towards 0; in addition the motion of the particle is resisted by a force equal to mk times the square of the velocity. If the particle is projected with velocity V from 0 along the line, obtain an equation to determine the distance X from 0 at which the particle first comes to rest. If k is so small that powers of kVIn above the first may be neglected, show that X is approximately 21e q

V n(

1

3n ) •

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PARTICLE DYNAMICS I

5. A particle of mass m hangs in equilibrium, suspended from a fixed point 0 by an elastic string of natural length a and modulus mg. Another particle of mass m is projected vertically upwards and strikes the suspended particle directly with speed V, the particles coalescing on impact. Show that, if the combined particle just reaches 0, then V -= I/14ga and in that case the time taken from the instant of impact to reach 0 is i/(2a/g) {1 + tan-117 -- n/4}. 6. A smooth circular wire of radius a is fixed in a vertical plane, and a bead of mass m is threaded on it. The bead is connected to the highest point of the wire by a light elastic spring of natural length a and modulus mg. The bead is projected from the lowest point of the wire with a horizontal velocity V. Show that in the subsequent motion the acute angle 95, which the spring makes with the vertical satisfies the equation 2 . 4a02 = V2 — 8ag sine 99 Prove further that if V2 < 4ag the bead comes to instantaneous rest after it has travelled a distance V2 4a sin-1 Bag along the wire; and that if V2 < 5ag the reaction between the bead and the wire changes direction during the motion. 7. A particle of mass m is suspended by a light flexible string of length 3na from the end of a horizontal diameter of a cylinder of radius a whose axis is horizontal. The particle is projected with speed V horizontally so that it moves in a plane perpendicular to the axis of the cylinder and winds the string round the cylinder. Prove that the string will not be slack at the top of the path if V2 12n g a. If V2 = 127tga find the tension in the string when the particle is first moving vertically upwards. 8. A bead is free to move on a smooth curve in the shape of the cycloid

x -= a(0 + sin 0) ,

y -=- a (1 + cos0),

where Ox is horizontal and Oy is the upward vertical. Write down the energy equation in terms of 0 and 0. If the bead is released from rest at 0 = s near the highest point of the cycloid, show that it arrives at the lowest point 0 = n after time (4a/g)2 log cot does this tend to infinity as e tends to zero? 9. A rough fixed wire is in the form of one bay of a cycloid

8.

Why

s = 4a siny (-17r < < In) with its plane vertical and its line of cusps horizontal and uppermost, and a small bead is threaded on the wire. If the bead is projected from the lowest point with speed U in the direction of increasing y, and the coefficient of friction is unity, show that the condition that the bead just reaches the cusp is

U 2= 2ag (en — 1) .

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10. Two particles of masses m, and m2 are connected by a light spring which is such that when in, is held fixed, m2oscillates harmonically in the line of the spring with period 2n/n,. Show that if m2is held fixed then ni, will oscillate harmonically n 1/ in, with period and find also the period of oscillation in the line of the n m, 2n spring when both particles are free to move. 11. A particle of unit mass is projected vertically upwards with a velocity u in a medium whose resistance varies as the product of its density and the square of the velocity of the particle. If the density at a height x varies inversely as (a + x), and if the initial resistance is u2/2 a, find the greatest height h attained, and show that the maximum velocity on descending is [2g (a + h)/e]E. 12. Two particles whose masses are m and 3m are attached to the ends of an inextensible string which hangs over a small smooth pulley. To the first of these another particle of mass 2m is attached by means of a light elastic string of natural length a and modulus of elasticity 4mg. The system is let go from rest, the strings being vertical where not in contact with the pulley, in a position in which the elastic string is of length a. Prove that the particle of mass 3m performs simple harmonic oscillations of period 2n 1/(a/3g) about a point distant a below its initial position. 13. A light elastic string, of natural length 1, lies in contact with a line of greatest slope of a smooth plane inclined at an angle a to the horizontal. The upper end A is held at rest, a particle of mass m is fastened to the lower end, and the extension of the string is c when the system is in equilibrium. The end A is now drawn up the line of greatest slope with constant velocity V. Prove that, if V2 < gc sin a, 1 after equal intervals the string never becomes slack and that its length is c of time n[c I (g sin a)]2. Prove also that the particle will overtake the end A if V2 > (c + 2 /) g sina . 14. One end of a light inextensible string, of length na, is fastened to a point on the highest generator of a circular cylinder, of radius a, which is fixed with its axis horizontal. At the other end of the string is attached a particle of mass m which is held at rest in contact with the cylinder at a point on the lowest generator and then released. Prove that, when the length of the straight portion of the string is a 0, the tension in it is mg (30 sin° — 4 sin2 20)/0. 15. A bead of mass m slides on a rough circular wire of radius a in a vertical plane, the coefficient of friction being 1/Y2. The bead is projected from the lowest point; show that in order that it may just reach the highest point the velocity of projection must be exp [112 a)]} where a is the acute angle tan-112.

CHAPTER VI

PARTICLE DYNAMICS II: MOTION WITH TWO DEGREES OF FREEDOM 6:1 Introduction If a particle is not constrained to move on a curve, then, unlike the motions considered in the last chapter, there are no unknown forces of constraint. On the other hand two parameters, usually coordinates, are required to specify the position of the particle, and the vector equation of motion has two component equations which, on integration, determine the coordinates as functions of the time ; these are the parametric equations of the trajectory (or orbit). The reader is already familiar with a simple case of this in the motion of a projectile under gravity; without air-resistance this trajectory is a parabola and the force on the particle, its weight, is constant in magnitude and direction. In this chapter we consider the motion of a particle in a resisting medium and the motion of a particle under the action of a force directed towards a fixed point (a central force). In these motions the resultant force on the particle varies; in the case of resisted motion the force depends upon the velocity and is not conservative ; in the case of the central force the force is almost always conservative and the energy equation can usually be written down. Example. A particle of mass m moves in the xy-plane under an attraction F towards the origin 0 varying directly as the distance and a resistance R proportional to the speed. At unit distance from 0 the attraction is of magnitude m}2 and, when the speed of the particle is unity, the resistance is of magnitude 2m2. If the particle passes through the point (a, 0) with velocity k Aa in the direction of the y-axis, show that the path is given by y

kx—

y — log (

ka kx—

y .

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The forces acting on the particle are shown in Fig. 73 where R is opposite to the velocity. The magnitudes of F, R are mA2 r, 2mAv respectively (OP = r). We resolve the vector equation of motion (5.1) parallel to the coordinate axes and obtain (1) mx = —2mAX — mA2 x, 924 = —2mAY — mA2 y, i.e.,

+ 2AX + A2 x = 0,

-I- 22y A2 y = O.

The general solutions of these equations are respectively x = (C,

Ay

P

C2 t)

y = (D,e-"

The constants of integration Cl, C2, Di, D2 are determined from the initial conditions x = a, = 0, x = 0, y =- kAa, when t = O. These lead to x = all + At) e-At,

0 Fm. 73.

y = kAate-",

(2)

from which At = y/(kx — y). We take logarithms of either of eqns. (2) and substitute the expression X for At. This eliminates t and leads to the equation of the trajectory as stated.

Exercises 6:1 1. If the motion of a particle is defined by the equations X = ay, Y = bX, where a, b are positive constants and 2C, Y are the components of the force in the positive directions of the x and y axes respectively, prove that the particle deb )1/2 scribes a hyperbola whose eccentricity is (1 + — a . 2. A particle P whose cartesian coordinates at any time t are x, y is projected from a point A (a, 0) with velocity na at right angles to OA, 0 being the fixed origin and n a constant. In the subsequent motion the acceleration of P is always n2 OP in the direction OP. Find x and y in terms of t and hence show that the path of P is a rectangular hyperbola. Show also that OP = a (cosh2nt)112 and that the angular velocity of the line OP is n sech 2nt. 3. A particle is projected on a smooth horizontal plane from the point (0, 2a), referred to rectangular axes in the plane, with velocity Ku/a) parallel to the xaxis. The particle moves under a force ,u/y2 per unit mass, directed towards and perpendicular to the x-axis. Show that its path in the first quadrant is the curve 82 -= 8a(2a — y), where s is the length of the path measured from the starting point. 4. A particle of mass m, moving in the x, y plane, is attracted towards the axis of x by a force /em/y3when at a point (x, y). Show that, if y is constant and the particle is projected from the point (0, k) with component velocities U, V parallel to the axes of x and y, it will reach a highest point if ,u > k2 V2. Find the coordinates of the highest point if this condition is satisfied.

§6:2

PARTICLE DYNAMICS II

171

6:2 The motion of a projectile in a resisting medium The projectile is a particle subject to gravity, which acts. vertically downwards and is constant; the resistance to the motion of the projectile is a force acting in a direction opposite to that of the velocity of the particle relative to the medium in which the particle moves. (Since both gravity and the resistance act in the same plane, the motion takes place in the vertical plane which contains the velocity of projection.) Usually the magnitude of the resistance is assumed to vary as some power of the velocity, and the method adopted to determine the trajectory depends upon the relation between resistance and velocity.

Linear law of resistance In this case we assume that the resistance is a force R = — mk v; the mass of the particle is m and its velocity of projection from an origin 0 is V at an angle oc, to the horizontal. We find the coordinates (x, y) of

the particle, at time t after projection, referred to horizontal and vertical axes Ox , Oy; by eliminating t we find the equation of the trajectory. Suppose the direction of motion of the particle when at P(x, y) makes an angle y with Ox, Fig. 74. Then the resistance, R, has components mkv cosy, mkv siny parallel to x0, y0 respectively. Hence the equation of motion resolved parallel to Ox, Oy gives 997J = —mkv cosy, = — mkv sing — mg.

(6.1) (6.2)

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A COURSE OF MATHEMATICS

But, if s is the arc distance along the trajectory (measured from 0), ds v = dt

""P =

dx dy ds siny = ds .

Hence (6.1) and (6.2) can be written k = 0.

(6.3)

+ kg g.= 0.

(6.4)

These equations, integrated by stages or as ordinary differential equations with constant coefficients, subject to the conditions x = 0 = y, = V cos a , = V sin a at t .= 0, give X

y=

------

V cos k

(17 sine'

(1

0 —kt)

(6.5)

/1

e —Ict)

(6.6)

k2

The equation of the path, obtained by eliminating t from (6.5) and (6.6), is kx y = x (tancx f g log ( 1 . (6.7) kV cos a k2 V cosa ) We note the following points: (1) from eqn. (6.5), x increases as t increases and x --> (V cosx)/Ic as t . Hence the path has a vertical asymptote; (2) the equations for an unresisted projectile can be obtained by letting k —> 0 in (6.5), (6.6) and (6.7). Examples. (i) Show that the effect of the resistance, if k is sufficiently small, is to reduce the horizontal range, R, of the particle by (4 k Y R sing)/3 g, approximately. Expanding the logarithm term in eqn. (6.7) gives

y= x Lana

g x2 2 V2 cos2 a

g z3 3 -173 cos3

+ (k2) .

Neglecting terms of 0 (7c2) we find that when y = 0, x satisfies the equation

x=

2 V2 sina cosa

2kx2 3 V cos a •

§ 6:2

173

PARTICLE DYNAMICS II

Successive approximations to the positive root of this equation are (see Vol. I § 5:8) 2 V2 sina cosa x — + 0(k), g

X

=-

2 V2 sina cosa

8 k V 3 sin2 a cosa 3g2

0(k2).

This second approximation gives the required result. (ii) Discuss the motion of a projectile moving under constant gravity and a linear resistance by vector methods. The equation of motion can be written dv dt — —gi — kv,

f

(1)

where z is a unit vector directed vertically upwards and f is the acceleration. Differentiation w.r. to t gives df (2) kf

whence f — foe—kt,

(3)

where f, is a constant vector. Hence the acceleration is always parallel to the (fixed) direction of the initial acceleration fo and tends to zero as t oo . Hence, from (1), as t 00

g

Y--> — Z ,

(4)

and this gives the limiting or terminal (vertically downwards) velocity. Note also that the hodograph of the motion is a straight line. Since v = dr/dt, where r is the position vector referred to the point of projection as origin, (1) integrates to give v = —gt — kr + v0 , (5) where v, is the initial velocity. Let i be a unit horizontal vector. Then the horizontal displacement

x=i•r=i• (vo — v)/k from (5). Hence as t given by

(6)

oo , we find from (4) and (6) the limiting horizontal range

lim i • r

i • vo /k.

Equation (5) can be written d dt

(rekt) = — gtekti+ vo ekt

(7)

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A COURSE OF MATHEMATICS

which integrates to give kg2 z -

r= k(1 — e-kt) dr dt

v

v e-kt °

kt

(8)

(9)

- e-k t).

The time of flight T is given by e-k T

v • z - v, •

i2

e-k

0

whence T= 1log {1 v k

-k( g o • i)}.

(10)

Resistance varying as vn When the resistance per unit mass is kvn, where v is the speed and n 1, we adopt a different procedure. The equation of motion resolved along the tangent and inward normal to the path (Fig. 74) gives vdv ds 2 -

— g siny — kvn, V v2

Clip

=

ds

g cosy,

(6.8) (6.9)

respectively, the path being taken as concave upwards in the usual manner (Vol. I § 4:5). Here we have used equation (4.21), for the tangential and normal components of acceleration. Dividing these equations gives kvn 1 dv -=- tany sect') v dye which is of Bernoulli's type (Vol. II § 1:4) and can be expressed in the form d 1 n tany nk secy dy v n ) vn g After multiplication by the integrating factor secny we obtain secny vn

^

f secn+ly dip,

nk g

(6.10)

§ 6:2

175

PARTICLE DYNAMICS II

where A is a constant determined by the conditions of projection. Equation (6.10) leads to the intrinsic equation s = f(y) of the trajectory. Then integration of the equations dx d dy dy,

dx ds ds d

==

cosy •

(— v2) g cos y

V2

g

dy ds (— v2) = siny • g cosy ds dy,

v2 tan y

(6.11) (6.12)

gives x and y in terms of p and hence the cartesian equation of the path can be determined. Finally, since dy dy ds g cos y = = dt ds dt from (6.9),

t

v dy' g cosy)

=

B,

(6.13)

where B is a constant. If n = 2, so that the resistance varies as the square of the velocity, or if the motion is almost horizontal, it is convenient to use the arc length along the trajectory as a parameter. Example. A particle of unit mass is moving under gravity in a medium which offers a resistance k w2, where w is the velocity. If u is the horizontal component of velocity, s the length of arc, and y the inclination of the velocity to the horizontal, show that the equations of motion are du ds

—

ku,

dy ds =

u2 cc's' V

The particle is projected from a point in a horizontal plane in a direction making a small angle 0 with the horizontal. Show that the range x is given approximately by g(eakx g), 1) = 2kx(2ku,10 where u, is the initial horizontal velocity. Here we resolve horizontally and along the normal and use s as a parameter. The equation of horizontal motion is du

dt i.e,

— kw2 cosy = — k du

ds

=

ku,

ds

dt

u,

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A COURSE OF MATHEMATICS

with solution

u Ne-ks .

(1)

As before the equation of motion along the inward normal is w2 = -g cosy. But w = u secy and

d2 y 1 = cos3 y. e dx2 d2 y dx2 dy ds

-

g u2 ' (2)

2 c°s3 V

[Note that this result is true whatever the law of resistance.] Equations (1) and (2) give d2 y g elks dx2 u;')

(3)

so that, if the trajectory is nearly flat (as in this case), s x and (3) can be written d2 y dx2

g akx

(4) approximately. Integrating eqn. (4) and choosing the constants of integration so that y = 0, dy/dx = 0 when x = 0 gives g)x — g (e2k x

41c2 u o2 y = 2k(2ku20 0

1) ,

approximately. The required range is the value of x obtained by putting y = 0. Exercises 6:2 1. A projectile moves under gravity in a medium whose resistance per unit mass is mkv, where v is the velocity and k a constant. If the velocity of projection has horizontal and vertical components U and V respectively prove that the highest point of the trajectory is at a horizontal distance U V l(g k V) from the point of projection. 2. A particle is projected under gravity with velocity V at an inclination a to the horizontal in a medium whose resistance per unit mass is lc times the velocity. Show that the particle is moving in a direction at right angles to the direction of 1 Vk projection after a time — log (1 lc g sins 3. A particle moves under gravity g in a medium that produces a retardation equal to k times the velocity. The particle is projected with speed u at angular elevation a and its path is observed to meet the horizontal plane through the point of projection at an angle /3 after time T/k. Show that /3 satisfies the equation tana

tant3 =

ku

secoc(eT — 1),

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177

and hence prove that sin (/3 — a) sin (fl + a)

sinhT — cosh -r — 1

Deduce that the angle of descent exceeds the angle of projection. 4. A particle of unit mass is projected horizontally with velocity Vin a medium the resistance of which is kv, where v is the velocity and k is a constant. Prove that when the direction of the velocity makes an angle y with the horizontal, then 1/v = (k/g) siny + (1/V) cosy. Prove that throughout the motion the velocity is never less than

V g 1 (g2

V 2)2.

5. The propulsive force on a heavy rocket moving in a vertical plane is so adjusted that its speed v is constant. Rectangular axes Ox, O y are taken in the plane of the motion with origin 0 at the highest point of the path, Ox horizontal and O y vertically downwards. If any change of mass is neglected, prove that the equation of the path is v2 g. x. y = — log sec

v2

Hence deduce that the horizontal range cannot exceed xv2/g, and that to approach this range the direction of projection must be nearly vertical. 6. A heavy particle moves under gravity in a resisting medium. Show that, if v is the speed of the particle and cp the angle its direction makes with the upward vertical, 1 dv +cot(p+.-0, vd g sin g) where f is the resisting force per unit mass. Find the form of v in terms of y when f = Cv 3, where C is a constant. 7. The resistance of a medium varies jointly as the density and the square of the speed. If the density at a height y is proportional to e-kY, prove that the horizontal resolute u of the velocity of a projectile is connected with the inclination q of this velocity to the horizontal by the equation

k sing du g u dy

d cosy du dry u 3d y

6:3 Motion under a central force— general theorems When the resultant force acting on a particle is always directed towards, or away from, a fixed point 0, the force is said to be a central force, the path a central orbit, and the point 0 the centre of force. Polar coordinates, with the pole at the centre 0, are particularly appropriate for the solution of the dynamical problems associated with central

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orbits, and the general theorems given below are an essential guide to the solution, whatever the law of force may be. In particular,the angular momentum, introduced in eqn. (6.15), is an important and fundamental concept. 1. A central orbit is a plane curve At any point C of its motion, where the velocity is directed along CD, the acceleration is along CO (see Fig. 75). Hence the velocity at the next instant remains in the plane OCD, and the particle continues subsequently to move in this plane.

Suppose that r is the position vector of the particle referred to 0 and that F is the resultant force on P. The equation of motion, (5.1), is mi = F. d Since — (r xmi)=i x mid-r x mi -= r x mE, the equation of dt motion, when multiplied vectorially by r, gives dt

(r x mi) = r x F.

(6.14)

The vector r xmi is defined to be the moment of momentum of P about the point 0 (c.f. the definition of the moment of a force about a point

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179

in § 2:1) and eqn. (6.14) shows that its rate of change is equal to the moment about 0 of the force acting on the particle. In particular, if F acts along OP and so is parallel to r, then r x F = 0 and r x mi = h, (6.15) where h is a constant (vector), the moment of momentum of P about 0. Taking the scalar product of eqn. (6.15) with r gives h • r = 0 so that r (= OP) always lies in the plane through 0 perpendicular to the fixed vector h.

2. The angular momentum integral If the force on P is mf (r, 0) in the direction P 0 , so that f (r, 0) is the force per unit mass, then the equation of motion resolved along and perpendicular to the radius vector OP gives respectively [see § 4:3 (b)] m(r — 9'62) = —mf(r,6),

(6.16)

• md (2.2 6 = 0 . dt

(6.17)

)

Equation (6.17) integrates to give mr2 0 = mh ,

(6.18)

where h is constant. But mr2 0=r•mr0 is the moment about 0 of the momentum of the particle localised at the particle and is the (scalar) moment of momentum or angular momentum of P about 0. (The vector h of eqn. (6.15) has magnitude I h = mh.) Hence when a particle moves in a central field of force its moment of momentum, h per unit mass, about the centre of force remains constant. If v is the speed of P, directed along the tangent to the path, and p is the length of the perpendicular from 0 to its tangent, Fig. 75, then the moment of momentum of P about 0 is mpv, i.e., h = pv. A counter clockwise moment is usually given a positive sign.

3. The Energy Equation If the force f (r, 0) is independent of 0, we write eqn. (6.16) in the form — r62) =

-M99 (r).

(6.19)

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A COURSE OF MATHEMATICS

Provided that 99 (r) can be integrated, we can always integrate this equation after using eqn. (6.18). Multiplication of (6.19) by r and substitution from (6.18) gives, after cancelling m,

h2 r3

—

Each of the terms in this equation can be integrated w. r. to t, giving h2

z r2 + 2 r2 = —f,99(r) dr + constant.

(6.20)

—f 99(r) dr + constant,

(6.21)

.•. (r2

r2 62) =

where we have substituted h = r2B . We can interpret eqn. (6.21) as an energy equation (after multiplicar2 0 2) , which is the kinetic tion by m). The 1.h. side becomes i m(i.2 energy of the particle. In Chap. II we introduced the potential energy of a system subject to the action of forces. The potential energy of the particle is the work done by an agency against the forces of the field when the agency moves the particle from a standard position to the point (r, 0). When the force of attraction is m99 (r) this work is

V = m f (s) ds,

(6.22) ro where the 'standard position' is (r0 , 00). It is usual to take r0 as infinite in orbital theory; this implies that a particle at an infinite distance from the centre has zero potential energy, so that, in the case of an attractive force, V is negative. From eqns. (6.21) and (6.22) we can write the energy equation in the form

T+V=

r2 62)

Mf 99(S)

ds = constant.

To

An analytic discussion of the motion represented by eqn. (6.20) gives variations in r similar to the variations in x for the rectilinear motion given in § 5:2, and will not be repeated here. Provided h 0, so that 0, the possible motions are : 0 (i) r may tend to infinity, so that the particle diverges to infinity; (ii) a limitation motion in which r -- a, so that the motion tends to become motion in a circle;

§6:3

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181

(iii) a libration motion in which r varies between fixed limits, so that the particle moves in an orbit which touches each of two fixed circles alternately. 4. The rate at which the radius vector sweeps out area in a central orbit is

constant (= 1h) . Let A be the (horizontally shaded) area swept out by OP in time t, Fig. 76. In time at, OP sweeps out the (vertically shaded) sector P 0 P,

of area 6 A = 1r26 0 ± 0 (6 02) . Hence the rate at which 0 P sweeps out area is d Ar 2 a0 r2 d0 1 60 li m 0)1 = +0c dt at 2 dt • 2 at -> o{ Hence, using eqn. (6.18), dA dt = 2u

(6.23)

This result is useful in determining the time between two points on a central orbit, particularly when the law of force is the inverse square law. The following examples illustrate the use of polar coordinates. Examples. (i) Two particles, A, B, each of mass m, are connected by an elastic string of natural length a and modulus pm a ai2 . They are placed inside a smooth horizontal tube of small cross-section so that their distance apart is a, and the tube

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A COURSE OF MATHEMATICS

is rotated with constant angular velocity w about a vertical axis through a point 0 of its length, not between the particles. If 2y > 1, show that AB varies between a and (2,a + 1)a/(2y — 1). Suppose that at time t after the start of the motion the particles A, B are at distances r, r x respectively from 0 (Fig. 77) and that the string remains taut, i.e., x a. Then the equations of radial motion are — r co2) = m a w2(x — a)/a,

A3) — (r

x)

= —,amaco2 (x — a)/a,

where we have used the relation 0 = wt. Subtraction and simplification gives — co2 x = —2,tcco2 (x — a), i.e., X = — (2y — 1) co2 x

2/2 a co2. (1)

If 2y — 1 > 0, the solution of (1) for which x = a, X = 0 at t = 0 is (2y — 1)x = a(2y — cosnt), where n2 = (2y — 1) co2 . Thus FIG. 77.

a < x < a(2y

1)/(2y — 1).

(ii) Two particles A,B, of masses m1, m2 respectively are attached to the ends A, B of a light inextensible string AB of length a + b which passes round a small smooth peg 0 on a horizontal table. When A0, OB are straight and AOB is a right angle with OA = a, OB = b, A is given a velocity V parallel to OB. Show that B reaches 0 with speed Vfm, b (2a + b) V 21(m1 +m2) (a + b)2} after a time /'{(91/1 m 2 )b(2 a + b)/m1V2} . Find the polar equation of the path of A. Suppose that the string remains taut, that, at time t, OA makes an angle 0 with its original direction and the tension in the string is T (Fig. 78). Then the equation of radial motion of A, the equation of motion of B along OB and the angular momentum integral of A are respectively 9/7.1(j"

- r 62 ) = — T ,

(1)

d2 m2 dt2

(a

b

—

r2

r) = — m2r =

—

a V.

T,

(2) (3)

Equations (1) and (2) give, on elimination of '1' (MI

+ M2) i" -Mir 62 = 0

or, using (3), (nzi

112, a2 V2 7)12 )

r3

= 0.

(4)

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183

Multiplying eqn. (4) by the integrating factor I- and integrating we find (m1 m2) r2 = m1V2

mia2 V r2

(5)

where the constant of integration has been chosen so that r = 0 when r = a. Equation (5) is the energy equation (expressing in this case the fact that the kinetic energy of the system remains constant) 2 m1

(7:2

+ r262) + _1,727:2

+7,1

V2.

In general, if the (unknown) tension T is not required, use of the energy and angular momentum equations gives #2 at once. Note that (2) and (4) give T as a function of r. From (4) we see that i > 0 for all r. Also r = 0 for r = a and hence r is a strictly increasing function of t. Hence from (5) . dr dt

1/ m, V2 (r2 — a2) t (m, ni,)r2 1

(6)

Substitution in (6) gives the speed with which B reaches 0, i.e., the value of when r = a + b, as stated. To find the time a at which B reaches 0 we separate eqn. (6) and integrate from r = a to r = a + b and obtain a+b 1Vm 22 ( m1M+

f

1(r2r

drat) — [1(r2 — a2)]aa +b = }/{b (2a + b)}

a

giving the required time. To find the equation of the path of A , we divide eqn. (6) by the square root of eqn. (5) and obtain 1 dr 1 m1 (r2 — a2) t r2 d 0 a V (m1 + m2 )7.2 J •

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A COURSE OF MATHEMATICS

Separation and integration gives 1 /( in, rdo a V m, + m2 ij 0

r

1 /r sec-' a —a)

dr

j rY(r2 — a2)

a

so that the equation of the path of P is r = a sec {0

v( m,+ mi )}.

(iii) A particle P, of unit mass, is attracted by a centre of force at 0 according to the law co2 r co2a3/r2where r is the distance OP. If P is projected from a point distant a from 0 with velocity 4 a co/0 perpendicular to OP, show that subsequently the distance of P from 0 varies between r = a and r = 2a, and find the speed of P when r = 2a. The equation of radial motion of P is — r 62 = — co2 (r

a3Irz)

(1)

4a2corv3.

(2)

The angular momentum equation is

r2 = a • 4aco/1/3 Eliminating 0 between (1) and (2) gives -= — co2

a3 r2

16a4 3r3

(3)

•

Multiplying by the integrating factor r and integrating we find = — co2 (30 — 13a2 r2 — 6a3 r

16a4)/3r2 ,

(4)

where the constant of integration has been chosen so that r = 0 when r = a. Alternatively eqn. (4) can be obtained by elimination of 0 between (2) and the energy equation a3 r2 62) f co2 2 dr = constant, the constant, 13 co2 a2 /6, being chosen so that r = 0, r 0 = 4a w11/3 when r = a. The discussion of eqn. (4) is similar to that of the rectilinear motion problems examined in § 5:2. In fact 1"2 = (.02 (r

But

— a) (2a — r) (3r2

9ar

8a2)/3r2.

(1) r2.„>. 0, (2) 3r2 -F Oar + 8a2 is a positive definite quadratic form in r and a, (3)

> when r = a, <

when r = 2a,

(5)

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185

Hence P oscillates between the circles r = a and r = 2a in a libration motion; the time of travel from one circle to the other is 2a

r dr I w j 1/{(r — a) (2a — r) (3r2

1/3

9ar

8a2)}

a and this integral converges. When r = 2a, r = 0 and (2) gives (r 0)r = a =2a 013 which is the required speed.

Exercises 6:3 1. If the curve r = a cos20 is described under a force to the origin r = 0, show that the force must be proportional to (8a2 — 3r2)/r5 . 2. A particle A of mass m is held at rest on a smooth horizontal table. To A a light inextensible string is attached which passes smoothly through a small hole H in the table, where AH = a, and carries at its other end a particle B, also of mass m, hanging freely. The particle A is projected horizontally with a velocity i/(2gh) in a direction perpendicular to the string AH. If r is the distance of A from H after time t, show that = g h (1 — a2Ir2)

g(a— r) .

Show also that the particle B will reach the table if the length of the string is less than {lc + 1/(h 2 4ah))/2. 3. P, Q are two particles, each of mass m, connected by a light inextensible string of length 21, which passes through a small smooth hole 0 in a smooth horizontal table. P is free to slide on the table, Q hangs freely. Initially, OQ is of length 1, and P is projected, from rest, at right angles to OP with velocity (8g O)2. Show that in the ensuing motion Q will just reach the hole. 4. A particle P is moving in a plane under an attraction of magnitude 2/r3 per unit mass towards a fixed point 0 in the plane, and at t = 0, r = 2 and the radial and transverse components of velocity are )/3/2, 1 respectively. Show that = 2/r3, and find as a function of t. 5. A straight smooth wire is forced to rotate in a vertical plane with constant angular velocity w about a point 0 of its length. A bead can slide on the wire, and originally the bead is at rest at 0 and the wire is horizontal. Show that after an interval of time t, the bead is at a distance from 0 equal to

g 2 co2

(mill t

sM t).

6. A particle of mass 2m, free to move on a smooth horizontal table, is connected by a light inextensible string, which passes through a hole in the table, to a particle of mass m which hangs freely below the table. When the system is at rest and the first particle is at a distance 3a from the hole, it is projected along the table at right-angles to the string with velocity 11/(ga) . Show that in the subsequent motion the hanging particle will move up and down through a distance 2a, and mg. that the tension of the string varies between * mg and

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A COURSE OF MATHEMATICS

7. A smooth thin horizontal straight tube contains a particle of mass m connected to a point A in the tube by means of an elastic string of natural length a and modulus }, the particle being at rest at a distance a from A. The tube is made to rotate with constant angular velocity co about a vertical axis 0 z in the sense which will cause the string to extend; OA is the shortest distance between the tube and 0 z and is of length a. If A > am w2, prove that the string will become slack after a time 2 A w2 . — {zr — tan-' — , where n2 = co n am

6:4 The differential equations of central orbits In this and the two following sections we consider the motion of a particle under a central force which is directed towards a fixed point and is a single valued function, F(r) per unit mass, of the distance, r, from that point. Using the angular momentum integral, r2 6 = h, we change the independent variable to O. Then

h d

d dt

r2 dO

= hu2dO

(6.24)

and we use u( = 1/r) as the dependent variable instead of r. The equation of radial motion — r 62 = —F(r) becomes ho \( \I h2u3 ( ul (6.25) {hu2 dO dO u)J which reduces to

d2u d 02

F(-117-) u =

(6.26)

h2 u2 •

This is the basic equation of motion. Equation (6.26) can be integrated after multiplication by the integrating factor du/d 0, giving du

) 2

1 2 =

f F(11u)

+2 1i

h 2'U2

d u + constant.

The substitution r = 1/u into the integral on the r. h. side and the use of the relation r = — h(du/d0), from eqn. (6.24) show that 1 (.2

r

h2 )

=

2

I F (r) dr

h2

constant.

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PARTICLE DYNAMICS II

187

This is equivalent to the energy eqn. (6.21). Example. Find the p — r equation (Fig. 75) of a central orbit. Since h = p v, the energy equation gives the p — r equation h2

p2 +

2 f F (r) dr = constant.

Alternatively, since p12

2 + r14 d dOr ) 2

d )2' d

he first integra lof eqn. (6.26), obtained above, gives 1

2

f F (r) dr = constant.

p2 — h2

This is the required p — r equation.

In a central orbit, described under a force depending on r only, a point K where the velocity is perpendicular to OK is called an apse, the line OK is an apse-line and OK is an apsidal distance. If we take this apse-line as the initial line, then changing 0 into — 0 in eqn. (6.26) leaves that equation unaltered. Hence, since the particle crosses an apse-line at rightangles, u is an even function of 0 and the orbit is symmetrical about any apse-line. Let L, M, N be the apses next after K. Then, since OL is an axis of symmetry for the orbit, OK = OM, KO -L = LOM. Similarly OM = ON, LOM = MON. It follows that there are only two apsidal distances and one apsidal angle, (the angle between successive apse-lines). When r1and r2are consecutive zeros of the function of r giving t2 in eqn. (6.20), the apsidal distances are r1and r2 . In general, the distances ri , r2 (ri > r2) are the maximum and minimum values of r and accordingly the orbit lies in the annulus r, r ,›- r2 and touches the circles r = rl , r = 9..2alternately. If and only if the apsidal angle cx is a rational multiple of 7 , the orbit is closed. (Special cases arise when r vanishes or diverges to infinity.) Example. As an illustrative example we discuss the stability of a nearly circular orbit and find the apsidal angle when the orbit is stable. Suppose a particle P moves with angular velocity w in a circular orbit, of centre 0 and radius a, under a central attraction f(r) per unit mass. Then, in the steady circular motion, w2 a f (a) •

(1)

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A COURSE OF MATHEMATICS

If this steady motion is disturbed, so that r = a + x, 0 = cat y, then substitution in the radial and transverse equations of motion — 9'62 = —f(r), 2r0 ± re = 0

gives respectively — (a + x) (co + ?)))2 = —f(a 2(co

x),

tp) + (a ± x) y = O.

(2) (3)

Assuming that x and ip (and hence their derivatives w.r. to t) are small, eqns. (2), (3) give, correct to the first order in x, y and their derivatives, — w2 a — xcu2— 2a (pip = —f(a) — xf' (a),

aip = 0,

(4) (5)

where we have used Taylor's Theorem to obtain the r.h. side of (4). Using (1) in (4) and integrating (5) w. r. to t we find — xoi2 — 2a (alp + xl (a) = 0, 2 cox + a fp =lc,

(6)

(7)

where k is constant. Eliminating ip between (6) and (7) we find + ft (a) + 3 0} x = constant.

(8)

The solution of (8) is periodic if

n2 =I(a) ± 3(02 = f(a) + 3If(a)/a12 > 0

(9)

in which case the period is 27c/m. The apsidal angle j3, i.e., the angle through which the radius vector turns between two consecutive apses, where x = 0, is given by

13

n co n

I

if

11 3f (a) f+ (a)af' (a)}

(10)

The orbit is said to be stable if, when slightly disturbed, it never deviates far from its original shape. Equation (9) gives the condition for stability. If f (r) = y/rm then for stability we find m < 3. When m = 2, the apsidal angle is n. Exercises 6:4 1. A particle of mass m is projected from a point A, at a distance a from a fixed point 0, with a velocity (yit)la, in the direction A P where the angle OAP is 45°. If it is subject to a force fl m I 7.3directed towards 0, where r is its distance from 0, show that it describes the curve r = ae-°. Show that the successive distances from 0 at which the particle crosses OA, form a geometric progression, and that the direction of crossing is, in each case, parallel to the original direction of projection.

§ 6:5

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PARTICLE DYNAMICS II

2. A particle describes a circular orbit of radius a about a point 0 which is the centre of an attractive field of force 99 (r) /r3per unit mass, where r denotes distance from 0 and 99(r) is a monotonic increasing function of r at r = a. Show that, if it is subjected to a small radial disturbance, then the period of small oscillation about the original circular orbit is 2or 1/{a3/9)' (a)} . Prove that, if q2(r)/r3is a monotonic decreasing function of r, and the orbit of small oscillation is closed after one revolution whatever the value of a, then 99(r) must be proportional to r. 3. A heavy particle, A, of mass m, is lying on a smooth horizontal table at a distance a from a small smooth hole, 0, in the table, and is connected with another particle B of mass 3m by an inelastic string passing through the hole. The particle A is held at rest supporting B, also at rest; find the velocity with which A must be projected along the table at right angles to Oil in order that it may move in a circle, centre 0. If the circular orbit be slightly disturbed, prove that the apsidal angle of the nearly circular path of A is 27E/i/3. 4. One end of an elastic string of natural length a is tied to a fixed point on the top of a smooth table and a particle, attached to the other end of the string, can move freely on the table. If the particle is slightly disturbed from a circular orbit of radius b, show that the apsidal angle of the new path is n V {(b — a)/(4b — 3a)} .

6:5 The direct law of force When the only force acting on a particle is an attraction n2 r per unit mass towards the pole, the equation of motion is with solution

r = — n2 r

(6.27)

r = A cosnt B sinnt

(6.28)

(c.f. § 5:3). The resolutes of this equation can be written x = alcos nt

b1 sinnt, y = a2 cos nt

b2 sin nt ,

(6.29)

which are the parametric equations of the ellipse (aiY

612x)2

(b1y

b 2x)2 = (al b2 — a2 b1)2 .

(6.30)

Equation (6.29) shows that the motion in each coordinate is simple harmonic with period 2cr/n and hence this motion is known as ellipticharmonic motion. If a simple pendulum of length 1 swings so that it does not move in a fixed vertical plane, the motion of the bob is approximately elliptic-harmonic in a horizontal plane for small oscillations, the period being 27t. I (11g) .

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Examples. (i) The particle of eqn. (6.27) is projected from the point (c, 0) with velocity (0, V). Then, at t = 0 , x = c, y = 0, = 0, = V. .•. al = c,

a2 -=- 0,

.•. x = c cosnt,

b1=0,

b, ------- V In.

y = (V/n) sinnt.

These are the parametric equations of an ellipse with semi-axes c, V In. Note that the eccentric angle q, = nt increases at the constant rate n. (ii) A particle P of mass m is describing an ellipse of major and minor axes 2a and 2 b respectively about a centre of force at the centre. When it reaches the end of the major axis, it strikes and coalesces with a particle of mass nm which is at rest. The central attraction per unit mass is unchanged. Prove that the new orbit is an ellipse of major and minor axes 2a and 2 b/(n + 1) respectively. Choosing the principal axes of the ellipse as coordinate axes, the collision takes place at time t = 0 at the point (a, 0), i.e., where r = ai + Oj. If the attraction is k2 r per unit mass, then for t< 0, we can write (6.28) in the form r = ai coskt + bj sinkt for the motion of P. Since the attraction per unit mass is unchanged, (6.28) gives the motion of the combined particles for t> 0, as r = A coskt + B sinkt, where A, B are constant vectors to be determined. At the time t = 0 the positions given by the two orbits are the same.

A _= ai. When the collision takes place the (linear) momentum of the two particles taken together is unaltered. (n + 1)mkB = mbkj.

B

bj n+1

so that after coalescence the (vector) equation of the orbit is r = ai cos kt

b

n+ j sinkt 1

which gives the required result.

Exercises 6:5 1. A particle of mass m moves in a horizontal plane under a force mn2 r directed to a fixed point 0, where r is the length of the radius vector from 0 to the particle. If when r = a, the particle is moving perpendicular to the radius vector with

§6:6

PARTICLE DYNAMICS II

191

velocity V, prove that, if V < na, the particle will describe an ellipse of major

V'

axis 2a and eccentricity 1/(1

n2,2. ) •

Show also that, when the particle is at a distance r from 0, (i) its speed is 1/( V2 + nza2 — n2 r2), (ii) its angular velocity about 0 is Va/r2 , (iii) its radial velocity is 1 _

,/f

r v

(

2

a

r

V2)}

2) (n2 r2

2. A particle is moving under the action of a force directed towards a fixed point 0 and equal to pr per unit mass, where r is the distance of the particle from 0. Show that the particle describes an ellipse with centre 0. If the particle is projected from a point P, when OP = c, with a velocity Op at an angle of 30° with OP, show that the eccentricity e of the ellipse is given by e2 = 4 j/3 — 6. 3. A particle of mass m describes the ellipse b2 x2 +a2 y2 = az b2 under the action of a force towards the centre of the ellipse. If x =.a cos 99, y = b sin cp is any point on the ellipse, show that q9 is constant and that if this constant value is w, the force towards the centre is mrw2. If two equal particles P and Q describe this ellipse under this law of force, show that the rate at which the straight line joining P and Q rotates is inversely proportional to the square of the distance PQ.

6:6 The inverse square law In this case the function F(r) in the equation of radial motion is given by

F (r)

= ,a1r2-=

pu2;

(6.31)

11± h2

(6.32)

the differential equation (6.26) becomes d2u d02

u

where the angular momentum h is given by (6.18). The differential equation (6.32) gives, on integration, u = A cos (19 — a) + ,u/h2 ,

(6.33)

where A, cc are arbitrary constants. If we write 1 = h2/p, and e = Al, eqn. (6.33) becomes e u = — cos (0 — a) ,

1

r

1

= 1 + e cos (0 — a) .

(6.34)

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A COURSE OF MATHEMATICS

This is the polar equation of a conic whose focus is at the origin, whose major axis lies along the direction 0 = a, and whose latus rectum is 2/. Equation (6.32) can also be integrated, after multiplication by the integrating factor d u/d 0 to give ( du )2 d

2,a2u + constant. h

u2 =

From (6.24) r = — h d u/d 0; therefore v2 =- h2

2a +C i( dd uo )2+

(6.35)

u21 =

where C is constant. This is a form of the energy equation. From the polar equation of the conic (6.34)

Since h2

du d0

,al, v2 =

e2

tt2 = 2u

2

)

2rh2(e2

1)

1 12

/2

r

12

e2

)

(6.36)

1

This shows that the sign of the constant C in eqn. (6.35),

C =- ,u, (e2 — 1)/l,

(6.37)

determines the eccentricity of the orbit. This constant can be calculated if the initial speed of projection is specified (or if the speed at a known distance from the centre of force is specified). If

C > 0, then e > 1 and the orbit is a hyperbola.

If

C = 0, then e = 1 and the orbit is a parabola. C < 0, then e < 1 and the orbit is an ellipse.

If

For a hyperbola whose major axis is 2a the semi-latus rectum is l = a (e2 — 1); for an ellipse of major axis 2a the semi-latus rectum is l = a(1 — e2) . Hence the energy equation (6.36) becomes v2 =ja (2 + ) for a hyperbolic orbit; r a

(6.38a)

v2 = 2/L for a parabolic orbit;

(6.38b)

2 v

=,"

2 r

1 a

for an elliptic orbit.

(6.38 c)

§ 6:6

PARTICLE DYNAMICS

II

193

Since the speed which the particle would acquire in a "free-fall" from infinity to a distance r from the centre of force under the attraction jalr2 is V =11(2,u1r), the orbit is a hyperbola, parabola or ellipse according as the speed at any point of the orbit exceeds, equals or is less than the appropriate free fall speed V. If the force acting on the particle is a repulsion ,u/r2 from the centre of force (e.g., the repulsion between two similarly charged particles), the equation of the orbit is obtained by changing the sign of p (and therefore of 1 and e, since 1 = h21,a, e = Al). The polar equation is, using the positive numerical values of 1, e in this case, /

= —1 + e cos (0 — cc);

(6.39)

and the energy equation becomes

v2 = v2

2p

_ 2p

r

+ constant. h2 (e2 — 1) /2 —(

2 r + e2

1)•

(6.40)

In order that v2 shall be positive the constant must be positive and, therefore, e > 1. Hence, the only possible orbit under the action of an inverse square law repulsion is a hyperbolic orbit; the particle traces out that branch of the hyperbola remote from the centre of force. The period of revolution in an elliptic orbit is obtained from eqn. (6.23). Since the area is swept out by the radius vector at a uniform rate, the time required to complete one circuit of an elliptic orbit whose semi-axes are a, b is na b

T=

2

h

Since b2 = a2 (1 — e2) and h2 =/l = pa (1 — e2) , T2

= 4700(1 — e2) pa(1 — 52)

T = 2or

4n2

(6.41)

Before Newton discovered the inverse square law of gravitation, Kepler stated three laws concerning planetary motion which he had dis-

194

A COURSE OF MATHEMATICS

covered by studying the observations made by other astronomers, notably Tycho Brahe. The laws were : (i) a planet moved in an ellipse having the sun at one focus; (ii) the radius swept out equal areas in equal times; (iii) the square of the time for one revolution was proportional to the cube of the mean distance of the planet from the sun. The results of this section, and eqn. (6.23), show how Newton's Law of Gravitation explains Kepler's Laws. The use of the p —r equation In general, the nature and size but not the orientation of a conic described under an inverse square law of force can be determined by comparison with the standard p—r equations of conics. (See Vol. I § 4:8.) For the energy and angular momentum integrals v2

=C,

pv -= h

give the p — r equation of the orbit in the form. h2 2,a — p2 = r

2C,

(6.42)

and comparison with the p—r equations of the conics (with focus as pole and in the usual notation), viz., b2

2a

p2

p2 = ar b2

2a

p2

r

b2

p2 = 1

1 for the ellipse, for the parabola,

(6.43)

+ 1 for that branch of the hyperbola enclosing the focus, 2a for that branch of the hyperbola r not enclosing the focus,

gives the nature and size of the orbit [see examples (iv) and (v) below]. In the following examples the relations in most frequent use are the polar equation of the orbit, the relation h2 =,a/, the energy equation(s) (6.38), and the definition of h in the form h = pv (Fig. 75).

§6:6

195

PARTICLE DYNAMICS II

Examples. (i) show that the velocity of a particle moving in an elliptic orbit about a centre of force at the focus is compounded of a constant velocity ylh perpendicular to the radius, and a constant velocity yell?, perpendicular to the major axis. The equation of the orbit is 11r = 1 e cos 0, where h2 = ,u/. The radial and transverse components of velocity are (f, re) where

—r = — e0 sine. ye sin 0

eh sin

h

1

•

re =

h

h

— r

(1 + e cos0) = — h (1

e cos0).

1

Resolving the two given velocities in the transverse direction gives

ey — + — cos0 = — (1 + e cos0) = r , h h h and resolving the given velocities in the radial direction gives /,te

— sin0 h

Thus the statement is verified. (ii) A particle describes an ellipse under a force u/r2towards the focus. If it is projected with velocity V from a point at a distance c from the centre of force, 27r 2 V2 —312 show that the time for a complete period is ( — It 1//z c From the energy equation 2 1

V2 =/4 1 a

2

c

—— 1, V2

Hence the period is

T ----

27r

3, 27E 2 a ' 2— 1/y c 1/kt

V2 —312 —

y

(iii) Show that, if a body is projected from the earth with a velocity exceeding 7 miles/s, it will not return. (We assume that the radius of the earth is a = 4000 miles, that, at the surface of the earth, the acceleration due to gravity is g = 32 ft/s2. We neglect frictional resistances and the gravitational forces due to other astronomical bodies.) Since the body is subject only to the attraction of the earth it will not return unless the orbit is an ellipse. We use the energy equation for the point of projection on the earth's surface; this shows that the orbit is not an ellipse if V2 2y/a .

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A COURSE OF MATHEMATICS

But the attraction on a unit mass at the earth's surface is g body will not return if

,1/a2 . Hence the

V2 a2ga = 2 x 32 x 4000 x 5280, i.e., the escape velocity is 7 miles/s approximately. It is worth noting the following point about hyperbolic orbits. When a particle describes a hyperbola under an attractive force it moves on that branch of the curve which includes the centre of force. If it is moving towards the apse of this curve, the angle between its velocity and the radius is an acute angle. This means that the particle cannot be projected from the surface of the earth so as to move towards the apse of a hyperbolic orbit—it would have to penetrate the earth to do so. A particle can therefore only be projected away from the apse of a hyperbolic orbit. (iv) A particle of mass m is describing a parabolic orbit of latus-rectum 4a under an inverse square law attraction to the focus. When the particle is at one end of the latus-rectum it meets and coalesces with a particle of mass nm at rest. Show that the composite body will trace out an elliptic orbit of eccentricity e given by the equation (n + 1)4 (1 — e2) = 2n(n + 2). In the parabolic orbit the velocity at the vertex is 1/(2a/a) and the velocity at the end of the latus rectum is U = 1/(/z/a). Hence the angular momentum is mh = ma 1/(2,a/a), .•. h = Y(2,aa). As a result of the collision the combined particles, of mass (n

1) m, have a speed

U 1(n ± 1). In the new orbit the speed v is given by v2 = 2a

C,

where C is constant. But when r = 2a, v = Ul(n + 1) <11(,ala). Hence C< 0 and the new orbit is an ellipse. If the major axis of the ellipse is 2a1,

U2 + 1)1 U2 (n + 1)2 a1 —

a

a1

a(n, + 1)2

(1 1 /4 a— a, '

a(n + 1)2 n(n + 2) •

But the angular momentum of the combined particles in the ellipse is the same as before the collision, m(n + 1)/i1 =mh.

§6:6

197

PARTICLE DYNAMICS II

if 11is the latus rectum of the ellipse, (n + 1)2 h7 = (n + 1)2

= (n + 1)2 ya1(1 — e)

= h2 2ya. 2a (1 — e2) (n + 1)2 •

••• al-

The result follows on equating the two expressions for a1. Alternatively this problem can be solved by using the p — r equation of the orbit. For the combined particle, the energy equation V2

-

r

—

pin (n + 2) a (n 1)2

/1/

a (n + 1)2

a

and the angular momentum equation pv—

1/(2,ua) n+1

give 2ya (n + 1)2 p2

2y r

2(n + 1)2 a n(n + 2)r

2a2 n(n + 2)p2

i.e.,

yn(v, + 2) a(n + 1)2 1 •

Comparison with the first of eqns. (6.43) shows that the new orbit is an ellipse of semi-axes a1, b1where

(a + 1) 2a —22 (n + 2)

, b2

1

2a2 -

n(n + 2)

-

(1 —

This again leads to the required result. (v) A particle A of mass m carrying a charge q moves towards a fixed atom, 0, which carries a charge q'. At a great distance from the atom the particle has a velocity v, in a line whose perpendicular distance from 0 is p0 . Show that the deviation produced in the path of the particle by the electric repulsion is 2 cot-1

(mpo v ,2

. (Neglect the gravitational force between the particles.)

41' Since the force on A is a repulsion, the path followed by A is a hyperbola, the initial direction being along one of the asymptotes and the final direction being the other asymptote. The deviation is that angle between the asymptotes which does not include the focus (the centre of force). The force of repulsion is m2/1 = qq: , and the energy equation gives /1 1)2

2\ )•

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A COURSE OF MATHEMATICS

Initially r is infinite and v = vo ; a

8,2

qq'

a ma

44, = ,a/ = ya (e2

h2 =

But

..

!APO

a

-=

-

a (e,2-- 1) ,

i.e., p = b2 so that I), = b. The angle between the asymptotes which includes the focus is a = 2tan-1(b la) = 2 tan-1(pola) . The angle of deviation is

P°

Po

13=n— a= 2 {-- — tan-1 (P°)1 — 2 cot-1 ( a 2

.

/3 = 2 cot-1 ( 192' 7)9v' I . qg

Alternatively the p — r equation of the orbit, derived from the energy equation v2 --- v2 — °

2y and the angular momentum equation pv = p0 v0 , can be written r 2y 13° 1 p2 vor

so that, by comparison with the last of eqns. (6.43) the orbit is the branch of a hyperbola with A as outer focus and a = y/v 2o , b = Po whence the result follows as before. (The result of this example is the basis of Rutherford's experiment on the scattering of charged particles, which confirmed the theory that an atom consists of a small, heavy, positively charged nucleus around which the electrons move in various orbits.) (vi) A planet P moves under the attraction of the (fixed) sun S in an ellipse of small eccentricity e, and periodic time T. If A is the perihelion and 0 = ASP, prove that the time taken by the planet to go from A to P is t where

2nt = T [0 — 2e sint9

e2 sin20 + 0 (0)].

Show that the difference between 0 and 2nt/T is greatest when

(The apse of an elliptic orbit nearest to the centre of force is called a perihelion. The other apse is called an aphelion.)

§6:6

199

PARTICLE DYNAMICS II

Since •2 =h and 11r= 1 + e cos 0 , d0 dt = k(1

where k = h112 .•.

d 99

e cos0)2 ,

I (1 — 2e cos + 3e2 cos2 + 0 (e")) dry).

(1 + e cos 99) 2 3 e2

•.

k t = 0 — 2 e sin 0 ±

2

(0 ± 1z sin 2 0) + (e3).

The periodic time T is given by

lcT =

i (1

do) e cosq2)2

3e2 gz) + 0 (0).

(27t

e2 (0 + Isin 2 0) + 0 (e3) 0 — 2e sine 27r (1 + e2 + 0 (e3))

T

.•. 27rt =- T[0 -

2e sin0 -I

3 e2

2 (0 + sin2 0) [1

301 2 + 0(e3)

= T (0 — 2 e sin 0 + 10 sin 2 6) + 0(e3). From this result

0— 2rrtIT = 2e sin° — *0 sin2 0 = ez (say). The stationary values of z occur when dz/d0 vanishes, i.e., when 4 cos0 = 3e cos20 cos 0 = le (2 cost 0 — 1).

or

We solve this equation by successive approximations since e is small. The approximations 00, 01, 02, ... are given by cos Or

= 'ie(2 cos2 0r — 1) .

The approximations are

cos00 =0, cos 0, = —

3e

3e , cos 02 = — • + 0(e3),

and, for the degree of accuracy required, we need go no further. We write

0 = 2 + x, so that 3e cos + X) = — sin X = — X + 0(X3) = — 4 + 0 (0) 3e

7r.

3e

200

A COURSE OF MATHEMATICS

d2 z Since d — —2 sin 0 + 3e sin 2 0 = —2 cosx — 3e sin 2 x < 0, this value of 0 gives a maximum value of z. The corresponding value of

t is

3 4

t= 2 n {0 — 2e sin0 + — e2 sin2 01 + 0 (63)

T 1gt 27r 1 2

3e 4

2e + 0(e3)} —

T14— 87r 5e

+0 (es).

Exercises 6:6 Questions 1-7 and 9-10 refer to the motion of a particle P moving under an attraction a/r2 per unit mass directed towards a fixed point S. 1. The particle P is describing a parabola about S as focus when it suddenly divides into two portions of masses m and (M — m) respectively. Assuming that there is no loss or gain of kinetic energy, and that the portion of mass m proceeds to describe an ellipse with S as a focus, prove that the path of the portion of mass (M — m) will be a hyperbola. 2. If the path of P is an ellipse, and w1, w2 are the greatest and least angular velocities of SP, show that the mean angular velocity of SP is 2(0)1(02)3 /4 + 0,33/2

coi l/

3. A particle which is describing a circle under this law of force collides and coalesces with an equal particle which is at rest. Show that, if the composite particle moves under the same law of force, it will describe an ellipse of eccentricity 3/4. Show also that the periods of description of the circle and the ellipse are in the ratio 7 1/7: 8 . 4. A particle of mass m moving in a circle of radius c under such an attraction collides and coalesces with a particle of mass Am which is at rest. Show that the orbit of the combined mass is an ellipse with major axis c cosec2 a , latus rectum 4c cos2 a, and eccentricity — cos 2 a , where see2 a = 2(1 + A)2. 5. A particle of mass m is describing an ellipse whose principal axes are of lengths 2a and a, under the action of a force ,u/r2 per unit mass, to one of its foci S. When the particle is at one end, A, of the minor axis, it receives an impulse and its orbit becomes a parabola with vertex at A. Prove that the magnitude of the impulse is m ±l (3 ± V2)}1/ 2 a 6. A particle describing an ellipse under a force to a focus reaches an extremity of the minor axis with velocity v and at that instant it collides and coalesces with another particle of equal mass but having a velocity v 1/7 . Show that the new orbit is an ellipse or hyperbola according as the angle between the directions of motion of the two particles is greater or less than a right angle. 7. A particle is describing a parabola about a centre of force which attracts according to the inverse square of the distance. If the speed of the particle is halved without change of direction of motion when the particle is at one end of the latus rectum, prove that the new path will be an ellipse whose eccentricity is (5/8)1/2.

§ 6:7

PARTICLE DYNAMICS II

201

8. Two particles of equal mass are moving in coplanar ellipses under a central attraction 1a/r2 per unit mass. The major axis of each orbit is 2a and the latera recta are 211, 212 respectively. The particles collide at right angles at a point distant c from the centre of force, and coalesce. Show that the semi-latus rectum of the orbit of the combined particles is

(yi,

1/12)2 /4.

9. The particle is projected from a point A, distant c from S, with velocity 1/(a/2c) in a direction perpendicular to AS. When moving parallel to AS it is given a blow, in the plane of the motion, perpendicular to, and away from A 5, such that the subsequent path is a parabola. Find the latus rectum of this parabola. 10. A particle is describing an ellipse of major axis 2a and eccentricity e under this force directed to S. If, when the particle is at one end B of the minor axis, the speed of the particle is halved but its direction of motion is not altered, prove that the new path will be an ellipse of major axis 8 a/7 and eccentricity 4 y(9 + 7 e2). Prove also that the angle between the major axis of the new path and SB is cos-

3+ e2 1/(9 + 7e2)

6:7 Newton's Theorem of the revolving orbit Suppose a particle P is describing a central orbit [Fig. 79 (i)] whose polar equation is 1 = r = 91(9). The attraction on the particle at any point is, from eqn. (6.26), / h2u2

d2u ±it) • d 02

Consider now a second orbit [see Fig. 79 (ii)] described by a point P1 whose polar coordinates are (r1, 01), where ul = 1/r1. If the second orbit is such that, for all values of r, 9, = r, 01= (1 + a) 9, where a is a constant, the equation of the second orbit is 01: a ) u1= ( 11 In the second orbit the angular momentum is hl = el r2 (1 +a)

= h(1 +a),

202

A COURSE OF MATHEMATICS

and the attraction acting on the particle P1 is = hiu]2L(ut

-c12 TOuT1-)

_1 a y 2h2 u2 iu + (1 1 = (1 + 0-)2h2,0 =/

h2 u3{(1

h2u2

d2u a)2 d 02 j

d2 //, d 02

a)2— 1)

=

I

h2(a + 2)a u3 .

In Fig. 79(i) and (ii), A.,g/3,.= 0 = A SrP, SP = SP„ and the line SA, is rotating with angular velocity a 0. Hence the second orbit has the

same shape as the first orbit but the apse-line SA/is rotating relative to the fixed apse line SA. In order to cause this rotation of the orbit an additional attraction of amount a (a + 2) h2 u3must act on P1, i.e., an inverse cube attraction must be added to the inverse square law. This is Newton's theorem of the revolving orbit. Example. The above result has been used to confirm the modifications suggested by the general theory of relativity to Newton's law of gravitation. Einstein's theory of gravitation gives the differential equation of the orbit of a planet (referred to the sun as pole) in the form d2 u d82

u=

3Mu2 ,

where M is the mass of the sun and units have been chosen so that the constant of gravitation and the velocity of light are both unity. Since the orbit of a planet

203

PARTICLE DYNAMICS II

has a small eccentricity (for the planet Mercury e '= 1/5), in the differential equa1 tion we put u = — + i.e., we suppose that the orbit differs slightly from a a circle of radius a. The additional central force arising from the term 3M u2 is \2

3 /0 mu4 = 3 h2 m u 2( 1 [ a = 310 Mu2( 2u a

1 3h2Mu2 a2 7i-

2 a

a2 ) '

where we have written = u — 1/a . The additional force is, therefore, 3 h2 m u4

6h2 11110

3h2 Mu2

a

a2

310 M u2

The effect of the second term

a2

whieh is proportional to 1/12 , is to

modify the Newtonian attraction Mug and so merely alters the periodic time in the orbit. The effect of the first term is to cause a rotation of the apse line .This rotation gives an angular advance 27t a per revolution where

h2 cr(v + 2) = 6 h2 M/a Since, in the units being used, M

is small, we can neglect a2 and take

3M1a, giving an advance of 3M/a revolutions per revolution in the orbit. In the case of Mercury the advance is 43" per century and agrees closely with observations.

Miscellaneous Exercises VI 1. A particle of mass m moves in a plane under an attraction µm/r3 towards the origin 0, where r denotes distance from 0. It is projected from a point K, at a distance a from 0, in a direction at right angles to OK, with velocity V, where V2 < ,a/a2. Write down the integrals of energy and of angular momentum, and prove that a' — r2 = U2 r2 where U' =— — V2. a2 Hence establish the relation connecting r and t, namely, for t < al (I,

r2 =a2 — U2t 2. Using the conservation of angular momentum, establish the relation connecting

0 and t, namely

Ut = a tanh(UO/V),

where the polar angle 0 is measured from 0 K. Find the polar equation of the orbit.

204

A COURSE OF MATHEMATICS

2. A particle of mass m is constrained to move inside a horizontal smooth straight tube OA which is made to rotate with constant angular velocity w about a vertical axis through 0. The particle is connected to 0 by an elastic string of natural length a and modulus of elasticity A, and initially the particle is at rest relative to the tube and at distance a from 0. Show that if A > maw2, the particle will perform simple harmonic oscillations relative to the tube and that the greatest length of the string during the motion is a (A maw2)/(A — maw2). 3. A particle P, of mass m, moves under the action of a force of magnitude peOP per unit mass directed to a fixed point 0. Show that, in general, the path of the particle is an ellipse, and find the time of a complete revolution. If P is projected in any direction with velocity u from a point distant c from 0, prove that the mean value of the kinetic energy of P in a complete revolution is m(u2 pc 2)/4. 4. A particle is projected from a point L with a velocity v(a/a), where OL = 2a, in a direction LP, where the angle OLP is n14. The particle moves under a force ,u1r2per unit mass towards 0. Show that the path of the particle is a parabola of latus rectum 4a. At the vertex of the parabola the particle strikes and coalesces with a particle of equal mass which is at rest. Show that the subsequent path is an ellipse of eccentricity 1/2. 5. Two particles of equal mass describe equal elliptic orbits of eccentricity e and latus rectum 2/ in the same sense about the same centre of force in the common focus S, the major axes of the two orbits being collinear and their centres on opposite sides of S. If the particles collide and coalesce, show that the angle between the directions of motion at impact is 2 tan-1 e and that the orbit of the joint mass is a circle of radius 1 and centre S. 6. A particle moves under gravity in a medium that produces a retardation always directly opposing the motion. If at a height y above the ground the velocity v makes an angle 0 with the horizontal, show that la

cos 0 = exp -

g

f

dy v

where h is the greatest height. Show that, whatever the law of resistance, the angle of descent on a horizontal plane through the point of projection cannot be less than the angle of projection above the horizontal. 7. A particle is projected in vacuo from a point A with velocity w, and elevation a. Prove that its range on the horizontal plane through A is w0 sin 2 a. g

Prove that, to obtain the same range in a slightly resisting medium, the elevation must be increased by approximately 2kwo sin a tan 2 a , 3g

PARTICLE DYNAMICS II

205

where k is the constant ratio of the resistance to the momentum of the particle. It may be assumed that a is not too near to r/4. 8. Two masses m, M. are connected by a light flexible inextensible string of length 2a, which passes through a fixed smooth ring 0 on a smooth horizontal table. Originally the masses lie on the table, the string being straight and just taut and the ring at its middle point. The mass M is then projected along the table transversely to the string with speed v. Show that the mass m reaches the a / M ring after a time – t 3 (1 — , and that the angle turned through by 0 M is then m /( m 3 V 1 + ) . 9. Two equal particles A and B are connected by a fine string of length 2 c which passes through a small hole in a smooth horizontal table. When the system is at rest, with A on the table at a distance c from the hole and B hanging freely, A is projected horizontally with velocity v perpendicular to the string. Find the initial tension of the string, and prove that the initial radius of the curvature of the path of A is

2 cv2 I(g c

v2).

10. Particles A and B, whose accelerations relative to an origin of reference 0 are 2,0A and µ/0B2respectively, both towards 0, are projected simultaneously from the same point U with the same velocity v perpendicular to 0 U. If the orbit of B is a parabola and the orbits cross at one end L of its latus rectum, prove that v2 = 47,a2 and ,u = 22,a3, where a = 0 U. Show also that the interval between the times when the particles pass through a ( 8

L is — v

— — ) 3

•

11. An elastic string of natural length 2l is stretched between two fixed points of coordinates (0, ± 41/3) referred to rectangular cartesian axes on a smooth horizontal table. A particle is fixed to the mid-point of the string and is drawn aside and released from rest at the point (c, d) a small distance from its equilibrium position. Prove that it vibrates in the parabola

c'y

d(2x2 — c2).

CHAPTER VII

SYSTEMS OF PARTICLES 7:1 The motion of two interacting particles When two particles interact with one another, the Law of Action and Reaction states that the forces on the particles are equal and opposite and act along the line joining them. Suppose that P1, P2(Fig. 80) are two interacting particles, of masses m,/ , m2 respectively and position vectors r1, r2 referred to a Newtonian

origin 0. In general the force F between the particles depends on the distance P1 P2and may even depend on the direction of P1/32 . In the absence of forces other than F, the equations of motion of the particles are

(r2 — ri)F ri =

(r1— r2)F m2 r2

ra

(7.1)

§ 7: 1

207

SYSTEMS OF PARTICLES

where F = IFI. From these equations we derive two important results. By adding eqns. (7.1) we find ml rl +

m2.112 = 0 But the position vector of the centre of mass G is r where (m1 +m2) r = m, r, m2 r2 . = 0. r = u,

(7.2)

where 1.1 is a constant velocity. Hence G moves with constant velocity. This is a special case of a general theorem proved in § 7:4. Also from eqns. (7.1) we find mi m2 (r2

=

(ml + m2) (r2 I Ire

r1) 1 F.

(7.3)

, Denoting r2 — r, = P1F2 by r so that I r2 — ri. I = I r I = r = P1-P2 , eqn. (7.3) becomes m.r.= — F ir , (7.4) where 1 1 1 r + and r = m I rI mi. m2 Since —Fr is a force directed along P2 P1, eqn. (7.4) shows that the motion of P2relative to P1is a central orbit with P1as centre of force. m2 r Since GPl = and GP2 = "21r themotions of P1 m1 + m2 ml + m2 and P2relative to 0 are also central orbits similar to but smaller than that of P2relative to P1. The results derived above are independent of the particular law of force between the particles. Since F in eqn. (7.4) in general varies with r, this equation has to be solved by the methods developed in Chap. VI for central orbits. Although the angular momentum and the energy of the relative motion are not strictly the angular momentum and energy of any actual motion, they are first integrals of eqn. (7.4). The motion described by the variation of r is sometimes called the 'internal' motion, and that described by r is sometimes called the 'external' motion. A double star is a good example of such a system. It consists of two stars of roughly equal mass reasonably close together, which rotate relative to each other; they trace out similar orbits about their centre of mass, which is moving with uniform velocity. A hydrogen atom, con-

208

A COURSE OF MATHEMATICS

sisting of an electron moving around a positively charged nucleus, forms another example. In both these examples the force of interaction F varies as 1/r2 which is the most important case of such an interaction. If we multiply eqns. (7.1) scalarly by r1, i2 respectively and add, we obtain • r*, m2 r2 • i2 = Fr • — r2)/r = — F(r • i)/r • But, since r2 = r • r = r2

, r •r = r dr/dt.

m1r1 •

m2i2 • r2 =

F dr/dt.

This equation can be integrated w.r. to the time between arbitrary limits to give t, dr m dt. (7.5) F + 2 m2r2, — — [2 1 dt In general, since F tends to decrease r, F = d V I dr , where V (r) is the potential function giving the force of interaction. Hence, eqn. (7.5) becomes + V = E, (7.6) where E is a constant. This is an energy equation for the system of two particles and shows that, in general, the internal forces in a system do some work as the system moves. Equations (7.2) and (7.6) have important applications in connection with 'collision problems' in astronomy, in atomic physics, and in the theory of gases. 7 : 2 The encounter of two interacting particles A 'collision' or 'encounter' takes place when two bodies, initially well separated, approach one another, interact, and finally separate. They may interact through a force F, as introduced in eqn. (7.1), or they may not affect each other until they come into physical contact. This latter case is the limiting one for which F = 0 until the bodies meet when F becomes infinite; this situation (for elastic collisions) is considered in Chap. X. The law of force may be an inverse square law, e.g. gravitating bodies, electrically charged ions, or it may be more complicated as in the encounter of two gas molecules. If we assume that V = 0 when the two particles are at a large distance apart, then both the 'momentum equation' (7.2) and the 'energy equation' (7.6) applied to the particles, when widely separated after the encounter, are independent of the law of force.

§7:2

209

SYSTEMS OF PARTICLES

In all these cases, with a change of notation, we can prove one important relation. Suppose that the initial velocities of P1, P2 , G are u1, u2 , u respectively and the final velocities are v1, v2 , v respectively. (The words 'initial' and 'final' here mean that the particles are so widely separated that both F and V may be taken to be zero before and after the encounter.) We introduce also the relative velocities U12, V12 , etc., by (7.7) V21 • U12 = u2 ul = - U21 v12 = v2 - vl = It is easy to show that - m2 U12/(ml m2), u2 = 171 - m1 U21/ (M1 + m2)

with similar relations for the velocities after the encounter. When we substitute these expressions into the energy equation we find + m21.4 = (m, + M2) U2 2 MI_ VT.

MA = (M1 + m2) v2

muh,

(7.8a)

M VT2 .

(7.8b)

But eqn. (7.2) implies that u = v since the velocity of 0 is unaffected at any stage during the encounter. Hence eqn. (7.6) with V = 0, implies that uy2 = V12, (7.9) i.e, the magnitude of the relative velocity of the particles is unaltered by the encounter. The difference between U12 and V12 is therefore simply one of direction. This difference of direction is dependent upon the law of force, as is shown in the examples which follow [see also example (v) p. 197]. This result is conveniently represented on a velocity diagram. In Fig. 81 0U1, 0 U2 represent the velocities u1 u2 respectively, 0 V1, ,

0 V2 represent v1, v2 respectively and 00 represents the constant velocity u = v of G. In the diagram U1 U2 = V1 V2, U10/0 U2 = = G/G V2 = M27M1 , and 0 is the angle through which the relative velocity has been turned. When this angle is known, Fig. 81 gives all the velocities concerned in the encounter. We emphasise again that all these results, except the value of 0 are independent of the law of force between P1and P2 . Examples. (i) Two ions P and Q, of masses m, and m2 and carrying like charges q, and q2 respectively, move freely under their mutual electrostatic repulsion. Prove that the orbit of P relative to Q is a branch of a hyperbola of which Q is the outer focus.

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A COURSE OF MATHEMATICS

Prove that, if V is their relative speed before encounter, p the perpendicular distance from Q to the asymptote of the orbit and 6 the deflection of the direction of their relative velocity by the encounter, then

1

4142 ( 1

) cots S. m2

P = V2

The first section of this problem has been discussed in § 6:6. The equation giving the relative motion is eqn. (7.4) with the sign of the r.h. side changed since the particles repel each other, where

0 FIG. 81. Hence, in the notation of the theory of central orbits,

_ 4m m

, h=--pV.

The energy equation for the relative (hyperbolic) motion is

/ 2

v2 p

+

1\

so that V2 = pia. Also, for a hyperbola 1 = b2/a. .• h2 =

p2 V2 =pl

= pb 2 la

.• p = b

= V'.

§7:2

211

SYSTEMS OF PARTICLES

If the angle between the asymptotes which contains the orbit is 2a, then tana = b/a. Also the angle of deflection is 6 = n — 2a, since the relative motion is initially an approach along the asymptote and eventually becomes a recession along the other asymptote.

pia

.•. cot+6 = tana

.•. p = v2 cot46 = ( 1 V2 'm

p V 2111. 1 --) cot16 m,

(The angle 6 here is the angle B of Fig. 81).

(ii) Find the angle through which the relative velocity V is turned when the law of force is an inverse cube attraction, the conditions of projection, etc., being the same as in example (i). nal With the usual notation of central orbits the force of attraction is F —

?.2

•

The differential equation of the orbit of relative motion is d2u d 02 i.e.,

d2 u d 02

4 (mi ± m2) u h2

mi m2 u m h2

u —

,a(m,

m2)} h2

u=

d2 u dO2

n2 u = 0,

where n2 = 1 — u(m1 + m2)Ih2is assumed > 0. The solution of this differential equation is u = A cosnO B sin220 . Since, initially, 0 = 0, u = 0, r = — V (see Fig. 82), A = 0. Also, initially du — V -= = h . d0

V

T, = nB• V . • . u = nh sin n 0 .

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A COURSE OF MATHEMATICS

After the encounter u,—>- 0 and 0

a; hence — n

since h = p V . [Note that since we have assumed 92' > 0, V2 > ,u(mi m2)/P2. If this condition is not satisfied the approaching particle is captured; it spirals about Q eventually coinciding with Q.] (iii) Find the deflection in space of the particle P in the above example (i), assuming Q is initially at rest. The final velocity of P relative to Q is V making the angle (5 with the initial direction. Suppose the particle Q acquires velocity components u, v, as shown in V

Q • m2 v V P • m Fm. 83. Fig. 83, after the encounter. Then the velocity of P in space finally has the components u + V cosa to the left, and V sin (5 — v upwards. Hence the angle of deflection in space is co, where V sino — v tang) — (1)

V cosa + u •

But, since the centre of mass moves with constant velocity, m, V — m2u + m,(u + V cos6), 0 = m2 v — ml (V sinS — v). m„ V(1 — coso) m1 + m2

,v—

m, V sin(' m2

Substituting these values into (1) gives tang) —

m2 sin + m2 cos a •

(iv) Find the modification that occurs in Kepler's Third law when the motion of the sun is taken into account.

§7:2

SYSTEMS OF PARTICLES

213

Kepler's laws were derived on the assumption that the sun was fixed, but § 7:1 shows that, in general, both bodies (planet and sun) move, and that each one describes a central orbit relative to the other and relative to their centre of mass G. We assume that 0 is at rest. Taking G as origin (see Fig. 84) and the planet P, of mass m, at the point (r1, 0) the sun S, of mass M, is at (s, B + 7c), where

r1

8

M m

(1)

M m,

and r = SP. Newton's Law of Gravitation states that the force of attraction between two particles, of masses m and M, is y Mm/r2 when they are at a

distance r apart. The constant y is called the constant of gravitation. Therefore, the equations of motion of P are d

.

dt

(1'2 0) — 0, m(i^1

2-1 02) —

yMm r2

or, using (1),

r2 = h,

— 9.02 —

Y(M r2

(2)

This shows that the motion of P relative to S is a conic with focus at S. If the relative orbit is an ellipse of major axis 2a, eqn. (2) shows that the periodic time is 2x

a3

lv(m + -211)

. This implies that Kepler's third law is only approximate:

if two planets, of masses m1and m2 , describe ellipses relative to the sun with respective major axes 2a1, 2a2 , their periodic times T1 and T 2 satisfy

Ti 17 2

a': (M m,) ct2,(M mi )

(Kepler's Third Law would give al'Ict; on the r.h. side of this equation). It should be noted that Kepler's first two laws still apply to the relative motion of planets and the sun.

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A COURSE OF MATHEMATICS

Exercises 7:2 1. Two gravitating particles of masses m1and m2 are initially at a great distance apart, and m, > m2 ; m2 is at rest and m1has a velocity u1in a line which passes m, at a perpendicular distance p. Prove that after the encounter, when m1 has again receded to a great distance, the angle between the directions of motion of the two particles will tend to the limit 6, where tan S —

Pu ∎ Y(mi. — m2)

and y is the constant of gravitation. 2. Two particles of masses m, and m2 move under the action of their gravitational attraction so that their relative orbit is an ellipse. Prove that the angular momentum of the system about the mass centre is 7rm,m2 (d 1 + d2) ' (d T(m,+ m2) 1d 2) 2' where T is the periodic time of description of the relative orbit and the greatest and least distances between the particles during the motion are d1and d2 . 3: Two gravitating particles of masses M and m are initially at a great distance apart ; M is at rest and m has a velocity v in a direction which passes M at a perpendicular distance a. Prove that after the encounter when m has again receded to a great distance, M will have acquired a velocity 2v Gm ylm(m

m)2

0-2v4)

where G is the constant of gravitation. 4. Two particles of mass m,, m2 respectively are connected by an elastic string of natural length land modulus A . Initially they are at rest on a smooth horizontal table at distance 1 apart. The mass m1is then suddenly projected at right angles to the string with velocity V. Prove that, in the subsequent motion, and before the string becomes slack, the greatest distance x between the particles is a root of the equation m1m2 V2 1(x + 1) = A (m, + m2) x2 (x — l). 5. Two particles of masses m1and mi are initially at a great distance apart. Initially m2 is at rest and m1is projected with velocity U1in a direction which passes m2 at a perpendicular distance p. After the encounter the final velocities of m, and m1make angles B and p with the initial velocity of in„. On the assumption that the mutual potential energy of the particles is a function of r, their distance apart, and tends to zero as r tends to infinity, prove that 1112

mi

sin sin(20

p)

Show further that the ratio of the final energy of m2 to the initial energy of 211, is 4m,m2 cost 0. (mi I-'m2)2

§7:

SYSTEMS OF PARTICLES

215

6. A swarm of small meteors is moving in a plane through the sun. When approaching the sun, and still at a great distance from it, they all have the velocity V in the same direction, and if undeflected would pass the sun at distances between c and c 6c, where &lc is small. Show that, when the meteors have receded to a great distance, their directions of motion will all be comprised within e2 V4) an angle approximately equal to 2 p V2 6,/(t42 7. Show that the results of § 7:1 concerning the relative motion of two particles apply if there are also 'external' forces m1P acting on P1 and m2 P on P2 .

7 :3

The motion of connected particles

In this section we give a few examples of the motion of particles which are connected together by means of strings. In these cases the solution is best found by obtaining the equation of motion of each particle separately. This usually produces a pair of simultaneous, second order, differential equations which are solved by standard methods. The reader should note the points of contrast, and of similarity with the systems discussed in § 7:4. Examples. (i) A particle P of mass

M is tied to one end of a light elastic string, of modulus Mg and natural length a, and the other end is fastened to A, the centre of gravity of a smooth board of mass M. The board is placed on a horizontal plane. If the system is initially held with P at a distance 2a from A, and is then released, prove that the particle will return to its original position relative to the board after a time 2(n + 2)/n, where n2 =g(M m)/am, assuming that the board is of such dimensions that the particle remains on it during the motion. Find the distance the board has described relative to the plane when the particle first passes over A.

x

P A FIG. 85.

Let 0 be the position of A at time t = 0 and let A, P be at distances x, x y respectively from 0 at time t (Fig. 85). Then the length of the string is y and, provided the string is taut, i.e., provided y n a, the equations of motion of P and the board are m(i — Mg(y — a)/a, (1)

Mx

Mg(y — a)/a.

(2)

Hence

g(M m) g (y — a) —

am

(3)

216

A COURSE OF MATHEMATICS

The solution of eqn. (3) which satisfies y = 2a, y = 0 at t = 0 is y = a(1 where n2 =g (M

cosnt),

(4)

m)/am. Also (1) + (2) gives

(M + m) ± mil = 0 which integrates to give

(M

m)th

= O.

(5)

[Equation (5) expresses the conservation of linear momentum for the whole system; see § 7:4.] Equations (1)–(4) hold for 0 C t < n/2n after which the string becomes slack. At t = n/2 n, the velocity y of P relative to the board is given by y = — n,a. Hence P passes over A after a further time a/an = 1/n. Thus the total time before P first passes over A is T = + 2)/2n. By symmetry the total time before P returns to its original position is 4T as required. For 0 < t < n /2n, eqns. (2) and (4) give Since = 0, x = 0 at t

= g cosnt.

0,

x = g(1 — cosnt)/n2. Hence the distance described by the board relative to the plane before the string becomes slack is g = m a l(M m) . When the string is slack the speed of the board is [from (5)] mna/(M m) and in time 1/n (the further time before P passes over A) the board moves a further distance man/(M m) n. Hence the total distance moved by A is 2 mal(M m) . 0

Y

ri

A

Yo

X0

B

FIG. 86. (ii) A light string of length 4a is stretched under tension 2 mn,2a between two fixed points A and B on a smooth horizontal table. Particles of masses 2m and m are attached to points X and Y of the string respectively, where AX = X Y = a. The system performs small oscillations in which the displacements x and y of X and Y, respectively, are horizontal and perpendicular to the string. Show that x and y satisfy the simultaneous differential equations th n2 (2x — y) = 0, If at time t

n,2 (3y — 2x) = 0.

0, x = 0, y = 0, = V and = 0 show that, at time t, 3ny = V(2 sinnt — sin2nt).

If X0 , Yoare the equilibrium positions of X, Y, so that X0 X = x, Yo Y = y and AX, = X, Y0 = a, BY0 =2a, the length of the stretched string is (see Fig. 86) y(a 2 + x2) + }/{a2 (y — x)2} 1/(4a2 + y2) = 4a + 0 (x2).

§ 7:3

217

SYSTEMS OP PARTICLES

Hence, correct to the first order in x, y, the tension T of the string remains equal to its undisturbed value 2 mn2 a. Also the inclinations 0,99,ip of AX, X Y , BY respectively to A B are, correct to the first order in x, y, X =

a

y -x

y

(1)

= 2a •

a '

The equations of motion of X and Y are respectively 2mx = T(sing9 — sine) 2mn2 a(q) — 0),

my = —T (sing, -1- sirp) = — 2m n2 a (99 + y). The component of force acting on X parallel to AB is T cos 99 T cos0 = 0(x2). Hence, correct to the first order, we neglect any motion parallel to AB, both for X and Y. Using (1) the equations of motion become

0

—

+ n2 (2x — y) = 0, n2 (3y — 2x) = O.

(2) (3)

Subtracting these equations gives — + 4n2 (x — y) -= 0.

3mg

Solution of this equation (in x — y) subject to the conditions x — y 0, x— y= V at t 0 gives

V x — y = — sin2nt. 2n

(4)

Substitution for x from (4) in (3) gives

p

n2 y = n V sin2nt.

The solution of this equation for which y = 0 = at t 0 is 3ny = V(2 sinnt — sin2nt).

mg FIG. 87.

(iii) A light inextensible string 0 AB of length 2a has a particle of mass 3m attached to its mid-point A and a particle of mass m attached to the end B. It hangs freely under gravity from the fixed end 0. If the system performs small oscillations in which the particles move in the same vertical plane, show that the equations for the displacements x and y of the upper and lower particles are (3 D2 + 5n2) x n 2 y, (D2 + ,2) y n2 x where D = d/dt and n2 =g/a.

218

A COURSE OF MATHEMATICS

Suppose that 0A , AB make angles 0, cp respectively, at time t, with the downward vertical, (Fig. 87). Then the vertical heights of A, B above their lowest positions are a(1 — cos0) = 0(02),

a(1 — cos0)

a(1 — cos(p) = 0(02) + O(q 2)

respectively. Thus, correct to the first order in 0, co we can neglect the vertical motions and put the tensions T 1, T2 in the strings equal to their equilibrium values 4mg, mg respectively. Further the (approximately horizontal) displacements x, y, of A and B are related to 0, 99 by the equations

x

0=— a'

—

y— x . a

(1)

The equations of horizontal motion of A, B are respectively 3mx = — T1 sin + T 2 sin 97

my = —

sin (p

mg(— 4 0 + (p),

— mg(?).

Therefore using (1) we find (3D2 + 5,2) ,

2 y, (D2 +)l2) y 91

where n2 =g/a.

Exercises 7:3 1. An elastic string, AB, of length 4a, is stretched to a constant tension (= 2man2) and lies on a smooth horizontal table with its ends A, B fixed. It carries a particle of mass m at its mid-point C and another of mass 2m midway between C and A. The particles are given small velocities v at right angles to the string and towards the same side of it. Find the displacements of the particles at any time t in the ensuing vibrations. 2. Two particles A and B, each of mass m, are placed on a smooth horizontal table and joined by a light spring of natural length 1 and modulus 2. Initially the particles are distant 1 apart and B is projected in the direction AB with speed v. In the subsequent motion x and y are the distances of A and B from the initial position of A. Show that x = n2 (y — x — 1) = where n2 = )/ml and dots denote differentiation with respect to the time. Show that the greatest extension of the spring is v (m1/2 A)V2 and that it first takes place after time 2r (m//24112/2. 3. A light elastic string, of natural length 3a and modulus mn2 a, is stretched between two fixed points A, D on a smooth horizontal table so that AD = 4a. Two particles of equal mass m are attached to the points of trisection B, C of the stretched string. The particle at B is displaced a small distance b in the direction AD and released from rest. In the subsequent motion the displacements of the particles from their equilibrium positions at time t are x and y measured in the direction AD. Show that = — 71,2 (2x

y) , y = - y2(2y

x).

§7:4

SYSTEMS OF PARTICLES

219

y and x y are periodic with periods Hence find x and y and show that x 2n/n and 2n/(nV3) respectively. 4. A light string of length 7a has two particles of equal mass in attached to one end and to a point distant 4a from that end. It hangs freely from the other end under gravity. If the system performs small oscillations in which the particles move approximately horizontally in the same vertical plane, show that the equations for the displacements x, y of the upper and lower particles are —

(12D2

11n2) x = 3n2 y, (4D2 +n2) y = n2 x

when D d/dt and n2 =g/a. Prove that x and y are each the sum of two oscillations of periods 2n/n, 2n3/6/n. Find x and y in terms of t, given that, at t = 0, Dx = Dy = 0, x y = 0.

7:4 The general theory of systems of particles In § 7 :1 and § 7 :2 certain results concerning the motion of two interacting particles were shown to be independent of the law of interaction. This section is devoted to a generalisation of these ideas to systems containing any number of particles and to obtaining those important results which are independent of the law of interaction of the particles. First we introduce the notation used. The system consists of n particles P1, ..., Pi , P„ having respective masses m1, ..., mn . The positions, velocities and accelerations are specified by the vectors : positions

r1

velocities

v1,

accelerations

fl

,

,

...,

r,,

rn

v,,

vn ;

;

• • • , fi , • • • , fn •

All these vectors are referred to a Newtonian frame of reference, with origin 0. Although d2 r, dv, dt2 = dt we shall, in general, keep the separate symbols for velocity and acceleration when referring to a particle of the system. When we refer to a point A, not necessarily coincident with a particle of the system, we give its position by the vector rA and use the differential coefficients = d rA/d t , 1"A = d2 rA/dt2 to denote its velocity and acceleration.

cg

220

A COURSE OF MATHEMATICS

The centre of mass G (which does not necessarily coincide with a particle) of the system plays a most important role. Its position vector is given by _Hi = m„r„, M = nt,. (7.10) (Here and in the following I denotes a summation over all the particles of the system.) By differentiation, the velocity and acceleration of G are given by My = mi vi , Mf m,fi, (7.11-12) assuming that the masses of the particles do not vary with the time. We also introduce the position vector, Ri , of P, referred to G, and its derivatives, by

=

R, , vi = ±

f, = f

(7.13-14-15)

The importance of the vector R and its derivatives lies in the relations mi It, = 0,

mi Ri =- 0, Imi Ri = 0. (7.16-17-18)

The new feature of systems of particles is that the momentum vector m, v, of a particle is localised at the point P,. When similar localised vectors occurred in connection with sets of forces, we introduced, in Chap. II, the idea of the moment of a force to take account of localisation. We adopt a similar procedure below with momentum, and now give a list of dynamical quantities and results which apply to systems of particles. 1. Kinetic energy. This is defined to be the sum of the kinetic energies of the constituent particles of the system, T=

v7.

This is modified by substitution from eqn. (7.14), giving + 2v • it, +

T=

= ivrv2

• If m,

Since / m, Ft, = 0 , T = 2111v2

(7.19)

This result illustrates the Principle of Independence of Translation and Rotation. The kinetic energy is made up of one term depending on the motion of G only, and of another term which involves only motion relative to G. [In eqns. (7.8) there is a similar result fort wo particles.]

§ 7:4

221

SYSTEMS OF PARTICLES

2. Linear momentum. This is defined to be the vector sum of the momenta of the individual particles regarded as non-localised vectors; p = fm,v, = Mv.

(7.20)

This shows that our definition makes the linear momentum of a sytem identical with the momentum of a particle of mass M moving with G. 3. Moment of momentum (also called angular momentum). It is here that we take account of the localisation of the individual momenta by the device of taking moments. Hence the name. The moment of momentum of a system of particles about a point A is defined to be h (A) --=

(r, — rA ) x m,v,.

(7.21)

By using eqn. (7.13) we find

- rA ) x mi v,

h(A) =

= (i — rA) x

R, x mi v,

it

R, x m,

Since / R, x mi v = (fm,R,) x v = 0 by eqn. (7.17),

h(A) = — rA ) x My — + 2R, x mi ni .

(7.22)

This is another example of the independence of translation and rotation, the first term of the r.h. side concerning motion of G only, the second concerning motion relative to G only. Since R, is not a velocity referred to a Newtonian frame, I R x ?NEC, is not strictly an angular momentum as defined in (7.21), but is a relative angular momentum. If in (7.21) we use, not v„, but the velocity of P relative to A, viz., v, — •, then

hr (A) =

(r — rA ) x m, (v,

This relative angular momentum, (A), is important only in the one case for which A coincides with G. Then

h, (0) =

x mi ni = (r% — r) x m, (vi — v)

= 2' (r, — i) x m„ v, — (r, — i) x

miv

The first term is h(G), by comparison with (7.21), and the last term vanishes since 2' (r, — x mi v = 2' R, x mi V = (2'm, Ri) x v = 0.

(G) = h(G),

(7.23)

222 and

A COURSE OF MATHEMATICS

h(A)

(r — r A ) x 114 V-

(7.24)

h(G) .

Although these results are quite general, for a distribution of particles which are moving in a plane all the vector products above are vectors perpendicular to this plane, and a typical moment of momentum vector may be written h = hk, the quantity h = IhI being, effectively, a scalar. 4. The equations of motion. We suppose that the particle Piis subject to a force F„ from external causes, e.g., gravity, electrical action, contact from a body outside the system, etc. In addition there is a force Fi acting on P, which is the resultant of the 'internal actions' from other particles of the system. In general F„ is given but Fi is unknown. The only fact we know concerning comes from the Law of Action and Reaction. Suppose that one element of force contributing to riis due to the action of particle P,, then the law tells us that there is an equal and opposite element acting on Pi which contributes to F;. Hence, if we consider the whole set of forces C. , , F;, , they must be in equilibrium. Hence = 0,

(ri — rA ) x

—0.

(7.25)

Now each particle moves according to the equation of motion mi ff = F„

(7.26)

(i = 1, 2, ..., n).

By using eqns. (7.25) we can eliminate the internal forces and obtain results which are true for any law of interaction between the particles of the system. First, by addition, / = I Fi

=F

.

Mf = F Since Mt = M d v/d t,

dp = F. dt

(7.27)

This is the equation of linear motion of the system. Second, by taking moments about A , and summing (ri— rA ) x mif, = I (r, — rA ) x F, =

r (A),

where F (A) is the resultant moment of the forces Fi about A.

(7.28)

§ 7:4

223

SYSTEMS OF PARTICLES

This equation has some particularly important forms which are obtained by modifying the 1.h. side as follows: (r, -- TA ) x mi ff = f

R, — rA ) x miff

=(i— ri,) x

4-

R, x mi f,.

But, by using eqn. (7.15), R, x m,f, =

y R, x mi(f

iti) =_-

X (r, — rA ) x mi f, = (-- — rA ) x Mf

I R, x V R, x m,

This is the first important form and illustrates, once again, the principle of independence of translation and rotation. The equation of angular motion, (i — rA ) x Mf ER, x mi Ri = T(A) , (7.29) is applicable for moments about any point A, moving or stationary. The second important form applies when A is stationary. The vector product v, x mi v, is identically zero in (it v""

(r;— rA ) x mt vil =

vi x m,v,

d -- — h(A) dt

(r, — rA ) x m,f,.

T(A).

(7.30)

The third important form occurs when A is chosen to coincide with G. In this case — r) x m,f, = R, x m, (f = R, x mi Ri . But d dt (. I R, x -= ft, x mi + R, x mi Ri = f R, x m, dt

h ((;') — ---h(G) = dt

r (G) .

(7.31)

The forms (7.30) and (7.31) bear a close analogy to eqn. (7.27) in that the 1.11. sides are rates of change of a momentum. Examples. (i) A light rod, of length 2a and carrying particles of mass m at each end, is projected so as to move in a vertical plane. Discuss the motion. Since the mid-point of the rod is the centre of mass, this point moves as if it were a particle of mass 2m under the action of a downward vertical force 2m g. Hence, this point traces out a parabola, the usual motion of a 'projectile'. This follows from eqns. (7.20) and (7.27).

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A COURSE OF MATHEMATICS

Since the sum of the moments about G of the applied forces, the weights of the particles, is zero, F (G) = 0. Hence d

(G) 0. c.t hr =

If 0 is the angular velocity of the rod, hr(G)

= 2ma2 0k.

.•. 0 = constant. Hence the angular velocity of the rod remains unaltered during the motion. (ii) A system of particles is such that every particle attracts every other particle with a force which is proportional to the product of their masses and to the distance between them, the attraction between two particles of unit mass at unit distance apart being ju. Show that the motion of any one particle relative to the mass-centre G of the system, or relative to any other particle of the system, is elliptic-harmonic of time period 2nCuMH, where M is the total mass of the system. At time t = 0, n such equal particles are situated in a ring at the vertices of a regular polygon of n sides inscribed in a circle of centre C and unit radius, each particle then moving tangentially to the circle, and in the same sense round the circle, with speed V. Describe the subsequent motion of each particle, and show that the particles always lie within the region between the unit circle and the concentric circle of radius V (uM)–+. If the particles are P1, P„ P„, of masses m1, m„ the force acting on P1 is F', where = sumilm2 P1P2

2n3 P1P3 + • • • +

= mi (mi Pi Pi + m 2 P1P2 + = It (m1 + m2 • • • +

mn respectively, then

mnP Pn)

m3 P1 P3 + • • • + M11 P1 P7i)

mn) P1 G = MP1 G

by the centroid theorem of Vol. II p. 171. But the centre of mass G moves with constant velocity since there is no 'external' force on any particle of the system. If 0 is any fixed origin,

GP, = OP,— OG. d • • d t2

OP

—

d d t2

1

= f1

and the equation of motion becomes 2

dt2 G-1;'1) = muMGP1'

(1)

§7:4

225

SYSTEMS OF PARTICLES

By the results of § 6:4, P, describes an ellipse about 0 with period 2n111(4M). A similar result holds for each particle of the system. The equations of motion of the rth and sth particles are d2 (GP dt2 r— Therefore by subtraction

—>

d2 dt2 Tr Ps )

d2 (GP ) dt2 '

= —

mMOZ.

M PrPs

from which it follows that the motion of P, relative to Pr is elliptic harmonic of period 27411/(4M). In the special case considered each particle describes an ellipse (the centre of mass of the system is initially at rest and hence remains at rest). The appropriate solution, see example (i) of p. 190, is X

= COS V, t ,

V

ysinn,t,

where n = 11(mM). Hence the particles always lie between the circles r = 1 and r = V11/(mM).

5. Conservation of linear momentum. If the resultant F of the applied forces is zero, then, since eqn. (7.27) is exactly the same as the equation of motion of a particle, p is a constant vector and G moves with uniform velocity, or remains at rest. If F always acts in a fixed direction, the component of linear momentum in a direction perpendicular to F is constant. [See also eqn. (7.2).] 6. Conservation of angular momentum. If F (A) = 0 and A is a fixed point, eqn. (7.30) shows that h(A) is constant. (In plane motion there is no conservation law for directions perpendicular to r (A) since F (A) , like h (A) , is a vector perpendicular to the plane of motion and is, effectively, a scalar). Similarly, if r (G) = 0, the angular momentum about an axis through 0 is constant. 7. Conservation of energy. If we multiply eqn. (7.26) scalarly by v, and sum for all particles, the 1.h. side becomes d dt

dT dt •

(7.32)

• v,. This expression is The r.h. side of eqn. (7.26) gives / F„ • v, the rate at which all the forces, both applied and internal, are doing work. If the internal forces are such that they do no work in any motion of a system (as in a rigid body), then dT = (rate of working of applied forces). dt

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A COURSE OF MATHEMATICS

If the forces, applied and internal, are conservative and so can be derived from a potential function V [see also eqn. (7.6)], then

EF, • v,

ri • v, —d

T

V = E,

Vld t (7.32a)

where E, the total energy, is constant. Examples. (i) A bead A, of mass m, slides on a smooth horizontal rail, and a particle B, also of mass m, is attached to the bead by a light inelastic string of length 2a. The system is let go from rest with the string taut and in the vertical

plane through the rail, and with AB making an acute angle a with the downward vertical. Prove that, if the inclination of the string to the vertical at time t is 0, then g (cos0 — cosa 4. 0•2 a 2 — cos2 0 • By considering the acceleration of the bead, or otherwise, prove that the tension in the string at any time during the motion is cos 0 (6 — cos2 0) -- 4 cos a (2 — cos2 0)2 If A0B0 (Fig. 88) is the initial position and AB is the position at time t, the coordinates of G are then (x + a sin 0, a cos e) . The velocity components of G are a0 cos 0 , — a 0 sin 0) . Since the rail is smooth there is no horizontal force on the system and G therefore has no horizontal velocity. (The horizontal component of linear momentum of the system is zero.) .•.

0,0 cos° — O.

(1)

§ 7:4

227

SYSTEMS OF PARTICLES

The velocity components of A are (.i!, 0) and those of B are (.i; — 2 a 6 sin 0) ; hence

T = im±2 4m {ct + 2a 0 cos o>2 = Im a' 62 coo o

2a0 cos 0,

4a2 02 sin2 0}

lm {a2 02 cos' 0 + 4 a' 62 sin' 0}

on using (1). The potential energy is V = —2mga cos0. Hence the energy equation is ma2 62 (2 — cos' 0) — 2mga cos0 = E. From the initial condition, 6 = 0 when 0 = a. .•. +a02 (2 — cos2 O) — g cos0 = —g cosa.

(2)

This leads to the required expression for 02, from which 0 may be obtained by differentiation. To find the tension S we consider the motion of A only. d dt

S sin0 = mx = m — (— a 6 cos()) = — ma (9 cos0 + ma 6' sin 0. Substitution of the expressions for 92 and 0 obtained from (2) leads to the required value of S. (ii) A light inextensible string, of length 2a, has equal particles, each of mass m, attached to its ends and a third particle of mass M attached to its mid-point. The particles lie in a straight line on a smooth horizontal table with the string just taut and M is projected along the table with velocity V perpendicular to the string. Show that, if the two particles at the ends collide after a time T when the displacement of M from its initial position is x, then

(M

2m) x= MVT+ 2ma.

Show also that the tension in the string just before the collision is mM2

(M

2m)2 a} .

Suppose the position at time t is as shown in Fig. 89. Then the equation of conservation of linear momentum along the original direction of motion of M is d 112- + 2m— d — (x — a cos0) = M V . t Integration gives

(M

2m)x = MVt

2ma cos°,

which gives the required result when 0 = 0. The energy equation is d +M±2 m dt (x

a cos 0) Y + (a 0 sin O)2] = IM V'.

(2)

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A COURSE OF MATHEMATICS

Also the equation of motion of 111 is - 2 7' cos 0. But by differentiating (1)

(M + 2m) x = — 2 ma (62 cos 0 + 0 sin 0) . Therefore just before the collision 0 = 0, T == To where

T o Mmael(M + 2m). But from (1) and (2) a20 =

m v21(m + 2m) and the required result follows. M

x

tv

•

Fia. 89. x

0

FIG. 90. (iii) A light rod of length 2a is free to rotate in a horizontal plane about its mid-point 0 which is fixed. Two small smooth equal rings, P, Q, are free to slide on the rod, and are initially equidistant from, but on opposite sides of, 0. The rod is given an angular velocity Q, when P, Q, are distant 3a/5 from 0 and at rest relative to the rod. If the system is now left to itself, show that at the instant when the rings leave the rod, the rings are moving with speed 3 af2/5, and the angular velocity of the rod is 9D/25. In the position shown (Fig. 90) the angular velocity of the rod is w. Since the rod rotates freely about 0, the angular momentum about 0 remains constant. Since the action of the rod on a ring is perpendicular to the rod this internal force does no work when the ring slides on the rod, and the work done on the particles in the rotation by this force is exactly cancelled by the work done on the rod by the reactions. Hence, the kinetic energy remains constant. In the position shown the (scalar) angular momentum is

h(0) -- 2m x2co

2m

9 a2 25

(1

)

§7:4

SYSTEMS OF PARTICLES

The kinetic energy is

m (±2 +

T

x2)

m 9aD2 25

229

(2)

From (1), w = 9Qa2 /(25x2). Hence using (2) gi th2

9 a 2 02

Q2 a4

625 x2

25

•

When x = a the rings leave the rod and at this instant 9 a2 D2 ( 1 25

9\ 25

9 16• a2 S22 625

i.e., X = 12aD/25. Also, when x = a, w = 9D/25. The speed v of a ring is given by v2 =V 2 + CO2 X2 and so, when x = a, v2 —

144 625

81 a2 D2 625 J

225 a2 D2, 625

3 v= — a D. 5

Exercises 7:4 1. A light rod of length a can turn freely about the end A, which is fixed, and to the other end B is attached a particle of mass m. The end B is joined by a string of length a to a small ring C of mass m, which can slide on a smooth horizontal wire passing through A. The system is released from rest when AC is 2a. Show that, when the angle CAB is 60°, the angular acceleration of the rod is — g/(16 a) , and find the tension in the string at this instant. 2. A ring A, of mass m, can move freely along a smooth horizontal wire, which is fixed just above a smooth horizontal table. The ring is attached by an inextensible string of length a to a particle B, of mass m, resting on the table. Initially the string AB is taut and alongside the wire, and then B is given a horizontal velocity u perpendicular to the wire. Find the angular velocity of AB when it makes an angle 8 with the wire and show that the tension of the string is then 2 m u2 a (1 + cost B)2 • 3. A light rod AB, of length a, has a small ring of mass vi fixed to one end A, and a particle of mass ni fixed to the other end B. The ring is threaded on a smooth fixed vertical wire. Initially A is at rest, AB is horizontal, and B is moving vertically upwards with velocity a w. Prove that after a time t, before the rod hits the wire, A has moved a distance x downwards and AB has turned through an angle 8, where — ant + a sin 0) x and wt

= (2 -- cos2 0)1/2 dd.

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A COURSE OF MATHEMATICS

4. A light inextensible string ABC has its end A fixed, it passes over a small smooth peg P, and it carries at its end C a particle of mass m. The line joining A to P is horizontal and is of length 2a. Another particle of mass 2m is fixed to the string at a point B, the length of the string from A to B being a. Initially the part ABP of the string is horizontal and C is hanging vertically below P. If the system is released from rest in this position, show that it will reach a position of instantaneous rest when AB has turned through an angle 0, given by sin 0 + cos 0 = cos2 0 Show also that the speed of B, when the angle ABP is a right angle, is 1/(2g a/3). 5. A block of mass M at rest on a smooth horizontal table has a smooth-walled cylindrical hole of radius a with its axis horizontal, and a small bead of mass m is at rest in the hole, in the vertical plane through the centre of mass of the block. If the block is then suddenly given a velocity V along the table in a direction normal to the axis of the hole, show that the bead will just rise to the level of the axis if V2 =2ga(M m)/M Prove also that when the bead is next at its lowest level the velocity of the block is (M — m) V 1(M m). 6. A cylinder of mass M, whose cross-section is bounded by the cycloid s = 4a sine and the straight line joining the cusps, is placed with its flat side on a smooth horizontal plane. A particle of mass m is placed at the middle point of the uppermost generator and slightly disturbed. Show that the particle leaves the cylinder at the point of the boundary of the cross-section for which .

.

cos 2 v., = — TIA,2 — 1 , where A =1 2M/m. 7. A wedge of mass M is at rest but free to move on a smooth horizontal plane and the inclined face of the wedge makes an angle a with the horizontal plane. A smooth particle of mass m is projected from the foot of the wedge up the inclined face along the line of greatest slope lying in the vertical plane through the centre of mass of the wedge. The speed of projection V is such as to cause the particle to attain a height h above the horizontal plane. Show that 2

V

2qh(M m) M m sin 2 a •

Show also that when the particle has returned to the foot of the wedge the horizontal distance traversed by the wedge is 4mh cota + M •

7:5 The motion of a rigid lamina We now consider the application of the results of the previous section to the case of a rigid lamina (or a solid body each of whose particles moves parallel to a fixed plane). We regard such a body as a system of particles in which the internal forces hold the particles always in the

§7:5

SYSTEMS OF PARTICLES

231

same relative positions. No matter how the body moves or what external forces act upon it, the distance between any pair of particles never alters. This condition has two consequences. First, since any two particles P,, Ps remain always at the same distance apart, their mutual interactions do no net work in any motion of the body. This is true for every pair of particles. Hence in any motion the internal forces of a rigid body do no work, and we may ignore these forces in the application of the energy equation. Second, the velocity of any point P of the lamina is given by ,

vi = v, + co (k x BP,)

(7.33)

where w is the angular velocity of the lamina and B is an arbitrary particle of the lamina, i.e., B is a point fixed in the lamina. (See § 4 : 6). We now use eqn. (7.33) in the results of the previous section to obtain special forms applicable to the motion of a lamina. Unless there is some good reason to do otherwise, we usually choose B in eqn. (7.33) to be the centre of mass G, which in the case of a lamina is fixed in the lamina. The reader should also bear in mind that the angular velocity co is independent of the choice of the point of reference B (see § 4: 6). Hence we can modify eqn. (7.14 — 15) as follows: v, -= v

= v ± co (k

x

R,) .

= co (k x R1) .

(7.34)

Here k is, as usual, the unit vector perpendicular to the plane of motion. By differentiating (7.34) and substituting for R% again we obtain f

=f

6 (k x R„) + co (k x 11,)

= f + ci> (k x R,) + co2 [k x (k x Rd] . In evaluating the triple vector product we remember that k • R, = 0 and k R, 0 f, = f + (k x R%) co2 [k (k 112 ) — R, k2] = f + 6 (k x R,) — co2 R,.

Ri = (k x R,) — co2 R,.

(7.35)

(This result is equivalent to the acceleration components in polar coordinates). Equations (7.16 — 18) still apply to a rigid lamina. Now we list the modifications in the dynamical quantities.

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A COURSE OF MATHEMATICS

1. Kinetic Energy. In eqn. (7.19) we substitute (0 2[10 it! — (k • 11 )2] 1

ft2 = (02(k x (02

(02/4

T = 2 MV 2

(Ern1 _TO)CO2 = M -)2 -I-

(G)w2 .

(7.36)

(0) E mil4 which we have introduced here is the moment of inertia of the lamina about an axis through G perpendicular

The quantity

to the plane of motion. It is sometimes written

(0) = Mg, where ko is the radius of gyration about the axis through G. If one point, B, of the lamina is stationary, the velocity of an arbitrary particle is v, = w k x (r„ — r„„). Hence the kinetic energy is, by definition,

T = E'm,„17,2 = 0)2 Em,(r„ — rB)2 = 2 5(B)w2 ,

(7.37)

since mi (r, — rB)2 = 5(B) is the moment of inertia about an axis through B. The theorem of parallel axes gives

(B) = (0) M • B02 . T=

(0)(1)2 +

co2B G2 .

But HI = co • B G

,*. T =

(0)0)2 + 2Mv2 .

This shows that the theorem of parallel axes, in this case, is equivalent to the principle of independence of translation and rotation. The formula (7.37) can be used when B is the instantaneous centre of the motion of the lamina. 2. Linear momentum. No modification has to be made to eqn. (7.20) when the system is a rigid lamina. 3. Moment of momentum. In eqn. (7.22) the term ER„ x m, R, is modified by the use of eqn. (7.34), thus, ER, x mi ni = co Em„[R, x (k x R,)] = co Em,[k 11,2, — R, (k • RO] = kw Em, .*. ER„ x m1 R1 = k (G)co = h,.(G) = h(G) ,

(7.38)

§ 7:5

SYSTEMS OF PARTICLES

233

by eqn. (7.24). Once again the moment of momentum is strictly a vector, but in uniplanar motion it is directed perpendicular to the plane. Hence h(A) = (i — rA ) x My + kJ(G)co. (7.39) If B, a particle of the lamina, is fixed, then, by definition, h(B) = E(ri — r,) x m,co[k x (r, — r,)] = w Ern, (r, — rB) x [k x (r, — rB)] = kco Emi (r, — rB)2 . h(B) =

(B)co.

(7.40)

The relation between (7.39) and (7.40) is again equivalent to the theorem of parallel axes. There is an analogy between eqns. (7.38) and (7.40) giving the angular momentum of a rigid body and the expression my for the linear momentum of a particle : the moment of inertia replaces the mass and the angular velocity replaces the linear velocity. (However, especially in three dimensions, this analogy can be deceiving and must not be pushed too far). 4. Equations of motion. The motion of the centre of mass of a lamina is given by eqn. (7.27) without modification. We consider the changes in the equation of motion obtained by taking moments. The last term on the 1.h. side of eqn. (7.29) becomes, using eqn. (7.35), ER, x mi Ri = ER, x mi ki)(k x Ri) — cot Ri] = Em,ci) [R, x (k x Ri)] = k(EmiRDth

ZR, x mi Ri =kJ(G)h,

(7.41)

since the vector product R, x R, is identically zero. Hence the equation of motion obtained by taking moments about A (which is not necessarily coincident with a particle of the lamina) becomes (r — rA ) x Mf +

(G)th =

r(A).

(7.42)

The parallel between Mf and (G) (k th) is another aspect of the analogy mentioned above. This equation of motion is one of the most useful forms; the other forms of the moment equation can, if required, all be obtained from it.

234

A COURSE OF MATHEMATICS

Equation (7.30), in which A is a point fixed in space, but not coincident with a particle of the lamina, is modified by the use of eqn. (7.39) thus : dt

h (A ) = (1* — rA) x Mf+v x “, k (0)6 =_ (r — rA) x Mf +kf(G)cb,

since (0) does not vary with the motion of the lamina. Hence cqn. (7.30) leads to the form (7.42). If the particle B of the lamina is fixed (i.e., has neither velocity nor acceleration) the equation of motion obtained by taking moments about B is obtained from eqn. (7.42) by substituting the appropriate form for f. Since B is fixed, = co (k x .B0) .

f=

th

(k x B

=th(k x B G)

+ co (k x Tv) w2[k x (1(

x

)[

= to (k x BG) — co2B G

(7.43)

(c.f. eqn. (7.35) where, in contrast, the origin of R, is not fixed). — r.„) x f=Bb x f=th [BG x (k x

= kodBG2.

Therefore the equation of motion (7.42) becomes kth [M • B G2 +

(0)] =

r (B) .

By the theorem of parallel axes this is

k f (B)th = r (B)

(7.44)

This is the most convenient form of equation of motion to use for the rotation of a rigid body about a fixed axis, not passing through G. It is important to note that B is fixed and not just instantaneously stationary. If B is instantaneously stationary, it may have an acceleration and in this case an additional term would be necessary in eqn. (7.43) to take account of the acceleration of B. Equation (7.41) shows that, for a rigid lamina, the eqn. (7.31) for moments about G becomes kf (G)th = T (G) (7.45) whether G is moving or fixed.

j7:5

235

SYSTEMS OF PARTICLES

We can sum up these results for taking moments as follows : (a) if A is moving or if A is fixed, but no one particle of the body is always coincident with A, use eqn. (7.42); (b) if A is fixed and is always coincident with one particle B of the body, use eqn. (7.44); (c) for moments about G, whether it is moving or fixed, use eqn. (7.45). When a lamina is subject to a set of forces in equilibrium Leqns. (2.2), (2.3)] we see from eqn. (7.27) that f = 0; then eqn. (7.42) shows that /V A FIG. 91.

co = 0 also. Hence if the lamina is stationary it remains stationary, as we assumed in Chap. II. Examples. (i) A uniform rigid rod, of mass m and length 2a, rotates with angular velocity w about a perpendicular axis through one end. Then its kinetic energy is a2 (02 m a2 w2 . (ii) A uniform solid sphere, of mass m and radius r, moves on a horizontal plane so that its centre 0 has speed V and it has an angular velocity Q about its horizontal diameter (see Fig. 93). Then its kinetic energy T = }2m V2 + s mr2 122 by eqn. (7.36). If the sphere rolls so that V = r S2 , then T = 7 mr2 122 /10. This latter result follows directly from eqn. (7.37). (iii) The end points of a uniform rod AB, of mass m and length 2a, are moving in a plane with velocities u, v. Prove that the kinetic energy is

1(, m (u2

u•v

v2) .

The velocity of G (Fig. 91) is 1(u v). Also, by the result of § 4:6, the angular velocity is w k, where k is a unit vector perpendicular to the plane of the motion and

v — u = wk x AB.

(I)

ma' T = m{z(u v)}2 +z 3 O.

(2)

Hence by eqn. (7.36)

But from (1) (v — u)2 = co2 (k x A B)2 = co2 A B2 =4a2 w2 , since k is perpendicular to AB. Therefore substitution for co2 in (2) and some reduction gives T = 6 m (u2 u • v + v2).

236

A COURSE OF MATHEMATICS

(iv) Two uniform rods A B ,BC, each of mass m and length 2a, are freely jointed at B and are constrained to move in a plane. ABC is a straight line, A has velocity components u, v along and perpendicular to ABC respectively, and the rods have angular velocities w1 , w2 (see Fig. 92). Find the moment of momentum about A.

Vt

GI

G2

v+ 2 awl +aco2

tv+awi

I

u

u

FIG. 92.

In uniplanar motion, such as this, 'moment of momentum' is interpreted to be the scalar magnitude of the vector which is directed perpendicular to the plane. (Anticlockwise senses of rotation are taken as positive). Elementary kinematics leads to the velocity components for 0, and G2 shown in Fig. 92, 01 and 02 being the centres of mass of AB and BC respectively. We use the principle of independence of translation and rotation and find the moment of momentum as the sum of contributions: from AB, motion of G„ motion about 01,

+ a • m(v

awl), ma2 • wi ;

from BC,

A Flo. 93.

motion of 02 , -1-3a • m (v + 2 a co, + a c02) motion about G2 , + mat • (02. The total moment of momentum is ;

h(A) = flta(12v + 22a co,

lOaco2).

(v) Find the moment of momentum of the sphere of example (ii) (a) about A, the point of contact, (b) about G, the centre, (c) about B, the highest point (see Fig. 93). Here again we use the independence of translation and rotation, applying eqn. (7.39);

h(A) =

1^ •

mV — 5mr2 S2

,

h(G) — 59nr2S2 ,

h(B) = +r • mV — 5mr2Q

.

§ 7:6

237

SYSTEMS OF PARTICLES

7: 6 Applications of the principles of conservation of linear momentum, angular momentum and energy to the motion of rigid bodies The use of these principles greatly facilitates the solution of certain classes of problems in rigid dynamics. Here we give some straightforward applications of the principles and defer to Chap. VIII the more elaborate problems of rigid dynamics. Examples. (i) A uniform rod AB, of length 2a and mass m, is freely jointed at A to a small ring of mass m. The ring is free to slide on a smooth fixed horizontal wire, and it is initially at rest with the rod vertical and rotating with angular velocity 1/(3 gla) in a vertical plane through the wire. Show that, when the rod is inclined at an angle 0 to the vertical, its angular velocity is 4 cos0) 3g(1 a(8 — 3 cos2 0) This problem is similar to example (i) of p. 226 (see Fig. 88). Here the horizontal linear momentum equation and the energy equation are respectively d j i x' + dt (a sin 0) = m V(3ag), 1

2

1

17?

4 d t (a sin 0)1

j.2 H 2

— m g a cos° — These equations give

(

1 4ma2 3g a 2 3

a

6 sin 0)2]

ma2 • 02 2 3

1

mg a .

6 as required.

(ii) A uniform circular disc of mass M and radius a rotates freely about a fixed vertical axis through its centre perpendicular to its plane, and carries a particle of mass +, M which is free to move along a smooth radial groove. Initially the disc rotates with angular velocity w and the particle is at rest at the centre. Prove that, when the particle, after being slightly disturbed, has moved a distance r along the radius, the angular velocity of the disc is 4a2co/(4a2 r 2) , and find the radial velocity of the particle at the moment when it reaches the edge of the disc. (See also example (iii) p. 228). If S2 is the angular velocity of the disc and v the radial velocity of the particle relative to the disc, then the equations of angular momentum and energy give 1Ma2 • m es-22 + Equation (1) gives S2 = 4a2co/(4a2 = 2a w/1/5.

Mr2 S2 = -1Ma2co , (r2

+ a2)

•

(1) M a2 w2 .

(2)

r2). Then eqn. (2) gives, when r = a,

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238

(iii) An insect P of mass m rests on a uniform circular ring of mass 111 and radius a which can rotate about a fixed vertical axis through its centre 0 perpendicular to its plane. The system being initially at rest, the insect begins to move along the ring, and when OP has turned through an angle 0 relative to the ring, the latter has turned backwards through an angle 0. If the system is acted upon by a retarding couple A 00 once the motion has started, show that subsequently

(M m)cp -1- Aq2=m0. In Fig. 94 we suppose that OX was the original position in space of OP, and that at the moment of starting (t = 0) the point A of the ring coincided with P at X. The angular velocities of OP and the ring are Jp and— 0 respectively (in the positive sense). Regarding the ring and insect as one system, the only external force is the retarding couple about the axis through 0. Since 0 is a fixed point both in space and in the system, we can use eqn. (7.30) with A=0. The equation of moments about 0 is

M a2(___,p) .•.

X

ma2

2,a2

— MO = Aq2,

after integrating and substituting the initial conditions. Since 0 + p = 0, this gives

- 9,) - m

FIG. 94.

= 9),

(1)

which leads to the required result. If A = 0, eqn. (1) is equivalent to the equation of conservation of angular momentum about the axis; the angular momentum is zero since its initial value is zero. Note that, for this and similar cases in which an 'intelligent' being does work by using its muscles, we cannot apply the energy equation. In fact, when A = 0 the energy of the system at any stage is the measure of the work done by the insect ; this is a case where the internal forces do work in the course of the motion. Exercises 7:6 1. A uniform rod of mass m and length 2a has a small ring of mass m pivoted at one end. The ring slides on a smooth horizontal wire and the system is released from rest when the rod is inclined at 60° to the downward vertical. Show that, when the rod makes an angle 0 with the vertical in its subsequent motion in a vertical plane through the wire, a02 (8 — 3 0000)

6g(2 cose — 1).

2. A uniform circular disc of mass m and radius 2a is free to rotate about a fixed vertical axis through its centre. An insect of mass m, initially at rest at a point P on the rim of the disc, begins to crawl along a chord through P subtending

§7:6

SYSTEMS OF PARTICLES

239

a right angle at the centre. If g) is the angle turned backwards by the disc in space after the insect has moved a distance ax along the chord, show that (4

(x — y2)2) =- X 1/2.

3. A uniform rod, of mass m and length 2a, is free to rotate in a horizontal plane about its mid-point 0 which is fixed and at a height a above the ground. Two small smooth rings, P, Q, each of mass m/2 are free to slide on the rod. The rings are initially equidistant from, but on opposite sides of, 0. The rod is given an angular velocity (27g1a)1/z, when P, Q are distant a/3 from 0, and at rest relative to the rod, and the system is then left to itself. Show that the angular velocity of the rod when the rings are about to slip off is (3 g/a)1/2, and that the distance between the rings when they strike the ground is 2 4. Two equal uniform rods AB,A C, each of mass m and length 2 a, are smoothly jointed at A, and a light elastic string of modulus 3m a w2and natural length 2a connects B and C. The rods lie on a smooth horizontal table with B and C in contact when each rod is given an angular velocity co about A, but in opposite senses. Show that the rods are instantaneously at rest when the angle 2 0 between the rods is given by 6 sin 0 = 5. 5. A uniform circular disc of radius a and mass m can rotate freely about its axis which is vertical. A thin smooth rod of length a and negligible mass is rigidly fixed to a point Pon the circumference of the disc so that it is inclined at an acute angle a to the upward drawn vertical at P, the rod and the vertical through P lying in a vertical plane tangential to the disc. A bead of mass M slides down the rod from rest at the top; prove that its speed v along the rod when it reaches P is given by m) cosa/(2M costa m). v2 = 2ga(2M 6. A uniform circular disc of radius a, initially at rest on a smooth horizontal plane, is free to turn about a fixed point 0 on its rim. An insect, whose mass is 1/n that of the disc, starts from the other end of the diameter through 0 and crawls with uniform speed v relative to the disc along a chord of the disc through this point which makes an angle of 60° with the diameter. Show that when the insect again reaches the rim the disc has turned through an angle 1,{2/(n + 2)) tan-' )12/(3n, + 6)); and that the angular velocity of the disc is then 2v a(n + 2) 1/3 7. A particle of mass m is attached by a light inextensible string of length a to the end A of a rod OA of length a. The system moves on a smooth horizontal table and the rod is kept rotating about the end 0 with uniform angular velocity a). At a given instant the rod and string are collinear and the particle is at rest. Prove that the time t taken for the angle between the string and the rod to increase to the value x is given by wt = log (tanx/2 seex/2)

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7:7 Moments and Products of Inertia of a plane system In § 7 : 5 and § 7:6 the moment of inertia of a lamina about an axis perpendicular to its plane arose. However, this moment of inertia is only one of a more general, complete set of constants associated with the inertia of a rigid body. At this point, therefore, we give a fuller account of the inertia properties of systems of masses which are distributed in a plane and include a lamina as a special case. The moment of inertia of a system about a line was defined in Vol. I § 6: 7 [see also eqns. (7.36) and (7.37) of this chapter]. For a continuous body the sum was evaluated by an integral, single or multiple, and some of the properties were developed in Vol. I and Vol. II. We defer a complete treatment of the inertia properties of a three dimensional system to Vol. VI of this series ; here we give an account for plane systems, but most of the concepts of the general case are included. Our notation, so far, has been to use ,l(P) to denote the moment of inertia about an axis through P in a direction perpendicular to the plane of motion. This is now replaced by the notation A, B, etc., given below.

Product of inertia If the system consists of n particles, a typical particle at P, (xi , y,), having mass m,, then the moments of inertia about Ox, Oy are A, B respectively where A = Em,0 , B = Em,4 (7.46) The product of inertia, H, of the system w. r. to Ox and Oy is defined by

H = Emi xi y„ Note that, whereas A 0, B (surface) density a (x, y),

(7.47)

0, H may be negative. For a lamina of

H= f f (x, y) xy d x dy taken over the area of the lamina. Clearly, if Ox and/or Oy is an axis of symmetry of the system, then H = 0 .

§ 7:7

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SYSTEMS OF PARTICLES

The parallel axes theorem Suppose that GX, G Y are axes through the centre of mass G y) of the system parallel to Ox, Oy and let He =Em,X, Y, be the product of inertia w.r. to GX, G Y. Then Pi is(X, Yi) where x, =

X„ y, = y

Y

Y,) H = Emi x i yi = Ern,(x X,) (y Y, Emi Xt y Em,X, Y. imiXy But Emi x Y, = x Erni Y, = 0, etc., by definition of G.

H = 11/1cg + Ho .

(7.48)

This is the parallel axes theorem for products of inertia, i.e., the theorem is formally the same as for moments of inertia. Examples. (i) PQRS is a uniform rectangular lamina of mass M and with PQ = 2a, PS = 2b. The product of inertia w.r. to the lines through the centre parallel to the sides vanishes (each line is an axis of symmetry). Hence the product of inertia w. r. to PQ, PS is Mab. (ii) The product of inertia of a semicircular lamina, of mass m and radius a, w.r. to its bounding diameter and the tangent at one end of its bounding diameter is 4Ma2 /3n.

The moment of inertia about any line in the plane Let 0 x', 0 y' be axes in the plane of the distribution which make angles 0, + 0 respectively with Ox, Fig. 95. Then, referred to the axes Oxy, Pi is (xi , y,') where

xi = xi cos 0 + y, sin 0, yi = — x, sin 0 + y, cos 0, (see Vol. I p. 143 and p. 273-4). If primes refer to moments and products of inertia w.r. to the axes Ox' y' ,

A' = Emiyi2= Ern, (y, cos 0 — x, sin 0)2 = Emi g cos2 0 — 2Emi x,y, sin0 cos0 + Emi x„! sin2 0

.*. A' = A cos2 0 — 2H sin° cos 0 + B sin2 0.

.

(7.49)

Similarly

B' = A sin2 0 + 2H sin 0 cos()

B cos20,

H' = (A — B) sin0 cos 0 + H(cos2 0 — sin2 0), i.e.,

H' = 1(A — B) sin2 0

H cos2 0.

(7.50) (7.51) (7.51 a)

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These formulae together with the parallel axes theorems enable the moments and products of inertia of the system w.r. to arbitrary lines to be calculated from the simple results already obtained. Example. The moment of inertia of the rectangular lamina of Example (i) p. 241 about a line /, which passes through P, lies inside the angle SPQ and makes the angle 0 with PQ is

eMb2 cos' — 2 Mab sing cos 0 + 4Ma2 sin2 0.

Ay

FIG. 95. Similarly the product of inertia w. r. to ll and a perpendicular line /2 is

M(b2— a2) sin20

Mab cos2e.

Principal axes of inertia If O x', Oy' are chosen so that H' = 0, then they are called the principal axes of inertia of the system at 0. In this case A', B' are called the principal moments of inertia at 0. Equation (7.51 a) shows that (i) if A B, H 0, the directions of the principal axes are given by tan2B = 2H/(B — A), (ii) if A = B, H 0, these directions are B = r/4, 344, (iii) if A+ B, H = 0, Ox and Oy are the principal axes, (iv) if A = B, H = 0, any pair of perpendicular lines through 0 can be taken as principal axes. Therefore, except in case (iv) [for which the moment of inertia about every line through 0 has the same value], there is one and only one pair of principal axes at any point of the plane.

7:7

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SYSTEMS OF PARTICLES

From eqns. (7.49), (7.50), (7.51) d A' — = dB' 2 H' d0 d0 and this leads to the conclusion that A' ,B' take stationary values when H' = 0, one having a maximum and the other having a minimum, i.e.,

the principal moments of inertia at 0 are the greatest and least moments of inertia at 0. Example. The lines /1and 12of the example of p. 242 are principal axes when tan2 0 —

3ab

2(a2 — b2) •

Then the principal moments of inertia are easily found to be -1M{2a2

2b2 ± V(4a4

01,2

4b4)}.

The momental ellipse Consider the conic whose equation is

A x2 — 2Hxy By2-= 11/c4 ,

(7.52)

where c has the dimensions of a length. This is cut by the line 01, y= x tan 0, at x= r cos 0, y= r sin 0 where r = r2

Mc4

Mc4 A cos2 0 — 2H sin° cos 0 + B sin2 0

S '

(7.53)

where 5 is the moment of inertia about 01. But, unless all the mass lies on 01, > 0 . Hence, excluding this special case, r2 > 0 for all 0 and so all lines through 0 cut the conic (7.52) in real points. Hence this conic is an ellipse which is called the momental ellipse at 0. Since 2r is the length of the diameter of the ellipse inclined at the angle to Ox, eqn. (7.53), written in the form =

M c4 r2

(7.54)

implies that the moment of inertia of the system about any line through 0 is inversely proportional to the square of the diameter of the momental ellipse lying along that line. In particular, if the momental ellipse

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at 0 is a circle (with all its diameters equal), all moments of inertia at 0 are equal. Note that sufficient conditions for an ellipse to be a circle are (1) the ellipse has three equal diameters, or (2) the ellipse has two pairs of equal perpendicular diameters. Example. A uniform lamina is in the shape of a regular polygon. Show that its moments of inertia about all axes in its plane passing through its centre 0 are the same. If the polygon is not a square, by symmetry the moments of inertia about 3 or more lines through 0 must be identical. Hence the momenta' ellipse has at least 3 equal diameters and it follows that it must be a circle. Hence the result follows from eqn. (7.54). When the figure is a square, by symmetry the moments of inertia about the diagonals are equal; so also are the moments of inertia about the perpendicular bisectors of the sides. Therefore case (2) above is applicable and the result follows as before.

Equimomental systems It is usually easier to find the inertia constants (i.e., moments and products of inertia) for a finite number of particles than it is to find them for a continuous body. Hence if two systems Si, and S2 , of which one is simpler than the other, have the same moments of inertia about an arbitrary line, then we can use the simpler one to find the inertia constants of either. Two such systems are said to be equimomental systems. One of the most useful equimomental systems is that given in example (ii) below for a triangle ; by its use any uniform lamina in the form of a polygon can be replaced by a finite number of particles. The necessary and sufficient conditions for two systems Si, and S2 to be equimomental are : (1) they have the same total mass and the same centres of mass ; (2) they have the same principal axes and the same principal moments at their common centre of mass. (These results are true in the case of general three dimensional systems, but, in the restricted circumstances of plane systems, we need only consider axes in the plane of the system. Moments of inertia about any axis perpendicular to the plane of the system are given by the perpendicular axes theorem, and we do not consider any axis inclined obliquely to the plane of the system.)

§7:7

SYSTEMS OF PARTICLES

245

Since the centres of mass coincide, condition (2) shows that they have identical momental ellipses at G [using the same value for c, in eqn. (7.52)]. Therefore, by eqn. (7.54) they have the same moments of inertia about any axis through G. The parallel axes theorem and the equal masses then imply that the moments of inertia about any axis in the plane are equal. Hence the conditions are sufficient. For axes in a given direction in its plane, the moment of inertia of any system is least when that axis passes through its centre of mass. A

Hence the axis for which the moment of inertia has this least value passes through GI and 02the centres of mass of Si and S2 respectively. But this is true for all directions. Hence GI and G, must coincide. The parallel axes theorem now implies that they have the same masses. Further, the equality of moments of inertia for all axes through G means that Si and S2 have the same momental ellipses at G and condition (2) follows. Therefore the conditions are necessary. Examples. (i) A uniform thin rod A B, of mass m and mid-point G, is equimo mental with three particles of masses m/6, m/6, 2m/3 placed at A, B, G

respectively. (ii) A uniform triangular lamina of mass M is equimomental with three particles, each of mass 1M, placed at the mid-points of the sides. To establish this result it is sufficient to show that the triangle and the particles have the same moment of inertia about any line in their plane through B, Fig. 96.

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[We use the fact that the moment of inertia of a triangle, of mass m and height h, about its base is mh2 /6 .] Let the surface density of the lamina be cr. Then

(B X) of AABC = f(BX) of AABX — (BX) of ABC X (1-qB Xa)q216 — (113B Xa)p2 I6 = B X a (q3 — p')/12. But M = lo-BX(q — p).

.•

(B X) of A ABC

M (qz q p p2) .

But f (B X) of the three particles is 1(5 2)2

q+ 2 P )2+ (E21 = `111(q2 (

" p2)

and the result follows. [The result follows similarly by subtraction when BX lies inside the angle ABC.] Exercises 7:7 1. If a lamina has the same moments of inertia about each of two perpendicular axes of symmetry Ox, Oy in its plane, prove that the moment of inertia about any line in the plane of the lamina through their intersection is constant. Hence, or otherwise, find the moment of inertia of a uniform square lamina of side 2a and mass m, about a line passing through a vertex and the mid-point of a non-adjacent side. 2. OPQ is a uniform triangular lamina of mass M whose vertices are at the points 0 (0, 0), P(2a , 0), Q (2 a , 2b), where a and b are positive. Find the moment of inertia of the lamina about the line y = x tan a, and show that the axes through 0 about which the moment of inertia is a maximum and a minimum make angles a and 7c/2 + a with Ox, where tan2a = 3ab/(3a2 — 3. A semi-circular lamina of radius a has AOB as its diameter, and 0 C as the radius perpendicular to AOB. Show that the radius of gyration of the lamina about AC is / 3 4 \ a 1/ k 4 37c ) • 4. A uniform wire, of mass 3m and length 3l, is bent to form three sides AB, BC, CD of a square. Find its moment of inertia about a line through B lying in the angle ABC and making an angle B with BC. 5. Prove that the radius of gyration of a uniform circular lamina of radius r about an axis through its centre making an angle 92 with its plane is zr j/ (1 + sin' 92). Hence or otherwise prove that the moment of inertia of a uniform solid right circular cone about a generator is L MR2 (1 + 5 cos' a), where R is the radius of the base, M the mass and 2 a the vertical angle of the cone.

SYSTEMS OF PARTICLES

247

6. Prove that a uniform rod, of mass m, is equimomental with three particles of masses m/6, m/6, 2m/3, respectively, at the ends and at the centre of the rod. Hence, or otherwise, show that a uniform lamina, of mass M, in the shape of a parallelogram, is equimomental with a particle of mass 4M/9 at its centre, four particles of mass M/36, one at each of the vertices, and four of mass M/9, one at each of the middle points of the sides. Obtain an equimomental system of particles for a uniform solid in the form of a parallelepiped. 7. Prove that a uniform triangular lamina ABC of mass m is equimomental with three particles each of mass -km placed at the middle points of its sides. Prove that the moment of inertia of this lamina about any line in space is

m [h2, +

+11,-P 9700 ]

where h1, h2, h3 and ho are the distances of A, B, C, and the centre of gravity of the triangle from the line. Miscellaneous Exercises VII 1. Two equal particles A, B lie on a smooth horizontal table, joined by a taut inextensible string. A third equal particle C moving on the table at right angles to the string hits it at its midpoint. Prove that in the subsequent motion C does not leave the string before A, B collide. Show also that at the time of this collision C has only one-ninth of its original kinetic energy. 2. Two particles, of masses m1and m2 , are connected by a light elastic string of natural length a and modulus 2, and are placed on a smooth horizontal table with the string at its natural length. The particle m2 is projected with velocity V along the table at right angles to the string. Show that the greatest length of the string during the subsequent motion is the value of r given by the equation rz (r — a) =

m1m2 V2 a(r + a)

(m + m2)

3. AB, BC, CD are three uniform rods lying in a straight line in a smooth horizontal plane, BC being freely jointed to AB at B and to CD at C. The mass of BC is M, the masses of AB and CD are each m and they are of the same length 2a. A constant horizontal force F is applied to BC at its middle point at right angles to its length and continues to act in this direction. Show that, when each of the rods AB and CD has turned through an angle 0(<17c) and the centre of BC has moved a distance x, {(M 2m)x — 2ma0 cos0}2 = 2F {(M 2m)x — 2ma sine} and a {2M + m(1 + 3 sin2 0)}62 = 317 sin°. 4. A uniform circular disc of radius a and mass m is free to rotate in a vertical plane about a smooth horizontal axis through its centre C. An insect A of mass w/20 is at the lowest point of the disc and the whole system is at rest. The insect

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suddenly starts to crawl along the rim of the disc with uniform speed V relative to the disc. Show that the initial spin of the disc is V/(11 a) . If in the subsequent motion 0 is the angle between CA and the downward vertical, prove that gsin0 -= 0. Show that the component of the force exerted by the insect on the disc in the direction perpendicular to the radius is (mg/22) sin 0. 5. Two particles, A and B, of equal mass m are attached to the ends of a light spring which exerts a tension of amount s per unit extension. Initially the particles are at rest on a smooth horizontal table with the spring just taut, and a constant force of magnitude sa is then applied to particle B in the direction AB. Obtain the differential equations for the displacements, x and y, of the particles A and B respectively at time t, and show that y x = +aw2t2, y — x = +a(1 — cos cot), where m co2 = 2s. 6. A uniform lamina is bounded by the parabola y2 = 4ax and the latus rectum of the parabola. Show that, if a principal axis of inertia at an extremity of the latus rectum is inclined to the x-axis at an angle 0, then 20 tan20 + 7 = 0. 7. The perimeter of a uniform lamina, of mass m, is a regular hexagon ABCDEF of side 2a. Find a system of particles equimomental with the lamina and hence, or otherwise, show that the moment of inertia of the lamina about AD is 5 ma2/6. Show also that the moment of inertia of the lamina about any line in its plane passing through the intersection of AD and BE has this same constant value. 8. A uniform rod AB and a string BC lie in a straight line on a smooth horizontal table. The rod AB, of mass M and length 2a, can turn freely about A which is fixed. The string BC, of length 2a, is fixed to the rod at B and has a particle of mass m attached at C. If the particle is projected along the table with velocity V perpendicular to the string, prove that the angle between BC and AB (produced) will lie between M < 12m . + 2 sin–1 M 12 m ' if 9. Two light rods, OA, AB, each of length a, are freely jointed together at A, and a particle of mass m is attached to the rod AB at B. The end 0 is attached to a fixed point of a horizontal table on which the system moves, and the rod OA is made to rotate with constant angular velocity w. If at time t the rodA B makes an angle yi with OA produced, establish the equation co2 sintp = 0. Prove that the tension in the rod AB is

ma (ip2 +2 co fp + w2

w2cosy)) .

10. A system of n particles, each of mass m, moves in a plane, the force between any two of the particles being a repulsion in the line joining them of magnitude ton2 times their distance apart. Initially the particles are held at rest and are symmetrically arranged around the circumference of a circle of radius a and centre 0. The particles are released simultaneously. Find the equations of motion of the rth particle referred to rectangular axes with origin at 0, and prove that, when the distance of this particle from 0 is 4, its speed is V{,unm — a2)).

CHAPTER VIII

THE UNIPLANAR MOTION OF A RIGID BODY 8:1 Introduction In this chapter we discuss the general motion in its own plane of a lamina of constant mass. The methods and tesults are applicable also to a three-dimensional rigid body which moves so that each of its constituent particles moves parallel to a fixed plane. We summarise here the basic theorems of Chap. VII which are most useful in the solution of these problems. 1. The motion of the centre of mass G of a body of mass M is the same as that of a particle of mass M coincident with G under the action of all the external forces transferred to act at G. The motion of G is therefore reduced to particle dynamics [eqn. (7.20)]. 2. The principle of independence of translation and rotation enables motion relative to G to be treated separately from the motion of G. [See eqns. (7.19), (7.22), (7.29), (7.31), (7.42), (7.45)]. As with particle dynamics, in many cases it is also possible (and frequently easier) to use the energy equation first and then use the equations of motion to find unknown forces. The subsequent sections of this chapter illustrate these points.

8:2 The motion of a rigid body about a fixed axis When a rigid body rotates about a fixed axis, so that every point of the body moves parallel to a fixed plane, then the equation of motion about that axis is eqn. (7.44), viz.,

(G)co = T(G) . If, however, the forces are conservative, we can employ the process, described in § 5:4, in which we write down the energy equation, differentiate it to obtain the angular acceleration and then (if necessary)

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write down the equations of motion of the centre of mass in order to determine the internal reactions in the system. The following examples illustrate four fairly typical problems concerning the motion of a rigid body about a fixed axis. Examples. (i) A rigid body can turn freely about a fixed horizontal axis distant h from G, its centre of mass. If the radius of gyration of the body about a parallel axis through G is k, show that the length of the equivalent simple pendulum is (k2 h 2)/h. A uniform rod AB, of mass m and length 2a, has a particle of equal mass m attached to it at the end B. The system executes small oscillations in a vertical plane about the end A. Find the period of these oscillations. Let 0, Fig. 97, be the point in which the axis of rotation intersects the vertical plane in which G moves. When 0Gmakes an angle B with the downward vertical the equation of motion of the lamina about the (fixed) axis is [see eqn. (7.44)] J1(0)0 = —Mgh sine. (1) By the parallel axes theorem 5(0) ---M (h2 k 2) and hence (1) reduces to (k2

FIG. 97.

h2) ••

0 —

g sin()

which is the equation of motion of a simple pendulum (the ESP, see § 5:4) of length (k2 h2)/h. If the body is the rod AB with the particle attached at B, taking moments about the end A when the rod makes the angle 0 with the downward vertical gives ma2 4ma2)o = — mga sin0 — mg 2a sin0, 160.; = —9g sing.

i.e., When 0 is small, so that sing

0, this equation reduces to

6

9g 0 16a

approximately. Thus the motion is simple harmonic with period

2

r /( 16a 9g j

87r a a\ 3 / g

§8:2

THE UNIPLANAR MOTION OF A RIGID BODY

251

(ii) A uniform circular hoop, of mass M, radius r and centre 0, can turn freely in its own plane, which is vertical, about a fixed horizontal axis passing through a point A on its circumference, and is making oscillations in which the diameter through A makes an angle a with the downward-drawn vertical in each of its extreme positions. Show that the least value of the magnitude of the reaction at the axis is iMg V(1 + 3 cos2 a) or 1M (1 — 15 cos2 a), according as a is acute or obtuse. When AO makes the angle 0 with the downward vertical (Fig. 98) the energy equation is (A)62= M g r (cos 0 — cosa). But J (A) = 2Mr2 and hence r02 = g(cos0 — cosa). Differentiating w. r. to t and cancelling 0 gives 2r0 = —gsin0.

(2)

The equation of motion of the centre of mass of the hoop resolved along and perpendicular to A 0 gives the components of the force exerted by the axis on the ring thus: — Mg cos° = Mre2, S— Mg sin0 = Mr (3) Substitution in (3) for 0 and 02 from (1) and (2) gives

R = Mg (2 cos0 — cosa),

S =1-mg sine.

Hence the resultant reaction is of magnitude

P = V (R2 + 52) ---- Mg 11(15 cos2 0 — 16 cos0 cosa + 4 cos2 a + 1) d(P2)

(M g )2

d0

2

sin 0 (16 cosa — 30 cos 0) .

Hence P2 has a maximum value when 0 = 0, and d(P2)/d0 < 0 for 0 < 0 < cos-1(8 cosa/15). But —a < 0 < a and hence, if a is acute so that cos-1(8 cosa/15) > a, the least value of P2 occurs when 0 = a in which case P = zMg 1/(1 + 3 costa). When a is obtuse P2 has a minimum for cos 0 = (8 cosa)/15 and the minimum (least) value of P is then zMg V(1 — 15 cos2 a). (iii) A heavy rod AB, of length 2a and mass m, whose centre of mass G is at its mid-point has radius of gyration k about an axis through G perpendicular to its length. The rod stands vertically on a perfectly rough horizontal plane with A in contact with the plane and is very slightly displaced. Show that if k2 < A will leave the plane before the rod becomes horizontal.

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Show further that if the rod is uniform, and when in contact with a plane of limited friction A slips when the inclination to the vertical is tan-4 (3/4), the coefficient of friction must be 18/49. Let AB make an angle 0 with the upward vertical at time t (Fig. 99). Then assuming that A remains in contact with the plane and does not slip on the plane, so that the rod rotates about A , the energy equation is 5(A)0' = mga(cose — cose), where e is the initial (small) inclination of AB to the vertical.

R

A F FIG. 99.

But

(A) = m (a' + . .•. (a' +

= 2ga(cos 8 - COS 0) .

Differentiating and cancelling 0 gives (a' + k2)0 = g a sine.

(1) (2)

If F and R are the horizontal and vertical components of the force exerted by the plane on the rod, the equations of horizontal and vertical (upwards) motion of G are d' F m do (a sine) = ma (6 cos 0 — 0.2 sine), (3) d'

R—mg—m do (a cos 0) = — ma Co sine + 0.2cos 0) .

(4)

Substitution from (1) and (2) for 62 and O in (3) and (4) gives

F —

raga' sin (3 cos 0 — 2 cos e)

a' ± k2 m g (k2 3a2 cos2 0 — 20 cos 0 cos e) B— a' + dR 2mga'sin0(cose — 3 cose) a' + kz d0

(5)

(6)

§8:2

THE TJNIPLANAR MOTION OF A RIGID BODY

253

Hence 1? has a minimum value when cos 6 = cos s and

mg (k2 — Rmin

a2

a2 cos e) k2

But Rmin cannot be negative and so A leaves the plane (when R vanishes) before the rod becomes horizontal if 1c2 < ja2 cos e. Letting e ---> 0 gives the required result. When the rod is uniform, V = s a2 and (5) and (6) give

F R

3 sin 0 (3 cos 0— 2 cos e) 1— 6 cos 0 cos e+ 9 cos2 0 •

(7)

If A slips when tan 0 = the coefficient of friction u (= FIR when tang =i) is given by letting e-. 0 and putting cos 0 = s in (7). This gives = 18/49. Note. In this example we have assumed that the rod falls from the position 0 = e. The time T taken to reach the position 0 = 13 is obtained by separation and integration of eqn. (1) in the form

T

(12

k2

2ga )

I

d0 f(cos e — cos 0) •

8

N As e —3- 0 this integral diverges at the lower limit like

d0 implying that, as the 0

initial disturbance is made smaller, the time required for the rod to reach a given inclination increases indefinitely, the time being infinite if e 0. This is generally the case when a system is given an infinitesimal disturbance from a position of unstable equilibrium. (iv) A uniform circular disc, of radius a and mass m, is free to turn about a horizontal axis through its centre 0 perpendicular to its plane. Over the rim of the disc hangs a light string which carries particles, A, B, of masses m and 2m respectively, at its free ends. The system is released from rest, and a restoring couple of moment mg a0 acts on the disc in its plane, where 0 is the angular displacement of the disc. Assuming that the string does not slip on the disc, find the acceleration of B when it has descended a distance x, and show that it describes a simple harmonic motion of period n V(14 a/g) . Find also the ratio of the tensions in the two vertical parts of the string in terms of x. Show that to prevent slipping the coefficient of friction ,a must not be less than 1 ( 18 — 5)

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Let T1, T2 be the tensions in the two vertical portions of the string as shown in Fig. 100. Then the equations of motion of the particles A, B and of the disc (about 0) are respectively

T1 — mg —

6,

(1)

2mg — T2 = 2mae,

(2)

(T 2— T1)a — mga0 =

(3)

where Co 0 C = B is the angle through which the disc has rotated Eliminating T,, T 2we find Co 7ct0 = 2g(1 0). The distance through which B has descended is x where 0 = x/a; hence the acceleration, &, of B is given by = 2g(a — x)/a. (4) The solution of eqn. (4) for which x = 0 = z at

t = 0 is x = all — cosn,t),

(5)

where n2 = 2g/7a. Hence B describes S. H.M., of period 27r/n = grV(14a/g) and amplitude a, about a point at a distance a below the initial position of B. From (1), (2), (4) we find T2

T1

10a + 4x 9a — 2x

But, from (5), 0 s x < 2a. Hence 18 10 T <—. < T2, 5 9

(6)

In order that the string shall not slip on the disc the ratio of the tensions must not exceed etu,. 18 017, 5 ' 1 — log (18/5). i.e., Fti 7t Exercises 8:2 1. A thin uniform rod AB, of length 2/ and mass m, is fixed at A and can rotate freely about A in a vertical plane. A circular disc, of mass 4m and radius //3, is clamped to the rod so that its centre is on the rod and it lies in the plane of rotation

§ 8:2

THE UNIPLANAR MOTION OF A RIGID BODY

255

of AB. Show that for different positions of the disc the length of the equivalent simple pendulum varies between 1.951 and 0.84 1 approximately. 2. Prove that the moment of inertia of a uniform circular cylinder of radius a, height 2h and mass M about an axis through its centre of mass perpendicular to the axis of symmetry is Maa2 A compound pendulum of uniform density comprises two cylinders having the same axis and rigidly joined base to base. The upper cylinder is of radius a and height 2h, while the lower cylinder is of radius 2a and height +h. If the pendulum swings about a horizontal diameter of the end of the upper cylinder, prove that the equivalent simple pendulum is of length (15a2 + 77 /0)/39 h 3. A uniform circular disc of mass M and radius a is suspended in air from one end of a diameter so that it can swing (a) about an axis perpendicular to its plane through the point of suspension and (b) about the tangent at this point. In (a), air resistance is negligible, but in (b) it causes a couple resisting the motion of magnitude 2Ma2 0, where 0 is the angular velocity. Show that when 0 is small the equation of motion in case (b) takes the form 50+ 426+ 4(g la)0 = 0. Prove that small oscillations occur in case (b) only if a < 5gI22 , and that the periods of oscillation in (a) and (b) are equal when a = 5 gI6 22 , and find the period of oscillation. 4. A uniform circular disc of mass m and radius a can turn freely in a vertical plane about a horizontal axis through a point 0 on its rim. A particle of mass m is attached to the point of the rim diametrically opposite to 0. The system is disturbed from rest with the particle vertically above 0. Show that the angular velocity of the disc after it has turned through an angle 0 is given by 11a62 = 12g(1 — cos0). Find the magnitude of the force exerted by the disc on the particle when = 703. 5. A uniform rodABof length 2a and mass M is free to turn about a fixed point in AB distant +a from A and has a particle of mass M attached to the end B. The rod is held in a horizontal position and then released. Find the angular velocity when AB is vertical and prove that the force on the point of support is then N Mg. 6. A uniform rod AB, of mass m and length 6a, is held horizontally and in contact with a fixed rough horizontal rail C where AC = 4a. The rod is released from rest. At time t after release and before slipping takes place AB makes an angle 0 with the horizontal. Show that 2a 02 =g sin0. Show also that the rod begins to slip when 0 attains the value tan-1(ta12), where /.1 is the coefficient of friction between the rod and the rail. 7. Two gear wheels A and B, of radii a and b and moments of inertia I, and I 2 respectively, are mounted on parallel axes and run permanently in mesh. When in motion, there are constant frictional torques P and Q on A and B respectively. A constant torque G is applied to A. Find

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(i) the tangential force between the wheels, (ii) the angular acceleration of A, (iii) the number of revolutions made by B in acquiring from rest a speed of N revolutions per unit time. Prove also that, for motion to be possible, G must be greater than P Qa/b. 8. A uniform circular disc of mass nm and radius a is free to turn about a smooth horizontal axis through its centre and perpendicular to its plane. The rotation of the disc is controlled by a spring which is such that a couple 2mga0 is required to twist the disc through an angle 0 radians. A light string, partly wound round the circumference of the disc, to which one end is fixed, supports at the other end a particle of mass m hanging freely. If in is displaced vertically a distance a/3 from its equilibrium position and released from rest, show that the period of its oscillations is 2 (n + 2) al(29)}112. Find the greatest speed of the particle. (It may be assumed that, during the motion, the particle does not reach the level of the centre of the disc.)

8: 3 Systems without a fixed axis of rotation In this section we show the use of the equations of motion, and of the equation of energy, for some simple systems for which there is no fixed axis of rotation about which to take moments. Examples. (i) A railway wagon of total mass M is fitted with four wheels each of mass m, radius a, and radius of gyration lc, freely mounted in pairs on axles whose masses and radii are negligible. A constant horizontal force P is applied to the coupling, and draws the wagon along rough horizontal rails. The wheels do not slip. P is parallel to the rails. Prove that the acceleration of the wagon is Pa2 Mae 4mk 2 • We suppose that F„ F2are the frictional forces exerted by the rails on the front wheels, and that F3 , F4 are the frictional forces on the rear wheels (see Fig. 101). We consider first the motion of the two front wheels, which together with their (light) axle form a rigid system under the action of F1, F2, and other forces which act through the axle. We assume that the centre of mass of each wheel lies on its axle and use eqn. (7.45) for moments about the centre of mass. If f is the acceleration of the truck, the angular acceleration of the wheels is f/a. Therefore, 2m k2 (Ea) = (F1 + F2)a. Similarly, for the rear wheels 2 m k2(Ha) = (F3 + F4) a.

§8:3

THE UNIPLANAR MOTION OF A RIGID BODY

257

Next, we consider the horizontal resolute of the equation of motion of G, the centre of mass of the whole system. This is

4) 111f = P — (F1 + F4 ,1- F2 F = P — 4mk2(f/a2).

f

P a2 111a2 4mk2

Alternatively we can use eqn. (7.32). When the wagon is moving forward with speed I), the applied force P is working at a rate Pv ; the frictional forces are not

doing work since the particles on which they act are instantaneously stationary. None of the other forces do work, being forces on smooth bearings or internal forces of a rigid body. The kinetic energy of the truck, by the principle of independence of translation and rotation, is 2. M f)2 + 44 m k2 (1 a

d

dt

( Ma2 4mk2 _ v 2} = P v.

- d f.) Since f = dt , this will give the required result. (In this problem we cannot find F1, F2, F3 , F4separately without further information about the distribution of mass in the truck. It then becomes a 3-dimensional problem.) (ii) The door of a stationary railway carriage stands open and perpendicular to the length of the train. The train starts off with acceleration f and at the same

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A COURSE OF MATHEMATICS

time the door is given an angular velocity 12 in the direction towards the front of the train, so as to shut the door. Show that, if the door can be regarded as a uniform rectangular plate of width smoothly hinged along a vertical axis, then 22/a,3f Q must be at least of magnitude V(- in order to close the door. 2a The forces which act upon the door in a horizontal plane are the components of force X, Y at the hinge; there are no others; see Fig. 102. (There are, of course, forces in a vertical plane which are in equilr librium.) Since we do not need to know X, Y it is most convenient to take moments about X the line of hinges and so keep X, Y out of the equations. Since this axis has an acceleration we must use eqn. (7.42). The acceleration of G, as shown in Fig. 102, is the sum of two contributions, one the acceleration, f, of the train, and the other from the angular acceleration of the door. Equation (7.42) gives

.M (a • ab+ a cos° • f)

-!,Ma2 0 =0,

Ma (4a 0 + 3f cos 0) = 0.

i.e.,

(1)

Equation (1) is integrated, after cancelling Ma and multiplying by 0, to give

FIG. 102.

2a02

3f sine -=-

where we have used the initial condition 0 = Q when 0 = 0. The door will close if 62 > 0 for 0 a 0 < Zx, i.e., if D > v(3f/2a). Exercises 8:3 1. The total mass of a car, including the wheels, is M. Each wheel is of mass m, radius a and radius of gyration k about its axis. The car is driven on horizontal ground by a torque T (not necessarily constant) applied to the back axle. Find the acceleration of the car, and the frictional force between each wheel and the ground, assuming there is no slipping. 2. A truck, which has four wheels each of mass m, is of total mass M. The radius of each wheel is a and its radius of gyration about its axis is k. The truck is allowed to run down a slope inclined at an angle a to the horizontal, which is rough enough to prevent the wheels from slipping. If axle friction causes a couple L to act on each wheel, prove that the acceleration f of the truck is given by

f

Mg sina — 4L/a 4mk2 /a '

and that the force exerted by the truck on each wheel, parallel to the slope, is

k2) L

ct2 f — mgsina. m(1+ —

§8:4

THE ITNIPLANAR MOTION OF A RIGID BODY

259

3. A uniform rod AB of mass m and length 2a lies at rest on a smooth horizontal plane when a constant horizontal force P is applied to B in a direction making an angle a (< 7c12) with BA. Prove that when the rod makes an angle 0 with the direction of P ma02 = 6P(cosa — cos0). Prove, also, that the rod oscillates with amplitude 2 (n — a) . 4. A uniform straight rod of length 2/ moves on a smooth horizontal plane, one end of the rod being constrained to describe a circle of radius a with constant angular velocity w. Prove that the angular motion of the rod relative to the radius through the point of constraint is the same as that of a simple pendulum of length 4g//3a ca2 performing oscillations under gravity in a vertical plane.

8:4 Conservative systems of forces In this section we illustrate the use of the energy equation for motions in which there is no fixed axis of rotation. (The method was suggested on p. 249 for bodies rotating about a fixed axis). It is a general method, which was first introduced in § 5:4 for particle dynamics. In order to write down the energy equation we need the expression for the potential energy and can use any of the expressions derived in § 7:5 for the kinetic energy of a lamina, viz., + .1,5(0)(02 , 0/( i2 + logo) , 15(0(.02 . Examples. (i) A uniform rod AB, of length 2a and mass m, is held at rest with the end A in contact with a smooth vertical wall and the end B in contact with smooth horizontal ground. The inclination of the rod to the vertical is a. (i > a > 0). If the rod is released, show that the rod leaves the wall when its inclination 0 to the wall satisfies cos 0 = 3 cos a . Discuss the subsequent motion. Suppose that at time t after release the ends A, B of the rod remain in contact with the wall and ground respectively and AB makes an angle 0 with the vertical, Fig. 103 (i). Then the instantaneous centre of AB lies at the intersection of horizontal and vertical lines through A, B respectively and the kinetic energy is 5 (I) 02 ma2 02. Hence the energy equation is 49na2 02 = mga(cosa — cos0), i.e.,

2a02= 3g(cosa — cos 0).

(1)

Differentiation w. r. to t and cancellation of 0 gives 4a0 = 3g sin0.

(2)

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A COURSE OF MATHEMATICS

If the normal reactions at A, B are R, S respectively, then the equations of motion of the centre of mass G of the rod are

R=m S — mg = m

d2

dt2

(a sine) = ma (0 cos0 — (52 sine) ,

d2

(a cos0) = — ma (0 sin 0 + 62cos 0) . dt2

Substituting from (1) and (2) we find

R= m g sin 0 (3 cos — 2 cos a), S =

(3)

(1 — 6 cosa cos 0 + 9 cos2 0).

(4)

Equation (1)—(4) hold only for the period during which A and B remain in contact with the wall and ground respectively. But

R > 0 for cos a „,>, cos 0 > cos a , = 0 for cos 0 = cosa , < 0 for cos0 < cosa. Also S =

g{ (3 cos 0 — cosa)2

sin2 a} > 0.

Hence, when cos 0 =1 cos a , A leaves the wall but B remains in contact with the ground. At the moment when A leaves the wall the angular velocity of the rod is w, where, from (1), a we = +g cosa. ( 5) Also, at this instant the horizontal component of the velocity of 0 is 2

V = {— (a sine)} = a 6 cos o evaluated when cos° = cosa. dt V=

cosa.

(6)

In the subsequent motion, Fig. 103 (ii), the only external forces acting on AB are its weight, acting vertically downwards through G, and the normal reaction, T, at B. Then the equation of horizontal motion of 0 implies that the horizontal component of velocity of 0 remains constant (= V). This result also follows from the principle of conservation of linear momentum (§ 7:4 and § 7:5). Suppose the rod now makes the angle ry with the vertical at time t and that B remains in contact with the ground. Since in the transition from one motion to the other (R vanishing) no work is done on the system, the total energy in the new motion is the same as in the former motion. Since in the former motion the rod started from rest, this total energy is equal to the initial potential energy. 1 ma2 2 .•. m [V2 + ddt (a cos q))} ± 2 3 (p2 ]

mga cos (p = mga cosa.

§8:4

261

THE TJNIPLANAR MOTION OF A RIGID BODY

Using (5) and (6) we find 3a(1 + 3 sin' g)) 02 =2g(9 cosa — 9 cos()) — cos3 a)

( 7)

and, by differentiation w.r. to t,

+ 3a sinq9 cos 99 GP = 3g sinT. (8)

a(1 ± 3 sin2 0

The equation of vertical motion of G is

T—mg—m

d2

(a cosy) = — ma (0 sin 0 + 192 cos q9) ,

B FIG. 103 (ii).

FIG. 103 (i).

whence, after use of (7) and (8) and some reduction,

T=

mg (12 — 18 cosy cosa + 27 cos2 +2 cos y cos3 a) 3(1

+ 3 sin2 02

ma {9 + 27 (cos 99 — A cosa)2 + 3 sin2 a + 2 cos y cos3 a) 3 (1 + 3 sine

>0

and the rod remains in contact with the plane for cos a > cos 92 0. When A strikes the plane, i.e., when y = In, the angular velocity of the rod is Q. where, from (7), g(9 cosa — cos3 a). 6a S22 We do not discuss the subsequent motion of the rod. (ii) Two uniform rods AB, BC of masses m and m' respectively are smoothly hinged to one another at B and to a horizontal table at C, so that they can move in a vertical plane. Initially BC is on the table and AB is set in motion (in a sense tending to increase the angle ABC) when it is vertically above B, with angular velocity Y(3 nag), where 1 is the length of AB. Show that so long as BC remains on the table

102= 3g (a

1 — cos0) ,

where 0 is the angle made by BA with the vertical.

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A COURSE OF MATHEMATICS

Show also that the vertically downward component Y of the reaction at B on the rod AB is Y= mg {6(1 n) cos0 — 1 — 9 cos2 0}. When n = 2/3, show that the condition that BC does not rise from the table is m' Sm/9. Suppose that BC remains on the table and AB makes the angle 0, where 0 < 0 < in, with the upward vertical at time t. [Fig. 104 (i).] Then AB rotates about a fixed (horizontal) axis through B and the equation of energy for AB is 1 m 12 ._

1

3 02 + g 2 cos0 —

1 m12 3ng

2

3

+ mg — 2

which reduces to 162 = 3g(n + 1 — cos0).

(1)

The equation of motion of the centre of mass of AB, resolved vertically upwards, is d2 ( 1 — — Y —nagg — m 1 cos0) nz/ (Osin0 62 cos0) (2) dt 2 2 2 [Fig. 104 (ii) shows the internal forces acting at the hinge B.] But differentiation of (1) w.r. to t gives

210 = 3g sinO, and so substitution from (1) and (3) in (2) gives after some reduction Y= mg(6(1

+ a) cos° — 1 — 9 cos2 0}.

(3)

§8:4

THE UNIPLANAR MOTION OF A RIGID BODY

263

The rod BC will rise from the table if the moment about C of the vertical component of reaction exceeds the moment about C of the weight of BC. Hence the condition that BC remains on the table is m' g

Y

,

(4)

where l' is the length of BC. When n = ,2 , Y = img(10 cos° — 1 — 9 cos2 0) = 1 mg {V — (3 cos0 — 021. Hence cos 0 = 5/9 gives the greatest value of Y for 0 e 0 < 17C and Ymax = 4m g/9 . In this case the condition (4) that BC remains on the table becomes 9m' a 8m. (iii) A uniform solid circular cylinder, of mass m and radius a, rolls without slipping on the inside of a fixed rough hollow circular cylinder of internal radius b (b > a) which has its axis horizontal. The angular velocity of the moving cylinder in its lowest position is .0. Show that it will roll completely round the inside of the hollow cylinder if 3a2 D2 11 g (b — a) . Find the force of friction between the two cylinders. Suppose that the moving cylinder remains in contact with the fixed cylinder and that C in Fig. 105 is a point of the moving cylinder originally in contact with a lowest point A of the fixed cylinder. Then, as shown on p. 125 the angular velocity of the moving cylinder is w where (A o B = 0) w-= —

= (b — a) o/c,.

Since B is the instantaneous centre of the moving cylinder, the kinetic energy of this cylinder is (B) co2 = m (b — a)2 02. 1 3ma2 But the kinetic energy of this cylinder at its lowest position is2 02 and 2 since there is no slipping the forces are conservative; the equation of energy gives lm(b — a)2 0.2— mg (b — a) cos 0 = Ima2 02 — mg (b — a), which reduces to (b — a)2 02 = a2 02 — g (b — a) (1 — cos 0) .

(1)

If R is the component along B 0 of the reaction between the cylinders, the resolute along B 0 of the equation of motion of the centre of mass of the moving cylinder is B— mg cos 0 = m(b — a) 02. (2) From (1) and (2) we find a2 02 R—mb— a

1 g (4 — 7 cos 0)}. 3

(3)

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From (1) and (3) we see that 02 and R both decrease as 0 increases from 0 to n. Hence, for 0 < 0 < n,

62

0 if 3a2 122

R > 0 if 3a2 i22

7 g (b — a), 11g(b — a).

But the moving cylinder will roll completely round the fixed cylinder if 02 .> 0 and R > 0 for 0 < 0 < n. Hence the required condition follows. [Note that it is

A FIG. 105.

not sufficient that the cylinder should have enough energy when in its lowest position merely to raise the centre of mass a height 2 (b — a).] In the position shown in Fig. 105 the resolute perpendicular to 0 D of the equation of motion of the centre of mass of the moving cylinder is

F — mg sin0

m(b — a) 6.

But differentiation of (1) w.r. to t gives 3(b — a) o = —2g sinO, and so substitution in (4) gives

F = m g sin 0 .

(4)

§ 8:4

THE UNIPLANAR MOTION OF A RIGID BODY

265

(iv) A uniform rod AB, of mass M and length 2a, freely pivoted about the end A, is released from rest in a horizontal position. Show that, when the rod is inclined at an angle 0 to the horizontal, the tension T and shearing force F in it at a point P at a distance x from the pivot are given by

T—

(2a — x) (10a + 3x) 8a2

F

(2a — x) (3x — 2a) Mg cos0. 16a2

Mg sinO,

Since the rod (Fig. 106) is freely pivoted, there is no frictional resistance and the energy equation can be used. Hence 1

2

4Ma2

•

3

0.2 = Mga sin O.

.•. 2a02 = 3y sin0.

(1)

4a0 = 3g cos0.

(2)

Differentiation w. r. to t gives

The mass m of the portion PB is given by m =M(2a — x)/2a.

(3)

The equations of motion of the centre of mass G of PB along and perpendicular (in the sense of 0 increasing) to GA are

T — mg sin0 = mGA02,

(4)

mg cos0 F = mGAO,

(5)

where GA =1(2a x). Substitution from (1), (2) and (3) in (4) and (5) gives the required results.

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Exercises 8:4 1. A uniform solid circular cylinder, of radius a and mass m, is placed in contact with the highest generator of a fixed rough circular cylinder of radius 2a, whose axis is horizontal, and is slightly displaced. If 0 denotes the angle which the plane containing the axes of the cylinders makes with the vertical at time t after the start of the motion, show that, so long as the motion is one of rolling, 9ao2 = 4g(1 — cos8). If the coefficient of friction between the cylinders is µ, show that slipping begins when sin 0 -= p (7 cos 0 — 4) . 2. A uniform sphere of radius a is placed on a fixed rough sphere, also of radius a, and is released from rest when the line of centres makes an angle a with the vertical. If the surfaces are sufficiently rough to prevent slipping, show that in the ensuing motion the moving sphere will leave the fixed sphere when the line of centres makes with the vertical an angle cos—'x, where 17x = 10 cos a, and that the corresponding angular velocity of the moving sphere is then 1/(2 gxla). 3. A uniform circular disc of mass 2m and radius a has a particle of mass m fixed to the circumference. The disc is projected, with its plane vertical and the particle initially in its highest position, so as to roll without slipping on a horizontal rail. Prove that, when the radius to the particle makes an angle 0 with the upward vertical, a02 ={7a..(22 + 2g(1 — cos 0))/(5 + 2 cos()) where .S2 is the angular speed of projection. Hence, or otherwise, find the vertical reaction on the rail when the disc has turned through one right angle. 4. One end A of a uniform rod AB of mass 3m and length 2a is freely pivoted at a fixed point on a smooth horizontal plane. The other end B of the rod rests against a face of a smooth uniform cube of mass m free to move with one face in contact with the plane, the vertical plane through AB being a plane of symmetry. If the system is released from rest, show that in the subsequent motion 2a02 (1 + sin2 0) = 3g(sina — sine), where 0 is the inclination of the rod to the horizontal and a is the initial value of O. Show also that the initial reaction at B is 3mg sina cosa 2(1 + sin2 a) 5. A heavy uniform solid sphere of radius ; a rolls without slipping on the inside of a fixed rough hollow cylinder of radius a, whose axis is horizontal, so that a diameter AB of the sphere always lies in the same vertical plane. Initially the centre C of the sphere is at the lowest point and is moving with a velocity 11(nag). If C comes to instantaneous rest when the perpendicular from C to the

§ 8:4

THE UNIPLANAR MOTION OF A RIGID BODY

267

axis of the cylinder is inclined at an angle cos-1(8) to the downward drawn vertical, prove that n, = Find also the value of the frictional force at that moment. 6. A uniform circular hoop, of radius a, mass m and centre 0, has a particle, also of mass m, attached to a point A of its rim. It rests on a rough horizontal plane with its own plane vertical and with A as the lowest point. If it is set in motion so as to roll in its own plane with an initial angular velocity co , prove that, when 0A makes an angle 0 with the downward vertical, (2 — cos0) ct02= a co' — g(1 — cos0)

.

Assuming that contact with the plane is maintained, prove that, if a co2 > 2g,

A can rise to a position vertically above 0 and that it will then have an acceleration (a co' — 2g)/3 downwards. 7. A uniform rod is held inclined at 30° to the vertical with its lower end in contact with a rough horizontal plane; if the rod is released from rest and falls under gravity, show that it will not immediately slip on the plane if the coefficient of friction exceeds (27)1/2 /13. 8. A uniform rod AB, of mass m and length 2a, has a particle of mass m fixed to the end A and is placed on a smooth horizontal table. If the rod is given an angular velocity 21l(g/a) about a horizontal axis passing through A so that the end B rises, prove that when the rod is inclined at an angle 0 to the horizontal and A is on the table a 52 (8 — 3 sin2 0) = 4g (8 — 3 sin 0) . Prove that the end A will leave the table when sin0

(4 — V10) .

9. A smooth solid cube of mass M and edge a rests on a smooth horizontal table. In the plane through the centre and parallel to two vertical faces of the cube, a uniform rod, of mass 1r M and length 4a, is placed with one end on the table and the other end leaning against a vertical face of the cube just at its highest edge. If the system is released from rest in this position, prove that at a subsequent instant when the rod makes an angle 0 with the horizontal and is still in contact with the cube,

02

3g(1 — 4 sins) a(16 — 3 sin2 0)

Show also that the rod and cube separate when sin0 has the value determined by the equation 3 sin3 0 — 48 sin 0 + 8 = . 10. A uniform straight rod of length 1 and mass m is free to turn about one end. Show that the angular velocity co with which it can rotate, making a constant angle a with the downward vertical, is given by 21w2 cosa = 3g,

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and that the tension T, shear S and bending moment M at a point distant from the fee end are given by

T

mg {1 + 2 3 (1

S

mg x (1 /

M

— mg x2 4/

I1

—

x 21

x

tan'al cos a ,

3x )sina, 2/

2

/ sincx.

8 :5 Rolling and sliding motion In this section we discuss the motion of rough bodies such as spheres, cylinders and hoops on rough planes. We consider only the motion of a body of circular cross-section with its centre of mass at the centre of a circular section made by a plane of symmetry. In such motions, where there is relative slipping of rough surfaces in contact, the energy equation cannot be used; on the other hand, when rolling takes place, the energy equation can be used. The type of motion we consider is well illustrated by a billiard ball moving in a straight line on a billiard table. The ball may be projected with or without an angular velocity about a horizontal axis perpendicular to the plane of motion; further, the rotation may be back-spin or top-spin. (We do not consider rotation about any other axis.) The friction of the table may increase or decrease the angular velocity as the motion proceeds, and in most cases a state is reached when the speed of the centre and the angular velocity are such that the ball is instantaneously rolling. Does it continue to roll? Or does it start to slip in a different manner? The examples below indicate a fairly systematic method of solving problems of this nature by considering the various possibilities which arise depending on whether the ball continues to roll or whether it slips once again but in a different manner. If this `billiard-table' type of motion takes place on an inclined plane, gravity is an additional factor to be included. When the character of the motion changes from sliding to rolling, or vice-versa, in general the frictional force alters discontinuously. The main features to be borne in mind for the solution of such problems can be summarised thus. l.. If the body rolls, then there is a condition for rolling and, in this case, the frictional force, F, is unknown; F must be determined from

j S:5

THE UNIPLANAR MOTION OF A RIGID BODY

269

the equations of motion. Further, since rolling takes place, F must not exceed limiting friction. 2. If the body slips, there is no condition for rolling to be satisfied, but the frictional force takes its limiting value. The direction of the frictional force acting on one of the surfaces in contact is opposite to the direction of the motion of that surface relative to the other surface. 3. The conditions of projection enable the nature of the motion to be determined in the initial stages. Example. A uniform hoop with centre 0, of mass m, radius a, and moment of inertia mk2about its axis, is projected with its plane vertical along a rough horizontal table. The coefficient of friction between the hoop and the table is a . The initial spin of the hoop is ,S2 and its centre has an initial speed V. Discuss the subsequent motion in the three cases (a) V = aS2, (b) V > aS2, (c) V < aS2. [By using mk2as the moment of inertia the results can be made to apply to other bodies with circular section by choosing the appropriate value for k, e.g., for a sphere k = a V-:2, .] Without loss of generality we suppose that V a 0 throughout this example. (a) In Fig. 107 (i), since V = aS2 the hoop is rolling at the instant of projection. Two possibilities arise; either the hoop will continue to roll or slipping will commence. Suppose that rolling continues and that at time t the situation is as shown on the right of Fig. 107 (i). Since we have assumed rolling, the condition for rolling is = a 0. (1) Taking moments about the point of contact gives

ma

mk2 6 = 0.

ma(1

k2 /a2) = 0.

Using (1) this gives .•. = ao = O. The equation of motion of the centre 0, when resolved horizontally and vertically, gives (2) F = mth -- 0, R — mg = O. Hence the assumption of continued rolling implies that the hoop does so with constant velocity V (x = 0) and that F = 0. The only alternative assumption, i.e., that slipping commences, implies that F has the limiting value pi? = ktmg. But the laws of friction (§ 2:1) imply that the magnitude of the frictional force brought into play is the least possible to prevent relative slipping. Hence we must conclude that rolling continues indefinitely. If at any stage of the motions to be discussed in cases (b) and (c) rolling takes place instantaneously, the above result allows us to assert that rolling must continue from that point onwards.

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(b) In this case, V > a9, the point of contact with the plane initially slips to the right. Hence the initial frictional force on the hoop is limiting and has magnitude ,aR (= ,amg) and acts to the left. After time t the situation is shown in Fig. 107 (ii). During the period for which the point of contact is slipping to the right, i.e., whilst u — ay > 0, the equations of motion of 0 and of moments about 0 are my = mumg, mkt 0 = ,umga. (3) Using the initial conditions z) = V, = 9, after integration we find

aS2

= V— u j g t,

,ug(a21k2)t.

(4)

••• j— a q1=V —aS2 — pg(1+ a21k2)t.

(5)

Equations (3), (4), (5) hold only so long as j — aq7 > 0 and (5) shows that — a ip decreases as t increases. The equations cease to hold when zj — a (p = 0, i.e., when t = T1, where

T1— 1

V — a S2 ,ug(1-1- a2/k2) •

(6)

At this instant = V1 —

a S2 + V (a21k2) 1 ± a2 /k2

(7)

'

and the hoop rolls instantaneously. The analysis of case (a) then implies that for

t> T1the hoop continues to roll with velocity V1. At the instant t = T1 the frictional force changes instantaneously from acting to the left to zero. A further integration of eqns. (4) shows that when rolling commences

1 (V — aS2) [V(1 +2a2 /k2) y = VT 1 — — ,agT ,2 = 2 2ag(1 a2/k2)2

aS2] •

(8)

If the hoop starts with a back-spin, S2 is negative, and if it has a sufficiently large numerical value so that V1< 0, i.e., aS2 + V (a2/k2) < 0, when rolling commences the hoop is rolling to the left. Further, if a S2+ V(1+ 2a2 /k2) < 0, the hoop is to the left of its starting point when rolling commences. (c) In this case, V < aS2, the point of contact with the plane slips to the left initially, and, whilst t — aye < 0, the frictional force on the hoop is y mg to the right and the situation is as shown in Fig. 107 As in the above cases the equations of motion are

mz

= ,a mg, m7c2 ip = —,umga.

(9)

.•. I = V + ,ugt, afp = aS2 — pg(a21k2)t.

(10)

.•. alp — I = aS2 — V —,ug(1± a2 /k2) t.

(11)

Equations. (9), (10), (11) are only valid whilst a instant t= T2, where

T,—

aS2 — V ug(1+ a2/k2)

— I > 0, i.e., up to the (12)

§ 8:5

THE UNIFLANAR MOTION OF A RIGID BODY

FIG. 107 (iii).

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272

A COURSE OF MATHEMATICS

when z has the value V2 —

a 52

V (a2/k2)

1 +colk2

.

(13)

At this point the hoop rolls instantaneously and therefore continues to do so with velocity V2 > 0, to the right. As before, the frictional force is discontinuous at this point.

The above example indicates that whatever be the conditions of projection of an axially-symmetric body with circular cross-section on

FIG. 108. a rough horizontal plane, the body eventually rolls with uniform velocity. Suppose now that such a body rolls on an inclined plane. If at time t the point of contact is at a distance x up the plane (see Fig. 108) measured from an arbitrary origin, the energy equation is lm ±2 plik2(t/a)2

mgx sing = E.

On differentiation this gives g sinoc + k2/a2

(8.1)

The equations of motion of the centre of mass are R = mg cosoc,

F — mg sinoc = mX.

F—

mg sin g a2/k2 1 4•

(8.2)

§ 8:5

THE UNIPLANAR MOTION OF A BODY

Rum)

273

But since rolling is taking place tan a 1 + a2/102

R

Hence it is possible for a body to roll for more than an instant provided that tan g < p,(1 a2/102). (8.3) When this condition is satisfied the body rolls on the plane, its centre having the uniform downward acceleration given by (8.1). When condition (8.3) is not satisfied the body can only roll instantaneously on the plane. The combination of a steep plane and a high moment of inertia is such that the maximum (limiting) frictional force cannot produce the required angular acceleration. (The situation is similar to that in particle dynamics in which a particle can only rest in equilibrium on an inclined plane for which tan oc s .) Any motion on a plane whose inclination satisfies condition (8.3) eventually becomes one of rolling on the plane, the centre of the body having the uniform downward acceleration given by eqn. (8.1). This result will not be proved since it involves the tedious detail of considering all the possible conditions of projection. The methods used for the horizontal plane (p. 269) should be used in any given problem; condition (8.3) determines the nature of the eventual motion as illustrated in the first two of the following examples. Examples. (i) A uniform solid sphere, of mass m and radius a, is projected down a line of greatest slope of a rough plane of inclination a. Initially the sphere has no angular velocity and its Centre has speed V. If p = s tan a, prove that the sphere will slide at first and that after time 4 V/(3g sin a) has elapsed it will roll down the plane. Initially, and as long as y — a q9 > 0, (see Fig. 109), the point of contact of the sphere slips down the plane. The lower position of the sphere in Fig. 109 shows the position after time t whilst y > a I). The equations of motion are

R= mg cos a, mg sin a — pR -= When ,u = +tan ,

g

sina,

.•.

— a

a cp =

=V—

pRa =-;.='; ma2 g5.

(1)

g sina.

(2)

sin a .

(3)

This shows that y — a cp decreases as t increases until at time T = 4 V 1(3g sin a) the sphere rolls instantaneously and eqns. (1) cease to hold. Condition (8.3), with k2 = a2, shows that rolling can take place if tana < 7, p =

tana

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A COURSE OF MATHEMATICS

(in this case). Since sliding requires a (limiting) frictional force +mg sina and rolling requires a frictional force Ong sina (<4-mg sina), [see eqn. (8.2)], rolling takes place when t> T. (ii) The sphere of example (i) is projected up a line of greatest slope with angular velocity S2, in the sense which would take it up the plane, and linear velocity V, where V > aS2. Determine the motion if ,a = + tan a. If Fig. 108 (with 0 instead of and F = pR acting down the plane) represents the state of affairs after time t whilst ic > a0, i.e., whilst the point of contact is

slipping up the plane, the equations of motion give, as in the previous example (i), =— sina, a 0=1 4 g sina. — ao = V — aS2 — gtsina.

(1)

The equations of motion, and eqn. (1), apply until time T1where 2 ( V — aS2) T1— 3g sina

when the point of contact ceases to slip. The sphere rolls instantaneously at this point; since, in this case, (1 + a2 /k2) = -itan a condition (8.3) is not satisfied and rolling cannot continue. If the limiting frictional force continued to act down the plane, eqn. (1) would continue to hold; this would imply that ,aR acted on the sphere in the same direction as the slipping of the point of contact of the sphere relative to the plane. Hence the frictional force must act up the plane (i.e., the force of friction changes its direction discontinuously at t = T1.) If we use y and 92 for the (upward) linear displacement, and angular displacement after the instant t = T1, the equations of motion are

mij = —mg sina

2 G)5 = —pRa,

(2)

§8:5

THE UNIPLANAR MOTION OF A RIGID BODY

275

where, R - - va y cos a .

= — g sin a , a (1) = — 14 y sin a .

(3)

Hence, for t > T1,

= V,— g(t — TO sin a , aq9 = a [2, — ti g(t — T,) sina, —

= V,— an, —

— 111) sina,

where V1 and Q1are the linear and angular velocities when t = T1, at which instant V1 = a Q1. (4) sina. .•. — aq7 = Equation (4) shows that the velocity of the point of contact relative to the plane is downwards and that its magnitude increases indefinitely as t increases. A 7 V1 after the closer investigation shows that the sphere ascends for a time 6g sin s instant of rolling, and then descends. (iii) A long plank of mass 3m rests on a smooth horizontal table. A rough sphere, of radius a and mass m, spinning about a horizontal diameter with angular velocity Q, is placed gently at one end of the plank so that its centre is initially at rest. The spin tends to take the sphere along the length of the plank and the coefficient of friction is ,u. Prove that, when slipping ceases, the angular velocity of the sphere is 8 Q/23 and find also the velocities of the plank and the centre of the sphere. Prove also that the sphere has turned through an angle 93a Qz/(529,u g) .

0•

• —:* x)F

Fie. 110. Initially the lowest point of the sphere has a velocity a Q to the left and hence in the early stages of the motion the frictional force F between the sphere and the plank is limiting, is of magnitude u j mg and acts to the right on the sphere and to the left on the plank as shown in Fig. 110. Suppose that at time t the plank has moved a distance y to the left, the centre 0 of the sphere has moved a distance x to the right and the sphere has rotated through an angle B about the given horizontal diameter. Then the equations of motion of the plank and sphere are 3 mi/ = itmg,

(1)

= ,umg ,

(2)

m

;ma2.6 = —,umga.

(3)

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A COURSE OF MATHEMATICS

Using the initial conditions -- 0 y, 0 z -S2 at t 0, eqns. (1)-(3) integrate to give = pat, t = pgt, ae = a S2 — 5pat12. (4) But the velocity of the lowest 'point of the sphere to the left relative to the plank is v =a0—x—y=aS2- 23,agt/6

(5)

and eqns. (1)-(5) hold only for v > 0. Hence slipping ceases when v = 0, i.e., when t = T where 6a T— (6)

23,ag

As before it can be shown that for t > T the sphere rolls on the plank with constant speed V and angular velocity to; the force of friction vanishes. The plank also moves with constant speed U. From eqns. (4) and (6)

V = pgT —

6a S2

23

1 2a S2 8 5pgT , U = pgT — 23, w = S2 — — . 2a 23

Integration of the last of eqns. (4) gives

ctO = a S2t — 5pgt 214 so that when slipping ceases (t = T) the sphere has rotated through an angle 93a122 /(529pg). Note that despite friction the linear momentum of the whole system is constant (zero) but the total energy has decreased. Exercises 8:5 1. A uniform solid sphere is projected along a rough horizontal plane with velocity V and no spin. Show that it skids for a distance 12 V2 /49µg, where p is the coefficient of friction, and that it subsequently rolls with speed 5 V/7. 2. A uniform solid circular cylinder of radius a is projected along a rough horizontal table; the coefficient of friction between the cylinder and the table is p. The initial velocity of the axis of the cylinder is V and the cylinder has a backward spin 9. Show that slipping ceases after a time ( V + aS2)13pg and the cylinder then rolls with constant velocity (2 V — a 9)/3 . Show also that if the cylinder has returned to its starting point when slipping ceases, then 5 V = a 9. 3. A uniform disc, of mass m and radius a, is rolling with speed V and with its plane vertical, on a rough horizontal rail. If a braking couple ipamg is applied, prove that skidding will follow and that rotation will cease at time 3 V 14pg while the disc is still moving forward. The coefficient of friction between disc and rail is 14. 4. A uniform solid rough sphere of radius a and mass m rests on a uniform rough board of mass which lies on a smooth horizontal table. If the board is projected with a velocity v along the table, show that the sphere will slip on the board for a time 2 M vl{pg (7 M 2m)}, where p is the coefficient of friction be-

§8:5

THE UNIPLANAR MOTION OF A RIGID BODY

277

tween the sphere and the board. Find the velocity of the board at this instant and the velocity with which the sphere starts to roll. 5. A uniform circular cylinder of mass m and radius a rests on a rough horizontal plane with its axis horizontal, and is in smooth contact along a generator with a vertical face of a uniform rectangular block of height 2a and mass lm which also rests on the plane. The cylinder is removed and gently replaced with a spin S2 in the sense tending to maintain contact between it and the block. If ja is the coefficient of friction at either contact with the plane, show that the cylinder slips on the plane for a time 3a S217 u j g and that its velocity and that of the block is then a S217 . Show also that the cylinder subsequently rolls on the plane and that the motion ceases after a further time 4a S217 pg 6. A uniform rough solid cylinder of radius a, rotating with angular velocity w about its axis which remains horizontal, is placed gently at the foot of a rough plane inclined at an angle /3 to the horizontal, the direction of rotation of the cylinder being such as to cause it to move up the plane. If the coefficient of friction between the cylinder and the plane is (5 tan(3) /3 , show that the cylinder ceases to slip after a time a w/(4g Find the distance moved by the axis of the cylinder before slipping ceases. 7. A uniform solid circular cylinder of mass m and radius a moves with its axis horizontal up a rough inclined plane of inclination a by means of a constant couple L acting so that the cylinder rolls up the plane. Show that if the coefficient of friction between the cylinder and the plane is a then for rolling to take place tana

(2L sec a)/(3 mag).

If this condition is satisfied and L = 2mag sina and moreover, the cylinder starts from rest at time t = 0, find its displacement at time t. 8. A uniform circular disc, of radius a, is projected, with its plane vertical, along a line of greatest slope of a plane of inclination a to the horizontal. The initial velocity of the centre of the disc is V down the plane and the initial angular velocity is S2 (in the sense in which the disc would roll up the plane). If the coefficient of friction between disc and plane is 2 tan a, show that slipping ceases after a time ( V + a S2)I(5g sina) , and that, if the disc has returned to its original position, V = a S2/9. Show also that the disc then rolls on the plane, coming to instantaneous rest after a further time 3 V/(2g sin a) . 9. A uniform solid sphere is projected up a line of greatest slope of a plane inclined at an angle a to the horizontal. Initially the sphere has no angular velocity and its centre is moving with velocity V up the plane. If the coefficient of friction between the sphere and the plane is kt, show that slipping ceases after a time 2 V1{g (2 sina + 7,u cos a)) and that the sphere subsequently rolls without slipping if y > tana. Show in this case that the sphere will begin to move down the plane after a total time V 1(g sin a) . 10. Two circular cylinders, both of radius a, are held in equilibrium on an inclined plane of inclination a so that their axes are horizontal and their surfaces are in contact along a common generator. The upper cylinder is of uniform density

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and of mass m, and the lower cylinder is of the same density, but hollow, with internal radius la. The contact between the cylinders is smooth and the other contacts are perfectly rough. If the cylinders are simultaneously released, show that they will roll down the plane with the same linear acceleration (56 g sin a) /87, and that the reaction between them is (mg sin a)/29 . 11. A uniform hollow circular cylinder of mass m and radius a is placed with its axis horizontal on a plane inclined at an angle a to the horizontal. A uniform solid circular cylinder of mass 2m and radius a is placed at a higher level on the plane so that it touches the hollow cylinder along a generator. The system is released from rest. If the contact between the cylinders is smooth but the contact between each cylinder and the plane is rough with coefficient of friction µ, show that motion of the cylinders in contact down the plane is possible such that the hollow cylinder always slips while the solid cylinder always rolls, provided tana > > Lana. 12. A uniform sphere of radius a is projected with velocity V down a rough inclined plane of angle a and at the same time is given an angular velocity Q tending to make it roll up the plane. Prove that the sphere comes to a position of instantaneous rest if the coefficient of friction µ = n, where n

tana, (1

5V 2a122

and 5 V < 2aD. Show further that it will turn back ifµ > n. 13. A wedge, of mass M, rests with one face on a horizontal table and one of its faces inclined at an angle a to the horizontal. A uniform sphere, of mass m, is projected without rotation down a line of greatest slope of the inclined face. The angle of friction between the wedge and the table and between the wedge and the spheres is 2. Prove that, if 2m cos a sin(a — 22) > M sin22, then the wedge will move and that its acceleration, while the sphere is sliding, is {2m cos a sin(a — 22) M sin22} g 2 { M cos2 2 m sin a sin(a — 22)}

8: 6 The use of the Instantaneous Centre: Initial Motions In uniplanar motion the instantaneous centre, I, is important for the description of the motion. Sometimes advantages are to be gained by using this point to investigate the dynamics of a rigid lamina, since the particle of the lamina instantaneously at / is stationary. Because of this fact it is easy to write down the kinetic energy and hence the energy equation. But since I is, in general, a moving point (see § 4 : 7) it is less convenient for obtaining the equation of motion by taking moments. Nevertheless, we obtain below a form of the equation of motion which is useful in certain cases.

§ 8:6

279

THE TJNIPLANAR MOTION OF A RIGID BODY

From eqn. (7.37) we can write down the kinetic energy, since I is a stationary point, T = i 5(I)co2. (8.4) This can be used, by differentiating it, to obtain the equation of motion, as, for example, the equation of motion (8.1) where the previous energy equation could have been obtained from eqn. (8.4). We can also use eqn. (7.42) and take moments about I to obtain the equation of motion. In eqn. (7.42) we use the fact that I is the instantaneous centre to obtain a different form for the term (i — rA) x (A -= I.) Since I is stationary = co k x (i — r„) f = k x (v — ii) + k X (I. — r„) . . (r — ri) x f w — ri) x [k x (v — ri)] +

— r,) x [k x

-I=wk

-

t

— ri)]

ri)2]

th k(r— r,)2

Hence the equation of motion is d _ co— d t [1M (r

— ri)2 +5(G)] =

+ k

(/)•

= 5(G) M —

But dt

[5(I)] = — [M (r — r,)2] , dt

since 5 (0) does not vary with time. kdif (/) + + ko — [ (1)1 = r (I)

dt

(8.5)

is the equation of moments about I. If the second term of the 1.h. side is zero, or is negligible, the equation takes the simpler form kth .1 (I) = (I) .

(8.6)

This simplification takes place in three cases. 1. When 5 (I) is constant. This occurs, for example, when a sphere rolls on a fixed surface.

280

A COURSE OF MATHEMATICS

2. For small oscillations about a position of equilibrium. In this case d

dt

[5 (/)] = 0 (co) and the second term on the 1.h. side of eqn. (8.5) is

0(co2) and is negligible in comparison with the remaining terms (see also Chap. IX). 3. Initial motion. When the body starts from rest w = 0 and eqn. (8.6) therefore gives the initial acceleration. The fact that eqn. (8.5) is simplified by inserting w = 0 is an example of a simplification which frequently occurs when a stationary system starts to move. Such mo-

B FIG. 111.

tions are called initial motions. The equations of motion of the system are obtained by any of the usual methods and are simplified by inserting zero values for any velocities which occur. The equations of motion, with the coordinate values corresponding to t = 0, then give the initial accelerations. Examples. (i) A uniform thin hollow cylinder, of mass m and radius a, is cut by a plane through its axis and one half is placed with its curved surface on a horizontal plane, slightly displaced from its equilibrium position and released from rest. Calculate the period of a small oscillation (a) when the horizontal plane is smooth, (b) when the horizontal plane is sufficiently rough to prevent slipping. (a) Since the plane is smooth, no horizontal forces act on the body and G moves vertically (Fig. 111). Hence I is the instantaneous centre. Also, correct to the first order in 0,

f (I) --- ,f (G) = m {a2

( 2a )12

and so the equation of small oscillations f(I) 0 = —mg GI becomes m (n2 — 4) a2 n2

— — mg

2a

§8:6

THE UNIPLANAR MOTION OF A RIGID BODY

281

which reduces to =

2ng 0 (7t2 — 4) a

representing S.H.M. with period 2}/{(n2— 4) gr a/(2 g)} . (b) When the plane is sufficiently rough to prevent slipping, B is the instantaneous centre. Also, correct to the first order in 0,

J(B) = J(A) = m Ice — ( rct )2 ±

\ 2t

7C

RO

FIG. 112 (i).

FIG. 112 (ii).

and the equation of small oscillations 5 (B) B = —mg GI becomes after some reduction —gO

a(gr — 2)

which represents S.H.M. with period 2gt Yfa (yr — 2)10. (ii) A uniform hemispherical shell, of mass M and radius a, is released from rest with its plane base against a smooth vertical wall and its lowest point on a smooth horizontal floor. Find the initial angular acceleration of the shell and the initial thrusts on the floor and the wall. At the moment of release (denoted by the suffix zero) the situation is as shown in Fig. 112 (i). At time t, whilst the shell is still in contact with the wall and the

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A COURSE OF MATHEMATICS

base makes an angle 0 with the vertical, as shown in Fig. 112 (ii), the equations of motion of G are d2 R=M (a sin0 + za cos0), (1)

dt2

8 — Mg — M

d2

dt2

(a

(2)

-la sin0).

Carrying out the differentiations and putting 00 =0, 00 = 0 we find B0

M a 0.0 ,

(3)

So Mg-12-Ma00 .

(4)

Using Fig. 112 (i), by taking moments about G the equation of motion is, initially 1

.11Ia2 60

a.Ro .

( 5)

Solution of eqns. (3), (4) and (5) gives ••

3g

00 — 10a

3My

, R,

10

80

17 Mg 20 .

(iii) A uniform rod AB, of mass In and length 2a, is held at inclination fi(> 0) to the vertical with B resting on a horizontal table and released from rest. Calculate the initial reaction between the rod and the table: (a) when the table is smooth; (b) when the table is so rough that the rod does not slide on the table. In case (b) show that, if the coefficient of friction u exceeds +, the rod will not slip initially whatever the value of 13. (a) Since the plane is smooth [Fig. 113 (H)] the reaction S at B is vertical and the equation of moments about I gives (4) tho = mga where tho is the initial angular acceleration of AB. .•. a G-

sin2[3) tho = y sin /3 .

But the equation of motion about G gives

(G) tho = S a sin/i. ilonatho = S sin . S=

1+ 3 sine /l

§8:6

THE UNIPLANAR MOTION OF A RIGID BODY

283

(b) In this case [Fig. 113 (i)] the rod rotates about B and we suppose that the rod makes the angle 0 with the vertical at time t after release. Then the equations of motion of G and the equation of moments about B are respectively d2

1 = nt dt2 (a sin 0),

(1)

d2 mg = m dt2 (a cos 0),

1?

(2)

raga sin 0 = m a' 0.

B

(3)

B FIG. 113 (ii).

FIG. 113 (i).

Again using suffixes to denote initial values so that 00 = 13, from (1), (2) and (3)

60 = 0, we find

Fo =ma4c cos , B0 =mg — m C4 .60 sinfi, a 00 =

= mg sin(3 cos (3, B0 =mg (1 — F0 • • Ro

3 sin0 cos (3 4 — 3 sine(

d( F0 \ • • dfl Ro )

sinfi.

sin2 jl).

3 sin 213 5 + 3 cos 2(3 •

6(5 cos 2 (3 ± 3) (5 + 3 cos 2(3)2 •

Hence Fo rk has a maximum (and greatest) value for 0<

< -In when cos 213 =— -;,-; and (F0 /R0),,,a„ =

Therefore, whatever be the value of (3, the initial frictional force is less than limiting if kt > and in this case the rod will not slip.

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Exercises 8 :6 1. AB, BC are two uniform rods, each of length 2a and mass m, freely hinged together at B. AB can turn freely about A, which is fixed, and C moves towards A with constant velocity u. Find the instantaneous centre for the rod BC. Prove that the body centrode and the space centrode are both circles and find their centres and radii. Show also that when both rods make an acute angle 0 with AC the kinetic energy of the rod BC is

m

24

(cosec2 0 + 6).

2. Two uniform rods AB, BC, each of mass m and length 2a, are freely hinged together at B. The rod AB can turn freely about A, which is fixed, and C is free to move below A along a smooth vertical wire which passes through A. The system is released from rest when both rods make an angle a with AC. At time t after release each rod makes an angle 0 with AC. Show that a (1 + 3 sin2 6) 62 = 3g(cos 0 — cosa). 3. The cross-section of a uniform right cylinder is a semicircle of radius a. Prove that the centre of mass is at a distance 4a/3n from the plane face and that the radius of gyration about the generator which is farthest from this face is rti/[(97E — 16)/6n]. Show that, if the cylinder is placed with its curved surface in contact with a rough horizontal plane, the period of small oscillations about the position of equilibrium is nV[(9n — 16) a/2g]. 4. A uniform solid is in the form of the portion of the paraboloid y2 ± z2 = 4ax between the planes x = 0, x =h (h < 3a). It is placed on a rough horizontal plane with its vertex on the plane and its axis of symmetry vertical. Prove that, if it is slightly disturbed from this position, the time of a small oscillation is nV[h (4a + 3h)/(3a — h) g]. 5. A square frame, made of four equal uniform rods each of length 2a, rigidly connected at their ends, hangs over two parallel smooth horizontal rails at the same level with which the two upper sides of the frame are in contact. In the position of equilibrium the plane of the frame is at right angles to the rails and a diagonal of the frame is vertical. If the rails are at a distance c (< a/1'2) apart, show that the length of the equivalent simple pendulum for small oscillations due to a slight angular displacement of the frame in its own plane is 10a2 — 6ac v2 + 3c2 3(a V2 — 2c) ' 6. A uniform rod AC D B, of length 2a, is supported by two smooth pegs at C and D, each distant a/2 from an end of the rod. The peg at D is suddenly removed. Show that the initial angular acceleration of the rod is 6g/7a and find the instantaneous alteration in the reaction at C.

THE UNIPLANAR MOTION OF A RIGID BODY

285

7. A uniform hemispherical shell of mass M and radius a is held with its rim in a vertical plane and the lowest point of the rim in contact with a rough horizontal table. The shell is then released. Prove that it will start to roll or slide on the table according as the coefficient of friction is greater or less than 6/17. 8. A uniform circular disc of radius a is suspended from a point on its rim. The disc is divided along a horizontal diameter and the two halves are kept together by a weightless smooth hinge at one end of this diameter and a weightless clasp at the other. The system being initially in equilibrium the clasp suddenly breaks. Show that the initial angular acceleration with which the halves begin to separate is 12ga/(9a2 12ac — 4c2), where c is the distance of the centre of mass of the upper half from the point of suspension. 9. A uniform circular hoop is suspended from a fixed point by means of three equal light inelastic strings, each of the strings making an angle a with the vertical and the points of attachment to the hoop being the vertices of an equilateral triangle. If one of the strings breaks, prove that the tensions in the others are instantaneously reduced in the ratio 6/(5 + sec2 a). 10. Two uniform rods AB, BC, each of length 2a, freely jointed at B, are suspended from smooth joints A, C in the same horizontal line so that the inclination of either rod to the horizontal is n/4. The end C is set free. Show that the initial angular accelerations of the rods are 9g/16a1/2 and 3g/4a 12. 11. A square lamina is suspended by vertical strings tied to two adjacent cornners, two edges of the square being vertical. If one string is cut, show that the tension of the remaining string is instantaneously diminished to `-; of its former value. 12. The middle points of two equal uniform thin rods AB, BC, freely jointed at B, are connected by an inextensible string. The system hangs in equilibrium from A and the angle ABC is a right angle. If the string is cut, show that the initial angular accelerations of the rods are in the ratio 1 :4, and that the direction of the reaction at B is instantaneously turned through an angle tan-16. Miscellaneous Exercises VIII 1. A piece of thin uniform wire is bent into the form of an open arc of a circle, and placed on a knife edge so that it can oscillate in its own plane about its mid-point. Show that the length of the equivalent simple pendulum depends only on the radius of the circle and not on the length of the wire. 2. A uniform circular disc of radius a has a concentric hole of radius b drilled out of it. The disc oscillates as a compound pendulum about an axis perpendicular to its plane through a point on its circumference. Show that the period of small 2)/(2ga)}1/2. oscillations is 27r{ (3a2 b If the effect of drilling the hole is to increase the period by 4% over that of the solid disc, find b/a.

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A COURSE OF MATHEMATICS

3. A thin uniform rod of mass m and length 2a is free to rotate in the vertical plane about a horizontal axis through one end. Each element of the rod is subject to a resistance per unit length of amount mk/a times its velocity. Obtain the differential equation of motion if the rod is released from rest in a position making a small angle with the downward vertical. State the condition for the motion to be oscillatory and solve the differential equation when this condition is satisfied. If a torque of magnitude L and frequency the same as the natural undamped frequency of the rod is applied, prove that the resulting final oscillation (assumed small) is of amplitude L 13 4mka ga • 4. A uniform circular disc of mass M and radius a is pivoted so that it can turn without friction about its axis, which is horizontal. Its rotation is controlled by a spring and a couple C is required to twist it through a radian. A light string, partly wound round the circumference of the disc, to which one end is fixed, supports at the other end a mass m hanging freely. If m is displaced vertically a small distance from its equilibrium position and let go, show that the period of its oscillations is 27ta v'{(2m

M)/2C}.

5. A uniform horizontal circular disc of mass m and radius a can rotate about a vertical axis through its centre, and the rotation is controlled by a spring which exerts a restoring couple of moment kmgaO when the disc is turned through an angle 0 from its equilibrium position. A particle of mass m rests on the disc and is connected to the axis of rotation by a light horizontal straight rod of length-la which can turn freely round the axis. The coefficient of friction between the particle and the disc is u. The disc is turned through an angle 13 from its equilibrium position, and is released from rest. If 13 > 3p/2k, show that initially the particle slips on the disc, and that if slipping first ceases after a time t, and if p2 = 2kgla, then pt is the root between 0 and 7r of the equation sinp t

Pt

13 — -fp •

6. A uniform square trapdoor of mass M and side 2a is initially at rest in a horizontal position with a particle of mass m lying at the centre of its upper surface which is so rough that the particle cannot move on it. The trapdoor can turn freely about hinges on one edge, and it is urged downwards by a force of constant magnitude F applied at the mid-point of the opposite edge and acting always perpendicularly to the plane of the door. Show that, if 6F < Mg, the particle will leave the surface when the inclination of the trapdoor to the vertical is sin-1(6F/Mg). Find also the velocity of the particle at this instant. 7. A symmetrical wheel, of radius a and radius of gyration k about its axle, with its plane vertical, is projected horizontally along a rough horizontal plane with linear velocity V in its own plane and angular velocity D. 0 is in the opposite sense to the rotation that the wheel would have in rolling. Prove that the wheel will return to the point of projection if Q k2 > a V.

THE UNIPLANAR MOTION OF A RIGID BODY

287

Show that a circular hoop thus projected with Q = 2 V la will return to the starting-point in time 9 V 14ktg, where ,a is the coefficient of friction between the hoop and the ground. 8. A smooth uniform cubical block of mass m and side 2a is held on a plane inclined at an angle a to the horizontal. A uniform solid circular cylinder of mass m and radius a is given a spin S2 about its axis in the sense to make it move up the plane and is then placed on the plane with its axis horizontal so that it touches the lower face of the cube along a generator. The system is released. If all contacts are smooth except between cylinder and plane where the coefficient of friction is it, show that the cylinder will never roll if 5,a < 2 tan a . If It 3 tan a , show that the block moves up the plane a distance a2S22 /(169 g sin a ) before the cylinder ceases slip. 9. A uniform solid circular cylinder, of mass M and radius a, rolls on the inside of a fixed cylindrical tube, of radius c + a, whose axis is horizontal. If the inclination to the vertical of the plane containing the axes of the cylinders is 0, measured so that 0 = 0 when the line of contact is the lowest generator of the tube, prove that, so long as the motion is one of pure rolling,

0+

2g 3c

sin 0 = O.

If the cylinder starts from the lowest position with a motion of pure rolling, the initial angular velocity being w, where w2 —

4cg 3a2

and if the coefficient of friction is e, prove that pure rolling ceases and slipping begins when tan 0 = 7µ. 10. A rough symmetrical circular cylinder of radius a, and radius of gyration z about its axis, is free to roll on a horizontal platform. Initially both the cylinder and the platform are at rest. The platform is then moved horizontally in a direction perpendicular to the axis of the cylinder. Show that, whatever the motion of 52

the platform, the displacement of the axis of the cylinder at any instant is

x2 + a2 times the displacement of the platform, provided that the acceleration of the platform never exceeds µg 1 + (

a2

in absolute value, it being the coefficient of

friction between cylinder and platform. If the total displacement of the platform is always small, show that the cylinder may be regarded as rotating about a fixed axis at height 5 ,7 : above its central axis. 11. A smooth wedge of mass M and angle a (< n/4) rests with a face adjacent to the angle a on a horizontal table. A solid circular cylinder of mass m and radius a is given a spin Q about its axis and then gently lowered on to the table so as to make contact along generators with the table and the slant face of the wedge simultaneously. All contacts are smooth except the contact of the cylinder with the table, where the contact is rough with angle of friction a. If the cylinder

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A COURSE OF MATHEMATICS

remains in contact with both wedge and table, show that the cylinder slips on the table for a time a.Q(m + 2M) g tana(3m + 2M) and that the magnitude of the subsequent steady velocity of the system is independent of a . 12. A thin hollow uniform circular cylinder, of mass 2M and radius 2a, open at both ends, is mounted so that it can turn freely about its axis, which is horizontal. A solid uniform sphere, of mass M and radius a, lies within the cylinder and the system is initially at rest. A constant couple of moment lMga about the axis is applied to the cylinder. Assuming that no slipping occurs and that at time t the sphere is still in contact with the cylinder, prove that 8a l2 =(20 — 24 sing-6)g, where 0 is the angle between the vertical and the plane containing the axis of the cylinder and the centre of the sphere. 13. A uniform rod AB of mass m and length 2a swings freely in a vertical plane about the end A, which is fixed. If 0 is the inclination of the rod to the downward vertical, find 6 and 62 in terms of 0 and the amplitude a of the oscillations. Let P be a point of the rod distant 2x from the free end B, where 0 < x < a. The reaction of the portion AP on the portion PB of the rod consists of a tension T, a shearing force S acting transversely at P and a couple G. Calculate T, S, and G when AB makes an angle 0 with the downward vertical. At what point P of the rod is the magnitude of the bending moment G greatest? 14. A square framework of four equal uniform rods, each of mass in and length 2a, freely jointed at their ends, is rotating with angular velocity a) about a vertical line through the centre 0 of the square perpendicular to the rods. If Q is the mid-point of one of the rods AB, prove that the tension T, the shearing stress S, and the bending moment M, at a point P on AB are given by

T=

w2 (3 — tan2 0) ,

S = lmaco2 tang, M=

(02(1 — tan2 0),

where 0 is the angle POQ. 15. A uniform circular disc of radius a and mass m is smoothly constrained to lie in a fixed vertical plane. Initially it is above and in contact with a rough horizontal plane (the coefficient of friction is equal to ,a) and the disc has an angular velocity co, and its centre of gravity has a horizontal velocity vo , the senses of these being such that the highest point of the disc has a horizontal velocity vo + a co, . Show that: (i) at all times the point of the disc momentarily situated at a height kt above the point of contact has the horizontal velocity vo +i a con ; (ii) if coo — 2 vo /a , then the disc will ultimately roll on the plane with a constant velocity. What happens if a), = —2vo/a?

THE UNIPLANAR MOTION OF A RIGID BODY

289

Show that the total amount of work done by friction during the motion is equal to m (a wo — v0)2. 16. A uniform circular hoop of mass m and of radius a hangs in a vertical plane on a small smooth peg A, and initially the hoop is held at rest so that the horizontal chord through the peg lies above the centre 0 of the hoop and subtends an angle 2a at it, where 0 < a < In. The hoop is then released. Calculate the initial reaction B, at the peg, and show that the vector AO vibrates like a simple pendulum of length a. If the hoop had been freely pivoted at A show that the initial reaction would have been B0 1/(1+ !f tan2 a), where B, is the reaction calculated above. 17. A uniform rod AB, of length 2a and mass m, moves in a vertical plane under the action of gravity and a force applied to its upper end A constrains that end to move in a horizontal line LM with uniform acceleration a. Find an expression for the horizontal component of the force acting on the rod in terms of 0, the angle between AL and AB. 18. A uniform rod is held nearly vertically with its lower end resting on an imperfectly rough plane. It is released from rest and falls forwards. The inclination to the vertical at any instant is 0. Prove (i) that if the coefficient of friction is less than a certain finite amount the lower end of the rod will slip backwards before sin2 40 = ; (ii) that however great the coefficient of friction may be the lower end will begin to slip forwards at a value of sin2 +0 between 4 and I. 19. A uniform rod, of mass m and length 2a, is placed with one end on a smooth plane inclined at an angle a to the horizontal and with the vertical plane through the rod containing a line of greatest slope of the plane. The rod is released from rest when it makes an angle 2 a to the horizontal and an angle a with the plane. Prove that, when the rod makes an angle q9 with the plane, a 422 (1

+ 3 cos2 q)) = 6 g cos a (sin a — sin 99) .

Also show that the reaction of the plane is mg cos a[1 + 3 costa + 3 (sin a — sin 0941 + 3 cos2q9)2. 20. A uniform solid hemisphere, of radius a, is initially held so that its curved surface is in contact with a smooth vertical wall and with a smooth horizontal floor, the plane face being parallel to the wall. If the solid is released from rest, show that it remains in contact with the wall until the plane face is horizontal and find the velocity of the centre of mass in this position. Show that, when the angular velocity subsequently vanishes, the plane face makes with the horizontal an angle cos-1(45/128) . 21. A rough circular cylinder, fixed with its axis horizontal, is surrounded by a thin uniform circular hoop which is initially held at rest in contact with the cylinder, its plane being perpendicular to the axis of the cylinder. Prove that, on being released, the hoop will begin to roll without slipping on the cylinder provided that the coefficient of friction is not less than ltan a, where a is the angle between the upward vertical and the radius to the initial point of contact.

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A COURSE OF MATHEMATICS

22. A particle A, of mass m, is fixed to the surface of a uniform solid sphere, of mass M and radius a, whose centre is the point C. The sphere is placed on a rough horizontal table and is released from rest when CA is horizontal. Prove that the sphere will begin to roll if the coefficient of friction between the sphere and the 5m2). If the sphere is rolling table is greater than 5m (M m)/(7 M2 17Mm when CA has turned through an angle 0, prove that

a(7M

10m— 10m sin0)62 = 10mg sine.

23. A uniform thin circular hoop, of radius a, surrounds a rough circular cylinder, of radius b, which is fixed with its axis horizontal. Initially, the hoop hangs freely at rest with its plane perpendicular to the axis of the cylinder. If it is given an angular velocity AA (a — b)g a9 about the generator through the point of contact, show that it will make complete revolutions, without slipping or losing contact with the cylinder, provided that is not less than 3 and that the coefficient of friction is not less than 1/[2 1/(22 — 22, — 3)]. 24. Two uniform rods PA, BC are smoothly hinged together at P on BC . The end C is smoothly hinged to a fixed point ; BC is held horizontal, and supports PA which rests on BC in the same line, with P between A and C. PA is of length 2a and mass m; BC is of length 2b and mass M, and P is at a distance c from C. BC is now released, and smooth constraints compel the rods to remain in a vertical plane. Show that PA and BC will begin to turn in opposite directions if 3c> 4b, and that they will begin to move as a single rod if

c<

4Mb(b — a) 3Mb ± ma

25. A heavy uniform rod is at rest in a vertical position when its upper end is made to move down a straight line of slope a with constant acceleration a, the rod being free to rotate in a vertical plane about its upper end. Prove that in the subsequent motion it just reaches a horizontal position if a = g/(cos a + sina). 26. A smooth uniform sphere of radius a and mass m is balanced on a smooth uniform sphere, of radius b and mass M, that rests on a smooth horizontal plane. Show that if the upper sphere is slightly displaced then the spheres are about to separate when the angle cbetween the line of centres of the spheres and the vertical satisfies 3(M + m) cos q) — m cos3 q9 (M m) . 27. A reel of mass M is made from three uniform circular cylinders by placing one of length 4a and radius a between two others of length a and radius 2a, the three cylinders being co-axial. Show that the moment of inertia about the axis is

-:!Ma2 . The reel moves on a horizontal table. A light inextensible thread unwinds without slip from a groove on the midsection of the inner cylinder, passes horizontally from the top point of the groove in a direction at right angles to the axis of the reel, over a smooth fixed pulley, and carries at its free end a mass 11M, which hangs clear of the table. If the coefficient of friction between the table and reel is less than s, i prove that when the system is released from rest, the reel commences

THE. UNIPLANAR MOTION OF A RIGID BODY

291

to slip on the table and continues to slip with constant linear and angular accelerations. Show further that if ft = 16, the distance travelled by the reel in a given time is Z1,1 , of that travelled in the same time when pc > 28. A uniform rectangular block rests on a horizontal table; a thin uniform square plate has one edge resting on the table and the parallel edge resting symmetrically against a vertical face of the block. There is no friction at any of the contacts. Show from general considerations of momentum that, if the system is released from rest, the two bodies must separate before the plate becomes horizontal. If the plate is of side 2a and is initially inclined at an angle a to the table, and if the masses of the two bodies are equal, prove that separation takes place when the plate makes an angle 0 with the table, where 3 sin3 0 — 24 sin 0 + 16 sin a = 0 .

CHAPTER IX

STABILITY 9:1 The concept of stability Everyone has a fairly clear idea of the difference between a stable and an unstable structure, or between a stable and an unstable state of motion. Roughly speaking, a heavy object hanging from a hook is stable, whereas a bottle standing upside down on a horizontal plane is unstable; a slight disturbance in the first case causes little deviation from the steady position, but a slight disturbance of the bottle causes it to fall over. A cyclist riding quickly along a straight line is stable because a small disturbance will not upset him, but, if he is just balanced moving very slowly, a small disturbance will upset him; he is unstable when moving slowly. The essential point is that the application of an arbitrary, small disturbance is followed either by a slight deviation from the steady state, or by a deviation which eventually becomes large. The first case is the stable case, the second is the unstable case. If there is any small disturbance which causes the system to deviate largely from the steady state, that state is unstable. It may be possible to find certain disturbances which will not upset an unstable state; hence it is essential that the disturbance applied to a steady state is arbitrary. Since the subsequent motion of a system after a disturbance is the criterion of stability or instability, we must investigate stability by means of the equations of motion. Stability is therefore a matter of dynamics and not strictly of statics, although associated closely with positions of equilibrium. For systems having one degree of freedom the distinction between stability and instability depends upon the sign of one term in the equation of motion. The differential equation

n2 y = 0

(9.1)

§ 9 :1 has a general solution

293

STABILITY

y= a cos (nt

e),

(9.2)

where a, 8 are arbitrary constants. If a is small, y and all its derivatives remain small for all values of t, y performing an oscillatory, simple harmonic, variation with time. On the other hand the differential equation n2 y = 0 (9.3) has the general solution y = A ent B e-nt . (9.4) For arbitrary, small values of A, B (or arbitrary, small initial values of y and y does not remain small for all values of t. Although y is small initially the factor en t shows that eventually y increases without limit. (There are some solutions for which y is small and remains so, i.e., when A = 0 , but such solutions cannot satisfy arbitrary initial conditions.) Hence, if the equation of motion of a system approximates to eqn. (9.1) for small deviations from the steady state, that state is stable; if the equation of motion approximates to eqn. (9.3) for initially small deviations from the steady state, that state is unstable. Consider now a system with one degree of freedom whose configuration is specified by a single parameter q which may be an angle, a length, etc. If the forces acting on the system are conservative, the energy equation takes the form A (q) 42+ V (q) = E, (9.5) where V (q) is the potential energy, 1- A (q) 42is the kinetic energy, and E is a constant, the total energy. Since the kinetic energy must be positive, A (q) > 0 for all possible values of q. The equation of motion of the system, obtained by differentiating eqn. (9.5) and cancelling through by 4, is (q) 42 + V' (q) = 0 . A (q)q (9.6) This equation may also be obtained in many cases by direct application of the Law of Motion. Since all the terms in eqn. (9.6) are continuous functions of t, it holds when 4 = 0 [despite the fact that (9.6) was obtained by cancelling 4 from an equation]. The position, or 'motion', given by q = qo is a possible equilibrium, or steady, state if q = 4 = 0, q = qo satisfy the equation of motion (9.6). Hence

V' (go) = i.e., V has a stationary value when q = q0.

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A COURSE OF MATHEMATICS

To investigate the stability of this position we substitute q = q, into eqn. (9.6), obtaining

A (qo

y)y

+A' (qo

y) y2 + V' (qo

y

y) = 0.

We assume that y and its derivatives are all small quantities of the first order (of smallness) and approximate this equation correct to the first order, using Taylor's expansion where necessary. Then

A (go

Y)

.1 A' (go

= V (go) + A (go) y

y)Y2 ± T7' (go ± y)

+ v" (go) y + 0(Y2) = O.

(9.7)

The term z A' (qo y) y2 is included in 0 (y2) because y2is of the second order. Since the condition for equilibrium is V' (q0) = 0, the equation of motion (9.7), correct to the first order, becomes

A (g0)y + V" (qo)y = 0.

(9.8)

Since A (q0) > 0, eqn. (9.8) is of the form (9.1) if V" (qo) > 0, and is of the form (9.3) if V" (q0) < 0 . We can therefore sum up : (i) if V' (qo) = 0 and V" (q0) < 0, the equilibrium is unstable ; in this case V has a maximum at q = q0 ; (ii) if V' (q0) = 0 and V" (q0) > 0, the equilibrium is stable ; in this case V has a minimum at q = q0 . The case for which V" (q0) = 0 requires closer investigation; each particular problem is best treated on its merits.

9:2 Stability of equilibrium For any system with one degree of freedom under the action of conservative forces the potential energy is a function, V (q), of a single parameter. Excluding the case V" (q0) = 0, the results of § 9:1 show that the positions of stable and unstable (statical) equilibrium can be found by determining the minima and maxima, respectively, of the potential energy. Any of the usual methods may be used for distinguishing between maxima and minima. (See Vol. I § 5:6). Examples. (i) A uniform rod of weight 12w and length 2a can turn freely about one extremity A, which is fixed. A light inextensible string attached to the other extremity passes through a small smooth ring fixed at a point C, distant 2a from A and at the same level as A, and carries at its other end a weight w. Show that the system is in stable equilibrium when 8 cos 6 = 1, where 6 is the inclination of AB to the horizontal. (See Fig. 11 with W = 12w.)

§9:2

295

STABILITY

The potential energy is

V = — 12wa sine + 4wa sin+0. We determine the stationary values of V and the nature of these stationary values. dV — 12wa cos0 + 2wa cos+0 = —2wa(12c2 — c — 6) dB —2wa(4c — 3) (3c + 2), where c = cos+ 0 . But d V/d 0 vanishes where c = 4 and c = — . As B increases through the value 2 cos-l(4) = cos-1(-`e ), so that c decreases through the value d V/dB changes from negative to positive: hence V has a minimum and c gives a position of stable equilibrium.

=a

(ii) Find the distance from the centre

C of a uniform solid sphere, of radius a, to the centre of mass G of the smaller of the segments into which it is divided by a plane distant +a from its centre. This segment rests with its plane face horizontal and uppermost and its curved surface in contact with the highest point of a fixed sphere of radius a. Assuming that the surfaces are sufficiently rough to prevent sliding, show that the segment may be rolled through an angle 2 cos-1(20/27) into a position of unstable equilibrium. Integration shows that the centre of mass of the segment is at a distance 27a/40 from the centre of the sphere from which it is cut. In the oblique position shown in Fig. 114 the segment has been turned through an angle 20. If W is the weight of the segment, the potential energy, V, is given by

V

1 dV W d0 Also

27a cos20. 40 27a 27a 27a/ —2a sine + sin 20 = sine 20 10 2a cos 0 —

—

1 d2 V 27a = 2a cos0 + 10 cos 20. W d02

Hence d V/d 0 vanishes for 0 = 0 and cos0 = 20/27. When 0 = 0, 27a d2 V —WI-2a I10 ) d02

> 0.

20

27

•

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A COURSE OF MATHEMATICS

Also when cos 0 = 20/27,

d2 V d 02

( 40 = Wa — 27

27 400 5 (27)2

27 10

< 0.

Therefore the position 0 = 0 is a position of stable equilibrium (V has a minimum there) and the oblique position, cos 0 = 20/27 , is a position of unstable equilibrium ( V has a maximum there).

There is also an 'intuitive' method of determining the stability of a system without finding V explicitly. In § 2:5, where the function V was introduced, we showed that — d V/dq is the 'force tending to increase q'. Since V' (q0) = 0,

V' (q,

y) = y V" (q0) + 0 (y2) ,

and the resultant force on the system in the slightly displaced position is, correct to the 1st order, — y V" (q0) . If V" (q0) > 0, this force changes sign with y and acts towards the position y = 0 implying that in all cases the force tends to destroy the displacement y. [This is shown by the sign of 1) in eqn. (9.8).] If the first non-vanishing derivative of V at q = q, is VP) (q0), then

v'(qo +

=

(q0

(p — 1)!

)

"P)

and so the force always tends to restore the system to the position

q = q0 if p is even and V(P) (q0) > 0, i.e., if V is a minimum at q = q0. Hence, if in the displaced position the resultant force on the system tends to restore the position of equilibrium, the position is stable. This is the 'intuitive' method used when stability is regarded as a matter of statics instead of dynamics. Examples. (i) A cylinder, not necessarily of circular cross-section, rests in equilibrium on the top of another fixed, rough cylinder. The generators of both cylinders are parallel. If the radii of curvature of the sections of the upper and lower cylinders are e , e' respectively at A , the point of contact, and if AG = h, where G is the centre of mass of the upper cylinder, find a sufficient condition for stability. (This is the 'rocking stone' problem in two dimensions.) In the equilibrium position [Fig. 115(i)] the common normal CAC' is vertical and passes through G. In the displaced position [Fig. 115 (ii)] the weight of the upper cylinder acting through G will tend to restore equilibrium if 0 is to the left of the vertical through B. Since the displaced position is obtained by rolling the upper cylinder, arc A'B = arc AB and so, correct to the 1st order in 0, 99, e'f9 = e97•

297

STABILITY

§ 9:2

The horizontal distances of B and G from the vertical C' A' are, respectively,

P

sine

(p + e') sin0 — ( e — h) sin (0 ± q))

e' 0, g')0 —

(g

(e —

h) (0 +

Hence the equilibrium of Fig. 115 (i) is stable if

P' 0 > ( e + e')0 — (e — h) (0

99)-

Fm. 115 (ii). Utilising the relation e'

e = e 9, the inequality gives

e e' > (e + e') e — (e — h) + V), i.e.,

e e' > (e — e + h) (e + e'), i.e., 1

1

1

— h>— e —e

'),

s a more detailed investigation is the condition for stability. If h = e e'l( e is required. In the limiting cases where e or e become infinite the upper cylinder or the lower cylinder becomes a plane, and the condition determines the stability of a thick plank resting on the highest point of a cylinder or the stability of a cylinder resting on a plane.

'

(ii) A uniform chain of length 2l and weight 2 w l is placed symmetrically over two small smooth pegs, at the same level and distance 2a apart, with equal portions hanging vertically and with the portion between the pegs hanging in the form of a catenary with parameter c. The ends of the vertical portions are each

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A COURSE OF MATHEMATICS

pulled with downward vertical forces F just sufficient to maintain equilibrium. Show that

F

w(ceale —1).

Hence show that, if ceaia = 1, the chain hangs in equilibrium under gravity alone and that this equilibrium position is stable for symmetrical disturbances in a vertical plane if a < c. With the notation of example (i) p. 47 and Fig. 14 p. 48 the equilibrium of

HB gives T H =wHB+ F. But TH

a wyh- = wc cosh (— c

.•. we cosh (— a = wHB F.

(1)

But

HB

ac =1. c sinh (—

(2)

From (1) and (2) we fin d

F = w(cealc —1) as required. The force F = 0 when ceak = 1 and in this case the chain hangs in equilibrium under gravity alone. If the length of the chain between the pegs is L, then

L = 2c sinh (a/c), dL( = 2 sinh — a cosh — 7 < 0. d c c c1 Hence, if L decreases, c increases. This implies that, if the ends of the chain are pulled down, then the parameter of the catenary in which the central portion hangs increases. But dF . dc = weak — Hence, if a < c, dF/dc > 0 and F increases with c. In this case, if the ends are pulled down below the equilibrium position, a (positive) downward force is required to maintain equilibrium. This implies that, when the chain is disturbed from the equilibrium position, the gravitational forces tend to restore the chain to its equilibrium position. Thus the equilibrium is stable when a < c. We note that, since in the equilibrium position / = cealc, dl -cTC-

a

eale (1

§9:2

299

STABILITY

and / has a minimum when c = a. Hence the least length of the chain for which equilibrium is possible is tae. From the graph of c eai° shown in Fig. 116 it is clear that when / > a e , two equilibrium positions are possible; the position for which c is greater (corresponding to a shorter length hanging in the catenary) is stable and the other position is unstable. Positions in which V' (q0) = 0, V" (q0) = 0 are sometimes called positions of neutral equilibrium, but, in fact, they are usually stable or unstable. A position of neutral equilibrium is such that on receiving a

y rce"c

C=0

C

FIG. 116.

small disturbance the system neither tends to return to the equilibrium position nor to move further away but is in equilibrium in the disturbed position. Cases of strictly neutral equilibrium are rare, being typified by a uniform sphere resting with its curved surface on a horizontal plane. (However, if the sphere is given a small velocity it will continue to roll indefinitely; it can then hardly be said to 'deviate slightly' from its original position, and so such a position could be called unstable.) Since the determination of stability of equilibrium is effectively the same as the investigation of the nature of the stationary points of the function V (q), those cases in which V" (q0) = 0 are the mechanical counterparts of those cases of stationary points of functions for which some derivatives of higher order than the first vanish. These cases often arise from the coincidence of a number of stationary points. (See example (ii) p. 301.) It remains true that, if V is a minimum at q = q0 , then this position is one of stable equilibrium. For suppose the system is given a small disturbance from the equilibrium position so that its energy in the sub-

300

A COURSE OF MATHEMATICS

sequent motion is V (q0) e, where e is the energy of the disturbance; then the energy equation T E becomes

T = e + V (go) — V (q).

(9.9)

Since V has a minimum at q = go , V (q) > V (go) and therefore 0 --<_T-<.e; it follows that V (q) — V(q0) = 0 (c) and, since V is a continuous function, I q — go t —› 0 as e —> 0. Hence, the equilibrium is stable because q — qo remains small. When V has a maximum at q = qo then this position is unstable. To establish this result it is sufficient to show that there is one disturbance which causes the system to move away from the equilibrium position, i.e., which causes I q — q0 1 to increase. Suppose that the system is released from rest at a position near to, but not coinciding with, the equilibrium position. The system will begin to move and acquire kinetic energy, so that, by eqn. (9.9), V (q0) — V (q) increases and I q — qo I increases. Examples. (i) We consider the 'rocking stone' problem [example (i) p. 296 Fig. 115] in the special case when the cylinders have circular sections and for which

1 h

1 g

1 (1)

P '

In these circumstances e , e' do not depend upon 0 and the relation

{(e + e')

V

e is strictly true. Here all the odd derivatives of V vanish when 0 = 0 and

1 d2 V W de2

—

cos°

(0 ± V)

(e +, e') (6,

h) cos

{

(e + e') {— cos e + cos{ (e

°)

r

L

e

e' )6

1} J

]}

d2 V when the condition (1) is used. Henced 02 = 0 when 0 = 0, and the equilibrium is apparently neutral. Further differentiations give

1

di

v

LV d 04

(e +

{cos°

e" )2 cos [+ c

Hence V has a maximum (since d4V/d 04 < 0) when 0 = 0 and the apparently neutral equilibrium is unstable.

§9:2

301

STABILITY

(ii) A uniform square lamina of side 2a rests in a vertical plane with two of its sides in contact with horizontal smooth pegs distant b apart, and in the same horizontal line. Find the positions of equilibrium and discuss their stability. Show in particular that, if a/1/2 < b < a, a non-symmetrical position of unstable equilibrium is possible in which b(sin0

cos0) = a,

where B is the inclination of a side of the square to the horizontal.

Fro. 117. Fig. 117 shows that the height of G above the fixed level PQ is

a V2 sin(8 ±

— x cos0,

where x = AQ = b sine.

V — a v2 sin (0 + In) — b sin 0 cos 0 . •• W 1 dV = a 1/2 cos (0 + • W de

— b cos20

(cos0 — sine) {a — b (cos 0 + sin0)} , 1 dV 1/2 sin (0 + 7)1. W dO = (cos 0 — sin 0) fa — b Equilibrium positions occur when d V Id0 = 0, i.e., when cos0 = sin0 which gives the symmetrical position, 0 = r, and when sin(0

a

= b V2

b (sin 0 + cos 0) = a.

(1)

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A COURSE OF MATHEMATICS

This latter value gives an unsymmetrical position. The corresponding value of 0( x/4) is real if b > a/1/2. Moreover 0 > 0 for equilibrium to be possible and a 1 > , i.e., b < a. Therehence for the existence of unsymmetrical positions b 1/2 1/2 fore unsymmetrical positions occur when a < b < a. I/2 As 0 is made to increase through 1-7r, d V/d 0 changes from positive to negative if a .„›.- b y2. Hence if b < a/I/2 there is only one position of equilibrium and this position is unstable. On the other hand, if b > an/2, d V Id0 changes from negative to positive as 0 increases through in and the symmetrical position is stable. In this case, too,

V^

V

7\ FIG. 118 (i)

a

112


< a.

Fin 118 (ii) .

12

= b < a.

the unsymmetrical positions exist (provided b < a) and, as 0 increases through the values given by eqn. (1), d Vid 0 changes from positive to negative showing that these positions are unstable. If a = b j/2, all three positions of equilibrium coincide in the position 0 = . Since 1 d2 V a -1/2 sin(0 a) 2b sin20, W cif/2 d2 V we see that, in this case, d 02 = 0 in the equilibrium position and the equilibrium is apparently neutral. However, we have shown above that this is, in fact, an unstable position. This example illustrates two features which occur frequently when V is a function of a single parameter. Fig. 118(i) shows that in the graph of V maxima and minima alternate, so that two positions of unstable equilibrium are separated by a

§9:2

STABILITY

303

position of stable equilibrium, and vice-versa. This is generally the case in physical problems. The situation shown in Fig. 118 (ii) is the limiting case in which the three points A, B, C, of Fig. 118 (i) coincide in one stationary point D. Although d2 V/d 02 = 0 at D the stationary value is nevertheless a maximum. Exercises 9:2 1. A heavy rod AB can turn freely in a vertical plane about one end A which is fixed. To the other end B is tied a light elastic string of natural length IAB and of modulus equal to half the weight of the rod. The other end of the string is attached to a light ring which can slide on a smooth horizontal bar, which is fixed at a height equal to twice AB above A and in the vertical plane through AB. Find the equilibrium positions of the rod and discuss the stability in each case. 2. A smooth ring P of mass m is free to slide on a smooth fixed vertical circular wire of radius a and centre 0. A light elastic string, of natural length 2a and modulus kmg, passes through the ring, its ends being fixed to the ends A and B of the horizontal diameter of the wire. Find the potential energy of the system when OP makes an angle 2 0 with AB, the ring being below AB. Show that if k> 2-1/2 the ring is in unstable equilibrium when at the lowest point of the wire. Investigate the stability in this position when k= 2 +1/2 . 3. A uniform lamina in the shape of a rhombus consisting of two equilateral triangles of side a rests in a vertical plane with two adjacent sides on two smooth I/3 pegs in a horizontal line, distance a — apart. Prove that in equilibrium either a 4 diagonal or a side is vertical, and find which of these is a position of stable equilibrium. 4. A ring A of weight W slides on a smooth circular wire of radius a fixed with its plane vertical. A light inextensible string attached to A passes over a small smooth pulley at 0, the end of a horizontal diameter, and supports a weight n W (n < 1) hanging freely. If, when the system is in equilibrium, OA makes an angle 0 above the horizontal prove that cos 26 = n sin 0 . Prove that there are two positions of equilibrium and that the position with the ring on the upper half of the circle is unstable. 5. A uniform rod A B, of weight W and length 2 a, is free to rotate in a vertical plane about the point A. A light elastic string, of modulus kW and natural length a, has one end attached to B and the other to a fixed point 0 which is vertically above A. If OA = 2a show that, when AB makes an angle 0 with the downward vertical, the potential energy of the system, when the string is stretched, may be expressed in the form W a I (4 k — 1) cos 0 — 4 k cos (0/2)} + constant. Deduce that, if k> the equilibrium position in which the rod is vertical with B below A, is unstable, and that there is an oblique position of equilibrium which is stable. 6. A uniform solid hemisphere of radius a rests with its plane face horizontal and its curved surface in contact with a fixed rough sphere of radius 2 a . Prove that the hemisphere can be rolled into a position of equilibrium with the plane

304

A COURSE OF MATHEMATICS

face inclined to the horizontal and find the angle between the common normal and the vertical in this position. Determine whether this position of equilibrium is stable or unstable. 7. A uniform plank of thickness 2h rests in equilibrium horizontally across the top of a rough cylinder whose cross-section is a cycloid with its vertex upwards, and is at right angles to the generators of the cylinder. If the radius of curvature of the cycloid at the vertex is a, prove that the horizontal position of the plank is stable if h < a, and unstable if h r a. Prove also that the plank can be rolled from this position into an oblique position of equilibrium if h < a, but that this position is unstable. 8. A uniform solid consists of a right circular cone of height h and a hemisphere with their bases, each of radius a, in contact and with their circular boundaries coinciding. The solid stands with its axis vertical and vertex upwards on the top of a rough sphere of radius 3a. Prove that the equilibrium is stable if h is less than Ey5 — 1)a, but is unstable if h is greater than or equal to this value. 9. A homogeneous cylindrical body whose cross-section is bounded by an arc of a parabola of latus rectum 4a and a double ordinate to the axis rests on a rough horizontal plane. Find the length of the intercept of the double ordinate on the axis in order that the position, in which the axis of every cross-section is vertical and the vertex is downwards, may be one of neutral equilibrium to a first approximation. Show that the equilibrium is then in reality stable.

9:3 Oscillations about stable equilibrium In § 9:1 we used eqn. (9.8) merely to distinguish stability from instability by means of the sign of the second term on the 1.h. side. We can, however, obtain more information by a comparison with eqn. (9.1) which implies that the system, when disturbed from the steady state, performs small oscillations about the position of stable equilibrium; this fact can be confirmed by a multitude of examples in everyday experience. All such examples are cases of systems in positions of stable equilibrium which receive disturbances and start oscillating. If the accurate equation of motion of some system is = —n2 sing,

(9.10)

then the equivalent simple pendulum has length 1 = gln2and is equivalent for oscillations of any amplitude. Nevertheless, in practice the period of small oscillations is usually required. One of the most convenient ways of obtaining the period of a small oscillation is to derive the approximate equation of motion by differentiating the energy equation, following the method of § 9:1. The equation of motion must be obtained correct to the 1st order in the dis-

§9:3

STABILITY

305

placement y and its derivatives, which are all taken to be of the same order of magnitude. It may however be simpler to make the approximations before obtaining the energy equation; if this is done the energy equation must be obtained correct to the 2nd order. (This is necessary because, after differentiating the energy equation, a common factor 4 is removed to give the equation of motion). Whatever method is used for any system, the equation of motion must be obtained for the arbitrarily disturbed motion and put into the form (9.1) [or the form (9.3) if the system is unstable]. A The period of oscillation is then 2z/n, which, in the case given by eqn. (9.8) and for V"(q0) > 0, is 27r 1/{ V" (q0)1 A (go)} . Examples. (i) A bead P of mass m can slide on a smooth wire which is bent in to a semicircle of radius a and fired Fm. 119. with its plane horizontal. Two similar elastic strings, each of natural length c(< a V2), have one end attached to the bead; their other ends are attached to the ends A, B of the wire. Prove that the symmetrical position of the system is stable. If the bead is allowed to make small oscillations about the position of stable equilibrium, the strings never becoming slack, prove that the period of oscillation is the same as that of a simple pendulum of length mga 1121 (cf), f being the force that would give unit increase of length in either string. From Fig. 119, AP = 2a cos°, BP = 2a sine, and so the potential energy of the stretched strings is

V = lf (2 a cos 0 — c)2± -if (2 a sin 0 — c)2 = f{ 2 a2 c 2— 2 a c (cos 0 + sin 0)} . The velocity of P is 2 ae, so that the kinetic energy is 2ma2 02 , giving the energy equation 2ma 2 62 f{2a2 c 2 — 2ac(cos0 sine)} = E . Hence, the equation of motion is

4ma6 — 2fc(cos 0 — sine) = 0,

e

fc ma V2 sin(0 — +v-c) = 0,

showing that the symmetrical position, 0 = in, is a position of equilibrium. If we write 0 = + 99, the equation of motion becomes

+

fc . ma V2 sin 99 = 0 ,

306

A COURSE OF MATHEMATICS

which we compare with the equation of motion of a simple pendulum

115

(g 11) sin 99 = 0 .

Hence the pendulum of length 1 = mga I/21(f c) is exactly synchronous with the oscillations of P, provided the strings remain taut. (The condition c < a 1/2 implies that both strings are stretched in the equilibrium position.) Since P oscillates about the symmetrical position, that position is stable. (ii) A heavy particle B of mass M is attached to one end of a light inextensible string that passes over a smooth peg A, and carries at its other end a particle C of mass m which can slide freely on a smooth fixed vertical rod at perpendicular distance a from the peg. Show that if m < M there is a position of equilibrium in which the inclined portion of the string makes with the horizontal an angle a given by sin a = m/M, and prove that the period of small oscillations in which each particle moves vertically is 2n 11(a tan a)/ {g (1 — sin a)}] . From Fig. 120, OC = a tan 0 , AB = — a sec0, and the potential energy is given by

V = —mga tan0 — Mg(l — a sec0). The velocity of C is a 0 sect 0, and the velocity of B is a 0 sec 0 tan 0 . Hence the energy equation is lma2 62 sec4 0 ± 1Ma2 62 sec2 0 tan20 — mga tan0

Mga sec° — Mgl -= F.

Differentiation gives (mat sec4 0 + Ma2 sec2 0 tan20)e + ± (2 m a2 sec4 0 tan 0 + Ma2 sec2 0 tan3 0 Ma2 sec4 0 tan 0) 02

mga sect 0

Mga sec 0 tan() = 0 .

There is an equilibrium position 0 =

(1 )

6 - 0 if

m sec2 a = M sec a tan a , i.e.,

m = M sina,

so that there can be equilibrium only if m < M. To investigate the stability, in eqn. (1) we put 0 = a

y, where y and its

derivatives are small. Since 0 = y is already small of the first order we substitute the approximate value 0 = a into the remaining factor of the first term of eqn. (1).

9:3

307

STABILITY

Ij2 is already small of the 2nd order we can neglect the second term altogether. In the remaining terms we use Taylor's theorem ;

Since

see2 (a + y) = sec2 + 2y sec2 a tan a+ 0 (y2), sec (a

y) tan (a

y) = seca tana

y (sec3 + seca tan2 a) + 0 (y2 ) .

Hence, the equation of motion correct to the 1st order in y is

(m at sec4 + a 2 sec2 a tan2cc)

— ga(m sec2 a — M seca tana) —

— g ay (2m sec2 a tana — M sec3 — M seca tan2 a) = 0.

(2)

The condition for equilibrium m = M sina removes the second term of eqn. (2) and reduces it to Ma2 sec2 a tana(seca tan a)Y

Y

g

a seca tana(seca

(Mga sec a) y = 0,

tan a)

— 0,

or

g 1 — sin a y — O. Y+ a • tan a Since (1 — sin a)/tan a > 0, this shows that the equilibrium is stable and gives the required period for the small oscillations. Exercises 9:3 (a) 1. Two small smooth rings A and B, each of mass m, can slide on a fixed vertical circular wire of radius a, with centre 0. They are connected by a light elastic string of natural length 2a/3 and modulus 2 mg/V3. If they are slightly displaced from rest at the highest point of the wire, so that they separate, show that they will come instantaneously to rest when (3 sin — 1)2 = 3 113(1 — cos0), where 20 is the angle AOB. Show also that the rings will be in equilibrium if placed in the same horizontal line so that the angle AOB is 60°, and that, when making small oscillations about this equilibrium position, the length of the equivalent simple pendulum is 213 a/5. cos0), the 2. A fixed smooth wire is in the shape of the cardioid r = a(1 initial line being the downward vertical. A small ring of mass m can slide on the wire and is attached to the point r = 0 of the cardioid by an elastic string of natural length a and modulus 4mg. If the particle is released from rest when the string is horizontal, show that

a02 (1

cos0) — g cos0(1 — cos0) = 0 .

Show further that 0 = x/3 is a position of stable equilibrium and that the period of small oscillations about this position is 27r 1/(2a/g).

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A COURSE OF MATHEMATICS

3. A small ring of mass m can slide on a smooth circular wire of radius a which is fixed in a vertical plane. The ring is attached by a light elastic string, of natural length 3a/2 and modulus of elasticity mg, to a fixed point distant 3a vertically above the centre of the wire. Show that 6, the angle which the radius through the ring makes with the downward vertical, satisfies the differential equation ao

g sin0{3 (10 + 6 cosh)-1/2 — 1} = 0.

Deduce that cos 0 = —1/6 gives a possible position of equilibrium, and hence or otherwise show that the period of small oscillations about this position is

4x 27a t+. 35g 4. To the ends of a light inextensible string which passes over a small smooth fixed pulley are tied respectively a particle of mass M and a bead of mass m. The bead moves on a smooth straight horizontal wire which is fixed in a vertical plane through the pulley at a depth a below it. If the bead is displaced slightly from its equilibrium position, prove that it will return to this position after a time

I I( ma 2 Y \ Mg) 5. A ring of mass M, which slides on a smooth rod inclined to the vertical at an angle a, is attached by means of an elastic string to a point so situated that its least distance from the rod is equal to the natural length 1 of the string. Prove that, if 0 is the inclination of the string to the rod when in equilibrium,

cot° — cos° = Mg cosa/A, where A is the modulus of the string. Show that, if the ring is slightly displaced, the time of a small oscillation is

Ml 1 2n A 1 — sin3 0

The following examples concern systems which contain one or more rigid bodies. The methods used to solve the problems are the same as those used above. Examples. (i) ACB, CD are two uniform rods, each of mass a per unit length ; AC = CB = CD = a. The rods are smoothly jointed together at C and a light ring, attached at B, can slide smoothly on a fixed vertical rail BD to which the rod CD is smoothly hinged at D. A light spring, of natural length a and modulus 5 aag, connects B and D. Show that the small oscillations of the system about its position of stable equilibrium are synchronous with those of a simple pendulum of length 48a/275. In this problem we must determine first the equilibrium positions and then find the period of oscillation. Both these objects can be achieved by the use of the equations of motion.

§ 9:3

309

STABILITY

The potential energy is made up of the energy of the spring and the gravitational potential energy of the rods (Fig. 121). Therefore V —

5Grag (2a cos° — a)2 aagla cos° 2a 5a2 ag (4 cost 0 — 3 cos 0 2

a2aga cos0

1).

The kinetic energy of the rods is 1 a2 1 as 02 ± • 2o-a ia2 612 2 3 2

a2

3 6,21 = — o'ct3 02. 2

Hence the energy equation is aa2 g (4 cos2 0 — 3 cos0

.:;-aa3 62

1) = E,

and the equation of motion is 6a0

5g sin0(3 — 8 cos0) = 0.

The equilibrium positions are given by

(1)

d = 0, i.e.,

sin° = 0 or 3 — 8 cos0 = 0, giving 0 = 0 or 2T, and 0 = a = cos-1(3/8). yin eqn. (1) To investigate the stability of the position 0 = a we put 0 = a and obtain the equation of motion correct to the first order in y and its derivatives. By Taylor's theorem, from eqn. (1), 6a

5g(sina

y cosa) (3 — 8 cosa + 8y sina) = 0.

310

A COURSE OF MATHEMATICS

Since 3 — 8 cos a = 0, this equation becomes 6ay

40gy sin2 a — 0.

Therefore the position 0 = a is stable and the period of small oscillations is synchronous with a simple pendulum of length /—

6a 40 sin2 a

48a 275

To show that the other positions of equilibrium are unstable, first we put

= y, where y is small, in eqn. (1) which becomes 0

6a

± 5g sin y(3 — 8 cos y)

0,

i.e.,

bay — 25gy = 0 approximately. Since the coefficient of y is negative in this equation, the position 0 = 0 is unstable. Similarly, putting 0 = y in eqn. (1) gives 6a Y+ 5gsin(n+ y) {3 — 8 cos (n y)) = 0, i.e.,

6ay — 55(0 = 0

approximately, so that the position 0 z– L-c is unstable also. These results illustrate the fact that two unstable positions of equilibrium 0 = 0,3z are separated by a stable position 0 = a. (ii) A smoothly jointed rhombus 0 ABC of four equal rods of mass m and length 2a hangs freely from the fixed point 0, with a FIG. 122. light elastic string of natural length a and modulus of elasticity mmg connecting 0 and B. If B is supported at a depth a vertically below 0 and the system is released from rest so that B oscillates between depths a and b (< 4a), prove that n > 4/3 and that the depth of B below 0 when the system is in statical equilibrium is b). +(a Show that this equilibrium configuration exists if n > 2/3 and is stable, and that when n = 2 the oscillations about the equilibrium position are synchronous with a simple pendulum of length 13a/18. B

From Fig. 122 the potential energy of the stretched string is

nnag 2a

(4a cos°

a)2 =

nmga 2

(16 cos2 0 — 8 cos() 4- 1) .

§ 9:3

311

STABILITY

Using 0 as the zero-level for gravitational potential energy, we obtain, for the whole system,

V = —2mga cos° — 2mg • 3a cos° + i-nmga(16 cos2 0 — 8 cosa + 1) mga {871 cos2 0 — 4(n + 2) cos0 + }n). The angular velocity of each rod is 0. The rods OA, OC each rotate about 0 1 4ma2 • 02. The centre of AB has velocity components and have kinetic energy 2 3 a0 cos 0 horizontally and — 3 a 0 sin° vertically downwards. Therefore the rods AB and BC each have kinetic energy, 1 m2 a 02 • + 1 m (a2 02 cos20 +9a202 sin2 0) 2 2 3 1 ( 4ma2 • 1 4ma2 • 02 + 8 ma2 62 sin' 0) 02 + 2 • 2 3 3

.•.T = 2 • 2 =

8

•

m a2 02 (1 + 3 sin2 0).

The energy equation is, therefore, ,7na2 62 (1 + 3 sin2 0) + mga (8n cos2 0 — 4(n + 2) cos° +

= E.

Initially 0 = 0 and 4a cos0 = a so that

mga(in — n, — 2 + 4n) = E = —2mga. a02 (1 + 3 sin2 0) + g{8n cos2 e — 4(n + 2) cos0 + 1n) = — 2g.

(1)

If B oscillates between OB = a and OB = b, 0 = 0 when cos 0 = b/4a Therefore 8n, cos2 0 — 4(n + 2) cos° + 1(n + 4) = 0

.

when cos 0 = and when cos0 = b/4a. But 8n cos' 0 — 4(n + 2) cos 0 + En + 4) = (2 cos° — {4n cos° (n + 4)) = 0.

b • • 4a

n+4 4n

•

But b < 4a. Therefore n + 4 < 4n, i.e., n > 4/3. Differentiating eqn. (1) gives the equation of motion l

ae(1 + 3 sin2 0) + 16a02 sin0 cos 0 — g {16n cos° sine — 4(n + 2) sin0} = 0. (2)

In a position of statical equilibrium 0 = 0 = 0, and therefore sin° {4n cos° — (n + 2)) = 0,

312

A COURSE OF MATHEMATICS

+ a (n + 2) i.e., excluding the case 0 = n, cos° = n4n 2 and 0 B — since b = a (n + 4)/n. Since

=

a+b 2

0 > 0, cos0 < 1, n + 2 < 4n, i.e., n > 2/3. To investigate the stability we put 0 = a + y, where cosa = (n + 2)/(4n), in eqn. (2). The equation of motion, correct to the 1st order in y and its derivatives, is 3sin2 a)— 4g (sin a + y cosa) x

-1 36

X {4n cos a — 4ny sing — (n + 2)} = 0. After using the condition for equilibrium this gives V- a (1 + 3 sin2 a) + y 16ng sin2 a =. 0. Y

3 ng sin2 a a(1 + 3 sin2 a) Y

1`).

Since the coefficient of y is positive the equilibrium is stable. If n = 2, cos a =1, sing = 1/3/2, and 13a .. 18 Y

gY = 0,

so that the length of the equivalent simple pendulum is 13a/18. FIG. 123.

(iii) The smooth horizontal axis of support of a swinging body is moved horizontally at right angles to its length with constant acceleration g tan A . Show that the plane containing the axis and the centre of mass of the body can maintain a constant inclination A to the vertical and that the period of small oscillations about this position is 27r i/{k2 cos2/lg}, where k is the radius of gyration of the body about the axis of support and 1 is the distance of the centre of mass from this axis. Since there is an external force acting on this system which does work on the system we cannot easily use the energy equation; consequently, we work with the equation of motion. The forces acting on the body are shown in Fig. 123. We take moments about 0, the axis of support, using eqn. (7.42), and find

M1c760 M 12 0 — g tanA l cos0) = —Mg 1 sine, (

(k + 12) o -= g 1 (cos° tan) — sin0), where k, is the radius of gyration about a horizontal axis through 64.

§9:3

313

STABILITY

For a steady state 0 = a, 6 = 0. .•. cosa tang, — sina = 0, Lana = tan2, a = 2. y and note that (by the To investigate the small oscillations we put 0 = theorem of parallel axes) k2 = ko +12. We find, correct to the first order in y,

y = gl {tang

(k 02 + /2)

(cos A — y sin2) — sin2 — y cos A} .

key = —gly (sin) tan). ± cos A) =

gl cost

y.

This shows that the steady state is stable and that small oscillations have a 2s

period 27r (

k ;

'

Exercises 9:3 (b) 1. A thin hollow cylinder of radius a has a particle of equal mass attached symmetrically to its inner surface. If the system is disturbed from its position of stable equilibrium on a rough horizontal table and then left to itself show that, when the radius to the particle makes an angle 0 with the downward vertical, a 62(2 — cos 6) — g cos0 = constant. Hence, or otherwise, prove that the period of small oscillations is 2x 1/(2a1g). 2. A uniform rigid rod AB, of mass 2m and length a, is smoothly hinged to a fixed point at A and has a particle of mass m attached to B, which is connected by a light elastic string of modulus mg and natural length 4a/5 to a point C vertically below A. If AC = 2a, show that the period of small oscillations about the position of stable equilibrium is 2zr{ 2 a/(3 g)}112. 3. A hollow right-circular cylinder of internal radius 7a is fixed with its axis horizontal. A rough solid circular cylinder of radius a can roll in the interior so that its axis remains horizontal during the motion, friction being sufficient to prevent sliding. When the plane through the two axes makes an angle 0 with the downward vertical, show that the angular velocity of the movable cylinder is 6 O. By means of the equation of energy, or otherwise, prove that, if the movable cylinder is slightly disturbed from its position of equilibrium, it will oscillate with period

4. A uniform cube A BCDEFGH of edge 2a rests in equilibrium with the face CDEF in contact with a fixed rough sphere of radius r. If the cube is disturbed so that it rolls without slipping and the face ABCD remains in the same vertical plane, show that when the face CDEF makes an angle 0 with the horizontal, (5a2 + 3 r2 02) 62 + 6g {r (cos 0 + 0 sin 0) + a cos 0} = constant. Hence, or otherwise, show that equilibrium is stable if r > a, and find the period of a small oscillation about the equilibrium position.

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A COURSE OF MATHEMATICS

5. Prove that the moment of inertia of a thin hemispherical cup of mass M and radius a about an axis through its centre of mass parallel to the plane of the rim is 5Ma2 /12. If the cup rests in equilibrium with the rim uppermost and the curved surface in contact with the highest point of a fixed rough sphere of radius 2 a, prove that the time of a small oscillation about this position of equilibrium is the same as that of a simple pendulum of length 4a. 6. A uniform rod of mass m and length 1 turns freely in a vertical plane about a fixed axis through one end 0. A light elastic string of modulus mg and natural length 11 has one end attached to the mid-point of the rod and the other to a fixed point A at distance -II vertically above O. If 0 is the angle made by the rod with the upward vertical OA, show that the equation of motion of the rod is

210 = 3g(1 — 2 siri40) cos 10. Show also that the period of small oscillations about the position of equilibrium

0 = zr is the same as that of a simple pendulum of length 81/9. 7. A uniform wire, of mass M, is bent into the form of a semi-circle of radius a and is mounted so that its ends can slide freely along a fixed horizontal straight line. Small smooth light rings attached to the ends of a uniform rod of mass m and length 2a sin a are threaded on the wire. Show that, when the system performs small oscillations in a vertical plane about the stable position of equilibrium, the length of the equivalent simple pendulum is a [3 M costa

(M

sin2 a] /[3 (M

m) cos a] .

8. The cross-section of a smooth fixed cylinder is a parabola with latus rectum 4a, axis vertical, and vertex upward. A pair of uniform equal rods, freely jointed together at their upper ends, rests symmetrically over the cylinder in the plane of a cross-section, and in their position of equilibrium the angle between the rods is 2a. Show that the length of each rod is 4a cosa cosec4 a, and that the period of small symmetrical oscillations, in which the joint moves vertically, is 2x cosa cosec2 a1/(2 a/3 g).

9 :4 Motion on a rotating curve A bead is threaded on a smooth wire which is bent into the form of a plane curve whose equation is y = /(x). If this plane and the wire carrying the bead are made to rotate with a constant angular velocity w about the vertical axis 0 y , the bead will, in general, move on the curve. Although strictly this is a three-dimensional motion we include it here because it gives an example of a state of steady motion whose stability we can investigate by the method of this chapter.

j 9 :4

315

STABILITY

If Q (see Fig. 124) is the foot of the perpendicular from P on to the axis Ox, then the polar coordinates of Q in the horizontal plane are (x, wt) at time t. Hence Q has a component of acceleration i — co2 x away from 0 (i.e., in the radial direction) at any instant. The acceleration of P therefore has a component — w2 x parallel to Ox and a component ij parallel to Oy. (There is also a component perpendicular to the plane Oxy, unless P is stationary relative to the wire). The forces acting on P in the plane Oxy are its weight, mg, and the reaction, R, of the wire perpendicular to the tangent. (There is also a component of the reaction which acts perpendicular to the plane of the diagram.) The equation of motion resolved along the direction of the tangent at P is (9.11) m( — w2 x) cosy, + my sirup = —mg sin y) . In this equation we write, X, = u, y = v, cosy) — dx/ds, silly, = dy/ds, where s stands for arc length, and also put =

du du dx du dv = = =u d x ' 'fi dt dx dt dt ( u du dx

—

ds 2

dv dy dv =v dy dt dy

dx dv dy )ds +v dy ds -=

w x

±1v2

—

dy

g ds ,

2w 2 x2 gy) = O.

Integrating this equation we obtain 2 1(112 + v ) :

- ICO2 X2

gY

(9.12)

where E' is a constant. This eqn. (9.12) resembles an energy equation except for the sign of the term — co2 x2. Since the curve is being made to rotate, the rotating agent is doing work on the system at an unknown rate. Therefore, we cannot write down an energy equation; eqn. (9.12)— a 'pseudo-energy equation' —must be used instead. Since the particle moves on the curve y = f(x), u = X, v = f' (x)X. [1 + {f' (x)}9X2 — co2 x2 +2g f (x) = 2E'.

(9.13)

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A COURSE OF MATHEMATICS

If the equation of the curve is given by a parameter q where x = x (q) , y = y (q) , so that

(q)

, y = y (4)4,

eqn. (9.12) takes the form, (x'

2 4_ y,2) 42

(02 x2

2gy = 2E'.

(9.14)

(The primes denote differentiation w.r.to q and x, y are both functions of q.) Investigation of relative equilibrium of the particle and the rotating wire, and of the stability of this equilibrium is conducted along lines

Fm. 124. exactly similar to those of § 9:3 using (9.12), (9.13) or (9.14) instead of the energy equation. Of course, the equation of motion (9.11) can be used directly if more convenient. The force R between the bead and the wire in the plane 0 x y can be found by the use of the equations of motion and the 'pseudo-energy equation' in one of its forms. From Fig. 124, by resolving horizontally

— co2 x) = — R siny,, and x may be found by differentiating eqn. (9.13). [See example (ii) below.] Similar methods can be used to solve problems involving the motion of a bead threaded on a smooth wire which is caused to rotate in its own plane about an axis perpendicular to its plane.

§ 9:4

STABILITY

317

Examples. (i) A smooth wire in the form of a circle of radius a rotates with constant angular velocity co about a vertical tangent. Derive the equation of motion of a bead of mass m sliding on the wire in the form

a0 + a co2 (1 + cos 0) sine — g cos 0 = 0, when the radius to the bead makes an angle 0 below the horizontal. If 0 = a in the position of relative equilibrium, deduce that the period of small oscillations of the bead about this position is that of a simple pendulum of length a sin a 1 — cosa cos2 The coordinates of the bead are (Fig. 125)

x = a(1

cose), y = — a sine.

The components of acceleration of the bead in the plane of the circle are

x — co2 x = — a (0.sin s + 62 cos0) — a co2 (1 + cos 0) in the direction Ox, and

= —a(6 cos0 — 62 sine) in the direction Oy. The equation of motion resolved along the tangent at P is

m(X — w2 x) sine + 994 cos° = —mg cos0. .•. a.o

aco2 (1

cos0) sine = g cos0.

(1)

For relative equilibrium 0 = a, 6 = 9 = 0. a co2(1 + cosa) sina =- g cosa. Put 0 =

(2)

y in (1). Then

a + a0(1 + cosa — y sina) (sina y cos a) = g (cos« — y sina)

(3)

approximately. Substituting for w2 from (2) in (3) and some reduction gives

ay sina gy (1 — cosa cos2 a) = 0 giving the length of the ESP as required. (ii) If the bead of example (i) above is released from rest relative to the wire at

A, find R in terms of 0. Resolving the equation of motion horizontally gives

R cos0 = — m(Fc — w2 x) = ma(e sin

02 cos 0) + ma w2 (1 + cos 0) .

Equation (9.14) in this case becomes

a2e2

a2 (1 + cos0)2

2ga sine = 2E'.

(4)

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A COURSE OF MATHEMATICS

The initial conditions give — 4a2w2 =211]'. .•. a(32 = 2g sin 0 + a ca2 (1 -F cos 0)2 — 41.

(5)

[This equation could also have been obtained by integrating eqn. (1) of example (i) after multiplying by the integrating factor B.] Using eqn. (5) and the equation of motion (1) for substituting into eqn. (4) gives, after some reduction,

B 3mg sine + ma w2 (2 cos2 0 + 3 cos° — 3). Exercises 9:4 1. A uniform circular ring of mass m and radius a is made of fine smooth wire. The ring is mounted so that it rotates freely about a vertical diameter. A bead of mass 2m slides on the wire. Prove that, if the bead is in relative equilibrium when the radius to the bead makes an angle a with the downward vertical, the corresponding steady angular velocity of the ring is w where awe = g sec a. The system is in relative equilibrium with a = a/3 when it is slightly disturbed. Prove that the period of the subsequent small oscillations is that of a simple pendulum of length si a . 2. A wire, in the form of an ellipse with axes 2 a, 2 b , is made to rotate with constant angular velocity co about the former axis, which is fixed in a vertical position, and a smooth heavy bead is free to slide on the wire. Prove that, if co2 > ga/b2, the bead can remain in relative equilibrium at a depth ga2 /b2 w2 below the centre of the ellipse, and that the period of a small oscillation about this position is 2n a /1 b4 — (a2 — b2) g2 1 b4 w4 — a2 g2 bw I_ j• 3. A small heavy bead can slide on a smooth wire in the form of the cycloid

s = 4a silly), (in > > —1n) the axis of symmetry being vertical and the vertex downwards. The wire is made to rotate with uniform angular velocity about the axis of symmetry, and the bead is at rest relative to the wire at the point given by q,o =arc. Show that, for small oscillations about this position, the length of the equivalent simple pendulum is 4a 2a . a

4. A bead, of mass m, is threaded on a smooth thin wire in the form of a parabola of latus rectum 4a. Initially the system is at rest in a horizontal plane with the bead at the vertex. The wire is then made to revolve with uniform angular velocity w about a vertical axis through its focus. When the bead is at a distance r from the focus, prove that the velocity of the bead relative to the wire is car, and that the horizontal component of the reaction of the wire on the bead is lmco2 [4r — 31,(ar)]. 5. A small bead is threaded on a rough straight rigid wire which is made to rotate in a plane with uniform angular velocity w about a point 0 of the plane. The perpendicular from 0 on to the wire is of length a and meets it at A. Show

§ 9:5

319

STABILITY

that the acceleration of the bead in space has components j — ya)2, 2p w -I-- a al respectively along and perpendicular to the wire, p being the relative displacement of the bead along the wire measured from A. The bead moves under no forces other than the reaction between it and the wire. Initially it is at A and has a velocity 2a co in space and a co relative to the wire. If the angle of friction is r/6, prove that the bead will ultimately come to rest relative to the wire at a distance a/1/3 from A.

9:5 Perturbation of an orbit In § 6:4 the stability of a circular orbit described under the action of a central force was investigated. We found that for the inverse square law of attraction the disturbed motion differed only slightly from the steady circular motion which was therefore,stable. This was a case of a steady motion being disturbed. Whilst motion in an elliptic orbit cannot strictly be considered as steady, nevertheless it is a periodic motion. Further, it is of some importance to know what changes occur in the orbit as a result of small disturbances. Such changes may occur in the motion of a satellite when it meets a meteorite, or when an artificial satellite is given a small impulse from its own rocket motors. We consider here two examples of the effect of small disturbances applied to a particle describing an elliptic orbit under a central force ,u1r2. Examples. (i) The particle receives a small tangential impulse which increases its velocity from v to v 5v. We suppose that the change takes place instantaneously so that the position 2 1 of the particle does not alter during the impulse. Since v2 = in any 117 — elliptical orbital motion, when v alters, it and r remaining the same, a must alter. Hence 2v O•t• -

pocc a2 •

This gives the change in the major axis of the ellipse. Since the focal distances arc equally inclined to the tangent, whose direction is unaltered in this change, the empty focus S, of the new orbit lies on the line PS' (see Fig. 126). Since SP + PS' = 2a, SP + PS, = 2(a Oa). S'S, = 2 Oa. The length of the perpendicular from S to the tangent is unaltered so that, since

h = pv, 6h, = p ov or

6hPh

6vIv.

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A COURSE OF MATHEMATICS

But h2 = as = iabz . •• 6b •.* b

2 6h

2 ab

h

b

6v 3a v + 2a

ba a• av3v

dv

v +

These results fix the position and size of the new orbit. (ii) A particle in an elliptic orbit receives a small impulse which imparts a velocity 6v along the outward normal.

Fm. 126.

FIG. 127.

Since the velocity V after the impulse is given by V2 = v2 + (Sv)2, 2 1\ , the length r aI of the major axis is unaltered. The direction of the velocity is turned through an angle 3v/v; hence S'PS,=- 2 6 vlv . (See Fig. 127). Since the sum of the focal distances is the same before and after the impulse, PS' = PS,. Hence the position of S, is determined and the new orbit is, in effect, known. Any other disturbance is the superposition of two disturbances such as (i) and (ii) above. we see that, correct to the 1st order, V = v . Because v2 =

The following examples show the effect of disturbances on other types of orbit. Examples. (i) A particle of mass m describes an ellipse under the action of a force mn2 r towards the centre of the ellipse. It receives a small impulse m 6v in the direction of its velocity v when the eccentric angle is . Show that the small changes in the lengths of the semi-axes are given by 6v Sa -=--sn2p, a v

6b

6v =

C"2

and that axes of the ellipse are turned through an angle ab dv sin 2cp. a2 — b2 v

'

§ 9:5

321

STABILITY

The equation of motion is

r

= —n2 r

which has a solution r= a i cosnt + bj sinnt, v = —na i sinnt + nb j cosnt, when the coordinate axes lie along the principal axes of the ellipse which are of lengths 2a, 2 b, and the eccentric angle 9) = nt. After the impulse the particle moves on an ellipse with axes of lengths 2 (a + 6 a), 2 (b + 6b) which lie along the directions i + Si, j + 6j, where

Oi = j 6a, 6j = —i 6a, (see § 4:2), inclined at an angle 6a to the original directions Immediately after the impulse the eccentric angle in the new ellipse is q + Sq) and the new vein-

6v

city, in the case in question, is v (1 ± , the position vector r being unaltered. Hence or = 0 = 6a i cosq) ± a oa j cosq) — a i 6q) sing) ±

Ob j sing) — b Oa i sing,

b j 699

COST.

Equating the coefficients of i, j to zero gives

6a COST — a 6co sing) — b ba sing) = 0,

(1)

ob sing) + b OT cos g) ± a Oa cos g) = 0.

(2)

The corresponding increment in v is

6v v

6v

na

6v

i sing) + nb

6v

j cos q)

=- —n 6a i sing, — n,a Oa j sing) — na i 6q) cos 92 ± n j cow — nb Oa

i COST -

nb j 6q) sing).

Equating the coefficients of i, j gives

6v Sa sing) ± a 699 cosq b oa cow = a — v

(3)

6v ob cos q) — b q, sinT — a 6 a sin99 — b v cosT.

(4)

From eqns. (1) and (3) we obtain

6v Oa = a -7 sin2 g), and from (2) and (4)

6v ob = b— cost v

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A COURSE OF MATHEMATICS

We also solve (1) and (2) for 6a and 69), obtaining

6a a 61) tang) .•.

—

q) b 6a cot q) a 6b tang) a2

1 b2 —

— a Oa cot()) — b 6b tan q)

b Oa cotq) b2

-

ab

Ov

a2 — b2 v

(ii) A smooth straight wire, of negligible mass, can rotate freely in a horizontal plane about its end 0 which is fixed. A heavy bead slides on the wire and is connected with 0 by a light spring of natural length 1. When the wire is at rest, the period of an oscillation of the bead on the wire is 2n/n. Find the position of relative equilibrium of the bead T with respect to the wire in the state of steady motion when the wire is rotating with uniform angular velocity w (n > a)), and prove that, if the particle is displaced slightly from this position, the period of a small oscillation of the system about the state of FIG. 128. steady motion is 2 n/(n,2 + 3 co2)1. This problem, although superficially of the type discussed in § 9:4, is in effect an orbital problem, for the wire carrying the bead is free to rotate, and is not constrained to rotate with a fixed angular velocity. In other words, the bead traces out an orbit in the horizontal plane under the action of a central force due to the spring (Fig. 128). The equations of motion of the bead are

— r02) = — T , m d (9.2 0) = 0, r dt 2 where the tension T = — - (r — 1), 2 being the modulus of elasticity of the spring. 1 Hence — 7.62 _

ml

(r

1), rb. ± 2r

o.

(1)

When the wire is stationary the equations of motion become _

2 ml

(r

/), 0 = 0.

The first of these corresponds to simple harmonic oscillations about the point

r= 1; hence n2 = Alml.

9:5

323

STABILITY

When the wire is free to rotate the conditions for steady motion are r = x, = 0, O. Hence — x co2 = — n2 (x — 1). x=

2121 n2 — w2

(2)

'

giving the position of relative equilibrium. To investigate the oscillations about this steady motion we put r = x 0 = w z. The equations of motion become (y and z being small) (x

y) (w2 + 2 co z (x

z2) =

n2 (x

/

y),

y,

(3)

y) 2g(co z) O. (4)

The zero-order terms in (3) cancel out because of (2), leaving, correct to the first order in y and z, y — w2 y — 2wzx n2 y = 0. (5) The first order terms of eqn. (4) give xi; + 2a4 = 0, which gives, on integration, xz = b — 2wy, where b is a (small) constant of integration. Substitution into eqn. (5) gives y + (n2 w2) y 4 w2 y = 2wb,

g + (n2 + 3 w2) y = 2 cob. 2 co b , hence the steady motn2 ion is stable the period of the oscillations being 2n/1/(n2 + 30). [The constant b is zero if the disturbance does not alter the angular momentum of the system.] This shows that y oscillates about the value y =

+3 w2

Exercises 9:5 1. A particle is describing an elliptic orbit about a centre of force in a focus. Prove the formulae, with the usual notation, h2 =y/,

v2 -= (21r — 1/a).

If, when the particle is at its greatest or least distance from the centre of force, it receives an impulse which communicates to it a small radial velocity Sy, prove that the apse-line is turned through an angle hôvleue. Show also that

h2 (6 ,2) /(210

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A COURSE OF MATHEMATICS

2. A satellite of negligible mass and under no atmospheric resistance is describing an orbit of semi-major axis a and eccentricity e under an inverse square force ,a/r2. It is possible, at some selected point of the orbit, to increase the tangential speed by a small amount (5v. Show that, if the object is to achieve the greatest increase in the pericentric distance all — e), the increase in speed must be effected at the apocentre, and that the amount of the increase in the pericentric distance is then 4 {a3 (1 — e)//4,(1 e)}2 ov. 3. A particle, of mass M, is describing an ellipse, whose major and minor axes are 2a, 2b, under a force to a focus S. Another particle, of small mass m, describing the same ellipse in the opposite sense under the same law of force, collides and coalesces with the first particle at one end of the minor axis. Prove that the new orbit is an ellipse. If terms of order m2 /M2 are neglected and Oa, 61) are the changes in a, b, prove that Oala = nib = —4m/M. Show also that the angle between the major axes of the two ellipses is 4mb1M1/(a2— b2) .

9:6 Systems with more than one degree of freedom In this section, but without proofs, we acquaint the reader with the generalisations of our results on stability and small oscillations to systems with more than one degree of freedom. In illustration we use the example (iii) of § 7:3 (p. 217). In that example the displacements x, y of A, B respectively were shown to satisfy the differential equations

(3D2

5n2) x = n2y5 (D2

+ 7i2) y

=Th 2 x.

(9.15)

To solve this pair of simultaneous differential equations it is convenient to adopt a somewhat different approach from the general method given in Vol. II § 2 :4. We multiply the second equation by A and add it to the first giving D2 (3x 4- Ay) ± n2 [(5 — 2)x + (-1 ± 2.)y] -= 0. We choose 2 so that the coefficients in the first term are proportional to the corresponding coefficients in the last term, i.e., 2 is chosen to satisfy the equation 3

5—2 —

—1 + 2'

22 — 3 = (A — 3) (2 1) = O. .'. 2 = 3 or —1.

§ 9:6

325

STABILITY

These values of 2 lead to the equations 3D2 (x + y) + 2n2 (x + y) 0, D2 (3x — y) + 2n2 (3x — y) = 0, or D2 +n 0, D227 + 2n2ii = 0 (9.16) after making the transformation of variables = x + y,

= 3x — y.

(9.17)

The new variables $, n are called normal coordinates. The solutions of eqns. (9.16) are = Clcos f-R nt +

C2 sin

nt, rl _=

C3cos

i/2nt + C, sinpnt,

and the solution for x, y is given by substituting these into x = 4( +97),

Y = -1- (3 e —

27) •

(9.18)

First we notice that the oscillations of x, y are compounded of two oscillations which have different frequencies; each normal coordinate oscillates with one of these frequencies. In general, a system with n degrees of freedom has n characteristic frequencies which occur in oscillations about a position of equilibrium. A suitable transformation of variables, such as (9.17), gives n normal coordinates each of which satisfies an equation of motion such as one of eqns. (9.16). Only if all these n equations of motion are of the simple harmonic motion type will the system remain near its equilibrium position, i.e., only then is the equilibrium stable. The potential energy of this system of two pendulums (Fig. 87) is V = —3mga cos° — mga(cos0 + cosc9) = — mga(4 cos() + cos(p). (9.19) But sin° = xla, since = (y — x)la; and since x, y are small, correct to the second order, (y — )2 cos° = 1 2a2 , COST= 1 a2 2 a2x 2 (y — x)2} V = mgai— 5+ 2x a2 + 2 a2

(9.20)

If V0 ( = — 5mga) is the potential energy in the equilibrium position and we make the substitution (9.18) into (9.20), we find mg (e2 ,2) (9.21) V — V, = . 4a

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A COURSE OF MATHEMATICS

The position given by x = 0, y = 0, or by e = o, = o, is the position of equilibrium and is the centre of the oscillations represented by eqns. (9.16). The form of V — Vo in (9.21) as the sum of squares with positive coefficients, shows that V has a minimum, in the strict analytical sense, at the position of stable equilibrium. This is an important feature which is true for a general system disturbed from a position of stable equilibrium. When the normal coordinates are substituted into V, the function V — Vo ( Vo again being the value of V in the equilibrium position) is a positive definite form being a sum of squares with positive coefficients. Only if all the equations of motion such as (9.16) are of the simple harmonic type do all the coefficients of V come out to be positive, i.e., only in the stable case does V have a strict minimum. Our general conclusion can be stated thus: When a system has n degrees of freedom the potential energy V is a function of n variables, which may be 'generalised coordinates' introduced in § 2 :5; the positions of stable equilibrium are those for which V has a minimum, in the strict analytical sense. When such a system is disturbed from a position of stable equilibrium, the resultant small oscillation is compounded of n different frequencies. A suitable general proof of these results, and a general method of finding the frequencies and normal coordinates require more powerful methods than lie within the scope of this book and are deferred to Vol. VI. We now give one example of the use of potential energy to examine the stability of equilibrium of a system with 3 degrees of freedom. Example. Two equal, uniform straight tubes AB, A C are rigidly joined at A so that CAB 2a. Each tube is of weight W. Inside each tube a smooth particle of weight W is attached to A by a light elastic string of natural length a and modulus A. Show that the symmetrical position is stable for all values of A when a < `7r, but that when a > i n this position is stable only if A > +W(seca — 2 cosa). The potential energy

A

— 2 Wa cosa cos° —

V — 2a (x .' —

W { (a + x,) cos (a — 0) + (a + x2) cos (a + 0)}

§9:6

327

STABILITY

is a function of three variables x1, x2 , 0 (see Fig. 129). The conditions for equilibrium are the conditions for a stationary value of V, i.e.,

aV

2x,

ax,

a

aV ao

— W cos (a — 0) = 0,

= 2 Wa cosa sin0

W{(a

aV

ax, —

Ax2 a

W cos (a + 0) = 0,

x1) sin (a — 0) — (a + x2 ) sin (a + 0)1 = 0.

FIG. 129.

The symmetrical position is a position of equilibrium if

0= 0,

x, = x2 —

Wa

Loss.

In V we write

x, =

Wa Wa 2 cosa yi , x2 = cos a + Y2,

where yi , y2 and 0 are small, and expand V correct to the second order in these variables. The result is

V = V 0 +–2— a (y2,

W(Y1 — Y2 )0 sing Wa (2

W cosa

) 02 cosa,

where V0is the equilibrium value of V. We can rearrange this in the form

V—

V0

2 (y, 2a

Wa sina y

Wa sina )2

+ 2a (Y2 + Wa cos a (2 +

W cosa ) 2 02

W2 a

02 sinza .

328

A COURSE OF MATHEMATICS

This last form of V makes V — Vc, a positive definite form ( a sum of squares with positive coefficients) if the coefficient of 02 is positive, i.e., if cos a (2 ± which reduces to

W cos a ) A

W sin2 a A

> W (sec a — 2 cos a).

The equilibrium is stable if this condition is satisfied. Hence, if a > 4.x, the symmetrical position is stable for all A ; if a <, this position is stable only if satisfies the specified inequality. When a= fna more detailed examination is required. Exercises 9:6 1. If x, y are the displacements of C, D respectively of question 1 of Ex. 7:3 prove that the transformations 3x = 2i, 3y = l — i give the normal coordinates

71. 2. For the system of question 3 of Ex. 7:3 show that the potential energy can be put into the form m n2 3,72} {8a2 V— 48 where the normal coordinates ri are given by =x

y,

— y.

3. A light string of length 7a has two particles of equal mass m attached to one end, and to a point distant 4a from that end, respectively. It hangs freely from the other end under gravity. If the system performs small oscillations, in which the particles move approximately horizontally in the same vertical plane, find the two differential equations for the displacements x, y of the upper and lower particles. Show that the periods for normal modes of oscillation are T and T y6, where T = 2x Valg), and find the corresponding values of the ratio xly.. Sketch the configurations in the two cases. 4. Two particles of equal mass m are attached respectively to the mid-point A, and a point B distant a from it, of a light string of length 4a fixed at the ends and stretched by a constant tension T. Find the periods of normal modes of transverse vibration and show that the ratios of the corresponding amplitudes of the masses at A and B are (1 ± V17)/4. 5. A smooth tube is pivoted at one end. Inside the tube and with one end also fixed to the pivot is a light spiral spring with particles of mass m attached at its mid-point and at its free end. Each half of the spring has stiffness A . When the tube is constrained to rotate in a horizontal plane with angular velocity, w, the particles move in a vibration relative to the tube. Find the periods of the normal modes of vibration, showing that for the motion to be necessarily vibratory a)2 <

3 — 1/5 2 m

STABILITY

329

6. A uniform rod AB of length 2a is suspended from a fixed point by a light string of length a attached to a point C of the rod so that AC = 2a/3. By using appropriate force and moment equations prove that if at time t the string and the rod make small angles 0, 92 with the downward vertical in the same sense, then a(3

cp) = —3g , a(p = g (0 — 92).

Hence show that the periods for normal modes of small oscillations in a vertical plane are 2nIco1, 2n/co2 where w3 , w2 satisfy

3a2 w4 — 7gaco2 3g 2 = 0.

Miscellaneous Exercises IX 1. Two equal uniform rods AB and AC, each of length 2 b , are freely jointed at A. The rods rest on a smooth circular cylinder of radius a with their plane perpendicular to the axis of the cylinder, which is horizontal. Prove that if 2 0 is the angle B AC when the rods are in equilibrium with A vertically above the axis of the cylinder then b sin3 0 = a cos°, and show that this equation has only one root in the range 0 < 0 < x/2. Show, also, that if the rods are constrained to move in a vertical plane and A is constrained to move vertically the position of equilibrium is stable. 2. A uniform bar AB of length 1 and weight W1is smoothly hinged to a fixed point A. The bar is supported by a light inextensible string attached to B and passing over a small light smooth pulley at a point C, which is vertically above A at a height a (> 1). The string carries a freely hanging weight W2 at its other end. Show that there are three positions of equilibrium, provided that 2a 2a W1 a + / < W2 < a — / If Wl = 2 W2 show that two of the positions of equilibrium are stable and one is unstable. 3. A thin hollow cylinder of radius a has a particle of equal mass attached symmetrically to its inner surface. If the system is disturbed from its position of stable equilibrium on a rough horizontal table and then left to itself, show that, when the radius to the particle makes an angle 0 with the downward vertical,

a 62 (2 — cos()) — g cos° = constant. Hence, or otherwise, prove that the period of small oscillations is 2ot Y(2a/g). 4. A uniform rod AB of mass M and length 2a rests in equilibrium on a perfectly rough fixed horizontal circular cylinder of radius a, with AB perpendicular to the axis of the cylinder. A particle of mass m is fixed to each end of the rod. Show that, if the rod rolls on the cylinder in a vertical plane, its kinetic energy is given by a2 02 {302 (M + 2m) + (M 6m)}/6,

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where 0 is the angle of inclination of the rod to the horizontal. Show that the period of small oscillations of the rod about its position of stable equilibrium is 27c.

/1 a(M ± 6m) g(3M + 6m) •

5. A small ring is free to slide on a smooth rigid wire in the form of a catenary. The wire is made to rotate with uniform angular velocity w about its axis of symmetry which is fixed and vertical, and it has its vertex downwards. If c be the parameter of the catenary, find the value of w when the ring is at a position of relative equilibrium such that the tangent to the wire at the ring makes an angle In with the horizontal. Show that the equilibrium is stable and that the time of a small oscillation about this position is 4x {c/(2g — cco2 )}÷. 6. A straight rod can turn freely in a vertical plane about a point A of its length, the point A being fixed. A light inextensible string has one end attached to another point B of the rod, passes over a small smooth pulley C, fixed in the plane in which the rod can turn, and carries at its other end a weight of such magnitude that there is equilibrium when the rod is horizontal. Show that the equilibrium is stable, if either of the anglesABC orACB is obtuse, and is unstable, if they are both acute. If either of these angles is a right angle, show that the equilibrium is apparently neutral but really stable. 7. A heavy uniform rod of length 2a and weight W carries light rings at its ends A and B. The ring A slides on the smooth horizontal constraint Ox, and the end .B (which is below A) slides on the smooth vertical constraint Oy. An elastic string of natural length a and modulus of elasticity W, fastened at one end to the ring B, passes through a smooth fixed ring at 0, and the other end is joined to the mid-point of the rod. If the angle OBA is 0, and 0 = a gives the position of stable statical eqUilibrium, show that when the rod is released from rest in a horizontal position it comes to rest instantaneously with 0 = (3, where cos/3 = 2 cos a = 4. 8. A uniform solid sphere, of mass M and radius a, rests on a rough horizontal plane with one third of its weight supported by an elastic string attached to the highest point A and to a fixed point at a height 2a vertically above A. A particle of mass m is rigidly attached to the sphere at the point A. Show that, if M > m and the system is slightly disturbed, the sphere makes small oscillations in the period 27c

[(7M

20m)a14•

5 (M — mg

CHAPTER X

IMPULSIVE MOTION AND VARIABLE MASS 10:1 Introduction The motions of particles, bodies and systems considered so far are motions for which the velocities vary as continuous functions of the time. We turn now to the consideration of the limiting cases of such motions in which for certain (short) intervals of time the accelerations are so large that the velocity, effectively, changes discontinuously. Of course, large forces must act on the bodies during the period of such changes in order to produce the large accelerations. The equation of motion, (5.1), of a particle under the action of a force F may be integrated from time t = 0 to time t = r to give m (v — u) = J Fdt= I,

(10.1)

where u, v are the velocities of the particle at the beginning and the end, respectively, of the time-interval (0, x). This equation defines the impulse, I, of the force F in the interval (0, x) and shows that the change of momentum of the particle is given by Zip

m(v — u) =- I.

(10.2)

Suppose, now, that motion occurs in a straight line and that F varies with the time according to

F = Fo sin (gttlx).

(10.3)

This force F is zero at the beginning and end of the interval and rises to a maximum, Fo, during the interval. On integration we find

= f F dt = 2F0 //Jr.

(10.4)

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A further integration of the equation of motion gives, for this case,

ms = MUT + F0 r2/7r

war +

,

(10.5)

where s is the distance covered during the interval. If we used some other form for F such that F = 0 when t = 0, T, e.g., F = Fo (t17)(1 — t/ r) we would obtain a result differing from (10.5) only in the coefficient of Jr. Now we suppose that the interval z is decreased and F0 is increased so that I remains finite. Equations (10.2) and (10.5) show that, in the limit r 0, the velocity (or momentum) changes discontinuously but that the position does not alter, i.e., s > 0 and the variation of F with time becomes a delta (or impulse) function (see § 3:8). This limiting case is an approximation to the state of affairs when a particle receives a 'blow' rather than a 'push'. This type of situation occurs when one body, such as a billiard ball, collides with another; when a shell is set into motion by an explosion ; when a body is jerked into motion by a string, etc. The processes which occur in practice require a short but finite time for completion; the billiard balls change shape slightly, and the string increases in length slightly. Nevertheless, for the purposes of mechanics we consider, as usual, idealised, limiting cases by making special, simplifying assumptions. We assume that, in impulsive motion: (i) the momenta, or velocities (linear and angular), in general, alter discontinuously; (ii) the positions of the particles of the system do not alter; (iii) the impulses act instantaneously. We emphasise two points arising from these assumptions. First, in —

the limiting case the impulse is I = lim

F dt, and eqn. (10.4)

0

implies that I F I becomes infinite ; therefore any forces whose magnitudes must remain finite cannot contribute to the impulse. Therefore gravitational forces, i.e., the weights of bodies, in particular do not occur in impulsive motion equations. Second, the equations of motion apply to one instant only. The equations of continuous motion apply for times up to the instant of impulse ; at that instant the equations of impulsive motion apply and the velocities change discontinuously; subsequently the equations of continuous motion apply again starting with the new velocities.

§ 10 : 2

IMPULSIVE MOTION AND VARIABLE MASS

333

The equation of impulsive motion of a particle is eqn. (10.2). To obtain the equations of impulsive motion of a system of particles we develop eqn. (10.2) in much the same way in which we developed the Law of Motion in § 7:4.

10:2 The impulsive motion of a system of particles We use a notation and treatment similar to that used in § 7 :4 and § 7:5. We assume that each particle Pi (i = 1, 2 , n) of the system is subject to an 'external' impulse Ii and to an 'internal' impulse Ii ; all these impulses act simultaneously. We also assume that the Law of Action and Reaction applies to the 'internal' impulses, as to the internal forces of § 7:4, so that, taken as a whole, the set of impulses Ii are in equilibrium. Therefore, 2' = 0, 2'(ri — rA) X = O.

(10.6)

Then, if ui, vi are the velocities of the particle Piimmediately before and immediately after the action of the impulses, mi vi —

= +

(10.7)

Adding these equations for every particle of the system, and using the first of eqns. (10.6) gives M — M = 2' I = J.

(10.8)

Here u, v are the velocities of the centre of mass before and after the impulse, M = Xmi and J is the resultant of the external impulses. This shows that the change in velocity of 0 is the same as the change in

velocity of a particle of mass M at 0 under the action of all the external impulses transferred to act at G. In particular, if internal collisions take place in a system they do not affect the total linear momentum. Similarly we can 'take moments' about an arbitrary point A and obtain, using the second of eqns. (10.6), (ri — rA) x mivi — f(ri —rA) x mi ui = Z(r, — rA) x I,. = K(A).

(10.9)

Equation (10.9) shows that the change in the moment of momentum of the system about A equals the moment, K (A), of the external impulses about A. Since we are assuming that no change of position occurs during the action of the impulses, any motion of the point A does not affect the

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result. In particular, if K (A) = 0, the moment of momentum about A is unaltered. We use the fact that there is no change in the position of the particles by using the same vector ri in both the terms on the 1.h. side of eqn. (10.9). In the following examples we apply eqns. (10.8) and (10.9) sometimes to the whole system, sometimes to a part of the system, and sometimes we use eqn. (10.2) applied to a single particle of the system. Examples. (i) A gun is mounted on a railway truck which is free to run without friction on a straight horizontal railway track. The gun and truck, together of mass M, are moving along the track with velocity u, when a shell, of mass m

Rio. 130. (not included in M), is fired from the gun with muzzle velocity v relative to the gun. If the gun barrel and track lie in the same vertical plane, and the former is inclined at an angle a to the direction in which the truck is moving, show that the shell has a horizontal range 2v sina g u

Mv cosa M m

Suppose that, when the shell leaves the gun, its velocity (in space) is V inclined at the angle 0 to the horizontal and that the speed of the truck immediately after firing is U. (See Fig. 130.) Then conservation of linear momentum of the truck and shell, considered as one system, in the (horizontal) direction of motion of the truck gives (1) m V cos° + MU = (M + m)u. Since the velocity of the shell relative to the truck is v inclined at a to the horizontal, (2) V cosg — U = v cosa,

V sing = v sin a.

(3)

Equations (1) and (2) give

V cos° - u

Mv cosa M 4- m •

(4)

§ 10:2

IMPULSIVE MOTION AND VARIABLE MASS

335

But the range of the shell is (2 V2 sin0 cos 0)/y and therefore the required result follows from (3) and (4). (ii) Four particles of equal mass m, are attached to a light inextensible string at points A ,B,C,D,where AB = BC = CD = a, and placed with the three parts of the string taut and forming three sides of a regular hexagon. An impulse is given to the particle at A so that it moves in the direction BA with speed u. Prove that the particle at D begins to move with speed u/13. Let w (Fig. 131) be the velocity of D, directed along DC since the impulse acting on D is along DC, and let the velocity components of Band C be as shown. [Since the string is inextensible, the component of velocity of B along BA must be the

same as that of A in this direction.] Equating the component of C's velocity along CB to that of B along CB, and similarly for D and C, we find

u cos

7C

3

cos = x cos — 3

y cos

w = — x cos 3 + y cos w

6

,

(1)

(2)

Since A moves along BA and the impulsive tension in AB acts along AB, the external impulse applied to A must act along BA. Hence no linear momentum perpendicular to AB is generated in the system as a whole. .•. mv — my — mw cos 73- = 0.

(3)

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Consider now the two particles C and D taken together; the only external impulse on these two is the impulsive tension in CB. This impulse has no moment about B and therefore the angular momentum of C and D about B remains zero. .•. ma (x cos — — y cos —) — maw cos — = 0. 6 3 6

(4)

Solution of eqns. (1) - (4) gives w = u/13 as required. (iii) Four equal particles A, B, C, D, each of mass m, rest on a smooth horizontal table at the vertices of a rhombus in which the angle BAD is 20 (0 < 45°). The sides of the rhombus are formed by four taut inextensible strings, each connecB

FIG. 132 W.

FIG. 132 (ii).

ting two of the particles, and of length a. A horizontal impulse P is given to the particle A in the direction CA. Show that, initially, the angular velocity of each string is P sin 0/(2 m a) . Clearly from considerations of symmetry A BCD will remain a rhombus and hence the angular velocities of AB, BC will be equal, co say, and in the directions indicated in Fig. 132 (i) which also shows the velocities of the particles in terms of u, the velocity of P, and co. To solve this problem we must introduce the impulsive tensions in the strings, Fig. 132 (ii). Then linear momentum of the whole system in the direction CA gives mfu + 2(u — aw sin0) u — 2aw sin01 = P. .•. — a co sin 0 —

4m

Also the impulsive equations for C--->- , B y , A --->- give 2T2 cos0 = m(u — 2aw sin0), (T1 + T2) sin0 = maw cos 0 , P — 2T1 cos0 = mu.

(1)

§ 10: 3

IMPULSIVE MOTION AND VARIABLE MASS

337

Elimination of T1, T2 from these equations gives

(„, _ Then (1) gives

P sin 0 2ma

P(1 + 2 sine B) • 4m

10:3 Collisions between particles In § 7 :2 we showed that in an 'encounter' the magnitude of the relative velocity of two particles was unaltered, provided that the potential energy V = 0 both before and after the collision and the interaction of the particles was through a force along the line joining them. The collision of two smooth spheres is a special, limiting case of such an encounter in which F = 0 at all times except when the bodies meet; the variation of F with time is an 'impulse function'. (See § 3 : 8.) In fact, in collisions between small bodies, or particles, the relative velocity is altered both in magnitude and direction; in an extreme case the relative velocity after a collision may be zero (an inelastic collision). This alteration of magnitude occurs because the bodies are deformed during the collision and not all the energy used to deform the bodies is released when they separate. Consequently, the value of V after the collision differs from its initial value, the difference depending on the elastic properties of the bodies. (Incidentally, we must also admit that a body is not strictly rigid during a collision). It is almost impossible to determine the variation of the force between two bodies during impact, particularly in the limiting case where the duration of the impact is infinitesimally short. We therefore need information from some other source to determine what will happen after the collision. This information is usually given in the form of Newton's experimental law of impact. This law states that after the impact the component of the relative velocity of the bodies along the common normal is — e times the component of the relative velocity before impact. The quantity e (> 0) is the coefficient of restitution, and its value depends on the materials of the colliding bodies. If e = 0, the bodies are inelastic ; if e = 1, the bodies are perfectly elastic ; otherwise 0 < e < 1. The ease of two perfectly elastic, smooth spheres in collision is the limiting case of the encounter considered in § 7 :2. [See also example (i) below.] A detailed account of the elementary problems of impulsive motion in particle dynamics is given in 'Theoretical Mechanics for Sixth

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Forms'. We can treat the case of smooth spheres colliding as a case of particle dynamics since the motions of their centres are the same as those of particles and the impulse between the spheres acts along the line of centres at the instant of impact.

Examples. (i) Two smooth spheres, of equal radii, moving on a table collide obliquely, as shown in Fig. 133 (i). The masses of the spheres are m1and 9n2 respectively, the initial velocities being u1, u2 and the final velocities being v1, v2 . (The radii of the spheres must be equal if the impulse between them is to be parallel to the table.) Since the spheres are smooth the impulse between them acts along the common normal at impact. Hence the momentum of each sphere perpendicular to this normal is unaltered. (1-2) .•. V1sin 01 = u1sin , r, sin 02 = 722 sin a2

-

§ 10:3

IMPULSIVE MOTION AND VARIABLE MASS

339

Since there is no external impulse on the system, the linear momentum of the whole system remains unaltered, in particular along the line of centres. .•. ml v1cos 01 + m2 v2COS 02 = MOGI cos al

m2 u2 cos a, .

(3)

The law of impact gives v1cos 01— v2 cos 92 = -e (u, cos a, — u2 cosa2) .

(4)

From eqns. (3) and (4) we find + m2) vi cos 0, = (n21— em2)u1cos a, + m2 (1 + e)u2 cosa2,

(5)

e) u1cos a, + (m2 — em1)u2 cosa2 .

(6)

(m1 + m2) v2 cos02 = m1 (1 +

Equations (1), (2), (5), (6) give the final velocity components for all conditions of incidence; special cases may be obtained by substituting special values for u1, u2 , oc2 , etc. The equations are simplified when expressed in terms of the initial and final velocities of G, the centre of mass of the spheres, ii, ecTtx , Tty),;i, (i,iy) respectively, and the initial and final velocities of A relative to B, U, (U x , Uy), V , (V x , V y) [see Fig. 133 (ii)]. In terms of u1, u2 , etc. + m 2) 26x = n21u1cos cc,

m2 u2 cosa2 ,

(m1 + 2n2) 2ly = n21u1sina,

m2 u2 sinaa,

u1 cos a, — 1t2 cosa2 ,

Uy = uisin a1— u2 sin a2 , with similar relations between v„, v1, etc., for the final velocities, eqns. (1) and (2) become VY = UY • (7-8) Y -v Y eqn. (3) becomes V, =U x , (9) and eqn. (4) becomes V x= — eU x . (10) The relative velocity is unaltered in magnitude if

Vx + VD =

U.

i.e., if ul = e2 U1 so that e = 1. The condition of the molecular encounter therefore only occurs when e = 1. In this case,

V, = — U x ,

Vy = Uy ,

so that AB bisects the angle between the directions of U and V, and the angle 0, through which the relative velocity is turned, is given by cos+0 = BN/AB. 0 = 2 cos-1(p/2a),

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where p is the perpendicular distance BN, and a is the radius of each sphere. This case of a perfectly elastic collision is unique in that 6 does not depend upon the magnitude of the relative velocity, (cf. the results of examples (i) and (ii), § 1:2). (ii) Two equal smooth spheres A, B, each of mass m, are joined by an inextensible string and lie on a smooth horizontal table with the string taut. A third equal sphere C moving on the table with speed V at an angle a with the string hits A directly, so that the string remains taut. Prove that C rebounds if the coefficient of restitution e is greater than z (1 + sin2 a).

Suppose that the velocity components of the spheres just after the impact are as shown in Fig. 134. Because the impulse on B is along BA and the string is inextensible, B must move along BA with speed u. Then conservation of linear momentum along and perpendicular to the line of centres of A and B gives mu ± mu + mw cosa = m V cosa,

(1)

my mwsina =- mVsina.

(2)

Also Newton's Law of impact along the line of centres of A and C gives (v sin a u cos a) — w — e(0 — V).

(3)

Equations (1), (2), (3) give

w—

(1 ± sin2 a — 2e) V 3 + sin2 a

and so w < 0, i.e., C rebounds, if e > z (1 + sin2 a). (iii) A smooth elastic sphere of mass m fits into a long straight tube of mass M with closed ends, so that it is free to move, without friction, along the length of the tube. The coefficient of restitution between the sphere and the ends of the tube is e. The tube is placed on a smooth horizontal table and is given an initial velocity V in the direction of its length. Find the velocity of the sphere after its nth impact in the tube.

§ 10 : 3

IMPULSIVE MOTION AND VARIABLE MASS

341

Suppose the velocities of the tube and the sphere after the nth and before the (n + 1)th impact are vn , un respectively, Fig. 135. Then Newton's Law of impact gives un — vn = — e(un _ i — Successive applications of this result give vn = ( e)n (uo — vo)

un But uo = V vo = O. ,

.•. un — vn = (—

V.

(1)

Also linear momentum parallel to the tube gives mun + Mvn = Solving (1) and (2) gives un —

fm + (— e)nMI V m+ M

(2) •

u„

•••■■ -,1.-

v, FIG.

135.

Exercises 10:3 1. The coefficient of restitution between any two of three indistinguishable smooth elastic balls, each of mass m, is e. Two are placed in contact on a smooth horizontal table, and are attached by smooth inextensible strings (parallel to one another and to the table) to fixed points, in such a manner that the strings are perpendicular to and on the same side of the line of centres of the balls, and lie in lines passing through the centres of the balls. The third ball is projected so as to slide along the table without rotating and with a constant velocity u in a line parallel to the strings and directed towards the balls and away from the fixed points ; it strikes both balls simultaneously. If the strings were taut before the impact, calculate the impulsive tensions T in them, and show that the third ball is brought to rest if e = b. If the strings were just slack before the impact, show that the impulsive tension in each of the strings is only ,÷7o T, and find in this case the value of e for which the third ball is brought to rest. 2. A, B, C, D are the corners of a smooth horizontal rectangular table, bounded along all four edges by a smooth vertical rim. From the mid-point of AB a particle is projected along the surface of the table in a direction making an angle a with AB, to strike in turn the rims BC, CD, DA. If the sides AB, BC are of length a, b, respectively, and if e is the coefficient of restitution between the particle and the rim, show that the particle will return to its starting point if tan a —

2 eb a (1 + e) '

and that if this condition is satisfied the particle will always pursue the same path.

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3. A number (n) of imperfectly elastic spheres are suspended by long threads so that their centres are in the same horizontal line and their surfaces very nearly touching. They are all made of the same material (coefficient of restitution = e), but their masses are 2/1/, nM. If a velocity u, in the line of centres, is given to the smallest sphere, prove that the largest will be set in motion with velocity

V = u(1

e)n -1 1 3

(n — 1)! 5 (2n — 1) '

whereas, if the velocity u is given to the largest sphere, the smallest will be set in motion with velocity n V. 4. Three particles A, B, C, of equal mass m, lie on a smooth horizontal plane and are smoothly connected by light rigid rods AB, BC so that the angle ABC = 7G - a, (a < n12). The whole system is moving parallel to A B with speed V, when C collides with a smooth inelastic wall fixed at right angles to AB. Show that the impulsive blow on the wall is m V(3 + sin2 a)/(1 + 3 sin2 a). 5. One end of an inelastic string is fixed to a point P on a smooth horizontal table. To the other end is attached a heavy particle R which is moving in a circle on the table with period T . At a point Q, where PQ : Q R = A:1, the string comes into contact with a smooth peg which is glued to the table so that it can withstand a force G. Prove that the peg will move if G < 2 T (1 + A), where T is the original tension in the string, and that it then moves off in a direction making an angle 0 with either part of the string, where 2 T (1 + A) cos0 -= . 6. Two particles A and B, each of mass m, are connected by a light inextensible string ; a third particle C, also of mass m, is attached to the middle point of the string. The particles are placed on a smooth horizontal table and are at rest, in a straight line, with the string taut. The particle C is projected horizontally with velocity V at right angles to the string. Show that, immediately before A and B collide, each has a velocity 2 V/3. If the coefficient of restitution between A and B is I-, show that, when the string next becomes straight, the velocity of C is zero, and find the velocity of A. 7. Particles A, B, each of mass m, are joined by an inelastic string of length 2a . Initially AB is vertical with B at a distance a below A. A is given a horizontal velocity V while B is simultaneously released from rest. Find the inclination of the string to the vertical when it becomes taut and the time which elapses before the string is horizontal. 8. A uniform solid triangular prism, of mass 2m, rests with one rectangular face on a smooth inelastic horizontal table. A small sphere, of mass m, drops vertically on the mid-point of a smooth face of the prism that is inclined at 45° to the horizontal. Assuming that the prism will not topple and that the coefficient of restitution between the sphere and the prism is I, find the velocities of sphere

§ 10 : 4

IMPULSIVE MOTION AND VARIABLE MASS

343

and prism just after impact in terms of U, the speed of the sphere just before impact. Show that, if the height of the point of impact above the table is a, the sphere will hit the prism face a second time after rebounding, provided that (73 < ga. 9. Three particles A, B, C, of masses m1, m2 , m3 , respectively, lie on a horizontal table, A, B, C being at the vertices of an equilateral triangle of side 2a. The particles A and B and the particles B and C are connected by light inelastic strings each of length 2a. If a horizontal impulse P is applied to B in a direction bisecting the angle ABC internally and such that impulsive tensions are at once set up in the strings, prove that the particle with mass m2 begins to move in a direction which makes an angle tan-1

1/3 (ml — 7,63) m, 4m2 + m3

with the direction of the impulse.

10:4 Collisions between bodies: energy changes From eqn. (10.1) we find the change in kinetic energy of the particle by multiplying scalarly by i(v u). ••• 1,0,2

u2) = II • (v

u).

(10.10)

The energy imparted by an impulse I is therefore given by the r.h. side. For a system of particles

.•. ZI T

— ui) =

I, • (v,

u,) ± if I', • (v, -

u„). (10.11)

We interpret the two parts of the r. h. side of eqn. (10.11) thus : 1/I, • (v1 u 1) is the energy imparted to the system by the external (applied) impulses; 22' Ii • (v, u.) is the energy generated by the internal impulses. In fact, the action of the internal impulses usually destroys kinetic energy or, rather, the bodies absorb some energy which is converted to a different form within the body, either as the potential energy of a permanent deformation, or as energy of thermal agitation of the molecules of the body. Exceptions to this are to be found in systems such as a gun and its shell where the internal explosion generates energy in a system initially at rest. (See § 10:5 on rockets and jets). If we assume that a body is rigid, we assume that the internal impulses inside any one rigid body generate (or destroy) no energy in that body. But, as we pointed out above, this assumption is inconsistent with our other assumptions, especially for imperfectly elastic colli-

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sions (0 < e < 1). Nevertheless, eqn. (10.11) often provides a convenient method of calculating energy changes. The direct application of Newton's experimental law to the collisions of bodies other than smooth spheres can be troublesome and is sometimes replaced by the following analysis of a collision process. The time during which two colliding bodies are in contact is divided into two periods, a period of compression and a period of restitution. During the compression the bodies are being deformed by their mutual action until their relative velocity along the common normal is zero ; during the restitution the bodies regain their original shape, or approximately so, and finally separate. Newton's law is replaced by Poisson's hypothesis that the impulse between the bodies for the period of restitution is the fraction e of the impulse for the period of compression. It is possible to calculate the impulse I' for the compression assuming zero relative velocity along the common normal at the end of the compression period; the impulse during an imperfectly elastic collision, compression and restitution, is (1 e) I'; we then use the latter value to calculate the velocities with which the bodies separate. If we adopt this view, we are in fact assuming a special form for the law of interaction between the bodies, but, without a much more detailed investigation according to general theories of the behaviour of elastic bodies, it is impossible to say whether the picture used in Poisson's hypothesis approximates at all closely to what occurs in practice. Clearly, unless e = 1, the force of interaction does work in deforming each body during the compression, but not all this energy is released during the restitution as kinetic energy of the separating bodies. In general, the internal impulses destroy kinetic energy in an inelastic or imperfectly elastic collision. In the cases we consider, Poisson's hypothesis leads to the same results as Newton's law of impact but involves the determination of all the internal impulses in a system of colliding bodies. Both hypotheses are generalisations from a limited experimental basis. However, they both give results in fair agreement with practice. Where energy considerations are involved in the following examples, we make the (doubtful) assumption that the internal impulses within a 'rigid' body do no work, and use eqn. (10.11) to find the energy imparted to a system by a set of applied impulses. Examples. (i) Calculate the energy imparted to the system of four particles by the impulse P in example (iii) p. 336.

§ 10 : 4

IMPULSIVE MOTION AND VARIABLE MASS

345

In this problem we assume that the strings connecting the particles are literally inextensible. Hence the internal impulsive tensions in the strings do no work and we may use eqn. (10.11) assuming no contribution from internal impulses. The only external impulse is P and its point of application has an initial velocity zero and a final velocity P(1 + 2 sin2 0)/(4m). Hence from eqn. (10.11) T = I • (v

P(1 +2 sin2 0) ui) = 1 -2,P

4m

P2 (1 + 2 sine ©) 8m

•

We leave the reader to verify that this is the sum of the kinetic energies of the particles after the impulse. (ii) Two uniform rods AB, BC, each of length 2a and mass m, are freely jointed at B and rest on a smooth horizontal table with ABC in a straight line.

A horizontal impulse I is applied to AB at A in a direction perpendicular to the line of the rods. Show that the initial velocity of A is of magnitude 71/(2m) and find the angular velocities of the rods. Find also the kinetic energy T, generated by the impulse. Suppose that A has components of velocity (u, v) and that the angular velocities of the rods are co,,, co2 so that the components of velocity of G1, B, G2 are as shown in Fig. 136. Then the linear momentum equations (for the whole system) along and perpendicular to AB give mv+ mv -= 0,

(1)

m (u — a co,) m (u — 2 a co, — a co2) = / .

(2)

After the impulse BC has no moment of momentum about B. .•. ma (u — 2a — a co2) — ma2 co2 = 0.

(3)

Also the change in moment of momentum of AB about B is 2a/ clockwise. .•. ma(u — awl)

I ma2 co1 = 2a/.

(4)

Equations (1)–(4) give u = 7//(2m), v = 0, wl = 9//(4ma), co2 = —3//(4ma).

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A COURSE OF MATHEMATICS

Note that here we do not introduce the impulsive reaction at B. Its resolutes could easily be determined by writing down the impulsive equation of motion of

Gior G2. We assume that the internal impulses in each rod do no work so that eqn. (10.11) gives

T = --4Pu = 7121(4m). (iii) Three uniform rigid rods AB, BC and CD, each of mass m, and length 2a, are smoothly jointed at B and C. The system rests on a smooth horizontal table with A, B, C and D collinear when a horizontal impulse P is applied to the midpoint of BC in the direction perpendicular to the line of the rods. Show that the angular velocity of CD immediately after the impulse is P/(2ma) and that the kinetic energy generated by the blow is P2 /(3 m) . Find the angular velocity of CD when it has rotated through 90°. u -a o.)

A

u,

B

tP

D.

C

FIG. 137.

The motion immediately after the impulse is shown in Fig. 137. Linear momentum of the whole system and conservation of angular momentum of CD about C give respectively m(u — au)) + mu + m(u — aco) P,

ma(u — aco) — Dna2 = 0, whence u = 2P/(3m), w = P/(2ma). Hence the kinetic energy generated is P2

T = 2-Pu =ant Suppose that, when the rod CD has rotated through 90°, BC has velocity V and AR, BC each have angular velocity D. Then conservation of linear momentum and energy give 3m V P, 2 lm {172

a 2 122 *a2522}

m V2 =

P2 3m

from which it follows that S2 = Pl(ma 2 V2). (iv) An infinite number of uniform rods Ao Ai , Ai A 2 , ..., each of length 2a and mass m, are freely connected at their ends and lie at rest on a smooth table in a

§ 10:4

IMPULSIVE MOTION AND VARIABLE MASS

347

straight line A5 A1A2 ... A horizontal impulse I is applied at Ao perpendicular to the line of the rods. If the velocity imparted to AT is yr , show that vr +s+ 4 vr+i 1), = 0 and find the kinetic energy T generated by the impulse. If the impulsive reaction at the hinge Ar+i isIr+I as shown in Fig. 138 then the angular momentum equations for AT Ar+i and Ar „.24„2 about AT , A„2 respectively are a • ml(vr vr +1) + flia2 (v74.1— vr )/2a = — 2a/r+i, a • m -Evr+i

9.910(vr+1 — vr+2)/2 a = 2 a/r+1•

vr+2)

•• • vr +2+ 4vr +1+ yr = 0.

IV°

i

A° I "

A,

I

V, +

V,

UMMIIIINIMINIMII.M. 0

A„ ,

Ir* FIG.

138.

The solution of this difference equation which remains finite as r oo is yr =C(— 2 +113)r, where C is a constant. But the linear momentum of the whole system resolved perpendicular to the line of the rods gives I = m(+ vo + G vr)• r =1

C=

121/3

m

Hence the kinetic energy generated is T =1/vo = +/C —

/2 4/3

(v) A uniform heavy rod AB, of length 2a and mass m, is held in a position inclined at an acute angle a to the vertical and is then released. It falls through a height h and strikes a smooth horizontal table. Show that, if the coefficient of restitution is zero, the lower end of the rod will leave the table immediately after impact if 18h sin2 a cos« > a (1 + 3 sin2 a)2. Before the end B of the rod strikes the table the centre of mass 0 descends vertically with acceleration g and so immediately before impact 0 has speed V = (2gh) vertically downwards. Further, AB does not rotate before the impact.

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A COURSE OF MATHEMATICS

Since the plane is smooth there will be no horizontal component of impulsive reaction with the plane and therefore the motion of the rod immediately after impact is defined by the downward velocity v of 0 and the angular velocity co, Fig. 139. There is no change of moment of momentum about B and so

A

mva sina + 3ma2 w = mVa sina.

(1)

Also, since the impact is inelastic, v — a w sina = O. 3 V sin2 a 1 + 3 sin' a '

a co —

(2) 3 V sina

1 + 3 sin2 a •

(3) To complete this problem we suppose that B remains in contact with the plane and that at time t after impact AB makes the angle q) with the vertical. Then following the procedure of example (i) p. 259 we find the reaction R between the plane and the rod is given by

R

—

mg

(1 + 3 sin2 q)) a2472 -

—

m

d2

(a cos g), dt2

9 V2 sin2 a 1 + 3 sin2,

6ga(cosa — cosq)),

(1 + 3 sin2 q)) a q, + 3a q)2 sing) cosq) = 3g since, so that

m(g a0 2cos(p) (1+ 3 sin2 99) • —

R—

But R cannot be negative and so B will leave the table immediately after impact if (a 4)2cos Ow , > g, i.e., if 9 V2 sin2 a cos a (1 + 3 sin2 a)2 a > g which leads to the required condition on substitution for V. (vi) A uniform rod AB, of length 2 a and mass m, can turn freely about one end A, and at time t = 0 is hanging from A in equilibrium under gravity. The end A is then set in motion in a horizontal straight line so that, when t 0,

OA = vt where 0 is a fixed point of the line and v, f are constants. Show that the initial angular velocity of the rod is 3v/(4a) and that it will make complete revolutions about A if

3v2 > 8 a [g

(g2

2)1j9 .

§ 10:4

IMPULSIVE MOTION AND VARIABLE MASS

349

Since A starts off with velocity v, the rod must be subject to an impulse acting at A. Ha) is the initial (backward) angular velocity of AB, the equation of angular momentum about A is ma(v — an) — ;ma2 a) = 0. .•. (4) = 3v/(4a). Suppose that at time t, AB makes the angle 0 with the downward vertical (Fig. 140). The equation of moments about A is

m(af cos° — a2 0) — )ma2 °= 9n,ga sine. .•

f cos0 — g sin0.

Bo Integration and use of the initial condition 0 = co when t = 0 = 0, gives

;lad' = f sin°

g cos°

F, a oi2 — g .

The rod will make complete revolutions provided 02 > 0 for all 0. But the least value of f sine gcos@ is — i'(f2 g2) and so the rod makes complete revolutions provided 2a w2 > 3{g Y(g2 f 2)} which leads to the required condition. (vii) A thin smooth circular wire, of mass m, radius a and centre 0, can rotate freely in a horizontal plane about a fixed vertical axis through 0 to which it is connected by light rods. A small ring, also of mass m, is threaded on the wire and is free to move between two stops lying at opposite ends of a diameter. The ring is projected from one of the stops with velocity v along the wire. If e is the coefficient of restitution between the ring and each stop show that the time to the

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A COURSE OF MATHEMATICS

nth impact is nag — en)/{v (1 — e) en -1} and that the velocity of the bead just after the nth impact is v(1 (— On} /2. Since the wire is smooth, between impacts the angular velocities of the particle and of the wire about 0 are both uniform. After r impacts the speed of the bead relative to the wire is er v . Therefore the time, Tr, r+1, between the rth and (r 1)th impacts is rca/(erv). Hence the time to the nth impact is n-1 rca n -1 1 na(1 — en). TT , 7+1 = v v(1 — e) en-1 r=0 If vn , con respectively are the velocity of the bead in space and angular velocity of the wire (measured in the same sense) after n impacts, then Newton's Law of Impact gives vn — awn = (

e)

— a con _ ,)

which leads to vn — awn = (— e)n (v0 — a wo) = (— e)nv. (1) Also conservation of angular momentum about the axis of the wire gives Fla. 141. mavn ma2 wn may.

(2)

Solution of (1) and (2) for vn gives the required result. (viii) A uniform hoop, of mass m, radius a and centre A, rolling upright on a horizontal plane with velocity v collides with a fixed perfectly rough inelastic rail which is perpendicular to the plane of the hoop and at a height h (< a) above the horizontal plane. Show that, for the hoop to surmount the rail without jumping, 2af{g(a — h)} 2a l'(gh) „.> v > 2a — h 2a — h • After impact the hoop will begin to turn about the rail; let the angular velocity just after the impact be w . Then, since the angular momentum of the hoop about the rail is unaltered by the impact, m(a — h) v maz — = 2mazco. a .•. a2 = 1(2a — h) v.

(1)

Let 0 be the point of the rail about which the hoop turns and suppose that the hoop remains in contact with the rail and that at time t after the impact 0 A makes the angle 0 with the upward vertical, Fig. 141. Then the energy equation for rotation about 0 is 2ma2 0.2 + mg (h + a cos0) = z 2ma2 w2 + mg a .

(2)

§ 10 : 4

IMPULSIVE MOTION AND VARIABLE MASS

351

Also, if B is the component along OA of the reaction between the rail and the hoop, the equation of motion of A resolved along AO, is

mg cos° — R = ma02.

(3)

a2 2 = a2 w2— g (h + a cos 0 — a), 0 R = mfg (2a cos() + h — a) — a2 w2}/a.

(4)

Equations (2) and (3) give

(5)

In order that the hoop should surmount the rail without jumping we must have both 02 > 0 and R > 0 for a— h < cos° < 1. a Therefore, from (4) and (5), a2 0,2 > g h, (6)

g (a — h) „.> a2 w2.

( 7)

Here (6) expresses the condition that the hoop has sufficient energy to surmount the rail and (7) expresses the condition that the hoop is not moving so fast that it leaves the rail immediately after the impact (when R is least). Substitution for w2 from (1) gives the required inequalities. Exercises 10:4 1. Two equal uniform rods AB, BC, smoothly jointed at B, lie in a straight line on a smooth horizontal plane. An impulse of magnitude P is applied at C, at right angles to ABC. Show that the initial angular velocities of the rods AB and BC are in the ratio — 1:3. Show also that the energy imparted to the system by the impulse is 7 P2/4 M, where M is the mass of each rod. 2. Four equal uniform rods each of mass m and length 2a, are freely jointed together at their ends and are at rest in the form of a square. If an impulsive couple Q is applied to one of the rods, show that the total energy generated in the system is 392 /16m a2. 3. Two equal uniform rods X Y, Y Z, each of mass m and length 2a are freely jointed at Y. X Y Z is a straight line and the rods are moving on a smooth horizontal table with velocity v perpendicular to X Y Z when X Y impinges at its midpoint on an inelastic stop. Show that the angular velocity of each rod just after the impact is 3 v/(7 a) . Find the impulsive force exerted on the stop and the loss of kinetic energy. 4. A uniform ball of radius r is at rest on a rough horizontal table. The ball is struck a horizontal blow in a vertical plane through its centre at a height r/2 above the table. Show that, when the ball stops slipping, its linear velocity is 5/14 of its initial linear velocity. 5. A uniform rod OA of mass m and length 2a turns freely in a horizontal plane about a fixed vertical axis through 0. A similar rod AB is hinged to the first at A so that it turns in a horizontal plane. At the moment when OA, AB are in line and rotating with respective angular velocities w, w' in opposite senses, the hinge

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A COURSE OF MATHEMATICS

at A is suddenly locked. If the rods come to rest prove that 5 co' = 11 co, and find the impulsive reaction on the axis through 0 in terms of cc. 6. A wheel of radius a mounted on an axle has its centre of mass at its geometrical centre, and the radius of gyration about the axle is k. The wheel is rolling along a horizontal line with velocity V and its plane is vertical. The highest point is then suddenly fixed by passing a smooth spindle parallel to the axle of the wheel through a small hole at the circumference. Show that the wheel will make complete revolutions about the spindle if V2 (a2 — k2)2 > 4g a3 (a2 Tk2) . 7. Show that a uniform rod of mass m is equimomental with two particles each of mass 1 m at its ends together with a particle of mass km at its mid-point. Six uniform rods form a regular hexagon smoothly jointed at the corner points. A blow is given to one of the rods at its mid-point in the direction perpendicular to the rod in the plane of the hexagon. Prove that the opposite rod begins to move with one-tenth of the velocity of the rod struck. 8. A uniform rigid rod of mass m and length 2a is free to rotate about a horizontal axis perpendicular to the rod and passing through a point a distance ia from one end. The rod is at rest in stable equilibrium when a horizontal impulse J perpendicular to the axis is applied to the uppermost point of the rod. Determine the greatest value of J for which the rod does not perform a complete revolution, and show that with this value of J the rod will be instantaneously horizontal at a time (4a/3g)2 log(Y2 + 1) after the action of the impulse. 9. A uniform lamina of mass M in the form of a square of side 2a is rotating with angular velocity co in its own plane about its centroid which is at rest. One corner of the lamina strikes a particle of mass m, which is at rest, and the particle adheres to the lamina. Find the angular velocity of the system after impact and show that the loss of kinetic energy of the system at impact is

M m a2 M + 4m

10. Two equal toothed wheels, each of mass M and radius a, which may be regarded as uniform circular discs, are rotating in the same sense with angular velocities co, and cue about two parallel spindles (of negligible inertia) at distance apart just greater than 2a. If the spindles are slightly moved towards each other so that the teeth suddenly engage, show that the loss of energy is f3- Mae (w1 + 0)2)2. 11. Two uniform rods AB and AC, each of mass m and length 2a, are freely hinged at A and held in a vertical plane with A above BC and each rod inclined at an angle a with the downward vertical. The system is released from rest and, after falling through a height h, B and C strike a smooth inelastic horizontal table. Show from energy considerations, or otherwise, that neither rod rotates before the impact. Show also that immediately after the impact each rod has an angular velocity 3 (2 gh sin2 a)112 /4 a and find the impulsive reaction at the joint A. 12. A uniform solid cylinder of radius a rolls down a rough plane of inclination (< 60°) . When it has rolled a distance 1 it impinges on a perfectly rough inelastic horizontal rail at a distance a/2 from the inclined plane. Assuming that the

§ 10 : 5

IMPULSIVE MOTION AND VARIABLE MASS

353

cylinder remains in contact with the rail after the impact, show that it surmounts the rail if 4/ sin a > 9a{ 1 — sin (a + 30°)). 13. A circular disc of radius a whose centre and mass-centre coincide can turn freely in a vertical plane about a fixed horizontal axle through its centre. The moment of inertia of the disc about the axle is pma2. The disc is initially at rest with an insect of mass m on it at the lowest point. If the insect suddenly starts to walk round the rim with speed V relative to the disc, show that his initial velocity in space is p V/(p + 1). Show also that if the insect continues to walk round the rim with speed V relative to the disc he will not reach the highest point if

2)2 V2 < 4ga(p + 1). 14. Three equal uniform rods AB, BC, CD, smoothly jointed together at B and C, lie at rest on a horizontal table with the points A, B, C, D in the same line. The system is set in motion by a horizontal blow applied at the point of trisection of BC which is nearer to C and in a direction perpendicular to the line of the rods. Prove that the rods AB,BC,CD begin to move with angular velocities of which the magnitudes are in the ratios 1:2:4. 15. A uniform billiard ball of mass m lies on a rigid horizontal table. A cue, striking at a point whose angular distance from the top of the ball is 0, delivers an impulse m U in a direction that intersects the vertical diameter and makes an angle pwith the downward vertical. The coefficient of friction between the ball and the table is y. Prove that the ball slips as it starts to move if tang) lies outside 5 sin° + 7y 5 sin 0 — the limits and 5 cos 0— 2 • 5 cos 0 — 2 16. A uniform solid spherical ball, of radius a, rolls on a level road directly towards a pavement of height h, where h < a. Initially the angular velocity of the ball is a, where

<

7Ogh (7a — 5h)2

Prove that the ball fails to mount the pavement. (It is assumed that the impact is inelastic, and that there is no slipping of the ball on the edge of the pavement.) Prove that the angular velocity with which the ball finally rolls away from the pavement is w where 5h \ 2 1 7a /I a ,

-

—

(It is assumed that the impact with the road is inelastic and without slip.)

10:5 The motion of a body with changing mass— Rockets The advent of the jet engine and the rapid development of the large rockets required to put earth satellites and space probes into orbit have given rise to a variety of problems involving bodies of changing mass or bodies which pick up and eject the material of the medium through which they move. Here we are concerned primarily with the rectilinear

354

A COURSE OF MATHEMATICS

motion of such a body but the methods are applicable to more complicated problems. Rather than apply the Law of Motion we solve these problems by applying eqn. (10.8), or (10.9), to such a system. The system in question consists of the body in motion, i.e., the aircraft or rocket, together with any matter which is absorbed and any matter which is ejected in a short interval of time at. At the beginning of this interval all these different parts of the system have their separate states of motion; at the end of the interval each part has, usually, a different state of motion. Equation (10.8) states that the change of linear momentum of the whole system in the interval S t is given by the impulse of the external forces acting on the whole system. Having written down the equation expressing this momentum balance for an interval of time at, we obtain a differential equation by dividing by at and then taking the limit as at ± 0 . In Chap. VII p. 238 we pointed out the effect 'an intelligent being' could have on a system by the use of its muscles. The process considered here is a generalisation of this idea. 1. A body falls under constant gravity picking up matter from rest as it falls, so that at time t its mass is m and speed v. Consider the system which consists of the body at a given instant and the material, of mass m, which it picks up in the subsequent interval t; this material is initially at rest. The initial momentum of this system is p =-- my + am

.

Finally the body and the additional material are moving with velocity v c5 v , and the final momentum is

p Sp = (m am) (v ay). The external force acting on the whole system throughout this interval is the total weight (m m)g.

(m Sm)g at = op = (m am) (v Sy) — my. (10.12) Correct to the first order this gives

mg at = v am m ay. This leads to the differential equation, in the limit as St -± 0, mg

= v

dm dv dt +m dt

d dt (my).

(10.13)

§ 10 : 5

355

IMPULSIVE MOTION AND VARIABLE MASS

If the matter picked up had been moving downwards with speed u, then in place of eqn. (10.12), the momentum balance would have been

(m + a m)g cSt = 6 p = (m + am) (v + 6v) — (my ± 6 m • u). .*. mg = m

dm dv +v dt dt

dm u dt

d dm (10.14) dt (my)— u dt .

2. A body moves vertically upwards under gravity so that at time t its mass is m and its speed is v. The body ejects material at the rate of k units of mass per second vertically downwards with a speed u relative to the body. In this case the system considered is the body and the material it ejects in the short interval at. At the start of the interval the body has mass m and velocity v; at the end of the interval it has mass m m (a mass — am = kat having been ejected) with a velocity v 5v, and the ejected mass has a velocity between v by — u and v — u. The external force acting on the whole system is the weight mg. The momentum equation is (m

am) (v

v — u) 0 (Om • v) — my

av) (— am) (v

—mg at. Correct to the first order this is m av + yam — yam u am = —mg at. .m

dv dt

u

dm —mg, dt =

(10.15)

where dm/d t = — k. 3. A jet aircraft, of mass M containing fuel of additional mass m , moves horizontally in a straight line under a resistance P(v), where v is its speed at time t. The aircraft picks up air which is moving with speed u horizontally in the direction of the motion of the aircraft, picking the air up at a rate a (v — u) units of mass per second, where a is a constant. This air and the heating agent (fuel) is ejected backwards with a speed V relative to the aircraft. In this case the system we consider consists of the aircraft of mass Ili, the fuel of mass m, and the air it will pick up in the interval at . Initially the air has a velocity u and the aircraft and fuel have velocity v; finally, after the interval 6t the air and a mass — b m of burnt fuel have a velocity between v — V and v + c5 v — V, the aircraft and the unburnt fuel,

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A COURSE OF MATHEMATICS

Ov. The force acting on of total mass M m Om, have velocity v the system is the resistance P (v) , and the mass of air picked up is a (v — u) t 0 (6 t2). The momentum equation is (M

{—om + a 6t (v — u)} (v

+ m) (v + 6 v)

v — V) —

— (M m)v — a 6 t (v — u)u = — P (v) t 0 (6 t2) (M

m)

dv dt

dm

±V

dt

+ a(v — u) (v — V — u) = —P(v). (10.16)

Here — d m/dt is the rate at which fuel is consumed. Examples. (i) A spherical hailstone, falling under gravity in still air, increases its radius r by condensation according to the law dr/dt = A r, where A is constant. If air resistance is neglected, prove that the hailstone approaches a limiting velocity g/(3 A). Let m be the mass and v the downward velocity at time t. Then dm dt

d ( 47tr3 g) dr — 42.tr2 e 71- = 47tAgr3, dt 3 t

where e is the density of the hailstone. 1 dm — 32. m dt

(1)

The momentum equation in the interval t to t St is

Sy) — my = mg St + 0(5t 2),

(m Om) (v which leads to dv dt

v dm m dt = g •

(2)

Substitution from (1) in (2) gives dv

dt

+ 3 A v = g.

(3)

The solution of this equation is

v=

3A

+ A e-3 At,

where A is a constant (determined by the initial conditions). Therefore, as t —>- co, the hailstone approaches its limiting velocity g/(3 A). In solving these problems it is advisable to write down the momentum equation for a short interval and derive the differential equation [(2) above], in each case rather than to quote eqn. (10.13) which may not apply exactly in a given problem.

§ 10 : 5

IMPULSIVE MOTION AND VARIABLE MASS

357

(ii) A particle whose mass increases through condensation of moisture at a constant time-rate motr, where m, is the initial mass of the particle and r is a constant, moves freely under gravity. The particle is projected from the origin of coordinates with a velocity whose horizontal and vertical resolutes are u and v respectively. Prove that the coordinates of the particle when its mass is m are given by x = UT log (m/mo), m2

y = ig-r2 (1

mot )

(VT + lg -c2) log (m/m0),

the y-axis being vertically upwards. Suppose that at time t the particle has coordinates (x, y) and horizontal and vertical components of velocity (U, V). Then the horizontal and vertical momentum equations in the interval t to t dt are (m

dm) (U + 6U) — mU = 0,

(m + dm) (V + 6V) — mV = —mg dt 0(6t2). These equations give respectively d (m U) = 0, dt

(1)

d V) =- —mg, dt (m

(2)

which, together with the equation for the variation of mass dm dt

m, (3)

give the solution of the problem. Equation (1) integrates to give m U = molt, i.e., dx m dt = mo u •

(4)

Dividing eqn. (3) by eqn. (4) gives 1 dm m dx

1 UT

which integrates to give x = u r log (m/m0). Now (3) integrates to give m = mo (t r)/r so that the integral of (2) is m V — mo v = —g f m dt =-

—

gmo 2x {(t +

r21

0 dy • • dt

gmr "70 2m, + m

(V

+ IgT)•

2 gmT + gmoT 2m, 7- 2

(5)

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A COURSE OF MATHEMATICS

Dividing (5) by (3) gives dy dm

gm T2 1 2m02 + m (VT + z9 T2)

which integrates to y =19, -0 (1

mo z

)

(v

M

-f gt2) log

mu

•

(6)

(iii) A Catherine wheel, of radius a, which can turn freely about a horizontal axis, is initially of mass M and moment of inertia /0 about its centre. A charge is spread along the rim and ignited at a time t = 0. While the charge is burning the rim of the wheel loses mass at a constant rate m, per second and, at the rim, a mass m2of gas is taken up per second from the atmosphere, which is at rest. The total mass m2 + m2 is discharged per second tangentially from the rim, with a velocity V relative to the rim. Prove that, if 0 is the angle through which the wheel has turned after t seconds, then 0 where A = mict2

V = a(u

A) [at — 1 +(1 — At)tilA],

—

,u = (m1 m2) a2110 .

In this problem we use eqn. (10.9) for angular momentum instead of linear momentum. Suppose that at time t the moment of inertia of the wheel about its axis is I and that the angular velocity of the wheel is w = d 0/dt . Then the angular momentum equation for the interval t to t dt is di) (co /

+

+ m2) ot (coa — V) a —

6(0)

du) d./ + co — dt dt

+ ma) (co a — V) a = 0.

But d/ dt =

Mia2.

.•. I = Io — mi a2t

Substitution from (2) and (3) in (1) gives d co

(1

At) dt

(It

A)

V/L a

This integrates to give d0 co = dt

Vy{1 —(1 — 2t)(m-42 } a Cu — A)

A second integration gives 0 —

=- 0(6t2).

V{,ut + (1 — At)/2/2— 1}. • a (y — A)

§ 10:5

IMPULSIVE MOTION AND VARIABLE MASS

359

Exercises 10: 5 1. A raindrop is observed at time t = 0 when it has a mass m and downward vertical velocity it. As it falls under gravity its mass increases by condensation at a constant rate r and a resisting force acts on it proportional to its velocity and equal to ry when the velocity is v. Show that the equation of motion is equivalent to d

at

Me g

where M = m rt, and find the velocity of the drop at time t. 2. A body consists of equal masses M of inflammable and non-inflammable material. It descends freely under gravity from rest whilst the combustible part burns at the uniform rate of AM per second, where 2 is a constant. If the burning material is ejected vertically upwards with a constant upward velocity is relative to the body, and the air resistance is neglected, show, from considerations of momentum, or otherwise, that d — [(2 — At) v] =- 2(u — v) g (2 — At) dt where v is the velocity of the body at time t. Hence show that the body descends a distance g/2A2 + (1 — log 2) u/A before all the inflammable material is burnt. 3. Show that the average value g of the acceleration due to gravity over a vertical rise H from the surface of the Earth (radius R) is gR/(R H), where g is the value at the Earth's surface. A rocket missile, starting from rest on the ground, attains a speed V after a vertical flight of duration T. The fuel is burnt at a constant rate, is ejected with velocity ve relative to the rocket, and is exhausted at time T . Neglecting air resistance, and assuming constant gravity q, prove that

V = —gT — velog {1 — (mo/MO} , where mo is the initial mass of fuel and Mo the total initial mass of the missile. 4. A gun of mass M stands on a horizontal plane and contains shot of mass M', The shot is fired at the rate of mass m per unit of time with velocity u relative to the ground. If the coefficient of friction between the gun and the plane is ,u, show that the velocity of the gun backward by the time the mass M' is fired is

M'

(111 + m,)2 m2

2mM

dug •

5. A small body is projected vertically upwards in a cloud, the velocity of projection being V(2gh). During the ascent the body picks up moisture from the cloud, its mass at height x above the point of projection being m0 (1 + ax),

360

A COURSE OF MATHEMATICS

where a is a positive constant, and the added mass is picked up from rest. Prove that the greatest height reached is h', where

h'

{j/(1

3ah) — 1)/a.

6. A railway engine, which together with its fuel and water has mass M(t) at time t, is subject to a total driving force F and a resistance R. Show that the appropriate equation of motion is

M

dv

at — F R,

where v is the velocity at time t. If the engine does work at constant rate P, while the resistance is always v and M = M, — mt where A and m are constants, prove that if v, is the velocity at time t = 0, then for motion on a level track, so long as these conditions hold, the velocity at any other time is given by V2 = V,; (M/11 / 0) +

2P [1(M/Mo)2]. Am

7. A rocket of initial total mass M propels itself by ejecting mass at a constant rate p per unit time with speed u relative to the rocket. If the rocket is at rest directed vertically upwards, show that it will not initially leave the ground unless flu> Mg, and assuming this condition to hold show that its velocity after time t is given by —u log(1 — ut/M) — gt. Show also that when the mass of the rocket has been reduced to half the initial value its height above ground level will be

uM 2p

(1 log 2 — Mg/4µu).

8. A rocket, initially of total mass M, throws off every second a mass fM with constant velocity V relative to the rocket. Show that it cannot rise at once unless f V > g and that it cannot rise at all unless f V > A.g, where 2 M is the mass of the case of the rocket. If the conditions are such that the rocket is just able to rise vertically at once, show that the greatest height it will reach is

V2

{1

A ± log 2 + 'z-(logA)2}.

9. A rocket is fired vertically with initial velocity V and is propelled upwards by projecting propellant material downwards at a constant rate and with a constant velocity u relative to the rocket. After a time T the propellant material is exhausted and the rocket still rising. Show that, neglecting air resistance, the maximum height attained is h, where

2gh =(u log C)2 — 2 Vu log 5 + V2 + 2guT {1 --IL (1 — C)-1log C},

IMPULSIVE MOTION AND VARIABLE MASS

361

and is the ratio of the mass of the rocket alone to the total initial mass of the rocket and propellant. 10. A raindrop is falling vertically with a speed 2 V when it overtakes a cloud which is falling vertically with constant speed V. While passing through the cloud the mass of the raindrop increases by condensation from the cloud at a constant rate k per second. The raindrop also experiences a resisting force equal to k times its velocity relative to the cloud. Obtain the differential equation relating the mass of the drop when in the cloud to the relative velocity and hence express the relative velocity in terms of the mass.

Miscellaneous Exercises X 1. A particle of mass m, moving in a straight line with velocity v0 , collides with an atom of mass M initially at rest. In the collision the atom is excited, absorbing an amount of internal energy E, and afterwards both particles move in the initial line of motion of the particle of mass m. Find the minimum value of v0 , for which the collision process is possible, and show that, if m = M, the initial kinetic energy of the colliding particle must be at least 2E . 2. A rocket driven car, of total initial mass M, loses mass at a constant rate in per unit time at constant ejection speed V relative to the car. If the total resistance to motion is icy when the velocity is v, show that the acceleration of the m V — kv car along a straight horizontal road is at time t from the start, and

M — mt

hence that the speed from rest is then mV

[1 —(1 m t M)09 .

3. Two equal uniform rods AB, BC, each of mass m and length 2a, smoothly hinged at B, lie in a straight line with the end A fixed. An impulse J is applied at the end C, perpendicular to the rods. Find the angular velocities with which the rods begin to move, and show that the kinetic energy imparted to the system is 12 J2 /(7 m) 4. A uniform circular disc of mass M, lying on a smooth horizontal table, is struck a horizontal blow at a point P of its circumference which causes P to move with a velocity of magnitude V in a direction making an angle a with the radius through P. Show that the normal and tangential components of the impulse are MV cos a and MV sin a and that the kinetic energy communicated to the disc is

mV2(1 + 2 costa). 5. A uniform square lamina A BCD of side 2a is falling under gravity without rotation, the plane of the lamina being vertical and the diagonal BD horizontal. When the velocity of the lamina is v, an impulse is applied at B which reduces B 3 V2 v to rest. Show that the lamina begins to rotate about B with angular velocity 8a If the point B is kept fixed, show that the lamina will make complete revolutions if 3v2 > 8gaV2.

362

A COURSE OF MATHEMATICS

6. Three equal uniform rigid rods AB, BC, CD, each of mass m and length 2a, smoothly hinged at B and C, lie in a straight line on a smooth horizontal table. The rod BC is struck a blow at its mid-point in a direction perpendicular to the line of the rods, which causes it to move with velocity u. Find the initial angular velocities of the other two rods, and show that the kinetic energy generated is }mug. Show also that a uniform tension is set up in the rodBC which is initially equal 9 m u2 to 16a 7. A coin, initially head upwards, is tossed from a horizontal position by a blow Jon its edge at an angle r/6 to the vertical in a vertical plane through the centre of the coin. Prove that if

J m(nnag13)1, where m is the mass and a the radius of the coin (which may be treated as a uniform disc), it will land flat on a table, head upwards, after n revolutions, the table being at the same level as the point of projection. 8. A uniform rod of mass m and length 2a is held at an angle a with the horizontal with its lower end at a height h above a perfectly rough and inelastic horizontal table. If the rod is allowed to fall freely under gravity, show that after striking the table it will not leave the table before it becomes horizontal provided that 3h(1 sina) < 4a. 9. Two equal uniform straight rods AB, BC, each of mass m and length a, are jointed together at B. When the rods rest on a smooth horizontal table with AB, BC in line, C is jerked into motion with a velocity v perpendicular to the length of the rods. If ABC remains a straight line after motion begins, prove that friction at the joint B must supply an impulsive couple of magnitude 8mva. 10. A uniform rigid straight rod AB, of length 2a, mass M and centre 0, has a small ring of mass m attached to the end A, the ring being free to slide along a smooth horizontal wire. A particle, also of mass m, is attached to the end B of the rod. The system is hanging at rest under gravity when a horizontal blow of magnitude P is applied to B in a direction parallel to the wire. Show that the velocity of B immediately after the blow is

4P(M 3m) (M ± 2m) (M 6m) and find the kinetic energy generated by the blow. Show that, if the rod just reaches the horizontal position in the subsequent motion,

132 =(2/3) (M + 2m) (M

6m) g a .

11. A uniform circular disc, of mass m, centre 0 and radius a, is rolling without slipping along a horizontal plane with velocity V when the highest point A is suddenly fixed. Show that the disc will execute complete revolutions about A if V2 > 24ag. If V 2 =24ag show that OA will be instantaneously horizontal at time (3a/2g)l-i2 log ()/2 + 1) after A is fixed.

IMPULSIVE MOTION AND VARIABLE MASS

363

12. A number of particles 131, P2,..., of masses m„ m„ ..., respectively, are connected by inextensible strings and placed on a horizontal plane so that the strings lie along the sides of an unclosed polygon, each of whose internal angles is equal to x — a. If an impulse is applied to P1in the direction P2Pi , prove that mr (Trd-i cos cc — T r )= mr+i(Tr — T r _i cos a), where Tr is the impulsive tension of the string Pr Pr „• If there are four particles of equal mass and the strings form three sides of a regular hexagon, show that the ratio of the initial speeds of the first and last particles is 13:1. 13. A rigid lamina of mass M is smoothly pivoted at a point P, and initially it hangs at rest under gravity. The pivot is suddenly set in motion so that P begins to move with velocity V in the plane of the lamina. Show that the initial angular velocity is (M//) R x V, and that the initial velocity of the mass centre G is

{/G V + M(R • V) R}/I, where I, IG are the moments of inertia of the lamina about axes through P, G, respectively, perpendicular to its plane and R is the position vector of P relative to G. If, after the point P has been set in motion, it is constrained to continue in motion with the same constant velocity V, show that the lamina will make complete revolutions about P if 11/(R x V)2 > 4/g1R.1. 14. A uniform rod AB of length 2a and mass m rests in equilibrium on two smooth planes facing each other and inclined at 45° to the horizontal. An impulse I is applied at the end A of the rod in a direction parallel to a line of greatest slope of the plane on which A rests and towards the other plane. Assuming A and B to remain in contact with the planes during the impulse, prove that the initial spin 3I1/2 of the rod is 4ma If contact continues after the impulse, show that initially the reactions at A and B are equal and that contact cannot continue if I> m(2ag)112. 15. A uniform inelastic sphere of radius a rolls without sliding down the rungs of a ladder inclined at an angle 13 to the horizontal. The rungs are thin, rough, horizontal, and a distance 2 a sin a apart, where a + < ln; the centre of the sphere moves in a vertical plane perpendicular to the rungs. Show that the sphere cannot strike each successive rung with the same angular velocity w before impact, unless co2 = 7g sinp /[a sin a (2 + 5 costa)] . If uis the coefficient of friction between the sphere and every rung, show also that in this case 1 < {17 g cos 0 — lOg cos(a 0) — 7a w2 }I(2g sin 0) for all 0 in the range (3 — a < < + a .

364

A COURSE OF MATHEMATICS

16. An infinite number of uniform rods are connected freely at their ends and lie at rest on a smooth table in a straight line A,B,A,Ba A r Br . .. The rods A,. B,. are of mass 9m while the rods Br Ar+1are of mass 5m. An impulse is applied at Al so that the point Almoves with velocity u1perpendicular to the line of the rods. If the velocity of A, is ur and the velocity of Br is vr , prove that 28u,. 28v, +

= 0, 5u7+, = 0.

Deduce that 15 ur+, — 226u, + 15u7 _1= 0, and hence show that ur = u1151-r. 17. A process by which a child starts up a swing is represented as follows. Amass M is able to turn without friction about a horizontal axis through its centroid G. This axis is suspended from a smooth fixed parallel axis through 0 by light rods of length 1 which are perpendicular to the two horizontal axes. The point 0 is vertically above the equilibrium position of G, and during the motion 0G makes an angle 0 with the vertical. Initially the system is at rest with 0 = 0. Then M is given a rotation by an internal impulsive torque L. When 0G comes to rest at an angle 01, M is given an opposite rotation by means of an internal impulsive torque —2 L. 00 comes to rest again at an angle — 02 , and M then receives an internal impulsive torque 2 L , and so on. Prove that at the nth point of rest the height of G above its initial position is (4n — 3) L2 /2 114-2 12 g 18. A particle falls from rest under gravity through a stationary cloud. The mass of the particle increases by accretion from the cloud at a rate which at any time is Inky, where m is the mass and v the speed of the particle at that instant and k is a constant. Show that after the particle has fallen a distance x

kv2 =g (1

e-gcx) ,

and find the distance the particle has fallen after time t. 17. A rocket starts from rest on the ground and rises vertically, the velocity of the expelled gas relative to the rocket being constant and equal to w. Prove that the velocity of the rocket when the fuel is exhausted is —w log(1 — a) — where a is the ratio of the initial mass of fuel to the total initial mass, and r the time of burning of the fuel. Find this velocity (i) if the rate of fuel consumption is uniform and the maximum acceleration is vg, and (ii) if the rate of fuel consumption is regulated so that the rocket has constant acceleration ng. 20. A raindrop moves into a stationary cloud, and thereafter its radius increases at a uniform rate through condensation of moisture. Show that, whatever the initial velocity of the raindrop, it will describe an arc of a hyperbola one of whose asymptotes is vertical, and that the acceleration of the raindrop tends to the limiting value 1,g, where g is the acceleration of gravity.

ANSWERS TO THE EXERCISES Exercises 2:1 (p. 16) 1. Along CP, where C is point of contact of the cylinder with the plane. 5. BC exerts a force zW vertically upwards on AB.

Exercises 2:2 (p. 21) 1. a cos-10 cot+0.

2.

G,(a2 b3— a3 b2) (a,b, — a3 b2)

G2 (a3 b3

b3) (a3bi — aib3)

03 (ai b2 a2bi) (ctib2 — a2bi)

4. 211 4 B AC (in sense of letters). Exercises 2:4 (p. 35) 5(l sin 0) ; n = 5/8.

1. P 8(21 cos() — nl cot()) —

4. 1(3 WI. ± 2 W2) tan 0 + -1(2 W1± W2) tang9 in BP; i tanv — 4(W 1472) tang) in CE.

10.-I W tany horizontally and outwards from D.

Miscellaneous Exercises II (p. 42) 2. CBB'C' slips first. 11. 1F cosec (:T/n) at an angle n/n with sides and outside the polygon.

Exercises 3:1 (p. 56) 1. 1 = c etc; 6. 13a.

8.

(sinh 1). 1 2 + 1 log ( 21 4

9. (/'' — b2) a/(l' +112).

3. w ( 3

4 11°

27c

±I2 )'

+ 7c).

2-1.2

11.31.

Exercises 3:4 (p. 65) 2(a) W e4 /17r

(01' n — 1)/(e4 tin — 1); (b) W (1 — e--H 7 ) Kew' — 1) .

+ 4 (00

—

12).

366

A COURSE OF MATHEMATICS

Exercises 3:7 (p. 80) 1. 61.

4. 11,

3. 7w 14 /1152E/.

-

ti30} from the stated end.

Exercises 3:10 (p. 94) 2. 3 w118 + 81 W/128. (b) lw(2/ - x)2 ;

5. 17 W/24.

6. (a) M -

wx2,

5w/4 /24E/. Exercises 3:11 (p. 99)

1. W(a

b - y);

W IE I, b = a(secn/ - 1);

if n2

b) = W a seem; W = n2E 1912 .

Mo = W(a

Miscellaneous Exercises III (p. 100) 6. w13 (5 - 6 log2)/12Ek. TV x2 8. R(x - a); 5 Wa3 /144E/. 7. (TV + 2 W')13148EI. 12a 8aE I 9. 15W1/128; 47 W/128. 12. sec In/. /2

3. {2 + 1/2

l'(2

41/2)}a.

2739wa2 /3529; wa3 /6.

21. 3a/8 < b < Za. Exercises 4:2 (p. 111)

3. u{3 sin a + 1(9 sin2 a - 8)}/2g. 4. V 2 I{q (1 ± cosfp sina)}, (+ up, - down).

8. 4ka3/(3u).

Exercises 4:5 (p. 121) 5. 4(h - y)2 (191)2 = 1 ± q2.

dy

6. y = vo (2a3

3a2 x - x3)/3a2 v;

4avo /3 V. Exercises 4:6 (p. 12S) -

4. a(w2 w1)

4w;)

Exercises 4:7 (p. 135) 1. Space centrode is circle centre 0, radius la (wi w)/w/I ; body centrode is circle centre C, radius (aw bcoi)lcoi . 4. The line parallel to the rod through the centre of the circle. 5. a w, where w is angular velocity of body.

367

ANSWERS TO THE EXERCISES

Miscellaneous Exercises IV (p. 136) 1. Radial r 62 (cot2 a - 1) + recota; transverse 2r e2 cot a + 2. b/(wa).

4. !, (52 + 2a)).

6. (-a tan2 0, 2a sec2 0 tame).

9. a co lb ; instantaneous centre is at highest point of upper disc.

14.

37),2 (a2 - b2)

4 a h2

r7

r5

Exercises 5:2 (p. 145) 2. 6.

8 /( kt\

4. z n2 (a2 - x2)2 /a2 ; x = a tanh nt .

V 7

,) -• .

1 (4a + but \ log 2a a + bu2

8. 31k log ( ) sec. Exercises 5:3 (p. 152)

1. 5a/4 below upper point.

7. }V (a - c)212 cm) .

4. (ue-kt sixty t)/kt . 8. X + kX

2. m(u co + yaco2).

to2 x = ku, .

10. m1k2 <82.

Exercises 5:4 (p. 160) 3. 3.

6. R = mg{(M + 3m) sine - 2M 6}1(M m);

T = M mg (1 + cos 0)1(M m);

1/{(7

- 3 m) g a/3 (M

m)} ;

(2n - 3)/9.

Miscellaneous Exercises V (p. 166) 4. (2 k2 V2 n2) e 2 kX = n2 (2kX - 1). 10.

2.7 / /

9

m, l (m -{- m2

7. 4mg(1 - 3n)/(5n).

)

Exercises 6:1 (p. 170) 2. (a coshnt, a sinhnt); x2 - y2 = a2 .

4. x -

k2U V

- k2 V2) '

Y

kr pc {1(p - k2 172)

Exercises 6:2 (p. 176) 1 6. =A sin3 T - C(3 sin2 y, cosy) v3

cos3q))Ig , where A is a constant.

368

A COURSE OF MATHEMATICS

Exercises 6:3 (p. 185) 4. Y{2 (t2

V6t + 2)). Exercises 6:4 (p. 188)

3. Y(3 ga). Exercises 6:6 (p. 200) 9. (4 + 2y3)c/3. Miscellaneous Exercises VI (p. 203) 1. r = a sech ( U0/ V).

3. 2 n/ya Exercises 7:3 (p. 218)

1. Both (v sinnt)/n.

3. x = -b(cosnt

cosnt1/3), y =

(cosnt - cosnt1/3).

4. x = b{9 cosnt + cos (nt/1/6))/10, y = 3b{cos(nt/1/6) - cosnt)/10.

Exercises 7:4 (p. 229) 2.

1. (3m0/3)/16.

u 2 a k 1 + cos2 ) • Exercises 7:7 (p. 246)

2. m(3a2 sin2 - 3ab sina cos a + b2 costa).

1. 8ma2 /15. 4.

4m12 (2 - 3 sine cos 0 + 2 sin2 0). 3

Miscellaneous Exercises V11 (p. 247) 7. m/18 at the mid-point of each 5. m.t = s(y - x); n2y = s a - s (y - x) . side and m/9 at the mid-point of each of the lines joining the centre to a vertex.

10. 4 = itninxr ,

= PmnYr • Exercises 8:2 (p. 254)

3.

(11/ 5)1A.

7. (i)

(G

4. (m0/43)/11.

5. 1/(24g/17a).

P)/2 a Q./l b (G - P)b2 Qab ; (ii) (= 2 say); 11b2 12 a2 11b2 12 a2

2n2 N 2/2.

8. 1/{2ga02 2/(n + 2)}.

369

ANSWERS TO THE EXERCISES

Exercises 8:3 (p. 258) 1. Acceleration x = T al(Ma2 4mk2); on each of front wheels mk2x1a2backwards; on each of rear wheels (Ma2 2mk2)2V2a2 forwards.

Exercises 8:4 (p. 266) 5. 71 mg.

3. m(68g - 7aS22)/25.

Exercises 8:5 (p. 276) 4. Board 7Mv/(7M + 2m); sphere 2Mv/(7 M + 2m). 6. a2 co2 /(48g sini3).

7. ij gt'sina.

Exercises 8:6 (p. 284) 1. Body centrode centre B radius 2a; space centrode centre A radius 4a. 6. Increased in ratio 8:7.

Miscellaneous Exercises VIII (p. 285) 2. 31/17/25.

3. 4a0

8kaO

3g0 = 0; 4k2 a < 3g.

6. VI 6(M m)gasin0 12aF0

3m + 4M

where cos0 =

6F

mg

13. 0 = -(3g sing)/4a; 62 = 3g(cos0 - cosa)/2a;

T = mgx {(8a - 3x) cos0 - (6a - 3x) cosa}/2a2 ; S = -{mgx(3x - 2a) sing}/4a2 ; G = {m g x2 (x - a) sing}/2a2 ; where

x =F, a.

15. The disc comes to rest after time vol(pg).

16. mg cosa.

17. 1,ma(1 - 3 cos6)2 + 7mg sing cos0.

3 /( 15ga 20. 16 1/ 1\ 2 ) horizontally. Exercises 9:2 (p. 303) 1. 11 B is angle between the rod and the vertical, 0 = 0, n, unstable; B = stable 2. Stable.

3. Diagonal vertical, stable.

6. sin-i (1/2 13), unstable.

Exercises 9:3(b) (p. 313) 4. 27i 1/1, 5a2 I 13g (r - a)

f

9. 3a.

370

A COURSE OF MATHEMATICS

Exercises 9:6 (p. 328) 3. A', = — 11 n2 x/3 n2 y, x/y--= —3 or

= n2 x — n2 y where n2 =g/4a;

.

4. 4n/a1n, 4n/a2 n, where n2 =Time and cq. , 5. 2m/z,, 2n/n2 where n;:,

= 7 ± f17.

= 1/{4(3 f }/5)p2 — a)2} and p2 = A,/m.

ExerciSes 10:3 (p. 341) 1. 3mu(1

e)I7; s.

6.1V.

7. n; a(in + 1/3)/V.

8. Sphere has velocity resolutes E U horizontally, horizontally.

U downwards; prism 3 U/8

Exercises 10:4 (p. 351) 3. 8mv/7; 4mv2 /7.

5. 4maw/5.

8. 2m Y(ga/3).

9. (M m) w/(M + 4m). Exercises 10:5 (p. 359) 1.

g lin+ rt 3r

m2 u

m3

(m + rt)2 } + (m ± rt)2

dv dv dm mg 10. v ( dt +k) +m =mg, equivalent to m dm + 2v = k , where v dt is the relative velocity and m is the mass; v =

n?,, V gm m2 +3k {1

where m, is the initial mass.

Miscellaneous Exercises X (p. 361) 1. }/{2E(M

m)/(mM)}.

6. 3u/4a.

1 18. —log cosh {t f(g k)} .

3. —3J/(7ma), 15J/(7ma).

a 19. (i) — w l0 1—a — a) + (1 — a) (1+ v)

I (11)

nw log(1 — a) n 1 •

mo)3 }' m

INDEX Absolute units 6 Acceleration 10511', 155 in polar coordinates 112, 179, 231 relative 118, 220 Action (and reaction) 8, 68, 206, 333 Amplitude (of SHM) 148 Angular momentum 178-9, 186, 191, 201, 207, 232, 23711 velocity 123ff, 23111 Apse 187, 201, 203 Archimedes 11 Atom hydrogen 207 Basic units 3 Beams 67f Bending moment 6711, 9611 Body centrode 12911 Catenary 47f,298 Central force 169, 177 orbit 143, 17711, 207 Centrode 129ff CGS units 3, 5, 7 Chain, flexible 2, 45 Characteristic frequency 325-6 Circle, motion in a 15411 Clapeyron's theorem 9111 Coefficient of restitution 337, 343 Collision 208-9, 33711 Conic 192, 243 Conservation, of momentum, energy 225, 237 g Conservative force 37, 39, 139-40, 154-5, 169, 226, 259, 293-4 Constraint 27, 139, 154 Couple 17, 18 work done by 23 Critical thrust (in a beam) 9711 Cycloid 162

6-function 8311, 85, 332, 337 Differentiation of a vector 105-6 Displacement, virtual 25ff Dissipative force 37 Double star 207 Double zero 144 Elastic body 2 string, heavy 60 Elasticity, modulus of 2 Ellipse 189, 192, 194 momental 243g Elliptic harmonic motion 189-90 Encounter, of two particles 208-9, 211-2, 33711 Energy 22, 226, 343 equation 140, 154-5, 161, 169, 179 to 180, 187, 192-3, 207-8, 226-7, 229, 231, 23711, 249, 256, 259, 268, 278, 293, 305 kinetic 140, 180, 225, 232, 278, 293, 34311 potential 37ll, 140, 155, 180, 227, 293ff, 326 Equilibrium 1011, 41 neutral 299 of chain 4511 of heavy elastic string 60 of particle 10, 25 of rigid lamina 1111, 17, 25 stability of 29411, 325 Equimomental systems 24411 Equivalent sets of forces 17g simple pendulum 160, 250 External motion 207 Flexible chain 2, 45 Flexural rigidity 75, 96 Flexure, pure 75, 76, 96

372

INDEX

Force 5 central 169 conservative 37, 39, 139-40, 154-5, 169, 226, 259, 293-4 dissipative 37 external 75, 222, 225-6, 231, 354 internal 24, 67, 222, 225-6, 230-1 of constraint 27 shearing 67ll unit of 5 Forces, polygon of 10 triangle of 10 FPS units 3, 5, 7 Frame of reference 4, 118 Newtonian 4, 107, 206, 219 Free fall 193 Friction 1211, 268 g Gravitation, Newton's inverse square law of 193, 202, 203 Einstein's theory of 202 Gravitational units 6 Gyration, radius of 232 Gyroscope 143 Heaviside's Unit function 8111, 85 Hodograph 116, 119, 173 Hooke's Law 2, 60 Hyperbola 192-4, 209-10 Impulse 331a, 343 Impulse function 8311, 85, 332, 337 Independence of translation and rotation 220-1, 223, 232-36, 249, 257 Indeterminate system 41 Inertia 4, 5 moment of 232-3, 24011, 279 principal axes of 24211 product of 24011 Inertial frame 4, 107 Initial motion 278, 280 Instability 29211 Instantaneous centre 129f, 232,278f Internal motion 207 Intrinsic equation 46 coordinates 46, 113, 175

Inverse cube law of attraction 202-3 square law 181, 191ff, 202, 208 Jet aircraft 353, 355 Kepler's Laws 194, 212-3 Kilogram (kg) 4 Kinetic energy 140, 180, 225, 232, 278, 293, 34311 Lamina 2, nu, 23, 23011 angular velocity of, 123-4 Lami's Theorem 10, 12 Law of Action and Reaction 8, 68 Law of Motion 611, 10 Length, unit of 3 Lever, Principle of 11-12 Libration 143, 181 Limitation motion 144, 180 Linear momentum 221, 225-6, 232-3, 23711, 333, 354 Load, on a beam concentrated 68-9, 84-5 distributed 68, 85 Mass 4 Macaulay's method 84 g Maximum 294, 300 Mercury (planet) 203 Metre 3, 4 Minimum 294, 296, 299, 326 MKS units 3, 5, 7 Modulus of elasticity 2 Moment, bending 67.11 of inertia 232-3, 24011, 279 of momentum 178-9, 221, 232-3, 236, 333-4 Moments, Clapeyron's theorem 91ll Momental ellipse 243 ft Momentum 7, 8, 190, 221, 232, 35411 angular 178-9, 186, 191, 201, 207, 232, 23711 conservation of 208, 216, 226, 334, 35411 Neutral equilibrium 299 surface 73, 75, 96

INDEX

Newton, the (N) 5 Newton's Law of gravitation 193-4, 202, 203 of impact 337, 344 of motion 6 g Newtonian frame 4, 107, 206, 219 time-scale 3 Normal coordinates 325-6 component 144, 161, 174 Orbit, central 143, 17711 eccentricity of 192 perturbation of 319 Parabola 55, 108, 169, 192, 194 Parabolic catenary 5311 Parallel axes theorem 232-4, 24111 Particle 1 Pendulum, simple 159, 160-1, 189, 250 spherical 143, 189 Planetary motion 193, 202, 213 Poisson's hypothesis 344 Polar coordinates 112, 179, 181, 186, 192, 201, 231 Potential energy 3711, 140, 155, 180, 227, 29311, 326 Pound (lb) 4 Poundal (pdl) 5 p-r equation 187, 194 Principal axes of inertia 24211 moments of inertia 24211 Product of inertia 232-4, 24111 Projectile 108, 169, 172 Pure flexure 75-6 Radial component 113 Radius of gyration 232 Relative motion 118, 220-1 angular momentum 221 Relativistic mechanics 3 Repeated zero 144 Resisting medium 169, 17111 Restitution, coefficient of 337 Revolving orbit 201-2 Rigid body (lamina) 1, 23011 Rocket 35311 Rocking stone 297, 300

373

Rolling 268 g Rotating curve 31411 Rotation, independence of translation and 220-1, 223, 232-36, 249, 257 Second, mean solar 3 Shearing force 67ll Sidereal day 3 Simple harmonic motion (SAM) 14811, 160, 162, 281, 293, 326 pendulum 159, 189, 250 equivalent 160-1 Sliding 268 g Slug 6 Small oscillations 149, 160, 216-8, 280,304f Space centrode 12911 Spherical pendulum 143, 189 Stability 29211, 325 Standard acceleration (ga) 6, 7 Statically indeterminate system 41 Statics 1011 Strain 67, 73 Stress 68, 74 String 2 Strut 9611 Suspension bridge 58 Systeme internationale 3 Systems of particles 21911 Tangential component 114, 161, 174 Technical units 6, 7 Tension in a beam 68-9, 74-5, 96 Thrust in a beam 96ll critical 97 11 Tie 96 Time interval 3 Torque 17-8 Translation 124 independence of rotation 220-1, 223, 232-6, 249, 257 Tycho Brahe 194 Units 311 Unit function (Heaviside's) 8111, 85 Unstable state 29211

374 Varignon's Theorem 11 Vector, differentiation of 105-6 Velocity 10511 angular 123ff, 23111 in polar coordinates 112 relative 118 Virtual work, Principle of 2511

INDEX

Wire 2 Work 22, 140 done by a couple 23 Principle of Virtual 2511 Yard 4 Young's modulus 74 Zero (of a function) 144, 187