A Course in Applied Mathematics, Vol. 1 and 2

A COURSE IN APPLIED MATHEMATICS

OLUMl!

V

I of this work is devoted to

the mathematical theory of mecha­ nics and Volume IT to the field theories of gravitation, electromagnetism and hydromechanics. The development is from first principles, but it is assumed that the reader has completed an Intermediate Course of mechanics and is studying for a First Degree, with applied mathematics as a pr incipal subj ect. The work is more than adequate as a text for the applied mathe­ ma tics section of Part TI of the London University General Degree (excluding the stat istics) and will also be found suitable for the course in natural philosophy leading to Part I of the Mathematics Tripos and mathematics courses for the applied arranged by other universities for their non-specialist mathematicians. Every effort has been made to assist those students who must depend largely upon their own resources when studying the subject. All arguments are given in full detail and the results are illustrated by numerous examples taken from examina­ tion papers set at Cambridge, London and the provincial universities. Each chapter

concludes with a comprehensive set of ex.ercises relating to its subject matter. The notation is modern and the techniques of vector analysis have been employed throughout. A knowledge of elementary vector algebra and calculus is assumed, but the more advanced ideas associated with the operators, 'div,' 'grad' and 'curl' have been developed ab initio as they are required. The fundamental laws of mechanics are carefully presented to be consistent with the special theory of relativity, and the difficulties associated with the definition of the electromagnetic field vectors in dielec­ trics and magnetic media arc also given

special consideration. Answers are provided for all the exercises and there is an index to each volume.

40/; net

A COURSE IN

APPLIED MATHEMATICS

PHYSICAL SCIENCE TEXTS Gmeral Editor

SIR GRAHAM SUT TON C.B.E., D.Sc., F.R.S. ,

Director-General, Meteorological Office, Formerly Dean of the Royal Military College of Science, Shrivenham, and Bashforth Professor of Mathematical Physics.

ADVANCED LEVEL APPLIED MATHEMATICS by c. G. LAMBE, B.A., Ph.D. APPLIED MATHEMATICS FOR ENGINEERS AND SCIENTISTS by c. G. LAMBE, B.A., Ph.D. ADVANCED LEVEL PURE MATHEMATICS by C. J. TRANTER, O.B.E., M.A., D.Sc. TECHNIQUES OF MATHEMATICAL ANALYSIS by C. J. TRANTER, O.B.E., M.A., D.Sc. GENERAL PHYSICS AND SOUND (To Advanced and Scholarship Level) by D. H. FENDER, B.Sc., Ph.D. HEAT (To Advanced and Scholarship Level) by A. J. WooDALL, O.B.E., Ph.D., F.Inst.P. LIGHT (To Advanced and Scholarship Level) by C. B. DAISH, M.Sc. ELECTRICITY AND MAGNETISM (To Advanced and Scholarship Level) by C. G. WILSON, M.Sc., A.Inst.P. EXPERIMENTAL PHYSICS (To Advanced and Scholarship Level) by C. B. DAISH, M.Sc. D. H. FENDER, B.Sc., Ph.D. A COMPENDIUM OF MATHEMATICS AND PHYSICS by DOROTHY S. MEYLER, M.Sc. SIR GRAHAM SUTTON, C.B.E., D.Sc., F.R.S. ELECTRON PHYSICS AND TECHNOLOGY by J. THOMSON, D.Sc. E. B. CALLICK, B.Sc. PRINCIPLES OF ELECTRONICS by M. R. GAVIN, M.B.E., D.Sc. J. E. HOULDIN, Ph.D. A COURSE IN APPLIED MATHEMATICS (Covering B.A. and B.Sc. General Degrees) by D. F. LAWDEN, M.A. PHYSICS FOR ELECTRICAL ENGINEERS by W. P. JoLLY, B.Sc.

In Preparation

REACTOR PHYSICS AND TECHNOLOGY by J. WALKER, Ph.D., F.Inst.P. D. ]AKEMAN, Ph.D., A.Inst.P. ELEMENTS OF MATHEMATICAL PHYSICS

A COURSE IN

APPLIED

MATHEMATICS By DEREK F. LAWDEN, M.A. Professor of Mathematics, University of Canterbury, New Zealand

VOLUME I Part I : Dynamics Part I I : Statics

THE ENGLISH UNIVERSITIES PRESS LTD 102

NEWGATE STREET

LONDON, E.C.l

Firstprinted 1960

© Copyright Derek F. Lawden, 1960

Printed in Great Britain for the English Universities Press, Limited, by Richard Clay and Company, Ltd., Bungay, Suffolk

GENERAL EDITOR'S FOREWORD

by

SIR GRAHAM SUTTON, C.B.E., D.Sc., F.R.S.

Director-General, Meteorological Office, Formerly Dean of the Royal Military College of Science, Shrivenham, and Bashforth Professor of Mathematical Physics

THE present volume is one of a number planned to extend the Physical Science Texts beyond the Advanced or Scholarship levels of the General Certificate of Education. The earlier volumes in this series were prepared as texts for class teaching or self-study in the upper forms at school, or in the first year at the university or technical college. In this next stage, the treatment necessarily assumes a greater degree of maturity in the student than did the earlier volumes, but the emphasis is still on a strongly realistic approach aimed at giving the sincere reader technical proficiency in his chosen subject. The material has been carefully selected on a broad and reasonably comprehensive basis, with the object of ensuring that the student acquires a proper grasp of the essentials before he begins to read more specialized texts. At the same time, due regard has been paid to modern developments, and each volume is designed to give the reader an integrated account of a subject up to the level of an honours degree of any British or Commonwealth university, or the graduate membership of a pro­ fessional institution. A course of study in science may take one of two shapes. It may spread horizontally rather than vertically, with greater attention to the security of the foundations than to the level attained, or it may be deliberately designed to reach the heights by the quickest possible route. The tradition of scientific education in this country has been in favour of the former method, and despite the need to produce techno­ logists quickly, I am convinced that the traditional policy is still the sounder. Experience shows that the student who has received a thorough unhurried training in the fun damentals reaches the stage of productive or original work very little, if at all, behind the man who has been persuaded to specialize at a much earlier stage, and in later life there is little doubt who is the better educated man. It is my hope that in these texts we have provided materials for a sound general education in the physical sciences, and that the student who works conscientiously through these books will face more specialized studies with complete confidence. v

PREFACE Mv intention has been to write a book on applied mathematics which will be of assistance to students reading for science degrees at a level lower than that of Honours. It is assumed that the reader has suc­ cessfully completed a course of study in the subject to the Inter­ mediate standard of his degree course or to the Advanced Level of the General Certificate of Education. It is also assumed that the reader will be attending a course of lectures in pure mathematics, during which he will study the theory of linear differential equations, both ordinary and partial, the definition and evaluation of surface and volume integrals and the notation and techniques of vector analysis. All these branches of the subject are now common to the courses arranged for this class of student by the various university bodies in the United Kingdom. Vector methods have accordingly been employed without comment at any point in the argument where it was felt they arise quite naturally. However, no previous knowledge of the properties of the vector operators " grad ", " div " and " curl " is assumed of the reader, these being developed ab initio as required. Since it is expected that the majority of the book 's readers will not be mathematical specialists, who could be expected to complete for themselves an argument not made fully explicit, very few links have been omitted from the chains of reasoning, with the result that the reader of average ability should be able to make considerable progress without outside assistance. This should commend the work to those many students who read for External General Degrees of London University by part-time attendance at technical-college classes. However, a serious attempt has been made to present a fairly complete logical structure. I have not therefore avoided diffi­ cult passages in the development of the subject, either by ignoring them, appealing to the reader's intuition or by offering an argument by analogy. The use of any of these stratagems may be justified in a book written for use by engineers or others who are primarily inter­ ested in acquiring the facility to use a tool. However, it is assumed that the readers of this book are motivated, at least partially, by a desire to study applied mathematics for its own sake as an �sthetic­ ally pleasing and elegant structure. Those engineers who appreciate that a proper understanding of the tools they employ will greatly assist them in their efficient use, should also find the book helpful. Another type of student to which the book may appeal is the man who is reading for Honours in mathematics but who, for one reason or vii

viii

P RE FACE

another, finds that the texts to which he is referred make too great a demand upon his abilities. The book should carry such a student a long way towards his goal, after which he may feel sufficiently con­ fident to study more comprehensive texts. The subject matter of the book divides into four parts, Part I Dynamics (Chapters 1-10}, Part II Statics (Chapters l l-13), Part III Field Theory (Chapters 14--18}, Part IV Hydromechanics (Chapters 19-21}. Part I includes a detailed discussion of the dynamics of a particle and of a rigid body moving in a plane, and an introduction to the theory of the general three-dimensional motion of a rigid body. This introduction explains how considerations of angular momentum may be employed to derive Euler's equations for the motion of a rigid body pivoted at a fixed point, establishes Lagrange's equations and illustrates their application by a number of easy problems on three­ dimensional motion. In Part II equations of equilibrium for a system of rigid bodies acted upon by a three-dimensional system of forces are derived as particular cases of previously established dynamical equa­ tions. This Part concludes with an introduction to the theory of elasticity. Part III gives an account of the Faraday-Maxwell theory of the electro-magnetic field together with an introduction to the Newtonian theory of gravitational attraction. The equations of the electro-magnetic theory are expressed in terms of an unrationalized Gaussian system of units, since the examination papers of the various university bodies which have been consulted indicate that such a system is normally employed by mathematical departments in their teaching. However, an appendix has been added at the end of this Part, in which an explanation of the advantages to be gained from rationalizing the unit system is given and the Giorgi system is described and related to the system employed in this book. In Part IV the equilibrium of a fluid in a general gravitational field is discussed. A chapter is devoted to the one-dimensional flow of an ideal fluid in a smooth pipe, the theory being illustrated by the problem, now very topical, of the rocket motor. The general equations of motion of an ideal fluid are obtained in a final chapter and the characteristics of a number of simple irrota­ tional flows and flows due to rectilinear vortices are investigated. By analogy with electrostatics, the velocity potential cp is related to the flow velocity by the equation q = -grad cp. Throughout the book important results and principles are illustrated by worked examples, some of which are original but many of which have been taken from examination papers set by the following bodies : Birmingham University (B.U.) Cambridge University (Mathematical Tripos) (M.T.) Durham University (D.U.) Leeds University (Le.U.)

P R E FA C E

ix

Liverpool University (Li.U.) London University (L.U.) Manchester University (M.U.) Nottingham University (N.U.) Queen's University, Belfast (Q.U.) Sheffield University (S.U.) Other problems, from the same sources, for working by the student, will be found in the sets of exercises at the ends of the chapters. The source of each such example or exercise is indicated according to the abbreviation scheme shown above. The author wishes to express his thanks to the bodies concerned for permission to make use of this material. The author is also indebted to Sir Graham Sutton, F.R.S., for a number of suggestions which have led to improvements in the text and to his brother, Mr. J. G. Lawden, who is responsible for the pre­ paration of the figures and diagrams. D. F. LAWDEN

University of Canterbury

CONTENTS PART I : DYNAMICS

l.

KINEMATICS OF A PARTICLE

.

l .l. The Velocity Vector 1 .2. Components of the Velocity Vector 1 .3. The Acceleration Vector 1 .4. Components of the Acceleration Vector .

3 3 5 10 12

2. NEWTON'S LAWS. RECTILINEAR MOTION . 2.1. Newton's Laws of Motion 2.2. Work and Power . 2.3. Rectilinear Motion of a Particle 2.4. Damped and Forced Oscillations

20 20 24 27 42

3. CARTESIAN AND INTRINSIC RESOLUTES OF ACCELERATION . 3.1. Resolving the Equation of Motion . 3.2. Cartesian Components. Projectiles 3.3. Cartesian Components. Elliptic Harmonic Motion . 3.4. The Energy Equation . 3.5. Tangential and Normal Components. Motion on Wires

55 55 55 66 69

4. PoLAR 4.1 . 4.2. 4.3. 4.4. 4.5. 4.6.

73

85 Central Forces 85 88 Inverse Square Law The Energy Equation 92 Time in an Orbit . 99 101 Kepler's Laws The pr-Equation of an Orbit. Elliptic Harmonic Motion . 103 4.7. Motion on Rotating Wires 106 4.8. Motion on Surfaces of Revolution . 1 10 RESOLUTES OF AccELERATION

5. IMPULSIVE MOTION OF PARTICLES 5.1. Impulse and Momentum 5.2. Newton's Third Law . 5.3. Impact of a Pair of Particles xi

118 118 121 124

xii

CONTENTS

6. MOTION O F A PARTICLE S YSTEM I . 6.1. Systems Involving a Small Number of Particles 6.2. Vibration of Two Particles. Normal Modes 6.3. Energy Equation for a Particle System . 6.4. Kinetic Energy of a Rigid Body . 6.5. Conservative Systems of one Degree of Freedom. Stability of Equilibrium. Small Oscillations 6.6. Stability of Rocking Bodies .

131 131 133 139 145

7. KINEMATICS OF A RIGID BODY. MOMENTS OF INERTIA 7.1. Instantaneous Centre. Space and Body Centrodes . 7.2. Motion of a Rigid Body about a Fixed Pivot 7.3. General Three-Dimensional Motion of a Rigid Body 7 .4. Moment of Inertia of a Lamina 7.5. Moment of Inertia of a Solid Body 7.6. Equimomental Systems

169 169 173 177 179 182 185

8. M OTION OF A PARTICLE SYSTEM-II 8.1. Equation of Motion of the Mass Centre . 8.2. Conservation of Linear Momentum 8.3. Isolated Systems. 8.4. Equation of Angular Momentum 8.5. Conservation of Angular Momentum 8.6. Angular Momentum of a Rigid Body in General Motion 8.7. Equation of AM for a Rigid Body in General Motion 8.8. Initial Motion of a Rigid Body 8.9. Motion with Variable Mass

191 191 197 201 203 205

-

.

152 159

208 210 214 215

9. IMPULSIVE M OTION OF A RIGID BODY 226 9.1. Equations of Linear and Angular Momentum for Impulsive Motion 226 9.2. Elastic Impact of Rigid Bodies 232 10. THREE-DIMENSIONAL M oTION. LAGRANGE ' s E QUATIONS 10.1 . Rotating Frames of Reference 10.2. AM of a Rigid Body Rotating about a Fixed Pivot 10.3. Euler's Equations 10.4. Kinetic Energy of a Rotating Rigid Body 10.5. Lagrange's Equations . 10.6. Small Vibrations. Normal Modes .

240 240 241 241 246 253 261

CONTENTS

xiii

PART II : STATICS

11.

EQUILIBRIUM AND E QUIVALENCE OF FORCE SYSTEMS

Equilibrium of a Rigid Body Equivalence of Force Systems Parallel Forces. Centres of Mass . Inertial Forces. Centrifugal Force The Principle of Virtual Work

271 271 276 282 284 285

12. E QUILIBRIUM OF STRINGS AND CHAINS 12.1. General Equations of Equilibrium . 12.2. String Hanging in Equilibrium under Gravity 12.3. String in Contact with a Cylinder .

296 296 297 303

13. DEFORMATION oF ELASTIC B oDIES . 13.1. Internal Forces and Deformations . 13.2. Shearing Force and Bending Moment 13.3. Extension of a Bar 13.4. Flexure of a Beam 13.5. Whirling Shafts . 13.6. Analysis of Strain 13.7. Analysis of Stress 13.8. Generalization of Hooke's Law 13.9. Simple Cases of Elastic Deformation

309 309 310 315 316 322 324 330 335 337

A NSWERS INDEX

348

11.1. 1 1.2. 1 1 .3. 11.4. 11.5.

PART I DYNAMICS

CHAPTER 1

KINEMATICS OF A PARTICLE 1.1. The Velocity Vector

A particle is defined to be a body whose dimensions are negligible to the degree of accuracy in linear measurement to which we choose to work. The Earth is a body of approximately 4000 miles radius, whereas the radius of its orbit about the Sun is over 93,000,000 miles. If, therefore, it is sufficient for our purpose to specify the orbit to an accuracy of 0·1 per cent, the Earth may be treated as a particle. This we shall do in Chapter 4. Since, in the sense just explained, a particle has no extension, it is conveniently represented by a Euclidean point. Rotation of a particle cannot be detected without improvement in the accuracy of measure­ ment. It follows that its motion is completely specified by that of the point. Let P (Fig. l.l) be a point representing a particle at some time t. Let 0 be any identifiable point in the neighbourhood of P (e.g., the z

X

FIG.

1 . 1 .-Position Vector of a Particle

centre of the Sun when we are discussing the motion of a planet). 0 is not necessarily " fixed" ; indeed, we have not yet stated the mean­ ing we shall attach to this term. Take 0 as the origin of rectangular Cartesian axes Ox, Oy, Oz in any identifiable, mutually perpendicular, directions (e.g., so as to pass through three known stars). These directions also are not necessarily " fixed". For example, it may be 3

4

A CO U R S E I N A P PLI E D M A T H E MA T ICS

[CH.

convenient to allow them to rotate with the Earth when we are dis­ cussing the motion of a projectile in its vicinity. These axes con­ stitute a frame of reference against the background of which the motion of P may be described. The position of P at time t may now be fixed

-+

by its coordinates (x, y, z) or, alternatively, by the vector OP. This vector is, accordingly, referred to as the position vector of P, and will be denoted by r. The quantities (x, y, z) are the projections of r upon the coordinate axes, and are therefore its components. r varies both in magnitude and direction as P moves ; i.e., r is a function of t. At time t + l>t, let the particle be at P', a point with position vector r + l>r. It follows-+from the triangle law of addition of vectors, that l>r is the vector PP'. Consider the vector l>rjl>t, i.e., a vector in the direction of l>r, but differing from this vector in magnitude by a factor ljl>t. As l>t -+ 0 this vector tends to a limit both in magnitude and direction. The limit vector is termed the derivative of r with respect to t and is denoted by drjdt. This derivative clearly provides us with a precise measure of the rate of change of the position of the particle P at the instant t relative to the frame of reference. It is accordingly accepted as the definition of the velocity of P relative to this frame and at this instant. Denoting the velocity vector by v, we have

dr V = -· dt

(1.1)

Before this definition of velocity is acceptable, we must show that it is in accord with the more or less vague ideas concerning this quantity commonly held. As l>t -+ 0, P' approaches P along the particle's path and l>r approaches the direction of the tangent to this track at P. The direction of the velocity vector is therefore along this tangent and specifies the direction of the motion of the particle in the frame of reference. Let A be any fixed point on the path of P and let s denote the length of the curve j oining A and P. s will be reckoned positive if the particle arrives at P after it has passed through A . Then the arc PP' may be denoted by l>s. As at -+ 0 and P' approaches P, the ratio PP'jl>s approaches unity. It follows that the limits of the ratios P P' jl>t and l>sjl>t are identical. But the first ratio is the mag­ nitude of the vector l>rjl>t and has limit JdrjdtJ. The limit of the second ratio is dsjdt = s. Hence

v = Jvl = ���� = s,

(1.2)

i.e., the magnitude of the velocity vector is the rate of increase of the distance of the particle from a fixed point on its path. v is termed the speed of P. Now suppose that each of two observers, moving relatively to one

1]

5

KI NEMATICS OF A PARTICLE

another, assesses the motion of P against the background of a reference frame which is fixed relative to himself. Let Oxyz, O'x'y'z' (Fig. 1 .2) be the two frames and suppose that corresponding axes always remain parallel, i.e., there is no relative rotation of the frames. Let r be the position vector of P relative to the first frame and r ' the position z

FIG. 1.2.-Relatively Moving Frames

vector of P relative to the second frame at the instant t. If p is the position vector of 0 in the second frame at the same time, we have r' = p + r.

Differentiating this equation with respect to the time or

r'

=

i.l + r,

v' = V + v,

(1.3)

t, we obtain (1.4) (1.5)

where v is the velocity of P relative to the first frame, v' is the velocity of P relative to the second frame and Vis the velocity of 0 in the second frame, i.e., of the first observer relative to the second. Equation (1.5) shows that, to calculate the velocity of a particle relative to a new frame of reference, to the velocity in the old frame we must add vectorially the velocity of this frame relative to the new. 1.2. Components of the Velocity Vector

The velocity vector, like any other vector, may be regarded as the vector sum of three vectors in any three non-coplanar directions. The magnitudes of these three vectors are the components of the velocity in these three directions. Consider the motion of a particle moving in a plane. If two of the

A CO U R S E I N A P P L I E D M A T H E M A T ICS

6

[cH.

component directions lie in the plane and the third does not, the velocity component associated with the latter direction will be always zero and we shall ignore it. Suppose that the remaining pair of directions are taken, one along the tangent and one along the normal to the particle's path at the instant t. Since v is a vector of magnitude s in the direction of the tangent, (s, 0) are the tangential and normal components of the velocity respectively. Let the axes Ox, Oy of the frame of reference be taken in the plane of motion and let (x, y) be the coordinates of P at time t. Then, if i, j are unit vectors in the directions of the axes, the position vector of P is given by r = xi+ yj. . (1 .6) Differentiating with respect to t, since i and j are constant in magnitude and direction, we obtain

v = drjdt = xi + yj. Thus, the Cartesian components of velocity are (.i, y).

(1.7)

This result does not depend upon the axes being rectangular, although it will generally be convenient to regard them so. If the axes are rectangular, squaring equation (1.7), we find that

(1.8) since i j = 0. Let (r, e) be the polar coordinates of P relative to a pole 0 and initial line Ox in the plane of motion (Fig. 1 .3 (a)). Then, if p, q are ·

(b)

(a) FIG.

1.3.-Polar Components of Velocity

unit vectors along and perpendicular to in the figure) respectively, we can write

r = rp.

OP

(in the senses indicated

(1. 9)

1]

KINEMATICS OF A PARTICLE

7

Differentiating, there results

v = rp + rp.

(l.lO)

Since p is not necessarily constant in direction, its derivative is not, in general, zero. In time l';t, e will increase by 1';6 and p will accordingly rotate through this angle into the position of p + 'i';p (Fig. 1.3 (b)). The magnitude of p + 'i';p is unity and hence, with p and 'i';p, forms an isosceles triangle with equal unit sides. From this triangle, we conclude that I 'i';p I = 86, to the first order. Hence .

'i';p

I I = 1"1m l';t = e.. 1';6

I p I = 1"1m &

But, as l';t ---+ 0, the angle between and thus the limiting direction of that

p and 'i';p approaches a right angle l';pjl';t is that of q. It now follows p = 6q. (1.11)

Similarly, the reader should prove that q=

-6p. (l.l2) Substituting from equation (1.11) into equation (1.10), we find that v = rp + r6q, (l.l3) i.e. the components of v along and perpendicular to OP are (r, r6) respectively. These are the polar components of velocity. Squaring equation (1.13), we obtain (l.l4)

6 is the rate of rotation of OP relative to the frame of reference. It is termed the angular velocity of P about the point 0. The velocity of P being known, its angular velocity may be calculated from the formula (l.l5) Example 1. A point A moves in a counter clockwise direction round a circle of centre 0 and radius a with angular velocity wv and a point B moves counter

clockwise round a concentric circle of radius 2a with angular velocity w2• If 6 is the angle OAB when the relative velocity of A and B is parallel to the line A B, prove that

provided that w1 � 2w2 •

tan2 6 -

"'•

"

- 4w•" 3wa•

Find also the speed of B relative to A at that instant.

(L.U.)

8

A C O U R S E I N A P PL I E D MATHEMAT I C S

[CH.

The velocities of A and B are aw1 and 2aw 2 respectively in the directions shown and relative to a frame within which 0 is stationary. Relative to an observer moving with A, this frame has a velocity aw1 in the opposite sense to that of A 's velocity. The velocity of B relative to A is therefore found by adding vectorially A 's velocity reversed to that of B. This is shown in the diagram. This rela­ tive velocity is in the direction BA , and hence, if it is resolved into components along and perpendi­ cular to AB, the latter component is zero. But this component is the sum of the components of the vectors of which it is the sum. We therefore have the equation 2aw 2 cos L OBA - aw1 sin (a - 90°) = 0 cos L OBA =

-

sin• L OBA = 1

�

2w 2

cos a

w• - 1 2 cos• a. 4w. Applying the Sine Rule to the triangle OAB, we obtain sin L OBA = t sin a, and then, from equations (i) and (ii), w• tsin• a = 1 - 1 2 cos• a. 4w. Whence, dividing by cos2 a, 2 w2 w ttan2 a = sec• a - -1-2 = 1 + tan• a - -1 , 4w 22 4w2 and the result stated follows by solving for tan2 a. The component of the relative velocity in the direction BA is aw1 cos (a - 90°) - 2aw 2 sin LOBA = awl sin a - aw. sin a (using equation (ii)) = a(wl - w.) sin a. -

(i) (ii)

A ship A is moving due East with constant speed u, whilst a second ship P is moving due North with constant speed 2u, and a third ship Q is moving North-east with constant speed 2y2u. 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 respec­ tively 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 l2au

Example 2.

9a2 + 16x2 (L.U.) Let 0, B, C be the initial positions of the ships A , P, Q respectively. Take axes Ox and Oy as shown. The Cartesian components of the velocities of P and Q are (0, 2u) and (2u, 2u) respectively. At time t after the ships are

1]

K I NE M ATICS O F A PARTICLE

9

collinear, the coordinates of P are (a, 2ut) and of Q are (2a + 2ut, 2ut) . If M is the mid point of PQ its coordinates are (ta + ut, 2ut), and hence its Cartesian velocity components are (u, 2u). A has velocity components (u, 0) and the velocity components of M relative to A are accordingly (0, 2u). If LMAx = 6 the resolute of this relative velocity perpendicular y

0

to A M is 2u cos a. Employing equation (1. 1 5), we find that the angular velocity of M about A is w, where 2u cos e. w = AM

But cos a A N{AM. Also, since A is the point (ut, 0) at time t, A N = ta and A M2 = :t-a• + 4u2t2 • Hence

=

3au N "' = 2uA = • :t-a + 4u2t2 = A M•

where x = ut = OA.

12au 9a2 + 1 6x2'

A n aircraft, moving along a straight track with uniform speed v, is under attack from a guided missile. The missile moves with constant speed nv, its motion being always directed towards its target. Show that the polar equation ofthe track of the missile, as observed from the aircraft, is

Example 3.

�r = sin a tan"!a.

where the aircraft has been taken as pole and its direction of motion as the initial line. Deduce that, unless the missile's speed is greater than that of the target, the latter cannot be hit.

Relative to the aircraft, a frame stationary relative to the ground has velocity v in a D 1 R E c11 oN oF direction opposite to that of the o L..:.:...--.l. A"" j_c_M_O_T_iO_N______.,. aircraft's own motion. Relative to the aircraft 0, the missile P accordingly possesses a velocity which is the vector sum of : (i) its velocity nv relative to the ground in the direction PO, and (ii) the velocity v of the ground relative to the aircraft.

10

A CO U R S E I N A P P L I E D MATH EMAT ICS

[CH.

Taking polar coordinates as shown in the diagram, we find that the polar resolutes of P's velocity are (i) r = -nv - v cos 6 r6 = v sin 6. (ii) Dividing (i) by (ii), we obtain 1 dr = -n cosec 6 - cot 6. rde Integrating with respect to 6, we find that log r = -n log tan !6 - log sin 6 + constant, and hence r = A cosec 6 cotn !6 :i = sin 6 tann !6, r

where A is a constant depending on the initial conditions. If the missile is to destroy the target, r must approach zero for some value of 6. This implies that A fr -+ oo. Now tan !6-+ oo as 6-+ 1r. Let 6 = 1r £. Then sin£ ,;, _ £_ = sin£ cotn !£ = = 2n£1 -n :i_ ' tann !£ (!£)n r 1 -n where £ is small. As £-+ 0, £ -+ oo if n > 1. This implies that the missile will reach the target if its velocity is greater than that of the target. -

·

1.3. The Acceleration Vector

The acceleration of a particle is defined to be the rate of change of its velocity. The acceleration vector is therefore derived from the velocity vector in the same way that this vector is derived from the

0

c

(b)

(a)

FrG. 1.4.-The Hodograph

position vector. Let v be the velocity of a particle P at time t (Fig. 1.4 (a)). Take v to be the position vector of a point Q relative to a frame having origin C (Fig. 1.4 (b)). As v varies with the time, the point Q will move so as to trace out some locus. This locus is called the hodograph of the motion of P. At time t + at, the particle will have

1]

K I N E M AT ICS O F A P A R T ICLE

11

moved t o P' and the velocity vector will have changed to v + av. � The point Q will have travelled along its locus to Q', where QQ' = ; as at� 0, is called the acceleration av. The limit of the vector av u vector of the particle at the time t and is denoted by f. We have, therefore, f

= dv dt .

(1.16)

Since v is the position vector of Q, f is clearly the velocity of this point and is in the direction of the tangent to the hodograph at Q. Now suppose that two observers assess the motion of P relative to two frames of reference, the first of which possesses a velocity V relative to the second and neither of which is rotating relative to the other. The velocities of P relative to the frames are related by equation (1.5) . Differentiating this equation with respect to the time t, we obtain the equation f'

=

F + f,

(1.17)

where f = v is the acceleration of p relative to the first frame, f' = v' is the acceleration of P relative to the second frame and F = V is the acceleration of the first frame relative to the second. Equation (1.17) shows that to calculate the acceleration of a particle relative to a new frame, we must add to its acceleration in the first frame the acceleration of this frame relative to the new. Calculate the acceleration vector of a particle which moves around a circle of radius a with uniform angular velocity w. The velocity vector is of constant magnitude aw and, being always perpen­

Example 4.

dicular to the radius through the particle, rotates with angular velocity

w.

The point Q accordingly describes a circle of radius aw (the hodograph) with angular velocity w and the magnitude of its velocity vector is aw2• This, then, is the magnitude of the acceleration vector of the particle. The direction of Q 's velocity is perpendicular to CQ, i.e., is in the direction PO. The direction of P's acceleration vector is therefore also along PO.

12

1.4.

A C O U R S E I N A P PL I E D M A T H E M A T I C S

[CH.

Components of the Acceleration Vector

In this section we shall obtain expressions for the components of the acceleration of a particle moving in a plane, in the three pairs of direc­ tions in which we have already resolved the velocity vector. Differentiating equation (1.7) once again with respect to t, we find that (1. 18) f = dvfdt = xi + yj. Thus the Cartesian components of acceleration are (x, y) . Also, a further differentiation of equation (1.13) yields f

= rp + rp + r6q + riiq + r6q.

Substitution from equations (1. 1 1) and (1.12) for p and q permits us to write the latter equation in the form

f = (r

r62)p

-

+ (rii + 2i6)q.

(1.19)

Equation (1. 19) indicates that the polar components of acceleration are r- r62 along OP, (1.20) and rii + 2i6 = (r26) perpendicular to OP.

}

��

Let t, n be unit vectors along the tangent and normal to the path of the point P, t being in the direction of motion and the rotation from t through a right angle to n being anti-clockwise (Fig. 1 .5). If t

t

/

/

" /

/;\

--���---�----------------- X

FIG. 1 .5.-Unit Tangent and Normal

makes an angle � with the x-axis, by a method similar to that used to derive equations (l.ll) and (1 . 12), the following equations may be obtained, (1.21)

K I N E M AT ICS OF A P A R T I C L E

1].

Employing the notation of Section

v = st.

Differentiating, this yields

f=

1.2,

13

(1.22) (1.23)

st + .st = st + .s�n.

Thus the tangential and normal components of acceleration are

(s, .s�).

Writing s = v, the tangential components of acceleration can be written in any of the equivalent forms , _ 11 -

_

dv ds dv d l 2 s - dv dt - ds di - v ds - ds ( zV ) • .

.

_

_

_

(1.24)

The normal component of acceleration can also be manipulated into a number of forms thus

In = s� = v� = v

�� �

= KV 2

= v2jp,

(1.25)

where K is the curvature and p the radius of curvature of the particle's trajectory at P. The thrust operating on a rocket in free space is directed so that the acceleration is always in a direction making a constant angle "' with the velocity . A ssuming plane motion show that, if v. is the speed at the commencement of the manreuvre, v is the speed of any later instant and .p is the angle through which the velocity vector has rotated from its initial direction at this instant, v = v.e.Pcota. Deduce that, if the rocket moves in a circle of radius a, v.a tan "' V= a tan "' - v.t ' where t is the time measured from the commencement of the manreuvre. The tangent to the rocket track makes an angle .p with the initial direction

Example 5.

of motion. Taking this latter direction as that of the x-axis of the frame of reference, we see from equations (1.24) and (1.25) that the tangential and normal components of the acceleration are v and v� respectively. Sine� the acceleration makes an angle a with the tangential direction

�v = tan

<X,

1 dv v d.f = cot"'· Integrating with respect to .p,

J

v d.f d.p = l dv

J

dv v

= .P cot "' + constant

log v = .p cot a + constant v = Ae.Pcota, where A is an arbitrary constant. But v = v., when .p = 0 and hence A =v Thus, (i) •.

14

A CO U R S E IN A P P L I E D M A T H E M A T ICS

[CH.

If s is measured along the track from the initial position of the rocket and the latter moves around a circle of radius a, then s = a.p. Hence v = s = a:.p and equation (i) yields di/J = dt dt di/JIntegrating with respect to .p,

� eo/1 cota.

a

!!_ e-o/1 cota.

v.

•

t = constant - � tan IX e-o/1 cot a. v.

(ii)

When .p = 0, t = 0 and hence the constant is !!_tan et . We can now v. solve for e.Pcot a in equation (ii) to obtain a tan et . e.Pcot<X= a tan IX- v.t Substituting in equation (i) for e.Pcot<X, we find v.a tan et v= a tan et- v.t Assuming that IX is a positive acute angle, so that tan et is positive, this •

equation shows that as t increases towards '

!!_

v.

tan a, v---+ oo. The circular

course cannot therefore be maintained for a time greater than

!!_

v.

tan a.

A particle, initially at rest at the origin, moves in a plane with an acceleration that is the resultant of a constant radial acceleration f outwards and a variable acceleration equal to 2w times the velocity in a v direction perpendicular to the velocity, where w is constant. Show that its angular velocity about the origin is equal to w f and find the polar equation of its path, taking the direction of f at the origin as the initial line. (M.T.)

Example 6.

Resolving the acceleration component 2wv radially and transversely, we obtain the components - 2wv sin cp and 2wv cos q, respectively. Equa­ tion ( 1 . 1 3} shows that these 0 components can also be written in the form - 2wr6, 2wr. The total acceleration in the radial direction is accordingly f - 2wr6 , and in the transverse direction is 2wr. This fact is expressed by the equations ;; - r() 2 = f - 2wr6 , (i)

� :e (r26)

=

(ii)

2wr.

From equation (ii), we obtain

:e (r26) = 2wrr

=

w

#t (r2) .

1]

K I N E M A T ICS O F A P A R T ICLE

15

Integrating with respect to it follows that •� wr• + constant � w + :i.

t, r = =

r• re -+

a:J as 0. But the particle Unless A = 0 , the transverse velocity 0. We conclude that A = 0. Hence � is at rest when w, i.e., the angular velocity of the particle about the origin is w. Substitution for� in equation (i) yields r+ w•r The general solution of this differential equation is

r=

r-+

=

= f.

r = B cos wt + C sin wt + wJ,., B and C being arbitrary constants. When t = 0, r = r = 0, i.e., f B + ;;;2 = 0, wC 0. B = -f/w2 and C 0. Thus Hence r wJ. (1 - cos wt). (iii) Since the angular velocity of P about 0 is w and 8 = 0 at t = 0, 8 = wt. Equation (iii) is therefore equivalent to r I.w (1 - cos 8), =

=

=

=

and this is the polar equation of the path (a cardioid).

0 is the centre of a circular wire and A is a point on its circumference. A bead P moves around the wire with uniform angular velocity w relative to the wire, whilst the latter rotates about A in the same sense with angular velo­ city !(1 + y3)w. If ON is perpendicular to A P and M is the mid point of ON, prove that the acceleration ofthe bead at any instant is -+ {4 + 2y3)w2 • PM.

Example 7.

p

Suppose the bead is at A', diametrically opposite to A, X at 0. Let AX be the position of A A ' at this in­ stant. At a later instant the wire will have rotated through the angle !(1 + y3)wt and the radius OP LAPO = through an angle wt, as shown in the diagram. Then LPAO

t=

t,

= Take A as pole and AX as the initial line of polar coordinates (r, 6) of the

}-wt.

bead. Then

where

a

6 = LPAX = !(2 + y3) wt, AP = 2AN = 2 cos }-wt, is the radius of the wire.

r=

a

(i) (ii)

16

A COURSE I N APPLIED MATHEMATICS

From (i),

&

From (ii),

=

[CH.

!(2 + y3)w.

r = -law2 cos lwt. The polar components of the acceleration of P are therefore,

!r

fo

=

rA•

r-

= -law2 cos lwt- !aw2(2 + y3)2 cos lwt = -aw2(4 + 2y3) cos !wt

= -(4

=

=

=

I d

r dt

I d

r dt

+ 2y3)w2• NP

(r•A) [a2 (4 + 2y3)w cos• lwt]

-a2(4 + 2y3)w2 cos lwt sin !wtf2a cos lwt -law2(4 + 2y3) sin lwt = -!w2(4 + 2y3)0N = -w2(4 + 2y3)MN. =

fr is therefore representable, both in magnitude and direction, by the --+ vector (4 + 2y3)w2• PN. fo is similarly representable by the vector --+ (4 + 2y3)w2• NM. The total acceleration of P is now given by the vector sum --+

--+

(4 + 2y3)w2 • PN + (4 + 2y3)w2• NM

=

--+

--+

(4 + 2v3)w2 • (PN + NM) --+

= (4 + 2y3)w2. PM.

EXERCISE 1

P and Q are moving around two coplanar circles in the same sense and with the same angular velocities w. If a is the distance

I. Points

between the centres of the circles and oc is the acute angle made by PQ with the line of centres at any instant, show that the angular velocity of PQ at this instant is

P2

oc

a1 a2 w1 w2, P2 P1 2 a1 -2 a2• n = t(w l + w .) + !(w l - w .) where P1P2• If P1, P2 represent two planets, it may be shown that w1 = fLlfa1�, w2 = fLlfa2�, where fL is a constant for the solar system. Deduce that, in this case, the motion of P2, as observed

2. Particles

P1

( 1 - ;Q cos )w.

and move around concentric circles of radii and in the same sense and with angular velocities and all respectively. Show that the angular velocity of about is given by r

r =

from Pv reverses its direction when the angle the planets is given by

6

between the radii to

v'ala2 a1 + a 2 - ya1a2 and prove that this equation always has a real root. cos 6 = ---�'---'-:== -'

I]

KINEMATICS

OF

A PARTICLE

I7

3. A point moves in the plane of axes Ox, Oy so that its x velocity com­

ponent is always k times its y coordinate and its y velocity component is always k times its x coordinate. Show that its path is a rectangular hyperbola.

4.

A particle P moves in the plane of axes Ox, Oy such that its velocity makes a constant angle with OP. Show that the path is an equiangular spiral. If P's angular velocity about 0 is constant, show that its speed is proportional to OP.

5. An aircraft moves with constant speed V along a straight line.

0,

is its distance from a fixed point

prove that

� (rr) = V2.

If r

6. A and B are two fixed points on a circle and P is a moving point.

If (u, v) are the components of P's velocity resolved in the oblique directions AP, BP respectively, show that u sin2

7.

ex;

= r s

-

r,

In Example

3, assuming that, at

r = a, 6 = ex:,

and that

BP =

v sin2 ex; =

and LAPB =

n > I,

when t

8.

ex;,

s cos

where AP =

=

ex:.

s

- r cos ex;,

Deduce that ur + vs = 0. (M.T.)

t = 0, the missile is at the point prove that the missile strikes the target

a(n - cos ex:) . v(n2- 1)

Prove also that the missile's acceleration is always perpendicular to OP and of magnitude nv2 sin 6fr. A particle moves around the equiangular spiral r = ae O co�"' so that its angular velocity about the pole is constant. Show that the accelera­ tion of the particle makes an angle 2cx: with the radius vector and is of magnitude v2fr, where v is the speed of the particle. (L.U.)

9. A bead is constrained to move around a wire in the shape of a cardioid r = a(1 + cos 6) in such a way that its angular velocity about the pole is a constant c:u. Show that the bead's speed v, when it is at a distance r from the pole, is given by v = 2ar and that the magnitude of its acceleration is then c:u(a2c:u2 + 2v2)t.

c:uv'

10.

y = c cosh:': with constant speed v. c Show that its acceleration has magnitude cv2fy2• 1 1 . A particle moves in the plane of rectangular axes Ox, Oy. If (x, y) A particle moves along the curve

are its coordinates at time t, show that (%, y) are the coordinates of the corresponding point on the hodograph with respect to parallel axes. Deduce that if x and y are quadratic functions of t, the hodograph is a straight line.

I2. A particle P moves in a plane and has polar coordinates

(r, 6) . Its velocity always makes an angle with the radial coordinate r. Show that its track is a straight line or a circle. In the latter case, show that, if the particle's acceleration is always in the radial direction and = 0 when t = 0, then t is proportional to - sin cos

e

e

e

e

e.

A CO URSE I N APPLIED MATHEMAT ICS

18 13.

[CH.

A particle moves in the plane of axes Ox, Oy so that its acceleration always makes the same angle .p with the tangent to its path that the tangent makes with Ox. Show that the speed v is given by v = sin .p, where v = v. when .p = rt/2. Deduce further that, if the tangent to the path rotates at a constant angular velocity w, the

v.

intrinsic equation of the path iss

= � (1 w

the path is a cycloid.

14.

-

cos .p) .

Hence show that

By further resolving the Cartesian components of the velocity of a particle, show that the polar components are

v,

vo

=

=

x cos6+ :Y sin6 -x sin6 +

:Y cos6.

Substituting x = r cos6, y = r sin6, obtain equation ( 1 . 13 ) . Show similarly, that the polar components of acceleration are given by .f,.=xcos6+ysin6

fo = -xsin6+ y cos6

and hence obtain equation

15.

A straight rod AB of length c has angular velocity w 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 rt/2 and 3rt/2. Assuming that w is less than rt/2 and that the eastward component of A's velocity is never zero, show that this component has alternate maxima and minima, provided/< cw2• Show also that the time interval between any maximum and the succeeding minimum is

and is less than

16.

( 1 . 1 8) .

rt w.

/

� cos-1 w

2 cw

(L.U.)

A ship leaves port and steers a straight course at 12 knots for a destina­ tion that is unknown. Six hours later a ship that can do 20 knots is sent from port in pursuit. It sails at top speed due northwards for 9 hours, and, on failing to find the first ship on this course, it proceeds to steer on such a curve that it would find it whatever the course taken by the first ship. Show that it describes the equiangular spiral r =

1 7.

(_1_)

1 80e!O.

(B.U.)

A particle moves in a plane so that the velocity components along and perpendicular to the radius vector from a fixed origin are c tan �6 and c respectively, where c is a constant. Show that the particle describes a parabola, and show that its acceleration is constant in magnitude and direction. (D.U.)

1] 18.

19

KINEMATICS O F A PARTICLE

A particle starts from the origin with velocity u i n the direction of the initial line and moves with constant angular velocity c:u about the origin and with constant radial component of acceleration uw inwards. Find its coordinates at time t from the start, and show that the equation of its path is wr = u(1 e-B) . (Li U.) -

.

CHAPTER 2

NEWTON'S LAWS.

RECTILINEAR MOTION

2.1. Newton's Laws of Motion

First Law of Motion may be rendered as follows : A body continues in its state of rest or of uniform motion in a straight line, unless acted upon by forces.

Newton's

We shall identify the " body " 'Yith the " particle " defined in the previous chapter. The motion of a particle can be assessed only against the background of a frame of reference and it is clear from the terms in which the law is formulated that this frame is supposed to be " at rest ". Newton assumed that it was possible to distinguish from among the infinity of possible frames of reference one which was in a state of absolute rest. Einstein's criticism of this assumption led him to develop the Special Theory of Relativity, and it is now uni­ versally admitted that the existence of such an absolute frame cannot be demonstrated. A more mature interpretation of the law must therefore be presented. If the centre of the Sun is adopted as the origin of a frame of re­ ference, the axes of which do not rotate against the background of the extra-galactic nebulce, then we may accept it as an observable fact that those stars of our galaxy which are widely separated from their neigh­ bours follow straight-line tracks at constant speeds over very long periods of time. Since the density of interstellar matter is small, it is a reasonable assumption that such stars are subjected to negligible external influences and their motion may be regarded as a verification of the First Law. If, however, we refer their motions to a frame which does not rotate relative to that based on the Sun, but whose origin moves with a velocity V relative to this body, it is clear from equation (1.5) that a particle which is in uniform motion relative to the first frame is also in uniform motion relative to the second. The First Law accordingly remains true in the second frame. Reasoning along these lines, we are led to expect that, in a hypothetical region of the universe populated by particles which do not interact, it would be possible to set up any number of frames of reference in each of which all the particles would follow straight-line trajectories at constant speeds. We shall interpret the First Law as asserting the existence of such frames of reference and shall term such a frame an inertial frame. Throughout the remainder of this book all motion will be supposed referred to such a frame unless otherwise stated. 20

[cH. 2]

NE WTON ' S LA W S .

RECT ILINEAR MOT ION

21

If mechanical phenomena within the Solar System are to be dis­ cussed, the inertial frame based on the Sun and described above may be employed. A frame which is fixed in the Earth is more con­ venient when the motions of bodies near the Earth's surface are to be treated. Such a frame is clearly not inertial. However, the rate of rotation of this frame due to the diurnal motion of the Earth and the acceleration of its origin on account of this motion, and of the motion of the Earth in its orbit, are so small that they can usually be neglected . For most practical purposes (except for long-range ballistic calculations) little error results from treating such a frame as inertial. Referred to an inertial frame, a particle which is not disturbed by external agencies will follow a straight-line path at a constant speed. If, therefore, relative to an inertial frame a particle is observed to follow a curved path, or if its speed is observed to vary, we conclude that the particle is being disturbed by the action of some extraneous agency. The disturbance of the particle's motion is said to be due to the forces which have been brought to operate on the particle by this agency. Thus, the speed of a falling particle steadily increases, and this phenomenon is regarded as being accounted for by the force of gravity acting upon the particle, the agency responsible for this force being the Earth. Again, a bead, spinning around a smooth circular wire fixed in a horizontal plane, is regarded as deviating from a straight path under the action of a force exerted by the wire upon it. Since inertial frames are not accelerated relative to one another, it follows from equation (1.17) that a particle whose motion is being disturbed by a force will have the same acceleration with respect to all inertial frames. A particle's acceleration vector is accordingly a satisfactory measure of the force being exerted upon it and we shall render Newton's Second Law in the form :

The force acting upon a particle in non-uniform motion relative to an inertial frame is defined to be a vector in the direction of the acceleration and proportional to it in magnitude. The constant of proportionality will depend upon the particle and will be called its mass. The second law has accordingly been reduced to the status of a definition of force. The " mass " of a particle occurs in our statement of the second law, but this scalar quantity is not known precisely until a technique for its measurement has been prescribed. This is the function of Newton's

Third Law: When two particles P and Q interact, the force exerted by P upon Q is equal in magnitude to the force exerted by Q upon P, but is in the opposite sense. The directions of both forces are in the line PQ.

22

A COURSE IN APPLIED MA THEMA TICS

[cH.

By observing the directions of the accelerations of P and Q, the final statement can be checked experimentally. However, the first part of the law cannot be subjected to a complete experimental verification, for we have not yet described how the masses of P and Q are to be measured and hence cannot find the magnitudes of the forces exerted upon them. In fact, this part of the law is partly a statement of how such measurements of mass are to be made. This can be made clear thus : If v is the velocity of a particle of mass m relative to an inertial frame, p = mv is termed the momentum of the particle. Applying the third law to two colliding particles, it follows that the sum of their momenta immediately after the collision is equal to the same sum before the collision. This is the Law of Conservation of Momentum (see Section 5.3) . Suppose then that two particles of masses m1 , m2 having velocities Uv u2 respectively come into collision and separate with velocities Vv v2 • Then or

mlul + m2u2 = mlvl + m2v2, m2 ml (v2 - u2) = ul - vi .

This result implies that the vectors u1 - Vv v2 - u2 are parallel, a prediction which has been checked experimentally. It is clear that if parallelism exists relative to one inertial frame, it will exist relative to all. However, such an experiment will clearly also provide us with a value for the ratio m2jm1 • Hence, if a certain particle is now adopted as the standard unit of mass (e.g. the pound (lb) or gramme (gm)) , the masses of all other particles can b e measured in terms o f this unit by permitting them to collide with the standard and measuring the appropriate velocities. Let F be the force acting upon a particle of mass m and let f be its acceleration. Then F = mf. (2. 1) If m is measured in lbs, f in ftjsec2, the units of F are called poundals (lbals). If m is in gm, f in cmjsec2, F is in dynes. If m is in kilo­ grammes (kg), f in mjsec2 , F is in newtons. Engineers employ the pound weight (lb wt) as a unit of force. This is the force of gravitational attraction acting upon a pound mass situated at the surface of the earth. With this unit of force, if f is in ftjsec2 , m is said to be measured in slugs. All particles in the vicinity of the earth's surface are observed to fall in vacuo with an acceleration of 32·2 ftjsec2 • It follows that the gravitational force exerted upon a pound mass is F lbals, where

F = l X 32·2 = 32·2.

2]

N E WTON'S LA WS .

But this force is, by definition, one pound weight. 32 ·2 lbals.

1 lb wt

Again, if m slugs is the mass of this particle, since it causes an acceleration of 32·2 ftjsec2, i.e.,

Hence (2.2)

1 lb wt acting upon

1 = m x 32·2 m = 1/32·2.

Hence, 1 lb or

23

RECT ILIN EAR MOT ION

1 slug

=

=

1 slugs 32 .2 32 ·2 lb.

(2.3)

Throughout the subsequent work, we shall always suppose that forces, masses and accelerations are expressed in terms of any set of consistent units. It will rarely be necessary to specify the par­ ticular set we are employing, since equation (2.1) is true for any such consistent set. Experiment reveals that, provided air resistance is eliminated or can be neglected, all particles near the earth's surface fall with the same acceleration 32·2 ftjsec2 or 981 cmjsec2 • If g denotes this acceleration in any convenient units, m denotes the mass of the falling particle in corresponding consistent units of mass, the force F of the Earth's attraction on the particle is given by

F = mg, .

(2. 4)

F being measured in units of force appropriate to those employed for mass and acceleration. Equation (2.4) reveals that the Earth's attraction on a particle at its surface is proportional to that particle's mass. This result is included in a further law due to Newton, which we shall state in Chapter 4 (p. 102) . If a number of forces act upon a particle the acceleration produced is found to be the vector sum of the accelerations each would cause if acting separately. This fact is an additional law of motion, whose validity can be checked only by experiment. It is termed the Principle of the Independence of Forces. If, therefore, forces Fv F2, Fn cause accelerations f1, f2, fn respectively in a particle of mass m, when acting individually, when acting in concert they will cause an + fn. .But acceleration f1 + f2 + Letting

+ fn) = mf1 + mf2 + = F1 + F2 +

(2.5) (2.6)

24

A COURSE

IN A PPL IED MATHEMATICS

[cH.

we see from equation (2.5) that a force F will also cause an acceleration in the particle of f1 + f2 + + fn. The forces Fv F2, F, acting together therefore have the same dynamical effect as the single force F. F is called the resultant of the n forces, and equation (2.6) shows that it is calculated by vector addition of these forces. It follows from equation (2.6) that \ \ the resolute of the resultant force F F ' \ in any direction is the sum of the \ F" taken in resolutes of Fv F2, ''----F;------./ the same direction. Fn are If the forces Fv F 2, FrG. 2.1.-Polygon of Forces represented by displacement vectors in the manner indicated in Fig. 2.1 the displacement vector representing their resultant F forms the closing side of a polygon. This diagram is accordingly called the Polygon of Forces. I

2.2. Work and Power

Suppose that a particle moves along a straight line from P to P' under the action of certain forces of which F is one (Fig. 2.2) . Suppose that F remains constant in magnitude and direction during the motion , F

L,, '�,,

p

p

(b)

(a)

FIG. 2.2.-Work Performed by Force

e being the angle made by the direction of the force with that of the displacement PP'. e is taken to be acute in case (a) and obtuse in case (b). The work done by this force during the displacement is defined to be the quantity F . P P' cos e, and is therefore negative if e is obtuse and zero if e is a right angle. F cos e is the resolute of F in the direction of the displacement PP', and hence the work done is equal to the resolute of the force in the direction of the displacement multiplied by the displacement. Alternatively, P P' cos e is the �

�

resolute of the displacement in the direction of the force, and the work �

done is this resolute of P P' multiplied by the force. If F is measured in dynes and P P' in em the units of work are called ergs. If F is in

2]

NE W TON 'S LA WS .

25

RE CTILINEAR MOTION

newtons and PP' in metres, work is in joules (= 10 7 ergs) . If F is measured in lbals and P P' in ft, the units are ft lbals. If F is in lb wt and PP' in ft, the units are ft lb (1 ft lb = 32·2 ft lbals). Employing the notation of vector algebra, if d represents the dis-+ placement PP', the work done by F can be written as the scalar product F d If F is the resultant of a number of forces F1 , F2, . . Fn acting on the particle, the work done by F is F d = (F1 + F2 + Fn) · d = F1 d + F2 d + + Fn d, having made use of equation (2.6). We have shown, therefore, that the sum of the works done by a number of forces is equal to the work done by their resultant. Now consider the general case when a particle moves along any curve from A to B under the action of forces of which F is one (Fig. 2.3) . ·

.

·

·

·

·

F B s

A

FIG. 2.3.-Work Done by a Variable Force

F will be supposed to vary both in magnitude and direction. Let P be any point on the curve and let the arc AP be denoted by s. Let the arc P P' be as. Let e be the angle between the direction of F when -+ -+ acting at P and the tangent PT to the curve at P. PT is taken in the same sense as the motion. As the particle moves from P to P', provided as is sufficiently small, we may assume that F does not vary appreciably in magnitude or direction. The work done by the force over the arc PP' is accordingly F cos eas, very nearly. The work done by F, as the particle moves from A to B, is

W = lim _2 F cos eas,

the summation being extended over all elements as into which the curve AB can be subdivided and the limit being approached by allow­ ing the number of these elements to tend to infinity, each tending to zero in its length. The limit of such a sum is written. w

=

1BF cos eds

and is calculated as a line integral.

(2.7)

26

A COURS E IN A P PL I ED MATHEMAT I C S

[CH.

If F is the resultant of a number of forces, since it remains constant in magnitude and direction over PP', provided this arc is small, the work it does over this element is the sum of the works done by the individual forces. Summing over all the elements as, we conclude that this statement is also valid for the whole arc AB. If P has position vector r and P' has position vector r + ar, the dis­ placement PP' along the curve A B is denoted by ar and the line integral (2. 7) may be written in vector notation thus :

W = LsF · dr.

Suppose that the curve AB and the force F lie in the plane of rect­ angular axes Ox, Oy and let (X, Y) be the components of F parallel to the axes. If P is the point (x, y) and P' is the point (x + ax, y + ay), ar has components (ax, ay). Hence F · ar = Xax + Yay and

W = LB (Xdx + Ydy) .

(2.8)

A particle can move in the plane of rectangular axes Ox, Oy. When its coordinates are (x, y), the components of a certain force applied to it are (ky2, kx2) . Calculate the work done by this force if the particle is constrained to move along the parabola y x 2 from the origin to the point (1, 1 ) . I n this case ky2, Y = kx2 • Since y = x2 at each point o f the path, k%4 and Y = ky. The work done is now given by equation {2.8) to be

Example 1.

X=

=

X=

W

=

1(1, 1) (0, 0)

(kx4dx + kydy) =

= tk + !k = 0·7k.

11 0

kx4dx +

i\

ydy

0

The power which is being developed by a force acting upon a particle at any instant is defined to be the rate at which the force is doing work at this instant. Consider again the particle which moves along the curve A B (Fig. 2.3). Let at be the time taken to move from P to P' The work done by F in this time at is F cos eas. Hence, the average rate of doing work over this time interval is

as F cos 6 -· at Letting

at -- 0, this expression approaches the value H = Fv cos e = F v, .

(2.9)

where v is the velocity of the particle at the instant under considera­ tion. H is the rate of doing work by F when the particle is at P and hence is the power developed by F at this moment. H is seen to be the resolute of F in the direction of motion multiplied

2]

N E W TON 'S LA W S .

RE CTILINEAR MOTION

27

by the speed or, alternatively, the resolute of the velocity in the direc­ tion of F multiplied by F. In the c.g.s. system of consistent units, equation (2.9) will give H in units of ergs/sec. In the f.p.s. consistent system, H will be given in ft lbalsjsec. In the m.k.s. system, there is also the practical unit of power termed the watt defined thus

1 watt 1 joulejsec

_

107 ergsjsec.

(2 . 10)

In the f.p.s. system, there are two practical units of power. They are defined as follows : (i) (ii)

1 ft lbjsec - 32·2 ft lbalsjsec, 1 horse power (H.P.) 550 ft lbjsec.

(2.11)

_

2.3. Rectilinear Motion of a Particle

Suppose a particle of mass m is in motion along a straight line. Its acceleration vector will be directed along the line and hence, by equation (2.1), so also will the vector representing the force acting upon it. Let 0 be a fixed point on the line and let P be the position of the particle at time t (Fig. 2.4). Let 0 P = x, x being reckoned positive 0

m

------- (}'

p

F

FIG. 2.4.-Motion Along the x-Axis

when P is on one side of 0 and negative when P lies on the other side . If v is the speed of P, v = dxjdt = i and v is positive if x is increasing. Let F be the magnitude of the force acting on P, F being given a positive sign if it acts in the direction in which x increases. We shall show how the motion of P may be determined in the three cases when F is given as a function of (i) t, (ii) x and (iii) v.

(1) F is a Function of t

The acceleration of P is dvjdt.

Hence, by equation

(2.1),

(2.12) F = m dv dt' m being a constant, this implies that dvjdt is a known function of t. Thus, dv = F = fl (t). (2.13) at :m Integrating with respect to t, we obtain (2.14) v = ffl (t)dt = f2 (t), .

28

[CH.

A COURSE IN APPL IE D MATHEMAT ICS

assuming that the integration can be carried out. v will usually be known at the commencement of the motion, i.e., at t = 0. The arbitrary constant of integration must be selected to ensure that equation (2.14) yields the value of v appropriate to this instant. Since v = dxfdt, by integrating equation (2.14) with respect to t, we find that

(2.15)

We shall normally find it convenient to choose 0 to be the position of the particle at time t = 0. If such is the case, x 0 when t 0 and the integration constant must be selected so that equation (2.15} is satisfied by these values. As a particular case, F may be constant. The acceleration is then also constant and equal to f, say. Thus, =

=

dv = f. dt Integrating with respect to t, we obtain

(2.16)

v = u + jt, where v

= u at t = 0.

A second integration yields

(2.17) x = ut + tft2 , having assumed that x = 0 at t 0. Eliminating t between equations (2.16), (2.17), we obtain the third well-known equation for constantly =

accelerated motion, viz.,

v2 = u2 + 2jx.

(2.18)

An electric train of mass M tons starts moving along a straight, level track from rest. The tractive force of the motors decreases uniformly with the time, over a period of T min, from P lb wt to R lb wt. The total resistance to motion is R lb wt. Show that the distance travelled in miles during the T min is

Example 2.

(P - R)P. 308M

(Take g = 32.)

The tractive force F lb wt diminishes at a rate (P - R) fT lb wtfmin. t sec after the commencement of motion, the tractive force is therefore given by F=

p

_

(P - R)t. 60T

The resultant force acting on the train at this instant is accordingly p

_

(

) � ) lbals. 6 T

t (P - R)t R = (P - R) l - __ lb wt 60 T 60T = 32(P

(

- R) I

-

2]

N E WTON 'S LA WS.

RECTILINEAR MOTION

The mass of the train is 2240M lb. Hence, if at time t sec,

29

v is the velocity in ftjsec

dv ( - t) 60T 1 dv = P - R ( 1 t ).

2240M dt = 32(P - R)

or Integrating, we obtain

dt

70M

-

60T

t2 )

(

P-R t 70M 120T ' 0 when t the constant of integration being zero, since A second integration yields

v

=

v=

=

0.

the integration constant again being zero if we agree to measure x from the starting position. When t = 60T sec, 120(P - R) T2 X = ' 7M and the appropriate units of x are feet. Converting to miles, we obtain the result stated in the question.

(2) F is a Function of x

In this case, the force acting upon the particle depends on the latter's position on the line of motion. We transform the acceleration thus,

dv - dv dx v dv - !_ ( 1v 2) dt dx dt - dx dx 2 • The acceleration will be a given function of x, i.e., d dx (!v2) gt(x). Integrating with respect to x, we obtain !v2 jg1 (x)dx = g2 (x) . Solving for v, we find that . 1v = dx dt = v 2g2 , _

_

_

=

=

and hence,

l dt -2 = g3 (x). dx = ·v. ;2g

(2.19) (2.20) (2.21) (2.22)

Integration of this last equation with respect to x, yields finally

t = j g3 (x)dx = g4(x).

(2.23)

30

A CO URSE

IN APPL IED MATHEMAT ICS

[cH.

It will generally be convenient to solve for x in terms of t in this equation, if this is possible. In any case, equation (2.23) specifies the motion. Thus, suppose that the particle is attracted towards a fixed point on the line of motion, with a force whose magnitude is proportional to the particle 's distance from the point. Taking the fixed point as the origin of x, the force, and hence the acceleration also, will be proportional to x. Thus,

(2.24) where cu2 is the constant of proportionality. Integrating equation (2.24), we obtain

fv2 = - !cu2x2 + constant

(2.25) It is clear from equation (2.25) that x can never exceed A!/ cu in magnitude, for otherwise v2 would be negative and v imaginary. Putting a = Aljw, this implies that the particle's motion is confined to that part of the x-axis lying between the points x = ±a. In terms of a, equation (2.25) reads as follows (2.26) Solving for v, v = cuva2 - x2 dt 1 or (2.27) dx - cuva2 - x2· Integrating with respect to x, we obtain t = cu1 sm. - aX + constant, or x = a sin (cut + �), (2.28) where � is an integration constant. The main characteristics of the motion are easily discerned by con­ sideration of equation (2.28). As t increases over an interval of length 2rt/cu, wt + � increases by 2rt and the sine function increases and de­ creases once through all values between ± L x therefore oscillates in value between ±a, one complete oscillation being performed in time 2rt/cu. The graph of x against t is shown in Fig. 2.5. The motion is called Simple Harmonic Motion (SHM) and the time of one complete oscillation is the period. Denoting the period by T, we have T = 2rtcu . (2.29) or

-

1

-

2]

NE WTON 'S LAWS .

RE CTIL INEAR MOT ION

31

The number of complete oscillations taking place in unit time is called the frequency f. Thus (>)

1 ! = T = 2"·

(2.30)

The extreme distance from the point 0 achieved by the particle, viz., a, is termed the amplitude of the motion. (wt + cf>) is called the phase

FIG.

2.5.-Graph of Simple Harmonic Motion

of the motion at the instant t and, hence, cf> is known as the initial phase of the motion. cf> can always be made zero by choosing t = 0 to be an instant when x = 0 and is increasing. If we choose t = 0 to be an instant when x = 0 and is decreasing, then cf> = " and x = -a sin wt. Very often, t = 0 is selected to be an instant when x = a ; then cf> = !" and x = a cos wt. If cf> takes none of these values, we may expand equation (2.28) thus, x

= a sin wt cos cf> + a cos wt sin cf> (2.31) = A sin wt + B cos wt, where A = a cos cf>, B = a sin cf> and are therefore also of the nature of constants of integration. Equation (2.31) represents an alternative general solution of the equation of motion (2.24). Equation (2.28) indicates that SHM may be regarded as the pro­ jection on a straight line of a uniform circular motion. Thus, consider a point Q which moves around a circle of radius a and centre 0 with

32

A COURSE

IN A P PLIED MATHEMAT I C S

[CH.

uniform angular velocity w in a clockwise direction (Fig. 2.6). Let A A ', BB' be perpendicular diameters of the circle and let QP be

FrG.

B'

2.6.-Geometrical Generation o f SHM

perpendicular to BB'. If 5 is the position of the point Q at time t = 0, LQOS = wt. Let LSOA = if>. Then x

= OP = OQ sin LPQO = a sin (wt + ¢>). (2.32) It follows that the point P executes SHM on the line BB'. The velocity of Q is of magnitude aw and is in the direction of the tangent at Q. This tangent makes an angle (wt + if>) with BB'. The resolute of Q's velocity in the direction of BB' is accordingly aw cos (wt + ¢>). Only this resolute of Q's velocity is communicated to P and, hence, P's velocity is

(2.33) a result which is also obtainable by differentiation of equation (2.28) with respect to t. It appears from this last equation that the maximum value of v is (2.34) = aw. v

v

= aw cos (wt + if>), Vmax.

takes this value when the phase is zero, i.e., when P passes through

0 in the positive direction.

The circular generation of SHM will be found very useful when problems concerning the passage of time between two phases of a SHM are to be solved (e.g., see Example 3 below) . Practically, the simplest way of causing a body to execute SHM is by connecting it to a fixed point by an elastic support. Thus, suppose that a particle P of mass m hangs freely from a fixed point A by an

2]

NE WTON ' S LA W S .

RECT IL INEAR MOTION

33

elastic string of natural length l (Fig. 2.7). We shall assume that the string obeys Hooke's Law, viz., the extension of an elastic string is ,

- - - - - - -

r

- - - - - -

A

l p- - - - - - - -- - - -- - -r 0 m9

.X

mg

t

T p

mg

FIG. 2.7.-Vertical Oscillations on an Elastic String

Then, if e is the extension beyond the natural length and T is the tension produced,

proportional to its tension.

T

= [1-e,

(2.35)

where lL is Hooke 's constant. lL is a constant for a particular piece of string, but is not independent of the length of the string. We often write

[L = ).,r

(2.36)

where :t.. is the modulus of elasticity of the string. :t.. depends only upon the material and cross-sectional area of the string. When the particle hangs in equilibrium (Fig. 2.7 (a)), the tension in the string must balance the weight force mg acting on the particle, i.e., T = mg. If c is the extension, therefore, by equation (2.35)

mg = [LC.

(2.37)

If the particle is now displaced vertically from its equilibrium position and released, it will oscillate in a vertical line. Let 0 re­ present the equilibrium position of the particle on this line and x the distance of P vertically below 0 at time t (Fig. 2.7 (b) ) . At this instant the total extension of the string is x + c, and the tension T is accordingly given by c

(r)

T

= tL (x + c) = [LX + [LC = [LX + mg,

(2.38)

34

A CO URSE

IN APPL IED MATHEMAT ICS

[CH.

use having been made of equation (2.37). The resultant downward force acting on the particle is therefore

mg

-

(2.39)

T = - fl.X

and the equation of motion is

mx = -fl.X

(2.40)

i.e.,

where w2 = fLfm. Equation (2.40) implies that the particle's motion is simple harmonic of period 2rr:/w = 2rr:vmffL . Clearly, equation (2.40) is valid only so long as the string remains taut. Should it go slack during the motion, the particle will then be in a state of free fall under gravity, i.e., will have a constant accelera­ tion g downwards. This phase of the motion must be dealt with separately from the SHM phase. A particle of mass m is attached to a fixed point A by an elastic string ofnatural length l and modulus t mg. The particle is held at A and then released. Prove that it returns to the point A after a time + �) �i (1 and that its greatest depth below A is (3 + 2y2)l. The particle first falls from A with a constant acceleration g to a point B, where AB = l. The string then becomes taut and the particle's motion is

Example 3.

simple harmonic until it returns to B, when the string goes slack again. On the two occasions that the particle is at B, its distances from the centre of oscillation 0 are the same. It follows from equation that the particle's speeds at these instants are also identical, though the senses of its velocities are opposite. The particle's upward motion is therefore identical with its downward motion, apart from direction, and it arrives back at A with zero velocity. If is the time of free fall from A to B and V is the speed at B, equations and show that

(2.26)

t1 (2.16)

(2.17)

= gtl> l = tgtl·· tl = �· v = v2lg. Hooke's constant for the string is 1-' = mgf2l and hence, during the SHM, w2 = JLfm = gf2l. The centre of oscillation 0 is at a depth c below B, where c is given by equation (2.37). Thus c = mg = 21. /-' Measuring x downwards from 0, x = -2l and v = - V = - v 2lg at B. Substitution in equation (2.26) yields, 2lg = .!I_21 (a• 412) :. a = 2vf2l. Hence

v

-

2]

N E W TON' S LA W S .

RECTIL INEAR MOTION

The maximum depth below achieved is therefore sponding depth below is + O + = + + 2 -yl2 = + The SHM may be regarded as the projection of a uniform circular motion of a point Q having angular velocity w = V around a circle of radius This is shown in the diagram, C being the point of maximum depth. When the particle is at SHM com­ mences and continues until the point Q generating the motion arrives at Q', when the particle has returned to Now OB = OQ = and hence BOQ = -f7T. The reflex angle QOQ' through which OQ moves is ac­ cordingly The time taken by OQ to sweep out this angle is

AB B

2-yl2l.

0 a l 2l

A

2-yl2l. l (3 2-yl2)l.

35

The corre­

gf2l

B,

2l,

2-yl2l

B.

377/2.

. "'F2li -� �2 "'/�g '

�': __,_ 2 and this is the time of the SHM. The total time required is

t

2

=

c

2tl + t. = 2,J!j + i�� = �¥ ( 1 + �)Example 4. 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 the point P are mp.PA-2, mp.PB-2 respectively . Find the velocity with which it must be projected from the point C, where A C = 9GB, 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 " ' (Iog ,a - ' ) ( ap_ )t (L.U.) Let x be the distance of P from 0 at any instant. The resultant force acting upon the particle in the direction of OP is then mp. mp. 4mp.ax (a - x)2 - (a + x)2 = (a2 - x2)0 The equation of motion of the particle is therefore m iiXd (tv ) = (a•4mp.ax - x•)2 An integration with respect to x yields 1•v2 - 4p.aJ xdx (a• - x•)• J 2p.a (a• - u)-•du, where u = x• = 2p.a(a2 - ut1 + constant, and, hence, v• = a 4p.a - x + constant. 2

•

2

_

=

2--2

•

36

A C O U RS E I N A P P L I E D MAT H EM A T I C S

[CH.

We require that when x = 0, then v = 0. The integration constant must therefore be chosen to be - 41-'fa, and then, �aa• - x•

v• =

_�= a

When P is at C, AP = 9PB, i.e.,

4/Lx• a(a• - x2) •

(i)

x + a = 9(a - x)

x = ta.

For this value of x, equation (i) shows that v2 =

641-' 9a

the negative sign being taken, since the velocity of projection from C is clearly towards 0. Putting v = dxfdt, equation (i) may be written � dx

=

-

(1-'�)

t (a• - x•Jl

2x

.

When P is at D, where 50D = 30B, x = 3aj5. (ii) over the motion from C to D, we obtain

(ii) Integrating equation

tD - tc = -! (JL-a)tl� a (a• -x x•)i dx a a) ll�a w• = !( dw, where w = + (a2 - x•)i -2a -w2 Ra ( a )il�a ( a• - 1 ) dw = ! w• p. a• la ! a + w - wlta = ! ( � ) �� log l

J.£

_

p.

2

= !(log t

(3) F is a Function of v

a -w

- tl(�t

�a

The mass being constant, the acceleration of the particle is a function of v, so that we may write

dv dt = hl (v)

or

(2.41)

dt = 1 h1 (v) · Integrating with respect to v, dV

{2.42)

2]

N E WT O N ' S LA W S .

RECTI L I N E A R M O T I O N

37

Assuming that we are now able to solve for v in terms of t, equation (2.42) may be put in the form

v = h3 (t), dx = ha(t). or (2.43) dt A final integration with respect to t now yields x as a function of t. Alternatively, equation (2.41) is equivalent to

v dv dx = h1 (v), dx = v or dv h1 (v)" Integrating with respect to v, we find that x = Jhv dv = r (v) h2 (v).

(2.44)

(2.45)

Solving for v in terms of x, we have

v = dx dt = h3 (x) dt l dx = h3 (x) ' An integration with respect to x now yields t as a function of x and the problem is solved.

(2.46) T

�

The methods of this section are particularly applicable to problems concerning the motion of a tr body in a vertical line under the action of gravity and air resistance. For small speeds v (up to m'J 20-30 ftjsec) it is found that the force of air re­ sistance is proportional to v. For larger speeds mkv (up to 800-900 ftjsec) this force is found to vary as v2 • As the velocity of sound is approached, the resistance increases very rapidly indeed, and its :X. variation cannot be represented satisfactorily by any power of v. At supersonic speeds the resist­ ance again increases approximately like a linear function of the speed. We shall discuss the cases when the resistance is proportional to v and to v2• Suppose that a particle of mass m is projected vertically upwards with a speed V from a point 0. 2.8.-Resisted When the speed is v, let mkv be the air resistance FrG. Motion in a Verti­ and let x be the body's height above 0. The cal Line

tp

fa

- - - - -

_ _ _

_

- - -

_ _

__[

38

A C O U R S E I N A P P L I E D M A T H E MA T I C S

[CH.

forces acting on the particle P are shown in Fig. 2.8. The equation of motion is

dv = -mg - mkv. mv dx

(2.47)

This equation is also valid for motion in the downwards direction, for then v will be negative and its sign serves to reverse the sense of the resistance mkv so that it acts upwards and continues to oppose the motion. From equation (2.47), l gjk v dx = dv g + kv = -k + g + kv' -

and, integrating, we obtain

x = ]i + � log (g + kv) + constant. (2.48) When x = 0, v = V and hence, g v constant = k fi2 log (g + k V). Equation (2.48) may now be written (gg ++ kVkv )· K log (2.49) x = !k (V - v) - k2 When v = 0, the particle has achieved its maximum height and is at the vertex of the motion T. If h is the height of the vertex above 0, equation (2.49) shows that V g log (l + kV ) h= k g . (2.50) Ji2 When the particle returns to 0, let U be its speed. Then, when x = 0, v = - U and, since equation (2.49) is valid for the downward trajectory, 0 = kl (V + U) - Ji2g log gg + kV kU g + kV. U + V = �k log g - kU (2.51) or This latter equation determines U. Alternatively, we may write equation (2.47) thus m dv dt = mg - mkv l dt dv = -g + kv' -

-

-

_

-

N E WT O N ' S LAWS .

2]

R E CT I L I N E A R M O T I O N

39

Integrating,

t = -k1 log (g + kv) + constant. When t = 0, v = V and hence, 1 constant = k log (g + k V) t = k! log gg ++ kV kv . At the vertex, v

= 0.

i log (1 + � v}

(2.52)

{2.53)

It follows that the time from 0 to T is

(2.54)

Solving for v in terms of t from equation

As

t -+ CXJ ,

e-kt -+

0,

(2.53), we obtain v = �� = (f + V) e-kt - f· {2.55) and hence

v -+ - f·

This implies that, after

some time has elapsed, the particle will be falling downwards with a speed gjk. This quantity is called the terminal velocity of the body. When v = -gjk, the resultant force ( - mg - mkv) acting upon the particle is zero, i.e., air resistance balances gravity and the particle's velocity is accordingly uniform. Integrating equation (2.55) with respect to t, and remembering that x = 0 when t = 0, we find that x

= Hf + V) (1 - e-kt) - f t.

(2.56)

mv ddvx = -mg - mkv2 •

(2.57)

This equation completely determines the motion. If now we take the resistance to be mkv2, the equation of motion corresponding to equation (2.47) is

This equation is obtained under the assumption that the motion is in the upwards direction. If the motion is in the opposite direction, v has a negative sign, but the sign of mkv2 remains positive. This implies that the sense of the resistance force is not automatically reversed when the sense of the motion changes. The equation of downwards motion must therefore be written d mv dxv = -mg + mkv2 •

(2.58)

A C O U R S E I N A P PLI E D MATH EMAT I C S

40

Arranging equation

(2.57) in the form dx - v dv g + kv2

[cH.

(2.59)

v, we obtain (2. 60) x = 2lk log (g + kv2) + constant. Since v = V, when x = 0, the constant is easily determined and then we find that x = 2lk log gg ++ kkV2 (2.61) v2 When v = 0, x = h and the particle is at the vertex. Thus the height of the vertex is given by h = ;k log (l + � V2} (2.62) and integrating with respect to -

Integrating equation downward motion

(2.58) in the same manner, we obtain for the

x = 2lk log (g

kv2) + constant. (2.63) When x = h, we know that v = 0. The integration constant is accordingly h ;k log g = ;k log H (l + � V2)} · -

-

(2.63) now yields (2.64) x = ;k log {(l � v2)(l + � V2)} When x = 0, the particle has returned to the point of projection . If U is the speed of return, equation (2.64) shows that 0 = 2� log {( l � U2)(l + � V2)} , Equation

-

-

or or

(1 � u2)(1 + i V2) = 1, U2 = V2� 2. l +gV _

To calculate the time of the motion, we rewrite equation the form

m dv dt = -mg - mkv2.

(2.65) (2.57) in (2.66)

N E WT O N ' S LA W S .

2]

R E CT I LI N E A R M O T I O N

I

Hence dt

dv

41

(2.67)

and an integration shows that t

=

)kg [tan-1 J� V - tan-1J� J,

(2.68)

- Ikg tan-1 J-gk V

(2.69)

v

the integration constant being chosen so that t = 0 when v = V Putting v = 0 in equation (2.68), we find that the time t0 to the vertex is given by t0 - .

1_

v

For the downward motion, we manipulate equation (2.58} in a similar fashion, obtaining instead of equation (2.68}, the equation

-

t - t0

+

I Ji 2 k log Ji vg

- v

�+

k

'

(2.70)

v

the integration constant being chosen so that t = t0 when v = 0 (i.e., at the vertex) . When v = - U, the particle has returned to 0. Let t = t1 at this instant. Then, from equation (2.70), t1 =

t0

U + I Ji . log Ji + 2 v kg 1_

�

_

k U

+ V2 + V J I k i V + log �===--Jg 2vkg f + V2 v Vkg J = � [tan-1 Jt V + log { I + V2 + J � J� v}} vg =

I

�

tan-1

�

_

(2.71}

equations (2.65) and (2.69) having been employed in the manipulation. Equation (2. 71) gives the time which elapses before the particle returns to the point of projection. The reader should show that in this case, the terminal velocity is

vgfk.

42

A C O U R S E I N A P P L I E D MATHEMAT I C S

(CH.

The engine of a motor car produces a power P when the car is running on the level at velocity v, where p = A (nvfu)� , (nvfu)s where A and u are constants, and n is the gear ratio which is being used. A s­ suming that the driver changes gear as soon as a higher power is obtainable from the next gear, and neglecting resistances, find the time to reach top gear from rest, if n = and respectively for bottom, second and top gear h.p., u = m.p.h., and the mass of the car is I ton. (B.D.) and if A = If n, n' represent two consecutive gear ratios, the velocity v at which a

Example 5.

I+

30

2·5, I·6

30

I,

change from one to the other should be carried out satisfies the equation A

Whence we find that

I + (nvfu) 3 = A I + (n'vfu) 3 (nvfu)�

(n'vfu)�

•

u ' v' (nn')

v=

When n = n' = I·6, since u = m.p.h., then v = m.p.h. When and n' = I, the appropriate velocity for the gear change is found to be m.p.h. Assuming that A, v, u are measured in absolute units of the f.p.s. system, the tractive force due to the engine is Pfv lbals. The equation of motion of the car during any phase of its motion when no change of gear is made is accordingly

I·6 I5y'I0/2

n=

where

2·5,

30

M dvdt

M is the car's mass in lb.

=

A

I5

(nfu) �vi

I + (nvfu) 3'

Separating the variables we obtain

� dt = [(E)

-4 v-i

+ (S) � v�] av.

Integrating over the phase being considered, we show that �

2M

( ) " J t J t't, = ( nvu ) t + ('!!!!) � I .l,.

?! u

7

u

� v,, v,

where JtJ :' = t1 - t0, etc., and t0, t1 are the times at the beginning and end of the phase respectively. In absolute units, A = X X ft lbal sec-1, u = ft sec-1, = lb. Thus 0

30 550 32 44 1232 t + ( '!!!!) ��v'. t = t1 75n2 J (�) u u Taking n = 2·5, v0 = 0, v 1 = 15 m.p.h., we find for the time spent in bottom gear 3·76 sec. With n = 1·6, v0 = 15 m.p.h., v1 15'\1'10/2 m.p.h., we calculate that the time spent in second gear is 3·15 sec. The total time required to reach top gear is accordingly 6·91 sec.

M 2240

_

],_

0

7

v,

=

2.4. Damped and Forced Oscillations

We have proved in Section 2.3(2) that a particle of mass m, which moves along the x-axis under the action of a force m(i}2x directed towards the origin, executes SHM of period 27tf(J). We will now investigate the effect of introducing, in addition, a resistive force

N E WT O N ' S LAWS .

2]

R E C T I L I N E AR M O T I O N

43

proportional to the particle's speed. A frictional force which varies as the speed of the body to which it is applied is said to be viscous. x being the speed of the particle at any instant, we shall take the �--- x 0

------�

mk:i:

mw'x

m

P

FrG. 2.9.-Forces for Damped Oscillatory Motion

frictional force to be mkx. The forces acting on The equation of motion is

2.9.

m are shown in Fig.

mi - mkx - mCil 2x, {D2 + kD + Cil2)x 0, =

or (2.72) where D dfdt. Equation (2.72) is a linear differential equation with constant co­ efficients, and we shall solve it by the standard procedure which is available for such equations. The characteristic equation is =

=

)_2 + k'A + Cil2 = 0

(2.73) and there are three cases to be considered, viz., (i) k > 2Cil, (ii) k 2Cil and (iii) k < 2Cil. If k > 2Cil, i.e., the frictional force is comparatively large, the roots of the quadratic equation (2.73) are both real and distinct. These characteristic roots are also both negative, for a positive value of 'A makes the left-hand side of equation (2.73) greater than zero. Let the roots be -'A1 and - 'A2 ('A1 > 'A2). The general solution of equation (2.72) may then be written x = Ae-A,t + Be-A•1• (2.74) Suppose that at t = 0, the particle is stationary at x X0• Then A and B must be chosen so that {2.75) =

=

Thus

(2.76) and, hence,

- 'Ale-A•t. X = Xo 'A2e-A,t "-2 "-r It is clear from equation (2.77) that x can only be zero if 'A2e-A,t = "-re-A,t

or

(2.77) (2.78)

44

A C O U R S E I N A P PLI E D M A T H E MAT I C S

[CH.

But the right-hand side of this equation is less than unity, whereas the left-hand side is greater than unity if t is positive, for then the exponent of e is also positive. It follows that the motion subsequent to t = 0 does not cause the particle to pass through the origin. How­ ever, as t -+ oo, e->.,l -+ 0, e->.,1 -+ 0 and, hence, x -+ 0. As time elapses, the particle therefore approaches the origin, but never passes through it. If x is plotted against t, the graph of Fig. 2.10 results. It is clear

FIG. 2.10.-Aperiodic Motion

that the motion, which is oscillatory in the absence of friction, has had its character entirely altered by the presence of the new force. The motion which now occurs is termed aperiodic motion. If k = 2w, the characteristic roots are both - w and the general solution of the equation of motion is accordingly x=

(A + Bt)e-<»1.

(2.79)

The reader should verify that the solution appropriate to the initial conditions x = x0, i = 0 at t = 0 is

X X0(1 + wt)e-<»1 (2.80) and that the graph of x against t takes the form shown in Fig. 2.10. The motion is again aperiodic. If k < 2w, the frictional force is relatively small and, as might be =

expected, the motion retains its oscillatory character. The character­ istic roots are complex and, as the reader may easily verify, have negative real parts. Denoting them by -ct ± i�, the general solu­ tion of equation (2.72) is

e-o:1(A cos �t + B sin �t). The initial conditions x X0, i = 0 at t = 0, require that X0 A 0 = -etA + �B. x=

(2.81)

=

=

,

(2.82)

2]

Hence

N E WT O N ' S LAWS .

RECTILINEAR MOTION

A = X0, B = rxX0/f'> and x = X0e- 1 (cos fjt + sin fjt) ( rx2)! = X0 1 + � e"""'1 cos (f'>t - g,) , "'

rx

45

(2.83)

where tan g, = rxff'>. Since Ieos (f'>t - g,) l <; l for all values of t, the graph of

x against t must lie between the two exponential curves x = ±xo( l + �y e-"'1 • Both the graphs and the bounding curves are shown in Fig. 2 . 1 1 .

It is

X:

FIG.

2. 1 1.-Damped Vibrations

now clear that the motion is similar to SHM, but that the amplitude of the oscillations steadily decreases towards zero. x is zero with cos (f'>t - g,). It follows that rt/f'> is the time interval between zeros and that a complete oscillation takes place in time T, where

T = 2f'>rt = 2rt (1 _ 4�)-!• (1)2 (J)

(2.84)

i.e., the friction has increased the period of the oscillations by a factor

(1 - 4k�2)-t

An oscillatory system which is modified by the introduction of a frictional or other force having the effects described above is said to be damped. In case (i) we say that oscillations have been completely damped out and the particle (system) is overdamped. In case (iii) the

46

A C O U R S E I N A P P L I E D MAT H E M A T I C S

[CH.

particle (system) is underdamped and oscillations occur if it is displaced and released. In case (ii) the damping is just sufficient to prevent oscillations occurring and the damping is critical. We shall now suppose that the particle is subjected to an additional force which is a given function of the time. This force will be pre­ sumed applied by some external agency. If F(t) denotes the force acting in the direction x increasing, the equation of motion (2. 72) must be modified to read

mx = -mkx - mc1h + F (t) (D2 + kD + cu2)x = J(t),

or

(2.85)

where j(t) = Ffm. Such an equation is well known to possess a general solution which is the sum of two expressions (a) the comple­ mentary function (CF) and (b) a particular integral (PI). The CF is the general solution of equation (2.72). The forms which may be taken by this general solution have already been given, and a reference to previous work will show that, in all circumstances, this solution approaches zero as t � w. This portion of the general solution of equation (2.85) is therefore only of significance during the early phase of the motion and plays no part in determining the motion during the later stages. It is accordingly referred to as the transient, and we shall disregard it. The PI, on the other hand, will generally comprise terms which remain finite for large t. This part of the general solution therefore completely specifies the motion, except during a period immediately subsequent to the initial instant. It is referred to as the steady state. The constants of integration are associated with the CF and do not appear in the PI. The initial conditions therefore affect the transient, but have no influence on the steady-state motion. Thus, no matter what the motion of the particle is when the force F (t) is first applied, it will eventually settle down into a state of motion which is prescribed entirely by the form of the function F (t) . We will illustrate our general remarks by the case F(t) = A sin pt, i.e., the applied force is itself oscillatory and of period 2-rrfp. The equation of motion is (D2 +

kD + cu2)x = a sin pt,

where a = Afm. The PI may be calculated by as the imaginary part of aeiP1 (i = v' - 1). Then x

= PI = � D2 + k� + cu2 aeiPt iPt = � (ip) 2 +aekip + cu2 t iP ae = � (,)2 - p2 + ikp"

(2.86) regarding a sin pt

(2.87)

2]

N E WT O N ' S LA W S .

RECTILINEAR MOTION

47

Putting the complex quantity (1)2 - p 2 + ikp into its polar form rei8, where r2 = ((1) 2 - p2) 2 + k2p2 and tan 6 = kP/((1)2 - p2), equation (2.87) becomes

aeipt = f a- ei(pt - 8) = a sin (pt - 6) . x = PI = f � r re'8 r-

(2.88) equation (2.86).

This equation defines the steady-state solution of It reveals that the particle's motion is simple harmonic, of frequency pj2rt. The particle's oscillations are therefore synchronous with those of the applied force and are termed forced oscillations. The function j(t) on the right-hand side of equation (2.85), which represents the cause of these oscillations, is correspondingly called the forcing Junction. The phase of the particle's motion lags behind that of the applied force oscillation by a constant angle e. The amplitude of the particle's motion is b, where

(2.89) b clearly depends upon the frequency P/2rt of the applied force. +b will take a maximum value for variation of p when P ((1)2 - p2) 2 k2p2 =

takes its minimum value.

b

a w

�: �;

Differentiating P, we obtain

- 4p( (1)2 - p2) + 2k2p = (2k2 - 4 (1)2) + 12p 2 . =

k

<

(2.90)

-l2 w

FrG. 2.12.-Amplitude Response Characteristic

Hence, dPfdp = 0 when p = 0 or ((1)2 - tk2) t. Assuming k < y2(1), the latter stationary point is real. If such is the case, when p = 0,

A C O U RS E I N A P P L I E D M A T H E MAT I C S

48

[CH.

d2PJdp2 is negative and this stationary point is a maximum of P and so corresponds to a minimum of b. When p = (w2 - !k2)!, d2Pfdp2 is positive, and hence P is a minimum, i.e., b is a maximum and (2.91) As p -+ b clearly approaches zero. The graph of b against p is therefore as sketched in Fig. 2.12. k < y2w implies that the damping must be less than critical if the relationship between b and p is to be of the form shown pictorially in Fig. 2.12. oo ,

If the frequency of the applied oscillatory force is slowly varied, its amplitude being maintained constant, from what we have proved, we know that the amplitude of the particle's oscil­ lation will rise to a maximum when the frequency of A',--r--, the force is (w2 - !k2)lj210. At this frequency the particle is said to be in a state of resonance with the _t - .. A ';---t-.... force . It will be noted that if the damping is small, so that k2 can be neglected, the resonantfrequency is wj210, i.e., is equal to the natural frequency of oscillation of the particle under the action of the force - w2x alone. The cases when k > y2w and k = y2w will be left to the reader for analysis. l+c As an example of the manner in which forced oscillations may be induced in a mechanical system, we will investigate the motion of a particle which moves vertically whilst hanging from an elastic string whose upper end also moves vertically in some prescribed manner. In Fig. 2.13 A 1 represents the 0 +---�--. position of the upper end of the string when t = 0 and T � A its position at some later instant t. A 1A = F(t), P+-----Lt this function prescribing the motion of A. P is the position of the particle of mass m at time t. 0 represents the equilibrium position of the particle my when the string is motionless with its upper end at FIG. 2.13.-Forced A 1 OP = x. Letting l denote the natural length Oscillations on an Elastic String of the string and c its extension when the particle hangs in equilibrium, A 10 = l + c and equation (2.37) determines c, fL being Hooke's constant for the string. At time t, the string's extension is

:

f�t)

AP - l = x + c - F(t),

(2.92)

and its tension T is accordingly given by T

=

fL (X + c - F(t)) = fLX + mg - ILF(t).

(2.93)

Assuming that the particle is moving in a medium which gives rise

2]

N E WT O N ' S LAWS .

to a force of viscous friction of motion is

RECTILINEAR MOTION

mkx opposing the motion, the equation

mi = mg - T - mkx = - fLX + fLF(t) - mkx mi + mkx + fLX = fLF(t).

or

49

(2.94)

This equation is of the form of equation (2.85), and the motion of the particle is therefore of the type we have been discussing. In particular, if A executes SHM about A ', the particle will, in the steady state, perform a SHM about 0 at the same frequency. Provided the viscous friction is not too great, there will exist a frequency of oscillation of the point of suspension at which the particle will give a maximum response. At this frequency, the system is in the state of resonance. If F(t) = 0, i.e., A is stationary, the equation of motion (2.94) corresponds to equation (2.72) . If, therefore, the particle is displaced from 0 and released, it will execute damped oscillations of the type described at the beginning of this section.

A particle of mass m is hanging in equilibrium under gravity on the end of an elastic string having a Hooke's constant m. At time t = 0, the upper end of the string commences to move downwards with SHM of amplitude unity and frequency 1j27T. If is the downward displacement of m, from its initial position, at time t, show that until the string goes slack, = !(sin t - t cos t). Friction may be neglected. Employing the notation of equation (2. 9 4), k = 0, p. = m, F(t) = sin t. The equation of motion is therefore (D 2 + 1 ) = sin t. (i) The characteristic equation is >.2 + 1 = 0 and the characteristic roots are ±i. This implies that the CF is A cos t + B sin t. A PI is found by a standard procedure to be -!t cos t. Hence, the general solution of equation (i) is x = A cos t + B sin t - !t cos t. (ii) Differentiating, it follows that 7: = -A sin t + B cos t - ! cos t + !t sin t. (iii) But, at t = 0, x = 7: = 0 and so, from equations (ii) and (iii), we find that A = 0, B - !- = 0. It is now clear that x = !(sin t - t cos t), as stated. Because of the t cos t term, the particle will make ever-larger excursions

Example 6.

x

x

x

either side of its equilibrium position and, eventually, the string must go slack, and equation (i) is then no longer applicable. This is the form the phenomenon of resonance takes in the absence of friction (see Ex. 15 below) .

EXERCISE 2

1. The combined weight of a motor cyclist and his machine is 500 lb.

When free-wheeling down an incline of I in 40 their maximum speed

50

A C O U R S E I N A P P L I E D MATH EMAT I C S

[CH.

is 25 m.p.h. ; and 40 m.p.h. down an incline of 1 in 20. Assuming that air resistance varies as the square of the speed, and other resist­ ances are constant, find the h.p. which the motor must develop to maintain a speed of 50 m.p.h. on the level. (L.U.) 2. Let it be assumed that the effective rate of working of the engine of a motor car is represented by the expression hn(n1 - n), where n is the number of revolutions per second and h, n1 are certain constants ; also that the velocity v is equal to kn, where k is a constant. If there is a fixed resistance to the motion of the car equal to A times its weight (w), find the speed with which the car will ascend an incline of angle et.. If the value of k is capable of adjustment, show that to secure the best speed k must be equal to hn1f2w(A. + sin et.) , the best speed being hn12/4w(A. + sin et.) . If the maximum power of the engine is 20 h.p., the weight of the car is 1 ton and the best speed up a slope of 1 in 10 is 20 m.p.h., show that the best speed on the level is nearly 50 m.p.h. and find the best speed up a slope of 1 in 5. (M.T.) 3. A pile of mass M is struck by a hammer which exerts a force increasing linearly from zero to a maximum F in time T, and then decreasing linearly to zero in an equal time. The ground exerts a constant resistance R in excess of the weight of the pile, whilst the pile is moving, within the limits given by (2 - y2)F < R < F. Find the time during which the pile moves. (L.U.) 4. The attraction of the Earth for a particle of mass m distant r from its centre is mfLfr2, fL being constant. A particle is projected vertically with velocity V from a point on the surface If a is the radius of the Earth, show that, neglecting air resistance, the particle will fall back to its point of projection unless V2 > 2fLfa. If the latter condition is satisfied, prove that the particle will be at a distance r from the Earth's centre after a time t given by

(�3t t = v'c2x2 + 2cx - v'c2 + 2c + cosh-1(c + 1) - cosh-l(cx + 1),

a V2 where x = rfa and c = - - 2. fL

5. A particle P of mass m moves along a straight line through a point 0, and, at any instant, the distance OP is x. When x > a the particle is attracted towards 0 by a force mkfx2, and when x < a the particle is repelled from 0 by a force makfx3 • If the particle is released from rest at a distance 2a from 0, show that it will come to rest instantaneously when x = afy2, and find the time the particle takes to travel from (L.U.) x = a to x = afy2. 6. A particle is suspended under gravity from a fixed point by a light elastic string and, when the particle hangs in equilibrium, the string is extended to a distance c beyond its natural length. Show that the

2]

N E WT O N ' S LAWS .

RECTILI NEAR MOTION

51

period o f small vertical oscillations about the position o f equilibrium is 2rt(cfg)t. If the particle is released from rest at a distance 3c below the position of equilibrium, prove that it will rise for a time {2y2 - cos-1 ( 1/3) + rt} (cfg) !. (L.U.) 7. A particle of mass m moves from rest on a rough horizontal table under an attraction to a fixed point of the table proportional to its distance from the point. Show : (a) that there are two positions of limiting equilibrium on the line of motion ; (b) that if the initial dis­ placement is large enough the particle describes half-cycles of simple harmonic motion about these two positions alternately ; and (c) that it completes a whole number of half-cycles before coming to rest. (L.U.) 8. A particle P of unit mass is acted on by an attractive force k 2 OP towards a fixed point 0 of a line and also by a small disturbing force in the line of amount + fl. . OP2. If the particle is released from rest at the point A , where OA = + a. prove that the particle will come to instantaneous rest at points OP = x given by the equation �fl. (aa - xa) . k 2 (a2 - x2) I f after leaving A the particle first comes to rest at B , prove that =

AB

=

(

2a 1

-

r:2)•

where powers of fl. higher than the first are neglected. (L.U.) 9. A railway engine weighing 32 tons exerts a constant horse-power of 448 and the motion of the engine is opposed by a constant force equal to £- tons weight. Prove that the equation of motion of the engine is dv = v2 £- (88 v), dx where v is the speed in ftfsec, x is the distance travelled in feet, and g has the value 32 ftfsec2 Find the maximum speed of the engine in m.p.h. and prove that the engine attains a speed of 30 m.p.h., starting from rest, in approximately 423 ft. (L.U.) 10. A particle of mass m is projected vertically upwards from the surface of the Earth, and the resistance of the air is mv2fc, where c is a constant and v is the velocity. If the velocity at height cf4 is Vf2 and at height cf2 is V/4, prove that V2 = 16gc(ye - 1)/(4 - ye), and find the maximum height attained in terms of c and e . (L.U.) 1 1 . A particle of mass m moves in a straight line against a resistance av + bv 2, where v is the velocity and a and b are positive constants. Prove that the particle cannot move a distance greater than -

•

� log ( 1 � u) from the point at which its velocity was u and that it +

cannot attain this distance in any finite time. Confirm, by other considerations, the result that the particle cannot come to rest in a finite time and show that the result is also true for negative values of b. (L.U.)

52

A C O U RS E I N A P P L I E D M A T H E M A T I C S

[cH.

12. A particle of mass m moves through a medium which offers resistance av + bv3, where v is the velocity and a, b are constants. If this be the only force acting and if the particle be given an initial velocity u, prove that it will come to rest after traversing a distance

7a

( J�) ( )

tan-1 u · v b) Show also that the velocity becomes !u after a time t given by 2at = 4a + buz log · m a + bu2 (L.U.) 13. A particle of unit mass moves in a straight line under a constant power H and a resistance !kv, where v is the speed of the particle. Prove that v tends to the limiting value V, where V = (2Hfk)t. In the absence of the resistance, the particle's speed increases from u to 3u(3u < V) in moving through a distance a. Prove that, when the resistance is also acting, the time required for the particle's speed to increase from u to 3u is 1 (L.U.) k log {(52u - 3ak)f(52u - 27ak)}.

14. A car of mass m moves on a level road. The engine works at a constant rate R, and there is a constant frictional resistance, the maximum velocity attainable being w. Prove that the distance in which the car, starting from rest, acquires a velocity V is mws ___!!!___ - _!:::" - vz log . w- V w 2w2 (M.T.) R 15. A particle P of mass m moves along a straight line under the action of a force mw2 OP directed towards a fixed point 0 of the line and a force mF sin pt, t being the time variable. Show that, if friction is sufficient to damp out the transient but may otherwise be neglected, in the steady state F sin pt OP = � ' (p =/= w) . w - pz Calculate the steady state solution when p = w and graph the ampli­ tude of the particle's oscillation as a function of p. 16. A ship of mass m moves from rest under a constant propelling force mf and against a resistance mk times the square of the speed. Prove that, when the ship has travelled a distance a, its speed v will be given by kv 2 = /(1 - e-zka) . If now the engines are reversed, show that the additional distance b travelled before the ship comes to rest will be given by e2kb + e-zka = 2 . (S.U.) 17. A particle of mass m can move in a straight line and is subject to a force rx.t at time t such that the force acts along the line of motion. The motion is opposed by a resistance of magnitude kv, where k is a constant and v the velocity of the particle. If the particle starts from rest show that at time t it has moved through a distance s given by rx.m2 (1 - e--ktfm) + ocez - rx.mt. s = k3 2k k2 (D.U.)

{

•

}

2]

N E W T O N ' S LAW S .

R E CT I L I N E A R M O T I O N

53

18. A particle moves from rest under a force which works at a constant p power P. Prove that the distance moved in time t is ta where m m is the mass of the particle. (Q.U.) 19. A light elastic string A B, of natural length l and modulus /.., lies slack and at rest on a smooth horizontal plane. A particle of mass m, also at rest, is attached to one end A of the string. The other end B of the string is pulled along the plane with a constant velocity V away from A along a straight line through A . Determine the time that elapses after the string first becomes taut until it becomes slack again. Show that the particle will have travelled a total distance 2l + Vrt(mlfJ..) t by the time it catches up with B. (B.U.)

(: Y,

20. A particle of unit mass moves along the x-axis under the attraction of an attractive force of magnitude 4x towards the point x = 0, and a resistance equal in magnitude to twice the velocity. If the particle is released from rest at x = a, find its position at any time, and show that its velocity is given by

-

J ae-•

sin y3t. 3 Show that when the particle is instantaneously at rest on the positive side of x = 0, after completing n full oscillations, its coordinate is ae-2n"'fv'3 • (Le.U.) i

=

21. A spring lies at rest on a horizontal plane and has one end fixed. A particle of mass m attached to the free end of the spring is projected with a velocity V directly towards the fixed end. The motion of the particle is opposed by a frictional resistance which is of magnitude mkv when the velocity is v, and the spring exerts a restoring force of magnitude mk2x when compressed through a distance x. Find the time during which the particle travels in the direction of projection before first coming to rest, and show that the distance traversed in this time is (Le.U.) Ve-"'13v'3fk. 22. In the case of damped vibrations governed by equation (2.72) , show that the natural logarithm of the ratio of two consecutive extreme distances from the origin is rtkfn, where n2 = 4w 2 - k 2 • This constant is called the logarithmic decrement. 23. A light elastic string, of natural length 2a and modulus Smg, is stretched between two fixed points A and B of smooth horizontal table, distant 4a apart. A particle of mass m is fastened to the mid-point of the string. Show that the period of small oscillations of the particle in the line of the string is !rty(afg) . Show also that, if the particle be pro­ jected from its equilibrium position in the direction AB with a velocity -yi(56ga), it will just reach B in a time

1J� {cos-1y� + y2 cos-1%}·

(S.U.)

54

A C O U R S E I N A P PLI E D M A T H E M A T I C S

[CH. 2]

24. A light spring AB is coiled round a fixed smooth vertical bar. If the lower end A of the spring is fixed and a bead of mass m threaded on the bar rests on the upper end B, the extent to which the spring is com­ pressed is b. Show that the period of small vibrations of the bead is 27tfw, where w2 = gfb. The end A of the spring is now made to perform a vertical simple harmonic motion of period 1tJ w and amplitude a, and at time t 0, A is in its lowest position and B is momentarily at rest at its original equilibrium position 0. Show that at time t the displacement of the bead below 0 is ta(cos wt - cos 2wt), provided the bead remains in contact with the spring. Find the great­ est downward acceleration of B and show that the contact will cease at some stage of the motion if 5aw2 > 3 g. (S.U. ) =

CHAPTER 3

CARTESIAN AND INTRINSIC RESOLUTES OF ACCELERATION

3.1 Resolving the Equation of Motion Since, by equation (2.1), the resultant force acting upon a particle of mass m is, at all times, represented by a vector F, which is parallel to the acceleration vector f for the particle and m times it in magnitude, the components of F in any three directions will be m times the com­ ponents of f in the same three directions. These directions need not be mutually perpendicular, but, since it is easier to calculate the various components when the directions satisfy this condition, it is usual to restrict them in this way. In particular, if the motion takes place in a plane, we can resolve the two vectors in any one of the three pairs of perpendicular directions considered in Section 1.4 and for which we have expressions for the resolutes of f. In this chapter, we shall study problems which may be solved by the resolution of f and F into (i) Cartesian components, and (ii) tangential and normal components.

3.2. Cartesian Components. Projectiles

Consider the motion of a projectile P of mass m, which moves under the Earth's attraction. We shall, at first, neglect air resistance, the variation of the attraction with the height above the Earth's surface and the curvature of the latter. Suppose the projectile is shot from a point 0 on the surface, with speed V making an angle rt. with the horizontal. Taking rectangular Cartesian axes Ox, Oy, such that Ox is horizontal and Oy is vertical (upwards) , let (x, y) be the coordinates of the projectile at time t after projection. The only force acting on P is that due to the Earth's attraction, and this is of magnitude mg in a direction parallel to Oy (see Fig. 3.1). The com­ ponents of the acceleration of P in the directions of the axes are (x, :Y) (Section 1 .4). The components of the force acting on P in the same pair of directions are (0, -mg). Accordingly, we have the equations of motion

mx 0, my = -mg, x = 0, y = -g. =

or

55

(3 . 1 )

A COURSE I N APPLIED MATHEMATICS

56

.X = 0 implies that x is constant. But velocity and is initially V cos r�:. Hence

x=

V cos

[cH.

x is the x-component of the

r�: .

(3.2)

Integrating again with respect to t,

X = Vt cos r�:, the integration constant being zero, since x =

(3.3)

0 when t = 0.

y DI R E CT R I X

y

' ' ' ' '

I ' ' I I

FIG. 3.1 .-Parabolic Motion under Gravity

Integrating y

= -g with respect to t, we obtain y = V sin r�: gt, -

G

(3.4)

V sin a. being the constant of integration, since y = V sin r�: at t = 0. Integrating equation (3.4) with respect to t, since y = 0 when t = 0, y = Vt sin

r�:

-

tgt2 •

(3.5)

Suppose the trajectory meets the horizontal plane through 0 in the point G. OG is called the horizontal range. At G, y = 0 and hence, from equation (3.5), 2V t = ta = g sin r�:' .

(3.6)

3]

R E S O L U T E S O F A C C E L E RAT I O N

57

the alternative root t = 0 corresponding to the point 0. Equation (3.6) gives the time of flight to G. Substituting from equation (3.6) into equation (3.3), we calculate that, after this time has elapsed, V2 2V2 x = OG = - sin IX cos IX = - sin

g

g

21X.

(3.7)

Equation (3.7) shows that the horizontal range which will be achieved with a given velocity of projection V, is a maximum when IX = rt/4 , and is then V2fg. When the projectile is at the highest point H of its trajectory (the vertex), its velocity will be horizontal and, thus, y = 0. Equation (3.4) shows that this occurs when

t tH vg Sin. IX. =

=

(3.8)

-

We note that tH = tt6. Substituting from equation (3.8) into equa­ tions (3.3) and (3.5), we obtain the coordinates of H, viz., V2 . V2 . X = XH = g Slll IX COS IX = Slll

2g

y = YH = V2g2 sin2 IX.

21X,

(3.9) (3.10)

We note that XH = txa. Equations (3.3) and (3.5) are parametric equations of the trajectory. Eliminating the parameter t between them, we obtain the equation of the trajectory in the form

g x2 sec2 IX. y - x tan IX 2V2

(3.11)

The expression on the right-hand side of this equation being a quad­ ratic in x, the trajectory must be a parabola. To obtain the equation of this parabola in standard form, we shift the origin to the vertex H, where we erect new axes HX, HY, HX being parallel to Ox and HY being vertically downwards. Let (X, Y) be the coordinates of P relative to these axes. Then X = X + XH,

y = YH Y. -

(3.12)

Putting xH and YH equal to their values as given by the equations (3.9), (3.10) and substituting in equation (3.11) for (x, y), we find that (X, Y) satisfy the equation

2V2 X2 = - cos2 IX g

. Y.

(3.13)

58

A C O U R S E I N A P P L I E D MATH EMAT I C S

[CH.

The parabola is now seen to have latus rectum of length 2V2 cos2 rx.fg and focus at the point 2 (3.14) X 0, Y V cos2 rx.. =

=

2g

The corresponding (x, y)-coordinates of the focus are found from equations (3.12) to be V2 . (3.15) x sm 2rx., =

2g

V2 2 - sin rx.

2

2

V V - cos2 rx. = - - cos 2rx. . - 2g (3.16) 2g 2g The directrix is the line Y = - V2 cos2 rx./2g, i.e., it is the line V -2g2 s1n. 2 rx. + 2gV-2 cos2 rx. = V-2g2 y (3.17) Equation (3.17) shows that the height of the directrix above the point of projection is independent of the angle of projection. If v is the speed of the projectile at time t, equations (3.2), (3.4) and (3.5) show that y

=

=

v2 = x2 + y 2 V2 cos2 rx. + (V sin rx. - gt) 2 V2 - 2Vgt sin rx. + g2t2 = V 2 - 2g(Vt sin rx. - !gt2) = V2 - 2gy. =

=

(3.18)

The depth of the projectile below the directrix at this instant is V2 - y. If, therefore, the particle were allowed to fall from the

2g

directrix to this point, it would acquire a speed u, where u2

=

2g(�; - y) = V2 - 2gy = v2,

(3 .19)

using equation (3.18). Hence u v and the speed of the projectile, at any instant, is equal to the speed it would acquire in falling from the directrix of the trajectory to its position at this instant. Let us now consider the anti-aircraft gunnery problem. Referred to the axes through 0, let (x0, Yo) be the coordinates of a target A. The velocity of projection V being given, it is required t o choose the angle of projection rx. so that the trajectory of the missile passes through A. Let T be the time of flight to A. Equations (3.3) and (3.5) show that the requirement is satisfied if =

rx.,

VT COS = o VT sin rx. Y X0

=

-

!g T2.} ·

(3.20)

R E S O L U T E S O F A C C E L E RAT I O N

3]

a:.

V and g, these two equations fix T and Eliminat­ Yo = X0 tan gx2Vo22 sec2 = X0 tan gx2Vo22 ( l + tan2 ) 2 2V 2V2y2 o + l = 0, tan2 - tan + -(3.21) gxo gxo (x0, Yo)

Given and ing T, we obtain

a:

or

59

a:

-

a:

-

a:

a:

-

a:

i.e., a quadratic for tan a:. If this quadratic has two distinct real roots there are two possible angles of projection. If the roots are complex there is no angle of projection which will cause the missile to strike A ; this implies that A is out of range. If the roots are real and equal, A is just within range and there is a unique angle of projection with the required property. For equal roots, we must have

V4 2V2y g2xo2 = gxo2 o + l , i.e., the point (x0, Yo) lies on the curve 2V2 (V2 -y) . x2 = g 2g

(3.22)

(3.23) This is a parabola with its vertex at the point (0, V2j2g), axis vertically downwards and latus rectum of length 2V2Jg (see Fig. 3.2). D I � E C T � I )(

OF

PAR

B O LA

OF

SAFETY

FrG. 3.2.-The Parabola o f Safety

V2j2g

The focus is therefore a distance below the vertex, i.e., is at 0. The directrix is at a height above the vertex, i.e., is at a height above 0. This is twice the height of the directrix of any tra­ jectory. Any target on this parabola is j ust within range from 0. Any target above this parabola is out of range from 0. The parabola is accordingly known in ballistics as the parabola of safety. It is the

V2Jg

V2J2g

60

A C O U R S E I N A P P L I E D MATH E MA T I C S

[CH.

envelope of all the possible trajectories emanating from 0 with constant velocity of projection Suppose, then, that the point (x0, is below, or on, the parabola of safety and hence equation (3.21) has real roots. Let these be tan and tan cx2 • Then

V

Yo)

<%1

(3.24) and hence, tan ( <%1

<%1 + tan <%2 = - � = -cot e, + <%2) = ltan- tan <%1 tan <%2 Yo

where e = LAOx (the angle of sight to the target from deduce from equation (3.25) that

<%1 + <%2 = t7t + e <%1 - (!7t + te) = (!7t + te) - <%2·

or

(3.25) 0).

We

(3.26)

Thus, there are two possible directions of projection by which we can reach any target A from 0, and they are equally inclined to the line If A lies on the parabola of safety, through 0 of slope (!7t <%1 = <%2 and equation (3.26) shows that the angle of projection must be !1t Eliminating between the equations (3.20) as follows :

+ te).

+ }-e. <%

Xo ' sm <% = Yo +VT-}-gT2, <% = VT 02 (y T2) 2 ' l = cos2 <% + sin2 <% = X + o2 +2 2 g VT 2 4 4 V T4 + g- (Yo - -g ) r2 + g (Xo2 + Yo2) - (3.27) we obtain a quadratic for T2 • If the point A is below the parabola of safety, this equation must have real positive roots. Let them be T12, T22, where Tv T2 are the times of flight to A along the two possible paths. Then T12T22 - g4 (Xo2 + Yo2) -- 4d-g22' where d2 = X02 + Yo2 = OA 2• Hence ·

cos

.1

2

-

o

'

•

2

(3.28)

R E S O L U T E S O F A C C E L E RAT I O N

3]

61

Suppose a gun is at a point 0 on an inclined plane and let the vertical plane of fire through 0 intersect this plane in a line of slope e (Fig. 3.3). The parabola of safety corresponding to the gun's velocity of projection V intersects this line in a point E, which is at extreme range in this direction along the plane. Let EN be the perpendicular from E on to the directrix of the parabola of safety. From a fundamental property of the parabola it follows that OE = EN. Thus, if OE = R,

V2 - R sin e NM - EM = g 2 V R - --;(-o-----,----.------=g 1 + sin e)

R or

=

(3.29)

This last equation specifies the extreme range along a line of slope e.

N

0

FIG. 3.3.-Maximum Range on an Inclined Plane

A particle is projected under gravity with velocity V in a direction making an angle £1 with an upward line ofgreatest slope of a plane of inclination the point of projection being on the plane. The motion takes place in the vertical plane through the line ofgreatest slope. Show that if the particle strikes the plane at an angle with the horizontal, then e is given by the equation (1 + 3 tan• ) tan e = 2 tan In this case show that the velocity with which the particle strikes the plane is Vfy'(l + 8 sin2cx). (L.U.)

Example 1. ex,

ex

C<

ex.

Relative to axes Ox, Oy through the point of projection 0 (see diagram), the parametric equations of the trajectory are

Vt cos (() + ) y Vt sin (e + ex) - tgt•.} x = =

ex ,

(i)

62

A C O U R S E I N A P P L I E D MAT H E M A T I C S

[cH.

At the point o f striking the sloping plane A , yfx = tan "' and hence Vt sin (a + ct) = tan "' Vt cos (fl + ct) 2V t= [sin (a + ct) - tan "' cos (a + ct)]

- !gt2

g

2 V sin a = - · -- · g cos "'

(ii)

!I

X

Differentiating the equations (i), we obtain x = V cos (a + ct) , y = V sin (a + ct) Hence, at A, x = V cos (a + ct) , y = V sin (a + ct)

- gt. 2 sin a cos "'

(iii)

But the velocity at A makes an angle "' with the horizontal and hence, at this point, -yfx = tan "'· Thus, by the equation (iii), 2 sin a - sin (a + ct) cos "' (iv) = tan "' . cos "' cos (a ct) Whence, expanding sin (a + ct) , cos (a + ct) and dividing the numerator and denominator of the left-hand side of equation (iv) by cos a cos• ct, we find that 2 tan 6 sec• "' - tan a tan "' = tan "' I - tan a tan e< tan 6 2 tan a tan• "' - tan "' = tan "'' I - tan fl tan e< (I + 3 tan• ct) tan (l = 2 tan "' (v) i.e., From this equation, it is easy to deduce that I + 2 sin• "' . 2 sin "' cos "' , cos a = sin a = V I + 8 sin2 "' V I + 8 sin2 "' Hence, from the first of equations (iii), V cos e< a (vi) X = V(COS ua COS C< - Slll u Slll e<) = _ 1 -v I + 8 sin2 "'

+

-

+

·

·

·

•

3]

R E S O L U T E S O F A C C E L E RAT I O N If is the velocity of arrival at

v

A, :i: = v cos "'

·

63

It follows that

v = x. sec "' = VI + v8 sin2 "'. Example 2. A particle is projected with velocity V from a point A, and moves under uniform gravity along a parabolic trajectory having focus S. Show that V2 = 2g . SA. A particle slides down a smooth inclined plane along a line ofgreatest slope from a point P to a point Q where it leaves the plane and falls along a parabolic trajectory under gravity. If S is the focus of the parabola and the particle starts from rest at P, show that LPSQ is a right angle. (L.U.) The point of projection A being on the parabola, its distance from the focus S is equal to its distance from the directrix. This latter dis­ tance has been proved to be V2f2g. Hence SA = V2f2g, or V2 = 2g. SA. The forces acting on the particle of mass m as it slides down the plane are the weight mg and the smooth reaction R, normal to the plane. If f is the particle's acceleration down the plane, the equation of motion resolved in this direction is mg sin "' = mf, or f = g sin "' being the angle of slope. The acceleration is therefore constant and we may apply equation (2.18). If V is the speed at Q, (i) V2 = 2g . PQ . sin "' = 2g . PN, . PN being the vertical height of P above Q. V is the velocity of projection into the parabolic trajectory at Q, and hence the directrix of this curve is a height V2f2g = PN above Q. Thus, the directrix passes through P. The direction of motion at Q is along PQ. This implies that PQ is tan­ gential to the parabola at Q. Now, since the segment of a tangent to a parabola between the point of contact and the directrix subtends a right angle at the focus, LPSQ = 90°. Consider a particle of mass m moving in a vertical plane under the action of its weight mg and air resistance proportional to its speed v and in a direction always opposite to that of the motion. We shall take the air resistance to be mkv. Erecting axes Ox, Oy as before through the point of projection 0 (see Fig. 3.4), let (x, y) be the co­ ordinates of the projectile at time t after projection. If if; is the angle made by the direction of motion with the x-axis at time t, the x- and y-components of the resultant force on the particle at this instant are ( -mkv cos if;, - mkv sin if; - mg). But (v cos if;, v sin if;) are the a,

x- and y-components of the velocity, viz., (i, y).

Hence the force

64

[CH.

A C O U R S E I N A P PL I E D MAT H E M A T I C S

components may be written ( - mkx, - mky - mg) . The corresponding components of the acceleration vector are (.X, y). Thus,

mi -mkx, my = - mky - mg, x -kx, :Y - ky - g, =

or

(3.30)

=

=

are the equations of motion. X=

f

cos Ol

FIG. 3.4.-Resisted Motion under Gravity

Writing the first of the equations (3.30) in the form dtjdx we can integrate with respect to x to obtain

or

=

t = - � log x + constant,

- 1/kx,

(3.31)

where A is a constant of integration to be determined by the initial conditions. Suppose the missile is projected from 0 with a velocity V, making an angle a: with Ox. Then, when t = 0, x = V cos a:. Thus A = V cos a: and

x

Integrating with respect to we now find that

=

t,

Ve-kt cos

a: .

and observing that at

v cos a: x- -k

(1 - e-kt) .

(3.32)

t = 0,

X =

(3.33)

0,

RESOLUTES O F ACCELERATIO N

3]

65

dt = - I (3.34) ky + g" we can integrate with respect to y. The result is = - � log (ky + g) + constant or (3.35) -- Be-kt - fi,k B being an integration constant. When = 0, y = V sin rx and hence B V sin rx + gjk. Equation (3.35) is therefore written (3.36) y = ( V sin rx + i) e-k1 - { Integrating this equation with respect to t and using the fact that y 0 at t = 0, we find that y = � ( V sin rx + f) (I ) �(3.37) Writing the second of the equations (3.30) in the form

dy

t

. Y

t

=

=

- e-kt

-

Equations (3.33), (3.37) constitute parametric equations for the trajectory. If t is eliminated between these equations, we calculate the equation of the trajectory to be

( rx + lv sec rx ) x + � log ( I � x sec rx ) . -

y = tan oo ,

(3.38)

As t-+ e-kt-+ 0 and hence equation (3.33) shows that x-+ V cos rxjk. For large values of t, x remains steady at Vthecosvalue rxjk. V cos rxjk, i.e., the particle falls vertically along the line x This line is asymptotic to the trajectory as indicated in Fig. 3.4. Equations (3.32) , (3.36) show that, as t-+ x -+ 0, y-+ -gjk. This implies that the particle is eventually falling vertically with a speed gjk. This is the terminal velocity at which the weight force =

oo,

is exactly balanced by the air resistance. Let OQ be the line of projection (Fig. 3.4), and let QP be the vertical through the missile P at time t. Putting OQ = �. PQ = 1), (�, 1l ) are termed the of P. The relationship between these coordinates and the Cartesian coordinates of P is easily shown to be given by the equations

Littlewood coordinates �=

x sec rx,

1) =

x tan rx - y.

v (I - e-k'), = � - � ( I - e-kt) .

(3.39)

Substituting from equations (3.33) and (3.37) , we find that �= k 1l

D {I)

(3.40)

66

A C O U R S E I N A P P L I E D MAT H E M AT I C S

[CH.

These are the most convenient equations for specifying the trajectory. It will be observed that (�. lJ ) depend upon t only and are independent of the angle of projection. If, therefore, the muzzle velocity V of a gun at 0 is known, tables of the functions (�. "IJ ) can be computed from which the trajectory for any angle of projection may be quickly determined. Practical ballistics is concerned with the motion of a shell moving through the atmosphere, a medium which does not lead to the simple resistance law we have been assuming. However, it is found that Littlewood's coordinates are only weakly dependent upon oc, with the consequence that, when tables of (�. "IJ ) corresponding to a few values of oc are available, a process of interpolation permits the calcula­ tion of these functions to a high degree of accuracy for any other value of oc and hence the determination of the trajectory.

A particle of mass m is projected under gravity with a velocity whose horizontal component is u in a medium in which the resistance acts along the tangent and is equal to mk times the velocity . Show that the resultant force acting on the particle is constant in direction, and that when the particle has travelled a distance x horizontally, its magnitude is equal to (l - kxfu) times (M.T.) its initial value. Employing our previous notation, u = V cos o:. Differentiating equations (3.32), (3.36), we find that .i' = -Vke-k' cos o:, y = -k ( V sin o: + !) e-k•.

Example 3.

The Cartesian force components are accordingly

- Vmke-k• cos o:, -mk ( V sin o: + i) e-k•

(i)

and, if q, is the angle made by the resultant force direction with the x-axis, tan q, = tan o: +

ffk sec o:,

indicating that q, does not vary with Let (X, Y) be the Cartesian force components at = in equations (i), we find that

t.

X=

t 0.

Putting

-Vmk cos o:, Y = -mk ( V sin o: + i)

t =0

and hence, the components of force at time t are (Xe-k', Ye-kl) . The resultant force at this instant is therefore e-ki(X2 + Y2)l, i.e., times the magnitude of the force at = Since, by equation (3.33), = 1 this proves the result stated.

t 0.

e-k'

e-k<

-

3.3. Cartesian Components. Elliptic Harmonic Motion

kxfu,

We will now consider the motion of a particle P which moves in the plane of fixed axes Ox, Oy, under the action of a force which is pro­ portional to the particle's distance from O(OP = r) and is always directed towards 0. The rectilinear motion of a particle subjected to such a force has been analysed in Section 2.3(2) and has been shown to

3)

R E S O L U T E S O F A C C E L E RAT I O N

67

possess an oscillatory character. Letting mfl.r denote the magnitude of the force, we see from Fig. 3.5 that the Cartesian components of force are (-mfl.r cos 6, -mfl.r sin 6) , or ( - mfLX, -mfl.y), if (x, y) are

FIG. 3.5.-Elliptic Harmonic Motion

the coordinates of P. The equations of motion of the particle are accordingly

mx - mfl.X, my = - mfl.J, x + w 2x = 0, y + w2y = 0, =

or

(3.41)

where w2 = fl.· These equations imply that the coordinates (x, y) oscillate in a simple harmonic manner during the motion, i.e., that the projection of the motion on to either axis (and therefore on to any straight line, for the axis directions are open to choice) is simple harmonic. We have already determined the general solutions of equations such as (3.41) (see equation (2.31)). They are x

= A sin

wt + B cos wt, y C sin wt + D cos wt, =

.

(3.42)

A , B, C, D being constants of integration determined by the circum­ stances of projection at t = 0. Solving equations (3.42) for sin wt and cos wt, we find that (AD - BC) sin (A D - BC) cos

wt = Dx - By, wt Ay Cx. =

-

.

(3.43)

68

A C O U R S E I N A PPLI E D M A T H E MA T I C S

[cH.

If AD = BC, clearly yjx = CJA = DjB and the trajectory is a straight line through the origin. The motion is rectilinear, and hence must be simple harmonic. This type of motion has already been discussed. If AD -=1=- BC, by squaring and adding equations (3.43), we eliminate t to obtain

(C2 + D2)x2 - 2(AC + BD)xy + (A 2 + B2)y2 = (AD - BC) 2 .

(3.44)

This is the equation of the trajectory. Being of the second degree, it represents a conic. Having no terms of unit degree, the conic has a centre at 0. Finally, since (A 2 + B2)(C 2 + D 2) > (AC + BD) 2, the conic is an ellipse. Suppose, now, we choose the directions of the coordinate axes to be those of the ellipse's axes and the instant t = 0 to be that at which the particle is at the positive end of the conic axis along Ox. As­ suming the motion to be in the counter-clockwise sense, at t = 0 the particle will have a velocity parallel to Oy. Let the magnitude of this velocity be V and let a be the particle's distance from 0 at this instant. The initial conditions are then x = a, y = 0, i = 0, y = V. Hence, from equations (3.42), and

a = B, O = A oo, O = D, V = Coo, x

= a cos oot,

y

= -v sm. oot. (1)

(3.45)

Equations (3.45) are the parametric equations of an ellipse of semi­ axes a and Vjoo. It also appears from these equations that the ec­ centric angle of P at time t is oot, i.e. , the eccentric angle increases at a uniform rate oo during the motion. The eccentric angle increases by 21t and the particle makes one complete circuit of the ellipse in time

T = 21t" .

(3.46)

(1)

Relative to such axes, we see from equations (3.45) that the ampli­ tude of the x-oscillation is a and of the y-oscillation is b = Vjoo. Let (u, v) denote the Cartesian components (i, y) of the velocity of P at time t. First integrals of the equations (3.41) are then (see equation (2.26)), u2 = oo2 (a2 x2), v2 = 002 (b2 y2) . (3.47) _

_

Adding these equations, we have where w

w2 = oo2 (a2 + b2 - r2),

= (u2 + v2) ! is the magnitude of P's velocity and r = (x2 + y2) ! = OP.

(3.48)

3]

R E S O L U T E S O F A C C E L E R AT I O N

69

A particle P, of mass m, is moving in a plane under an attraction 2m . OP towards a fixed point 0 in the plane and a resisting force 3mv, where v is the speed of P. Show that the equations of motion in rectangular Cartesian coordinates with origin 0 are x + 3x + 2x = 0, ji + 3y + 2y = 0. Deduce that, in general, P describes a parabola passing through 0 and that it reaches 0 after an infinite time. (L.U.) The velocity of the point P has components (x, y) parallel to the axes and, hence, the resisting force has components ( -3mx, -3my). The equations of motion are accordingly mx = -3mx - 2mx, mji = -3my - 2my,

Example 4.

which are equivalent to those stated. Both the equations of motion are linear with constant coefficients. They have the same characteristic equation, viz., ,\2

+ 3:\ + 2 = 0.

The characteristic roots are - 1 , -2 and the general solutions for are therefore,

x and y

x = Ae • + Be-••, y = Ce-• + De-21 (i) e-• and e-••, we obtain (AD - BC)e-• = Dx - By, (AD - BC)e-•• = Ay - Cx. (ii) If AD = BC, yfx = DJB = CJA and the trajectory is a straight line through the origin. In general, however, by eliminating e-• between the equations (ii), we obtain the equation of the trajectory in the form (iii) (Dx - By) 2 = (AD - BC)(Ay - Cx). •

-

Solving in these equations for

Equation (iii) clearly represents a conic and, since the second-degree terms form a perfect square, it is a parabola. That the curve passes through the origin is also clear from its equation. As oo, and equations (i) show that i.e., the particle approaches the origin, but does not arrive at this point in a finite time.

(x, y) -+ (0, 0},

t-+ e-'-+ 0, e-21 -+ 0

3.4. The Energy Equation

Suppose that a particle P of mass

m moves in a plane under the action

F

of a resultant force F (Fig. 3.6). Let the curve DAB be a part of its tra­ j ectory. If e is the angle made by the force with the direction of motion at any instant, resolving the equation of motion in this direction, we obtain s

d m ds (lv2)

= F cos e,

(3.49)

being the distance moved by P along its track from some fixed point D, and v being the speed of the particle at time t.

D

FIG. 3.6.-Plane Motion of a Particle

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A P P L I E D MAT H EM A T I C S

[CH.

Suppose that the particle moves from A to B during the time in­ terval Quantities which are to be given their values at will be distinguished by a suffix and those which are to be given their values at by the suffix l. Integrating the equation with respect to over the interval we obtain

(t0, t1). t = tv s

t = t0 (3.49)

0

(s0, s1),

(3.50) tmv2

Kinetic Energy v.

m

is called the (KE) of a particle of mass moving with a speed It is a scalar quantity. The left-hand side of equation therefore represents the increase in the particle's KE as it moves along its track from A to B. The right-hand side of equation may be interpreted by re­ ference to equation where it is shown to represent the work done by the force acting on P as the particle moves from A to B. Accordingly, we have proved the result :

(3.50)

(2.7),

(3.50)

Increment in KE = work done by the applied force.

(3.51)

Although this equation is a direct consequence of the tangential equation of motion, it is essentially simpler, for, unlike the equation of motion, it does not involve a second-order derivative. The KE de­ pends upon the speed of the particle, rather than its acceleration, and is therefore expressed in terms of first-order derivatives only. On this account, it is frequently convenient to replace one of the equations of motion by an equation expressing the principle As a simple example, consider the problem of the projectile moving in a vertical plane under gravity alone. Employing the notation described in Section as the projectile moves from the point of projection to the point (x, the work done by gravity is

(3.51).

(0, 0)

3.2,

y), { -mg dy = -mgy,

(3.52)

!mv2 - tmV2 = -mgy or v2 = V2 - 2gy which is equation (3.18).

(3.53)

v !mv2 - !mV2.

and, if denotes the speed at the latter point, the increment in the KE Hence is

Provided frictional forces are absent, i t frequently happens that the work done by the forces acting on a particle as it moves from A to B is independent of the mode of motion between these points. This is clearly the case, for example, in the problem of the projectile just considered. The work done by gravity as the particle moves between the points (x0, (x1 , will be no matter how the

y0),

y1)

-mg(y1 - y0),

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71

transference between these points is effected. I f such is the case, the forces are said to be and we can define a quantity termed the (PE) of the particle. The PE of the particle when at a point is defined to be the work which would be done by the forces acting upon it, if it were to move from to some arbitrarily chosen datum point D. Thus, in the case of the projectile at the point choosing the point of projection 0 as the datum point, the work which would be done by gravity if it were moved from P to 0 is This is the PE of the projectile at A problem which will often occur in our later work is that of calculat­ ing the PE associated with the force acting on a particle due to a stretched elastic string. Suppose the particle is at some point and is con­ nected to a fixed point 0 by an elastic string of 0 modulus "A and natural length (Fig. 3. 7) . Let the datum position D be selected such that when FIG. 3.7.-Potential Energy due to an Elastic the particle is at this point String the string is its natural length. We shall prove that, if the particle is taken from to D by any path, the work done by the tension T of the string acting on the particle is always the same. Let Q be any intermediate position of the particle on its path and Let cp be the angle between the let OQ = At Q, T = ).. tangent to the path and the string. Let the arc DQ be denoted by Then cos cp The work done by T as the particte moves from to D is the negative of the work done as the particle moves from D to ; i.e., it is

conservative Potential Energy P

P

P(x, y}, mgy.

P

P.

P

l

P

r.

P P

(r - l)Jl.

=

s.

drjds.

1: T cos cp ds 1: f (r - l) �: ds P = J � (r - l)dr l P = J ")...__ (r - l)2 / 2l =

D

)..x2

= u'

/)

(3.54)

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[CH.

where x is the extension of the string at P. This result is observed to be independent of the path, and therefore represents the PE due to the stretched string when the particle is at P. It will be noted that we have assumed that the string does not go slack during the particle's motion from P to D. If this is not the case, the reader should show that the argument may be modified to yield the same result as before. In general, taking D to be some point on the path of the particle (Fig. 3.6), if represents the PE of the particle at any point on its path, the work which would be done by the forces acting on the particle if it were to be moved from B to D is, by definition, the PE of the particle at B, viz., Similarly, the work which would be done by these forces if the particle were moved from A to D is We can now conclude that the work which would be done if the particle were moved from B to A is In the actual motion the particle moves from A to B and the work done by the forces is therefore Equation (3.50) can therefore be written in the form

V

V1•

V0•

V1 - V0•

-(V1 - V0).

tmv12 - !mv02 = -(V1 - V0), tmv12 + V1 = tmv02 + V0• (3.55) or The quantity ( tmv2 + V) is termed the total energy E of the particle at any instant. Equation (3.55) indicates that E does not change its

value as the particle moves from A to B. Since B is an arbitrary point on the particle's track, this implies that the total energy remains constant throughout the motion. Thus

E = constant. This is the Energy Equation.

(3.56)

KE and PE are mechanical forms of energy. The molecules of a piece of material are well known to be in a state of vibration. The mechanical energy associated with this motion is the heat energy possessed by the material. There are other forms of energy which are not reducible to the mechanical forms, e.g., the energy associated with wireless waves. It is a fundamental law of modern physics that the total energy of all forms present in the universe is a constant, so that, as any phenomenon develops, there is a transformation of energy from one form into another, but no net loss or gain of energy. This is the It follows from this law that if a mechanical system neither receives energy from nor communicates energy to any external agency, and if there is no conversion of energy within the system into non-mechanical forms, then the sum of the kinetic and potential energies of all its particles will remain constant. This is the content of equation (3.56). If energy is lost from the system, as, for example, when friction is present and heat is generated, it will prove impossible to define a PE function and the energy equation, in the form (3.56) , is no longer applicable.

Law of Conservation of Energy.

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3.5. Tangential and Normal Components. Motion on Wires

If a particle is in motion in a plane, by resolving the resultant force acting upon it and its acceleration vector in the direction of motion, equation (3.49) is obtained. It will frequently be convenient to replace this equation by the equation of work (3.51) or the equation of energy (3.56). Resolution in the direction of the normal to the path yields a second equation, viz.,

2 mv - = F sm e -

•

p

(3.57)

being the radius of curvature of the trajectory. These equations are particularly appropriate when we are presented with the problem of the motion of a bead on a smooth plane wire of specified shape and under the action of given applied forces (e.g., gravity) . The only unknown force involved is the reaction of the wire upon the bead and, since the wire is smooth, the direction of this force will always be that of the normal to the wire. This unknown reaction will have no resolute in the tangential direction, and hence will not appear in the tangential equation of motion or in the equations of work or energy. Any one of these latter equations will therefore be sufficient to determine the motion of the bead. The normal equation of motion will involve the unknown reaction and is accordingly employed to calculate this force when the motion has been established. As an example, consider the motion of a bead on a smooth vertical circular wire of radius a. The bead is acted upon by two forces : (i) its weight mg, and (ii) the reaction R of the wire acting along a radius. At any instant, let e be the angle made by the radius 0 P through the bead with the down­ ward vertical (Fig. 3.8). Taking D, on the same level as the centre of the wire, as the datum point, the work which will be done by the forces acting on the bead if it is FIG. 3.8.-Motion of a Bead on a moved from P to D is -mga cos e, Vertical Circular Wire R doing no work, since its direction is everywhere perpendicular to the direction of motion. If v is the speed of the bead at any instant, its total energy is fmv2 - mga cos e. The energy equation is therefore p

or

tmv2 - mga cos e = constant, v2 = constant + 2ga cos e .

(3.58)

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[CH.

Suppose that the bead is projected from the lowest point of the wire with velocity V Then, when e = 0, v = V and, by substitution of these values in equation (3.58), we find that the constant is V2 - 2ga. Hence, in this case,

(3.59) v2 = V2 - 2ga(1 - cos e). It is clear from equation (3.59) that, as e increases, v2 decreases, as we should expect. When e = v2 takes its minimum value given by v2 = V2 - 4ga. (3.60) v will only be real, and this value of e can be attained only if V2 ;;,: 4ga. If V2 > 4ga, the bead will arrive at the topmost point of the wire with a non-zero velocity and will describe a complete circle, returning to the lowest point of the wire, where e = 2TC, with its initial velocity V. Complete circuits of the wire will now be made indefinitely. If V2 < 4ga, the bead cannot rise to the highest point and its velocity TC,

becomes zero when

e satisfies the equation cos e

V2 = 1 - 2ga

(3.61)

- ·

Having arrived at this point, the direction of its motion is reversed and it returns to the lowest point with speed V, after which it rises to the same height on the opposite side of the wire. The motion is oscillatory. The normal equation of motion is

2 -· - mg cos e = mv a Substituting for v2 from equation (3.59), we find that mV2 + mg(3 cos e - 2). R= a R

-

(3.62) (3.63)

an equation which determines R at all points on the particle's path. The motion of a particle which is connected to a fixed point 0 by an inextensible string of length a and which is free to move in a vertical plane through 0 may be discussed in a similar fashion. R is now re­ placed by the tension T in the string. T cannot take negative values (unless the string is replaced by a light rigid wire) and hence, if the particle is to be able to describe complete circles, must be positive for all values of e. Equation (3.63) indicates that T will take its mini­ mum value when e = 1t, and that then

T

2 T = mV a - 5mg.

(3.64)

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75

This must be positive, and thus V2 � 5ga. This condition is stronger than the condition V2 > 4ga, under which a bead can complete the circuit of a vertical circle and therefore includes it. Referring to Fig. 3.8, we see that the tangential equation of motion is

m dv dt = -mg sm e. .

Since

v ae, this equation is equivalent to e + !I sin e = 0. a

(3.65)

=

(3.66)

If the particle is making small swings in the vertical plane, e will remain small throughout the motion and we may replace sin e by e to obtain

(3.67) as an approximate equation of motion. We deduce that the mode of oscillation of e is simple harmonic, with period given by

T 2rrJf =

(3.68)

A more complex problem which may be attacked successfully by resolving along the tangent and normal to the path is that of the motion

FIG. 3.9.-Resisted Motion under Gravity

of a projectile in a vertical plane under gravity and air resistance proportional to the square of the speed. If m is the mass of the pro­ jectile P and v is its speed at time t, let mkv2 be the air resistance (Fig. 3.9). Taking axes Ox, Oy through the point of projection 0, let ifl be the angle made by the tangent to the trajectory at P with Ox.

[CR.

A C O U R S E I N A P PLI E D MATH E MAT I C S

76

}

Resolving along the tangent and normal at P, we obtain the equations of motion.

mv �� = -mg sin !f; - mkv 2 (3.69) mKv2 = mg cos lj;, where s is measured along the trajectory from 0 and K is the curvature. The path being concave downwards, dlf;fds is negative and, since is always taken to be positive, K = - d!f;jds. Equations (3.69) are there­

;s (tv2) = sin lj; - kv2 v2 �� = -g cos lj;.

fore equivalent to

-g

K

}

(3.70)

Dividing the first of these equations by the second and putting

v2 = u, we obtain

l du k 2u dlj; = tan !f; + g u sec lj;.

(3.71)

This equation can be linearised by substituting w =

dw + 2w tan !f; = -g 2k sec lj;. dlj;

lfu to yield (3.72)

The integrating factor is found to be sec2 lj;. Multiplying both sides of the equation by this factor, we can then write it in the form

d 2k 3 do/ (w sec2 lj;) = -g sec lj;.

(3.73)

An integration with respect to !f; now gives

w sec2 !f; = -gk- [log(sec !f; + tan lj;) + sec !f; tan lj;] + constant. (3.74) If V is the velocity of projection at an angle to Ox, then, when !f; = w = lfVZ. Substituting these values for w and !f; in equation (3.74), we calculate that sec2 k constant = +- [log(sec + tan ) + sec tan ] oc

oc,

oc

g

- � - g log

(sec sec

�

and hence that sec2 !f; sec2 oc �

_

k[

oc

oc

)

oc

oc

oc ,

+ tan oc + sec oc tan oc - sec lj; tan lj;l . + o/ tan o/ J

v2

(3.75)

Equation (3.75) determines in terms of the angle lj;. If this expression for v2 be substituted in the second of equations (3.70),

3]

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77

dsjdif; may be calculated as a function of if;, and then an integration with respect to if; expresses s as a function of this angle. This is the intrinsic equation of the trajectory. The integration cannot be per­ formed analytically, but may be treated by numerical methods. =

A fine smooth cycloidal tube, whose intrinsic equation is s 4a sin .p, is fixed in a vertical plane with its vertex uppermost. Two equal particles, each of mass m, are connected by a light inextensible string of length 2a and are placed inside the tube so that the upper particle is at the vertex and the string is fully extended inside the tube. If the upper particle is released, show that the tension of the string remains constant at t;mg, and that the time which elapses before the lower particle reaches a cusp is 2 (i) t cosh-1 3. (L.U.) Let T be the tension in the string. This force acts on each particle along a

Example 5.

tangent to the cycloid as shown in the diagram. Let R 1, R 2 be the reactions of the tube on the particles. These forces are normal to the cycloid, since the tube is smooth. At time after the motion commences, suppose the particles are at and Let rfov be the angles made by the tangents to the cycloid at and with the tangent at the vertex 0. Then, if and are the lengths of the arcs = sin = sin (i)

s2

P P

t Q. r/12 Q OP, OQ, s1 4a r/11, s 2 4a rp2 .

s1

}

Resolving along the tangent to each particle's path, we obtain the equations of motion (tv ) sin r/J 1 (ii) sin

T + mg = md�1 2 , mg r/12 - T = mdds,(!v2),

where i s the common speed o f the particles. But = + and hence, = The right-hand sides of the equations (ii) are therefore equal, implying that sin r/11 sin = ! sin sin

v s 2 s1 2a

ds2 ds1. T + mg = mg r/12 - T, T mg( rp2 - rp 1) = ?a,cs. - sl) (by (i)) t;mg. =

A C O U R S E I N A P P L I E D M A T H E MA T I C S

78

[cR.

Addition of equations (ii) yields =

1{11 + sin 1{12) 2dds1 (!v2) g d ds1 (v2) 4a (s1 + s2) = 2ff...a(s1 + a). Integrating with respect to s1, we obtain (iii) v2 = fa(s12 + 2as1), since v = 0 when s1 = 0. Since v = ds1fdt, equation (iii) may be written in the form .!!._ )t 1 . (iv) ds1 = 2 (�g · v' s12 + 2as1 When the lower particle is at a cusp, 1{1 = trr and hence s = 4a. Thus, s1 = 2a. Integrating equation (iv) with2 respect to s1 over2 the interval (0, 2a) we therefore obtain the time required, viz., 2 (i) t I cosh-1 s1 t a I :· = 2 (i) t cosh-1 3. Example 6. A bead slides on a rough circular vertical wire. Starting from rest at the level of the centre, it comes to rest again at the lowest point. If fL is the co­ efficient offriction, show that (M.T.) Let A, B be the initial and final positions of the bead and let P be its position at time t. Let a be the radius of the wire and LAOP = e. If R g (sin

=

,

is the normal component of the reaction of the wire, fLR is the frictional

8

component in the direction of the tangent. If is the mass of the bead and s is the arc the tangential and normal equations of motion are

AP,

m mg cos e -J.
R E S O L U T E S O F A C C E L E RATI O N

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79

Eliminating R between these equations, we find that

�� + 2J.LU

=

2ag(cos 6

-

J.L sin

6} ,

where u = v2. This is a linear equation having CF Ae-2�-'o. A PI is found thus : l iJ.L) eiO 3J.L cos 6 + (1 2J.L2) sin a. /?4 ___ (eiO + iJ.LeiO) = !?4 i!_j2J.L + i = 4J.L2 + I D + 2J.L The general solution is accordingly -

When e =

Hence

0,

u_ - v2 _ - A e_2�'0 + 3J.L cos a + (I - 2J.L2} sin e . 4J.L2 + I v = 0 and consequently 3J.L 0=A + 4J.L2 + t' v2 - 3J.L cos _

a + (I - 2J.L2) sin 6 - 31-Le-2�'0 4J.L2 + I

When a = t7T, v is zero again.

0

•

It follows that

I - 2J.L2 - 3J.Le-J.L" = , 4J.L• + I

and this is equivalent to the result stated in the question.

EXERCISE 3

1 . A projectile is launched with velocity V, and moves in vacuo under the action of a constant gravitational force. Show that its Littlewood coordinates at time t after projection are � = Vt and 'Y) = !gt2• 2. A projectile is fired from a point 0 on an inclined plane with velocity V. The plane of its motion intersects the inclined plane in a line making an angle � with the horizontal. Taking this line as x-axis, the positive direction being up the plane, and Oy in the plane of motion and perpendicular to Ox in the upwards direction, write down the equations of motion of the projectile. Hence prove that the range R of the projectile on the sloping plane is given by 2 V2 sin (tX - �) cos tX sec2 �. R = g where tX is the angle made by the velocity of projection with the horizontal. 3. Employing the notation introduced in the text, show that for variable angle of projection and constant velocity of projection the locus of the focus of parabolic motion under gravity is the circle x2 + y2 = a2, where a = V2f2g. Show also that the locus of the vertex is the ellipse x2 + 4y2 = 4ay. -

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[CH.

4. 0, A and B are three collinear points on a piece of ground which is horizontal, and A lies between 0 and B. A vertical wall of height 2h is erected at A , and a parallel vertical wall is erected at B, both walls being perpendicular to the line OAB. A particle is projected from 0 under constant gravity in a vertical plane through the line OAB. It just clears the wall at A and strikes the wall at B at a height h above the ground. If OA = 2d, AB = d, prove that the particle was projected at an angle tan-1(7h/3d) to the horizontal. (L.U.) 5. Any number of particles are projected from the same point at the same instant and fall under uniform gravity. Prove that : (i) at any sub­ sequent instant the directions of their velocities are concurrent ; (ii) the point of concurrence rises with constant acceleration. (L.U.) 6. A particle is projected with given velocity up a line of greatest slope on a finite smooth plane inclined at a fixed angle to the horizontal. After passing the top edge of the plane, the particle describes a para­ bola under gravity. If the point of projection remains fixed whilst the length up the plane is varied, prove that the locus of the focus of the parabola is a straight line. (L.U.) 7. An aeroplane is travelling with constant speed v round a horizontal circle ABC of radius a at a height h above the ground. An anti­ aircraft gun is situated on the ground at a point 0 vertically beneath the centre of the circle A BC. The gun is fired when the aeroplane is seen in the direction OA , and the shell just hits the aeroplane when it is at the point B. If the muzzle velocity of the gun is 3(hg) l, prove that a = (3y7)h, and that the length of the arc AB is 4v(hfg) i. (L.U.) 8. A particle of mass m moves in a vertical plane in a medium whose resistance is [Lm times the velocity. Show that the resultant accelera­ tion is in a fixed direction and that the vertical distance of the particle from a line through the point of projection parallel to the direction of the resultant acceleration is gtj[L. (L.U.) 9. An aeroplane is flying with uniform speed v in an arc of a vertical circle of radius a, whose centre is at a height h vertically above a point 0 of the ground. If a bomb is dropped from the aeroplane when at a height Y and strikes the ground at 0, show that Y satisfies the equation k ¥2 + Y(a2 - 2hk) + k (h2 - a2) = 0, (L.U.)

10. A particle of unit mass is projected with a velocity whose horizontal and vertical components are (u, v) in a medium in which the resistance is k times the velocity. Show that when the horizontal distance traversed is uf2k, the particle is at a height v g (log 2 !) 2k

-

above the point of projection.

k2

-

(M.T.)

3]

RESOLUTES OF ACCELERATI ON

81

1 1 . A thin smooth wire is bent into the shape of a parabola of latus rectum 4a and is fixed in a vertical plane with its vertex A downwards and its axis inclined at an angle 6( < !7t) to the vertical. A heavy bead of mass m can slide on the wire and is released from rest at the point B where the vertical through the vertex intersects the wire again. Find the speed of the bead when it reaches the vertex and prove that the reaction of the wire on the bead is then mg cos 6(1 + 4 cosec2 6). (L.U.)

12. A smooth wire bent into the form of a catenary of parameter c is fixed with its axis vertical and vertex upwards. A smooth bead which is threaded on the wire is projected from the vertex along the wire with velocity y'(2gc). Prove that, when the bead reaches the point at a depth h below the vertex, it has a velocity y2g(h + c) and find the reaction of the wire on the bead at this instant. (L.U.) 13. A smooth wire bent into the form of a parabola is fixed with its axis vertical and its vertex downwards. A particle of mass m oscillates on the wire coming to rest at the extremities of the latus rectum. Show that the pressure of the particle on the wire when passing the vertex is 2mg. (L.U.) 14. A smooth tube is in the shape of the catenary y = c cosh xfc and is fixed in a vertical plane with its vertex downwards and axis of sym­ metry vertical. A particle of mass m moves in the tube under the action of gravity and of an attractive force towards the x-axis equal to mgfc times its distance from it. If the speed of the particle at the vertex is (4gc)l, show that it will come to instantaneous rest at points distant (2y'2 - 1)c from the x-axis. Find also the reaction of the tube on the particle at the vertex. (L.U.) 15. Two particles, P, Q, each of mass m, are joined by a light inextensible string. The particle P is held at rest at a point on the highest generator of a smooth circular cylinder, of radius a, having its axis horizontal. The particle Q hangs freely (the length of the string being greater than a1tj2) and lies in the vertical plane through P at right angles to the axis of the cylinder. If P is now released, show that it will leave the surface of the cylinder when the radius through P makes with the vertical an angle oc, where oc is given by the equation 2 cos cc = (1 + cc) . Show also that the tension in the string at the instant when P leaves the surface is tmg(1 - sin cc) . (L.U.)

16. An elastic spring of small cross-section, natural length 1taf2 and modulus of elasticity 1tmgy'2 lies inside the upper half of a smooth, hollow tube which has been bent into the shape of a circle with centre 0 and radius a, and fixed with its plane vertical. The spring carries particles of mass m at its ends A , B. If the spring has its natural length initially and A , B occupy symmetrical positions, show that if the

82

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A C O U R S E I N A P PLI E D M A T H E M A T I C S

particles are released from rest, the maximum value 2y of the angle AOB is given by 1 - y2 cos y = (2y - rt/2) 2. Show that, if the system is symmetrically disturbed from its equili­ brium position, it will perform simple harmonic oscillations of period T given by gP ( 4y2 - cos ct) = 4rt2a, where 2ct is the equilibrium value of the angle AOB. (L.U.) 17. A smooth wire bent into the form of a catenary s = c tan is fixed with its plane horizontal. A small ring of mass m can slide on the wire and is attracted towards the directrix with a force proportional to its distance from the directrix. Show that the motion of the ring is simple harmonic and that the normal reaction varies inversely as the square of the distance from the directrix. (L.U.) 18. A smooth particle is projected along the inside of a vertical circle whose centre is 0 and radius is a. The particle leaves the circle at the point P on the upper half and meets it again at the point Q ; the point R is the highest point of its trajectory. Show that the time from P to R is (a sin 6/g) t cos 6 and that the time from P to Q is four times this, 6 being the angle between the horizontal and OP. (L.U.) 19. A particle is suspended from a fixed point A by a string of length l. Find the velocity with which it must be projected horizontally from its lowest position if it is to strike A . 2 0 A small ring can slide o n a thin smooth wire which is in the form of a circle of radius a. The wire is fixed in a vertical plane, the ring being attached to the lowest point of the wire by a light elastic string of natural length a and modulus equal to the weight of the ring. The ring is projected from the highest point of the wire with speed (!ga) t . Find its speed when it reaches the lowest point. Show that the re­ action of the wire on the ring at first acts outwards and begins to act inwards when the string has turned through an angle cos-1 ( 1 1/12) . (L.U.) 21. A particle P of mass m moves on a smooth horizontal plane and is attracted to a fixed point 0 in the plane by a force !J.m OP, where !J. is a constant. A fixed smooth vertical wall is erected on the plane at a perpendicular distance a from 0. The particle is projected from 0 with speed (2!J.} ta cosec ct and strikes the wall at an acute angle ct. If the coefficient of restitution at the impact is e, prove that, at time t after the impact, x = a( cos !J.lt - e sin !J.lt), y = a cot ct (cos !J.tt + sin !J.lt) , where the origin is at 0 and the x-axis is perpendicular to the wall. (L.U.)

.p

•

22. Two particles of unit mass P and Q perform coplanar elliptic harmonic motions of equal period 2rt/n, about a centre of force 0. Initially, 0, P and Q are collinear, OP = a, OQ = 2a and P and Q are moving

RESOLUTES OF ACCELERATION

3]

83

in the same sense perpendicularly to the line OPQ with speeds v and J..v respectively. Prove that the path of P relative to Q is an ellipse described in time 2rr:fn and show that this path becomes a circle if J.. = 1 ± nafv. (L.U.) 23. A particle P, of mass m, is attracted towards the origin 0 of rectangular coordinates x, y by a force mn20P. Initially, the particle is projected from the point (a, 0) with velocity u parallel to the axis of y. Prove that the particle describes the ellipse (x;a) 2 + (nyfu) 2 I, and that the eccentric angle o f P o n the ellipse increases at a constant rate. Show that the hodograph of the motion is an ellipse which is similar to the orbit. (L.U.) 24. Two particles P and Q are both attracted to a fixed point 0 of their plane of motion by a force fLY per unit mass, where r is the distance from 0. They are projected simultaneously from a point A , at a distance a from 0, in a direction at right angles to OA . If the initial velocity of P is byfl. (where b < a) and that of Q is ayfL, show that as P describes its elliptical orbit, Q will describe the auxiliary circle in such a manner that PQ is always perpendicular to OA . (L.U.) 25. A bead slides on a smooth wire which is bent to a circle and fixed with its plane horizontal. The circumference is divided into n equal parts by points A 1, A 2, A n - The bead, when at any point B of the wire, is simultaneously attracted to these points by forces whose magnitudes kA nE, where k is a constant, acting along the are kA 1B, kA 2B, chords BA 1, BA 2 , BA n respectively. Prove that the bead moves round the circle with constant speed. (L.U.) 26. A particle is attached to the end of a light inextensible string and hangs at rest with the upper end of the string attached to a point on the upper half of a fixed horizontal circular cylinder of radius a, so that the straight part of the string is of length arr: and is a vertical tangent to the cylinder. The particle is projected horizontally with velocity V at right angles to the axis of the cylinder so that the string begins to wrap round the cylinder. Show that the tension in the string, when the length of the straight part is acfo, is given by Tacfofm = V2 + ga(2 sin cfo 3cfo cos cfo 2rr:) . Hence show that if the string becomes slack when cfo = �. then � � �. where � is the acute angle given by cot � = 3�. (M.T.) 27. A bead whose weight is negligible by comparison with the other forces acting on it is set in motion with velocity V round a fixed rough horizontal circular wire of radius a on which it is threaded. Show that the angular distance 8 travelled by the bead after time is fL log + where fL is the coefficient of friction between the bead =

-

�

(I

:t),

-

t

and the wire. (B.D.) 28. A projectile moves under the action of gravity and of an air resistance proportional to the nth power of the speed. If the velocity of the

84

A COURSE I N APPLIED MATHEMATICS

[cH. 3]

projectile at any time is inclined at an angle .p above the horizontal and has a horizontal component u, prove that n u

1

=

1 -n n

n Uo

w

1"' o

secn+l.pd.p'

where u, is the speed at the highest point of the trajectory and w is the speed at which the magnitude of the resistance equals the weight of the projectile. (Li.U.) 29. The normal cross-section of a cylindrical surface is a complete arch of a cycloid with vertex uppermost whose intrinsic equation is s = 4a sin .p. The surface of the cylinder is smooth, and it is held fixed with its cusps resting on a horizontal plane. A particle is projected from a point on top of the cycloidal surface on the highest generator, the direction of projection being along the tangent to the normal cross-section through the point of projection. Show that, if the speed of projection is y(2ag), the speed of the particle while still in contact with the surface is y{2ag(1 + 2 sin2 .P)} and that the particle will leave the surface when it has descended a vertical distance ta. (Le.U.) 30. A smooth curve in the form of the parabola y2 = 4ax is fixed in a vertical plane with the x-axis horizontal. A particle on the concave side is projected vertically upwards from the vertex with velocity y(2gh) . Show that it will leave the surface at a point P whose y­ coordinate, Yv is a root of the equation ya + 12a2y - 8a2h = 0. Show also that the horizontal velocity component at P is (gy13f4a2) ! and deduce that the greatest height of the particle in the subsequent free motion is 3y1/2. (S.U.) 31. A smooth tube, in the form of the catenary y = c cosh xfc, is fixed in a horizontal plane. A particle of unit mass slides in the tube and is attracted to the x-axis by a force equal to k times its distance from it. Show that the motion is simple harmonic. If the speed of the particle at the vertex of the curve is cyk, show that the reaction of the tube is then 2kc. (S.U.)

CHAPTER 4

POLAR RESOLUTES OF ACCELERATION

4.1. Central Forces In this chapter we shall study a class of problems which receive their solution by resolving the equation of motion along and perpendicular to the line j oining the position P of the particle at any instant to a fixed origin 0, in the plane of motion. We shall denote 0 P by and the angle made by OP with a fixed reference line Ox in the plane of motion by 6. Thus 6) are polar coordinates of P. The polar resolutes of the acceleration of P have been given at (1 .20). If (Fr. F0) are the corresponding resolutes of the resultant force acting on the particle, then the equations of motion are

r

(r,

m(r - rfl2)

=

� :e (r28) = Fo.

Fr,

(4.1)

In the special case when the resultant force acting on the particle is always directed along PO and is of magnitudef per unit mass (Fig. 4.1)

FIG. 4.1.-A Central Force

these equations of motion reduce to

-J, it (r2(l) 0. (4.2) Such a force is termed a central force and, in general, it will be variable in magnitude, being dependent upon the particle's position relative to 0 and Ox. Thus f will be a function of r and 6. We shall only be concerned with the case when f is a function of r, this being the case r - r(l2

=

=

which more often arises in practice.

85

A C O U R S E IN A P P L I E D M A T H E M A T I C S

86

[cH.

From the second of equations (4.2), we deduce that (4.3) r26 = h (a constant) . If A is the area of the plane of motion which has been swept out by the radius OP since the instant t = 0, the area swept out as e increases to e + de and the particle moves to P' (Fig. 4.1) is the sector OPP' and may be denoted by dA. To the first order of small quantities, we have dA = !r2de, and hence, dividing by dt, we obtain (4.4) A = tr26 = fh. h may therefore be interpreted as twice the rate of description of area by the radius 0 P, and we have proved that this rate is a constant if the force is of the central type. An alternative physical interpretation of h, which we shall find useful in the sequel, may be derived as follows : If v is the velocity vector of P and p = ON is the perpendicular from 0 on to the line of motion of P (Fig. 4.2) the moment of the velocity about 0 is

pv = r sin {> v .

=

r

.

v sin {> = r . r6

=

r26 = h,

(4.5)

;.

r

0

!

- -p -

---

I I I I

I

I

I

I

- cf

FIG. 4.2.-Moment of the Velocity

where {> is the angle indicated in the figure. h accordingly represents the moment of the velocity about the centre of force. Putting = equation (4.3) implies that

r lju,

Hence

(4.6)

r. = u2l dudt = u2l dude dedt = h dude · -

-

-

(4.7)

POLAR RESOLUTES O F ACCELERATION

4]

(

h de ) - de ( - h de )dt - - h d62 6. - - h2u2 dd62u2 ·

Differentiating again with respect to t, we obtain .. d d2u du de du d _

r - di

-

_

_

_

87

(4.8)

It now follows from equations (4.6) and (4.8) that r -

y(l2 - h2u2 �:� - h2u3 . =

(4.9)

(4.10) h2u2 ( ddf)2u2 + u ) -- j. Given I as a function of r, and therefore of u, equation (4.10) is a differential equation determining u as a function of 6. This latter

The first of equations (4.2) can therefore be written in the form

equation is the polar equation of the trajectory of the particle. A trajectory described under a central force is usually referred to as an orbit. Alternatively, if the orbit is known from observation of the motion of the particle, u will be specified as a function of 6, and equation (4.10) then permits a calculation of the manner of variation of I with the particle's position in space. This latter application is the simpler and we shall illustrate it first.

A planet is observed to describe an elliptical orbit with the sun at a focus. Show that this is consistent with the assumption that the sun is attracting the planet with a central force varying as the inverse square of the distance between the two bodies.

Example 1.

The sun being a massive body compared with a planet, its motion is little influenced by the latter. We shall accordingly take the sun to be at rest at the origin 0. The correction necessary to allow for this erroneous assump­ tion will be discussed in Section 8.3. The polar equation of the planet's orbit may be taken in the form r=

=

l lu 1 + e cos a, . (4.11) where l is the semi-latus rectum and e the eccentricity of the ellipse. Differentiating equation (4.11) twice with respect to a, we find that (4.12) ldda•2u e cos a. Substituting from equations (4.11) and (4.12) into equation (4.10}, we deduce that f - h•u· (-!.1 cos v + 1 + el cos a ) - �1 . u2. (4. 1 3) Thus f u2 1/r2, and the result is proved. A point on an orbit at which the particle is moving at right angles to the radius vector is called an apse. At such a point r = 0 and hence r is a maximum or a minimum. Also, at an apse, dr r.16. 0 du = I dr = 0 (4.14) de -� de · de = -

oo

0

=

=

=

'

[cH.

A C O U R S E IN A P P L I E D M A T H E M A T I CS

88

The distance of an apse from the centre of force is termed an apsidal

distance.

4.2. Inverse Square Law Physical attractions between particles are almost invariably governed by an inverse square law. Thus, gravitational, electrostatic and mag­ netic attractions all conform to this law, i.e., the force of such an attrac­ tion is inversely proportional to the square of the distance between the attracting particles. It is therefore of practical importance to investigate the type of motion which will be executed by a particle under the attraction of another when the latter is fixed in position and the attraction is according to the inverse square law. Taking the position of the fixed particle to be 0, let f1./r2 = fLU2 be the attraction per unit mass on the moving particle. Equation (4.10) shows that the orbit is governed by the equation (4.15) This is a linear equation of the second order having characteristic roots ±i and CF = A cos 6 + B sin 6 = C cos (6 - ex) , where C and ex are arbitrary constants. A PI is found by inspection to be f1./h2 • The general solution is therefore u =

x2 + c cos (6 - ex) .

(4.16)

C and ex will depend upon the initial circumstances of projection of the particle. Suppose C and have been determined in some particular case. Without loss of generality, C can be restricted to be positive, ex

for if C is originally negative, C cos (6 ex) = -C cos (6 - ex n-) = -C cos (6 ex ') , where ex ' = ex + n-, and -C is positive. Let Ox' make an angle ex with Ox in the sense indicated in the diagram. If 6' is the angle made by OP with Ox', then 6' = 6 - ex and equation (4.16) is equivalent to -

-

u=

�+

c

cos 6'.

-

4]

P O L A R R E S O L U T E S O F A C C E L E RA T I O N

89

If, therefore, we replace the reference line Ox by Ox', and measure 6 from Ox', the equation of the orbit will be

� + C cos 6

u= or

l = 1 + e cos 6,

(4.17)

h2 l = -,

(4.18)

e = Ch2/[.L.

(4.19)

r

-

where

[.1.

and

Equation (4. 17) is the equation of the orbit and contains one inte­ gration constant e, which is positive. It is well known to represent a

0

e<

I

0

I?> I

E L L I Ps E

HYPERBOLA

0

FIG. 4.3.-Types of Conic Orbit

conic with focus at 0 , of semi-latus rectum l and eccentricity e. If e < 1, the orbit is an ellipse. If e = 1 the orbit is a parabola. If e > 1 the orbit is a hyperbola. The three types of orbit are illustrated in Fig. 4.3, from which it will be noted that the reference line is the axis of each conic. Equation (4.18) shows that l is determined immediately the moment of the velocity about the centre of force can be calculated.

[cH.

A C O U R S E I N A P P L I E D M A T H E M AT I C S

90

Since r2e =

h, the transverse resolute of the particle's velocity is . h J� (1 + e COS (4.20) Ve

= r8 = r =

l

8) ,

use having been made of equations (4.17) and (4.18) . Also, differentiating equation (4. 17) with respect to

· 1.e.,

Vr

=

· a r· = el sm

y_!_2 r = -e sin ee' heT sm = J!/V: sm r2e =

t, we obtain

_

v .

·

·

a v

·

a v, .

(4 . 21)

an equation which determines the radial component of velocity in the orbit.

"'---

It

e

sin e

FIG. 4.4.-Components of the Velocity Vector

Equations (4.20), (4.21) indicate that v may be regarded as the sum

J�e sin e along OP ; (ii) J�e cos e per­ pendicular to OP, and (iii) J� of three vectors : (i)

in the same direction. The vector sum of the vectors (i) and (ii) is the vector w (Fig. 4.4), which is of magnitude

FrG. 4.5.-Hodograph of the Orbital Motion

J�e and whose

direction makes an angle 8 with the perpendicular to OP. w is accordingly perpendicular to Ox. v can therefore be resolved into

4]

P O L A R R E S O L U T E S OF A C C E L E R A T I O N

two components o f constant magnitude, viz., direction and

91

J� in the transverse

J�e in a direction perpendicular to the conic axis.

This

resolution is shown in a velocity diagram in Fig. 4.5. As the motion proceeds, the vector CG remains fixed in magnitude and direction and the vector GQ rotates about G, maintaining its magnitude constant. Q therefore traces out a circle having centre G. This circle is the hodograph of the motion. If < 1 the orbit is an ellipse and the hodograph encloses the point C. If > 1 the orbit is a hyperbola and the hodograph excludes C.

e e

A particle is projected from infinity with velocity V so as to pass a fixed point at a distance c if undisturbed. If it is attracted to the fixed point with acceleration l-'fr2, where r is the radius vector, find its orbit and show that its line of motion is deflected through an angle 21X, where v p tan IX = .!:._. v•c The initial moment of the velocity is Vc. Hence h = Vc and the semi-latus rectum of the hyperbolic orbit is l = v•c•fl-'· Taking Ox, the axis of the hyperbola, as reference line, initially when the particle is at infinity r = r = - V and e = -LxOZ = -{3, OZ being the line through 0 parallel

Example 2.

:z

oo,

to the initial line of motion. Substituting these initial values into equations (4.17) and (4.21), we find that 0 = 1 + o cos f3 -V = -

�e

- -frc sin {3.

sin {3 =

This pair of equations determines and {3. Thus

e V• e• - 1-'"c_2 + 1' tan f3 = v•c _

_

- -·

1-'

The equation of the orbit is accordingly

+ ( v;�· + 1 ) t cos e 1 1-' cos e �-<" i r = v•c• + -c- ( 1 + v•c• ) . When the particle arrives on the axis Ox it is moving in the direction of the perpendicular to this line. Its line of motion has accordingly been rotated through the angle NOx = IX. But LNOx = {3 - !1r and hence tan IX = -cot {3 = L . v•c

�

=

1

The hyperbola being symmetrical about its axis, the particle will experience a further deflection of its line of motion through an angle IX as it recedes to infinity. The total deflection is therefore 2cx.

A COURSE I N APPLIED MATHEMATICS

92

[CH.

4.3. The Energy Equatio-n

Consider a particle P of mass m acted upon by an attractive force f(r) per unit mass directed towards a fixed point where OP = r (Fig. 4.6). If the particle is taken from a point A , such that OA = r,

0, 0

FIG. 4.6.-Work Done by a Central Force

to a point D, where r = d, the work which is done by the attractive force is -

D J A

mf(r) cos ds

=

-m

D J

dr f(r) - ds = -m dS A

J

d

r

f(r)dr,

(4.22)

where s is arc length measured along the path from A and is the angle between OP and the tangent to the path. It is clear from equation (4.22) that this work depends only upon the form of the function f(r) and not upon the shape of the path between A and D. Taking D as the datum point, the PE of the particle at A is therefore given by v = -m

r r

f(r)dr

=

m

fd

f(r)dr.

(4.23)

We can now write down the energy equation of the motion of P under the influence of the attraction. If v represents the speed of the particle when it is distant r from 0, then tmv2

or

+ m [ f(r)dr = constant,

v 2 = constant - 2

J: j

f(r)dr

= constant - 2 f(r)dr,

(4.24)

absorbing the contribution of the lower limit to the integral in the constant.

4]

POLAR RESOLUTES O F ACCELERATI ON

93

In the particular case of the inverse square law of attraction,

f(r) = fL/r2 and equation (4.24) leads to v2 = constant + �­ r

(4.25)

To be able to calculate the constant we must know the speed at one point on the path. Suppose that the particle moves in an ellipse of eccentricity e, semi-latus rectum l and semi-major axis a. When it is at the apse nearer to the centre of attraction, putting 8 = 0 in equation (4.20), we find that

(4.26) At this instant r = lj(l + e) (equation (4.17)). Substituting this pair of corresponding values of v and r in equation (4.25), we obtain constant

= y (1 + e)2 - ¥ (1 + e) = -] (1 e2) (4.27) = - !:-a. -

The energy equation is accordingly

v2 = fL (� - D (ellipse). (4.28) In the case of parabolic motion, e = 1 and equation (4.27) indicates

that the constant is zero. Thus

v2 = �r (parabola) . In the case of hyperbolic motion, semi-axis of the orbit.

Hence

(4.29)

-y (1 - e2) = �' where a is the

v2 = fL a + D (hyperbola) . (4.30) The velocity from infinity V at any point in the vicinity of a centre of

attraction is defined to be the velocity which a particle would acquire in moving from a great distance to the point in question under the action of the force alone. The particle is assumed to be at rest initially, and hence v = 0 when r = <XJ . Equating the increase in the KE of the particle as it moves between r = <XJ and r = a to the work done by the central force, we obtain the equation

(4.31)

A C O U R S E I N A PPLI ED M A T H E M A T I C S

94

[CH.

r a. r a.

V being the particle's velocity at = This latter equation specifies the velocity from infinity at a point where = Clearly, this velocity exists only when is of such a form that the integral in equation converges. Suppose that a particle is projected from the point at which = with speed v0 in any direction. Let v be the speed at any later instant when the particle's distance from the centre of attraction is Equat­ ing the increase in KE to the work done by the central force, we obtain the equation

f(r)

(4.31)

r a

r.

!mv2 - !mv02 = -mff(r)dr.

(4.32)

This may be written in the form

v2 v02 =

-

or

"" 2[rf(r)dr = V02 2[ f(r)dr + 2j"" f(r)dr, (4.33) v2 = V02 - V2 + 2["" f(r)dr. -

a

a

r

V0 < V, the right-hand side of this equation will be negative when and hence v will be imaginary for this value of r. This implies that r cannot approach infinity and the particle will not recede in­ definitely from the centre of attraction. If, however, v0 � V, r can If

r

= �XJ,

tend to infinity and the particle will escape from the centre of attrac­ tion. The velocity from infinity is therefore the minimum

velocity.

f(r) tL/r2, V2tL/a

escape

V2[Lja

the particle will If = V= Thus, if V0 � escape to infinity along a hyperbolic or parabolic trajectory. If Vo < the particle will describe an ellipse and cannot escape from the centre of attraction. It will be proved in Chapter that a spherical distribution of matter such as the Earth attracts any external particle with a force varying inversely as the square of its distance from the Earth's centre. In c.g.s. 0 units, fL = X 10 2 • The escape velocity at a point on the Earth's surface, where = Earth's radius = x 108 em, is therefore A particle projected with this speed in V= = 11·2 kmjsec. any direction will, neglecting air resistance and providing its trajectory does not intersect the Earth's surface, recede to infinity along a para­ bolic or straight path.

V2[Lja.

14

3·991 a V2tL/a

6·379

A particle describes an ellipse of major axis 2a and eccentricity e under a force directed to a focus S. When the particle is at the end of the latus rectum through S, its velocity is suddenly turned through a right angle without change of magnitude. Prove that the particle will now describe 1an ellipse whose major axis is still of length 2a, but is inclined at an angle tan- (1 - e2)fe to the major axis of the original orbit. Find the eccentricity of the new orbit.

Example 3.

(L.U.)

4)

POLAR RESOLUTES O F ACCELERATION

95

Since the particle's speed and its distance from the centre of attraction are unaltered, it follows from equation (4.28) that there is no change in the value of i.e., the particle continues to describe an ellipse of major axis 2a.

a,

Let 5, 5' be the foci of the original orbit and let 5, 5" be the foci of the new orbit. Let L5 be a semi-latus rectum of the original orbit. The normal to the original orbit at L makes the same angle
+

a

tan L5"5N = L 5"5N =

a( - e2), ae,

- e•, e e" tan-1 c -;; ) . 1

This is the angle made by the new major axis with the old. Also 55"2 = 5N2 + N5"2 =

4a2e2 + 4a2 (1 - e2) 2 4a2(1 - e•2 ++ e') e•)t. 55" = 2a ( l - e But 55" = 2ae', where e' is the eccentricity of the new orbit. Hence e' = (1 - e• + e•)t. Example 4. An electron of mass m moves in an orbit under the action of a central attractive force Ze2fr2, where r is the radius vector. If the total energy of the electron is such that the velocity tends to zero as r tends to infinity, show that the sine of the angle which the radius vector makes with the direction of the tangent to the orbit is inversely proportional to yr. (B.D.) The total energy of the electron is !mv2 - Ze2fr, and, since v -+ 0 as r -+ the constant value of this quantity must be zero. Thus 2 v• 2Ze mr =

oo,

=

-- ·

A C O U RS E I N A P P L I E D M A T H E M AT I C S

96

[CH.

If cp is the angle between the radius vector and the tangent to the orbit, by equation (4.5) rv sin cp = h. sin cp

Hence

=

proving the result stated.

hfrv

=

�J�rl.

Equations (1.14) and (4.24) together imply that the energy equation for the motion of a particle under the action of a central force may be expressed in the form But r28

= h and hence

.yz

+

r2(l2 = g(r).

(4.34)

(4.35) r2 = g(r) - hr22 = G(r). Thus, if r is to be real, G(r) must take positive values. Suppose that G(r) is plotted against r as shown in Fig. 4.7. Then a range of r' •

G(t)

FIG. 4.7.-Graph of G(r)

possible values of r is from r = r1 to r = r2 • Suppose that initially r = ro within this range and that r is increasing. Then r is initially positive and r must continue to increase until r begins to take negative values. Before this can happen r must pass through the value zero. Hence r will increase until it reaches the value r2 at which r vanishes. r cannot then increase further, since to do so would result in G(r) becoming negative. Hence r becomes negative and r begins to de­ crease. r must persist in decreasing until it reaches the value r1

POLAR RESOLUTES OF ACCELERATION

4]

97

at which r = 0, and hence r can once again change sign. It follows that r will oscillate between the extreme values ri> r2 and the whole orbit will be contained between two circles of these radii.

An elastic string has one end fastened to a smooth horizontal table, the other end being attached to a particle. The string is stretched to twice its natural length and the particle is projected along the table at right angles to the string. Prove that in the subsequent motion the string will never attain its natural length if more than a quarter of the total energy is initially kinetic. If one-fifth of the total energy is initially kinetic and if at some instant the string attains its natural length, prove that it will remain slackfor a time �,Jrr;,_a, where m is the mass of the particle, a is the natural length of the string and i\ is the modulus of the string. (L.U.) Let V be the speed of projection and let r be the length of the string assumed taut, at time t. The initial length of the string is 2a, and hence the constant moment of the particle's velocity about the point on the table 0 to which the string is attached is 2aV. If (l is the angular velocity of the string at time t, we have the equation (i) r26 = 2aV The PE of the stretched string at time t is i\(r - a) 2f2a (equation 3.54). The KE of the particle at this instant is im(r2 + r2S2). It follows that the energy equation is !m(f2 + r2S2) + ia, 3mV2 • This latter inequality implies that !mV2 < ti\a,

Example 5.

i.e., initial KE < ! (initial PE) or less than one-quarter of the total energy is initially kinetic. If more than one-quarter is initially kinetic the string cannot attain its natural length. If one-fifth of the total energy is initially kinetic four-fifths is potential and hence = i.e., = Substituting this value of into equation (v), we find that when the string first goes slack,

V2

V2 i\af4m.

!m V" !(!i\a),

r

=

-H�r·

(vi)

the negative sign being taken, since the string is contracting. At this instant = (i), = The radial and

r = a, and hence, from equation E (I)

re 2V (i\afm)l.

98

[CH.

A C O U RS E IN A P P L I E D M A T H E M A T I C S

transverse velocity components of the particle are accordingly respectively, where The particle is now at the point P in the diagram and proceeds with constant velocity to the point Q, at which the string becomes taut again. If LOQP LOPQ then tan

( -!u, u)

u = (>.afm)l.

=

= cr.,

a

cr. = uf!u = 2.

a

The particle's constant speed will be cosec and the distance PQ is cos The time taken by the particle to move from P to Q is therefore sin cr. cos i 5 ---:.\

u

cr.

2a

cr..

= 4af5u = 4 (ma) Example 6. A particle P, of unit mass is attracted by a centre offorce at 0 according to the law w2r + w2a3fr2, where r = OP. If P is projected from a point distant a from 0 with velocity 4awfv'3 perpendicular to OP, show that sub­ sequently the distance of P from 0 varies between r = a and r = 2a, and find the velocity of P when r = 2a. (L.U.) Since f(r) = w2(r + a3fr2), the energy equation for the motion is v2 = constant - 2w2J (r + ::) dr = constant - w2 (r2 - 2:" ) When r = a, v = 4awfv'3 and hence the constant is l3a2w2/3. Thus ) v• = r• + r262 = w2 ( 313 a• - r• + 2a3 r . The initial moment of the velocity is 4a2w/v'3 and thus r•(J = 4a2w/y'3. Accordingly ) r• = w• ( 133 a• - r• + 2a3r - 16a' 3r2 w• (- 3r4 + l3a2r2 + 6a3r - 16a'). = 3r2 (i) The velocity of projection being in a direction perpendicular to the radius vector, we know that r = 0 when r = a. Hence (r - a) is a factor of the right-hand side of equation (i). By inspection, we also find that r = 2a is a zero of this expression and hence (r - 2a) is a factor. Dividing these factors out, it is easy to show that w• (r - a)(r - 2a)(3r• + 9ar + Sa•). r• = - 3r• But 3r2 + 9ar + Sa• = 3 ( r + ia ) + �a•, and hence is always positive. 2a

·

·

•

2

cr.fu

P O L A R R E S O L U T E S O F A C C E L E RATI O N

4]

99

The sign of f2 is accordingly determined by that of the expression This expression takes positive values only when Hence f is real for such values of only and the motion is confined between the circles = and = When = r = 0 and, since = = At such an instant, therefore, the particle is moving in the transverse direction with a speed

-(r - a)(r - 2a). a :::;;; r :::;;; 2a. r 2a, 2awfy3.

4.4. Time in

an

r r a 2 r 2a. r 6 4a2w/y3, rli 2awfy3.

Orbit

Having calculated the equation of the orbit of a particle, it may be required to find the time which elapses as the particle moves from one point on its path to another. If is the area swept out by the radius vector from the centre of attraction during the motion between these points, and if is the time of transit between the points, the constant rate of sweeping out of area is Thus

A

t

A ft. At = ih, or t 2A h =

(4.36)

.

In the case of an elliptical orbit described under the influence of an inverse square law of force per unit mass, let be the semi­ major and semi-minor axes respectively. The semi-latus rectum is and hence = Let be the time necessary to complete one circuit of the orbit. During this time the radius vector sweeps over the whole area of the ellipse and = Hence

f.'./r2

b2Ja,

h bv';{a.

a, b

T

A rr:ab. T 2rr:ab _ 2rr:at (4.37) T is called the periodic time in the orbit, and is here observed to be _

- -----;;- - f.'.!

.

proportional to the three halves power of the major axis of the orbit. A more difficult problem associated with this type of orbit is to calculate the time of the motion from the nearer apse to any other The problem is essentially that of point on the path (Fig. calculating the area of the sector of the elliptical path. Let be the point on the auxiliary circle such that is perpendicular to the major axis ', meeting it in N. If is the centre of the ellipse we shall denote the angle by If 0 represents the Sun and is any planet, is called the and then = e Each element of the sector such as YZ is the (or Y'Z'), is times the corresponding element XZ (or X'Z') of the area of the auxiliary circle. Hence

P

A

4.8).

AA

E real anomaly. bja OAQ b area OAP = a

OAP PQ C QCA E. eccentric anomaly

Q

OAP,

LPOA

OAQ ab (sector ACQ - t:.. OCQ) = � (ia2E - i;aOC sin E) a (4.38) = iab(E - e sin E). -

X

area

=

-

P

100

A C O U R S E IN A P PL I E D M A T H E M AT I C S

[CH.

We have already proved that h = If t is the time required for the radius vector to sweep over the sector OA P, equation (4.36) shows that

bv'fL/a.

;!_

t This is

Kepler's Equation.

a• (E - e sm. E) .

(4.39)

= --"1

fL"

FIG. 4.8.-Time in an Elliptical Orbit

E

t may be calculated from Kepler's equation only when has been determined. It is the real anomaly e which is usually available, however, and we therefore require a formula giving in terms of e. Now r cos e = ON = CN - CO = cos (4.40) Hence, employing equation (4.17), l cos e 1 + cos e = cos (1 e2)cos e cos or (4.41) l + cos e Putting 1 tan2 te ' cos E = 1 tan2 tE' cos e = 1 + tan2 te 1 + tan2 tE we find that

E a E - ae.

-

e e

_

-

-

tan

a( E - e), E e. -

!E R, tan te.

(4.42)

=

This equation is convenient for the calculation of

E.

P O L A R R E S O L U T E S O F A C C E L E RA T I O N

4]

101

A particle moves in an elliptical orbit of eccentricity e under the action of an inverse square law of force. If the major and minor axes are AA', BE' respectively, A being the apse nearer the centre offorce, show that the times taken to traverse the arcs B'AB, BA'B' are in the ratio 2e : + 2e. (L.U.) The ratio of the times of description of the arcs B'AB and BA'B' is the same as the ratio of the shaded area to the unshaded area in the diagram. If a, b are the semi-major and semi-minor axes respectively, shaded area = area semi-ellipse ABE' - area !'::,. OBB' = i1rab !BB' OC = !1rab - abe = !ab(1r - 2e).

Example 7.

1r -

1r

-

B

A

c

a'

Similarly, unshaded area = + The ratio of the areas is accordingly

!ab(1r 2e).

1r

(1r - 2e) : ( + 2e).

4.5. Kepler's Laws

Analysing the results of observation of the motions of the planets, Kepler came to the following conclusions :

(a) The planets describe ellipses with the Sun at a focus.

(b) The area swept out by the line joining the Sun to a planet is proportional to the time of its description. (c) The squares of the periodic times of the planets are pro­ portional to the cubes of their mean distances from the Sun.

Kepler's Laws.

These are With the aid of the analysis developed in this chapter we may deduce from each of these laws one important characteristic of the forces acting on the planets due to the Sun. Thus, taking the Sun as pole of polar coordinates 6) and employing the notation introduced in the first section of this chapter, the second = constant. This implies law is equivalent to the equation A =

(r,

that

�� (r28)

!r2B

=

0, and hence that the force acting on a planet has zero

[CH.

A COURSE I N APPLIED MATHEMATICS

102

transverse resolute, i.e., is directed along the line joining the planet to the Sun. The first law now implies that this force is one of attraction and has a magnitude varying inversely as the square of the distance between the planet and the Sun. The details of this deduction have been explained in Ex. l. Knowing that the force is o f the inverse square type, we can now derive equation (4.37) for the periodic time T in an orbit. Let Tv T2 be the periodic times of two planets moving in elliptical orbits having semi-major axes and respectively. Then, if and are the respective attractions per unit mass acting on these planets due to the Sun,

a1

f!.1/r2

a2

f!.2/r2

(4.43)

i.e.,

e1

a1 ae1) a1

If is the eccentricity of the orbit of major axis 2 v the least and greatest distances from the Sun in this orbit are (1 and (1 + respectively. The mean distance from the Sun is accordingly Similarly, is the mean distance of the second planet from the Sun. By the third law therefore, -

a2

T_12 = a___!_s_ . T22 azs

This equation, in conjunction with equation

e1) a1.

(4.44) (4.43),

implies that

= !-Lz· f!.1 The fundamental content of the third law is therefore that the attraction per unit mass due to the Sun at a distance r is of magnitude !-L/r2, being a constant for the solar system independent of the body !-L

being attracted. All of the above conclusions were summarized and generalized by Newton in his This states that

Law of Gravitational Attraction. any two particles attract one another with equal forces acting along the line joining them and that the magnitudes of these forces are proportional to the masses of the particles and inversely proportional to the square of the distance between them. The equality of the forces acting on the two particles is a consequence of Newton's third law of motion (p. 21). If m1, m2 are the masses of the particles and r is the distance between them the magnitude F of the force acting on either particle is given by the equation F = ymrl2m2, (4.45) where is a constant of proportionality known as the Gravitational Constant. In c.g.s. units, = 6·66 I0-8. It follows that if S is y

y

X

POLAR RESOLUTES OF ACCELERATION

4]

103

the mass of the Sun and P the mass of a planet, the force of attraction acting on the latter body when at a distance from the Sun is i.e., a force per unit mass of the planet. Thus, for this planet, flo = and flo is seen to be a constant for all the planets of the Solar system as deduced from Kepler's third law. There will, however, be an equal force acting on the Sun, causing it to experience an acceleration This acceleration will be small by comparison with the acceleration experienced by a planet, since P is always very small compared with (for Jupiter, the most massive planet, 5/P = 1050). The assumption, implicit in the above argument, that the Sun can be regarded as a stationary body is therefore a valid approximation, but an accurate discussion of planetary motions must make an allowance for the Sun's motion. This discussion we shall defer until Section 8.3, where it will appear that, assuming the funda­ mental validity of Newton's law of gravitation, the first two laws of Kepler are accurate consequences, whereas the third law is only approximately so.

y5

r

y5jr2

y5Pjr2,

y5Pjr2 yPjr2• 2 y5jr 5

4.6. The pr-Equation of an Orbit.

Elliptic Harmonic Motion Consider again the plane motion of a particle of mass m subjected

f(r)

to an attractive force per unit mass acting towards a fixed point 0 (Fig. 4.9). If cf> is the angle between the radius vector OP and

FIG. 4.9.-The pr-Equation

the direction of motion, by resolving along the normal to the orbit we obtain the equation of motion

mf(r) sin cp mKv2, =

(4.46)

where K is the curvature of the path at P. Let p be the perpendicular from 0 to the line of motion through P. Then sin cp = pfr and

104

A C O U R S E IN A P PLI E D M A T H E M A T I CS

K = dpjrdr. Also, from equation (4.5), v = hjp. in equation (4.46), we obtain h2 dp = f(r), · pa dr

[cH.

Hence, substituting

(4.47)

an equation which permits the calculation of f(r) when the relationship between p and r for the orbit is known. Integrating equation (4.47) with respect to r, we find that h2

j)2

=

constant

This is the pr-equation of the orbit. tion (4.48) yields the equation v2

=

constant

-

2fj(r)dr. .

Substitution of -

2Jj(r)dr, .

(4.48) h

=

pv in equa­

(4.49)

which is the energy equation (4.24). Conversely, we are provided with an alternative mode of derivation of the pr-equation from the energy equation. y

FIG. 4.10.-pr-Equation of an Ellipse

As a particular case, we will suppose that j(r) = w2r, i.e., the attrac­ tive force is proportional to the particle's distance from 0. This is the problem already solved by another method in Section 3.3. Equa­ tion (4.48) takes the form

(4.50 )

4]

105

P O L A R R E S O L U T E S O F A C C E L E RA T I O N

We shall show that this is the pr-equation of an ellipse relative to its centre as pole. Let the Cartesian equation of an ellipse be expressed in the standard form (4.51) The equation of the tangent at the point P(X, Y) (Fig. 4.10) is Xx + Yy = 1 b2 a2 '

(4.52)

and the perpendicular p from the centre (0, 0) to this line is accordingly given by the equation (4.53) X2 Y2 Now 2 + b2 = 1 and hence, a

where X2 + Y2 = OP2 = r2 . The pr-equation of the ellipse is accordingly (4.54) Comparing this equation with equation (4.50), we observe that they are identical if a and b are chosen so that

h = wab, C = w2 (a2 + b2) .

(4.55) (4.56)

It follows that the orbit represented by the equation (4.50) is an ellipse with its centre at the centre of attraction, as previously deduced in Section 3.3. Putting h = pv in equation (4.50) and substituting for C from equa­ tion (4.56), we obtain the energy equation for this type of motion, viz., v2 = w2 (a2 + b2 y2 ) . (4.57) _

This is equation (3.48) again.

A C O U R S E IN A P PLI E D M A T H E M A T I C S

106

[CH.

During the periodic time T, the radius vector sweeps out the whole area of the ellipse rrab. Hence

rrab y - zlh - 21 wab ,

from which it follows that

T = 2rr.

(4.58)

(J)

This equation reveals that the time taken to make a complete circuit of an orbit depends only upon the constant w governing the attractive force and is independent of the particular orbit being described.

A particle moves in a circular orbit under the action of a central force directed towards a point on the circumference. Prove that the force is one of attraction varying inversely as the fifth power of the distance. Let 0 be the centre of attraction (see diagram) and let OA be a diameter of the circular orbit. If P is the position of the particle at any instant and ON is the perpendicular from 0 on to its line of motion, LOPN = LOAP = q,. Hence p = r sin q, = r . OPJOA = r2f2a, where a is the radius of the orbit.

Example 8.

Thus

dp = � and h'J_• r!P_ h2r = 8a"h" dr a p dr ap• r5 and, by equation (4.47), f(r) = 8a2h2fr5• Thus, the attractive force must vary inversely as the fifth power of r. =

4.7. Motion on Rotating Wires

Suppose that a smooth straight wire is compelled to rotate in a horizontal plane about a vertical axis passing through a point 0 of its length with constant angular velocity w and suppose that a bead P

4]

POLAR RESOLUTES O F ACCELERATION

107

of mass m is free to move along the wire. At time t, let r be the distance of the bead from 0 and let e be the angle made by the wire with a fixed direction. The reaction of the wire upon the bead will act in a direction perpendicular to the wire and, since the bead has no resolute of acceleration vertically, the component of the reaction in this direction will exactly balance the weight force mg. Let R be the R

FrG.

4.11.-Bead on a Rotating Wire

resolute of this reaction in the plane of the motion and perpendicular to the wire (Fig. 4.1 1) . Resolving along and perpendicular to OP, we obtain the equations of motion m(r

But

- r62) = 0, '; :e (r26) = R.

(4.59)

- w2r = 0,

(4.60)

e = w (a constant) and hence, these equations are equivalent to r

R = 2mwr.

The first of the equations (4.60) determines the motion of the bead on the wire and the second specifies the reaction R at any instant. The first of equations (4.60) is a linear equation with constant coefficients having characteristic roots ± w . The general solution of this equation is accordingly (4.61) where A and B are arbitrary constants to be determined by the initial conditions. Assuming that the bead is stationary on the wire and distant a from 0 at time t = 0, the initial conditions are r = a, r = 0. Hence A + B = a, wA - wE = 0, (4.62) yielding A we have

=

B = fa.

Substituting these values in equation (4.61), (4.63)

A C O U R S E IN A P PL I E D M A T H E M A T I C S

108

[cH.

determining the manner in which the bead recedes from 0 along the Wire. Substituting for r from equation (4.63) into the second of equations (4.60), we now obtain (4.64) R = 2maw2 cosh wt . The following two examples illustrate more difficult problems in which additional forces act upon the bead.

A smooth wire is constrained to rotate in a verticalplane about a point 0 of its length with constant angular velocity w. At time t = 0 the wire is horizontal and is moving downwards. At this instant a bead, threaded on 0 ...--::-.----­ the wire, is released from 0. If r is the distance of the bead from 0 at time t, prove that r = /w• (sinh wt - sin wt) . Show also that the reaction of the wire at this time is given by R = mg(cosh wt 2 cos wt).

Example 9.

-

If 6 is the angle made by the m9 wire with the horizontal at time t, then 6 The forces acting on the bead are the reaction of the wire perpendicular to the wire, and the vertical weight Resolving radially and transversely, we obtain the equations =

wt.

R, mg. d (r2o)" = R + mg cos 6. r62) = mg sin 6, m m (r r lit Since 6 = wt, we have (i) r - w2r = g sin wt, R = 2m wr - mg cos wt. The first of these equations is a linear equation forr, having CF Aewt + Be-wt. A PI is found to be 2!,2 sin wt. The general solution is accordingly _ff_ sin wt r = Aewt + Be-wt _ 2w2 (ii) Initially, t = 0, r = r = 0. Hence A + B = 0, Aw - Bw - fw = 0, from which we deduce that A = !', • B = 4!',2 Substitution of these 4 2 values for A and B in (ii) yields the required solution for viz., r = 2gw2 ( ewt -2 e-wt - sm. wt) = !,. (sinh wt - sin wt). 2 -

-

•

-

r,

It now follows that

r

=

fw (cosh wt

-

cos

wt).

Whence, from the second of equations (i), we obtain the stated formula for

R.

4]

POLAR RESOLUTES O F ACCELERATION

109

A narrow straight tube of length 2a is rotating in a plane about one end with constant angular velocity w. Inside the tube is a small bead of mass m which is instantaneously at relative rest at the mid point. The coefficient of friction between the bead and the tube is f· Neglecting the effect of gravity, calculate the normal reaction between the tube and bead at time t after the instant of relative rest. Show that the particle reaches one end of the tube after a time 2 log 2·5/w approximately. (M.T.) If R is the normal reaction, !R is the frictional component of the reaction. Let r be the distance of the bead from the centre of rotation 0 at time t. With a = wt, the equations of motion are m(� - w2r) = -!R, 2mwr = R. (i) Eliminating R between these equations, we obtain {2D2 + 3wD - 2w2)r = 0, where D dfdt. The characteristic roots of this equation are found to be !w and -2w. The general solution is accordingly r = A elwt + Be-2wt. (ii)

Example 10.

==

The initial conditions at = 0 are given by = from equation (ii) that these imply A + = !wA - 2wB = 0. Thus A = 4 5, B = f5 and hence,

af

a

r a, r = 0.

t B a,

r = �(4eiwt + -e-2wt) .

We deduce

(iii)

Suppose that the particle reaches the end of the tube at = T. At this instant = Hence, employing equation (iii),

t

r 2a.

2a = �(4etwT + e-2wT) 5 4elwT + e-2wT = 10. (iv) Since is positive, e-2wT will be less than unity, whereas 4e!wT will be greater than 4 . Assuming, for the moment, that e-2w T is negligible by com­ parison with 4elwT, equation (iv) is approximately 4eiwT 10 and hence = log 2·5/w very nearly. For this value of e-2wT = (2·5)-• = 0·0256, whereas eiwT = 2·5. Thus e-2wT is only 1 % of eiwT and the above approximate procedure is justified.

T

=

T 2

T,

1 10

A C O U R S E I N A P PLI ED M A T H E M A T I C S

[cH.

Substituting for from equation (iii) in the second of equations (i), we calculate the required formula for R, viz., R=

r

tmaw2 (elwt - e-2wt).

We conclude this Section b y solving a problem i n which two particles are involved. The method is to deal with each particle separately, writing down appropriate equations of motion in each case. A smooth straight wire AB is made to rotate about A in a horizontal plane with constant angular velocity w. A small bead of mass m slides on the wire. The bead is attached to a light inextensible string, which passes over a pulley at A and then hangs vertically supporting a particle ofmass 3m. Initially the bead is at distance a from A and its radial velocity is zero. Show that it reaches A after a time w2a) _! cosh-1 ( 1 3g w (B.U.) has elapsed. What happens if a > 3gfw2 ? Let r be the distance of the bead from A at time t and let T be the tension in the string. Then the radial equation of motion of the bead is (i) m(r - w2r) = -T. The acceleration of the mass 3m will be ii in an upward sense and hence its equation of motion is (ii) T - 3mg = 3mr. Eliminating T between equations (i) and (ii), we deduce that ·,y !-w2r = -!g. This equation has the general solution r = A cosh iwt + B sinh iwt + w3�. Initially t = 0, r = a, r = 0 and thus r = w3g2 - (w3g2 - a) cosh iwt. (iii) Assuming that a < 3gfw2, this indicates that r decreases steadily. When r = 0, 2a -1 cosh iwt = ( 1 � ) g 2 2 1 t = w cosh- ( 1 - w3ga ) If a > 3gfw2, equation (iii) shows that r increases indefinitely and thus the

Example 11.

-

-1

-

-

-

-

1

bead never reaches A .

4.8. Motion on Surfaces of Revolution

Suppose that a smooth surface of revolution is fixed with its axis Oz vertical and that a particle P of mass m moves upon it under the action of gravity and the reaction R of the surface alone (Fig. 4.12). Let (r, e, z) be the cylindrical polar coordinates of P. By consideration of the motion of the projection of P on the xy-plane, we conclude that

4]

POLAR RESOLUTES O F ACCELERATION

Ill

the components of the particle's velocity along and perpendicular to NP are respectively. In addition, the particle has a velocity component of z in the direction due to its vertical motion. We can now write down the energy equation in the form

(r, re)

Oz tm(r2 r282 z2) mgz +

+

z

R

+

=

constant.

(4.65)

FIG. 4.12.-Particle on Smooth Surface of Revolution

The particle's components of acceleration comprise the familiar polar components in the horizontal plane along and perpendicular to NP and a vertical component z. Hence, resolving horizontally and in a direction perpendicular to NP, we obtain the equation

h

m- -d (r26) r dt

=

0 or

r26 h' =

(4.66)

where is a constant equal to the moment of the particle's velocity about the axis Eliminating e between equations (4.65) and (4.66), we find that

Oz.

r2 �r .i2 +

+

=

c

-

2gz.

(4.67)

Suppose that the equation of the surface is given in the form

r = f(z). Then r = f' (z)z and hence equation (4.67) is equivalent to z• 2 C 2gz f-'2h2/f2 -_ G(z) . _

-

-

l

+

(4.68)

(4.69)

A C O U R S E IN APPLI E D M A T H E M A T I C S

112

[CH.

an argument identical with that which was applied to equation (4.By 35), we now deduce that, if z1, z2 are two consecutive zeros of G(z) and the initial value of z lies between them, then during the subse­

quent motion z will oscillate between these values indefinitely. Thus, the particle will rise and fall between two horizontal circles which are the sections of the surface by the planes = and = As an example, consider the case of a particle moving on the interior of a smooth sphere of radius Taking 0 at the centre, the equation of the sphere is

a.

z z1 z z2•

r = (a2 - z2)!-.

Hence, = may be written

j(z) (a2 - z2)!, f'(z) = -zj(a2 - z)!- and

(4.69) (4.70)

equation

G(z) are given by the roots of the cubic equation (C - 2gz)(a2 - z2) - h2 = 0. (4.71) If = z0 initially, z2 must be positive for this value of and hence also must be the left-hand member of equation (4.71). This member is accordingly negative for z = -a, positive for z z0, negative again for z = a and positive for z = Hence the roots of equation (4.71), viz., zv z2 , z3, lie in the intervals (-a, z0), (z0, a), (a, ) respectively and the particle will oscillate between the planes z = zv z = z2 • It has been tacitly assumed in the preceding theory that R does not The zeros of Z

Z

SO

=

C¥J.

oo

vanish during the motion, for if it does the particle will leave the surface. To check this we require a formula for R. Let !f; be the angle made by the normal to the surface with the z-axis (Fig. Then, resolving vertically, we obtain = R cos t/; -

4.12).

mi

mg.

(4.72)

(4.69) with respect to t yields = m(g + fG' ) sec !f;. (4.73)

But, differentiation of equation G ( z. Hence R= '

z)

2zi

Since 1/J can be obtained in terms of z, this latter equation specifies R in terms of this variable. Thus, in the case of the spherical motion, cos 1/J = (negative sign since 1/J is acute when is negative) and hence

z

R=

m(C - 3gz)ja. We observe that R is positive provided z < Cj3g.

-z/a

(4.74)

A particle moves on the inside of a smooth sphere of radius a. It is projected with speed y(3ga) along the horizontal great circle. Show that the motion is confined to the part of the sphere between the horizontal great circle and a horizontal circle with centre at a depth !a below the centre of the sphere.

Example 12.

(S.U.)

4]

POLAR RESOLUTES O F ACCELERATION

1 13

In this case = Hence equation

C 3ga, h = ay'(3ga). (4.7 1) takes the form z(2z + a)(z - 2a) = 0 and the roots are 0, -!a and 2a. It follows that the particle moves as stated provided R is positive for values of z in the interval (-!a, 0) . But, by equation (4.74), R = 3mg(a - z)fa and R is positive for all z < a. Example 13. A paraboloid of revolution having equation r2 = 4az is fixed with its axis vertical and vertex downwards. The inner surface is smooth and a particle of mass m is projected horizontally from a point on it at the level of the focus with speed 2(ag)l. Show that the particle will move between two circles of radii 2a and 2y'2a. Prove also that the reaction R of the surface is given by the equation � R = 6mg ( 1 + �r The conditions of projection require that C = 6ag, h = 4a(ag)l. Also, f(z) = 2(az)t. Hence, equation (4.69) shows that .z.• 2g(2a - z)(z - a) G(z). = z+a G(z) is zero when z = a or 2a and hence the particle moves between the two horizontal circles at these levels. Also, rp is the angle made by the tangent to a vertical section of the surface through its axis and the horizontal. Hence tan rp = Equation

dz = r = (zfa)i, 2a r ( sec rp = l + � · dr

(4.73) now yields the stated formula for R.

EXERCISE 4 I. A particle is describing an elliptic orbit about a centre of force at one of its foci. At a point P on the orbit, distant c from 5, the centre of force, the tangent to the orbit is inclined at an angle sin-1 (3/5) to 5P, and at a point Q on the orbit, distant 3cf2 from 5, the speed of the particle is one-half of its speed at P. If the intensity of attraction on unit mass at unit distance is fl., find the speed at P, and prove that the major axis of the orbit is 9c/5 and find its eccentricity. (L.U.) 2. A comet is describing a parabolic orbit about the Sun and at its closest approach to the Sun the velocity V in the orbit is suddenly decreased to n V (n < 1), without change of direction of motion. Show that subsequently the comet describes an ellipse and that the ratio of its least to its greatest distance from the Sun is l/n2 I. (D.U.) 3. A particle P is describing a circle under a law of force [L/OP2 per unit mass directed towards the centre 0. It 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 ec­ centricity 3/4. Show also that the periods of description of the circle and the ellipse are in the ratio 7y7 8. (L.U.) -

[cH. A C O U R S E I N A P PLI E D M A T H E M A T I C S 114 4. A particle of mass m is describing an ellipse whose principal axes are of lengths 2a and a, under the action of a force fLfr2 per unit mass, to

one of its foci. 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

{�

(3 ±

y2)r

(L.U.)

5. A particle P moves under the action of a force fLfSP2 directed towards a fixed point S. 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 with respect to the time is

2(w1w 2) !/ w1t + w

(L.U.)

6. A particle is describing a circle of radius under an attractive force per unit mass directed towards the centre. An outward radial velocity is suddenly impressed on it in such a way as to leave the transverse velocity unaltered ; prove that it will describe an ellipse and find the periodic time. (L.U.)

a

fLfr2

(fLf5a)t

7. A particle is describing an ellipse of eccentricity e about a centre of force at a focus. When it is at one end of the minor axis its velocity is, doubled. Prove that the new path will be a hyperbola of eccentricity y(9 (L.U.) -

8.

8e2).

and under A particle of unit mass is describing an ellipse of axes an attraction fLY to the centre C when it is at a distance from C. When the particle is at a distance 3b/2 from C the intensity of attrac­ tion is suddenly increased in the ratio Prove that the axes of the new orbit are (y79 ± (L.U.)

4b

y15)bf4.

r

2b

1 : 4.

9. A particle of mass m is describing an ellipse of major and minor axes and respectively about a centre of force at the centre. When it reaches the end of the major axis it strikes and coalesces with a which is at rest. The central attraction per unit particle of mass mass is unchanged. Prove that the new orbit is an ellipse of major and minor axes and + respectively. (L.U.)

2a

2b

nm 2a 2bf(n 1)

10.

Particles P and Q of equal masses describe the same elliptical orbit under an attractive force directed towards the centre. If, at any instant, OP and OQ are conjugate semi-diameters, show that they will always be conjugate semi-diameters and that the sum of the KEs of the particles is constant. (L.U.)

1 1. A comet describes a parabola about the Sun as focus.

Show that its velocity perpendicular to the axis of its orbit varies inversely as the radius vector from the Sun. (L.U.)

12.

A particle of mass moves in a given plane about a fixed point 0 under the action of a force which is always normal to OP( = r) and of magnitude where is a constant. Initially the particle is at

m

mkifr,

k

4]

POLAR RESOLUTES OF ACCELERATION

1 15

rest at r = a. Prove that, in the subsequent motion, the speed V of the particle when at a distance r from 0 is given by a V 2 = 2k 2 + log !:: � . r aJ (L.U.) 13. At a point A on a smooth horizontal table is a particle attached to one end of an elastic string of natural length a. The other end of the string is attached to a point P on the table, the distance A P being ta. The particle is projected along the table, perpendicular to PA . In the subsequent motion c is the greatest length attained by the string. Prove that the initial KE of the particle is 2fc2 (c - a) 2/(4c2 - a2), where f is the force that gives unit increase of length to the string. (L.U.) 14. A particle P of unit mass is repelled from a fixed point 0 by a force fl.fr2, where r = OP. Prove that the orbit of P is a branch of a hyper­ bola of which 0 is the outer focus and that the speed v of the particle is given by v• =

{ -r

fl.G - �)·

where 2a is the major axis of the orbit. The particle is projected from infinity with velocity V along a line which is at a perpendicular distance p from the centre of force. Prove that the particle will eventually return to infinity along a direction making an angle 2 tan-1(f1./P V2) with the initial direction of motion. (L.U.) 15. A particle of unit mass is attracted towards a centre of force S with a force equal to a(u2 + 2cu3) where 1/u is the distance of the particle from S. It is projected from a point P at a distance c from S with a velocity 2a c at an angle of 60° with PS. Show from the formulre for the acceleration components in polar coordinates that d 2u 3 + tu = de• 8c' where 8 is the angle between SP and the radius vector to the particle, and find the distance of the particle from S when 8 = 120°. (M.T.) 1 6 . A particle P is moving in a plane under an attraction of magnitude 2fr3 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)! and 1 respectively. Show that r = 2fr3, and find r as a function of t. (L.U.) 17. A particle P, of mass m, is moving in a circle of radius a and centre 0 under a force [Lm(r + 2a3fr2) directed towards 0. If P is acted on by an impulse tangential to the path and of magnitude (3[Lm2a2) t , show that the velocity of P is immediately doubled and that the greatest and least distances from 0 in the ensuing motion are a and 3a. (L.U.)

,

:( t

116

A C O U R S E I N A P P L I E D MATH EMAT I C S

[CH.

18. If (lfu, 8) are polar coordinates of a particle which moves in a plane under the action of an attractive force fLU3 per unit mass towards the pole 0, and the particle is projected from a point P with velocity 2yfL/ay3 at right angles to OP where OP = a show that the particle describes the curve u = cos !8fa. 19. A light rod AB of length 4a is rotating in a horizontal plane about its fixed end A , when a smooth ring of mass m is projected towards A along the rod from the end B with speed V relative to the rod. Throughout the motion of the ring, the rod is constrained to rotate with constant angular velocity w. Prove that the ring will come to instant­ aneous rest at the mid-point C of AB if V = 2y3aw, and find the force exerted by the rod on the ring at time t during its motion from B to C. (L.U.) 20. A fine smooth tube rotates about a point 0 of itself with uniform angular velocity w in a horizontal plane. The tube contains a particle of mass m attached to 0 by a fine elastic thread of natural length a and modulus 2maw2• Initially the thread is just taut and the particle is at rest relative to the tube. Show that the particle makes one com­ plete oscillation in the tube in each rotation, and that the greatest horizontal pressure of the tube on the particle is 2maw2• (M.T.) 21. Employing the usual notation, show that the energy equation (4.24) can be written in the form ,

h2[(��r

+

J = constant - 2 jJ(r)dr.

u2

Hence, by differentiating with respect to 8, obtain equation (4. 10) . 22. A comet describes a parabolic path. Determine the time it spends within the Earth's orbit (regarded as circular) in terms of c, the least distance of the orbit from the Sun, a, the radius of the Earth's orbit, and T, the length of the Earth's year. (B.U.) 23. If a stream of water is moving with uniform angular speed in such a way that the force on each individual particle acts at right angles to the radius vector from the centre of rotation, show that the path of the particle is of the form r = aeO + be-0, and prove that the force at any point is proportional to y(r2 - c2), where c is a constant. (B.U.) 24. A particle of mass m on a smooth horizontal table is attached to one end of a string which passes through a small hole C in the table. When the particle is moving, the tension in the string is maintained by a force applied to the other end of the string. Initially the particle is moving with angular velocity 0 in a circle of radius a and centre C. The string is then pulled so that the distance of the particle from C decreases at a constant speed u. Obtain expressions for the variation with re­ spect to the time of : (i) the angular velocity of the particle about C, and (ii) the tension T of the string. Deduce the time taken for the radius vector from C to the particle to make one complete revolution starting from the instant when the radius first begins to decrease. (B.U.)

4]

P O L A R R E S O L U T E S O F A C C E L E RATI O N

1 17

25. A particle, subject to a central attractive force !J.U4 per unit mass, is projected from a point at a distance b from the centre of attraction with a velocity (2!J./b3)t at right angles to the radius. Show that the other apsidal distance is !b(yl3 - 1 ) . (Hint : Use the result of Ex. 21.) (S.D.) 26. Two particles of masses m, M, connected by a light inextensible string of length 2a, can slide in a long, smooth, straight tube. The tube is rotated in a horizontal plane about a point 0 of itself with constant angular speed w. Show that in the ensuing motion the tension of the string is constant ; and prove that if initially the particles were at rest with the string taut and with its middle point at 0, in a time t they will move along the tube a distance a(M m) (cosh wt - 1) (Li.U.) M+m 27. A particle, of mass m, is projected horizontally with speed V along the inner surface of a fixed smooth spherical shell of internal radius a. The point of projection is at a depth a/2 below the level of the centre of the spherical surface. Obtain the energy equation and show that the particle rises initially if V2 > 3agf 2. If the greatest and least depths below the level of the centre are aj2 and af4 respectively, show that when the depth of the particle is 3af8, the magnitude of the normal reaction of the surface is 2 1mgf8. (S.D.) 28. A right circular cone of angle oc is fixed with its axis vertical and vertex downwards. A particle of mass m is projected with velocity v horizont­ ally along the smooth inner surface of the cone from a point at a height h above the vertex. Derive the energy equation and from this show that, if v2 < gh, the particle will begin to fall below its original level. Taking v2 ghf6, find the height above the vertex of the cone of the lowest point of the path. (S.D.) 29. A smooth surface of revolution is formed by rotating the parabola y 2 = 4ax about the tangent at the vertex. The surface is fixed with Oy vertical and a particle is projected with velocity (2agf15) l horizont­ ally from a point vertically above a focus of a section. Show that the particle moves between two horizontal circles of radii a and ta. -

=

CHAPTER 5

IMPULSIVE MOTION OF PARTICLES 5.1. Impulse and Momentum

m

Consider a particle of mass which moves under the action of a force Let be its velocity vector at time Its equation of motion is therefore (5. 1 )

F.

v

t.

F = mv. . The vector mv is termed the momentum o f the particle. Denoting this quantity by p, it will be observed that equation (5.1) is equivalent to F = p, (5.2 )

i.e., the applied force is equal to the rate of change of momentum. This statement is an alternative interpretation of Newton's Second Law. Integrating equation (5.2) with respect to over some time interval we obtain

t

(t0, t1},

(5.3) p0 and p1 being the values taken by the momentum vector at the instants respectively.

t0, t = t1 ,Ft dt ist =a vector quantity called the impulse of the force F calculated f t,

{t0, t1).

over the time interval Equation (5.3) accordingly asserts that this impulse of is equal to the increase in momentum experienced by the particle during this time. In the application of equation (5.3) to particular problems, it will be convenient to equate the components of the impulse in any three directions to the corresponding components of the momentum vector. Thus, if represent the components of the velocity vector in three assigned directions (usually mutually perpendicular) and (X, Y, Z) are the corresponding components of the force vector,

F

(u, v, w)

will be the components of p the components of the impulse. the three equations

(mu, mv, mw) , t , dt, ftZ, dt) will be and (iX dt, itY to

to

to

Equation (5.3) is then equivalent to

fz dt = mwl - mwo. t,

118

(5.4)

[CH. 5]

I M P U L S I V E M O T I O N O F PARTI C L E S

119

The impulse equation (5.3) , though generally applicable, is par­ ticularly useful in circumstances where a large force acts upon a particle for a very short time, e.g., when the particle is struck a blow or collides with another particle. In such a problem, the magnitude of the force acting will rise very rapidly to a maximum value and then de­ crease to zero in an equally short time, as illustrated graphically in Fig. 5.1. It will usually be very difficult to specify precisely the F

FIG. 5.1.-Graph of Impulsive Type Force

manner of variation of F in such a case. If, however, the equation of motion (5.1) is to be employed to calculate the effect of the force, F should be a known function of the time. This difficulty is avoided by having recourse to equation (5.3). If t0 is the instant when the force is first applied and t1 is the instant when this application terminates, the left-hand side of this equation represents the impulse of the blow applied to the particle and is easily specified in terms of three numbers, its components. In the case of an impulsive force which varies after the manner indicated in Fig. 5.1, its impulse is measured by the area under this graph. It is clear, therefore, that although the maximum magnitude of the force may be very large, its period of application will be small so that its impulse will be a conveniently finite quantity. Given the impulse, equation (5.3) permits us to calculate the increment in momentum it causes and hence the change in velocity. Alterna­ tively, from an experimental determination of the velocity change, we can infer the impulse which must have acted. The specification of a blow by means of its impulse is clearly not so detailed a description as one in terms of the force and its mann('}r of variation with the time during the period of its application, Less information is therefore available from the impulse equation than can

120

A C O U R S E IN APPLI E D M A T H E M AT I C S

[CH.

be obtained from the equation of motion (5.1). Thus, although the impulse equation will determine the velocity change caused by the blow, it will not permit a calculation of the particle's displacement during the blow. This can be assessed only from the fundamental equation of motion, since it depends upon the manner of variation of the force constituting the blow. It is a common experience, however, that the action of the blow is very nearly instantaneous, so that the displacement observed during the infinitesimal time interval of its application is very small and can usually be neglected. We shall accordingly regard the effect of an impulsive force acting on a particle to be an in­ stantaneous change in velocity of the particle with no concurrent change in its position. A particle of mass m is attached to a fixed point 0 by an inelastic string of length a. Initially it is on the same level as 0 and at a distance ia from this point when it is projected vertically downwards with velocity U. Show that if the particle eventually describes complete vertical circles about 0, u• � 3ga(4 + v3). The string becomes taut when the angle made by the 0 string with the vertical is 30° as shown in the diagram and the particle is at P. At this instant the particle has fallen through a distance h = iv3a and its velo­ city is where u• = u• + 2gh = u• + y3ga.

Example 1.

u,

The string being inelastic, the particle's velocity is now transformed instantaneously from u vertically to a value at right angles to OP. This velocity change is caused by an impulsive tension set up in the string. Resolving along and perpendicular to OP and employing equations we have cos sin (i) It will be noted that the impulse of the gravitational acting on the particle has been omitted from force these equations. If -r is the duration of the impulse, the impulse of is and this will be negligible, since -r is very small. In general, in such problems U the impulses of ordinary finite forces may all be disregarded. The first of equations (i) defines The second of these equations shows that and hence + Let V be the particle's velocity when it is vertically below 0. The loss of PE in falling from P to this point is - cos From energy considerations it therefore follows that

v

I

(5.4), I = mu 30°, 0 = mv - mu 30°. mg mg mg-r,

I. v2 = t{U2 y3ga). mga(l 30°) = mga(l - iv3). tm v• - imv• = mga(l - h/3) v• = t(U"• + -yl3ag) + 2ga(1 - tv3) = tu + ga(2 - iv3). (ii) On p. 75 we have shown that the particle will now describe complete vertical circles about 0 if v• � 5ga. The condition required is accordingly iU" +• ga(2 - iv3) :;;?: 5ga, u :;;?: 3ga(4 + y3). or v = iu

5]

I M P U L S I V E M O T I O N O F PARTI C L E S

121

5.2. Newton's Third Law

Hitherto we have been principally concerned with the motion of a single particle. If a number of interacting particles are involved in a problem an additional hypothesis concerning the forces which are exerted between the particles must be introduced before we are in a position to arrive at a solution. Such an hypothesis is provided by Newton's Third Law, which has been stated on p. 2 1 . It is a direct consequence of this law that if two particles in motion are connected by an inelastic string which suddenly becomes taut, for a short period of time equal and opposite forces acting in the direction of the string will be applied to the particles. These will accordingly subject each other to equal and opposite impulses. We shall make use of this fact in the examples which follow.

Two particles A and B, each of mass m, are attached to the ends of a light, inextensible string of length l, and B moves in a smooth straight groove in a smooth horizontal table. Initially A is at rest in the groove, AB being perpendicular to the groove and of length il. If A is projected with velocity v along the table and parallel to the groove, show that, when B begins to move its velocity is 3vf1. Find the impulsive tension in the string and the impulsive (L.U.) reaction of the groove. When the string becomes taut, AB makes an angle of 30° with

Example 2.

the groove. The groove being J smooth, it can only exert a force on B in a direction perpendicular to its length. Let ] be the im­ pulse of this force when the string becomes taut. Let be the im­ pulsive tension in the string exerted on both particles as shown in the diagram. Let be B's speed when its motion commences. Resolving A 's velocity into components along and perpendicular to A B, the component along BA must equal the component of B's velocity in this direction, viz., u cos Resolving the change in the IL c os 30• momentum of A along and perpendicular to AB and equating the component along AB to the corresponding component of impulse, we obtain the equation cos cos (i) Resolving along and perpendicular to the groove for the impulses acting on B, we find that (ii) cos = (iii) ] sin Elimination of between equations (i) and (ii) shows that Then, from (ii), Hence, from (iii),

I

u

30°.

I = m(v 30° - u = imv'3(v - u) .

mu = I 30° !Iy3, = I 30° = !f. I I = 2y'3mvf1.

30°)

J = y'3mvf1.

u = 3vf1.

A C O U R S E IN A P P L I E D M A T H E M AT I C S

122

[cH.

Three particles A, B, C, of masses 3m, m, 2m respectively, lie on a smooth, horizontal table, and the angle ABC is IX (< !,.). The particles A and B and the particles B and C are connected by light, inextensible strings which are just taut. A horizontal impulse P is applied to B in the direction AB and the initial velocities of A and C are u and v respectively. Show that the com­ ponents of the initial velocity of B are u along AB and (u cos IX - v)fsin IX perpendicular to AB, and prove that + 2 sin2 IX) . u - P(l 6m(l + sin2 ) (L.U.)

Example 3.

_

IX

Let I, 1 be the impulsive tensions in the strings AB, BC respectively. Since I is the only impulse to act upon the particle A, it commences to move in the direction AB with velocity Similarly, C begins to move in the direction CB with velocity Let be the com­ ponents of B's velocity along and per­ pendicular to AB. The component in the direction AB must equal the com­ ponent of A 's velocity in this direction, i.e., = Similarly, if B's velocity is resolved along and perpendicular to BC the component along CB must equal Thus = cos IX - sin IX, or sin IX = cos oc - = cos IX v, U COS IX - V = n_IX_ -� si� Resolving the impulses acting on B A 3m along and perpendicular to AB and equating to the momentum increments in these directions, we obtain - I - 1 cos IX = (i) 1 sin IX = cos IX - ) si n IX (ii) Since the impulses I and 1 generate velocities and in the particles A and 1 = Hence equations (i) and (ii) and C respectively, I = are equivalent to = + cos IX, (iii) cos IX = + sin• IX) . (iv) Elimination of between these equations now yields the stated result for

u. (p, q)

v.

p u.

q

v.

v p p q

q v u

P

u.

v

-

m(u

3mu 2mv. P 4mu 2mv u v(l 2

mu, . vf u

v

.

Another type of problem in which equal and opposite impulses are applied between two interacting bodies is that in which a missile is fired from a gun. During the short time interval the missile is moving down the gun barrel the pressure of the propellent gases exerts equal and opposite forces on these two bodies. Equal and opposite impulses are accordingly applied to the missile and the gun, causing the former to be launched into its trajectory and the latter to recoil. 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

Example 4.

5]

IMPULSIVE MOTION O F PARTICLES

123

(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 to the direction in which the truck is moving, show that the shell has a horizontal range. 2v sin {u + Mv cos } g M+m and that the distance between the shell and truck when the former lands is independent of M and m. (L.U.) a

a

a

Let I be the impulsive reaction between the shell and the gun, acting at an angle {3 to the horizontal (see diagram) . The smooth track will exert an

�'--

J

impulse j on the gun in a vertical direction (since it is smooth) . Let be the velocity of the gun in a forwards direction after firing. Resolving horizontally for the impulses applied to the gun and truck, we obtain the equation I cos {3 (i) The horizontal and vertical components of the shell's velocity before firing are respectively and after firing are cos a, sin a . Resolving horizontally for the impulsive motion of the shell, we obtain the equation I cos {3 cos a (ii) Equations (i) and (ii) show that cos a.

V

= M(u - V).

(u, 0)

(V + v

v

)

= m(V + v - u). V = u _ mv M+m

The initial horizontal and vertical components of the shell's velocity are accordingly cos a

u + Mv M + m ' v sin a

respectively. Thus, if T is the time of flight of the shell to the point where it strikes the horizontal plane, = sin cc - T2, sin a. = i.e.,

0 vT T 2v g

!g

124

A C O U R S E IN A P PLI E D M A T H E M A T I C S

[cH.

The range o n the horizontal plane i s now found to be 2v sin IX + cos u (iii) g as stated. During the time the truck moves a distance cos IX 2v sin IX u (iv) = . g The difference between the expressions (iii) and (iv) is 2v2 sin IX cos 1Xfg and represents the distance between the shell and truck when the former lands . We have verified that this distance is independent of and m.

{ Mv IX} M+m

T,

VT

(

_

mv ) M+m

M

5.3. Impact of a Pair of Particles

If two particles are in collision with one another each exerts a large variable force on the other for a short period of time and, by Newton's Third Law, these two forces are always equal in magnitude but opposite in sense. It follows that the impulses of these forces are also equal in magnitude but opposite in sense, and · hence that the momentum increments for the two particles are identical apart from sign. If pi> p2 are the momenta of the particles immediately before collision and bop is the increment in momentum of the first particle due to the col­ lision, then - bop is the increment in momentum of the second particle and p1 + bop, p2 - bop are the momenta of the particles immediately after the collision. It will now be observed that the vector sum of the momenta after the collision is equal to the vector sum of these quantities before the collision, i.e., the total momentum is conserved in the col­ lision. This is the restricted form of the Principle of Conservation of Momentum which will be proved in a more general form in Chapter 8. If we take components of the momentum vectors in any direction it follows from the principle just enunciated that the sum of the com­ ponents before collision is equal to the sum after collision.

(9 (9 C9 C9 FIG. 5.2.-Impact of Smooth Spheres

In the particular case when the particles are small smooth spheres the forces between the spheres will be directed along the line of centres. It follows that the impulses will also be in this direction, and hence neither sphere experiences any change of momentum in a direction per­ pendicular to the line of centres, i.e., the components of p1 and p2 perpendicular to the line of centres are unaltered by the collision. Suppose the spheres are in motion in the same plane. Let (uv v1)

5]

I M P U L S I V E M O T I O N O F PART I C L E S

125

be the velocity components of one sphere before collision in directions along and perpendicular to the line of centres respectively. Let (u2 , v2) be the velocity components of the second sphere in the same directions and at the same instant (Fig. 5. 2) . Let (u1 ', v1 '), (u2 ', v2 ') be the corresponding velocity components immediately after collision. If m1, m2 are the masses of the spheres the above reasoning implies that (5.5)

or

Also, by considering momentum components along the line of centres and applying the principle of conservation of momentum, we obtain the equation (5.6) The velocities of the spheres before collision being given, the three equations (5.5) , (5.6) are insufficient to determine the motion after collision as specified by the four velocity components u1 ', v1 ', u2 ', v2 '. A fourth equation expressing the elastic properties of the materials of the spheres in mathematical form is clearly necessary. This is provided by Newton's Law of Restitution, which states that the ratio of the component of the relative velocity along the line of centres after collision to the same quantity before collision is a constant depending only upon the elastic properties of the spheres. Since these relative velocities are clearly in opposite senses, the ratio is negative in value and is denoted by -e. e is called the Coefficient of Restitution. Symbolically, we have (5.7) This law of Newton is not on the same fundamental plane with his other laws. Experimental evidence indicates that it is of an approxi­ mate nature only. However, it is sufficiently accurate for most purposes. The total KE of the spheres before collision is lm1 (u1 2 + v12) + fm2 (u22 + v22). Subtracting the corresponding expression for the total KE after collision and using equations (5.5), we find that the loss in KE due to the collision is

tm1u12 + tm2u22 - tm1u1 ' 2 - tm2u2 '2 1 [ + 2(mi + m2) mi2ui2 m22u22 - ml2ui '2 - m22u2 1 2 + mlm2 (u12 + u22 - ut ' 2 - u2 ' 2)J .

(5.8)

Squaring equation (5.6) , we obtain

m12u12 + m22u22 - m12ut '2 - m22u2 '2 = 2mJ m2 (ut 'u2 ' - ulu2),

(5.9)

A C O U R S E IN A P P L I E D M A T H E M A T I C S

126

(CH.

showing that the expression (5.8) may be written in the equivalent form

m1m - 2(m1 + 2m ) (ul 2 _

-

u2) 2 ( 1

-

e2 ) ,

(5.10)

having employed equation (5.7) in the final reduction. Since the loss in KE given by the expression (5. 10) must be positive, e is numerically less than unity. This is verified by experiment. If e = 1 , there is no loss of KE and the spheres are said to be perfectly elastic. If e = 0 the loss of KE is a maximum and the spheres are said to be inelastic. In these latter circumstances they do not separate after collision. The energy lost is dissipated mainly in the form of heat and sound. In the case of impact between two bodies of finite extension, i.e., one or neither of which can be regarded as a particle, it is usually assumed that Newton's law is applicable to the two points, one on each body, which are in contact during the collision, and relates the com­ ponents of the relative velocities of these points in the direction of the normal common to the two surfaces in contact. Thus, if a small sphere strikes a fixed plane wall the components of its velocity per­ pendicular to the wall before and after the collision,u and u' respectively, are related by the equation u' = - eu. A particle is projected from a point A on the inside of a circular ring which is fixed to a smooth horizontal plane. If is the angle between the radius of the circle through A and the initial direction in which it leaves A , and e is the coefficient of restitution between the particle and the ring for each impact, show that the condition that the particle returns to the point A after rebounding twice from the ring, at two points B and C, can be expressed in the form (1 e3)tan2 oc = e3(1 - e). (B.U.) Upon impact at B the velocity com­ ponent in the direction at right angles to OB is unaltered, whereas that in the direction OB is multiplied by -e.

Example 5.

oc

-

Thus

1 tan fJ = 8 tan oc.

A

Similarly, for the impact at C tan y = 8 tan fJ.

1

5]

I M P U L S I V E M O T I O N O F PART I CL E S But y = tn"

-

127

a. - {3. Hence 1 e tan {3. tan a. + tan f3 = tan a. tan {3

cot (a. + {3) tan {3 ·

1

=

-

e.

Eliminating tan {3 by use of the first equation, we obtain e tan a. + tan a. tan2 a. tan2 a. = + +

tan a.

•

e (1 e e2)

or

-

-

which is equivalent to the result stated.

=

2

e' e•,

Example 6. A

particle is projected at an angle of elevation a. to strike a smooth, vertical wall and after rebounding it passes through the point of projection. If is the angle of inclination to the horizontal at which the particle rebounds from the wall and e is the coefficient ofrestitution, show that e( 1 + e) tan rp = (1 e) tan {3. (M.T.) Let 0 be the point of projection distant a from the wall PN and let u be the velocity of arrival of the particle at the wall, inclined at an angle {3 to the horizontal (see diagram). Let v be the velocity of rebound from the wall -

y

at an angle to the horizontal. Since the wall is smooth, the component of the momentum of the particle in the direction is unaffected by the impact, and hence sin = sin {3. Also, by Newton's law of restitution we have

NP

v u

v cos = eu cos {3.

(i)

e tan = tan {3.

(ii)

Dividing this pair of equations, we obtain

128

A C O U R S E I N A P P L I E D MAT H E M AT I C S

[cH.

Taking horizontal and vertical axes Px, Py through the point of impact, the equation of the trajectory of the particle after rebound is (see equation (3. 1 1))

gx•

sec2 q,. y = x tan q, (iii) 2v• If a particle is projected from P with the velocity u reversed it will traverse the same track as did the particle projected from 0. The equation of this track is therefore

gx• ge2x2 sec2 q,, . y = -ex tan q, y = -x tan {3 - 2u• sec2 {3,

(iv) 2v• where we have made use of equations (i) and (ii). Differentiating equation (iv) with respect to x, we calculate that the gradient of this trajectory at O(x = is

a)

-e tan q, - ge2a """V2 sec2 q,, � sec2 q, = e-2 (tan - e tan rp) . v

-tan a =

a

(v)

Since 0 lies on both trajectories, equations (iii) and (iv) must yield identical values for y when we put x = Hence

a. ga2 a tan q, - 2v• sec• q, = -ea tan q, - ge2a2 2V2 sec• q,, :; (1 - e) sec2 q, = tan q,.

(vi)

Dividing equations (v) and (vi), we obtain tan q, �� ( 1 ) = tan IX - tan q,' which is easily verified to be equivalent to the relationship stated.

-e

e•

e

EXERCISE 5

I. Four equal particles A , B, C, D, each of mass m, are connected by equal light, inextensible strings AB, BC, CD, DA and lie at rest on a smooth, horizontal table with the strings taut, so that A BCD is a rhombus and angle BAD = 20( ((X < i-11') . The particle A is given a velocity V along the table in the direction CA (produced) . Find the initial velocity of the particle C and prove that the initial kinetic energy of the system is 2m V2f(l + 2 sin2 0() . (L.U.) 2. 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 = 71' - 0(, (0( < The whole system is moving parallel to AB 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 + sin20()/(l + 3 sin2 0() . (L.U.) 3. A smooth prism, whose cross-section is a right-angled isosceles triangle, is fixed with its largest rectangular face in contact with a horizontal

t11') .

5]

4.

5.

6.

7.

I M PU L S I V E M O T I O N O F PART I C L E S

129

plane. Two particles, whose masses are in the ratio 2 : I, are simul­ taneously projected with speed V from the feet of, and directly up, the opposite sloping faces of the prism. The particles move in the same vertical plane and coalesce on meeting at the top of the prism. Prove that the combined particle falls to the initial point of projection of the lighter particle if 4 V2 = 1 7gh, where h is the vertical height of the fixed prism. (L.U.) Four particles of equal mass are attached to a light, inextensible string at points A , B, C, D, where AB = BC = CD, 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 uf13. (L.U.) A small, smooth sphere falls freely from rest through a distance h and then strikes a fixed smooth plane which is inclined at an angle a to the horizontal. If the coefficient of restitution between the sphere and the plane is 0·5, prove that the distance between the first and fourth points of impact is 105h sin a/ 1 6, and find the impulsive reactions of the plane at these impacts. (L.U.) A particle is dropped vertically on to a smooth inclined plane and rebounds, the coefficient of restitution being e. If three suc­ cessive jumps are of lengths a, b and c respectively, prove that (be - c)f(ae - b) = e2• (L.U.) Two smooth spheres with equal" momenta approach each other from opposite directions, their coefficient of restitution being e. After impact, each moves perpendicular to its original direction. If, at the moment of impact, their common normal is inclined at an angle (L.U.) 6 to the original direction of motion, find 6.

8. One end of a light, inextensible string of length a is attached to a smooth, horizontal floor at a point distant b (< a) from a smooth, vertical wall. An elastic particle of mass m is attached to the other end of the string and rests on the floor with the string taut. The particle is given a horizontal velocity u perpendicular to the string and after striking the wall it rebounds. Find the impulsive tension in the string when it again becomes taut. Show that the particle strikes the wall again in the same place if e, the coefficient of restitution between the particle and the wall, is less than b2f(a2 - b2). Find the velocity of the particle just before it strikes the wall for the nth time. (L.U.) 9. A, B, C, D are the corners of a smooth, 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 IX with AB, to strike in turn the rims BC, CD, DA . If the sides AB, BC are of lengths a, b, respectively, and if e is the co­ efficient of restitution between the particle and the rim, show that the particle will return to its starting-point if tan IX = 2ebfa(1 + e) and that if this condition is satisfied, the particle will always pursue the same path. (L.U.) F (1)

A COURSE I N APPLIED MATHEMATICS 130 [cH. 5] 10. A smooth sphere A impinges on a stationary and equal smooth sphere B, the velocity of A making an angle 8 with the common normal at the moment of impact. Show that the angle through which A is deflected is a maximum when tan2 8 = !(I - e) and is then equal to !1t 28, where e is the coefficient of restitution between the spheres. Show further that in this case the fractional loss of KE of the spheres (L.U.) is ( 1 - e2)/(3 - e). -

I I . A smooth, narrow groove in the shape of a circle is cut in the surface

of a horizontal table. Two particles, A and B, lie respectively at opposite ends of a diameter of the groove. The mass of the particle A is that of B is and the coefficient of restitution between the particles is The particle B is projected along the groove with speed v and collides with A . Prove that, immediately after the second col­ lision between the particles, the KE of B has decreased by

m,

e.

2m

m9 (I - e2)(5 + e2) 2 v

12.

13.

•

Prove also that, if the speed of A is then one-quarter of the speed of B the value of is (L.U.) A ball of coefficient of restitution is dropped on to a level floor from a height h. Find how long it will continue to bounce. If it is given initially a horizontal velocity v and is subjected to frictional impulses fL times the normal impulse at each bounce, how far will it travel horizontally before it ceases to bounce, assuming that v is large enough for a forward velocity to be maintained throughout? (M.T.) Two particles A and B of equal mass lie touching one another in a smooth, horizontal, circular groove of radius a. A is projected away from B with speed V If is the coefficient of restitution for impact between the particles, show that after n collisions have taken place and B's speed is ! V[l - ( - e)n]. Cal­ A 's speed is culate also the time which elapses before the nth collision takes place. A particle is projected from a point 0 on a horizontal plane, to hit at a height h a vertical rigid pole, whose base A is at a distance h from 0. The coefficient of restitution between the particle and the pole is t. Find the horizo�tal and vertical components of the velocity of pro­ jection in order that the particle shall hit the plane again at a distance h from A , and prove that the maximum height of the particle above the plane is 9h/8. (The whole motion may be assumed to take place in the plane through the point of projection and the pole.) A particle of mass M is travelling uniformly in a straight line with energy E when it breaks up into two particles, each of mass tM. One of the final particles is observed to be moving at right angles to the original direction. If the release of energy in the process is Q, find the direction and energy of the other particle resulting from the disinte­ gration. Show that this situation is only possible if Q > E.

e (2/3)!.

e

e tV[l + (-e)n]

14.

(B.U.)

15.

(B.U.)

CHAPTER

6

MOTION OF A PARTICLE SYSTEM-I 6.1. Systems Involving a Small Number of Particles

In this and the succeeding chapters we shall be concerned with the motion of aggregates of interacting particles. We shall usually take the point of view that each of these systems involves an indefinite but very large number of particles, each of which is a molecule, or a small collection of molecules, of one of the bodies entering into the system. In such a case we shall make use of certain equations of motion which refer to the system as a whole and which assume no precise knowledge in their formulation of the number of particles present in the system. If, however, it is possible to regard a system as being composed of a few particles only, the analysis of its motion may be undertaken by considering the motion of each particle separately, making an allowance for the influence of each particle on the motion of the others according to Newton's Third Law.

The cross-section of a wedge, with three smooth plane faces and of mass 2m, is a triangle whose angles are 30°, 60° and 90° respectively. A small smooth pulley is fixed at the middle point A of the edge at which the angles of all cross-sections are 60°; a fine string is passed over the pulley and has particles P and Q, of masses 3m and m respectively, tied to its ends. The wedge is placed on a smooth, horizontal table with the face opposite to A in contact with the table and the particle Q held at rest vertically below A in contact with the vertical face of the wedge, the particle P lying on the inclined face. If Q is now set free, prove that the acceleration of the wedge is y3gf23, and find the tension in the string and the reaction of the wedge on Q. (L.U.)

Example 1.

A

(a) (b) In the diagram (b) we show the forces acting on the particles P and Q. The forces acting on the wedge ( + pulley) are shown in diagram (a). F and G are the forces of interaction between the wedge and P and Q respectively. T is the tension in the string. R is the reaction of the smooth, horizontal plane on the wedge. Let f be the acceleration of the wedge in the sense shown and let a be the acceleration of the particles relative to the wedge. The acceleration of P 131

A C O U R S E I N A P P L I E D M A T H E MA T I C S

132

(CH.

relative to a fixed observer will be the vector sum of its acceleration relative to the wedge and the acceleration of the wedge. Similarly for Q. Treating the wedge as a particle (the justification for this procedure will be found in Section and taking horizontal components of the forces acting upon it, we obtain the equation of motion (i) Resolving the forces acting on Q and its accelerations horizontally and vertically, we find that = (ii) = (iii) Resolving the forces acting on and its accelerations along and per­ pendicular to the sloping face of the wedge, we obtain (iv) (v) = Elimination of from equation (i) by the use of equation (ii) provides us with the equation

8.1)

tF - G - tTv 3 = 2mf.

G mf, T - mg ma. P - T = 3m(a - !Jy3), img mgy3 - F imf. !

G

F - Ty3 = 6mf. (vi) 2T 3mg - 5y3mf. a (vii) 8T 9mg 3y3mf. f = �:c. T = 2327 mg. Whence, from (ii), G = y3mgf23. Example 2. Three light, inextensible strings BA , A D, DC support in equilibrium two particles each of mass m, at A and D. B and C are fixed points at the same horizontal level. AD is parallel to BC, and AB and DC are each inclined at 45° to the horizontal. Find the tension in the string BA . If the string DC is cut, show that the tension in BA is immediately reduced in the ratio of 2 : 3. (L.U.) When the particle at A is in equilibrium, let S be the tension in the string AB. Resolving forces acting on the particle vertically, we find that S = mgy2. Adding this equation to (v), we obtain = Eliminating between equations (iii) and (iv), we find that = + Equations (vi) and (vii) now yield

B

8

T A

p

,

I

'"9

J/

/

A

/

m9

0

0

f

/

I>

/

,

,

0

I oQ I '

'

"'9 t

When the string CD is cut, let and Q be the tensions in the strings AB, AD respectively as the motion commences. A begins to move around a circle of centre B and radius AB. Initially, its velocity in this circle is zero, and hence the radial component of its initial acceleration is zero. A 's initial

T

M O T I O N O F A PART I C L E S Y S T EM-I

6]

133

acceleration is therefore in a direction perpendicular to AB. Let it be f. Relative to A , D moves around a circle of centre A and radius AD. Since initially D is stationary in this relative motion, its acceleration relative to A is directed at right angles to AD. Let it be a. D's initial acceleration relative to a fixed observer is the vector sum of this acceleration relative to A and A 's initial acceleration. Resolving horizontally and vertically for the forces acting on the particles at A and D and also for their accelerations, we obtain the equations I 1 T = y 2 mf, mg y' 2 1 1 T - Q = 2mf, .y 2 .y 1 Q = 2mf, .y mg = m

(J2 ! + a) ·

Adding the first three equations, we find that y2 f = a g.

Whence, from the first equation, T = 2 y' 2mgf3. = 2 : 3.

T:

S

6.2. Vibration of Two Particles.

This proves that

Normal Modes

A particular particle system whose motion can be discussed by the separate consideration of the equations of motion of the individual particles is that comprising a string of negligible mass stretched between two fixed points A and B and a number of particles fixed to the string at points along its length. We shall consider the case of two particles P and Q of masses m1, m2 respectively, free to vibrate in a horizontal transverse direction (Fig. 6.1). We shall assume that the vibrations p

A

a

- - - - -

I lx I

L

- - - - - - -

1

�

oV

c

- -

FIG. 6.1 .-Transverse Vibrations of Particles

8

are of small amplitude so that the lines of motion of the particles are both perpendicular to A B and the variations in the lengths a, b, c, of the three parts of the string are negligible. It follows that the tension T remains constant during the motion over the whole length of the string. If e, q,, if; are the inclinations of the three parts of the string to A B as shown in the figure, when the displacements of the particles from their equilibrium positions are x and y, the equations of motion are m/i = -T sin e - T sin q,, m2Y = T sin q, T sin if;.

-

134

A C O U R S E IN A P P L I E D MATHEMAT I C S

[cH.

= xfa, sin if> = (x - y)fb, sin tj; = yfc. Hence (D2 + p)x - ry = 0,} (6.l) -sx + (D2 + q)y = 0, where P = T(a + b)jm1ab, q = T(b + c)jm2bc, r = Tjm1b, s = Tjm2b. Eliminating y between these equations, we obtain [(D2 + p)(D2 + q) - rs]x = 0 . (6.2) But sin e

The equation for the characteristic roots of this linear differential equation is ).4 + (p + q) ).2 + pq (6.3) - rs = 0. Solving for J...2 , we calculate that

)_2 = - HP + q) ± !{(P + q)2 - 4 (pq - rs)}t, (6. 4) = -t(P + q) ± t{(p - q)2 + 4rs}t . (6.5) The form (6.5) shows that both these values for ).2 are real. Also, + b + c) > pq _ rs = P(am1m 0 2abc and hence from the form (6.4) it follows that both values of ).2 are negative. We shall write ).2 = - (1)2, - !1 2• The characteristic roots are then ±iffi, ±in and the general solution of equation (6.2) is x = A sin(ffit + ) + B sin( n t + �), (6.6) A, B, � being arbitrary constants fixed by the initial conditions. Substituting for from equation (6.6) in the first of equations (6.1), we now find that - Q 2 B sin(n t + �), - (1)2 A sin(ffit + ) + p-y = p-r r = (J.A sin(ffit + ) + vB sin ( n t + �) . (6.7) where !J. = (p - (1)2)/r, v = (p - !12)/r. ot ,

ot

x

ot

ot

If now the initial conditions are so chosen that B = 0, the oscilla­ tions of the system are determined by the equations

x = A sin(ffit +

)

ot ,

y = !J.A sin(ffit +

)

ot .

(6.8)

In these circumstances both particles oscillate with simple harmonic motions, each of frequency (1)/2rr:. Moreover, throughout the motion y = !J.X, i.e., the particles' vibrations are in phase or in anti-phase with one another according as !J. is positive or negative respec­ tively. Both particles pass through their equilibrium positions on AB at the same instants and arrive at the extremes of their paths to­ gether. Such a simple form of oscillation of the system is referred to as a normal mode. By choosing the initial conditions so that A = 0,

6]

M O T I O N O F A PART I CL E S Y S T EM-I

135

the system will be set oscillating in a second normal mode, this time of frequency D.f2rr:, viz., (6.9} x = B sin(nt + �), y = vB sin(nt + �).

Any other mode of oscillation for which neither A nor B is zero is governed by the equations (6.6), (6.7). It may clearly be regarded as the resultant of two oscillations conducted in the normal modes and proceeding simultaneously. In any particular case, having obtained the equations of motion (6.1), it is preferable to seek for a solution corresponding to a normal mode than to calculate the general solution in the manner described above. Thus, suppose that a = b = c, m1 = m2 = m. Then p = q = 2r = 2s = 2Tjma = 2n2 and the equations of motion are

(D2 + 2n2)x - n2y = 0,} (6.10) - n2x + (D2 + 2n2}y = 0. Let x = A sin(wt + ) y = C sin(wt + ) represent a normal mode. Substituting in the equations (6.10), we obtain (2n2 - w2)A - n2C = 0,} (6.11) - n2A + (2n2 - w2)C = 0. This pair of equations can only possess a non-zero solution in A and oc ,

oc

c if

(6.12)

i.e., if

(6.13)

Thus w2 = n2 or 3n2 and w = n or y'3n. Equation (6.12) accordingly determines the frequencies nj2rr:, y'3nf2rr: of the two normal modes. If w = n, either of equations (6.11) shows that C = A and the normal mode is given by the equations oc

(6.14) x = A sin(nt + ) = y. This mode of vibration is illustrated in Fig. 6.2 (a). If w = y'3n, we find that C = -A and the normal mode is deter­

mined by the equations

x = A sin(y'3nt + ) = -y. This mode is illustrated in Fig. 6.2 (b). oc

(6.15)

Any other mode of vibration can be represented as the sum of two normal modes. Thus, in the case we have been considering the general solution of the equations of motion is

x = A sin(nt + ) + B sin(y'3nt + �).} y = A sin(nt + oc) - B sin(y'3nt + �). oc

(6.16)

A C O U R S E I N A P P L I E D M A T H E MA T I C S

1 36

[CH.

the arbitrary constants A, a occurring in the equations (6.15) having been replaced by B, � respectively, since the amplitudes of the two component modes are not necessarily identical and neither are their phase angles related in any way (cf. equations (6.6), (6.7) ) .

8

A

(a)

B

A

(b)

FIG. 6.2.-Normal Modes of Vibration

A light string of length 7af5 is fixed at one end and carries a particle of mass 4m at the other. A second light string of length a has one end attached to this particle and a second particle of mass 3m is attached to the other end. The system makes small oscillations in a vertical plane under gravity. Dis­ cuss the normal modes of oscillation. Let 6, cp be the inclinations of the two strings to the vertical at any instant and let x, y be the dis­

Example 3.

placements of the upper and lower particles respectively from their equilibrium positions at be the tensions in the same instant. Let the upper and lower strings respectively. We shall assume that the vertical motions of the particles may be neglected. Hence, resolving vertically for the forces on each particle, we obtain the equations cos 6 = + cos cp, T cos q, = Since 6, cp are small, we shall approximate to the values of cos 6 and cos cp by unity. Thus r = = ' The equations governing the horizontal motions ;- --- - - - -1 If of the particles are I = sin q, sin 6, 3 mJ = sin q,. Since sin 6 = sin cp = these equations may be written (i) = + + + = (ii) where = Let a normal mode be determined by the equations = sin + a) , = sin + a) . Upon substitution of this solution into equations (i) and (ii), we obtain = + = These equations have a non-zero solution in and only provided = + = i.e., if

S, T

S

4mg T 3mg.

T 3mg, S 7mg.

4mi T - S 3mji - T 5xf7a, (y - x)fa, (4D2 8n2)x - 3n2y 0, -n•x (D2 n2)y 0, n2 gfa. x A

(wt y B (wt (8n2 - 4w2)A - 3n2B 0, -n2A (n2 - w2)B 0. A B (8n2 - 4w2)(n2 - w2) - 3n4 0, 4w4 - 12n2w2 5n4 0.

Q

6]

M O T I O N O F A PARTI CLE S Y S T E M- I

1 37

Hence w = or If w takes the first ,value, it will be found that corresponding normal mode is

A : B = 1 : 2, and the t x = A sin ( �� + •+ y = 2A sin ( � + 0<.) · 2 This mode is illustrated in diagram (a). nfy'2 ny'(5f2).

If

w

(a)

takes its other value, then

(b)

A : B = 3 : -2 and the normal mode is �f ( nt + �) . y = -2C sin ( �f nt + � ) · x= 3C sin This mode is illustrated in diagram (b)

0

If a system has two normal modes of vibration whose frequencies differ little from one another, when both modes are excited simultaneously their interaction can lead to the phenomenon of sympathetic vibration. Thus, consider the system built as follows. A light rectangular frame of sides a and b is smoothly hinged along its horizontal edge b. To its other edge b, two strings of the same length l are attached, and these carry particles both of mass m at their lower ends. The strings are set swinging in parallel, vertical planes perpendicular to the plane of the frame. Let e, 4> be the inclinations of the strings to the vertical at any I instant and let if; be the inclination of the frame to the T 1 vertical at the same instant (Fig. 6.3). These angles - - -! J will be assumed to remain small during the motion. For the purposes of this analysis, we shall suppose the my m'J frame replaced by a light string OK, of length a, to 6.3.-Sym­ which the pair of length l are attached at its lower FrG. pathetic Vi­ extremity K. A sufficient gap is supposed to exist brations _ _ _ X

138

A C O U R S E IN A P P L I E D MATHEM AT I C S

[cH.

between this pair of strings to permit their vibrating side by side with­ out touching. Let S, T be the tensions in the lower pair of strings and let P be the tension in the upper. The knot K is of negligible mass, and hence the forces acting on it must be in equilibrium. Resolving vertically for these forces, we obtain P = S + T, since cos e, cos ¢>, cos if; are all unity to the first order in e, ¢>, if;. Assuming the vertical motions of the two particles to be negligible, we also obtain T = S = mg. Hence P = 2mg. Taking the horizontal components of the forces acting on K, we find that

P sin if; = T sin () + sin ¢>, 2 sin if; = sin () + sin ¢>.

or

(6.17)

Let x and y be the horizontal displacements of the particles from the vertical through 0 and let z be the horizontal displacement of K from the same line. Then

l sin e, y z = l sin ¢>. (6.18) and hence, from equations (6.17) and (6.18), we find that . () x (2l + a) - ay sm. ,.�. - y( 2l + a) - ax (6.19) sm 2l(l + a) 'I' 2l(l + a) z =

a sin if;, x

-

-

z =

_

_

'

·

The equations of motion of the particles resolved horizontally are

(6.20) mx = - T sin e, my - S sin ¢>. In view of equations (6.19), these are equivalent to {D2 + g(2l + a)}x - ga Y - O, (6.21) 2l(l + a) 2l(l + a) + a)} (6.22) - 2l(lga+ a) x + {D2 + g(2l 2l(l + a) Y = 0 Let = A sin(wt + <X), y = B sin(wt + <X), (6.23) specify a normal mode. Substitution in equations (6.21) and (6.22) =

_

x

shows that

{g(2l + a) } ga 2l(l + a) - w2 A - 2l(l + a) B - O , - ga A + {g(2l + a) - w2} B 0 · _

2l(l + a)

For a non-zero solution in

2l(l + a)

A and

B,

}2 {g(2l + a) 2l(l a) - w 2 +

=

(6.24) (6.25)

w must satisfy =

g2a2 4l2 (l + a)2

(6.26)

6]

M O T I O N O F A PART I C L E S Y STEM-I

w2 = gjl or gj(l + a) . 1 -1 and 1 1 respectively.

139

The corresponding ratios A B are It is left as an exercise for the reader to examine the characteristics of these two modes. The general solution of the equations (6.21) and (6.22) is now seen to be (6.27) x = A sin (nt + IX) + C sin (pt + �),

Thus

(6.28) y = A sin (nt + IX) - C sin (pt + �), where n2 = gj(l + a) and p 2 = gjl. Suppose that initially one particle is displaced a distance h from the vertical through 0, the other particle lying on this vertical and the system is then released from rest. Then x = h, :X = y = y = 0 at t = 0 and the appropriate form of equations (6.27), (6.28) will be found to be x = !h(cos nt + cos pt) = h cos HP - n)t cos!(P + n)t, (6.29) y = !h(cos nt - cos pt) = h sin t(p - n)t sin t(p + n)t. (6.30) Suppose a is small compared with l. Then p-n=

JH1 - ( 1 + ]r1} = itJf

(6.31)

approximately and the difference between the frequencies of the normal modes is small. The sinusoidal functions cos HP - n)t, sin t(p - n)t are then of small frequency, i.e., oscillate only slowly in value. Equations (6.29), (6.30) accordingly indicate that x and y oscillate in SHM with angular frequency !(P + n) and slowly changing amplitudes h cos !(P - n)t and h sin HP - n)t respectively. When x's amplitude is at its maximum h, y s amplitude is zero, and vice versa. An observer will therefore witness a slow transfer of oscillation from one particle to the other and back again indefinitely. As cos t(p - n)t, sin t(p - n)t go through a complete cycle of values, four such transfers take place. The time for one transfer is therefore '

1t

21tl t l (6.32) p - n - aJi - a T' by the use of equation (6.31), T being the period of oscillation of a simple pendulum of length l. _

_

6.3. Energy Equation for a Particle System

For any particle of a system we can write down an equation of work (see equation (3.51)), viz., increase in KE

=

work done by the forces acting.

If we write down one such equation for the motion of each particle of the system over a certain period of time and then add all the equations

140

A C O U RS E I N A P P L I E D MATHEMAT I C S

(CH.

together we obtain an equation which refers to the whole system, viz., total increase in KE of the system = total work done by all forces operating on the system.

(6.33)

It is convenient to divide the forces operating on such a system into two classes. Firstly, there are those forces which arise from the action of agents external to the system. Thus, if the system comprises a number of interconnected particles falling under gravity the gravi­ tational forces are due to an agent external to the system, viz., the Earth. Such forces are therefore called external forces. Secondly, there are those forces which are caused by the interactions between the particles of the system. These forces are governed by Newton's Third Law and are referred to as internal forces. This classification is convenient, but is not imposed by any fundamental difference between the natures of these two types of forces. In fact, by increasing the scope of the system, it is always possible to transfer forces from the external to the internal category. Thus, in the case of a system falling under gravity, by enlarging the system to include the Earth the gravi­ tational forces become internal forces. The solar attraction is an external force if the system being considered is the Earth, but is an internal force relative to the whole solar system. It follows from Newton's Third Law that the internal forces of a system occur in equal and opposite pairs, and it might be expected that in any motion of the system the work done by one force of a pair would be cancelled by the work done by the .other force, so that the internal forces could be neglected when writing down equation (6.33). How­ ever, the motions of the particles to which such a pair of forces are applied are not necessarily identical, and hence neither are the works done by the forces. On the other hand, there is a case of great practical importance when we are justified in such a neglect of the internal forces. This is when the particles of the system form a rigid body, i.e., the distance between any particle pair remains fixed during the motion. Suppose P and Q are any two particles of the system. When the latter undergoes any displacement the motion of Q relative to P must be along an arc of a circle having centre at P. Thus Q's displacement is compounded of a displacement along the circular arc together with the displacement of P. We shall suppose that the equal and opposite forces F between P and Q act along the line PQ. The work done by F acting on Q is the sum of the works done by this force over the two displacements into which that of Q has been resolved. Since the motion over the circular arc is always at right angles to the direction of F, no work is done by F in this displacement. The remaining component of Q's motion is the same as that of P, and the work done by F acting on Q in this motion is therefore the negative of the work done by F acting on P. The total work done is accordingly zero.

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M O T I O N O F A PA R T I C L E S Y S T E M - I

141

Again, if two bodies of a system are in contact with one another they will exert equal and opposite forces upon one another at the point of contact and, in any displacement of the system for which there is no slipping between the bodies, the works done by these forces will cancel out, since their points of application are always identical and undergo the same displacements. It will often be the case that the work that is done by the forces of a system (internal and external) as the latter moves from one con­ figuration to another is independent of the mode of motion between the configurations, provided only that such a motion would not cause a rupture in the bodies forming the system. For example, consider the R

c

A

B

FIG. 6.4.-Simple Conservative System

system shown in Fig. 6.4. This comprises a rigid bar AB, at one end of which is a ring A sliding on a smooth, horizontal wire and at the other end is attached an elastic string BC attached to a point C of the wire. In any displacement of the system in which neither the bar nor the string is ruptured and the ring is not removed from the wire, the only forces which contribute to the total work done are the weight forces acting on the particles of the bar and the internal forces of the string. The reaction R of the wire on the ring acts vertically and, since the motion of the ring is horizontal, clearly performs no work. The in­ ternal forces of the bar do no work, since the bar is a rigid body. As the system moves from one configuration to another the work done by the internal forces of the string depends only on the difference between the string's extensions in the two positions and, if M is the mass of the bar, the work done by the gravitational forces is the product of the total

[cH.

A C O U R S E IN A P P L I E D MAT HEMA T I C S

142

weight Mg and the vertical displacement of the centre of mass of the bar G (see below for proof). Hence, the total work done by the forces of the system is independent of the mode of motion between the two con­ figurations. A system of this type is said to be conservative (cf. p. 71). The construction of a system may be such that during its motion certain points are caused to move along specified paths. For example, in the system of Fig. 6.4 one end of the bar A is forced to move along the horizontal wire. Such prescribed motions are said to be due to constraints, and the system, of which AB is a part, is said to be con­ servative relative to motions which do not violate the constraints. When dealing with a conservative system it is convenienient to introduce a Potential Energy function V The PE of a conservative system in any position is defined to be the work which is done by the forces of the system, internal and external, if it is moved from that position to some standard or datum position in such a manner as not to violate the constraints. Suppose that a conservative system moves under the action of certain forces from a configuration which we shall denote by A to an­ other we shall denote by B. Let VA, VB be its respective potential energies in these two positions. Let D be the datum configuration. If the system were moved from the configuration A to the configuration D the work done by the forces of the system would be VA· If this displacement were followed by another from D to B the work would be - VB · Hence the work done as the system moves from A to B in any manner is VA - VB· This will accordingly be the work done by the forces in the actual motion from A to B. Let TA be the KE of the system in the configuration A in the actual motion and TB the KE in the configuration B. By equation (6.33) TB - TA = vA B, TA + vA =:= TB + vB, (6.34) or i.e., the total energy of the system remains constant during the motion. We shall often require to know the contribution to the PE of a system of weight forces acting on the particles of a body included in the system. Suppose the body comprises particles of masses mv m2, mn, and suppose that at some instant these are situated at points at heights z1, z2 , Zn respectively above some fixed horizontal plane. The sum of the PEs of these particles relative to the reference plane as datum is -

v

•

•

•

•

+

mngZn

n

= g L m;Z;. i= 1

But, if z is the height of the CM of the body above the reference plane, Im;z; = Mz, where M is the total mass of the body (see eqn. (11.28)) . Hence

V = Mgz.

(6.35)

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M O T I O N O F A P A R T I C L E S Y S T E M-I

143

This equation indicates that the PE may be calculated by supposing the whole mass of the body is concentrated at the CM. We will now solve two problems by the use of the energy equation.

Ox, Oy are two smooth, straight wiresfixed perpendicular to one another at Oy being vertically downwards. To each end of a light rod of length a, a ring of mass m is attached and the rings slide freely oni the wires. The bead on Oy is projected downwards from 0 with velocity (2ga) . Show that a time (ga) l log tan 831T elapses before the other bead arrives at 0 and find its speed of arrival.

Example 4. 0,

mg 'f

Q

Let e be the angle made by the rod with at any instant. Let x, y be the distances of the particles from 0 (see diagram) . Then (i) sin e, COS (). The system is conservative and its PE is cos e. Differentiating the equations (i) with respect to the time, we obtain (ii) X cos M, sin ee. These are the speeds of the rings. The KE of the system is accordingly x• + • • (cos• e + sin2 e)62 •e•. Initially the PE is zero and the KE is the bead on being stationary. Hence the energy equation is cos e + cos e) = 4g cos• te, (iii) 6 1 cos te,

X=a

=a

tm

tmy = tma

Oy y=a -mgy = -mga y = -a = tma tm(2ga) = mga, = mga,

Ox

tma262 - mga a6• = 2g(l = - 2 (!) 6 being clearly negative. When e = 0, 6 = -2(gfa)t and hence, from equations (ii), x = -2(ga)l. The speed of arrival of the bead on Ox at 0 is therefore 2(ga)l.

1 44

[CH.

A C O U R S E I N A P P L I E D M A T H EMAT I C S Writing equation (iii) in the form

dt ( a) 11 de = - ! g sec !6,

and integrating with respect to time of the motion,

e

over the range (!7T, 0) we obtain for the

A particle A of mass m moves on a smooth, horizontal table and is con­ nected by a light, inextensible string passing through a smooth hole 0 in the table to a particle B of the same mass m, which moves along the vertical through 0. Initially B is at rest and A is distant a from 0, moving with velocity y'(gaf3) perpendicular to OA. Show that in the subsequent motion the distance r = OA lies between a and ta. Show also that the tension in the string is mg ( 3 + �) . 6 r3 The forces acting on the particle A are the tension T in the string, its weight mg and the smooth reaction of the table R. The last-mentioned forces cancel, there being no vertical motion of A. Resolving in the hori­ zontal direction perpendicular to OA for the particle A, we obtain the equation m d (r2o)" = 0 or r•e. = h (a constant) . (i) r lit Initially r = a, rO = (gaf3)i and hence h = a(gaf3) l .

Example 5.

R

0 T 8

•"9

l! L r

r

A

The system comprising the two particles and the string is conservative, the internal forces between particles of the string doing no work in any dis­ placement, since the distances between them remain invariant. Taking the datum position of the system to be that in which is at 0, the PE of the system when = will be If 6) are polar coordinates of its velocity is r + and its KE is accordingly + If is the length of the string, The vertical velocity of is = -r downwards. The KE of is therefore We can now write down the energy equation, viz., + + + constant, + 2r• + 2 = constant . . or (ii)

A OA r mgr. (r, A, ( 2 r2(J2)l !m(r2 r202). l OB = l - r. B d(OB)fdt B !mr2• tm(r2 r202) tmr• mgr = r202 gr

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M O T I O N OF A PART I C L E S Y S T EM-I

1 45

Initially = r = and = It follows that the value of the constant in equation (ii) is Eliminating from equation (ii) by the use of equation (i), we now find that _§_ (iii) '

r a, (J

0

r6 (gaj3)l. 7gaj3.

r2 = 6r" (7ar2 - a• - 6r3) (a - r)(2r - a)(3r + a). 6�2 But r2 must be positive and g(3r + a)j6r2 is certainly of this sign. Hence ta � r � a. Also, r is only zero when r = ta or a. It follows that r must oscillate between these two values (see p. 96) . Since OB = l - the acceleration of B downwards is r The equation of motion of B is therefore (iv) T - g mr. =

r,

-

m

.

=

Differentiating equation (iii) with respect to t, we obtain

2rr =

g (;;. - 1) r.

Whence r Substituting in equation (iv) and solving for the result stated in the question. .

T, we now obtain

6.4. Kinetic Energy of a Rigid Body

Before we can write down the energy equation for a system com­ prising one or more rigid bodies we must find a method of calculating

FrG. 6.5.-Rotation about a Fixed Axis

the KE of such a body when it is moving in any manner. We shall restrict our attention to the case of a rigid body, all particles of which are moving parallel to some fixed plane (i.e., the case of two-dimensional motion). As a preliminary to the general discussion, consider the case of a rigid body rotating about a fixed axis. Let P be any particle of the body of mass m and let PO = r (Fig. 6.5) be the perpendicular from P

1 46

A C O U R S E I N A P P L I E D MATH EMAT I C S

[CH.

on to the axis of rotation. If w is the angular velocity of the body at the instant under consideration, P is moving around a circle of centre 0 and radius r with this angular velocity, and its speed is accordingly wr. The KE of P is therefore �mw2r2• Summing over all the particles of the body, we obtain KE of body y

�w2Imr2 = tJ w2,

(6.36) where I = Imr2 is the moment of inertia (MI) of the body about the axis of rotation (Chapter 7).

=

Iimw2r2

=

As a particular example of the use of this formula, suppose the rigid body is free to rotate about a horizontal axis through 0. Let G be the CM of the body and let OG = h (Fig. 6.6). If the pivot at 0 is smooth the action of the bearings on the axle can be represented by a single force R at 0. Let ( Y, X) be the components of this force along and perpendicular to GO respectively. The only other external force acting on the body is the weight Mg acting at G. Since R does no work in any displacement con­ FIG. 6.6-The Compound sistent with the constraints, the system .is Pendulum conservative and, if the datum position is taken to be that in which OG is horizontal, the PE is - Mgh cos 6. Let I be the MI of the body about the axis through 0. Then the energy equation is

tJ62 - Mgh cos 6

=

constant.

(6.37)

Suppose that the body is displaced from its equilibrium position so that 6 = IX and then released from rest. It will clearly oscillate as a pendulum. The system is called a compound pendulum. Initially we have e = 0 and 6 = IX. Hence the constant in equation (6.37) is -Mgh cos IX and

62 =

2�gh (cos 6 - cos

IX .

)

(6.38)

If IX is small, then so is 6 throughout the motion and cos 6 - cos IX = t(1X2 - 62) to the second order of small quantities. Hence, in these circumstances the motion is governed approximately by the equation

(6.39) We have already proved (equation (2.26)) that a point executes SHM of amplitude a and period 2rr:f w along a straight line if its coordinate

M O T I O N O F A PA R T I C L E S Y S T E M - I

6]

147

.i2
X is governed by the equation = Equation (6.39) therefore shows that the angular motion of the compound pendulum is approximately simple harmonic of amplitude tx and period

T = 2rrJJgh .

(6.40)

Comparing this result with equation (3.68), we observe that the period of the compound pendulum is the same as that of a simple pendulum of length IjMh. A simple pendulum of this length is called the simple

equivalent pendulum.

The axle of a wheel is supported horizontally in smooth bearings. A chain of length l and mass m per unit length has one end fixed to the axle and is wound around it. To the free end of the chain is attached a particle of mass equal to that of the chain. Initially this particle is on the level of the axle and the wheel is stationary. The chain begins to unwind. If I is the moment of inertia of the wheel and axle and r is the radius of the axle (assumed small), show that the length of chain unwound after a time t is 2l siuh •>..t, where 4>..2(I + 2mlr2) = mgr2 • (L.U.) Let x be the length unwound after a time t.

Example 6.

The system is conservative. Taking the level of the axle as the datum for the purpose of calculating the PE, the CM of the hanging G part of the chain is at distant tx below the datum level. The weight of this portion of the chain is mgx. Hence its PE is - imgx•. The PE of the remainder of the chain and of the wheel is zero. The PE of the particle is -mlgx. The velocity of the particle and of all the links of the chain is %, and hence their ml contribution to the KE is mlx2• The angular velocity of the axle is xfr, and the KE of the wheel and axle is accord­ ingly tJ(xfr) 2 • The total energy is initially ze:ro. Hence the energy equation is

G,

F

(�)

2

+ mlx• - imgx2 - mlgx

or x• 4>..2 (x 2 + 2lx), where >.. is defined in the question. Writing (i) in the form

=

= 0,

dx = 2>..dt, {(x + l)" - l"}!

we integrate to obtain 2>..t

When t = 0, x or

=

= cosh-1 X-+-l + A .

0 and hence A X+l

l

= 0.

Thus

-- = cosh 2At, l x = l (cosh 2>..t - 1) = 2l sinh2 >..t.

(i)

148

A C O U R S E IN A P P L I E D M A T H E M A T I C S

[CH.

Before commencing consideration of the calculation of the KE of a rigid body in general two-dimensional motion, we shall deal with the general case of a system of particles executing any motion whatsoever. Let m1, m2, rn mn be the masses of the particles and let rv r2 , be their position vectors relative to a fixed origin 0. Also, let Pv p2 , Pn be the position vectors of the particles relative to G, the --+ mass centre. Then, if OG = r, •

•

(6.41) ri = Pi + r, and thus ri = Pi +r. But i:i is the velocity of the particle mi. Hence, the total KE of the system is given by

n

T = ti=�1mii'2i = t�m;(Pi + r). 2

= !�mipi2 + !�mii2 + �mipi i = t�miPi2 + tr2�mi + r �miPi· ·

·

(6.42)

Since the mass centre of the system lies at the origin of the position vectors pi, it follows from equation (11.29) that

�m,pi = 0,

(6.43)

and, by differentiation with respect to the time, that

�mipi = 0.

(6.44)

Hence equation (6.42) may be written

T = t�mipi2 + tMr2, (6.45) where M = �mi. Now p; is the velocity of m; relative to G. Thus the first term in the right-hand member of equation (6.45) represents the KE of the motion of the system relative to G. If we imagine the whole system con­ densed into a single particle of mass M at the centre of mass and given the velocity f of this point, its KE would be tMr2 • This provides us with an interpretation of the second term of the right-hand member of equation (6.45). We have therefore proved that the KE of a system

is equal to the KE of the motion relative to the CM together with the KE of a particle of mass equal to that of the whole system and moving with the CM. In the particular case of a rigid body in motion parallel to a plane, let V be the velocity of the centre of mass and w the angular velocity of rotation about G. The motion relative to G is a simple rotation about this point and, by equation (6.36), the KE of such a motion is tJ0w2 , where I is the MI of the body about G. Thus the total a

KE of the rigid body is given by

(6.46)

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M O T I O N O F A PA R T I C L E S Y S T E M - I

149

A

uniform circular cylinder, of mass M and radius a, carries a particle of mass m fixed in its surface. It is placed on a rough horizontal table with the particle in the highest position. If the system is slightly disturbed and the cylinder rolls without slipping, prove that when it has turned through an angle a

Example 7.

ae•{fM + 2m(l + cos a)}

= 2mg(l

-

cos 6) . (L.U.)

R is the normal component of the reaction of the plane on the cylinder and F is the frictional component. As the cylinder rolls, successive particles on its surface come into contact with particles of the plane. However, since there is no slipping, these particles do not move relative to one another during contact, and no work is done by the reaction forces between them. Thus neither R nor perform any work during the motion. The remaining external forces acting on the system " cylinder + particle " are the weight of the cylinder Mg and the weight of the particle mg. The system is con­ servative, with PE mga cos a (the datum configuration is with OP horizontal).

F

m

a.e

'

'

'

'

'

' '

'

'

The KE of the cylinder alone is the KE of a particle of mass M moving with 0 together with the KE of the rotary motion about 0. In the position a. 0 has moved a horizontal distance aa from its initial position and its speed is therefore ae . The contribution to the KE of the particle at 0 is accordingly !Ma• e•. The MI of the cylinder about its axis through 0 is !Ma2 and its angular velocity is e . The KE of the rotary motion is therefore !Ma•e•. The net KE of the cylinder is now seen to be !Ma•e•. Now consider the particle. Relative to 0 its velocity is ae perpendicular to OP. Adding O's velocity ae vectorially to this, we obtain P's velocity v relative to a stationary observer (see right-hand diagram) . By the Cosine Rule, v2 a•e• + a•e• + 2a•e • cos a 2a•e•(I + cos a),

= =

and hence the KE of the particle is ma262(1 + cos a) . We can now write down the energy equation of the system, viz., !Ma•e• + ma2t'l 2 ( l + cos a) + mga cos a = constant.

When a =

0, e = 0 and hence the constant is equal to mga. ae2{!M + m(l + cos 8)} = mg( l

which is equivalent to the equation stated.

-

cos a),

Thus

150

A C O U R S E IN A P P L I E D M A T H E MA T I C S

[cH.

A

wheel of mass M and radius a is caused to move along a horizontal straight line without slipping at a constant speed V under the action of a horizontal force P applied at its centre 0. The distance of G, its CM, from 0 is b. Show that when OG makes an angle 6 with the downward vertical, Mgb V" P= sin 6. + a ga (L.U.)

Example 8.

(1

)

It follows from equation (6.33) that the rate at which the forces of a system do work is equal to the rate of increase of its KE. During the motion of the system being considered, only the forces P and Mg do work. To find the rate at which these forces are working, we employ equation (2.9) . P is doing work at the rate P V. The angular velocity of the wheel is Vfa and hence, relative to 0, G has velocity b Vfa perpendicular to OG. The

'

F

\

\

\

\

\

\

\

\

\

\

velocity of G relative to a stationary observer is found by combining this velocity with that of 0, viz., V horizontally. This has been done in the right-hand diagram to give the velocity v. The vertical component of v is observed to be b V sin 6fa, and it follows from equation (2.9) that the rate at which Mg is doing work is -Mgb V sin 6fa. The net rate at which work is being done by the forces of the system is therefore PV -

Mgb V sin 6. a

(i)

Since the angular velocity of the wheel does not change, the rate of change of the rotary component of the KE is zero. We have b2 v• = v• + ii2 v•

- 2b v• cos e, a

so that the KE of a particle of mass M moving with G is tMV2

[1 + a•� - 2ba cos e]

and the rate of increase of this component of the KE of the wheel is clearly

.;

M V2 � sin 6 e = M V3 sin 6, a a

(ii)

since e = Vfa. Equating the expressions (i) and (ii) according to the principle previously explained, we obtain the result stated. Two equal rods AB, BC of length 2a are smoothly hinged at B. The rod AB is smoothly pivoted at to a point on a smooth horizontal plane in such a way that both rods are free to move in a vertical plane through the pivot. The end C of the rod BC slides on the smooth plane and is connected by a light, inextensible string which passes through a smooth hole in the table at A to a

Example 9.

A

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MOTION OF A PARTICLE SYSTEM-I

151

particle hanging vertically below A . If the mass of the particle i s half the mass of either rod and the inclination of AB to the horizontal is initially rr/6, show that C can only move to A if it is given an initial velocity v towards this point, where 6ga v2 > 5 (2y2 - y 3 1) . -

Prove also that if v is just sufficient to permit C to move to A , then its velocity of arrival at A is V, where v• = 2l ga(v'2 - 1 ) . Let 2 a be the length of the rods, 2 m their mass and let a be their inclination

to the horizontal at any instant. The system is conservative. The PE associated with the weight forces acting on the bars is 4mga sin a. The weight of the hanging particle P is mg and its PE is -mg . AP = -mg(l - A C) -mg(l - 4a cos a), where l is the length of the string. The net PE of the system is accordingly (i) 4mga(sin a + cos a), where we have neglected a constant component -mgl, since this can be absorbed into the constant appearing in the energy equation.

=

B

p

The bar AB rotates about a fixed axis and its MI about this axis is 8ma2f3. By equation (6.36), its KE is therefore (ii) Taking horizontal and vertical axes A x, Ay as shown, let the CM of the bar BC, have coordinates (x, y) . Then X = 3a COS a, y = a sin a and hence the velocity components of are % - 3a sin a y = a cos a (J, and its speed is therefore a6 (9 sin2 a + cos2 a)! = ail (l + 8 sin2 a)t.

G,

=

(J,

G

152

A COURSE I N APPLIED MATHEMATICS It follows that the KE of a particle o f mass 2m moving with G is

[CH.

ma262(1 + 8 sin2 e) . This is one component of the KE of BC. The contribution of the rotary component is ma262f3. The total KE of BC is accordingly tma•e•(l + 6 sin• e) . (iii) Now A C = 4a cos e and hence, by differentiation, we find that the velocity of P is - 4a sin e () upwards. Thus the KE of this particle is 8ma•e• sin2 e. (iv) Combining the expressions (i)-(iv), we now obtain the energy equation, viz., tma202(1 + 6 sin2 e) + 4mga(sin e + cos e) constant, or a202(1 6 sin2 e) + fga(sin e + cos e) = constant. (v) The velocity of C at any instant is the same in magnitude as that of i.e., -4a6 sin e. Initially, e = TT/6, and hence this velocity is then of magnitude 2a6. If V denoteS thiS initial Velocity, ae = tv When e = TT/6 and, by substitution of these values in equation (v), we calculate that the constant is given by constant = tv• + !ga(1 y 3). Hence a202(1 + 6 sin2 6) = fv2 + !ga(1 + v3) - fga(sin e + cos e) (vi) As C moves towards A , e takes all values in the interval (1r/6, TT/2) . Since the left-hand side of equation (vi) must always be positive, such a motion will be possible only if the right-hand side of this equation is positive for such values of e. This side of the equation takes its minimum value when e = t7T· This value is ;tv• - !ga(2y2 - y3 - 1) and it is positive provided v2 > tga(2y2 - y3 - 1). This then is the condition that C should be able to reach A. If v2 = 6ga(2y2 - y3 - 1)/5, equation (vi) may be written a202(1 6 sin2 e) fga( y2 - sin e - cos e), and hence, when (l i1r, a•()• 3ga( y2 1)/14. At this instant the velocity of C is 4a6 = V. Thus v• = 16a•e• = :l-,i·ga(v2 - 1).

=

+

P,

+

+ =

= =

-

6.5. Conservative Systems of one Degree of Freedom. Equilibrium. Small Oscillations

Stability of

The configuration adopted by a system of bodies, which can move subject to the limitations imposed by various constraints, at any instant during its motion, can usually be completely specified by stating the values taken at this instant by a finite number of independent quantities of a geometrical nature associated with the system and called its generalized coordinates. For example, the configuration of the system shown in Fig. 6.4 is determined by stating the magnitude of LCAB and the distance AC. This angle and this distance are independent quantities, since the value of one does not fix the value of the other,

6]

MOTION OF A PARTICLE SYSTEM-I

153

and they are accordingly convenient generalized coordinates of the system. Consider in this connexion Example 7, p. 149. If the angle e through which the cylinder has rotated is known, the position of every particle of the system is determined. e is a generalized coordinate of this simple system. Instead of e, we might choose the distance moved by the axis of the cylinder from its initial position as the coordinate of the system, since, this being known, the angular rotation of the cylinder can also be determined. We conclude that a set of generalized coordinates for a particular system can be chosen in many ways, but the number of such coordinates will always be the same and is called the number of degrees offreedom of the system. Consider a rigid body free to move in any manner parallel to a fixed plane. If G is its CM, two coordinates will be necessary to specify the position of G in its plane of motion. Having fixed the position of G, it will be necessary to state the rotation the body has undergone about G from some reference orientation before the position of the body is completely known. The number of generalized coordinates is accordingly three and the system has three degrees of freedom. Let e be the generalized coordinate of a conservative system having one degree of freedom. The PE of the system depends only upon the latter's configuration, and hence upon e. Thus, if V is the PE, then V = V(e) . Let m be the mass of any particle of the system and let (x, y, z) be its coordinates at time t referred to any fixed set of rectangular axes. The velocity components of this particle in the directions of the axes are (.X, y, z) and hence its KE is

tm(x2

+ y2 + z2).

But (x, y, z) are determined when the configuration of the system is known and hence x = x(e), y = y(e), z = z(e). Differentiating these equations, it follows that

.X = x'6, y = y'6, z = z'6, primes denoting differentiations with respect to e. Thus the particle's KE is lm(x'2 + y'2 + z'2)62 and the KE of the system is

!!m(x'2

+ y'2 + z'2)62 = !62!m(x'2 + y'2 + z'2) = !]62,

where ] = !m(x'2 + y'2 + z'2) is a function of e only. The energy equation of the system can now be written down in the form

!]62 + V = constant.

(6.47)

This equation is a differential equation of the first order determining e as a function of t, and hence determining the motion of the system. A conservative system having only one degree of freedom is therefore

1 54

A COURSE IN APPLIED MATHEMATICS

[CH.

of an especially simple nature, in that energy considerations alone suffice to determine its motion. Differentiating equation (6.47) with respect to t, we obtain the equation ]66 + U'63 + V'6 = o, or (6.48) Je + U'62 + V' = o. Suppose that an equilibrium configuration exists for which 6 takes the value ex. Then 6 = ex represents a possible ' ' motion ' ' of the system and hence satisfies equation (6.48). Substituting, and noting that 6 = e = 0, we find that

V'(ex) = 0.

(6.49)

This implies that the PE takes a stationary value in any position of equilibrium. Now suppose that the system rests initially in the equilibrium configuration 6 = ex and is then slightly displaced so that motion commences. Let 6 = ex + e: during the subsequent motion, where e: and E: are small quantities, at any rate during the early stages of this motion. Substituting for 6 in equation (6.48), we find that e: is governed by the equation

](ex + e:) e + U'(ex + e:) E:2 + V'(ex + e:) = 0.

(6.50)

In this equation we shall now neglect all small quantities of the second order. Thus, the second term of the left-hand member of the equation will be disregarded. Also, by Taylor's theorem,

(6.51) ](ex + e:) = J(ex) + e:J'(ex) + O(e:2), V'(ex + e:) = V'(ex) + e:V"(ex) + O(e:2), (6.52) = e:V"(ex) + O(e:2). To the first order, therefore, equation (6.50) is equivalent to ] (ex)€ + V"(ex)e: = 0, or (6.53) e: + V"(ex) J(ex) e: = 0 . Since tf62 is the KE of the system, J is positive for any admissible value of 6. The sign of the quantity V"(ex)/J(ex) is accordingly de­ termined by that of V"(ex). If V"(ex) > 0, equation (6.53) can be written (6.54) 2 where (i) = V"(ex)/J(ex) and, during the early stages, the motion of the ··

system is clearly approximately simple harmonic of period

T = 21tJ V" (ex)"

JM

(6. 55)

MOTION OF A PARTICLE SYSTEM-I

6]

155

During such a motion, e: remains small, and hence equation (6.53) remains valid indefinitely. In such circumstances, therefore, the system oscillates indefinitely about the equilibrium position, showing no tendency to move far away from the equilibrium con­ figuration. We say that the equilibrium is stable. For stability, we must have V"{oc) > 0. Taken in conjunction with equation (6.49) , this implies that the PE takes a minimum value at e = oc. Next, suppose that V"(oc) < 0, so that equation (6.53) may be written (6.56) where n2 = - V"(oc)/](oc) . The general solution of this equation is e: =

Aent + Be-nt ± as t -+

(6.57)

and hence, in general, e: -+ oo oo This implies that although e: is initially small, it does not remain so, and eventually equation (6.53) ceases to be a valid approximation to the equation of motion of the system. Clearly, in these circumstances, the system reveals a tendency to move away from the equilibrium con­ figuration when slightly disturbed, and we say that the equilibrium is unstable. In this case the PE takes a maximum value in the equili­ brium position. The intermediate case V" (oc) = 0 cannot be satisfactorily decided unless the effects of the second-order terms in equation (6.50) are taken into account. A , B, C, D are four points on a smooth, horizontal table which form the corners of a square whose diagonals are each of length 2d and whose centre is at 0. A particle of mass m is placed at 0 and is connected to A and C by two equal light, elastic strings, each of natural length l( < d), and connected to B and D by two equal light, elastic strings, each of natural length l'( < d), all four strings having equal moduli of elasticity A. If the particle is slightly displaced from rest at 0 towards C, show that the period of small oscil­ lations is 2m/A t. (D.U.) 'TT lfl + lfl' - lfd

Example 10.

If OP

Also

{

D;c------;lc

0

}

B

= x, the PE of the two strings AP and CP is A A A 21 (d + x - 1) 2 + 21 (d - x - l)2 = [(d - 1) 2 + x2] . I PD 2

(

= d2 + x2,

PD = d l +

where we have neglected terms O (x4) .

�t = d �' +

156

A COURSE I N APPLIED MATHEMATICS

[CH.

The PE o f PD o r PB i s accordingly

� ( d - l' + ;�r = 2�,[(d - l') 2 + ( l - �) x•}

2'

where we have again neglected a term O (x4) . The total PE of the system is now found to be V =

� [ (d - l)2 + x2] + � [(d - l') " + (r - �) x•}

The energy equation is therefore

,\ ( } + � - �) x• = constant, (! + .!:.l - !)d (constant - x2) x• = !mx• +

2,\ m l'

·

Comparing this with equation (2.26), we deduce that the vibration is approximately simple harmonic of period 27r/w where

(!

!)

+ .!:_ w• = � m l l' d ·

A uniform plank is of length 2l and thickness 2t. It is placed sym­ metrically across a rough cylinder of radius r, the axis of the cylinder being horizontal. Prove that if the plank is slightly disturbed so that it makes small oscillations (without slipping) the period of the oscillations is equal to the period of small oscillations of a simple pendulum of length 1• + 4t• " 3(r - t) Deduce the condition necessary for oscillations to be possible. (B.D.)

Example 11.

Let C be the point of contact of the plank with the cylinder before it is disturbed. Let A be the point of contact during the subsequent motion. If LAOB = e, since no slipping has taken place, A C = arc AB = r6. The height of the CM of the plank, above 0 is accordingly t cos e + re sin e + r cos e and the PE of the plank is V(6) = Mg{(t + r) cos e + r6 sin 6}, where M is the plank's mass.

G,

6]

MOTION OF A PARTICLE SYSTEM-I

157

At the instant under consideration the plank is rotating about A , the instantaneous centre (see Chapter with angular velocity e . The speed of G is accordingly AG e . The KE of a particle of mass M moving with G is therefore tMe • . A G• = tMe2 (r2B2 + 2 . The KE of the rotary motion about G is tM(l2 + •) e • Hence, the total KE of the plank is

7)

t .

tMe2(l2 +

It now follows that, in this case, But

t)

4t2 + 3r2B2) .

](B) = tM( " +

l 4t2 + 3r2B2).

V"(B) = Mg{(r

- t) cos B - rB sin B}. V"(O) = Mg(r - t), ](0) = tM(l2 + 4t2). Using equation (6.55), we can now compute the period of small oscillations about the equilibrium position e = 0. It is / l2 + 4t2 27rV 3g(r - t)' i.e., the period of oscillation of a simple pendulum of length (l2 + 4t2)/3(r - t) . If r < t, V"(O) is negative and B = 0 represents a position of unstable equilibrium. The condition for oscillations to be possible is therefore r > t. Example 12. A heavy, uniform chain hangs over a rough pulley consisting of a uni­ form disc of radius a, which can turn freely about a horizontal axis through its centre, and which has a particle of mass m attached to a point of its rim. The mass of a length 2l of the chain is equal to the mass of the particle. The particle is at the lowest point of the pulley when the lengths of chain hanging on each side of the pulley are the same. If the angu­ lar displacement of the wheel e is mea­ b-ae sured from this position, show that e = 0 is a position of stable or unstable equili­ btaS brium according as l is greater than or less than a, and discuss the stability of other positions of equilibrium that may occur if l > a. (M.T.) Let b be the lengths of chain which hang vertically in the position e = 0. In the displaced position the lengths of the hanging portions are b ae and b aB, as shown in the +diagram. The weights of these lengths are mg(b + aB)f2l and mg(b - aB)f2l respectively, and since their mass centres Hence

1 l

-

are at distances i(b + aB), !(b - aB) respectively below the centre of the disc, we can take the sum of their PEs to be

- mg (b + aB)2 - '!!.� (b � �

-

aB)2 =

-

mg (b2 + a2B2) �

0

158

(CH.

A COURSE IN APPLIED MATHEMATICS

Let P (independent of a) be the PE of the portion of the chain in contact with the pulley. The total PE of the chain is then p -

�f (b2 + a2a 2) .

When a = 0 , this PE i s P - mgb2f2l. The change i n PE o f the chain when the pulley rotates through the angle a is accordingly -mga2a2f2l. The change in PE of the particle as a result of this rotation is mga( l cos a), whereas there is no change in the PE of the disc. Measuring the PE from the datum position when a = 0, we obtain

-

mga2 a2 V = - --- + mga(l 2l

Differentiating,

we

���

find that

-

- cos a).

mga•a dV -- + mga sin a, = l da 2 m a = + mga cos a = mga cos a

�

(

- i)

.

(i)

When a = 0, d Vfda = 0, confirming that the datum position is one of equilibrium, and d2 V2/dfJ 2 = mga(l - a)fl. If l > a, d2 Vfdf:J 2 > 0 and this position is stable. If l a, d2 Vfdf:J 2 and this position is unstable. Other positions of equilibrium are given by roots of the equation d Vfd6 = i.e.,

0,

<

<0

sin a= 7 a.

(ii)

If l > a, the graph of aafl is a straight line through the origin of gradient less than unity. This curve therefore intersects the graph of sin a in points (see diagram) other than the origin 0. These points of P 1 , P 2 , P3 ,

intersection have absciss<e which are roots of equation (ii) and therefore correspond to equilibrium positions. The graph of sin a has gradient cos a at every point, whereas that of a6fl has gradient afl. The sign of the difference between these gradients at the points P 1, P2 , P3 , accordingly determines the sign of d2 Vfd6 2 and hence the nature of the equilibrium positions. Thus, it will be observed from the diagram that the points P1, P 2, P 3, correspond to unstable, stable, unstable, equilibrium positions respectively. There will also be a set of equilibrium positions in which a takes negative values.

6]

MOTION OF A PARTICLE SYSTEM-I

159

6.6. Stability of Rocking Bodies

A problem of stability of equilibrium which is conveniently introduced at this juncture is that relating to the equilibrium of a movable cylinder which rests under gravity with one of its generators in contact with a generator of a fixed cylinder. A right section of the two cylinders is shown in Fig. 6.7, C being a point on the generator of contact. Only two forces act upon the movable cylinder, viz., the weight force Mg through its centre of mass G and R, the reaction of the fixed cylinder. The movable cylinder can rest in equilibrium only if these two forces are equal and opposite and have the same line of action. Hence G is vertically above C. Suppose R makes an angle oc with the common normal at C. Then oc must be less than the angle of friction at this

FIG. 6.7.-Rocking Body

point, otherwise slipping will take place. We shall assume the co­ efficient of friction to be sufficiently large to permit the upper cylinder to rest in equilibrium. Consider the nature of this equilibrium relative to small rocking motions of the upper cylinder. If the latter is caused to roll on the lower cylinder without slipping so that the arc CA rolls on the arc CA ', A eventually making contact with A ', then arc CA ' = arc CA = ds. In the equilibrium position, let the tangent to the upper cylinder at A make an angle drp with the common tangent at the point of contact C and let the tangent to the lower cylinder at A ' make an angle drp' with this line. If p is the radius of curvature of the section of the upper cylinder at C, the arc AC is approximately an arc of a circle of radius Hence ds = pdrp. Similarly, p subtending an angle drp at the centre. ds = p 1 drp 1, where p' is the radius of curvature of the section of the lower cylinder at C. As the upper cylinder rolls, the tangent at A moves into coincidence with the tangent at A 1, and hence the angle of rotation is

drp + drp' = ds

G f,). +

(6.58)

A COURSE IN APPLIED MATHEMATICS [CH. For small angles of rotation, G moves along an arc of a circle having centre C (the instantaneous centre) and radius CG = h. The dis­ placement of G is accordingly h(dcf> + dcp 1 ) = hds (� + t�} (6.59) 160

and this displacement will be horizontal to the first order of small quantities. Let G1 be the displaced position of G. If G1 is to the right of the vertical through A 1, as shown in the right-hand figure, the weight will possess a moment about the new point of contact A 1 in a sense tending to roll the upper cylinder back towards the equilibrium position. This position is then stable. If G1 is to the left of the vertical through A 1, the weight will have a moment about A 1 in a sense tending to roll the upper cylinder farther away from the equilibrium position, which is accordingly unstable. Now the distance between the verticals through A and A 1 is approximately A 1 C cos ds cos Hence, for stability we must have

GG1 < ds cos

hds

or

(� + ;,) < ds cos !

i.e.,

<X,

p

+

� < cos p

h

<X.

<X =

<X.

<X,

(6.60)

Reversal of the inequality gives the condition for instability. If the lower surface is concave upwards, it is necessary to treat 1 p as a negative quantity. A uniform thin, hemispherical bowl of radius a rests in equilibrium on a fixed solid sphere with its outer curved surface in contact with the sphere so that the common normal makes an angle <X with the vertical. Show that cx must be less than 1r/6. Show also that, if the radius of the sphere is 3a, equili­ brium is stable only if sin ct < l f y'5.

Example 13.

The above argument is clearly applicable with only slight modification when the cylinders are replaced by spheres. If G is the CM of the hemispherical bowl and Q is its centre, then QG ta. G is vertically above the point of contact C, so that L QCG = ct . Applying the Sine Rule to 6 CGQ, we have

=

sin cx = But

g� sin L QGC = ! sin L QGC.

0 � sin L QGC � l and hence

i.e., cx � 1r/6.

O � sin cx � t,

6]

M O T I O N O F A PART I C L E S Y ST EM-I Let CG

161

Construct QN perpendicular to CG produced. Then CN - GN a cos "' - (!a2 - a2 sin2 <X)t, = a{cos "' - (! - sin• <X)l}. The condition for stability is l cos "' 1 < 3a a a{cos "' - (! - sin2 <X)l}' 4 cos "' - 4(! - sin2 <X)l < 3 cos "'• cos "' < 4(! - sin2 <X), - sin2 "' < 4 - 1 6 sin2 "'• sin2 "' < 3

= h. h

=

=

+ 1

15

Slll <X < .

V 5. 1

EXERCISE 6 1 . A triangular prism of mass M rests on a smooth, horizontal plane so that one of its faces is vertical and another inclined at an angle ex to the horizontal. At the mid-point of a horizontal line on the inclined face rests a rough particle of mass m, the coefficient of friction between the particle and the prism being tan ex. If the prism is pushed by a hori­ zontal force P applied symmetrically at right angles to the vertical face, show that the particle can never slide downwards. If M = 2m, ex 30° and P 4mgy3, show that the particle will slide up the inclined face with an acceleration of 2gf5 relative to the prism. (L.U.) 2. A rectangular block of mass M is provided with four light wheels and is free to move along two horizontal smooth rails. A particle of mass m1 slides on the horizontal smooth upper face of the block and is attached to a light, inextensible string which passes over a smooth pulley secured to one edge of this face and then hangs between the rails supporting a particle of mass m2 • The string lies in a vertical plane parallel to the rails. The string is drawn aside from the vertical through an angle "'· and the system is then released from rest. If the inclination of the string to the vertical remains constant at its initial value during the motion, show that M M . cosec ex + sm ex = - + - + 2, ml m 2 and that the acceleration of the block is g tan ex . 3. Three light inextensible strings AB, BC, CA are respectively of lengths a, a, ay2, and are knotted together at A , B, C. Masses m, M are carried at B, C respectively, and are moving on a smooth, horizontal table, so that ABC rotates freely about the knot A , which is pinned to a point of the table. If the string A C is cut, show that the ratio of the tensions in BC and AB immediately afterwards is Mf(m + M) . 4. Four equal light, inextensible strings are joined together to form the sides of a rhombus ABCD. A light bar forms the diagonal ED. Four particles of equal mass are placed at the vertices and the system hangs in equilibrium from a string attached to A . If the bar is G (1)

=

=

A C O U R S E I N A P PL I E D M A T H EMAT I C S

162

[CH.

suddenly removed without disturbing the configuration, show that the tension in this string is instantaneously reduced by a factor - cos L BAD) .

1/(2

5. Two particles whose masses are in the ratio 5 : 8 are attached to the points of trisection of a light, elastic string stretched between two points on a smooth, horizontal table. Find the periods of normal modes of transverse oscillation, and prove that in one of the modes the particles move in the same sense with amplitudes in the ratio : 5, and in the other mode the particles move in opposite senses with amplitudes in the ratio Three equal light springs AB, BC, CD, each of unstretched length and the same modulus, are joined end to end and carry masses at apart, on a B and C. The ends A and D are fixed points, distant smooth, horizontal table. Determine the normal modes of oscillation along the line AD and show that their frequencies are in the ratio : Show also that at the lower frequency the oscillations of the particles are identical, whereas at the higher frequency they are of the same amplitude but are in anti-phase. OAB is a light, elastic string of modulus :A hanging vertically from a support 0. A particle of mass is attached at A and a particle of mass at B. OA and AB The particles are set oscillating along the line of the string. Assuming that the string never goes slack during the motion, show that the equations of motion of the particles are 0, 0, where are the downward displacements of the particles from their equilibrium positions, and Show that two normal modes exist, one having double the frequency of the other. Calculate the ratio of the amplitudes of the particles' oscillations in each case. 8. A bead of mass slides on a fixed smooth circular vertical wire of radius and is connected by a light, inextensible string which passes over a smooth peg at the highest point of the wire to a particle of equal mass which hangs vertically. Calculate the minimum velocity with which the bead must be projected from the lowest point of the wire if it is to reach the highest point.

2 l.

6.

3a

4 (B.U.) a m

y'3 l.

7.

m

=a

(D2 + 3w2)x - w2y = 2w2x - (D2 + 2w2)y = w2 = :Af4ma.

x, y

a

2m = 2a.

m

A particle P of mass lies on a smooth table and is attached to a long, inextensible string which passes through a small smooth hole in the table. The other end of the string carries a particle Q of mass hanging freely. The particle P is projected from rest along the table at right angles to the string with speed when it is a distance from the hole. Prove that the particle Q will begin to rise if < 8, then and that, if the greatest distance of P from the hole is (L.U.) 10. A smooth, narrow tube, with open ends, in the form of one complete arch of a cycloid s sin .P is fixed with the tangent at the mid-

9.

m

(Sag)!,

2a,

= 4a

km a k k = 3.

6]

M O T I O N O F A PARTI CLE S Y ST EM-I

163

point A (rp = 0) of the arch vertical. A particle of mass is placed in the tube at A and is connected to a light, inextensible string. The string, whose length exceeds that of the arch, lies in the tube, and after passing out through the upper end of the tube is attached to another particle of mass which hangs freely. If the system is > show that released from rest and = sin sin and that if the reaction between and the tube vanishes when = t7t', then + (1t If < show that will = emerge from the lower end of the tube provided < 1tm. (L.U.) The period of small oscillations of a compound pendulum is It is hanging in equilibrium when it is suddenly set rotating with angular velocity Show that > if the body is to make complete revolutions in a vertical plane. Two simple pendulums are formed from equal light, rigid rods A BC, DEF of length suspended from the points A and D, which are on the same level. The bobs, of equal mass are placed at C and F. A light spring of stiffness and natural length AD joins B show that the normal periods of small and E. If AB = DE = oscillations about the position of equilibrium are and (M.U.) = = and where A light rod AB of length is freely hinged at B to a uniform rod BC of length and mass M. The end A is freely hinged to a fixed sup­ port, and C is attached to a small, light ring which moves on a smooth, horizontal rail passing through A . The system is released from rest with the rods horizontal, and B mid-way between A and C. If, after time t, the rod AB makes an angle 6 with the horizontal, prove that _ sin 6 ff. + sin2 6) (M.T.) A uniform rod of mass and length turns freely in a vertical plane about a fixed axis through one end 0. A light, elastic string of modulus and natural length :ll has one end attached to the mid-point of the rod and the other to a fixed point A at distance !1 vertically above 0. If 6 is the angle made by the rod with the upward vertical OA , show that the equation of motion of the rod is = sin !6) cos !6. Show also that the period of small oscillations about the position of equilibrium 6 = is the same as that of a simple pendulum of length (L.U.) One end of a uniform, rectangular plank of length and mass M is smoothly hinged to a smooth horizontal table. The plank rests against an upper edge of a smooth block of rectangular section, height and mass M', which lies on the table with the edge against which the plank rests parallel to the hinge. If the block is held at rest with the plank inclined at to the table, and then released, find the

m

11. 12.

13.

14.

m', m' m (m + m')s2 Sm'ga .p - 2mga 2,P - 4mga.p, m .p m m' jm (2y2 1) + 4) /14. m' m, 4m' T. w. w 47t/T 1

w12 gfl 2a

21mgfl 1/3, 3w 22 17gf1. 2a

() 2 - 2(1 3 6 m l

m,

27t/w1

a

mg

81/9.

15.

216 3g(1 - 2 7t/3

2a

ta

30°

27t/w 2

164

A C O U R S E I N A P P L I E D M A T H E MA T I C S

[cH.

angular velocity of the plank when it is inclined at an angle 6 to the table. Show also that the block has velocity

)t 2 ( M 6Mga + 48M'

16.

at the instant when the plank falls off. (L.U.) Two equal uniform rods A B, BC, each of length and mass smoothly hinged together at B, are held in a vertical plane with their ends A and C resting on a smooth, horizontal table, the angle ABC A light string attached to B passes over a small pulley being vertically above B and carries a mass at its other end. If the system is then released, prove that when the rods become horizontal (L.U.) the angular velocity of each of them is cos A uniform, solid sphere, centre 0 mass and radius has a uniform rigidly attached to a point A rod AB of length and mass of its surface so that 0, A , B are always collinear. Initially the system is at rest, with A B vertically above 0 and the sphere on a perfectly rough, horizontal table. Show that if AB is given a small displacement from the vertical and 6 denotes the angle between A B and the vertical, the value o f when B strikes the table is (L.U.) Two uniform rods A B, BC, freely hinged at B, are of equal mass and of lengths and respectively. AB is freely hinged at A to a fixed point on a smooth, horizontal table. The rods are held in a vertical plane, with C resting on the table and the angle ABC a right angle. If the rods are then released, prove that they will strike the table when the angular velocity of AB is V (L.U.) is fixed with A hollow, right-circular cylinder of internal radius its axis horizontal. A rough, solid, circular cylinder of radius 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 6 with the downward vertical, show that the angular velocity of the movable cylinder is If the movable cylinder is slightly disturbed from its position of equilibrium, show that it will oscillate with period (L.U.) A uniform circular hoop, of radius mass 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 prove that, when OA makes an angle e with the downward vertical, cos cos 6) . Assuming that contact with the plane is maintained, prove that, if > A can rise to a position vertically above 0 and that it will downwards. then have an acceleration (L.U.) A uniform cotton-reel has radius of gyration about its axis. A length of light cotton is fixed to the surface of the reel and wound

2a

2cc.

17.

7mf20

2a

m/3 y(2g m

ccf3a). a,

6

18.

3a

m,

(2gf3a)i.

4a

4gf5a.

19.

7a

a

66.

20.

61ty(a{g). a, m

(2 -

aw2 2g,

21.

w, 6)a62 aw2 - g(l =

(aw2 - 2g)/3

k

6]

M O T I O N O F A PA R T I C L E S Y S T EM-I

165

,around it in a circle of radius whose plane is perpendicular to the axis and contains the mass centre. The free end of the cotton is held fixed whilst the reel falls vertically from rest and the thread unwinds from it. During the motion, the unwound thread remains vertical. Equating the rate of increase of KE to the rate at which the forces do work, show that the downward acceleration of the reel is Hence calculate the length of thread which unwinds in time t. A uniform, isosceles, triangular lamina of height and weight can turn freely about its base, which is fixed and horizontal. One end of a light, elastic string of natural length is attached to the vertex of the triangle and the other end to a point at a height vertically above the mid-point of the base. If the modulus of the string is find the inclinations of the triangle to the vertical in the four positions of equilibrium, and show that the positions in which the plane of the triangle is vertical are unstable. (L.U.) . A wheel, of mass and centre 0, consists of a uniform circular disc of radius in which have been cut four circular holes. The radius of each hole is and their centres, which are at a distance from 0, form the vertices of a square. Show that the moment of inertia of the wheel about an axis through 0 normal to the plane of the wheel is The wheel is free to rotate about this axis. A light, elastic string of modulus and natural length joins P, a point on the circumference of the wheel, to a fixed point A in the plane of the wheel. If A O = show that the length of the equivalent simple pendulum for small oscillations of the wheel about its stable position of equilibrium is (L. U.) To the ends of a light, inextensible string which passes over a small, smooth pulley are tied respectively a particle of mass and a bead of mass The bead moves on a straight, smooth, horizontal wire which is fixed in a vertical plane through the pulley at a depth below it. If the bead is initially at rest at a distance from its equilibrium position, find an equation to give its speed when at a distance x. If the bead is displaced slightly from its equilibrium position, prove that it will return to this position after a time (L.U.) A smooth, rigid wire in the form of a circle of radius is fixed in a vertical plane. A second smooth, rigid wire forms the lower vertical radius of the circle. A uniform rod of length such that < l < is constrained by small rings at its ends P and Q, which can slide the one P on the circle and the other Q on the straight wire. Show that there are positions of equilibrium in which the rod is inclined to the vertical and find the angles which the rod makes with the vertical in these positions. (L.U.) A smooth, circular wire of radius and centre 0 is fixed with its plane vertical. A uniform rod PQ of length and weight has small, light rings at its ends, which slide on the wire. A light, elastic string of natural length and modulus has one end tied to C, the mid­ point of PQ, and the other end to A , a fixed point at a distance

a,

ga2f(a2 + k2).

22.

3a

W

a

4a

5Wf48,

23.

a

M

;!a,

ta

55Ma2/96.

2Mg

a

3a, 55af288.

24.

M

m.

a

2a

irty(mafMg).

25.

a

l,

26.

a

a

6Wf7

6af5

ta

a,

W

2a

166

A C O U RS E I N A P P L I E D MATH EMAT I C S

[CH.

vertically above 0. Show that when the angle is - 6, the PE of the system is cos 6 cos 6)] + a constant. lrr Deduce that there are four possible positions of equilibrium, and that the two in which the rod is horizontal are both unstable. (L.U.)

AOC 1t

Wa[5

27.

A uniform rod of mass and length is smoothly hinged to a fixed point at A and has a particle of mass attached to which is connected by a light, elastic string of modulus and natural length to a point vertically below If C = show that the period of small oscillations about the position of equilibrium with vertically below is (L.U.)

AB,

!a

28.

3y(29 + 20 2m

C A 1t(5af3g)i.

a, m

A.

A

B,

mg 2a,

B

Two small, light rings, joined by a uniform rod whose mid-point is M, can slide on a smooth, circular wire fixed in a vertical plane with its centre at 0. The weight of the rod is W and the angle it subtends at 0 is The rings are connected by a light, elastic string which passes over a smooth peg situated at the highest point of the wire the modulus being W and the natural length where a is the length of the radius of the wire. Find the PE of the system when makes an acute angle 6 with the downward vertical. Show that there is a position of equilibrium when and determine whether the equilibrium is stable or unstable. (L.U.)

120°.

2a,

OM

e 60° =

29. A uniform bar AB of length l and weight W1 is smoothly hinged to a fixed point A. The bar is supported by a light, inextensible string attached to B and passing over a small, smooth pulley at a point C,

which is vertically above A at a height a( >l) . The string carries a freely hanging weight W 2 at its other end. Show that there are three positions of equilibrium, provided that

� < w1 < � w 2 a - l·

a+l

If W1 2 show that two of the positions of equilibrium are stable and one is unstable. (L.U.) =

30.

2W

A thin, smooth wire, in the form of a circle of radius is fixed in a vertical plane, the horizontal diameter being A bead of mass can slide on the wire, and attached to it is a light, inextensible string which passes through a smooth ring fixed at and carries at its other end a mass hanging vertically below Prove that there is an equilibrium position in which the bead is above the diameter and show that this position is unstable. If the bead is held at A and then set moving upwards with velocity find the least value of for which the bead will pass to (L.U.)

m,

B.

V,

B.

AB. B,

a,

m

AB V

31. A cylinder of radius a, whose axis 0102 is always horizontal, can roll on a perfectly rough plane which is inclined at an angle IX to the hori­ zontal. The centre of gravity G of the cylinder is at a distance b from 0102• Prove that, if b > sin IX, there are two possible positions of equilibrium and that in each case the plane G0102 makes an angle

a

M O T I O N O F A P A R T I C L E S Y S T E M- I

6]

167

sin-1 sin with the vertical. Prove also that one of the two positions is stable, and indicate which. (L.U.)

(a

32.

ajb)

A uniform, hemispherical shell of mass and radius c rests with its curved surface on a rough, horizontal table. A particle of mass is fixed on the rim and the system is allowed to roll to rest. If it be slightly disturbed in the vertical plane through the particle and the

M

3Mj8

centre, prove that the frequency of small oscillations is

4� ( 1:f) t

(L.U.)

33.

A uniform rod of length and weight can turn in a vertical plane about a smooth, horizontal axis through 0, distant from the end and carries a bead of the same weight which can slide smoothly on The bead is attached to a light, inextensible string of length which passes through a smooth ring at 0, over a smooth, fixed peg at a height vertically above 0 and is attached to the end of the rod. If is the inclination of to the downward vertical, show that the PE of the system is = cos EJ (cos !8 + constant. Hence show that there is a position of equilibrium of the system in which is acute and is equal to Show also that this position is stable. (L.U.)

AOB

A,

7a,

e

34.

W

P

OA .

e

A

6a

W

2a

OA 1) (11 2y10)af3.

V 4Wa OP

A uniform, thin rod of length is suspended from two fixed points at the same level by vertical strings each of length attached to its ends. It is turned through an angle about a vertical axis through its centre, and let go, the strings being kept taut. Construct the energy equation for the subsequent motion. If a is small, determine the period of the oscillation. (L.U.)

2a

a

35.

2a

b,

A right cylinder rests in equilibrium under gravity on a rough plane. Show that the equilibrium is stable provided the centre of mass is below the centre of curvature of a right section. The base of a uniform, solid cone is of radius and is secured to the plane face of a uniform, solid hemisphere of the same radius. The body so formed rests in equilibrium with the curved surface of the hemisphere in contact with a rough, horizontal plane. If p, p ' are the densities of the cone and hemisphere respectively and is the height of the cone, show that the equilibrium is stable provided

r

h

3r2 h2

p -' < - · p

36.

A hollow, right circular cylinder is fixed with its axis horizontal. A uniform, solid, right elliptical cylinder rests in equilibrium on the outside surface of the circular cylinder with its generators horizontal. One of the equi-conjugate diameters of a right section of the elliptical cylinder is vertical. Show that the equilibrium is unstable. If the elliptical cylinder is transferred to the inside of the fixed cylinder so that it rests in equilibrium in the same position, show that the equilibrium is then stable.

168

A C O U R S E I N A P P L I ED M A T H E MA T I C S

[cH. 6]

of length 2a and mass m is smoothly hinged at 37. A uniform rod A light spring of natural length and stiffness is attached at one end to and at the other end to D, a point vertically above = D = 3af2, find whether or not the positions of equilibrium If are stable when 2mgf3a and when 2mgf3a. (M.U.) A particle of mass m is connected by a light, inextensible string of 38. length 2a to a bead. of mass m, which is threaded on a fixed, smooth, vertical wire ; initially the string is taut and horizontal, is at rest and is moving vertically upwards with velocity u. If is the inclination of the string to the horizontal in the subsequent motion, find expressions for the reaction between the bead and the wire in terms of 8 and its derivatives with respect to the time. Deduce that + sin2 8) = u , and show that the tension of the string is mu2 (S.U.) 39. A homogeneous solid, in the form of a segment of a paraboloid of revolution, latus rectum cut off by a plane perpendicular to the axis at a distance from the vertex, is balanced vertex downward on the top of a rough sphere of radius Prove that the equilibrium is stable if (S.U.) A uniform rod of mass m and length can turn freely in a vertical plane about a fixed, horizontal axis at the end An elastic string of unstretched length �a and modulus ). is attached to and to a = If 3:A mg, prove that point vertically over where there is a stable position of equilibrium given by sin !oc: = mg) , where Ct: is the angle the rod makes with the upward vertical. If the rod is slightly disturbed from this position, show that the period of small oscillation is mg)am (S.U.) 7t 3 (5:A - mg) (3:A - mg) ·

ABC A. B AB A k> A B A

4a262(1

a

B

2

4a,

a.

b < a. AB,

C

A.

k<

b

40.

k

a,

A,

4[

A.

>

AC a. :A/(4:A -

(4). -

J

t

B

e

CHAPTER 7

KINEMATICS OF A RIGID BODY.

MOMENTS OF INERTIA

IN this chapter we shall be concerned with certain geometrical con­ siderations relating to a rigid body and its motion. 'i .1. Instantaneous Centre.

Space and Body Centrodes

Consider a rigid lamina moving in its own plane in which Oxy is a fixed, rectangular, cartesian reference frame (Fig. 7.1). Let O'uv be a

FIG. 7.1.-Two-dimensional Motion of a Lamina

similar reference frame fixed in the lamina. It will be convenient to specify a point P having coordinates (x, y) relative to the stationary frame by the complex number z = x + iy and a point having co­ ordinates (u, v) relative to the moving frame by w = u + iv. If P represents a particle of the lamina, z will be a function of the time t and, differentiating, (7.1) The vector representing the complex number z will accordingly have components (x, y) relative to Oxy, and it follows that z specifies com­ pletely the velocity of P. w will be constant for such a particle. Let 0' be represented in the stationary frame by the complex number -+ z0• Relative to O'uv, the vector 0' P is represented by the complex number w. Thus, if ifi is the angle made by O 'u with Ox, relative to

169

170

A C O U R S E I N A P PLI E D MATH EMAT I C S

Oxy,

this vector is represented by the number frame, ---+ ---+ ---+ OP = 00' + O'P, and thus z = z0 + weio/1.

wei.P.

[cH.

But, in any

(7.2)

Differentiating this equation with respect to t, we find that z = z0 +

iweio/1�,

(7.3)

giving the velocity of the particle P. Since wei.P represents the vector . 0'---+P---+ relative to the frame Oxy, iwei.Pf represents a vector perpendicular . to 0' P and of magnitude f . 0' P. Equation (7.3) therefore states that the velocity of P is compounded of the velocity of 0' and a com­ ponent due to the rotati0n of P about 0' with angular velocity � · The particle P will be stationary if z 0, i.e., if =

w -- izo . e-io/1. f

(7.4)

Thus, provided � =F 0, there will always be a particle of the lamina (given by equation (7.4)) which is at rest at any instant. The position of this particle is called the instantaneous centre (IC) of the motion at the instant under consideration. If 0' is taken to be coincident with the IC at some instant t, z0 will vanish and the velocity of any particle P of the lamina at time t can be calculated by regarding it as being in rotation about the instantaneous centre 0' with angular velocity � · The reader is warned that the acceleration of a particle cannot be found in this way (see Ex. 2 at the end of the chapter). If � = 0, the lamina is not rotating and there is no IC. Alternatively, since w ---+ cc as � ---+ 0 in equation (7.4), we can say that the IC is at infinity in this case. If the directions of motion of two particles of the lamina are known, the position of the IC is easily found. For this must be such that rotation about it causes the particles to move in the known directions. Hence, it must lie on the perpendiculars through the particles to their respective directions of motion. The IC lies at the intersection of these perpendiculars. Exercise AB, BC, CD are three bars pivoted at B and C. AB and CD rotate in the same plane about pivots at A and D respectively. Show that the IC for BC lies at the intersection of and CD (produced if necessary) .

AB

As the motion of the lamina proceeds, the IC will, in general, take up different positions on the lamina. Thus, in the case of a disc rolling along a straight rail without slipping, the point of contact of the disc with the rail at any instant is the IC, since this point is clearly at rest.

7]

K I NEMAT I C S O F A R I G I D B O D Y

171

But all points on the circumference of the disc come into contact with the rail. Hence, an observer moving with the disc and rotating with it will note that the IC traces out a circle on the disc, viz., the circum­ ference. The locus of the IC relative to the lamina is a curve which can be imagined as drawn upon the lamina and is called the body centrode. A stationary observer will also note that the IC traces out a curve. This locus, fixed in space, is termed the space centrode. In the case of the rolling disc the space centrode is the line of the rail. Writing equation (7.4) thus, (7.5) we have an equation of the form w = w (t/J). As t/J varies, the vector w traces out the body centrode. Equation (7.5) accordingly defines this locus. Substitution for w from equation (7.5) in equation (7.2) yields an equation for the position of the IC relative to the frame Oxy, viz., (7.6) This equation is of the form z = z (t/J) and determines the space centrode. At any instant during the motion the body and space centrodes

FrG. 7.2.-Space and Body Centrodes

must have a point I in common, namely the IC at this instant, for it lies on both loci (Fig. 7.2). As the motion proceeds, successive points along the body centrode will take up their position on the space centrode and, as they do so, play the role of the IC. Suppose that the points of the arc IK of the body centrode take up positions on the arc IH of the space centrode, K coinciding eventually with H. If the arcs IH and IK are always equal, then the phenomenon we have been

1 72

[cu.

A COURSE I N APPLIED MATHEMATICS

describing may be regarded as the rolling of the body centrode on the space centrode. We proceed to the proof of this equality. Differentiat­ ing equations (7.5), (7.6) with respect to if;, we obtain

dw = ( dz0 + . d2z0 ) e--i.P, dif; dif; z dif;2 dz _ dz0 . d2z0 dif; - dif; + z dif;2 '

(7.7) (7.8)

the first equation referring to the body centrode and the second to the space centrode. It follows that

= ,;�, 1 �: 1 (u' 2 + v' 2) ! = (x' 2 + y' 2)!,

or

(7.9) (7.10)

primes denoting differentiations with respect to if;. But

={'(x'2 + y'2)idif;, ' arc IK = { (u' 2 + v' 2)idif;, .

arc IH

"'·

"'·

(7.11) (7.12)

if;0, if;1 being the values of if; when the IC is at I and H respectively.

Equation (7.10) now implies the equality of the two arcs. We have accordingly proved that the motion of a lamina in its plane can, in general, be represented as the rolling of a curve fixed in the lamina (the body centrode) on a curve fixed in space (the space centrode). Example 1.

A wire ABC encloses an angle {3 between the straight A parts AB and BC. The wire moves in its own plane so that A B passes through a fixed point P, whilst BC touches a fixed circle, centre 0. Prove that the fixed centrode is a circle which passes through 0 and P, and find the moving centrode. (L.U.) The particle of the bar AB which is instantaneously at P

c

must be moving along the line of AB. Hence I, the IC, lies on the perpendicular PI to AB. Let BC touch the fixed circle G at Q. Then the particle of BC instantaneously at Q is moving along the line of BC. Hence I lies on the normal OQ to the fixed circle, The position of I has now been determined.

K I NEMATICS O F A RIGID BODY

7]

173

Consider the quadrilateral BPIQ. The angles at P and Q are right angles. Hence LPIQ = 7T - {3, i.e., this angle is constant throughout the motion. It follows that the segment PO, fixed in space, subtends a constant angle at I and therefore that the locus of this point in space is a circle passing through P and 0. This is the space centrode. Construct OG parallel to CB, meeting AB produced in G. Then LBGO {3. Hence the quadrilateral PIOG is cyclic and G lies on the space centrode. Now GI subtends a right angle at P, and thus GI is a diameter of the space centrode. But the space centrode is a fixed circle, and hence GI remains constant in length throughout the motion. Also, if is the radius of the fixed circle, centre 0, BG cosec {3, and it follows that G is fixed relative to the wire ABC. Relative to the wire, therefore, I moves on a circle, centre G, radius the constant length GI. This is the body centrode. We note that its radius is double that of the space centrode. The motion of the wire can now be reproduced by rolling the larger circle around the smaller, the latter remaining within the former.

=

r

=r

'i 2 Motion of a Rigid Body about a Fixed Pivot .

.

In the case of a rigid body in general motion, there is no particle which is at rest at any instant, i.e., there is no instantaneous centre. If, however, the rigid body moves about a fixed pivot at any instant, all the particles which lie upon a certain straight line through the pivot are stationary, i.e., the body is instantaneously in rotation about this line as axis and an instantaneous axis of rotation exists. This we proceed to demonstrate. Let 0 be the fixed pivot and let x, y be the respective position vectors, relative to a fixed frame of reference with origin 0, of two particles of the body X and Y. x and y will be functions of the time t and completely specify the orientation of the body at any instant. Let z be the position vector of any other point Z of the body. Since the body is rigid, x, y, z will be constant in magnitude, and thus x2, y2, z2 will each be constant. Differentiating these quantities with respect to t, we deduce that X · i: =

Y

·y=

Z

·

Z = 0,

(7. 13)

i.e., the points X, Y, Z move at right angles to their position vectors. Also, the angles between the vectors x, y, z will not vary during the motion, and hence x · y, y · z, z · X are all constant in value. Differentiating, we obtain the conditions Y

·Z + y·Z=

Z•

X+

Z X = X · Y + X · Y = 0. •

(7.14)

It follows from equations (7.13) and (7. 14) that y.z

x.

i: +

x.

y(z . i: + z . x) = 0.

(7.15)

z) • X = 0,

(7.16)

Writing this equation in the form (y Z X + •

Z•

XY+X Y •

174

A C O U R S E I N A P PLI ED MATH EMAT I C S

[CH.

we deduce that either (7.17) U = Y · z x + z · x :V + x · Y Z = O, or u is perpendicular to x. Similarly, by consideration of the equations z . X: y . y + y . z(x · y + X: . y) = o, x · y z · z + z · :ic(y · z + y · z) = 0, we conclude that either equation (7.17) is valid or u is perpendicular to y and z. We shall assume, for the moment, that the general particle Z does not lie in the plane OXY. Then x, y, z are not coplanar and thus u cannot be orthogonal to all three. Hence u = 0. Equation (7.17) is of the form

ax + by + c.Z = 0, where a, b, c are scalars.

If these scalars are all non-zero, we have

z=

Ai:

+ fLY,

(7. 18)

where A, fl. do not vanish. This equation implies that the direction of motion of any particle Z of the body is parallel to a plane containing the vectors x, y. This plane is, of course, normal to the instantaneous axis of rotation. We now show that a vector w exists with the property that the velocity of any particle Z of the rigid body is given by the equation (7. 19) We first suppose that x · y -=1c- 0, i.e., the direction of motion of Y is not perpendicular to x. If, in the first instance, this is not so, we alter the direction of x suitably by choosing another particle for X.* We now define w by the equation

w = (Y X x)/x · y . .

(7.20)

Substituting in the right-hand member of equation (7.19), we obtain X

:ic) . X z = z · y :ic - . z · x y X·Y X·Y y . z x +. z · i Y (by equations (7. 14)), =X·Y Z = X Y . (by equation (7.17)), X·Y = Z,

W XZ=

(Y

'

•

--

proving the property stated. If the rigid body is a lamina, this may not be possible, since x is then restricted to lie in a plane. However, we shall then suppose that the plate has been extended by the addition of some rigid structure.

*

7]

K I N EMAT I C S O F A R I G I D B O D Y

175

This result has been obtained on the assumption that Z does not lie in the plane OXY. However, since it can lie arbitrarily near to this plane and since the velocity vector is continuous over the region occupied by the rigid body, it follows that equation (7.19) is true without restriction. It is now clear from equation (7. 19) that if Z lies on the line through 0 parallel to w, then z = 0. Thus, any particle on this line is station­ ary. Let OA (Fig. 7.3) be this line and Z any particle of the body. Construct ZN perpendicular to OA . By equation (7.19),

i z l = w . OZ sin LZON = w . ZN.

(7.21)

Also, by equation (7. 19), the direction of Z's motion is perpendicular to the plane ZON. It follows that Z is moving, at the instant under consideration, with angular velocity A w about the axis OA . This is true for any particle of the body, which is accordingly in rotation about an instantaneous axis of rotation OA with angular velocity w. The vector w, having direction that of the instantaneous axis and magni­ tude that of the angular velocity about this axis, is termed the vector angular velocity of the body about 0. Although, for simplicity, we found it convenient in the foregoing to regard the reference frame, relative to which the motion of the rl·g1'd body was measured as bei'ng FIG. 7.3.-Angular Velocity Vector at rest, it could, in fact, be in rotation about 0 without in any way affecting the argument. Thus, suppose that 51 is a moving frame rotating about 0 and that 52 is a stationary frame momentarily co­ incident with 51 . The frames being coincident at the instant under consideration, we shall not find it necessary to state in the case of each vector relative to which frame it is to be taken. Let w 1 be the angular velocity of the rigid body relative to 5 1 and let w2 be the angular velocity of 5 1 relative to 52 at the same instant. Then the body's angular velocity relative to 52 is w1 + w2 . For, if P is any point of the rigid body and r is its position vector, its velocity relative to 51 is w 1 X r, i.e., this is the velocity of P relative to the point of 51 momentarily coincident with P. But the velocity of this point fixed in 51 relative to 52 is w2 X r. It follows that the velocity of P relative to 52 is '

(7.22)

176

A C O U R S E IN A P P L I E D M A T H E MA T I C S

[cH.

This being true for all points of the rigid body, we deduce that it has an angular velocity w1 + w2 relative to 52 • More generally, we can suppose that 52 is rotating with angular velocity w3 relative to a third frame 53 and that this in turn rotates with angular velocity w4 relative to 54, and so on. Then the body 's angular velocity relative to the final frame is w1 + w2 + w3 + A rough right circular cone of semi-angle ex is rolling on a rough horizontal plane, the contact generator revolving round the vertical through the vertex with angular velocity W1 • Find the angular velocity of the cone and discuss the motion of the cone relative to the moving vertical plane through the (Li.U.) contact generator.

Example 2.

If OA is the contact generator, each particle of the cone lying on this line is instantaneously at rest. Hence OA is· the instantaneous axis of rotation and w, the angular velocity of the cone, is directed along this line. Let C

be the centre of the base of the cone and let OC = h. Then C is rotating about OA with angular velocity w. Hence its speed is wh sin ex. But the vertical plane OA C has angular velocity w1 about the vertical through 0. It follows that the speed of C can also be expressed as w1h cos ex. Thus wh sin ex w1h cos ex, or w w1 cot ex. Relative to an observer moving with the vertical plane OAC, the particles on OC are at rest. Thus, relative to this plane, the cone rotates about OC. Let w0 be its angular velocity about OC. Then w = W0 + Wv and it is clear from the diagram that w0 = w1 cosec ex.

= =

w,

K I N EMAT I C S O F A R I G I D B O D Y

7]

17 7

7.3. General Three-Dimensional Motion of a Rigid Body

Let 0 represent some arbitrarily chosen particle of the body and let V be its velocity at the instant under consideration relative to some reference frame which we shall regard as stationary (Fig. 7.4) . Relative

FIG. .4-.-General Motion of a Rigid Body

7

to a parallel, non-rotating frame moving with 0, the body can only pivot about this point. Let w be the angular velocity of the body about 0. The velocity of a particle P relative to the frame moving �

with 0 is then w X s, where s is the vector OP. relative to the fixed frame is accordingly v=V

+w

X s.

The velocity of P (7.23)

If P is stationary, s must satisfy the equation

w X s = -V.

(7.24)

This is possible only if V is perpendicular to w (this is the case for two­ dimensional motion and an IC then exists) . In general, all particles of the body will be in motion. We will now calculate the effect of transferring the base point 0 to �

some other particle of the body, e.g., 0'. Let 00' = r. Let V' be the velocity of 0' and w' the angular velocity of the body about 0' The position vector of P relative to 0' is ( s - r) . Hence, by analogy with equation (7.23), the velocity of P is given by

X (s - r) . V' = V + w X r. .

v = V'

But

+ w'

(7.25) (7.26)

IN A P PLI E D MATH EMAT I C S [cH. v = V + w' X s + (w - w') X r. (7.27) (7.23) and (7.27) must yield the same value of v for all

A COURSE

178

Hence Equations values of s.

We conclude that

(7.28)

w' = w,

i.e., the angular velocity of a rigid body is independent of the point about which it is measured. A fortiori, w2 is independent of the base point 0 to which w is referred and is a scalar invariant of the motion. A second scalar invariant is the product V · w, for

V' · w' = (V + w X r) · w = V w . ·

(7.29)

This result implies that the component of the velocity of any point of the body in the direction of the angular velocity is a constant. Although, in general, we cannot find a base point 0' which is station­ ary, it is possible to find any number whose velocity V' is parallel to the angular velocity w. For such a point, w X V' = 0 and hence, by equation (7.26),

w X (V + w X r) = 0, (7.30) or w X V + w rw - w 2r = 0, xV i.e., r = w-(7.31) w2 - + tw, where t = w rjw 2• If we substitute for r from equation (7.31) into equation (7.30), the latter equation proves to be satisfied for all values of t. It follows that t is an arbitrary scalar and equation (7.31) defines a line l of points 0', through the point r = w X V/w 2, parallel to w. The body moves so that any such point has its velocity in the direction of l and, at the same time, the body rotates about l with angular velocity w. We have accordingly reduced the general motion of a rigid body to a " screw motion ", the axis of the screw being the line l. The pitch of the screw motion at any instant t is defined to be the ratio of the distance the body moves along l in a short time interval St to its angle of rotation about l. Thus, if p is the pitch and 0' lies on l, V' . w' _ V . w V'St (7.32) P - w'St - V' w' - � - w2 · Thus p is the ratio of the two scalar invariants. ·

·

_

_

_

The point (1, 0, 0) of a rigid body is free to move only in the direction (1, 2, 2) . The point (0, 1, 0) of the body is free to move only in the direction (0, 1, 2) . The point (0, 0, 1) of the body is free to move only in the plane x = 0. Find the pitch of the screw motion that is possible for the body. (Le.U.) Take the point 0 to be at (1, 0, 0) . Then V = (q, 2q, 2q). Let { w1, w2, w3) . The position vector of the point {0, 1, 0) relative to 0

Example 3.

w=

7]

K I N E M A T I CS O F A R I G I D B O D Y

179

is given by s = ( - 1, 1, 0) . Thus, by equation (7 .2 3) , the velocity of the point (0, 1, 0) is given by (q, 2q, 2q) + ( - w 3, - w3, w1 + w 2) , = ( q w 3, 2 q w 3, 2 q + w1 + w 2 ) . But (0, p, 2p) . Hence w 3 = 0, 2q - w 3 = p, 2q + w1 + w 2 = 2p. q From these equations, we derive the results w1 + w 2 = 0, w 3 =P = q . The position vector of the point (0, 0, 1) relative to 0 is s ( - 1 , 0, 1) and thus the velocity of this point is given by equation (7 .23) to be

V=

v=

-

-

-

=

V == ((qq, +2q,w2q), 2q+ (ww2, --w1w -, 2qw3+, ww2) ), . 1 3 2 2 But v lies i n the plane x = 0 and its z-component is therefore zero. -

2q +

w.

= 0.

Thus

We can now deduce that w1 = 2q, w 2 = - 2q, w3 = q . = w1 2 + w 2 2 + w3 2 = 9q 2, Hence V • W = qw1 + 2qw 2 + 2qw 3 = 0 and the pitch is zero. This implies that the body is instantaneously in rotation about a fixed point.

w2

7.4. Moment of Inertia of a Lamina

Let Ox, Oy be perpendicular axes in the plane of the lamina and let m be the mass of a typical particle at the point (x,y). Then AI= !my2 , B = !mx2 are the moments of inertia of the lamina about the axes Ox, Oy respectively. If Oz is an axis perpendicular to the plane of the lamina, the distance of the point (x, y) from it is y(x2 + y2) and hence the lamina's MI about Oz is

C = !m(x2 + y2) = !mx2 + !my2 = A + B. . (7.33) This is the theorem of perpendicular axes, viz., the MI of a lamina about

an axis perpendicular to its plane is equal to the sum of the moments of inertia of the lamina about a pair of perpendicular axes in its plane and intersecting the first axis. Now let OA (Fig. 7.5) be any axis through 0 in the plane of the lamina. If q = PN is the perpendicular from the particle P of mass m at the point (x, y) on to OA , the lamina's MI about OA is I=

!mq2 •

Let OA make an angle 6 and OP an angle 4> with Ox.

(7.34)

Then

q = OP sin(¢> - 6) = OP sin 1> cos 6 - OP cos 1> sin 6,

= y cos 6

and hence

x sin 6, I = !m(y cos 6 x sin 6) 2 , = A cos2 6 2H cos 6 sin 6 + B sin2 6, . -

-

-

(7.35)

1 80

A C O U R S E I N A P PL I E D MATH EMAT I C S

[cH.

where A and B have been defined already and

H = Lmxy. (7.36) H is termed the product of inertia of the lamina with respect to the axes Ox and Oy. y

A

FrG. 7.5.-MI

about any Axis in the Plane of a Lamina

Consider the conic

Ax2 - 2Hxy + By2 = K (a constant), (7.37) with its centre at 0. Let OA intersect the conic in the point Q, where OQ = r. At Q, X = r cos e, y = r sin e and hence Ar2 cos2 e 2Hr2 cos e sin e + Br2 sin2 e = K, or (7.38) Equation (7.38) implies that the MI of the lamina about OA is in­ versely proportional to the square of the radius of the conic in the direction OA. Alternatively, since I = Mk2 , where k is the radius of gyration, k is inversely proportional to r. The conic must be an ellipse, since, by equation (7.38), r is real for all e. The conic is called the Momental Ellipse. Suppose we rotate the axes Oxy about 0 until they are principal -

axes for the ellipse. Then the equation of the ellipse will be

Ax2 + By2 = K,

(7.39)

where A and B are now the moments of inertia about the new axes. Thus, measuring 6 from the new position of Ox, equation (7.35) is re­ placed by I = A cos2 e + B sin2 e (7.40)

K I N E MA T I CS O F A R I G I D B O D Y

7]

181

and, clearly, H = 0. Perpendicular axes through 0 for which the product of inertia vanishes are termed principal axes of inertia and the moments of inertia about them are the principal moments of inertia at 0. Returning Ox, Oy to their original positions, if OB is perpendicular to OA , as indicated in Fig. 7.5, the product of inertia relative to the axes OA B is 1 = L.mpq, where p = ON. But

p = OP cos(¢> - 6) = OP cos ¢> cos e + OP sin ¢> sin e,

Hence

= X COS 6 + J sin 6.

1=

"'i.m(x cos e + y sin 6) (y cos e - X sin 6),

= (A - B) sin e cos e + H(cos2 e - sin2 6) , = t (A - B) sin 26 + H cos 26.

(7.41)

This will vanish if e is selected so that 2H (7.42) ·. B_A Hence, for such a value of e, OA, OB are principal axes. Although there are four values of e in the range (0, 21t) satisfying equation (7.42), these differ by multiples of t7t and thus do not represent essentially distinct sets of principal axes. If the axes Oxy are principal axes, H = 0 and consequently equation (7.41) reduces to 1 t(A - B) sin 26, (7.43} tan 26 =

=

e being measured from the new x-axis. Equations (7.40) and (7.43) permit the calculation of moments and products of inertia relative to any pair of perpendicular axes through 0 , when the principal axes and moments of inertia are known at this point. If the lamina is symmetrical about Oy, then Ox, Oy are principal axes, for to each term mxy of the sum H contributed by a mass m at the point (x, y) there will correspond a term -mxy contributed by a similar mass m at the point ( -x, y). Thus, all terms of H will cancel in pairs. If a lamina has the same moments of inertia about each of three con­ current lines in its plane, prove that the moment of inertia about any line through their intersection and in the plane of the lamina is constant. Hence 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.

Example 4.

(L.U.) Let 0 be the point of concurrence of the three lines and let Ox, Oy be principal axes at 0. Let A, B be the principal moments of inertia at 0. If I is the moment of inertia about any one of the lines and these make angles ex, {3, y with Ox, then I = A cos• ex + B sin2 ex, = A cos2 f3 + B sin2 {3, = A cos2 y + B sin2 y.

182

A COURSE I N APPLIED MATHEMATICS

[CH.

Subtracting the first equation from the second, we deduce that A (cos• "' - cos• {3) = B(sin2 f3 - sin• e<), = B(cos2 "' - cos• {3) . Thus A = B or cos2 "' = cos• {3. A = B or cos "' = cos {3. I.e., A = B or "' = n1T f3 (n an integer) . Therefore If "' = n1T + {3, the lines having slopes "' and f3 are identical. Hence, if A i= B, "' = n1T - {3. But similarly, if A i= B, "' = m1T - y and therefore f3 = (n - m)1T + y, i.e., the lines of slopes f3 and y are identical. We con­ clude that, if the three lines are distinct, then A = B. Then, the moment of inertia about any line through 0 making an angle e with is given by I = A cos• e + A sin2 0 = A , i.e., is a constant independent o f e. If 0 is the centre of the uniform square lamina, axes through 0 parallel to the sides are clearly principal axes, for they are axes of symmetry. The principal moments of inertia are both tma2 and it follows that the MI about any axis through 0 is also tma2 . But the distance between the axis l through a vertex and the mid-point of a non-adj acent side and a parallel axis through 0 is afy'5. It follows from the parallel axes theorem (see next section) that the MI about l is tma• + tma• = 186ma2 •

± ±

Ox

7 .5. Moment of Inertia of a Solid Body

Let 0 be any particle of the body and let Oxyz be a rectangular reference frame. The distance of a particle of mass m situated at the point (x, y, z) from the x-axis is y(y2 + z2). Thus, the body's MI about Ox is A = �m(y2 + z2) . (7.44) Similarly, its moments of inertia about Oy and Oz are B = �m(z2 +

x2), C = �m(x2 + y2 ),

(7.45)

respectively. Let l be any axis through 0 having direction cosines (:r., fl., '1) (Fig. 7.6) . z

FIG. 7.6.-MI of a Body about any Axis

7]

K I NEMAT I C S O F A R I G I D B O D Y

183

If u is the unit vector in the direction of l, u has components (:A, tJ., v) . Let r be the position vector of the particle of mass m at the point P(x, y, z). The vector r X u has magnitude r sin
V

-

3.

4.

P

(

A

(2

5.

6.

a

e

OX, OY OX

Y

a.

7.

e),

!a;

Ox, Oy A

y2 = 4ax

Oy, ay2 4x(a - x)2 2a e

e.

e.

8.

[cH.

A C O U R S E IN A P PLI E D M A T H EMAT I C S

188

is a point fixed in a lamina, and l is a line fixed in the lamina and passing through The lamina moves so that always lies on a line A fixed in the plane of motion, and l always passes through a point (M.T.) 0 fixed in the plane. Find the space centrode.

A

A.

A

9. A lamina is in motion in a plane and two of its particles describe straight lines. Show that the motion can be represented by the rolling of a circle on the inside of a second fixed circle of twice the size. Deduce that the particles which lie on a certain circle fixed in the lamina all de­ scribe straight lines. 10. A rigid body has simultaneous angular velocities about the lines = a, = -a; = a, = a; = a, = - a. If the resultant screw motion has a pitch a, show that its axis lies on the plane + y + z = 0. (M.T.)

z

11.

x

z

-

x

y

y

x

A body is said to have two motions m1, m 2 simultaneously, when it has the first motion in a certain frame and the frame has the second motion relative to a fixed frame. If m1 can be analysed into a velocity V1 of the point r1 and an angular velocity w 1 about this point and m 2 can be similarly analysed into a velocity V 2 of a point r 2 and an angular velocity w2 about it, show that the resultant motion relative to the fixed frame has pitch [(V1

+ V2)

•

(w1

+ w2) + (r1 - r2)

•

(w1 X w2)] f(w1

+ w2) 2.

12. A rigid body is given angular velocities cu and cu', respectively, about

a pair of non-intersecting perpendicular lines, together with linear velocities pcu and pcu', respectively, along these lines. The shortest distance between the lines is c. Show that the motion is equivalent to an angular velocity Q about a certain line together with a linear velocity w Q along that line. Calculate w and Q, and hence show that the velocity of any point of the body cannot be less than pcu2 ± ccucu' pcu'2 · 2 cu ' 2) ! (M.T.) ( cu

13.

I

+

1

+

The CM of a plane lamina of mass M is and are principal axes of inertia at are axes in the plane of the lamina parallel to respectively. Prove that the product of inertia of the lamina with respect to is Moc�. where oc, � are the coordin­ If the principal moments of inertia ates of with respect to and one of and Mq� 2 about of the lamina are Mpoc2 about the principal axes of inertia at 0 is inclined at an angle of 45° to prove that

G. Ox, Oy GX, GY Ox, Oy G Ox, Oy. GX p

-

1

q- 1

G

GX, GY

GY,

Ox,

(Li.U.)

14. Px, Py are coordinate axes in the plane of a lamina of mass M, H is the product of inertia about parallel axes at the centre of mass G, and (%, ji) are the coordinates of G. Show that Px, Py are principal axes of inertia at P if H + Mxy = 0. OA, OB are perpendicular axes in the plane of the lamina, OA being a principal axis at A and OB principal axis at B. Show that G must lie on OC, where OACB is a

a

7]

K I N E MA T I CS O F A R I G I D B O D Y

18

9

rectangle. If OGfOC = 4/3, show that the points on OC at which the principal axes are parallel to OA and OB are distant 20C and iOC from 0. (L.U.) A uniform plane lamina, of mass M, is in the shape of a rectangle whose sides are of length 2a, 2b and whose centre is 0. The perpendicular distances from a point P of the lamina to axes through 0 parallel to the sides are x and y. Find the moments and the product of inertia about axes through parallel to the sides. If one of the principal axes through P makes a constant angle ex; with the side of length 2a and x is measured parallel to this side, show that the locus of P, with respect to the axes through 0, is x2 - y2 - 2xy cot 2a = (b2 - a2)f3. (L.U.) 16. A homogeneous wire of length ). is bent to a semi-circular arc. Its MI is the same about all lines in its plane which pass through one or other, or both, of two points P, Q. Prove that the distance PQ is 4i.Jrt2• (L. U.) A uniform rectangular plate whose sides are of lengths 2a, 2b has a portion cut out in the form of a square whose centre is the centre of the rectangle and whose mass is half the mass of the plate. Show that the axes of greatest and least moment of inertia at a corner of the rectangle make angles with a side, where + tan 26 = �abf(a2 - b2) . (M.T.) (Hint : Consideration of the momental ellipse will reveal that the axes of greatest and least MI are the principal axes.)

15.

P

17.

e, f1t e,

18. A right circular cylinder made of uniform metal sheet, the ends being open, has height 2h, diameter 2a, and mass M. A is a point on one end rim, B the point farthest from A on the other end rim. Verify that the MI of the cylinder about A B is Ma2(3a2 + 8h2) 6(a2 h2) (Le.U.) Referred to axes Ox, Oy, Oz fixed in a body of mass M, the moments and products of inertia are A , B, C and F, G, H. The coordinates of the mass-centre are (x, ji, z). Show that, referred to parallel axes through the point P, coordinates (a, b, c) , the moments and products of inertia are A + M(b2 + c2) - 2M(bji + cz), etc., and F + Mbc - M(bz + cji), etc. If the principal axes of inertia at 0 and P are parallel and none of them coincide, show that the mass-centre is at the mid-point of OP. (S.U.) 20. OA , OB, OC, of lengths a, b, c respectively are three coterminous edges of a uniform rectangular block of mass M. Determine the moments and products of inertia of the block with respect to the edges. Deduce that the MI of the block about a diagonal of the face A OC is M(2a2b2 a2c2 2b2c2)/6(a2 + c2). (S.U.)

+

19.

+

+

190

A C O U R S E I N A P P L I E D M A T H E MA T I C S

[cH. 7]

21. Show that a uniform, triangular lamina of mass M is equimomental with three particles, each of mass M/3, placed one at the mid-point of each side. 22. In a homogeneous, triangular lamina A BC, the mid-point of AB is D and the foot of the perpendicular from C on to AB is E. Prove that AB is a principal axis at the mid-point of DE. (L.U.) 23. Show that if a line through the CM of a body is a principal axis, it is a principal axis at every point along its length.

CHAPTER 8

MOTION OF A PARTICLE SYSTEM-II 8.1. Equation of Motion of the Mass Centre

Suppose that a system of n particles is in motion in a space in which a set of rectangular axes forming an inertial frame. Let mn be the masses of the particles and let r; be the position vector of the particle of mass m; at time t relative to the frame. This latter particle may be subject to both internal and external forces. Let F'; be the resultant internal force and let F; be the resultant external force acting upon the particle. Then the equation of motion of the particle is (8 . 1}

Oxyz is mv m2,

•

•

Summing this equation with respect to system, we now obtain

i over all the particles of the (8.2)

But, as has been remarked before (Section 6.3) , the internal forces occur in equal and opposite pairs. It follows that the terms of the sum !F'; cancel one another in pairs and hence that the sum is zero. Equation (8.2} accordingly reduces to (8.3)

m;i:; is the momentum of the particle m;. The vector 1m;i:; is the sum of the momenta of the particles of the system and is termed the linear momentum of the system. Equation (8.3} can Now

p

=

be written

(8.4)

the rate of change of the linear momentum of a system is equal to the sum of the external forces. i.e.,

Let r be the position vector of the system's mass centre G at time t . r i s defined b y equation (11.29}, thus

Mr = 1m;r;,

where M = 1m; is the total mass of the system. differentiated with respect to t the result is

Mr = 1m;i:; = p, 191

(8.5) If this equation is (8.6)

A C O U R S E I N A P P L I E D M A T H E MA T I C S

192

[CH.

indicating that the linear momentum of a system may be calculated by assuming that the total mass within the system is condensed at the CM and is given the velocity of this point. Substituting for p from equation (8.6) in equation (8.4), we obtain (8.7) Equation (8.7) is a second-order differential equation for the quantity

r when the external forces are prescribed, i.e., it is an equation de­

termining the motion of the centre of mass. Consider a hypothetical particle of mass M situated at G and suppose that the points of applica­ tion of the various external forces are shifted so that they all act on this particle with the same magnitudes and directions they had originally. The equation of motion of such a particle will be precisely equation (8.7). We sometimes express this by saying that the CM of a particle system moves as though the whole mass of the system were condensed there and all the external forces acted there. This conclusion is our justification for treating extensive bodies as particles in some examples in earlier chapters (e.g., p. 132). As a simple example of the use of equation (8.7), we shall consider again the compound pendulum problem (p. 146) . The centre of mass G moves around a circle of centre 0 and radius h (Fig. 6.6). Its speed in this circle at the instant under consideration is h6, and hence its acceleration component in the normal direction GO is h2fJ 2/h = hfJ2 . Its acceleration component in the tangential direction perpendicular to GO is d(hfJ)jdt = h8. Resolving the external forces in these two direc­ tions and employing equation (8.7), we find that Y X

Mg cos e = MhfJ 2,

(8.8)

- Mg sin e = MM.

(8.9)

-

If the pendulum is swinging through an angle et on either side of its equilibrium position its motion is governed by the energy equation (6.38) Substituting from this equation for 82 in equation (8.8) , we obtain .

Y = Mg cos e

+ 2M12gh2 (cos e - cos et) . . -

(8.10)

If k is the radius of gyration about an axis through G parallel to the axis of rotation, by the parallel-axis theorem, ! = M(h2 + k2) . Hence, equation (8.10) is equivalent to y

=

�k2 [(3h2 + k2)cos e - 2h2 cos et].

h2

(8.1 1)

8]

M O T I O N O F A PART I CL E S Y ST E M- I I

193

Differentiating equation (6.38) with respect to t, we obtain 2M = -

�

2 gh

sin e e,

. gh . 6. = - 2 sm e. h + k2

or

(8.12)

Substituting for e from this latter equation into equation (8.9), we obtain an equation for the second component of the reaction at the axis of rotation, viz. , 2 X = hMgk (8. 13) 2 + k2 sm e . •

A uniform, solid cylinder of radius a rolls without slipping on the inside of a fixed, rough, hollow cylinder of radius 4a with axis horizontal, and the angular velocity of the solid cylinder when in its lowest position is w. Show that, if w > y(llgfa), it will roll completely round the hollow cylinder, but if w is small it will oscillate with period 37ry(2afg). (L.U.)

Example 1.

When the moving cylinder is in its lowest position suppose that the point A on its surface is in contact with the point B of the fixed cylinder. At some other instant let C be the corresponding point of contact be­ tween the cylinders. Since there is no slipping, arc BC = arc A C. If is the angle made by the plane containing the axes of both cylinders with the vertical at this instant, arc BC = Hence arc A C = and LAGC = If G V i s the downward vertical through G, L CGV = and hence LAGV = The rate of rotation of GA relative to the fixed direction G V is accord­ ingly This is the angular velocity of the solid cylinder. The CM of the solid cylinder G moves around a circle of radius Its speed in this B If is circle is therefore the mass of the moving cylinder, the contribution to its KE of a particle of this mass moving with G is The MI of this cylinder about its axis is and hence, the KE of its rotary motion about G is = The total KE of this cylinder is therefore

6

4a6.

4a6

46.

36.

6

38.

3a.

3a6.

tma• t(!ma2)(38)2 9ma262f4. 21ma•6•f4.

m

9ma262/2.

The only external forces acting on the solid cylinder are its weight and the normal and frictional components of the reaction at C, R and F respec­ tively. Since there is no slipping, the system is conservative, having PE cos We can now write down the energy equation, viz., H

- 3mga 6. •:ma•e• - 3mga cos e = 247ma2w2 - 3mga,

(I)

mg

194

A C O U RS E I N A P PL I E D M A T H E MA T I CS

[cH.

6 = 0, () = w. Solving for 62, we obtain ()2 = w2 - 9a4g (1 - cos 6) ' (i) indicating that 62 decreases as 6 increases, in accord with our expectations. If the solid cylinder has insufficient momentum to reach the position 6 = equation (i) will reveal this fact by giving a negative value for 62 (i.e., an imaginary value for 6) at this value of 6. Putting 6 = in equation (i), since initially

TT,

TT

we obtain

()2 = w2 - 9aSg_ w2 > Bg9a,

If

(ii)

the solid cylinder has been given sufficient momentum in the lowest position to permit it to climb to the highest position. However, it will reach this position without leaving the surface of the fixed cylinder only if R does not vanish during the motion. The component of G's acceleration along the radius GO of its path is = Resolving along GO for the motion of the mass centre G, we find that R cos = Substituting for from (i) and solving for R, R = + cos an equation which proves that R decreases as the solid cylinder rises. When e = TT, R takes its least value, viz., Rmin. = If Rmin. is still positive, the motion under consideration is possible. This will be so if

(3a6)2/3a 3a62.

62

- mg 6 3ma62.

3maw2 - tmg fmg 6, 3maw2 - 'fmg.

w2 > llg 9a _

(iii)

The inequality (iii) includes inequality (ii), and hence is the condition required. If is small, 6 will remain small during the motion and equation (i) may be approximated by

w

9aw2 ()2 = w2 - 9a2g 62 = 2g 9a ( 2g - 62) .

This is an equation of SHM of period

2TTJ;; = 3TTJ�· Example 2. A uniform, straight rod is placed with its ends in contact with a smooth, vertical wall and a smooth, horizontal plane, the vertical plane through the rod being perpendicular to the wall. If the rod is allowed to fallfreely under gravity, show that it will lose contact with the wall when its centre has fallen through one-third of its original height above the plane. (L.U.) Let 6 be the angle made by the bar AB with the wall at any instant Since LAOB is a right angle, a circle on AB as diameter will pass through 0. If G is the centre of the bar, GO is a radius of this circle. Hence, if AB = 2a, OG = a. Thus the locus of G is a circle of centre 0 and radius a. Since

8]

M O T I O N O F A PART I CL E S Y S T E M - I I

195

= = a, = = 6. The speed of in its circle is accordingly ae, as shown in the diagram. If m is the mass of the bar, the KE of a particle of this mass and moving with is !ma•e•. Relative to the bar rotates with angular velocity e. The KE of this rotary motion is ma262f6. The net KE of th� bar is therefore 2ma2fl2/3.

GO GB

LGOB LGBO

G

G

Q

'

'

'

'

'

'

'

G,

{)

'

0

The external forces acting on the bar are the smooth reactions P and Q and its weight mg. The system is clearly conservative, having PE mga cos 6. Hence the energy equation is tma•e• mga cos 6 = constant,

+

e• =

:! (cos

Cl

- cos 6),

(i)

where a is the initial value of 6 when e = 0. Let x be the distance of from the wall at any instant. The horizontal component of acceleration is then iii. But x = a sin 6. Hence, differentiating this equation twice with respect to t, we obtain

G's

G

iii = a cos e a - a sin 6 e• Differentiation of equation (i) with respect to t yields the result �-.·· 3g 2o6 = sin e }.o,

2a . . 3g 6 = 4,a sin 6.

(ii)

We can now substitute for e• and e in equation (ii) to obtain � = !g sin 6(3 cos 6 - 2 cos a) . The x-component of equation (8·7) may now be written down as !mg sin 6(3 cos 6 - 2 cos a) = P. The rod loses contact with the wall when P = 0, i.e., when 6 satisfies the equation cos 6 = t cos a. The height of above the horizontal plane at this instant is a cos 6 = fa cos a. This is two-thirds of its initial height above the plane, and hence the con­ clusion stated in the question has been verified.

G

196

A C O U R S E I N A P PLI E D MAT H E M A T I C S

[cH. Example 3. A uniform, solid, circular cylinder has a particle ofequal mass attached to the mid-point of one of its generators. It is placed on a rough, horizontal plane with the particle vertically above the generator in contact with the plane. It is slightly disturbed and rolls through a right angle before slipping occurs. Show that the coefficient offriction between the cylinder and the plane is 4/19. Referring to the problem solved on p. 149, by putting M = m in the energy equation, we obtain the equation appropriate to our present problem, viz., 4 1 - cos e) e• = 7 + 4 cos e '

a

g(

ali

22g

being the radius of the cylinder. Differentiating this equation with respect to t, we find that sin e . = (7 + 4 cos e)• Hence, when e = i1r, = de = (i)

a

aa• tg.

!!g.

mg

Relative to the axis of the cylinder, the mass centre of the system G describes a circle of radius ta with speed t 6 . The component of this acceleration in the direction of the radius GO is therefore t e•, and the component in the direction of the tangent perpendicular to GO is ta'e. The acceleration of G relative to a fixed observer is obtained by adding vectorially to these components the acceleration of 0. Since 0 moves along a straight line with speed ae, its acceleration is horizontally. This is the third com­ ponent of G's acceleration. These components are shown in the diagram for the case e = t7T. Resolving horizontally and vertically for the motion of G, we obtain the equations F= - R = t . Substituting from equations (i) into these latter equations, we find that F = 1 6 f49 and R = 76 f49. Hence FfR = 4/19. Since slipping now takes place, this ratio is the coefficient of limiting friction.

a

aS

mg

2m(a6 - tae•),6 2mg 2m . a mg

a

M O T I O N O F A PART I C L E S Y ST E M- I I

8]

197

8.2. Conservation of Linear Momentum

The equation associated with the motion of the CM of a system of particles is a differential equation of the second order. If, however, it is known that the sum of the components of the external forces in some direction vanishes throughout the motion, a first-order equation may be written down and the solution to the problem correspondingly simplified. Suppose that we choose the x-axis to be in this direction, so that, if X; is the x-component of F;, then �X; 0 for all t. Then, resolving equation (8.4) in the x-direction, we obtain Px = 0, where Px is the x-component of the linear momentum of the system. Inte­ gration now yields constant. (8.14) Px =

=

Equation (8.14) states that if the sum of the components of the external forces in any direction vanishes throughout the motion of a system, the component of the system's linear momentum in this direc­ tion is conserved. This is the general Principle of Conservation of

Linear Momentum.

By equation (8.6), the component of the linear momentum of a system in any direction may be calculated as the component in this direction of the momentum of a particle moving with the centre of mass and of mass equal to that of the whole system.

Two particles A and B are connected by a light, inextensible string of length 3a and A can slide along a smooth, straight groove cut in a smooth, horizontal table. The mass of B is twice that of A , and initially the string is just taut and lies along the groove. If B is projected along the table at right angles to the groove, prove that it describes an ellipse and find the lengths of the semi-axes of this ellipse. (B.U.) Let 0 be the initial position of A and let Ox be in the direction of the

Example 4.

groove.

(x, y)

8

0

Let (x, y) be the coordinates of B at any instant referred to the axes Ox, Oy (see diagram) . Let (X, 0) be the coordinates of A at the same instant. Consider the system comprising the particles A and B and the string. The only external force acting on this system and having a component in

198

[CH.

A C O U R S E IN A P PL I E D MATH EMAT I C S

the plane of motion is R, the reaction of the groove on A . Since this force is always perpendicular to linear momentum is conserved in the direc­ tion Thus + 0, (i) the initial linear momentum in this direction being zero. Integrating equation (i) with respect to the time, we find that X= - 2x, (ii) since = X = 0 initially. But, if e is the angle made by the string with the groove, X=X+ cos e = - 2x + cos e (iii) by (ii), or + COS e. = 3a sin e. (iv) Also Equations (iii) and (iv) constitute parametric equations for an ellipse of semi-axes and

Ox.

Ox, mX 2mx = 6a

x 3a,

X = 2a a y 3a

6a

3a

a 3a. Example 5. A uniform, straight rod of length 2a and mass 2m has a smooth ring of mass m attached at one end. The ring is free to slide on a smooth, horizontal wire. The rod is drawn aside from the vertical through an angle and the system is released from rest so that the rod oscillates in the vertical plane con­ taining the wire. Show that the reaction of the wire on the ring when the rod passes through its vertical position is 3mg(3 - 2 cos ) and that if IX is small, the period of the oscillations is the same as that of a simple pendulum of length 2af3. The configuration at any instant may be specified in terms of generalized coordinates (x, e), where x is the distance of the ring from a fixed point on IX

IX

the wire and e is the rod's inclination to the vertical. Two equations are

u.

therefore necessary to determine the motion. We shall obtain these by application of the principles of conservation of linear momentum and of energy. The external forces acting on the system " rod + ring " are the weight forces and the reaction R of the wire on the ring. These forces are always vertical. Hence linear momentum is conserved in the horizontal direction. The centre of mass C of the rod describes a circle of radius

mg, 2mg

a

8]

M O T I O N O F A PART I CL E S Y S T E M- I I

199

relative to the ring, and hence its velocity relative to the ring is per­ pendicular to the rod (see diagram) . Let = x be the horizontal velocity of the ring. The velocity of C relative to a fixed observer is the vector sum V of the velocities and The horizontal component of V is the sum of the horizontal components of its constituent velocities, viz., u cos e. The horizontal component of the linear momentum of the rod is accordingly cos e) . The corresponding component of the linear momentum of the ring is The linear momentum of the system is initially zero, since it starts from rest. But linear momentum is conserved in the horizontal direction. Thus cos e) + = 0, i.e., = cos e. (i) By the Cosine Rule, = + cos e, = - 8 cos• having employed equation (i). The KE of the rod is therefore = cos2 + = Adding the KE of the ring, cos• we obtain the total KE of the system, viz., sin• The depth of C below the wire is cos The PE of the system can accordingly be written cos The system is conservative so that an energy equation exists in the form cos = cos ex, + sin2 3g(cos cos ex) . (ii) + sin2 If ex is small, is always small, R and to the second order of small quantities this latter equation may be approximated by

a6

2m(u - a6

a6

u

u.

- a6

mu.

2m(u - a6 mu u taa v• u2 a262 - 2ua6 ta2B2(9 6),

t . 2mV2 t . -!2ma• . 6• �ma2B2(3 - 2 6). tmu• tma•6• 6, tma2B2(l + 6). a 6. -2mga 6. 6) - 2mga 6 - 2mga Jma•6•(1 (J• - a(l 6 - 6) 6 _

62 = �! (ex2 - tl2) .

�!;

This i s the energy equation o f a SHM of period 21T

. A simple

. %, cx 8

,

G l-----� u - % cx O pendulum of length oscillates with this period, thus proving one c statement of the question. The CM of the system G is dis­ tant from the ring 0 and hence, relative to the ring, it moves around a circle of radius The acceleration of G rela­ tive to the ring accordingly has a radial component along GO and a tangential component 2 6J perpendicular to GO. Adding vectorially to this relative acceleration the acceleration u of the ring, we obtain the acceleration of G relative to a fixed observer. The various components of G's acceleration are indicated in the second diagram for the particular configuration when the

2af3

2af3

2af3.

2aB2/3

a 3

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[CH.

rod is vertical. Resolving vertically in this position for the motion of G, we obtain R =

3mg 3m . ia62• a62 3g(l ) Hence, at this 3mg + 6mg(l ) 3mg(3 - 2 Example 6. A uniform, solid sphere of mass m and radius b roUs down the rough, curved surface of a uniform hemisphere of mass M and radius a, which is free to slide on its base in contact with a smooth, horizontal plane, starting from rest when the common radius makes an angle with the vertical. Show that, when this angle has increased to e, [t(M + m) - m cos2 6](a + b)62 = 2(M + m)g(cos - cos 6). When = 0, equation (ii) shows that = - cos 0< instant, R = - cos 0< = cos 0<) .

6

0<

.

0<

(M.T.)

Let be the speed of the hemisphere at any instant. Relative to the centre 0 of this body, the mass centre G of the sphere moves around a circle in a of radius Its velocity in this circle is therefore direction perpendicular to OG. The velocity of G relative to a fixed observer is found by compounding this velocity vectorially with the horizontal

u

(a + b)6

(a + b).

<£ ----.,

The horizontal component of this resultant velocity velocity of 0, viz., is cos and hence the same component of the linear momentum of the sphere is cos 6] . The linear momentum of the hemi­ sphere is horizontally. The applied forces being vertical, linear momentum is conserved in the horizontal direction. But initially the linear momentum is zero. Hence cos = 0,

u. u - (a + b)6 6 m[u (a + b)6 Mu

Mu + m[u - (a + b)S 6] u = M � m (a + b)() cos 6.

(i)

There being no slipping at the point of contact C between the two bodies, the system is conservative. Let A ' be the point on the sphere which would

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M O T I O N O F A PA RTI CLE S Y S T E M - I I

201

come into contact with the topmost point of the hemisphere if the former were rolled into its highest position without slipping. Then arc A 'C = arc A C = Hence LA 'GC = The anti-clockwise angle made by GA' with the downward vertical is accordingly = The rate of rotation of the sphere is now seen to be and the KE of its rotary motion is The KE of a particle of mass moving with G is f [u + cos whence the total KE of the sphere. The hemisphere has no rotary motion, so that its KE is fMu2• This KE has been generated by a loss in PE of b (cos IX cos We accordingly have the energy equation cos !Mu2 !m[u2 = b cos IX - cos Eliminating u from this equation by the use of equation i , we obtain the equation stated.

a6.

+

a6Jb.

6 + a6Jb (a + b)6Jb. (a + b)6jb m(a + b)262j5. m m 2 (a + b)2e2 - 2u(a + b)(J 6], mg(a + ) - 6). + (a + b)•(J• - 2u(a + b)(J 6] + lm(a + b)•(J• mg(a + )(

()

6).

8.3. Isolated Systems

A system which is not disturbed by any external agent, i.e., is subject to no external forces, we shall term an isolated system. It follows from the theory of the previous section that the component of the linear momentum of the system taken in any direction will be constant. In particular, the component in the x-direction is constant, i.e.,

Im;i; = MX = constant.

(8.15)

Thus x = constant. Similarly, all other components of the velocity of the mass centre are constant, and this point accordingly moves with uniform velocity. A non-rotating frame of reference which moves with the mass centre of an isolated system is an inertial frame as defined in Chapter 2. p

FIG. 8.1 .-Motion of Sun and a Planet

Newton's Laws of Motion will be valid relative to such a frame, which is accordingly the most convenient for use when studying the motion of the system. The Solar System, comprising the Sun, planets, comets, etc., forms an isolated system, since the influence of the neighbouring stars is, for most purposes, quite negligible. Consider a simplified form of this system consisting of a Sun of mass S and a planet of mass P (Fig. 8.1). If y is the gravitational constant, the forces acting on the

[CH.

A C O U R S E I N A P PL I E D M A T H E MA T I C S

202

two bodies are both yPSjr2, where r is the distance between them. The acceleration of P is therefore ySJr2 towards S and of S is yPJr2 towards P, both these accelerations being referred to an inertial frame such as that moving with G, the centre of mass of the two bodies. The acceleration of P relative to S is

(8.16) The motion of P relative to S is accordingly that which it would have if S were fixed and were attracting P with a force �J./r2 per unit mass, where !J. = y(S + P). P is small compared with S, so that (J. = yS approximately, and this value of (J. is independent of the planet being considered (this was the approximation accepted in Section 4.5). If this approximation is not made, !J. can no longer be regarded as a constant of the Solar System, but varies from planet to planet. Thus, if T is the periodic time of the planet in its orbit relative to the Sun, an ellipse of semi-major axis a , equation (4.37) indicates that

(8.17) i.e., the square of the periodic time of a planet is proportional to the cube of its mean distance from the Sun and inversely proportional to the sum of the masses of the Sun and the planet. This is the required

modification of Kepler's Third Law referred to in Section 4.5. We will conclude this section by solving a problem concerning an isolated system, where it is convenient to refer the motion to the centre of mass.

Two particles whose respective masses are m, M are attached to the ends of a light, elastic string of natural length l and modulus >.. The particles lie on a smooth, horizontal table with the string just taut and are then projected away from each other, in the line of the string, with speeds u and v respectively . Find the time that elapses before the particles collide. (L.U.)

Example 7.

The weight forces acting on the two particles A and B are neutralized by the smooth reactions of the table, so that the system is effectively isolated. Initially the velocity of G is in the direction of A 's motion (equation (8.14)). This is indicated in diagram Relative to G, the initial velocities of projection of the two particles are as shown in diagram Consider the motion of A relative to G. Let be the displacement of A from its initial position relative to G. The displacement of B must then be since BG : A G = at all times. The extension of the string is therefore ( 1 and the tension in it is The equation of motion of A is accordingly

(Mv - mu)J(M + m)

(b). Mxfm,

or

+ Mfm)x

M: m

Mi = -�� (m + M)x, + M)x -_ 0 x. + >.(mMml '

x

(a).

>.(m + M)xfml.

8]

M O T I O N O F A PA R T I C L E S Y S T E M-I I

203

i.e., the motion of A is initially simple harmonic of period

2'"VI>.(mMml + M) "

When one-half of a complete oscillation has taken place the particles are back in their initial positions relative to G and the string goes slack. The velocities of the particles relative to G are equal in magnitude to those shown in diagram (b), but are in the reverse senses. Now AG = and Mu/J.

mlf(m + M)

- m u.

f
m

B

(a)

M (uw)

M

"

A

G

m�u.-.u-) M -. m

M+m

8

(b)

A

G

the time taken for A to reach G after the string has become slack is therefore Clearly B must arrive at G at the same instant. Hence the total time which elapses before the particles collide is

lf(u + v).

l " '"VI>.(mMml + M) + u + v

8.4. Equation of Angular Momentum

Suppose that a particle P of mass m is in motion relative to an inertial frame Oxyz and let r be its position vector at time t. If v = i: is the velocity of the particle its momentum is mv. The moment of its momentum about 0 is therefore r X mv. This quantity is called the angular momentum h of the particle about 0. If F is the resultant force acting upon the particle, F = m\r and hence

dh d dt = dt(r X mv) = r X mv + r X mv = r x F, (8.18) since i: X v = v X v 0. But r X F is the moment of the applied force about 0. Equation (8.18} accordingly asserts that the rate of increase of the angular momentum of a particle about a fixed point is equal to the moment of the applied force about the same point. If, now, equation (8.18) is applied to each particle of a system moving relative to the frame Oxyz, we obtain by summation over these particles 1dh· d/ = the sum of the moments about 0 of all forces of the .

·

=

system,

(8.19)

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A COURSE I N APPLIED MATHEMATICS

h; being the AM of the ith particle about 0.

Since the internal forces occur in equal and opposite pairs, their net moment about 0 is zero. Hence, equation (8.19) is equivalent to the equation

dh = the sum of the moments about 0 of the external forces of the dt (8.20)

system,

where h = !11;, is the total AM of the system about 0. In words, the rate of increase of the AM of a system of particles about a fixed point is equal to the sum of the moments of the external forces about this point.

For the purpose of solving problems, it is convenient to resolve equation (8.20) along an axis such as Oz. We then have the equation

;t (angular momentum about a fixed axis Oz)

= sum of moments of external forces about Oz,

(8.21)

the component of the moment of a vector in any direction being equal to the scalar moment of the vector about an axis in this direction (see Section 11.1). When the motion is two-dimensional we shall speak of the moment of force and momentum about a point in the plane of motion, meaning thereby the scalar moment about an axis through this point per­ pendicular to the plane. Before we can make use of equation (8.21) we must calculate formulre from which the AM of a system of particles such as a rigid body may be found when its motion has been specified. We will first examine the case of a rigid body rotating about a fixed axis. Referring to Fig. 6.5, 0 is on the axis of rotation of the body and w is the angular velocity. A typical particle P of mass m and distant r from the axis has velocity wr perpendicular to OP. Its momentum is mwr in the same direction and the moment of this momentum about the axis of rotation is mwr2 • The total AM of the rigid body is then (8.22) where I is the MI of the body about the axis of rotation. If L is the sum of the moments of the external forces acting on the body about the axis of rotation, the equation of AM is

dw L ddi (Iw) = I di'

(8.23)

=

i.e., the moment of the external forces about the fixed axis gular acceleration.

=

MI

x

an­

M O T I O N O F A PART I C L E S Y S T E M - I I

8]

205

In the case of the compound pendulum (Fig. 6.6) the equation of AM is therefore -Mgh sin 6 = !6, (8.24) an equation which is equivalent to (8.12).

The flywheel of an internal-combustion engine has moment of inertia I and is being run up to full speed by the action of the motor torque which is inversely proportional to the angular velocity w of the flywheel. The bearings exert a frictional torque proportional to w. Show that a I dw dt = ;;; - bw, where a and b are constants. At maximum speed, w = n. If w = 0 when t = 0, prove that w n vl - e 2btfl. If afw is the motor torque and bw is the frictional torque, the first of the stated equations may be obtained by application of equation (8.23). When the speed is a maximum, dwfdt = 0, and hence w2 = afb. Thus a = b!l2 and the equation of motion can be written in the form 2wdw - ¥ dt. - n• w2 -

Example 8.

=

This integrates immediately to give log (!12 When

t = 0, w = 0, so that

Thus and hence

8.5.

y + constant. - w2) = - 2bt

constant = log n•.

•log n -no- = - ' T = !12(1 - e-2btfl) .

w2

w2

2bt

Conservation of Angular Momentum

If we can find a fixed axis Oz in the neighbourhood of a particle system about which the external forces applied to the system have zero moment throughout its motion, the rate of increase of the AM of the system about Oz will be zero. Thus

hz = constant, (8.25) and the angular momentum about Oz is conserved. This is the Principle of Conservation of Angular Momentum. Example 9. A rough, circular wire, of mass m and radius a, is free to rotate in its own plane about its centre 0. A bead P ofmass m moves on the wire. Initially, the wire is rotating with angular velocity 2w and OP with angular velocity w, both relative to a fixed observer. Show that when the bead is at rest relative to the wire the angular velocity of the latter has been reduced to 3w/2 (gravity is to be neglected).

A C O U R S E I N A P P L I E D M A T H E MA T I C S

206

[cH.

If 6 is the angle moved through by OP from its initial position at any instant bef�re the bead comes to rest on the wire and p. is the coefficient offriction between the bead and the wire, show that the angular velocity of OP at this instant is given by 6 = wei'O, Deduce that the angle turned through by OP before the bead comes to rest on the wire is I 3 - log -· 2 fL

Consider the system " bead + wire ". The only external force acting on it is the reaction Q of the bearings in which the wire is pivoted. This force has zero moment about 0, so that AM about this point is con­ served. Initially, the momentum of is perpendicular to Its moment about 0 is The MI of the wire about 0 is and hence its initial AM about this point is The total initial AM is Let be the angular velocity of bead and wire when there is no relative motion between them. The total MI of the system about 0 is Hence the AM of the system is then Since AM is conserved,

maw

OP. ma•w. ma2

2ma2w. n

P

3ma2w.

2ma2!1. 2ma2!1 = 3ma2w, n = tw.

2ma2•

Whilst is sliding on the wire, the forces acting on it are the normal reaction R and the frictional force p.R. Resolving normally and tangentially for the motion of we obtain = R, = p.R.

P

P,

maS

ma.62

Dividing the second equation by the first, we find that

Integration now yields

fL = ro1 d6dt = e1 aad6· p.d6 = ad(J- ·

log () = + constant, = or where A is a constant. When = 0, = and hence A = result stated. When (J = =

p.6 (J Aei'O, 6 (J w

n 3wf2,

w, proving the

3w wel'-0, 6 = �1 log 23 . This is the angle through which OP rotates before the bead comes to rest on

the wire.

2=

M O T I O N OF A PART I C L E S Y S T E M - I I

207 Example 10. A cube of edge 2a and mass m is free to rotate about a vertical edge. A smooth, horizontal groove is cut across the upper face along the diagonal which does not intersect the axis of rotation, and a particle of mass 2mf3 is placed at the mid-point of the groove. The cube is given an initial angular velocity 0. If the velocity of the particle relative to the cube is V when the particle is about to leave the groove and the angular velocity of the cube at the same instant is w, show that 4a(2w - 0), v v'2 and that w satisfies the equation (L.U.) 6w2 - 6w0 + 02 0. 8]

=

The diagram shows the particle about to leave the groove. This point of the groove has velocity in the direction shown at this instant. The particle partakes of this velocity and has also a velocity along the groove as shown. Each of these velocity components gives rise to a component of linear momentum of the particle in the same direction, and the sum of the moments of these components about the axis of rotation is

=

2aw

V

!maw . 2a - !m V y'2a.

This expression represents the angular momentum of the particle about the axis of rotation at this instant. The MI of the cube about its axis of rotation is Hence its AM about this axis is AM of the system is

8ma2f3.

8ma2w/3 and the net

'."ma•w - iv'2m Va.

The external forces acting on the system " cube + particle " comprise the weight forces and the smooth reactions at the pivots securing this axis. None of these forces have any moment about the axis of rotation, so that AM is conserved about this axis. Initially the particle is at rest and the AM of the cube is Hence = or (i) = When the particle is about to leave the groove, let be its resultant speed. Then = cos + Its KE is accordingly = + = + where equation (i) has been employed. The KE of the cube is then Initially the particle has zero KE and the cube a KE of Since energy is conserved during the motion, + + = + = 0. or

8ma20f3. ¥m,a•w - iv'2mVa tma20, v v'2 4a(2w - 0). v v• v• + 4a2w2 - 4 Vaw 45°, V2 4a2w2 - 2y'2Vaw.

}mv2 !m(V2 4a2w2 - 2y'2Vaw) tma2(5w2 - 60w 202), 4ma2w2f3.

4ma202f3. tma2(5w2 - 60w 202) tma2w2 tma202, 6w2 - 60w 02

208 8.6.

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[CH.

Angular Momentum of a Rigid Body in General Motion

Consider the particle system m1, m2, mn. Let r be the position vector of its CM relative to an inertial frame and let Pi be the position vector of mi relative to a parallel moving frame with the CM as origin . Then equations (6.41), (6.43) and (6.44) are valid. Hence, we have for the angular momentum of the system about 0. •

h = _2ri X m;i�i = _2(pi + r) X mi (Pi + "r) = _2p; x m;pi + _2r X mir + .2r x mipi + LPi x m;� = _2pi X mipi + r X f _2mi + r X _2m;f)i + (_2m;pi) X r (8.26) = ha + r x .Mf where ha = !Pi X mi()i is the AM of the motion relative to the CM and about this point and M _2m;. Now r X iff is the AM about 0 of a particle of mass M moving with the CM of the system. Accordingly, we have proved that the AM of a =

system about any point is equal to the AM about the point of a particle whose mass is that of the whole system and which moves with the CM, together with the AM about the CM of the motion relative to this point.

(Cf. corresponding result for KE on p. 148.) In the particular case of a rigid body in two-dimensional motion parallel to the plane of fixed axes Ox, Oy, let (u, v) be the components of the velocity of the mass centre G parallel to these axes and let cu be the angular velocity of the body about G. Then the particle moving with G has components of momentum (Mu, Mv) and, if (x, y) are the coordinates of G, the moment of the momentum of this particle about 0 is Mvx - Muy. From the point of view of an observer moving with G, the rigid body's motion relative to him is simply that of a rotation about G with angular velocity cu. It follows that the AM about G of the relative motion is Iacu, where Ia is the body's MI about G. Thus, the AM of the body about 0 is given by

h Mvx - Muy + Iacu. =

(8.27)

We can now solve more complicated problems by application of the Principle of Conservation of Angular Momentum. A smooth, straight, rigid tube is free to rotate in a horizontal plane about a vertical axis not intersecting the tube. A straight bar can slide freely inside the tube. The tube is set rotating with the of the bar at the point in the tube closest to the axis of rotation. Show that the equilibrium of the bar relative to the tube is unstable, and obtain an equation determining the motion of the bar inside the tube when the former is disturbed.

Example 11.

CM

Immediately after the bar has been disturbed, let its speed in the tube be V and let the angular velocity of the tube be Let x be the distance of the centre of mass G of the bar from its initial position A at any later

n.

8]

M O T I O N O F A PART I CL E S Y ST E M- I I

209

instant and let be the angular velocity of the tube at this time. 0 is the point of intersection of the axis of rotation with the horizontal plane con­ taining the tube. Let OA = OG = L A OG = 6. Relative to the tube the velocity of G is in the direction indicated. The velocity of the tube at the point occupied by G is in a direction per­ pendicular to OG. The resultant of these velocities is the velocity of G

w

a,

r,

x

wr

..__ X _., I I

I I

A 1 ::Jiiiiiii �1 G�ii�====�� x ::::::::::::::::::� � wr

0

relative to a fixed observer. The angular velocity of the bar is if is its mass and its MI about G, the KE of the bar is cos 6)

m

I !m(x2 + w2r2 + 2wrx

w.

Hence,

+ !Jw2 !m(.i-2 + 2wax) + t[m(a2 + x2) + I]w2. If ] is the MI of the tube about 0, its KE is tJw2• The system being conservative, there is an energy equation, viz., !m(x2 + 2wax) + ![m(a2 + x2) + I + ]]w2 = !m(V2 + 2QaV) + t(ma2 + I + ])02, (i) since initially x = 0, x = V, w = 0. A particle of mass m moving with G has linear momentum components mx, mwr along the tube and perpendicular to OG respectively. Its AM about 0 is accordingly max + mwr2 = m[ax + w(a2 + x2)]. The A M about G o f the motion o f the bar relative t o G i s Iw. The total AM of the bar about 0 is therefore m[ax + w(a2 + x2)] + Iw. =

The only external forces applied to the system which have any com­ ponent in the plane of motion are the reactions of the bearings on the axis of rotation. These forces have no moment about 0, so that AM about this point is conserved. The expression of this fact is the equation = (ii) Eliminating between equations (i) and (ii), we obtain the equation

m[ax + w(a2 + x2)] + (I + ])w maV + (ma2 + I + ])Q . w + k2 x. = v�p2x2 (iii) x2 + k2 ' where mk2 = I + ], p2 = [(V + a0)2 + k202]fV2. This equation indicates that the speed x of the bar in the tube steadily increases from V to p V = {( V + a0)2 + k2Q2}t as x increases from 0 to The sense of the bar's motion ao.

is accordingly never reversed, and its original position was therefore unstable. Equation (iii) determines the motion of the bar relative to the tube.

210 8.'i.

A COURSE I N APPLIED MATHEMATICS

[cH.

Equation of AM for a Rigid Body in General Motion

Consider again the particle system mv m2 , mn. Employing the notation of the last section, we have by equation (8.20)

dh = 1r X F;, ; dt

(8.28)

where F; is the resultant external force acting upon m;. Also, from equation (8.26),

h = ha + r x Mf.

(8.29)

Differentiating this equation and employing equation (8.28), we obtain

dha + r dt

_

But, by equation (8.7), Mr

··

X Mr = 1r; X F;. =

(8.30)

1F; and hence,

d!a = 1 (r; - r)

X F; = 1 P; X F;.

(8.31)

The right-hand member of equation (8.31) is the moment of the ex­ ternal forces about G. We have accordingly proved that the rate of change of the AM about G of the motion relative to G is equal to the sum of the moments of the external forces about G. Thus, when writing down an equation of AM for a system, provided moments are taken about the mass centre, we may refer the whole motion to this point and neglect the point's motion. In the particular case of a rigid body in two-dimensional motion, the motion relative to G is a simple rotation about this point with angular velocity w , say. Thus, if Ia is the MI of the body about G , the AM about G of the relative motion is Iaw. Equation (8.31) therefore takes the form

Iaw = moment of external forces about G.

(8.32)

This is precisely the equation we should obtain if G were a fixed pivot. We conclude therefore that the rotary motion of the rigid body is quite independent of its translatory motion. This is the Principle of In­

dependence of Rotary and Translatory Motions.

A uniform, circular disc of radius a rolls in a vertical plane without slipping down a line of greatest slope of a plane inclined at an angle ex to the horizontal. Show that the coefficient of friction is not less than t tan ex. Let (R, F) be the normal and frictional components of the reaction at the point of contact C between the disc and the plane. Let mg be the weight

Example 12.

force acting at the mass centre G. If w is the angular velocity of the disc at any instant the velocity of G is v = aw . Differentiating, we obtain v = aw, i.e., if f is the acceleration of G, ffa is the angular acceleration of the disc.

8]

M O T I O N O F A P A R T I C L E S Y S T E M- I I

211

Resolving along and perpendicular to the plane for the motion of G, we obtain mg sin IX - F = mf, R - mg cos IX = 0.

I

Taking moments about G and employing equation (8.32) yields the equation of AM, viz., Fa = !ma• . !., a F = J;;mf.

or

It is now easy to show that f = %g sin IX, F = }mg sin IX, R = mg cos IX, and hence that FJR = l tan IX, thus proving the result stated.

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 fm which also rests on the plane. The cylinder is removed and gently replaced with a spin 0 in the sense tending to maintain contact between it and the block. If p. is the coefficient offriction at either contact with the plane, show that the cylinder slips on the plane for a time 3a0f7p.g and that its velocity is then aOJ7. Show also that the cylinder subsequently rolls on the plane and that the motion ceases after a further time 4a0f7p.g. (L.U.)

Example 13.

The external forces acting on the cylinder and on the block during the first phase of the motion are indicated in the first diagram. f denotes the 1 -----

f

p

pQ

212

A C O U R S E I N A P P L I E D MAT H E M AT I C S

[CH.

acceleration of the CM of the cylinder and hence also of the block. Let be the angular acceleration of the cylinder in the sense indicated. By resolving horizontally and vertically for the external forces acting on the cylinder, the following equations of motion are found, 1'-R = f R = The same procedure applied to the block yields the pair of equations Q = g. 1'-Q = Finally, taking moments about the CM of the cylinder, we obtain its equation of angular momentum, viz.,

A

-

P

-

P m !mf,

mg. !m

,

a !ma2A. A

1'-R =

A

Solving these equations for f and we determine that f = 1Lgf3, = 21'-gfa. Let v be the velocity of the block and w the angular velocity of the cylinder at time t after the commencement of the motion. Since both the linear and angular accelerations are constant, we have the equations

n - 2a�'-g t.

w=

v = il'-gt,

When rolling commences v = aw and hence ip.gt

=

t=

an 3an .

-

21'-gt,

71'-g

At this instant v = a0f7. During the rolling phase the external forces acting on the two components of the system are as indicated in the second diagram. The block is now

f

Q

fL Q

L,

being retarded, so that w e shall denote its acceleration b y f in the sense shown. The angular acceleration of the cylinder is thenffa. The equations of motion of the mass centres of block and cylinder are now = Q = 1'-Q F = mf, R = g . Taking moments about the axis of the cylinder, we obtain its AM equation, viz., -

P

-

P !mf,

!mg, m

Solving for f, we find that f = il'-g. Let t' be the time which must elapse during the second phase of the motion

M O T I O N O F A PA R T I C L E S Y S T E M - I I

8]

213

before the system i s reduced to rest. Then, since the velocity o f the block at the commencement of the phase is we have an l7 i.e.,

a0.J7, = fLgt ' t' = 4aO.f7fLg. Example 14. A uniform, solid cylinder of radius a is free to rotate about its axis, which is horizontal and fixed. A second equal cylinder rests in equilibrium on it, touching it along a generator. Prove that, if equilibrium is slightly disturbed, the angular velocities of the cylinders are equal and, until slipping takes place, the angle a, which the plane through the axes makes with the vertical, is given by the equation 5a& = 2g sin e. (L.U.) Let 0. be the angular velocities of the upper and lower cylinders respec­ tively in the senses indicated in the diagram. The external forces acting on the upper cylinder comprise the weight force mg and the frictional and '

w,

(a) (b) normal components F, R respectively of the reaction at C. These forces are shown in diagram (a). Taking moments about G, we obtain the AM equation for the upper cylinder, viz., Fa = !ma2w. (i) The external forces acting on the lower cylinder are indicated in diagram (b). P is the smooth reaction of the bearings on the axis of rotation. Taking moments about 0, we find that Fa = tma•nn . (ii) From (i) and (ii) it is clear that w = and hence, by integration, that n, the integration constant being zero, since, initially, = n = 0. Let OA be a radius fixed in the lower cylinder which is initially vertical. Let BG be a radius fixed in the upper cylinder which is also vertical at the commencement of the motion so that at this instant A and B are in contact. Since there is no slipping, arc AC = arc BC, and hence LAOC = LBGC rp, say. At any time the angle made by OA with the upward vertical is (6 - rp) and the angle made by BG with the downward vertical is (6 + rp). The angular velocities of the lower and upper cylinders are accordingly given by the equations = 6 + �. (ii) n = 6 +· respectively. But = 0.. Thus rp = 0 and hence rp = 0, since this is its w =

w

=

-

w

w

initial value. This implies that the upper cylinder does not roll on the lower, but remains continually in contact with it along the same generator.

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A C O U R S E IN A P PLI ED MATH EMAT I C S

[CH. G moves around a circle of radius 2a and centre 0 with angular velocity 6.·e Its acceleration in the direction perpendicular to OG is therefore 2a .

Resolving the external forces acting on the upper cylinder in this direction, we obtain the equation of motion =

mg sin 6 F 2mal But F = !maw by equation (i) and w = (J by equation (ii) . fma6 = mg sin 6, --

Hence

which is equivalent to the equation stated in the question.

8.8.

Initial Motion of a Rigid Body

If the problem is to calculate the forces present within a dynamical system at the instant it commences its motion from rest, equations of linear and angular momentum can be written down for each rigid body according to the principles explained above. However, it is not necessary to find equations which are valid throughout the subsequent motion of the system. We need only write down equations which refer to the initial position. Such problems are therefore essentially simpler than the type we have already discussed.

Two uniform rods OA, AB, of masses m, 2m and of lengths 2a, 4a respectively, are freely jointed at A, and the rod OA is smoothly hinged to a fixed point at 0. The rods are held in a horizontal line, and then released from rest. Find the initial reactions at 0 and A. (S.U.) Let P, Q be the reactions required (see diagram). These will act vertically as shown. Let w, 0 be the initial angular accelerations of the bars.

Example 15.

p

0

Q

A

w

m'l

q

11

l

B

2m']

Taking moments about 0 for the bar we have the equation of AM (i) = The acceleration of the mid-point of relative to will be vertically downwards. The acceleration of is in the same direction. Thus the net acceleration of the mass centre of is Resolving vertically for the external forces acting upon we now obtain (ii) = Taking moments about the CM of the bar (iii) = Eliminating from equations (i)-(iii), we find that

OA, mga + 2Qa tma2w. AB A 2aD A 2aw AB 2a(w + D). AB, 2mg - Q 4ma(w + D). AB 2aQ !ma2D. w, D Q = -mgflO. Now, from (i), w = 3gf5a.

8]

M O T I O N O F A PA R T I C L E S Y S T E M - I I The acceleration of the CM of OA being mg + Q

we have the equation

maw.

P= P = 3mgfl0.

Thus

8.9.

aw,

215

-

Motion with Variable Mass

In this section we shall consider briefly the motion of a body, such as a rocket or raindrop upon which water condenses as it falls through a cloud, whose mass alters as the motion proceeds. Let I: denote the system of particles which comprises the body B at the instant t (Fig. 8.2). Let a denote the set of particles which

t

t

.,. S t

FIG. 8.2.-Body with Changing Mass

belong to B at the instant t + at, but not at the instant t, i.e., those particles which j oin B from some external source across the surface bounding this body, during the interval (t, t + at) . Let a ' denote the set of particles which belong to B at the instant t, but not at the instant t + at, i.e., those particles which leave B during the interval (t, t + Bt) . Then, if I:' denotes the set of particles comprising B at the instant t + at, I: + a = I: ' + a ' (8.33)

Let v, v + av be the velocities of the mass centre of B at the instants t, t + Bt respectively and let F, F + BF be the respective vector sums of the external forces applied to the particle system I: + a (or I:' + a') at these instants. Let u be the velocity of the mass centre of a at the instant t and let u' be the velocity of the mass centre of a ' at the instant t + at. M will denote the mass of I:, am the mass of a and 3m' the mass of a ' . Then the mass of I:' must be M + 3m - 3m'.

For a system of constant mass, the increment in the linear momentum over any period is equal to the impulse of the external forces (Section 9.1) . But the systems of particles I: + a, I:' + a ' are identical, and hence we can apply this principle over the interval (t, t + at) to obtain the equation Fat = (M + am - Bm') (v + av) + u'Bm' - Mv - uam,

(8.34)

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A COURSE IN APPLIED MAT HEMAT I C S

[CH.

where second-order quantities have been neglected. After the neglect of further terms of the second order, this equation may be reduced to

Fat = Mav + (v - u)am - (v - u')am'

(8.35)

Dividing by at and proceeding to the limit, we find that

F

=

M�; + 1X(V - u) - 1X'(v - u'),

(8.36)

where IX is the rate at which B is gaining mass from its environment and IX ' is the rate at which it is losing mass to this environment. Also, in the limit, F is the net external force applied to B at the instant t, not including the forces being exerted upon B at this instant by particles which are just leaving or j oining this body. u, u' are the limiting velocities of the mass centres of cr, cr ' respectively as at -+ 0. In the particular case of a body, such as a raindrop, which is only gaining particles from its surroundings and is not losing particles, IX' = 0 and a = dMjdt. The equation of motion (8.36) can then be written in either of the forms

F = �; + (v - u) d:{' d dM F = dt (Mv) - u llt .

If u

=

}

(8.37)

0, i.e., particles are being picked up from rest, then

F = dtd (Mv) , .

(8.38)

and the rate of increase of the linear momentum of B is equal to the sum of the external forces. Again, if u = v, i.e., particles have B's velocity just before combining with it, then (8.39) and the product of the mass and acceleration of B is equal to the sum of the external forces. In the case of a body, such as a rocket, which is only losing particles, IX = 0 and IX ' = - dMfdt. The equation of motion can again be written in either of the forms (8.37) (with u' replacing u). u' - v is the velocity of the particles being lost relative to B and, in the case of a rocket, this is the y"et velocity c . The first of equations (8.30) can then be expressed

dv F + e dM dt = M d( .

(8.40)

Thus, the product of the mass and acceleration of a rocket is equal to the sum of the applied external forces, if we include among these a

8]

M O T I O N O F A PA R T I C L E S Y S T E M - I I

217

force cdMfdt. This latter force is opposite in direction t o that o f the jet (since dMjdt is negative) and is called the thrust of the rocket motor. Consider a rocket moving along the x-axis under the action of its thrust alone. The scalar equation of motion is dM dv M = -c (8.41)

dt

Tt '

assuming that c is in the negative direction of the axis. Thus = dv. - c dM M Integrating this equation over a time interval during which v increases from v0 to v1 and M decreases from M0 to M1, we obtain (supposing c constant)

(8.42)

If all the propellant has been expended during this manceuvre, M0/M1 is called the mass-ratio of the rocket and equation (8.42) shows how this is related to the velocity increment which can be produced in free space. Thus, with a mass ratio of 6-about the maximum that can be attained at present (1959) in practice-we have v1 - v0

=

c log 6 = 1 · 8c.

(8.43)

I.e., the maximum velocity increment to be expected from a single­ stage rocket is about 1·8 times its exhaust velocity.

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 JV > g, and that it cannot rise at all unless JV > 'Ag, where 'AM is the mass of the case of the rocket. If conditions are such that the rocket is just able to rise vertically at once, show that, if the resistances are neglected, the greatest height it will reach is v• - {I - ). + log ). + !(log 'A)2}. (Li.U.) g We have dMfdt = -fM, and hence the motor thrust is flvlV The rocket will not move so long as the thrust is less than the weight. Hence, unless JMV > Mg, i.e., fV > g, the rocket will not move initially. When all the propellant is exhausted, the weight is >.Mg. Hence, unless fMV > 'AMg, i.e., fV > 'Ag, the rocket can never move. IffV g, the rocket can just move initially. At time t after commencing to rise, the rocket's mass is M - fMt = M(l - gtfV). Its thrust is Mg. Hence its equation of motion is M(l - gtJV) dtdv = -Mg(l - gtfV) + Mg, dV = -g + g . dt l - gtJV Integrating under the initial condition v = 0 at t = 0, we obtain v = -gt - V log (l - gtfV).

Example 16.

=

A COURSE I N APPLIED MATHEMATICS

218

[CH.

If

x is the height risen in time t, v = dxfdt and a further integration yields x = Vt - !gt2 + gv• (I gtfV) log ( 1 gtfV). The propellant is fully expended after a time (1 "A)/f = (1 "A) Vfg After this time v = v1 = V(l "A + log "A), x = x 1 = v•g [!(1 - /..2) + "A log /..] . The rocket will continue to rise under its own momentum for a further height v12f2g. Hence the total height attained is as stated. -

-

-

-

-

-

-

EXERCISE 8 of length 2a and mass M is free to turn about a 1 . A uniform rod fixed point in distant !a from and has a particle of mass M attached to the end The rod is held in a horizontal position and then released. Find the angular velocity when is vertical and prove that the pressure on the point of support is then 82Mgfl 7. (L.U.) 2. Two particles, and of masses 2m and m respectively, are attached to the ends of a light rod of length 3a and are initially at rest on a smooth, horizontal table. The particle is then projected vertically to the horizontal upwards with velocity u. If the inclination of during the time that is in contact with the table is denoted by show that 3a2 (cos2 + 2) (d6Jdt) 2 = u2 - 6ga sin Show also that the particle will leave the table when a sin 6(d6fdt)2 = g. If this occurs when e = 60°, show that the initial velocity must have the value given by 1 5y3 -- ga. u• 2 (S.U.) 3. A particle of mass m is fixed to a point P of the rim of a uniform circular disc, centre 0, mass m and radius a. The disc is held with its plane vertical, its lowest point in contact with a perfectly rough, horizontal table, and with OP inclined at 60° to the upward vertical and is then released. If the subsequent motion continues in the same vertical plane, show that when OP makes an angle e with the upward vertical a(7 4 cos 8) 82 = 2g(l - 2 cos 8). Show also that when OP is first horizontal the acceleration of 0 is 1 8g/49, and find the reaction between the disc and table at this instant. (L.U.) 4. A heavy, uniform, solid sphere of radius af5 rolls without slipping on the inside of a fixed, rough, hollow cylinder of radius a, whose axis is horizontal, so that a diameter of the sphere always lies in the same vertical plane. Initially the centre of the sphere is at the lowest point and is moving with a velocity y(nag) . If comes to instantane­ ous rest when the perpendicular from to the axis of the cylinder is

AB AB

A

B.

A

AB

B,

B

A e

AB

e,

e.

A

=

+

AB

C

C

C

8]

M O T I O N O F A PART I C L E S Y S T EM-I I

219

inclined at an angle cos-1 � to the downward drawn vertical, prove that n = 3/7. Find also the value of the frictional force at that moment. (L.U.) 5. A uniform rod is held so that one-third of its length rests on a hori­

zontal table at right angles to an edge, the other two-thirds projecting beyond the edge, and it is then released. Show that it will rotate about the edge through an angle tan-1 !-fl., where fL is the coefficient of friction between the rod and the edge before it begins to slip. (L.U.)

6.

A thin, hollow, uniform, cylindrical circular tube of mass M and radius a has a particle of mass m attached to its inner surface at a point mid­ way along its length. It is placed on a smooth, horizontal plane with the particle at the level of the axis, one generator of the cylinder being in contact with the plane, and released from rest. Write down the equations of linear momentum and of energy for the system in terms of the displacements x of the axis and e of the radius to the particle from their initial positions, and show that the maximum displacement of the axis is 2maf(M + m) and that its maximum velocity is 2maga (B.U.) M(M + m) (M + 2m)

(

)t

7. The cross-section of a right cylinder of mass M is bounded by a para­ bolic arc and the latus rectum (length 21) of the parabola. The surface of the cylinder is smooth, and it is placed with its rectangular face on a horizontal plane. A particle of mass m is placed at the mid-point of the highest generator of the cylinder and slightly disturbed. Find the speed of the cylinder and the speed of the particle relative to the cylinder when the particle has descended a vertical distance x. (L.U.) 8. A string of length 2a with three particles each of mass m attached to it, one at each end and one at its mid-point, is laid in a straight line on a smooth, horizontal table. The middle particle receives a blow in a direction at right angles to the string. Show that, at the instant when the two extreme particles collide, the tension of the string is P/9ma. (B.U. )

]

9. Two equal, uniform rods A B, A C, each o f mass m and length 2a are smoothly jointed at A , and a light, elastic string of modulus 3maw2 and 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 w about A , but in opposite senses. Show that the rods are instantaneously at rest when the angle 28 between the rods (L.U.) is given by sin e 5.

6

10.

=

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 perpendicular to the string. Show that, if the

V

220

11.

12.

A C O U R S E I N A P PLI E D M A T H E M A T I C S

particles at the ends collide after a time t when the displacement of M from its initial position is x, then (M + 2m)x = M Vt + 2ma. Show also that the tension in the string j ust before the collision is mM2 V2f(M + 2m) 2a. (L.U.) 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)fM. Prove also that when the bead is next at its lowest level the velocity of the block is (M - m) Vf(M + m). (L.U.) A rough, uniform sphere of mass M can turn freely about a horizontal axis through its centre 0, and a particle of mass M/5 rests on it at its highest point. The sphere is slightly disturbed, and the system allowed to move freely. Prove that when OF makes an angle e with the upward drawn vertical through 0, the angular acceleration of the sphere is (gf3a) sin e, so long as the particle does not slip on the sphere. Prove also that the particle will begin to slip when the sphere has turned through an angle ex, given by the equation (5 cos ex - 2)!L 2 sin ex, where {1. is the coefficient of friction. (L.U.) A uniform, cubical block of mass M can slide on a smooth, horizontal plane and a uniform rod, also of mass M, leans with one end on the plane and the other against a vertical cube face, which is smooth, the rod being in a vertical plane at right angles to the cube face and through its centre. The system is released from rest with the rod inclined at an angle ex to the horizontal. Show that contact between the rod and the cube ceases when the inclination e of the rod to the horizontal satisfies the equation 24 sin e + 16 sin ex = 0. (M.T.) 3 sin3 e A wheel of radius a can turn about a horizontal axis which passes through the centre of the wheel at right angles to its plane. The moment of inertia of the wheel about the axis is 3Ma2 • An inextens­ ible string passes over the wheel and carries freely hanging masses M and !M at its ends. The system is released from rest and moves under the action of gravity and of a frictional resistance whose moment about the axis is tMa2w2 when the angular velocity of the wheel is w. Assuming that there is no slip between the string and the wheel, prove that the speed of the mass M is {2ag(1 e--.z/9•)}! when it has fallen a distance x, provided that the mass !M has not then reached the wheel. Find also the tensions in the two parts of the string. (L.U.) Two equal, uniform rods AB and BC, each of mass m and length 2a, are joined by a light but rough hinge at B so that the angle ABC is always equal to a right angle. If the system is suspended so that it

P

=

13.

-

14.

-

15.

[cH.

8]

M O T I O N O F A PART I CL E S Y ST E M - I I

221

can oscillate freely about a fixed horizontal axis through find the length of the equivalent simple pendulum. Prove that when the makes an angle 6 with the down­ internal bisector of the angle ward vertical, the moment of the couple exerted by each rod upon the other is mag cos 6f-yl2. (L.U.) 16. Two smooth, fixed, horizontal, concentric circular wires are of radii a and 3a. A particle of mass m can slide on the inner wire, and another particle of mass 2m can slide on the outer wire, the two particles being connected by a light elastic string of natural length a and modulus :A. If the particles are slightly disturbed from the position of equilibrium in which they are collinear with the common centre but on opposite sides of it, show that when the particles are next collinear with the centre the velocity of the inner particle is 12-yi(:Aaf19m), and find the angle through which the radius to this particle has turned. (L.U.) 17. 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/8 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 4a2w/(4a2 + r2), and find the radial velocity of the particle at the moment when it reaches the edge of the disc. (L.U.) 18. 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 on the circumference of the disc so that it is inclined at an acute angle IX to the upward drawn vertical at the rod and the vertical through 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 is given by 2 v = 2ga(2M + m) cos 1X/(2M cos 2 IX + m) . (L.U.)

B,

ABC

P

P,

P

P

19. A bead of mass m is threaded on a smooth, uniform wire circle of mass m and radius a, which turns in a horizontal plane about a smooth, vertical axis through a point 0 of its circumference. If
P

COP,

OP

C

P

P

20. A straight, smooth wire of negligible mass is free to rotate in a hori­ zontal plane about a fixed point 0 of itself. Two beads of equal mass are threaded on to the wire and are joined by a light, inextensible

A COURSE I N APPLIED MATHEMATICS

222

[cH.

string of length 2a. P is the point mid-way between the beads and OP = x. Initially, the wire is rotating with angular velocity w, x = a, i = y2aw and the string is taut. By applying the principles of conservation of energy and angular momentum, show that during the subsequent motion 2awx . X = (x2 + a2) l -

•

Hence calculate the time taken for

x

to decrease from a to !a.

21. Two particles A and B, each of mass m, attract each other with a force kr, where is constant and r is the distance AB. At an instant when the particles are at rest at a distance a apart, B is projected with velocity na in a direction perpendicular to A B, where n = y(2kfm). Prove that the subsequent motion of the particles is the same as that of the ends of a diameter of a circular disc which rolls, with uniform speed, on a straight line. (M.T.) 22. A light, smooth tube, of length 4a, can rotate freely in a horizontal plane about its mid-point 0, which is fixed. A heavy, uniform rod, of length 2a, is placed inside the tube so that its mid-point M is slightly displaced from 0, and the system is given an initial angular velocity w about the vertical through 0. Find the radial and angular velocities of the rod when OM = r ( < 3a) . Show that, when the rod is about to leave the tube, its radial velocity is 3aw/y28 and its angular velocity (L.U.) is w/28.

k

23. A uniform, straight rod OA , of length l and mass m, is freely hinged to a uniform circular disc, of radius kl and mass m at one end A of a diameter AB. The system is free to rotate about the fixed point 0 on a smooth, horizontal table. If initially 0, A , B lie in that order along a line OX, and if at any subsequent instant the inclinations to OX of OA , AB are respectively e, ¢>, find expressions for the KE and the AM about 0 in terms of ¢>, 6, �' k, m. If initially 6 = 0, if> = w, and if � = 0 when OA and A B are perpendicular, determine the value of (M.T.) 24. A uniform sphere of radius r is projected with velocity V, but without rotation, up a line of greatest slope of a rough plane inclined at an angle � to the horizontal. If the coefficient of friction is tan t�. prove that the centre of the sphere will come to rest in time 1 7 V/(18g sin �) . (Q.U.) 25. A uniform, right circular cylinder of radius a which is rotating with angular velocity w about its axis is gently placed in contact with a rough plane inclined at an angle tan-1 (5/12) to the horizontal, its axis being placed in a horizontal position perpendicular to the lines of greatest slope and released. If the sense of rotation is such that the cylinder begins to move up the slope and if the coefficient of friction between the cylinder and the plane is t. find the time that elapses before the cylinder begins to roll up the plane and prove that it comes to instantaneous rest at time 13awj10g after it was placed on the plane. (L.U.)

e,

k.

l,

8]

M O T I O N O F A PART I CL E S Y S T E M- I I

223

26. A uniform sphere of radius a is projected with back spin along a rough horizontal plane. Show that it will not turn back in the course of its motion if 5u > 2aw, where u and w are the initial linear and angular velocities of the sphere. Show also that the angular velocity of the sphere will then change sign during the motion and that rolling takes place after time 2(u + aw)f7fLg, where fL is the coefficient of friction of the plane. (B.U.) 27. Two circular cylinders, both of radius a, are held in equilibrium on an inclined plane of inclination et so that their axes are horizontal and their surfaces are in contact along a common generator. The upper cylinder is of uniform density and of mass m, and the lower cylinder is of the same density, but hollow, with internal radius ia. 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 (56g sin et)/87, and that the reaction between them is (mg sin et)/29. (L.U.) 28. A uniform, circular cylinder of mass m and radius a rests on a rough, horizontal plane, with its axis horizontal and parallel to and distant l from a vertical wall. The two ends of the axis are joined to the wall by similar elastic strings which are horizontal and perpendicular to the axis. The coefficient of friction between the cylinder and the plane is fL, and the natural length and modulus of each string are l and iA respectively. The cylinder is now given a spin n in the sense causing motion away from the wall. Show that, so long as slipping occurs and the strings remain stretched, the motion of the axis is

( ��) from the wall, and

simple harmonic about a centre at distance l +

its displacement from the starting position after time t is x = ',Lg(l - cos nt)fn2, (L.U.) where n = y(A/ml) . 29. A smooth wedge of mass M and angle et( < t7t) rests with a face adjacent to the angle et on a horizontal table. A solid circular cylinder of mass m and radius a is given a spin n 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 et. If the cylinder remains in contact with both wedge and table, show that the cylinder slips on the table for a time a!1(m + 2M) g tan et(3m + 2M) and that the magnitude of the subsequent steady velocity of the system is independent of et. (L.U.) 30. A uniform, circular disc, of mass M and radius r, rolls without slipping on a rough, horizontal plane. A uniform rod, of mass M and length 2a( > r) , has one end freely attached to the centre of the disc and the

224

31.

A C O U R S E I N A P PL I E D M A T H E M A T I C S

other in contact with a smooth, vertical wall. The whole system moves in a vertical plane perpendicular to the wall. The inclination of the rod to the horizontal is denoted by e. and the system starts from rest when e = o:. Show that a62(2 + 9 sin2 e) = 3g(sin o: - sin e) . Prove also that the ratio of the reaction of the wall to the frictional force on the disc is independent of the value of e. (M.T.) A symmetrical cotton-reel, consisting of two uniform discs of radius a joined by a cylinder of radius ia. can roll on a rough, horizontal table. The total mass of the reel is M, and its radius of gyration about the axis of the cylinder is !a. A fine, light, inextensible thread is wrapped round the cylinder in a vertical plane through the CM of the reel. The thread leaves the cylinder at its highest point at right angles to the axis, and thence passes horizontally over a small, light pulley at a height 3af2 above the table. To the free end of the thread is attached a mass m which can move vertically under gravity whilst the thread unwinds without slipping from the cylinder. If the reel does not slip on the table, prove that the coefficient of friction must exceed (L.U.) mf(5M 9m . A thin, uniform, circular disc of mass M lies on a smooth, horizontal table. Cut in the upper face of this disc is a circular groove of radius a, with its centre at C, the CM of the disc, and in the groove a particle P of mass m can slide without friction. The disc is set moving with­ out rotation so that C has an initial velocity V towards the momentary position of P, and at the same instant P is given a velocity v relative to the disc. Show that the centre of mass of the system moves in a straight line, making an angle tan-1 [mvf(M m) V] with the initial direction of CP, and that the angular velocity of the disc relative to the table remains zero. (L.U.) A uniform, square lamina, of side c, is suspended in a vertical plane with two sides vertical by two equal strings attached to the upper corners. The strings are equally inclined at an angle � to the vertical so that their upper ends are at a distance apart greater than c. The one string being suddenly released, the tension in the other changes from T0 to T. Prove that T 2 + 2 cos 2� · T0 5 - 3 sin 2� (L.U.) A raindrop of mass 3k Vfg is falling vertically with speed 2 V when it overtakes a cloud which is falling vertically with constant speed V. During its passage through the cloud the mass of the raindrop increases by condensation from the cloud at a constant rate k, and the force of resistance is k times the relative speed. Find 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. (M.T.) A raindrop, falling freely under gravity in a straight line, grows by the condensation on it of vapour previously at rest. The drop falls

+ )

32.

+

33.

34.

35.

[CH.

M O T I O N O F A PART I C L E S Y S T E M - I I

8]

225

from rest at time t = 0 and its mass at time t is Mek', where M and k are constants. If T is the time required for the mass to increase to 2M, prove that the speed of the drop is then gTf(2 log. 2) and find how (L.U.) far it has fallen. 36. A rocket, with exhaust velocity c, is to rise vertically from the Earth's surface with constant acceleration ng. If g is the acceleration due to gravity (assumed constant) and A.v is the air resistance, prove that fuel must be expended so that its mass m at time t is given by A.nc -(n+ llgtfc )..n c gt , (n m - m0 (n + l) 2g (n l) 2g { m 0 being the mass at t = 0. 37. A uniform, circular disc is suspended horizontally by three vertical strings symmetrically attached to points on the circumference. If one string is cut, show that the tension in the other strings is instantaneously reduced in the ratio 3 : 4. (Q.U.) 38. AB, BC are two uniform rods of masses m, m' and lengths 2a, 2b respectively. AB is pivoted freely at a fixed point A and is initially held in a horizontal position, while BC hangs freely from B. When AB is released and is free to rotate round A , show that � 0, its initial angular acceleration, is given by 2m'. � 0 - 3g . m 4a m 3m' (M.T.) _

{

_

}e

_

I (r)

+ +

+ +

_

+ 1) }

CHAPTER 9

IMPULSIVE MOTION OF A RIGID BODY

Equations of Linear and Angular Momentum for Impulsive Motion Consider the system of particles mi> m2 , mn. As in Section

9.1.

8.1, we derive the equation

(9.1) Integrating this equation with respect to t over the interval (t1 , t2) , we obtain (9.2) where rl > r2 are the velocities of the CM at the beginning and end of the interval respectively. Let (9.3) Then � is the impulse of the external force acting upon the ith particle and calculated over the time interval (t11 t2) . Mf is the linear momentum of the system. Hence, equation (9.2) states that the increase in the linear momentum of a system over any time interval is equal to the sum of the impulses of the external forces over the time interval. For practical application it is, of course, convenient to resolve this equation in various directions. The principle we have just enunciated is generally applicable, but is particularly useful in problems where a system is subjected to blows at various points. We may usually assume that the effect of such blows is to cause an instantaneous change in the velocities of the different particles of the system without any appreciable change in their positions. Equation (9.2) may be employed to calculate the velocity changes involved. If h is the AM of the system about the origin 0 of the frame of reference, we have, by equation (8.20),

dh = !ri X Fi. dt

226

(9.4)

[ca. 9]

I M P U L S I VE M O T I O N O F A R I G I D B O D Y

227

Integrating with respect to t over the interval (tl> t2), we obtain

h2 - h1 = 1

t r; t,

X

F;dt.

(9.5)

Assuming that the system is being subjected to a number of blows, (tl> t2) being the very short time interval during which their effect is felt, the position vector r; of the ith particle will not vary appreciably during this interval, and equation (9.5) is therefore equivalent to

h2 - h1 = 1r;

X

t F;dt t,

=

Ir; X 1;.

(9.6)

The right-hand member of equation (9.6) is the sum of the moments of the impulses of the external forces (the blows) about 0. We may accordingly state the principle that the increment in the AM of a system about a point 0 caused by the simultaneous application of a series of blows is equal to the sum of the moments of the impulses of these blows about 0.

Three uniform rods AB, BC and CD, each of mass m and length 2a, are smoothly jointed at B and C and lie at rest in a straight line on a smooth, horizontal table. The rod BC is given a horizontal blow ] at a point 0 in a direction perpendicular to BC. Find the initial velocity of 0 and prove that the initial velocity of the mid-point Q of the rod BC is independent of the position of O. Find the KE generated as a function of the distance OQ. (B.U.) The impulses acting upon the separate rods are shown in diagram (a). It

Example 1 .

is evident that all particles are set in motion in a direction perpendicular to the initial line of the rods and that, therefore, the impulses have no com­ ponents along this line. The actions and reactions at the joints B and C are always equal but opposite in sense, and hence so are their impulses.

(a)

_j

A

-

A

d

0

8

1

14

(b )

K

J

I

0

8

I X ._ I

c

'I

L

·d ,(1

D

K

0

c

D

The motion generated in the system is represented in diagram The initial motion of each rod is specified by the velocity of its CM together with its angular velocity about this point. Since B is a point of both the rods AB and BC, its velocity may be obtained by consideration of the motion of either rod, and the results must

(b).

A C O U R S E I N A P P L I E D M A T H E M AT I C S

228

[CH.

be equal. The velocity of relative to is Combining this with the velocity of (which is in the opposite sense), we obtain a resultant velocity of of in the direction of Similarly, its velocity is + Hence (i) + Similarly, (ii) + By resolving in a direction perpendicular to the rods, we obtain the following equations of linear momentum : (iii) (iv) J (v) We will now calculate the angular momentum about its CM generated in each rod. Equation (8.26) will be employed for this purpose. The contri­ bution of the moment of the momentum of the particle of mass moving with the CM is zero in each case, since moments are being taken about the CM. This fact makes the CM a particularly convenient point about which to calculate the AM in all problems concerning motion under impulses. The AM of the motion relative to the CM is in the case of , for and for These quantities are accordingly the angular momenta about their CMs generated in the three rods. Applying equation (9.6) to each in turn, we obtain (vi) + (vii) (viii) Solving the equations (i)-(viii), we find that 2] 9x + 9x ]

B

P B (u - ap)

P ap.

u. u - ap = v aq. v - aq = w ar.

(v aq).

I = mu, - I - K = mv, K = mw .

m

BC

ma2rf3

ma2pf3

CD.

AB ma2qf3

-Ia = tma•p, Ka ]x - Ia = tma2q, Ka = tma•r. 5a - · m.]· w=� v = 3m' u = 5a � · ;:ii: 5a - 9x J ' + 9x J ' q = 6x J ' r = ----yoa p = - 5a� · ma 5a · ma ma Clearly v is independent of O's position. The velocity of 0 is v + qx = :� ( 1 + �::) . •

The KE generated is

tmu• + }ma•p• + tmv• + tma2q2 + !mw2 + tma2r2 3x2) . = E_ (!3 + 5a2 Example 2. (i) An impulse of mar,nitude P is applied at a point A of a body. Prove that the KE given to the body is !P(u + v), where u and v are the com­ ponents (in the direction of Pj of the velocity of A just before and just after the blow. (ii) A blow P applied at a point A of a body at rest in the direction l produces at another point B a velocity whose component in the direction l is v. Show that a parallel blow P applied at B to the same body at rest in the direction l produces at A a velocity whose component in the direction l is also v. (Q.U.) Let (p, q) be the components of the CM's velocity in P's direction and perpendicular to it respectively, immediately before the application of P. Let be the angular velocity of the body at this instant. In the same define the body's motion immediately after the impulse. way, let (p', q'), m

w

w

'

IMPULSIVE MOTION O F A RIGID BODY

9]

229

If is the mass of the body, we obtain the following equations of linear momentum, (i) p= 0 = (ii)

M

M(p' - p), M(q' - q). If I is the MI of the body about G, taking moments about G, we derive the following equation of AM, -Py = I(w' - w), (iii) where y is the dimension shown in the diagram.

The increment in KE resulting from the blow is

!M(p'2 + q'2) + !Jw'2 !M(p2 + q2) Fw2 !M(p' - p)(p' + p) + !M(q' - q)(q' + q) + !f(w' - w)(w' + w) = !P(p' + P) !Py(w' + w) , (iv) by the use of equations (i)-(iii). But the velocity component of A in the directionof P is u = p - wy before the impulse and v = P' w'y afterwards. Hence, the increment in the KE (iv) may be expressed in the form !P(u + v). If the body is initially at rest, p = q = w = 0 and we obtain from equations (i) and (iii) Py P P' = M, w' = - - r If B is a point of the body a perpendicular distance TJ from the line through G parallel to P, its velocity after the blow has a component in the direction of P given by v = P' - w,TJ = Mp + 1p YTJ· If y and TJ are interchanged in this expression, the value of v is unaltered. This implies that if P is applied at B, the component of the velocity generated at A in the direction of P is also v. Example 3. Two uniform rods OA, OB each of mass m and length 2l are smoothly jointed at 0 and lie on a smooth, horizontal table, in the same straight line. A particle of mass M, joined to 0 by a light, inextensible string, is projected from 0 with velocity u along the table. Show that, at the instant after the string becomes taut, the velocity v of 0 and the angular velocity of each rod are =

_

_

-

-

w

230

A C O U R S E I N A P P L I E D M A T H E M AT I C S

[CH,

given by w = 3vf4l and v = 2Muf(2M + m), and that the loss in KE is Mmu2f(4M + 2m). Show also that at the instant in the subsequent motion at which the rods come together, the velocity of 0 is Muf(M + 2m). (L.U.) Let I be the impulsive tension generated in the string when it becomes taut. To maintain the symmetry, suppose that impulses ii are applied

Mt � ( "t r t''1) -+I

I I I

A

-

w

-8.

to each rod at 0. Considering the change in linear momentum of the particle of mass we obtain the equation (i) The velocity of the CM of either rod immediately after the string becomes taut is in the direction of The equation of linear momentum of a rod is accordingly (ii) if Taking moments about the centre of either rod, we obtain (iii) Solving for and from equations (i)-(iii), we find that

M,

I = M(u - v). (v - lw) v. = m(v - lw). !Il = }ml2w . v w 3Mu 3v v = 2M2Mu + m' w = 2l(2M + m) = 4f The KE generated in a rod is l;m(v - lw)2 + i-ml2w2 = i-mv•. The KE of the particle M changes to l;Mv2• The total KE of the system is accordingly reduced to !mv2 + !Mv2 = i(m + 2M)v2 = 2MM2u2 + m' The initial KE of the system is !Mu2, indicating that a loss of amount Mmu2 4M + 2m has occurred. Let w be the velocity of 0 when the rods come together and let n be their angular velocity at this instant. Then the CMs of the rods both have velocity components (w, lQ)

L n.

n.

along and perpendicular to the string. The component of the linear momentum of either rod in the direction of the string is therefore and the total linear momentum of the system in this direction is But there are no external forces acting on the system in the plane of motion, and hence its linear momentum

mw,

(2m + M)w.

IMPULSIVE MOTION O F A RIGID BODY

9]

231

is conserved in all directions from the outset. In particular, linear momentum is conserved in the direction of the st>ing. Thus

(2m + M)w = Mu, w = 2mMu + M. Example 4. A uniform, circular disc of radius a rolls without slipping, with its plane vertical, on a rough, horizontal table, the speed of the centre being v. If the disc strikes a rough, inelastic step of height h( 12a2ghf(3a - 2h)2• (L.U.) Let m be the mass of the disc. The disc is subjected to an impulse I acting from the corner of the step 0. Since this impulse has no moment about 0, the AM of the disc about this point is unaltered by the impact.

Initially, this AM is equal to the moment of the momentum of a particle of mass moving with the centre of the disc, together with the AM of the rotary motion with angular velocity about i.e., it is equal to

m

C

vfa C, mv(a - h) + !ma2 .1:'.a = !mv(3a - 2h). Immediately after striking the step, the disc is rotating about 0 with angular velocity w, say. Since the disc's MI about 0 is �ma•, its AM about this point is then �ma2w. Since AM is conserved, !mv(3a - 2h) = �ma2w, w = v(3a3a2- 2h) . The KE of the disc immediately after striking the step is

fma•w• = mv2(3a12a2- 2h)".

During the subsequent motion, energy is conserved, KE being transformed into PE. The increase in PE necessary if the disc is to surmount the step is Thus the KE at the beginning of this phase of the motion must he larger than this in value, i.e.,

mgh.

mv2(3a� - 2h)2 > mgh, 1 v > (3a12a2gh - 2h) 2 --

2

A C O U R S E I N A P P L I E D MAT HEMA T I C S

232 9.2.

[CH.

Elastic Impact of Rigid Bodies

When two rigid bodies collide, there are a number of alternative possibilities for the conditions at the point of contact during the short period of the collision. These are as follows :

(a) Friction may be sufficient to prevent sliding throughout the period of contact. (b) Sliding may occur in one sense throughout the period of contact. (c) Sliding may take place in one direction at first and in the contrary direction later. (d) A sliding phase may be followed by another during which there is no sliding, or vice versa. If it can be assumed that, whatever the conditions during contact, at the conclusion of this phase sliding has ceased, we are thereby provided with one condition to be satisfied by the motion immediately after impact. This condition, together with the generalized form of Newton's Law of Restitution (p. 125), will be found sufficient to de­ termine the effect of the impact. The appropriate form of Newton's Law can be stated thus : When two bodies impinge at a point the relative

velocity of the points of contact immediately after the impact resolved along the common normal is -e times this component of the relative velocity before impact. This method of approach to the problem is adequate when conditions (a), or certain cases of (d), prevail. If conditions (b) are to be expected, the ratio of the frictional com­

ponent to the normal component of the reaction between the bodies during the collision will be constant, and hence the impulses of these quantities will be in the same ratio, viz., the known coefficient of friction between the two surfaces. This fact provides us with a condition which may be combined with that obtained from Newton's Law and the problem solved. In all other circumstances the impact must be divided into two or more phases and the change in the motion over each phase assessed separately. We shall not consider this more complex type of situation.

A uniform, rigid sphere of radius a is falling under gravity with angular velocity w about an axis perpendicular to its plane of motion when it impinges with velocity v upon a horizontal plane. If e is the angle ofincidence and is the angle of reflection, show that, if slipping takes place in one sense only during the collision, is independent of w, whereas, if the contact is sufficiently rough to prevent sliding, 5v sin e tan 'I' - 2aw7ev+ cos e . e being the coefficient of restitution. Taking w in the sense indicated in diagram (a), suppose sliding at the

Example 5.

.1.

-

point of contact C is to the right in this figure. Let (I, ftl) be the normal

9]

I M PU L S I V E M O T I O N O F A R I G I D B O D Y

233

and frictional components of the impulsive reaction respectively is the coefficient of friction) . Immediately after the impact let V be the velocity of the centre of the sphere 0 and let be the new angular velocity. The equations of linear momentum in the directions along and perpendicular to the plane are then = V sin sin a, (i) (ii) V cos cos a . Application of Newton's Law yields v cos cos e. (iii)

(p.

w'

p.I m I=m

Eliminating V and

mv + mv = ev -

I from equations (i)-(iii), we obtain e tan = tan e + p.(l + e),

determining the angle of reflection. rf> is clearly independent of Should slipping take place in the opposite sense to that just assumed, the direction of must be reversed. Changing the sign of has this effect, and then tan = tan e If no slipping takes place we replace by and then equations (i) and (ii) become sin rf> - sin e), (iv) (v) ( V cos cos 6). Also, taking moments about 0, we have the angular momentum equation (vi) The point of contact is brought to rest by the impact, and thus V sin (vii) From equations (iii)-(vii) we find that sin e tan "' cos e .

p.I

p.

e

w.

p.(l + e). p.I F, F = m(V v I=m +v Fa = fma2(w - w'). = aw'. = 2aw7ev+ 5v Example 6. A uniform, straight rod falls vertically without rotation on to a smooth plane inclined at an angle ex to the horizontal. If e is the coefficient of restitution between the plane and the rod, show that a fraction (l e2) cos2 l + 3 sin2 ex of the initial energy is lost in the impact. Let V be the velocity of fall of the rod on to the plane. The impulsive reaction of the plane I must act normally to the plane, since the latter is smooth. Let (u, v) be the components of the velocity of the centre G of -

-

ex

A C O U RS E I N A P PLI E D M A T H E M AT I C S

234

[CH.

the rod along and perpendicular to the plane after impact (diagram (b)). Let be the angular velocity of the rod after impact. Resolving along and perpendicular to the plane and taking moments about G, we obtain the impulse equations ( sin ex = 0, (i) (ii) + cos ex = I, (iii) = I sin ex,

w

mV u) m(v V ) tma•w a 2a being the rod's length and m its mass. I

G

(b)

(a)

The velocity of the lower end of the rod in the direction of the plane's normal is cos ex before impact and + sin ex) after. Hence, by Newton's Law, (iv) + sin ex = cos ex.

V

(v aw eV

v aw Solving equations (i)-(iv) for u, v, w we find that u = V sin ex, ex) V cos v = (e -13+sin•3 sin• ' ex 3 ( 1 + e) V sin cos ex. aw _- 1 + 3 sin• ex The KE before impact is !m V2• The KE after impact is ex + 4 sin• ex �m v•. �m(u2 + v") + •ma w - e2 cos• 1 + 3 sin• ex

1

1

2

2

_

The loss of energy is accordingly

ex

oc

•

1

�os• ex, !mV• (11 -+ e2)3 sm• ex

which proves the result stated.

Two uniform, circular discs with rough edges lie fiat on a smooth, horizontal plane; one of them, of radius a and mass m, is in motion along the line of centres with velocity V and angular velocity n so as to impinge on the other, whose mass is m'. Show that if the discs are inelastic and sufficiently rough to prevent slipping the line of motion of the centre of the moving disc is deflected through an angle whose tangent is m'a0f3m V. (M.T.)

Example 7.

Let (I, ]) be the components of the impulsive reaction along and per­ pendicular to the line of centres respectively. After the impact, suppose the centre of the disc of mass has velocity components in th�

m

(u, v)

9]

I M P U L S I VE M O T I O N O F A R I G I D B O D Y

235

directions shown in diagram and let be its angular velocity in the sense indicated. Let denote the corresponding quantities for the other disc.

u', v', w'

(b)

w

b

a

The discs being inelastic, the relative velocity of the two points of impact after collision in the direction of the common normal is zero. Thus (i) Since there is no slipping between the discs, the components of the velocities of the points of impact in the direction of the common tangent after collision are equal. Thus (ii) where is the radius of the disc of mass The equations of linear and angular momentum for the disc of mass are (iii) (iv) (v) The corresponding equations for the other disc are (vi) (vii) (viii) From equations (i}, (iii) and (vi), we deduce that

u = u'.

a'

v - aw = v' + a'w', m'. I = m(V - u}, ] = mv, . ]a = !ma2 (0 - w}.

m

1 = m'u', -] = m'v',2 -]a' = !m'a' w'. u = mmV + m''

and, from the remaining equations, that

v = 3(mm'aO + m') "

Thus, if
m

tan "'

= uv = m'an 3m v·

EXERCISE 9

1. A uniform rod AB of mass 2m is freely jointed at B to a second rod

BC of mass m. The rods lie on a smooth, horizontal plane at right angles to one another and an impulse I is applied to AB at A in a direction parallel to BC. Find the initial velocity of BC and prove that the KE generated is 5I2/6m. (L.U.)

236

A C O U R S E IN A P P L I E D M A T H EMAT I C S

[cH.

2. Three equal, uniform rods A B, BC, CD, each of length 2a and mass m, are smoothly hinged at B and C and laid out on a smooth, horizontal plane so as to form three sides of a square. An impulse I is applied to A in the direction AB. If the initial angular velocity of BC is w and the initial velocity of its mid-point is u, prove that I 7aw u=9 = ' 3m Determine the impulsive actions at B and C. (L.U.) 3. Two uniform rods AB, BC are smoothly jointed at B and lie in a straight line on a smooth, horizontal table. AB is of length 2a and mass M, BC is of length 4a and mass 2M. A horizontal blow is applied normally to the rod AB at a distance b from A . Find the ratio of the initial velocities of A and B, and show that these points begin to move in the same direction if 5a > 3b > 2a. (L.U.) 4. A uniform rod AB of length 2a is attached at one end B to the end of a light, inextensible string of length whose other end is attached to a fixed point 0. The rod and string are rotating freely in a horizontal plane about 0, the end A being at a distance 2a + l from 0. Find at what point an impulse should be applied to bring the rod to rest. (B.U.)

l,

5. A uniform, circular disc of radius a is rolling without slipping along a smooth, horizontal plane with velocity V when the highest point becomes suddenly fixed. Prove that the disc will make a complete revolution round the point if V2 > 24ag. (M.T.) 6. A uniform rod of length 2a and mass m starts moving so that the ends have parallel velocities u and v normal to the rod and no velocity along the rod. Prove that the KE is im(u2 v2 + uv). Find also by what impulses, acting at the ends, the rod can be so started. (M.T.)

+

7. A uniform ball of weight W and of radius r is at rest on a rough hori­ zontal table. The ball is struck a horizontal blow P 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. (M.T.)

8. The end B of a uniform rod AB of mass 2M and length 2a is smoothly

hinged to a point on the rim of a uniform, circular disc of mass M and radius r. The rod and disc are laid on a smooth, horizontal table so that the direction of A B passes through the centre of the disc. An impulse P is applied at A at right angles to AB. Prove that the KE produced is 9P2/10M and that the impulsive reaction at the hinge is (M.T.) P/5.

9. A uniform, rectangular plate with sides 2a, 2b is falling freely without rotation and with its plane vertical, with velocity when a string

V,

9]

IMPULSIVE MOTION O F A RIGID BODY

237

attached to the top corner becomes vertical and taut. Show that the velocity of the CM immediately afterwards is 3 Vp2 3p 2 + a2 + b2' where p is the length of the perpendicular from the centre of mass to the string, and find the loss of KE. (M.T.)

1 0. A uniform rod of length 2a is in motion in a plane with speed u at right angles to its length when it collides with a stationary, small, uniform, elastic sphere of equal mass whose centre lies in the plane. If the sphere is free to move, prove that the angular velocity acquired by the rod cannot exceed ( 1 + e)uy'6 ' 4a where e is the coefficient of restitution. (M.T.) 1 1 . Two uniform, thin rods AB, BC, each of mass m and length 2a, lie freely jointed together at B in a straight line of length 4a on a smooth, horizontal table. The rod AB is pivoted freely at its mid-point to a point in the table. A blow of impulse ] is applied horizontally and perpendicular to the length of the rods at C. Determine the initial angular velocities of the rods and the initial velocity of the mid-point of BC. Show that the energy generated by the impact is 12j2;7m. (B.U.) 12. A uniform, circular disc is lying on a smooth, horizontal table and rotating about a point A on its circumference. The point A is re­ leased and at the same instant another point B, also on the circum­ ference, is fixed ; the angular distance of B from A is oc. Find the ratio of the angular velocities of the disc immediately before and after the change in the motion occurs, and deduce that the angular velocities are in the same sense or in opposite senses according as oc is less than or greater than 120°. (B.U.) 13. A uniform, square lamina of side 2a is moving without rotation in its plane with uniform speed V in a direction parallel to a diagonal when a corner, not in this diagonal, is suddenly fixed. Find the angular velocity with which the lamina begins to rotate. (B.U.) 14. A uniform, square lamina ABCD lies on a smooth, horizontal table. The lamina is struck by a horizontal blow at the vertex A and the initial direction of motion of A then makes an angle oc with the direction of the blow. Prove that oc can never be greater than tan-1 f. (L.U.) 15. A lamina of mass m rotates on a smooth, horizontal table about a point A at distance h from its centre of mass G, its angular velocity being w. If 0 is any point in the plane of the lamina and ON is the perpendicular from 0 on A G, show that the moment of momentum of the lamina about 0 is mw(k 2 hx), where k is its radius of gyration about A and x is the distance A N. If the lamina is released at A and -

A C O U R S E I N A P PL I E D M A T H E MA T I CS

238

[cH.

at the same instant smoothly pinned to the table at 0, show that its subsequent angular velocity is tw when 0 lies on a circle of radius and centre A . (L.U.) Three equal, uniform, rigid rods AB, BC, CD, each of mass and length 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 KE generated is jmu2• Show also that a uniform tension is set up in the rod BC which is initially equal (L.U.) to A uniform rod OA , of mass and length is free to rotate about the end 0 which is fixed. From 0 a particle P, of mass is suspended at the end of a light, inelastic string of length If the rod is released from rest when OA is horizontal, show that the angular velocity of the rod immediately before it strikes P is and that there is no impulsive reaction at 0 as a result of the impact. Show, further, that if the impact is perfectly elastic and if the rod and the particle remain in the same vertical plane the particle will come instantaneously to rest when it has risen through a vertical distance (L.U.) A uniform rod AB, of length and mass m, rests on a smooth, hori­ zontal table, and the end B is constrained to move along a small, smooth, straight groove on the table. Initially the rod is at rest parallel to the groove. A horizontal impulse J is now applied to the end A of the rod in a direction at right angles to its length. Show that the angular velocity of the rod immediately after the impulse is Show further that when AB is at right angles to the direction of the groove the angular velocity of the rod is and find the reaction between the rod and the groove at this instant. (L.U.) Two uniform rods AB and A C, each of mass and length are freely hinged at A and held in a vertical plane with A above BC and each rod inclined at an angle o: with the downward vertical. The system is released from rest and, after falling through a height B and C strike a smooth, inelastic, horizontal table. Show from energy considerations that neither rod rotates before the impact. Show also that immediately after the impact each rod has an angular velocity sin2 and find the impulsive reaction at the j oint A. (L.U.) A uniform, circular disc of mass touches internally a uniform, cir­ cular ring of mass and they lie at rest on a smooth, horizontal table. An impulse is applied to the ring, directed towards its centre, at a point whose angular distance from the point of contact of the two bodies is o:(< ire) . Show that, if the bodies are rough, the disc will at first roll or slide according as the coefficient of friction is greater or less than + tan o: (M.T.)

k

16.

17.

18.

m

2a,

9mu2f16a.

m

48l/49.

2l,

m, 4lf3. (3gf2l) l,

2l

3]f2lm.

19.

3]flm,

m

2a,

h,

3(2gh

20.

o:) tf4a

M,

m

(M m) ' 3M + 2m

9) 21.

I M P U L S I VE M O T I O N O F A R I G I D B O D Y

239

A uniform rod of mass lies at rest on a smooth, horizontal plane. A uniform disc of mass and radius spinning with angular velocity slides on the plane so that its centre moves with velocity along the perpendicular to the rod through one of its ends. The disc collides with the rod, the sense of being such that it tends to cause the disc to roll along the rod. During impact, the disc slips on the rod. If the disc emerges from the collision without spin, show that the coeffi­ cient of friction at the point of impact is

m M

co,

a,

v

co

aco(4M + m) ' 2mv{l + e)

where is the coefficient of restitution. Show also that if the line of motion of the disc's centre is deflected through an angle if> and if IX is the angle made by the initial velocity of the point of impact on the rod with the line of the disc's motion before the collision, then

e

4M - em tan if>. IX = 4m(l + e) 22. A uniform sphere of radius a is given a spin co about a horizontal axis and released with its centre at a height h above a rough, horizontal plane. If no slipping occurs when the sphere impinges on the plane, show that its spin is multiplied by a factor 2/7 and that the distance between its first and second point of impact is 2(h a) 4' aeco � ; ' where e is the coefficient of restitution. tan

CHAPTER 10

THREE-DIMENSIONAL MOTION. LAGRANGE'S EQUATIONS 10.1.

Rotating Frames of Reference

Let 5 be any frame of reference determined by rectangular axes Oxyz (Fig. 10.1). Let 5' be another frame Ox'y'z' having the same z

z'

p

:x:. ' :X:

FIG. 10.1.-Frames of Reference

origin as the first frame, but rotating with angular velocity Q relative to it. At the instant t under consideration, suppose that the two frames are momentarily coincident. Let A be any vector quantity defined in the space in which the frames are set. A may be regarded as the position vector of a point P in this space. Then dAjdt will represent the velocity of P. But the velocity of this point will depend upon the frame from which observation is being kept. Thus (dAjdt)8 will denote the velocity of P relative to 5, and (dAjdt)s' will denote the velocity of P relative to 5'. If P were stationary in the frame 5', it would possess a velocity Q X A relative to 5. But P has a velocity (dAjdt)8' relative to 5', and hence its velocity relative to 5 is given by

(10.1) This last equation shows how we may convert the rate of change of a vector relative to one reference frame into a rate of change relative to 240

[CH. 10)

T H R E E - D I M E N S I O NA L M O T I O N

241

another frame momentarily coincident with the first, but rotating relative to it. 10.2. AM of

a Rigid Body Rotating about a Fixed Pivot

Suppose that a rigid body is pivoted at a point 0 of itself and possesses an angular velocity w about this point at some instant. If

m is the mass of a typical particle at a point P, where-+ OP = r, the velocity of this particle is w X r, and hence its momentum is mw X r. The AM of the rigid body about 0 can now be written down as h !r X (mw X r) !m(r2w - r · wr). (10.2) Let Oxyz be any rectangular frame of reference and let P be the point (x, y, z). Then, if i, j, k are unit vectors along the axes and ( wx, wy, wz) are the components of w along the axes of the frame r2 x2 + y2 + z2 , r xi + yj + zk, Wxi + Wyj + Wzk, r XWx + JWy + ZWz. =

=

=

=

•

W =

W =

Hence

h = ![m(x2 + y2 + z2)(wxi +

wyj + Wzk) - (xwx + ywy + zw.)(xi + yj + zk) ] = (Awx - Hwy - Gwz)i + ( - Hwx + Bwy - Fwz)j + (-Gwx - Fwy + Cw.)k, (10.3) where A , B , C, F, G, H are the moments and products of inertia de­ fined by equations (7.47) with respect to the frame Oxyz. It follows that the components of the AM in the directions of the axes are

A wx - Hwy - Gwz, -Hwx + Bwy - Fw,, - Gwx - Fwy + Cwz.

(10.4)

In the particular case when the axes of the frame of reference are the principal axes of inertia of the body through the pivot 0, the com­ ponents of AM are given by

h=

(Awx, Bwy, Cw,).

(10.5)

The AM about a point 0 of a rigid body executing any motion whatsoever can now be computed by use of the equation (8.26), for ha is the AM about the mass centre of the relative rotational motion about this point as pivot. 10.3.

Euler's Equations

Consider a rigid body rotating about a fixed pivot 0 under the action of certain applied forces and the reaction of the pivot. Let Ox'y 'z' be a rectangular frame of reference 5' fixed in the body with its axes in the directions of the principal axes of inertia at 0. Let Oxyz be a fixed (i.e., inertial) frame of reference S momentarily coincident

242

[cH.

A COURSE I N APPLI ED MATHEMATICS

with the moving frame. Then, if L is the sum of the moments of the applied forces about 0, by equation (8.20) (10.6) where h is the AM about 0. Employing equation (10.1), this equation of AM can be shown to be equivalent to

(��) 8,

+

w

X

h

=

L,

(10.7)

where w is the angular velocity of the body and therefore also of 5'. If A , B, C are the principal moments of inertia of the body and ( w.,, wy, wz) are the components of w in the directions of the axes of 5 and 5', we have already proved that h = (A w.,, Bwy, Cw,) . As the body moves, this formula continues to be valid within the frame 5', but not in the frame 5, for the axes of this frame soon cease to be principal axes. Hence (10.8) where the derivatives w.,, etc., are calculated relative to 5'. Also, the components of w X h in the directions of the axes of 5 or 5' are ( wyhz - w2hy, Wzhx - w.,h,, w.,hy - wyhx) = [ (C - B) wyw2, (A - C) wzw.,, (B - A) w.,wy] .

IJ

(10.9)

Hence, resolving equation (10. 7) in the direction of these axes, we obtain Euler's equations, viz.,

Aw., + (C - B) wywz = L.,, Bwy + (A - C) wzwx = Ly,

Cwz + (B - A ) w.,wy = Lz.

(10. 10)

In the case of a rigid body in general three-dimensional motion, equation (10.6) is valid if h is the AM about the mass centre G of the relative motion and L is the sum of the moments of the external forces about G (see equation (8.31)). But the motion relative to G can only be a rotation of the body about G as the pivot. Euler's equations are therefore derivable as before. Three other equations of motion are also available. These are obtained by resolving the equation of motion of the CM (8.7) in three directions. Since a rigid body, free to move in any manner, has six degrees of freedom, the six equations of motion we have obtained permit a complete solution of the problem of the motion of the body under the action of given forces. For further details and for the solutions to particular problems of this type, we refer the reader to more advanced textbooks of the subject.*

* E.g., Classical Mechanics, by D. E. Rutherford (Oliver and Boyd).

1 0]

T H REE-DIMENSIONAL MOTION

243

In the case of a body, such as the Earth, which has two of its principal moments of inertia about the pivot equal and which is moving under the reaction of the pivot alone, Euler's equations take an easily soluble form (for the Earth, we consider the motion relative to its mass centre, which assumes the role of the smooth pivot). Thus, if A = B, Lx = Ly = Lz = 0, they become A �x + (C - A ) wywz A wy - (C - A ) wzWx

cw.

0, 0,

=

=

}

o.

=

(10 . 1 1)

Hence Wz = n, a constant. For the Earth, Oz is the polar axis, and hence, in the absence of external forces, its angular velocity about this axis will be constant in magnitude. Putting n(C - A ) /A

=

v,

(10.12)

the first pair of equations (10. 11) become Eliminating wy, we deduce that wx + v2wx and thus

=

I

(10.13)

0. Hence

wx = a cos(vt + ex) , wy = a sin(vt + ex) J

(10.14)

These two components ( wx, wy) combine to give a vector a in the xy-plane of constant magnitude a and making an angle (vt + ex) with z

N FIG. 10.2.-Resultant Velocity of Rotation of the Earth

Hence, this vector rotates about 0 with constant angular velocity The angular velocity vector w is obtained by the vector addition of the third component Wz = n to this vector in the xy-plane. The result is shown in Fig. 10.2. It is clear that the magnitude of w takes Ox. v.

244

A COURSE I N APPLIED MATHEMATICS

[CH.

the constant value (a2 + n2)! and that the vector makes a constant angle e: = tan-1 (ajn) with Oz. However, since a is rotating, w must describe a cone about Oz as axis with angular velocity v. The AM vector h is given by equation (10.5) to be

)

Aa sin ( t + ) Cn]. (10.15) It is therefore the sum of a vector along ON (Fig. 10.2) of magnitude Aa and another parallel to NJ of magnitude Cn. It follows that h h=

[Aa cos(vt +

oc ,

v

oc ,

lies in the plane ON] and rotates about Oz, making a constant angle 3 with this axis, where tan 3 = AaJCn. (10.16)

In the case of the Earth, A < C, and hence 3 < e:. But, in equation (10.6), L = 0 and hence h = constant. Thus h is a constant vector relative to a non-rotating frame and, in particular, its direction is constant. We deduce that the axis Oz and the vector w rotate about a fixed axis, making constant angles 3 and ( e: - 3) respectively with it. In the case of the Earth, therefore, its axis of symmetry should generate a cone in space with semi-vertical angle 3, the period of rotation being

21t

21tA (10.17) _ A) ' For the Earth, (C - A)/A = 0 ·00327 and n = 27t radians/day. Hence the period should be 306 days. This is the phenomenon of Eulerian Nutation. It is found, however, that the period of rotation is 427 days. The discrepancy is usually explained as being due to the ---; = n(C

non-rigidity of the Earth.

Obtain Euler's dynamical equations, and show that for a body moving under no forces about a fixed point they possess the two integrals A wx2 + Bwv2 + Cw.• = 2T, A 2wz2 + B2wv 2 + C2w.2 H2, where T and H are constants and the other symbols have their usual meanings. If w• = wz• + wy2 + w.•. deduce that dw = ± v{(a: - w2)(f3 - w2)(y - w2)}, dt where BCa: 2T(B + C) - H2, and f3 and y are similarly defined.

Example 1.

=

=

(JJ

(Li.U.) Multiplying Euler's equations (10. 10), with Lz = Lv L. = 0, by wz, wy, w. respectively and adding, we obtain A w,Wz B wuwv + Cw.w. = 0. Integration with respect to now yields A wz2 + Bwy2 + Cw." = 2T. (i) It will appear later (Section 10.4) that this is the energy equation for the motion. Multiplying Euler's equations by A wz, Bw•• Cw. respectively and adding, we obtain =

t

+

1 0]

T H RE E - D I M E N S I O N A L M O T I O N Integrating with respect to

245

t, we now find that

A 2wx2 + B2w." + C2w." = H2• (ii) It is clear from equation (10.5) that H is the magnitude of the angular

momentum vector h. But, since L = 0, as remarked earlier h = constant and, in particular, its magnitude is constant. This is the physical significance of equation (ii). Taking w2 = Wx2 + wy2 + w.", (iii) i.e., w is the magnitude of the angular velocity w, and solving equations (i)-(iii) for Wz2, wy2, w.', we obtain w2 1 1

1I

2T B C H2 B2 C2 1 1

1

A B C A 2 B2 C2

I I

BC(w2 - o:) = A -B A C) ' ( )(

etc. Thus ± v{(o: - w2) (fl - w2) (y - w2)} = (B - C) (C - A ) (A - B) wxwvw,fABC. Multiplying Euler's equations by wxfA , wvfB, w,fC respectively and adding, we deduce that

d (!w2) = WzWx + WyWy + w,w,

dt

Hence

= (B - C) (C - A ) (A - B ) wxwvw,fABC.

d (!w") = w dw = ± v{(o: - w2) (fl - w2) (y - w2)}, dt

di

as stated.

A uniform rod XY, of length l, can turn freely about a horizontal axis at X, which is always perpendicular to the rod. This axis is made to rotate with uniform angular velocity w(w > y (3gfl)) about the vertical through X. The rod is initially rotating in a horizontal plane without vertical motion. Show that in1its subsequent motion the greatest inclination of the rod to the horizontal is sin- (3gfw2l). Also find the couple acting on the rod at X. (D.U.) Xx, Xy, Xz are principal axes for the rod through the pivot X. Xx is along the rod, Xy is horizontal and perpendicular to the rod and Xz lies in the same vertical plane as the rod and is also perpendicular to it. The forces at the pivot are equivalent to a single force R acting at X and a couple G (Section 1 1 .2) . Let G have components Gx. G, in the directions Xx, Xz. The component of G in the direction Xy is zero, since the rod is

Example 2.

free to rotate about this axis. At any instant, the rod possesses an angular velocity w about the vertical axis through and another about Hence, in Euler's equations, we may put Wx = COS 6, Wy = 6 , w, = sin 6. Also = 0, B = C= Accordingly, Euler's equations take the form 0 = Gx, sin 6, - w2 sin 6 cos 6) = iml2w cos 6 = G,.

X

6

w

A

!mt• (e

Xy.

!ml2, 6

w !ml2•

-imgl

A C O U R S E I N A P P L I E D MAT H E M A T I C S

246

[cH.

The second equation o f motion may be written

:a (ta•) = w• sin 6 cos 6 - M sin 6,

which, upon integration, yields

a• = constant

-

w2 cos• 6 )\

Now 6

=

z

0 when 6

=

+ ¥- cos 6.

f,-. Hence the constant is zero and a• = cos 6

( ¥- =

)

w2 cos 6 .

It is now clear that a is again zero when cos 6

3g/w2l

and 8 cannot take a smaller value than that given by this equation, since, for such a value, a• would be negative. The remaining pair of Euler equations show that the couple to be applied at the pivot to maintain the constant angular velocity w must have its axis along and be of magnitude !ml•wa cos 6.

Xz

10.4.

Kinetic Energy of a Rotating Rigid Body

Suppose that a rigid body is rotating about a fixed pivot 0 (Fig. 10.3) and that, at the instant under consideration, its angular velocity is w. Let r be the position vector relative to 0 of a typical particle of the body of mass m. Then the velocity of this particle is w X r and hence its KE is tm(w X r)2. Summing over all the particles of the body, we deduce that the KE of the motion is

T = l�m(w X r)2 = !!m(w X r) = t!mw r X (w X r)

·

(w

X r)

·

= t!mw · (r2w - r wr) = t!m{r2w2 - ;r . w)2}. ·

( 10.18)

10)

THREE-DIMENSI ONAL MOTION

247

Introducing rectangular axes Oxyz, let r have components (x, y, z) and w components (wx, wy, w.) along these axes. Then equation (10.18) may be written

2 T = :2m{(x2 + y2 + z2) ( w2x + wi + cu.2) - (x wx + ywy + zw.) 2} = A wx2 + B wy2 + C wz2 - 2Fwywz - 2GwzWx - 2Hwxwy,

(10.19)

where A, B, C, F, G, H are the moments and products of inertia of the body relative to the Cartesian frame of reference. It should here be z

X.

w

FIG. 10.3.-KE of a Rotating Body

noted that, unless the axes are rotating with the body, these quantities will vary with the time. If the axes are principal axes at the instant under consideration,

(10.20) As an example of the use of this last equation, we will solve the problem of the motion of a top or gyrostat free to rotate about a smooth, fixed pivot 0 under the action of its weight and the reaction of the pivot. Thus, if the top is spinning on a horizontal plane this is sup­ posed to be sufficiently rough to prevent lateral motion at the point of contact. Let OXYZ be fixed rectangular axes through the pivot, of which OZ is vertical (Fig. 10.4). Let Oxyz be a moving frame defined as follows : Oz is along the axis of symmetry of the top ; Ox is per­ pendicular to Oz and lies in the vertical plane containing Oz and OZ ; Oy is perpendicular to the plane Oxz and in such a sense that the system is right-handed. Thus Oy is horizontal. Let LZOz e and L YOy = if>. Then if> is the angle between the vertical plane through OZ containing the axis of the top and the fixed vertical plane XOZ. e and if> completely determine the position of the axis =

248

A COURSE I N APPLIED MATHEMATICS

[cH.

of the top. Let if; be the angle between a plane fixed in the top con­ taining the axis of the top and the plane xOz. Then the angles e, rfo, if; completely determine the position of the top at any instant. These angles are called the Eulerian angles. z

X .X

FIG. 10.4.-Spinning Top

If we take rectangular axes fixed in the top, one of which is along the axis Oz, these axes will be principal axes throughout the motion of the top and, on account of the symmetry, the principal Mls about the axes other than Oz will be equal. Also, the two external forces R (the reaction at the pivot) and Mg (the weight) have no moment about Oz. It accordingly follows from Euler's equations, as explained in the last section, that the component of the angular velocity in the direction Oz is constant. Now the top has the angular velocities 6, � if; simultaneously about the axes Oy, OZ and Oz. Hence, relative ' to the axes Oxyz, the components of its resultant angular velocity are given by = (-� sin e, 6 , � cos e + f) . (10.21) But, as already explained, � cos e + if; = n (a constant) (10.22) and hence (10.23) = (-� sin e, 6, n). Employing equation (10.20), we can now write down an expression for the KE of the top, viz., w

w

(10.24)

THREE-D IMENSIONAL MOTION

10]

249

it being noted that the axes Oxyz, though they do not follow the motion of the top completely, yet remain principal axes thoughout that motion. The system is clearly conservative, with PE Mgl cos 6. Hence there is an energy equation for the motion, viz.,

A(�2 sin2 e + 62) + tCn2 + Mgl cos e = E.

t

(10.25)

There being two unknowns 6 and �' a further equation must be found before the motion is specified. Now the external forces have zero moment about the fixed axis OZ. Hence, by equation (10.6), the component of angular momentum h in the direction OZ is constant. But, by equation (10.5), the components of h in the directions of the axes Oxyz are (-A� sin e, Ae, Cn) . (10.26) Resolving again, we find that the component along OZ is A� sin2 e + Cn cos e = H. (10.27) Equations (10.22), (10.25) and (10.27) completely determine the top ' s motion. Eliminating � between equations (10.25) and (10.27), we obtain a first order differential equation for e, viz., A62 = 2E - Cn2 - 2Mgl cos e - (H -A Cs� 2cos6 e)2• (10.28) Putting A = cos 6, this equation may be written m

i2

= (a -

- (c - dA) 2 - j(A) , (10.29) b = 2MgljA, c = HJA , d = CnfA and all

bA) (1 - A2)

where a = (2E - Cn2)jA , these quantities are positive. j(A) is a cubic in A. It will be shown later that a motion is possible with 6, and therefore A, a constant. However, for any other motion, i will be non-zero for some values of 6 and hence for some values of A in the interval ( -1, 1). Let A = A0 be such a value. Then j(A) is positive for this value of A, for otherwise � would be imaginary. Hence, it is easily verified that

0, J(+ oo) > 0. Thus, j(A) has a zero in each of the intervals ( - 1, A0) , (A0, 1), (1, ) and its graph is of the form indicated in Fig. 10.5. If Al> A2 , As are f(-1) < 0,

j(A0) >

0,

j(1 ) <

oo

the zeros, where A1 < A2 < As, then, since A cannot be numerically greater than unity and j(A) cannot be negative during the motion, A is confined to lie in the interval (Al> A2) . Also, by arguing as in Section 4.3 (p. 96}, it may be shown that A oscillates between the values

250

A C O U R S E I N A P P L I E D MATH EMAT I C S

[cH.

At and A2• It follows that the axis of the top moves between the cones 6 = ex1 , 6 = ex2, where cos ex; = A;. The exact mode of this motion can be determined only by solving the equation (10.29), a process which involves elliptic integrals. It may happen that At = A2, i.e., /(A) = 0 has a double root. Then A cannot change in value and the axis of the top remains at a constant angle to the vertical. In these circumstances equation (10.27) shows that c{> is constant, i.e., the axis of the top rotates about OZ with constant angular velocity. The motion is said to be one of

steady precession.

J (>.)

FIG. 10.5.-Graph of f(>.)

Suppose, then, that /(A) = 0 has a double root (J.. Then j'(A) = 0 also possesses a root (J.. Hence, for steady precession at an angle ex, where cos ex = (J., we have the equations

(10.30) ex

Suppose that we regard the angle and the spin n as specified. Then the quantities (J., b, d will be known and equations {10.30) will determine a and c. Hence E and H can be found and the initial conditions of projection for steady precession calculated. In fact, only the value of � remains to be determined and, from equation (10.27), we obtain Eliminating (a - b(.L) and (c - d(.L) between equations (10.31), we find that

(10.31) (10.30) and (10.32)

10]

THREE-DIMENSI ONAL MOTION

. d ± vd2 - 2bfl. 2fl. 2- '4A--M'-.---. g'l'c_o_s-oc . _ Cn ± yrrC"'2n'"

251

This equation yields

rp = -

(10.33)

2A cos oc 2 2 Unless C n � 4AMgl cos oc, (10.34) the equation has no real roots, and steady precession is impossible. Assuming this inequality to be satisfied, it will be noted that initially � may be given either of two values to cause steady precession. Writing ± cos oc ! , � = Cn 1 _ 4AMgl C 2n2 2A cos oc 1 and, assuming that C 2n2 is large by comparison with 4AMgl cos oc, as it generally will be in practice, we may employ the binomial ex­ pansion to give Cn Mgt (10.35) rp = Acos oc or Cn '

[ (

)]

·

where we have neglected all but the leading terms in the two expansions. It will be observed that one angular velocity of precession is large whilst the other is small. It is the small value which is generally observed. w

A flywheel rotates with constant angular speed about a horizontal axle, about which the moment of inertia is C. This axle is to be rotated about the vertical with constant angular velocity the centre of theflywheel remaining fixed. Find the magnitude of the couple which must be applied, the plane in which it must act and the sense in this plane. Find the magnitude of the couple in ft lb wt if C = 100 lb ft2, = 20 revfsec, = 0· 1 revfsec.

Example 3.

' w ,

w

w

'

(Li.U.) Let 0 be the fixed centre of the flywheel. The external forces which must be applied ( to cause the motion desired w may be taken to be equivalent to a force R at 0 together with a couple G. Since the CM of the flywheel is fixed at 0 and therefore has no ac­ celeration, R = 0, i.e., the weight force will be balanced by the reactions at the flywheel bearings. Taking rectangular axes being along the axis of the flywheel and along ::c the axis about which it is required that the flywheel shall precess with angular velocity w ', let (Gz, Gv, G,) be the components of G in the directions of these axes. The angular

Oxyz, Oy

Oz

252

A C O U R S E I N A P P L I E D MATH EMAT I C S

[CH.

velocity of the wheel has components w, w') . Hence, the axes being principal axes, the components of AM are given by Cw, A w') , h = where A is the MI of the wheel about Oz. The angular velocity of rotation of the axes is given by w'). n= the rate of change of AM relative to fixed axes instan­ By equation taneously coincident with Oxyz, is

(0,

(0,

(0, 0,

(10.1),

dt (�)

moving

a.xes

+n

x h.

But h is a constant vector relative to the moving axes, and hence the first By equation term of this expression is zero. Also, Sl X h = -Cww', therefore, = (Gx, Gu, G.) . - Cww', Thus the applied couple is in the plane Oyz, acts in the sense from Oz to Oy, and has magnitude Cww'. radsfsec, w' = 1r/5 radsfsec, this couple has If C = lb ft2, w = ft lb wt approximately. magnitude ft lbals = 57T2 ft lb wt =

(8.20),

0, 0).

0, 0)

(

100 8001r2

(

407T 2

250

Example 4. A uniform, solid sphere of radius a can turn freely about a fixed point 0 on its surface. The sphere is held with the diameter OA through 0 horizontal,

and given a spin n about OA . It is released whilst OA is stationary, and allowed to turn freely about 0 under gravity. Find the greatest inclination of OA to the horizontal in the subsequent motion, and show that the angular velocity of precession then has its greatest value 5gfna. (D.U.) This is the problem of the spinning top in the particular case when l = a, A = 7Ma2f5, C = 2Ma2f5. Also, initially () = !1r, G = if, = Hence, by = Ma2n 2f5 and = equations and In equation

(10.25)

(10.29) therefore

and consequently

H 0. (10.27), E a = 0, b = l0gf7a, c = 0, d = 2nf 7, 4n2A 35g f(A) = _ 49a {2an2 (1 A•) + A} .

0.

(i)

_

The zeros Av A 2 , A3 are now found to be Ar =

an• 35g

I a2n4 \/ 1225g2

+ I,

an 2 A 2 = O, A a = 35g

2n4 + l. + \1I"Ia225g2 --

It follows from the general theory that the axis OA will oscillate between a horizontal position and a position in which it is inclined at an acute angle IX to the downward vertical, where COS IX

Equation

=

a•n• \/I 1 22 +1 5g2

-

an• ' 35g

(10.27) takes the form �= (l - A2) + A = 0.

Since - A2) is increasing steadily for - I < A < I, if, takes its maximum numerical value when A assumes its largest negative value, viz., A1• Then if, is given by

A/(1

7¢> (I - A12) + A1 = 0.

2n

(ii)

T H RE E - D I ME N S I O N A L M O T I O N

10]

253

But ,\ = ,\ 1 makes the expression (i) zero. Hence 35g 2an.

(1 - ,\12) + ,\1 = 0.

From equations (ii) and (iii), we now deduce that �

10.5.

(iii) =

5gfna.

Lagrange's Equations

Generalized coordinates of a mechanical system have been defined in Section 6.5. Suppose that a system has n independent generalized coordinates qv q2, qn, so that to each set of values of these quantities there corresponds a unique configuration of the system. However, if the constraints (Section 6.3) of the system alter in some predetermined way with the passage of time, it may be necessary to know also the instant t at which the coordinates q; take their given values before the configuration is completely specified. For example, if the system comprises two bars OA, AB hinged at A , supported by a pivot at 0 and free to swing in a vertical plane, (6, if>), the angles made by the bars with the vertical are suitable generalized coordinates. If, however, the pivot 0 is made to oscillate horizontally with simple harmonic motion of given amplitude and frequency, values of e and if> will not alone determine the configuration. The instant t must also be stated in order that the position of 0 can be found. In general, therefore, if r is the position vector relative to some fixed reference frame of a typical particle of the system, then (10.36) Let F be the resultant force acting upon the particle. Then, if m is the particle mass, F = mr. (10.37) Consider now the configuration of the system at time t. Let us imagine that the constraints (if they are movable) are fixed in the positions they occupy at this instant and that the system is given a small displacement which is consistent with these constraints. In general, this displacement will not be the actual displacement under­ gone by the system during the interval (t, t + lit) and is therefore of an entirely imaginary character. It is called a virtual displacement. Let us further suppose that all the forces acting on the particles of the system, both internal and external, remain constant during this dis­ placement at the values they possess at the instant t. Then the total work performed by these forces is !F . ilr, where ar is the displacement of the particle m during the virtual displacement and the summation is carried out over all particles of the system. We have the equation

!F . ar = !mr . ar.

(10.38)

254

A C O U R S E I N A P P L I E D MATH EMAT I C S

[cH.

During the virtual displacement, let the generalized coordinates change their values to (q1 + 'Sq1, q2 + 'Sq2, qn + 'Sqn) · Then, by equation (10.36), retaining only terms of the first order, or 'Sq1 uq 1

+ uorq2 'Sq2 +

or u

(10.39) It should be noted that t is maintained constant in equation (10.36), 'Sr

= <>

qn . + <>'S qn

<:\

since the constraints are supposed to remain immovable during the virtual displacement. Thus, the work done in this displacement is (10.40) + Qn'Sqn, _!F · 'Sr = Ql 'Sql + Q2'Sq2 + where

(10.41) The Q; are termed the generalized components of force acting upon the system. The right-hand member of equation (10.38) may be expressed in the form where X;

or mr. · ­ oq;

� = k

.

Equation (10.38) is now seen to be equivalent to + Qn'Sqn Xl'Sql X2'Sq2 + Q l'Sql + Q2'Sq2 +

=

+

(10.43) + Xn'Sqn•

(10.44)

Since the q; are independent, so, in general, will be the 'Sq; (in certain special cases this is not so and the system is said to be non-holonomic). Hence we can give the 'Sq; any set of values we please. Taking n), we find from equation (10.44) 'Sq1 =t- 0, 'Sq; = 0 (i = 2, 3, that X1 Q1• In general, it may be proved that X; Q; (i 1, 2, n). (10.45) It remains to transform the X; into a form suitable for practical calculation, when the equations (10.45) will be Lagrange's Equations

=

=

=

of Motion.

For the actual motion of the system, the q; will be certain functions of t, and hence r can also be expressed as a function of the time variable alone. Assuming this has been done, r denotes its derivative with respect to t. Then, from equation (10.36), .

r

=

ar

.

q oql l

+ oorq2 q2. +

(10.46)

This equation shows that r can be expressed as a function of the quantities t, qv q2, qn, q1, q2, lJn, being linear with respect

T H R E E - D I M E N S I O NA L M O T I O N

1 0]

255

to the last n variables. In all that follows it will be assumed that partial derivatives such as are to be calculated from such an expression for r. Thus, from equation (10.46),

oi:foq;, oi:foq;, orjot ar ar oq; = aq.,·

Also,

+

8rj8q;

q;

(10.47)

82r 82r q. n + oq;ot" oq;oqn (10.48)

t,

Now is a function of the and and hence is expressible, when the motion of the system has been determined, as a function of alone. Supposing this has been done and differentiating with respect to t, we obtain

t

+ oq8 n( oq;8r ) + N8 ( a""i;8r )

(10.49) aq; , by equation (10.48) . Here, and throughout the argument, the quantity to which an operator djdt is applied, is assumed to be ex­ pressed as a function of the time variable alone. Let T be the KE of the whole system. Then (10.50)

and hence (10.51) Thus, by equation (10.47),

aT = "'mr. at = "'mr. ar . oq; � . oqi � . oq; It follows that d ( aT) "'mr. · ar + "'mr. · d ( ar ) itt oq; = � oq; � dt oq; ' = xi + 1mi: . ::/ (equation (10.49)), (10.52) = X; + 8T oq; . Hence X; � ( :r) - :�· =

Lagrange's equations can now be expressed in the form

� ( :�) - :� = Q;.

(10. 53)

[cH. To calculate the Q; in any particular problem, it is necessary to imagine that the system undergoes a virtual displacement in which the q; increase by 'iiq;, and to calculate the total work done by all the forces of the system, both internal and external. This will be of the form shown in equation (10.40) , and then the Qi can be picked out as the coefficients of the 'iiq;. Provided the relative distances between pairs of particles of the system are unaltered by the virtual displace­ ment, as has been proved in Section 6.3, the work done by the internal forces will be zero. If, however, elastic strings form components of the system the work done in stretching these during the virtual displace­ ment must be included. Since the displacement is consistent with the constraints, if these are smooth, their reactions upon the system will do no work, and these forces will not appear in Lagrange's equations. Since we are generally interested in the motion of the system alone and do not require to calculate such reactions, this is an advantage that Lagrange's equations have over other dynamical equations such as that of angular momentum. As an example, consider again the problem of a uniform disc of radius a , rolling down a line of greatest slope on a rough, inclined plane of inclination with its plane vertical (see Section 8.7). Let e be the angle through which the disc has rotated since the motion commenced. The system has one degree of freedom, and e is an appropriate general­ ized coordinate. If e increases by ae, the disc rolls a distance a'ii 6 down the plane and the weight force performs an amount of work mga sin 36. No other forces of the system, internal or external, do work during such a displacement, and hence the generalized com­ ponent of force corresponding to the coordinate 6 is Q8 = mga sin Also, since the linear velocity of the centre of the disc is a 6 , its KE is given by A C O U R S E I N A P P L I E D M A T H E M AT I C S

256

ex

ex

T

ex.

T = !ma2()2 + tma2()2

=

£ma2()2,

should be expressible as a function of t, 6 and 6. This has been done, though in this case t and 6 do not appear in the expression for T. Thus

aT

86 = 0, and Lagrange's equation for the system is d ma = mga sm dt .2 e .. 2 . ae = 3 g sm or as obtained previously. As another example, consider again the problem of the spinning top, solved in the previous section. Employing the notation intro-

(3 2 ' )

.

ex,

ex,

THREE-DIMENSI ONAL MOTION

10]

257

duced in that section, (6, if>, if;) are the generalized coordinates. Sup­ pose these quantities increase by a6, a¢>, aq; respectively in the virtual displacement. Only the weight force performs work, and this is of magnitude Mgl sin 6 a6. Hence Q0 = Mgl sin 6, Q.p = 0, Q.p = 0. The KE is given by T = tA (82 + �2 sin2 6) + !C(� + � cos 6)2. Thus . . . . aT = A¢>2 sin 6 cos 6 - qq; + if> cos 6)¢> sin 6, a6 aT aT = 0• = aq; a¢> . . . aT aT = Aif> sin2 6 + C(if; + if> cos 6)cos 6, 88 = A e, a� aT . . = C(if; + if> cos 6). a�

Hence Lagrange's equations are : . . . . d dt (A e) - A¢>2 sin 6 cos 6 + C(if; + if> cos 6)¢> sin 6 = Mgl sin 6, d [Aif>. sin2 6 + C(if;. + if>. cos 6) cos 6] = o, dt d . . & C(if; if> cos 6) = 0. +

From the last equation, we obtain � + � cos 6 = n (a constant), in agreement with equation (10.22) . From the second equation, we derive A� sin2 6 + Cn cos 6 = H (a constant) , in agreement with equation (10.27). The first Lagrange equation may now be shown to be equivalent, after integration, to the energy equation (10.25). A fine, uniform tube of mass M is bent to form a hoop of radius a. The tube has a smooth inner surface. A particle of mass m moves inside the tube which rolls, without skidding, in a vertical plane on a rough horizontal table. The radii to the particle and to a marked point of the tube make angles e and ¢ respectively with the upward vertical, both angles being measured in the same sense. The system is gently disturbed from rest when a and ¢ are both zero.

Example 5.

K

258

A C O U R S E I N A P PL I E D M A T H E M A T I C S

[CH,

Form expressions for the potential and kinetic energies of the system at any time, and use the energy equation and one Lagrangian equation to prove that m sin2 e) = 2g(2M + m) (l - cos e) . (M.T.) a6 • (2M

+

e and "' are the generalized coordinates of the system. The KE of the tube is Ma•¢• . The particle P has a velocity a6 perpendicular to OP relative to 0. 0 has a velocity a¢ horizontally. Thus the particle's velocity v relative to a fixed observer is the resultant of two velocity components a6, a(f, directed as shown. Hence v• = a26 2 + a•¢ • + 2a26 ¢ cos e and its KE is I ima2 (6• + :,• + 26 ¢ cos e) . The net KE of the system is accordingly given by (p• + 26� cos e) . T = Ma2f2 + tma2(62 The PE is given by v = mga cos e. If e, "' increase by se, ll,P respectively, only the weight force acting upon the particle performs work. The work done is mgalle sin e. Hence Q.f> = Qo = mga sin e. Lagrange's equations of motion can now be found, viz.,

+

0,

d

dt

• {2Ma2cp + ma2(cp + •

A o

cos e)}

=

0,

4t {ma2(6 + 1> cos en + ma•e;p sin e = mga sin e.

From the first of these equations, we deduce that 2M¢ m((f, + 6 cos e) = constant = the constant of integration being zero since initially = ¢ = The system being conservative, there is an energy equation, viz., T + V = Ma2f2 + ima2(f2 + + 2 6¢ cos e) + mga cos e = constant. Initially e = 6 = ¢ = Hence

+

(J

0,

(i)

0.

62

0,

0.

mg M¢2 + tm (6 2 + ¢• + 26¢ cos e) = a

(1 - cos e) . .

(ii)

Eliminating ;p between equations (i) and (ii), we now obtain the equation of motion given in the question. A truck has four wheels, each of which is a uniform disc of mass m, and apart from the wheels its mass is M. It rolls without slipping down an inclined plane which makes an angle oc with the horizontal, and the floor of the truck remains parallel to the slope. Meanwhile, a heavy solid spherical ball of mass m' rolls on the floor of the truck, also without slipping, parallel to a line of greatest slope.

Example 6.

T HREE-DIMENSI ONAL MOTION

10]

259

If x is the distance moved by the truck, and if y is the distance relative to the truck floor moved by the ball, show that the kinetic energy of the system is !-(M + 6m + m')x2 + m'xy + i'om'y2• Use Lagrange's equations to find the accelerations of truck and ball.

(Li.U.)

x, y are generalized coordinates for the system. x is the speed of the truck and xfr is the angular velocity of its wheels (r being the radius of a wheel) . Hence the KE of the truck and wheels is !Mx• + 3mx• .

The centre of the sphere has velocity (x + y) and its angular velocity is yfa, where a is its radius. Hence its KE is !m'(x + Y) 2 + tm'y2• The total KE of the system is now found to be T = !Mx2 + 3mx2 + !m'(x + Y)2 + tm'y2, = l(M + 6m + m') x2 + m'xy + 170m'y 2 • If x increases by ox and y by oy, the weights perform work (M + 4m)gox sin IX + m'g(ox + oy) sin IX . Hence Qz = (M + 4m + m')g sin IX, Qy = m'g sin IX. Lagrange's equations are accordingly

:e { (M + 6m + m')x + m'y} = (M + 4m + m')g sin IX, i (m'x + tm'y) = m'g sin IX,

(M + 6m + m')x + m'ji = (M + 4m + m')g sin IX, x + i:Y = g sin IX. Solving for x and ji, we obtain . .. 7M + 28m + 2m' . .. 10m sm x= g sm IX, y = 7M + 42m + 2m' 7M + 42m + 2m' g The acceleration of the truck is x and of the ball is 7M + 38m + 2m' . .. .. g sm IX . x +y 7M + 42m + 2m'

IX.

=

A smooth, circular wire of radius a is constrained to rotate with constant angular velocity w in its own plane, which is horizontal, about a point on its circumference. A bead is free to move on the wire. Show that the bead oscillates on the wire isochronously with a simple pendulum of length gf w2•

Example 7.

Let A A ' be the diameter of the wire through the point A about which it rotates. Let OP, the radius through the bead, make an angle 6 with A A '.

260

A C O U R S E I N A P P L I E D M A T H E MA T I CS

[cH.

The bead has a velocity a6 along the tangent to the wire at P and relative to the wire. At this point P the wire has velocity w . A P = 2aw cos !6

along the perpendicular to A P. The bead's velocity v relative to a fixed reference frame is the resultant of these two velocities and is accordingly given by v2 a2()2 + 4a2w2 cos2 !6 + 4a2w6 cos2 !6. =

Hence, the KE of the bead is T = tma2{S2 + 2w(6 + w) (l + cos 6)}.

In this case, our dynamical system comprises a single particle, viz., the bead. 6 is the generalized coordinate. Since the constraint (i.e., the wire) is in motion, the position of the bead is only com­ pletely fixed when t is known, for only then can the angle wt through which the wire has rotated since t = 0 be found. Letting 6 increase by 86 and keeping t fixed (i.e., the wire is not allowed to rotate during the virtual displacement), the work done by the forces acting on the bead (smooth reaction and weight) is zero. Hence Qo 0 . Lagrange's equation is there­ fore

2 a w c.os Yz B

a9

=

� [ma2{6 + w(l + cos em + ma2w(6 + w) sin e a + w2 sin 6 = 0.

=

0,

This is the equation of motion of a simple pendulum of length gfw2•

If the dynamical system for which q1, q , qn are the generalized coordinates is conservative, let V be its2 potential energy. Then V depends upon the configuration of the system alone, and hence, (10.54) During the virtual displacement, the qi increase by the 8qi and remains fixed. Hence V increases by 8V, where

t

(10.55) But, as shown in Section 6.3 (p. 142), the work done by the forces of the system is - 8V. Hence n n av � Q;8q; = -� - 8q;. i 1 i = 1 aq; =

(10.56)

10]

THREE-DIMENSI ONAL MOTION

261

The 'Sqi being independent quantities, it now follows that Q;

= - iW aq; '

(10.57)

and Lagrange's equations (10.53) take the form (10.58) These equations can be put into yet another form by introducing the function L, where L=T-V (10.59) Since v does not depend upon the (10.58) are equivalent to

q;, avIoq

!:_ ( oL ) aL o dt oq; - oq; = L is called the Lagrangian of the system.

i

=

0 and equations (10.60)

·

In the case of a conservative system, therefore, the dynamical behaviour is completely specified by a single function L. Thus, for the spinning top,

L = tA (62 + �2 sin2 e) + tC(� + ¢ cos 6)2 - Mgt cos e and its behaviour in all circumstances can now be ascertained. 10.6.

Small Vibrations. Normal Modes

In Section 6.2 we studied the behaviour of simple vibrating systems with two degrees of freedom and defined the normal modes. A system having n degrees of freedom, which is free to vibrate about a stable equilibrium configuration, possesses n normal modes, and these are most easily obtained from the system's Lagrange equations. We first select generalized coordinates q;, which are all zero when the system lies at rest in the equilibrium configuration. Then, provided the vibrations are small, the quantities q;, q; will all be small and we can write down approximate expressions for and V correct to the second order in these small quantities. Setting up the Lagrange equations, we seek a solution of the form

T

q; = A; sin (wt +

)

oc ,

(i = 1 , 2 ,

n),

(10.61)

specifying a normal mode. The procedure is now that already de­ scribed in the earlier section. Two examples will be given to illustrate the method.

A C O U R S E I N A P P L I E D M A T H EMAT I C S

262

[cR.

Two uniform rpds AB, BC have masses 5m and m respectively and equal lengths l. They are smoothly jointed at B and can swing freely in vertical plane about a fixed horizontal hinge at A. Show that for small oscilla­ tions the frequencies of the normal/ modes are 1 'V 3i 1 'V/21g( 27r 7' 27r 23 Also show that in one of the normal modes the rod BC turns through three times the angle turned through by AB, but in the opposite sense. (D.U.) Let e, cf> be the angles made by AB, BC respec­

Example 8.

a

tively with the downward vertical. These are the generalized coordinates of the system. Then the PE of the system is given by cos cos + cos v= = cos + cos ) = + + constant, provided and are small so that terms of order cf>3 can be neglected. The KE of is The rotational com­ ponent of the KE of is The centre of mass of has a velocity perpendicular to 'o l ¢ and relative to Compounding this velocity com- I(; -.-o:::,� 1 ponent with the velocity of viz, perpendicular to we obtain the resultant velocity of the centre of to be where + ()� cos ( + = c Hence the total KE of the system is given by + cos ( + + + = Neglecting terms of order higher than the second in we can put cos = l and then T= + Thus the approximate Lagrange equations are

AB, BC

-tmgl -tmgl(7 !mgl(7e• e cf> AB BC B.

e - mg(l e tl ) e cp2) f1n1282• BC -i4-m12�2• tf� B, !18

e•,

BC

v, - e)} . v2 12{82 t�· T %ml•e• -l-4-ml•�· tm12{82 !�· e� ( - e) !ml•(se• + �· 38¢).

d {!m12(16o" + 3¢)}. = -fmglfl, d dt {!ml2(2 t/> + 38) } = -!mglcf>, (16D2 + 21n2)fl + 3D2t/> = 0, 3D2fl + (2D2 + 3n2) = 0, where D = dfdt and n2 = gfl. Let the equations f) = A sin (wt + ) if> = B sin (wt + dt

- e)}. 8, �.

•

ot , ot) define a normal mode. Substituting in Lagrange's equations, we obtain = 0, (i) + = 0. These equations have non-zero solutions for and provided = 0,

i.e., if

(21n2 - 16w2)A - 3w2B } 3w2A (3n2 - 2w2)B A B, 1 21n2 - l6w2 - 3w2 -3w2 3n2 - 2w2 (w2 - 3n2)(23w2 - 21n2) I = 0.

THREE-DIMENSIONAL MOTION

10]

263

w = y3n or vHn Thus determining the frequencies of the normal modes. In the case w2 = 3n2, it will be found from either of the equations (i) that A : B = 1 : -3, i.e., the angle made by BC with the vertical is always three times that made by AB and is in the opposite sense. If w2 = 21n2/23, then A :B = 3:7 and the rods swing together on the same side of the vertical. A smooth, circular wire of radius a and mass Sm, swinging in a vertical plane, is suspended by an inextensible string of length a attached to a point A of it; a particle of mass m slides on the wire. Show that the periods of normal oscillations are (Li.U.) 2Tr y(8af3g), 2Tr y(8af9g), 2Tr y (af3g) . Take generalized coordinates a, ¢> .p as shown. Then the velocity of 0,

Example 9.

•

the centre of the wire, relative to the point A is a,P and of A is a8. Hence, provided a , q, are small, to the first order the velocity of 0 relative to a fixed reference frame is a(8 + ¢) . We can now write down the KE of the wire as l . Sma• . ,p• + l . 8ma2 (G + ¢) " = 4ma2 (8• + 284> + 22). The velocity of the particle P rela­ tive to 0 is �- Thus, again to the first order, the velocity of this particle relative to the fixed frame is a(6 + 4> + ,P) and its KE is my !ma2 (8 + 4> + ,P) • . It follows that the total KE of the system is given by T = 4ma•(6• + 28¢ + 2¢ 2) + tma2 (8 + ¢ + rfo)2 • The PE of the system is V, where V = - Smga(cos a + cos ) - mga(cos a + cos ¢> + cos t{l) = !mga(982 + 9¢>2 + t/J2) + constant, to the second order. Lagrange's equations for the system are now found to be 9ii + 9� + f + 9n28 = 0, + 17� + f + 9n2¢> = 0, fi + � f + n•.p = 0, where n2 = gfa. Assuming a normal mode solution in the form a = A sin ( wt + ot) , = B sin (wt + ot) , .p = C sin ( wt + ot) , � find upon substitution in the equations of motion that 9(w2 - n2)A + 9w2B + w2C = 0, 9w2A + (l7w2 - 9n2)B + w2C = 0, w2A + w2B + (w2 - n2)C = 0.

9l:i

+

}

(i)

1

A C O U R S E I N A P PL I E D M A T H E M A T I C S

264

[cH.

These equations have a non-zero solution in A , B , C provided =0 9(w2 - n2) 9w2 w2 i.e., if Hence

I

9w2 w2 17w2 - 9n2 w2 w2 - n2 w2 (8w2 - 9n2)(8w2 - 3n2) ( w2 - 3n2) = 0.

:

w = vtn. vin. y3n, giving the angular frequencies of the normal modes. In the case w2 = 9n2/8, by solving equations (i) for the ratio A B C, we calculate that A : B : C = 0 : 1 : - 9. Thus, in this normal mode the string remains stationary and the particle and wire oscillate in anti-phase. In the case w2 = 3n2f 8, we find that A : B : C = 2 : 3 : 3, i.e., cp = .p and the particle remains stationary on the wire, which oscillates in phase with the string. In the case w 2 = 3n2, A : B : C = -5 : 3 : 3, i.e., the particle again remains stationary on the wire, which oscillates in anti-phase with the string.

EXERCISE 10

I . A uniform, solid cube of mass and edge is spinning with angular velocity about a fixed diagonal of one face. Show that the magni­ tude of the angular momentum about one of the fixed corners is (Li.U.) A body moves about a point 0 under no forces, the principal moments of inertia at 0 being Initially the angular velocity of the body has components = n, about the principal = 0, = axes. Show that at any later time

2.

3.

w Ma2wy'(43/9).

M

A, 2A, 3A. w1 w2

2a

w3 3-tn,

(Li.U.) A rigid body is rotating about an axis, fixed in the body, with constant angular velocity n. If H is the angular momentum vector of the body, show that the couple exerted on the bearings is G = H x n. A uniform, solid, circular cylinder, of mass m, radius and length is rotating with constant angular velocity about a fixed axis through its centre of gravity and through a point of the circumference of one end. Show that the magnitude of the couple exerted on the bearings is

2l,

w

a

mal(3a2 - 4l2)w2 12(a2 + l2) (S.U.) 4. A uniform disc can turn freely in a vertical plane about a horizontal axis at its centre. A mass m equal to the mass of the disc is attached •

10]

THREE-DIMENSI ONAL MOTION

265

to a point on the rim and the plane containing the disc is made to revolve with uniform angular velocity about a vertical line through the centre of the disc. Prove that the inclination of the radius to the particle to the downward vertical is given by sin cos + sin = Find also the horizontal and vertical components of the couple required to maintain the motion. (D.U.) A body moves about a fixed point 0 under no forces, the principal moments of inertia at 0 being and Initially the angular = = n about velocity of the body has components = the corresponding principal axes. Show that, at time = tanh and that the body ultimately rotates about its mean axis. (S.U.) A uniform, thin, spherical shell of radius is free to turn about a fixed point 0 of its surface. It is set spinning about the diameter through 0 as axis, and that axis is set moving in the horizontal plane through 0. Show that the axis cannot rise through unless the angular velocity of the spin is at least If the spin has this value, find the initial angular velocity about the vertical which must be imparted to the axis so that it will rise through precisely in the ensuing motion. (S.U.) 7 . A particle of mass is attached by an inextensible, light string to one end A of a uniform rod AB of mass and length The end A rests on a smooth, horizontal table, and the string passes along the table, over a smooth edge and vertically downwards to the hanging particle. Both parts of the string are in a vertical plane perpendicular to the edge. The system starts from rest with AB vertical, and B begins to fall away from the edge. Show that if the rod has turned through an angle and the particle has descended a distance x after time then = + and + + 3M sin2 = sin + + cos 6) }. (S.U.) 8. A rod of length is suspended at one end from a light ring which slides on a smooth, horizontal rail. If is the distance of the ring from a fixed point on the rail and is the inclination of the rod to the vertical, prove that cos = constant, x + (Li.U.) + cos + sin 6 = 9. A particle of mass rn moves under gravity on the inner surface of a smooth sphere of radius By use of Lagrange's equations, derive the equations :

w

3a6 - 2aw2

6

5.

6 2g

3k, 5k

w2 (3n/y5)

6.

6 6 0.

6k. w1 n, w 2 0, w3 t, (ntfy5), a

y(15gf4a).

30°

30°

rn

M

2a.

6

{M 4rn

t,

(M rn)x - Ma sin 6 !rngt2 6}a62 6g{rn 6 (M rn)(l -

2a

x 6 a6 6 4a6 3.i' e 3g 0. a. h � sin2 6, 62 + h2 cosec2 6 + (2g cos 6)fa, =

E where h and E are constants. =

266

A COURSE I N APPLIED MATHEMATICS

[CH.

If initially e = 60°, e = 0, � = (gfa) l, show that the greatest value of e in the subsequent motion of the particle is about 1 57°. (Li.U.) 10. A top of mass M rotates about a point on its axis distant h from the centre of gravity, the principal moments of inertia at the point of suspension being A , A , C. Initially there is a spin n about the axis, the axis itself being held so that it makes an angle ()(( < !rt) with the upward vertical. If the axis is suddenly released, obtain the equation to determine the subsequent inclination of the axis to the vertical and deduce that the axis will become horizontal if 2A Mgh > C2n2 cos ()(. (S.U.) 1 1 . A smooth, circular wire of centre 0 and radius a is arranged with the radius to its lowest point A inclined at an angle ()( to the vertical. A bead of mass m can slide freely on the wire, which is made to rotate with uniform angular velocity Cil about the vertical through 0. If the radius to the bead makes an angle 6 with OA , prove that the kinetic energy of the particle is !m[a282 + 2a26Cil sin ()( a2Cil2(1 - cos2 e cos2 ()()], and hence use Lagrange's equations to find the differential equation governing e and prove that the torque about the vertical required to keep the wire revolving is ma2[8 sin ()( + 2SCil sin e cos e cos2 ()(] . (D.U.) 12. Two uniform rods AB, BC, of masses 3M and M, and lengths a and 4a respectively, are smoothly hinged at B, and AB is pivoted to a fixed pin at A . If the system is set moving in a vertical plane, under gravity, show that the kinetic energy of the system is given by Ma282 + �Ma2�2 + 2Ma28� cos (6 - cf>), T

+

=

where 6 and c/> are the angles which AB and BC make with the down­ ward vertical at any instant, and find the potential-energy function. Derive Lagrange's equations for the system and find the initial angular accelerations of the two rods if the system is set in motion with the rods in a line inclined at an angle ()( to the downward vertical. (S.U.) 13. A ring of mass m can slide smoothly on a hoop of radius a and mass M. The hoop is free to rotate about a vertical diameter. Obtain the Lagrangian equations of motion. Show that the hoop can rotate with constant angular velocity Cil, with the ring in relative equilibrium at angular distance ()( from the lowest point, provided cos ()( gfaCil2• Prove that, if this state is slightly disturbed, the period of a small oscillation of the ring is 2rtfp, where p {M + 2m(1 + 3 cos2 sin2 ()( · � M 2m sin2 ()( (S.U.) 14. A particle, of mass m, is attached by a string, of length b, to the lowest point of a circular disc of mass m and radius a, which is free to rotate =

_

+

()()}

10]

THREE-DIMENSI ONAL MOTI ON

267

about its centre in a vertical plane. Show that the Lagrange equations, for small oscillations about the position of equilibrium, reduce to Ha6 + = -g6 a6 + = -gcf>, where 6 and cp are the angular displacements of the disc and string respectively. Find also the normal periods of oscillation. (Le.U.) 15. Four equal uniform rods AB, BC, CD, DA are smoothly jointed to­ gether in the form of a rhombus and suspended from a fixed point 0 by means of a light, stretched, elastic string attached at its other end to A . 0, A , C are always in that order in the same vertical straight line and the distance from 0 to the mid-point of A C is x. The rods are inclined to the vertical at an angle 6. Express the kinetic and potential energies in terms of x, 6, i, 6, m, A, a, b, where m is the mass and 2a the length of each rod, A is the modulus of elasticity and the un­ stretched length of the string. Hence form Lagrange's equations of motion. (M.T.) 1 6. A uniform, circular disc of mass M rolls with its plane vertical on a fixed rod inclined at an angle IX to the horizontal. A mass m is sus­ pended by a light string of length l from the centre of the disc and the system is released from rest with the string vertical. If after a time t the disc has rolled down the rod through a distance x and the string has swung forward through an angle 6, show that (�M + m)x + ml[sin (6 + IX} - sin IX] = i(M + m)gt2 sin IX. (M.T.) 17. A rod AB of mass M hangs from two fixed supports A ', B' by light, inextensible strings A A ', BB', each of length a ; the line A 'B' is hori­ zontal, and the lengths AB, A 'B' are equal. A bead of mass m slides smoothly on the rod, and the system moves in the vertical plane through A 'B'. If at time t the strings make an angle 6 with the vertical, and the bead is at a distance ax from a fixed point on the rod, form the kinetic and potential energy functions for the system, using 6 and x as Lagrangian coordinates. Form also the Lagrangian equations of motion, and prove that M 6 + m sin 6 (sin 6 6 + cos 6 02) (M m) (gfa) sin 6 = 0. Find the period of a small oscillation about a position of equilibrium in which the strings are vertical. (M.T.) 18. A uniform, circular disc of mass m is constrained to move in a vertical plane with its centre in a smooth vertical groove. It is suspended by two similar vertical, light, elastic strings of length l and elastic modulus A attached at their lower ends to points at opposite ends of a diameter of the disc, and at their upper ends to points on the same horizontal level. Prove that, if the system is displaced so that there is both vertical motion of the centre and rotation of the disc, the frequencies of small oscillation will be

b�� b

b

+

+

(D.U.)

A C O U R S E I N A P P L I E D M A T H E M A T I C S [cH. 10]

268

19. A uniform rod of length is suspended from a fixed point by a light, inextensible string of length fastened at a distance from the centre of the rod. Show that the periods of two-dimensional oscilla­ tions are where = and = and that, to the first order, in one normal mode the top of the rod is fixed, and in the other the point of the rod distant from the bottom is fixed. (Li.U.) A uniform, circular disc, of centre 0 and radius has a point P on its circumference smoothly hinged to one end of a uniform rod, of equal mass and length which has its centre C fixed. The system makes small oscillations in a vertical plane. Find the periods of the normal modes, and the general expressions for the inclinations of CP, PO to the vertical at time t. Show that, if CP, PO are initially vertical, it is possible for the system to move : (i) so that OPC remains a straight line ; (ii) so that 0 remains in the vertical through C. (S.U.) Two equal, uniform bars AB, BC, each of mass and length are smoothly jointed together at B and suspended from a smooth pivot at 0, the mid-point of AB. Show that the kinetic energy of the rods during their small motions in a vertical plane is approximately

2a

�a P12 gf2a

2rcfp,

a/3

P 22 3gf2a,

2af3

20.

a,

3a,

8,

m

21.

e

22.

24.

tma2(262 + 2¢2 + 36¢),

where and are the small angles which AB and BC make with the vertical. Find the normal periods of oscillation. (S.U.) The kinetic and potential energies of a dynamical system are given respectively by = + = + + b being constants. Show that small vibrations about a position of equilibrium are possible provided > and find the periods of the normal modes. Show that = L sin cos qt is a possible motion for certain values of the constants q. Determine these, and find the corresponding value of (N.U.) A circular hoop of mass and radius is free to swing in its own plane about a point A of its rim. A small bead of mass is free to slide on the hoop without friction. Prove by means of Lagrange's equations that the periods of small oscillations about the position of stable equilibrium are

a,

23.

2a,

2T 62 �2, 2V a82 2b8rf> arf>2, a fbi, e pt p, rf>. M a

m

2rc�( 2; ). 2rc�( (M�am)g) ·

(Le.U.) ABCDE is a string stretched between two fixed points A and E. Particles of equal mass are fixed to the string at the points B, C and D. If AB = BC = CD = DE and the particles are free to vibrate in a plane along lines perpendicular to AE, show that the normal periods of vibration are of the form where takes one of the values

2rcfp, p2 2). � (l ml ( l a�). (2 ± y2) ml a�). A being the modulus of elasticity of the string, l its natural length, m the mass of a particle and a the distance between adjacent particles. _

-

PART I I STATICS

CHAPTER l l

EQUILIBRIUM AND EQUIVALENCE OF FORCE SYSTEMS 11.1.

Equilibrium of a Rigid Body

If it is found that when a rigid body is subjected to a certain system of external forces, it can remain at rest, the body is said to be in equili­ brium under the action of the system. Let F 1, F Fn denote the forces and r11 r , rn the position vectors of their respective points of application relative to the origin 0 of a fixed frame of reference. Then the equations of linear and angular momenta, viz., (8.4) and (8.20), are applicable and, since both the vectors p and h remain zero for all values of t, we obtain the equations 2

•

2,

•

n

L Fi = 0 , i=l n .L ri X Fi = 0,

i= 1

•

(11.1) (11.2)

as necessary conditions for the equilibrium of the body. In words,

if a rigid body is in equilibrium under the action of a system offorces, it is necessary (i) that the vector sum of the forces should be zero, and (ii) that the sum of their vector moments about any point 0 should also be zero. The theory of force systems in equilibrium is referred to as the theory of statics. The majority of statical problems may be readily solved

by application of the principle just enunciated, for it is generally the case in such problems that we are aware at the outset that we are concerned with a force system in equilibrium. For example, our problem may be to calculate the upward thrusts exerted upon a bridge by two piers supporting it at either end. It is clear that these thrusts will automatically adjust themselves so that the bridge rests in equili­ brium. Whether the two conditions already stated are sufficient of themselves to ensure equilibrium is therefore a question of somewhat academic interest, since we rarely, in practice, seek to apply it in this converse form. We shall, however, prove that if (i) the vector sum of

the external forces acting upon a rigid body is zero, (ii) the sum of the moments of these forces about some fixed point 0 is zero and (iii) the body is at rest at some instant, then the body is in equilibrium. To establish this principle, we first note that by conditions (i) and (ii), equations (11.1) and (11.2) are valid, and hence, from equations (8.4) and (8.20), we have (11.3) p = 0, (11 .4) h = O. 271

272

A C O U R S E IN A P PL I E D M A T H E M A T I C S

[CH.

Thus the linear and angular momentum vectors p and h are constant in magnitude and direction. But, at the instant when the body is known to be at rest, p = 0, h = 0. We deduce that p and h are zero at all times. By equation (8.6) therefore, (11.5) where r is the position vector of the body's CM. This equation shows that the CM is stationary. It now follows from equation (8.26) that ha = 0, (11 .6) i.e., the AM about the CM of the body's motion relative to this point vanishes. But the only motion the body can possess relative to its own CM is one of rotation about this point. Let ( wy, wz) be the components of its angular velocity of rotation about the CM, taken in the direction of the principal axes of inertia at this point. Then, by equation (10.5), (11. 7) where A, B, C are the principal moments of inertia of the body at its CM. Comparing equations (11.6) and (1 1.7), we deduce that Wx = Wy = Wz = 0, . (11 .8) i.e., the body does not rotate about its CM and is accordingly stationary. By resolving the vector equations (11.1) and (11.2) along three non­ coplanar directions, six scalar conditions of equilibrium may be derived. It is usually convenient to take these directions to be those of rect­ angular axes Ox, Oy, Oz. Then equation (11.1) indicates that, in equilibrium, the sum of the x-components of the external forces will be zero. Similarly, for the sums of the y- and z-components. Let F be a force whose point of application P has position vector r. Suppose F to have components (Fx, Fy, Fz) and let P have coordinates (x, y, z) relative to the rectangular reference frame Oxyz. Then, if i, j, k are unit vectors in the directions of the axes, the moment of F about 0 is r X F = (xi + yj + zk) X (Fxi + Fyj + Fzk) = (yFz - zFy)i + (zFx - xFz)j + (xFy - yFx)k, (1 1.9) i.e. , this vector moment has components c.>z,

yFz - zFy, zFx - xFz, xFy - yF., in the directions of the axes. It will be observed that each of these components is equal to the scalar moment of F about the corresponding axis as ordinarily defined. For the scalar moment of F about Ox can be computed as the sum of the moments of its three components

E Q U I L I B R I U M OF F O R C E S Y ST E M S

1 1]

273

about this axis, and this sum is equal to yFz - zFy (Fx is parallel to

Ox and therefore has no moment about it) .

By resolving equation (11.2) in three perpendicular directions, we can now deduce that the sum of the scalar moments of the external forces about any axis is zero. A uniform bar of weight W and length 2a is smoothly pivoted at one end and rests with its other end against a smooth, vertical plane. It is main­ tained in this position by the application of a horizontal force F parallel to the smooth plane and at a point on the rod distant x from the pivot. If ex is the angle made by the bar with the horizontal and (3 is the angle made by the vertical plane containing the rod with the smooth plane, prove that

Example 1.

Wa F = x cot ex cos (3, and find the reactions at the smooth plane and the pivot.

Take rectangular axes Oxyz through the pivot 0, Ox being horizontal and parallel to the smooth plane, Oy being perpendicular to this plane and Oz vertical. Let P be the reaction of the smooth plane. Taking moments about Oy, we obtain Fx sin ex = Wa cos ex cos (3. Hence

z

Wa F = x cot ex cos (3.

Taking moments about Ox, we find that P 2a sin cx = Wa cos cx sin (3. P = tW cot ex sin (3. Thus It will be found that the equation x of moments about Oz is now satisfied 0 identically. The reason for this is that none of the forces has any moment about an axis along the bar, i.e., the equation of moments about this axis is an identity. Thus only two out of a possible three equations of moments can be independent, no matter what axes are taken. Let (X, Y, Z) be the components of the reaction at the pivot in the direc­ tions of the axes. Then resolutions parallel to these axes yield the equations X = F, Y = P, Z = W, determining the reaction at the pivot.

When a rigid body is acted upon by forces in such a way that its surface 51 tends to slip upon the surface 52 of another rigid body, frictional forces will be brought into play at the point of contact P (Fig. ll.l) in such a manner as to maintain the equilibrium. If R is the force exerted by the surface 52 upon the surface 51 at P we may resolve this force into two components N and F, N being in the direction of the common normal and F in a direction perpendicular to this. N is called the normal reaction and F the frictional force. If the

274

A C O U R S E IN A P P L I E D MATH EMAT I CS

[CH.

applied forces responsible for the tendency to slip are steadily in­ creased, slipping will eventually take place at P. When the magnitudes of these applied forces take critical values such that any further increase will result in slipping, the frictional force at P is said to be limiting.

FIG. 1 1 . 1.-Forces due to Friction

It is usual to accept the following laws by experiment :

of friction as being justified

1. The direction of F is opposite to that of the motion of 51 relative to 52 at P which would take place if F were removed. 2. The magnitude of F is just sufficient to maintain equilibrium. 3. When F is limiting, FJN = fL , where fL is a constant of the surfaces in contact called the coefficient of friction between them. It must be appreciated, however, that these laws represent rough descriptions only of frictional phenomena which have been observed. For a more accurate account the reader is referred to texts concerned with the properties of matter.* The third law implies that the total reaction R must lie within or on the surface of a right circular cone with its apex at P, its axis along the normal to the surfaces 51 and 52 and its semi-vertical angle equal to A, where tan A = fL . A is called the angle of friction and the cone is termed the cone of friction. If 51 is in contact with 52 at three points or more, the hypothetical removal of the frictional force at one of these points will not necessarily destroy the equilibrium, since the frictional forces at the remaining points may be able to adjust themselves so that the equilibrium is maintained. In these circumstances the precise significance of the first law is not immediately apparent. If, however, we take into account the fact that the body cannot be perfectly rigid, it will appear that the removal of F will cause a slight reduction in the body's elastic deformation, with the result that a small relative motion between the * E.g., The Friction and Lubrication of Solids, by F. P. Bowden and D. Tabor (Oxford University Press) .

EQUILIBRIUM OF FORCE SYSTEMS

1 1]

275

surfaces at P will take place. The first law will then be taken to refer to this motion, and hence can be applied only when the elastic properties of the body are known. In the absence of a knowledge of these properties, the solution to the problem is indeterminate. However, when the equilibrium is limiting, so that the friction at all points of contact with the exception of one at most is limiting, then the body is on the point of rotating about an instantaneous axis of rotation, and it will be found that the position of this axis, and hence the directions of the tendencies to relative motion at all points of contact, are uniquely determined by laws 1 and 3. An example is given below. A uniform sphere of radius a and weight W is supported by three equally rough rods joined at their ends to form a horizontal equilateral triangle of side b. If p. is the coefficient of friction, show that a couple about a vertical axis applied to the sphere must have moment greater than

Example 2.

p.Wab y(l2a2 - b 2) if the equilibrium is to be disturbed.

The points of contact A , B, C of the sphere with the rods lie on the inscribed circle of the triangle formed by the rods. Let Q be the centre of

w

this circle and 0 the centre of the sphere. Let L OA Q = ct. The normal components of the reactions at A , B, C are all equal and will be denoted by R. When friction is limiting, the frictional component p.R of the reaction at A will be directed along the rod through A , opposing the tendency of this point on the sphere to move in the opposite direction under the action of the couple G. Resolving vertically, we obtain the equation 3R sin " = W.

A C O U R S E I N A P P L I E D M A T H EMAT I C S

276

[cH.

Taking moments about a vertical axis through 0, we find also that 3pRa cos "' = G. Hence G = p.Wa cot <X. But cot "' = A QJQO = bfy(12a2 - b2) , proving the result stated. A uniform, straight rod of length l lies on a uniformly rough horizontal table. The table supports the rod uniformly along its length. A slowly increasing horizontal force is applied to the rod at one end in a direction per­ pendicular to its length. Show that when slipping commences, the instan­ taneous centre of rotation is at a point upon the rod distant lf y2 from the point of application of the force, and calculate the magnitude of the force at this instant. Let P be the magnitude of the force when slipping commence� and let I be the instantaneous centre of rotation. It is clear from diagram (a) that if I does not lie on the line of the rod, then the frictional forces F v F2 , etc.,

Example 3.

applied to the elements of the rod will have components in the direction of the rod whose senses are all the same and hence that the sum of these com­ ponents is not zero. This contradicts our assumption that slipping is only just taking place and hence that P is in equilibrium with the limiting frictional forces. p

d

}+- oc I

0

f,

pwdx

(a)

0

(b)

We accordingly take I to lie on the line of the rod distant d from 0 (diagram (b) ) . Then, if p. is the coefficient of friction and w is the weight per unit length of the rod, the frictional force acting upon the element dx is p.wdx. For values of x within the interval (0, d), this force will be in the sense shown. For values of x within the interval (d, l), the sense will be reversed. Taking moments about 0, we obtain the equation

or

{ 0

1

1 p.wx dx - p.wx dx = 0,

d

!d2 - (!l2 - !d2) d = lfy2.

l

11.2.

=

0.

Thus Resolving at right angles to the rod, we obtain l d P + p.w dx = p.w dx, d 0 P = p.w(2d - l) = p.W(y2 - 1 ) .

f

Equivalence of Force Systems

Suppose that a rigid body is in equilibrium under the action of a system of forces S. Let 51 denote a sub-system comprising some of the

1 1]

EQUILIBRIUM OF FORCE SYSTEMS

277

forces belonging to 5. If the forces of 51 are removed and the equili­ brium can be maintained by the substitution for it of another sub­ system 51', then 51 and 51' are said to be statically equivalent systems. Since 5 is in equilibrium, we may write down equations of the types (l l . l) , (1 1.2). If, when 51' replaces 51 these equations remain true, then the equilibrium of 5 will not be disturbed. Thus 51> 51' are equivalent if their contributions to the left-hand members of these equations are the same, i.e., if (a), the vector sum of the forces com­ prising 51 is equal to the vector sum of those comprising 5/ and (b) the sum of the moments of the forces of 51 about 0 is equal to the sum of the moments of the forces of 5/ about the same point.* We will now consider some particular cases of this principle. Sup­ pose, firstly, that 51 comprises a single force F applied at a point P having position vector r. Let 51' be the same force F applied at the point P' having position vector r ' . These two forces are equivalent according to the principles of the previous paragraph if r X F = r' X F, i.e., if (ll.lO) (r - r') x F = 0 . -+

-+

But r - r' = P' P and hence P' P must be parallel to F. This implies that P' P is the line of action of either of the forces F and that 5/ is obtained from 51 by moving the point of application of F from P to another point P' on its line of action. We have therefore proved that in statics a force may be regarded as applied at any point on its line of action, all such points yielding equivalent systems. This is the -+

Principle of Transmissibility of Force. Now suppose that 51 comprises two forces F1, F2 whose lines of action intersect at P. Let 5/ be the single force F1 + F2 applied at P. Then it is easy to verify that the conditions (a) and (b) are satisfied and hence that 51 and 51' are equivalent. This result is equivalent to the Parallelogram of Forces for statics. It can clearly be generalized for the case of any number of forces acting at a point. Let 51 comprise a pair of equal and opposite forces F, -F whose lines of action are parallel but not coincident. Such a force pair is called a couple. Let r, s be the respective position vectors of the points of application P, Q of these forces. The vector sum of the two forces is zero. The sum of the moments of the forces about 0 is (ll.ll) r X F s X F = (r - s) X F = G. -

G is called the

moment of the couple and is independent of the origin

* It follows that two sets of forces which remain statically equivalent through­ out the motion of a rigid body are also dynamically equivalent, for they will yield the same equations of motion (8.4) and (8.20) of the rigid body.

A C O U R S E IN A P PLI E D M A T H E M A T I C S

278

�

[cH.

of position vectors, since r - s = Q P. Also, if P' , Q ' are any other points on the lines of action of F, -F respectively, then �

�

�

�

�

G = QP X F = (QQ' + Q' P ' + P ' P) X F = Q' P ' X F (11. 12) QQ' , P ' P being parallel to F. Thus G depends only upon the forces comprising the couple and their lines of action. Taking Q' P ' to be perpendicular to the forces, it is easy to see that the magnitude of G is Fp, where p is the perpendicular distance between their lines of action, and its direction is normal to the plane containing these lines. If 5/ is any other couple having the same moment G, then application of the conditions (a) and (b) shows that 51 and 51' are equivalent. I.e., couples having equal moments are equivalent. In statics, therefore, it is sufficient to specify a couple by its moment. If 51 comprises two couples of moments G1, G2 and 5/ is a single couple of moment G1 + G2, conditions (a) and (b) are satisfied, and hence these two systems are equivalent. This may be expressed by saying that the resultant of two couples is a couple having moment equal

�

�

�

to the sum of the moments of the two couples.

If 51 comprises a single force F acting at a point P, the addition to the system of a pair of forces F, -F acting at some other point 0 yields an equivalent system 51'. But the force pair, F acting at P and - F acting at 0, constitute a couple of moment r X F, where r = OP. Thus 51' comprises a couple of moment equal to that about 0 of the original force, together with a force F acting at 0. We have therefore shown that if the line of action of a force F acting at P be displaced parallel to itself to act through 0 and a couple of moment equal to that of the force acting at P about 0 be then introduced, the new system is equivalent to the old. Let us now consider the general case when 51 comprises forces . Fn acting at points Pv P2, Pn respectively. Let 0 be F1, F2, an arbitrarily selected origin and suppose Pi has position vector r; with respect to 0. According to the principle just explained, the line of action of F; may be displaced to act through 0, provided a compensating couple r; X F; be introduced. Displacing all the forces of 51 in this manner, we construct a system equivalent to 51 comprising forces Fv F2, Fn acting at 0 and couples r1 X F1, r2 X F2, rn X Fn. The forces acting at 0 are equivalent to a single force R acting at this point, where (11.13) The couples are also equivalent to a single couple G, where G = r1 X F1 + r2 X F2 + + rn X F,. . (11.14) �

•

•

•

1 1]

EQUILIBRIUM OF FORCE SYSTEMS

279

We have accordingly proved that any system of forces is equivalent to a force acting at an arbitrarily assigned point 0 together with a couple. The force is equal to the vector sum of the forces of the original system, and the couple has moment equal to the sum of the moments of the forces of the system about 0. There are three special cases to consider : (i) if R = 0, the system is equivalent to a couple ; (ii) if G = 0, the system is equivalent to a force ; (iii) if R = G = 0, the system is in equilibrium. If neither R nor G is zero it remains to investigate whether, by appropriate choice of the point 0, we can simplify the system still further, e.g., by showing that it is always equivalent to a single force. Let us displace the line of action of R to act through P, where OP = r. A couple -r X R must be introduced to maintain equivalence. Thus the new system comprises a force R acting at P and a couple G' = G - r X R. It is shown below that only in a special case is it now possible to choose r so that G' vanishes and the system has been reduced to a single force. However, we can always find P so that R and G' are parallel. A system consisting of a force and a couple whose axis is parallel to the force is termed a wrench. Thus R and G' constitute a wrench if R X G' = 0, i.e., if R X (G r X R) = R X G - R2r + R · r R = 0. (11.15) For this condition to be satisfied, r must be such that �

-

r

RxG

= -----w-

+ tR,

(11.16)

where t = R . rjR2 and is a scalar. If we substitute in equation (11.15) for r from equation (11.16), it will be found that the former equation is identically satisfied for all values of t. Hence equation (11.16) shows that the point P can lie anywhere upon a line l parallel to R through the point Q where r = R X GjR2 (Fig. 11.2). Clearly all such points P correspond to the same force system, viz., a force R acting along l and a couple G' whose axis is parallel to this line . l is called the central axis of the original force system and (11. 16) is its equation. The couple G' is given by G'

=

G-r X R=G-

=

R · G R.

(R� G + tR) X R,

R2

(11. 17)

If R · G = 0, i.e., R is orthogonal to G, the system reduces to a single force acting along the central axis. This is the case if the original forces lie in a plane for by choosing 0 to lie in we ensure that the 1t,

1t

280

A C O U R S E IN A P P L I E D M A T H E M A T I C S

[cH.

position vectors ri also lie in 1t and hence, by equation (11.14), that G is normal to But R lies in and hence R and G are orthogonal. A coplanar system of forces is therefore always equivalent to a force or a couple or is in equilibrium. If the coplanar system reduces to a rt.

1t

G R

c/

FIG. 1 1 .2.-The Central Axis

force its line of action is given by equation (11.16). Taking the vector product of both sides of this equation with R, we obtain r X R = G, (11.18) since R . G = 0. Taking rectangular axes Oxy in the plane let the coordinates of any point on the line of action of the resultant R be (x, y) and let (X, Y) be the components of R along the axes. Then r = (x, y, 0), R = (X, Y, 0), G = (0, 0, G) . Hence Yx - Xy = G. (11.19) This is the cartesian equation of the line of action of the resultant. It is clear from equation (11.13) that R is independent of the point 0. Thus R2 is an invariant of the original system. Also G' cannot depend upon 0. It follows from equation (11.17) that R . G must also be an invariant of the system. Since G' and R are parallel, equation (11. 17) indicates that R·G p G' (11.20) rt,

R = R2 = ,

where p is an invariant of the system called the pitch of the resultant wrench, i.e., the pitch is the ratio of the magnitudes of the couple and force components of the wrench. The choice of the term " pitch "

1 1]

E Q U IL I B R I UM O F FORCE SYSTEMS

281

is suggested by the parallelism which exists between the above theory of the central axis and the kinematical theory of rigid body motion (see Chapter 7) . Example 4.

=

=

Forces P, Q, R act respectively along the three lines

y c; z = -c, x = a ; x = -a, y = b. -b, z Show that they are equivalent to a wrench of pitch 2 (aQR + bRP + cPQ) pz + Q• + R •

Show also that if the system reduces to a single force, its line of action will lie on the surface ayz + bzx + cxy + abc = 0.

=

=

The three forces have components along the axes as follows :

(B.U.)

(P, 0, 0), F 2 (0, Q, 0), F3 = (0, . 0, R), F1 and may be taken as acting at the points

r1 = (0, -b, c) , r 2 = (a, 0, -c), r3 = ( - a, b, respectively. By equations (1l. l3), (11.14), R

=

0)

(P, Q, R) ,

G = (cQ + bR, aR + cP, bP + aQ).

R2 = P2 + Q 2 + R2, R G = P(cQ + bR) + Q(aR + cP) + R(bP + aQ) = 2(aQR + bRP + cPQ) and equation (11.20) shows the pitch to be as stated. If the system reduces to a single force R G = 0 and ( 1 1 . 18) is the equation of its line of action. Thus, if r = (x, y, z) is any point on the line of action

Thus

•

.

=

then

Ry - Qz cQ + bR, Pz - Rx = aR + cP, Qx - Py = bP + aQ. R G 2(aQR + bRP + cPQ) = 0.

=

Also, The last equation may be written •

a c b p+g+R= O and thus the first two equations are equivalent to a y z = o, P + Q -R X b + Z = o. -P + Q R.

Eliminating the reciprocals of P, Q, R from the last three equations, we obtain

! _; � -: I

=

o.

ayz + bzx + cxy + abc = 0.

I.e., the line of action lies in this ruled surface.

282

A C O U R S E IN A P P L I E D M A T H E MA T I C S

[cH.

Show that if a system of forces is reduced to a force through 0 togethel' with a couple, then the angle between the force and the couple vectol's is tan-1 (dfp), where p is the pitch of the equivalent wrench and d is the distance of 0 from the axis of the wrench. (Le.U.) Since R X G is perpendicular to R, OQ (Fig. 1 1 .2) is perpendicular to the central axis QR. Thus

Example 5.

I G.

R x G = R sm cx, R• where ex is the angle between R and G. R·G p = R2 = R cos ex. Also,

I

d = OQ =

tan ex = dfp.

Thus

11.3.

G

Parallel Forces. Centres of Mass

If the forces comprising the general system considered in the previous section are all parallel, let u be a unit vector in the direction of any one of them. Then we can write F; = F;u,

(i = 1, 2,

n) ,

(11.21)

where F; is the magnitude of F; prefixed by the appropriate sign. Reducing the system to a force R applied at an origin 0 and a couple G, we have from equations (11.13), (11. 14) (11.22) n

1

G = ! F;r; X u, i

=

(11.23)

respectively. Thus R is parallel to u, whereas G is perpendicular to u. If !F; = 0, the system is equivalent to a couple. Otherwise R · G = 0 and the system is equivalent to a single force R. The magnitude R of this force is given by the equation (11.24)

i.e., is the algebraic sum of the magnitudes of the forces of the system. The equation of the line of action of the resultant follows from equation (ll. l6), viz.,

r= =

I R

[u x (!F;r; x u)] + tRu

� !F;r; + t'u,

(11.25)

where t' = tR - u . !F;r;/R and is arbitrary. This equation re­ presents a straight line parallel to u through a point C having position vector (11.26)

EQUILIBRIUM OF FORCE SYSTEMS

1 1]

283

The_position of C is observed to depend upon the magnitudes of the forces and their points of application, but not upon the direction of the forces as governed by u. It follows that if the forces be rotated so as to remain parallel and their magnitudes and points of applica­ tion are not altered, the resultant will also rotate but will always act through C. C is called the centre of the force system. A particular case is that of a system of particles of weights Wv w2, Wn, rigidly connected together. Provided the dimensions of the system are small by comparison with those of the Earth, the weight forces will be very nearly parallel and their resultant will act through a point C, which is fixed in position relative to the particles no matter what their orientation with respect to the Earth. The point C is called the centre f gravity (CG) of the particles, and its position is given by the equation •

o

!

w;r; r= � · ,kWi _

-

(11.27)

If a rectangular reference frame Oxyz is defined and ( y;, z;) are the coordinates with respect to this frame of the particle of weight w;, the coordinates (.X, y, z) of the CG are given by x;,

Wx =

!w;x;,

Wy =

!w;y;,

Wz =

!w;z;,

(11.28)

where W = !w; is the total weight. These equations follow from equation (11.27) by taking components of both members in the directions of the three axes. The equations can also be obtained by equating the moment of the resultant weight force about any one of the axes with the sum of the moments of the individual weights about the same axis. These two quantities are equal, since the resultant is equivalent to the original system and hence conditi on (b) on p. 277 is satisfied. Since weights are proportional to masses, equation (11.27) is equiv­ alent to the equation (11.29) Mf = !m;r;, the notation being obvious. Even in the absence of gravity when the CG is non-existent, this equation defines a unique point r with respect to the particle system. This point is referred to as the centre f ma (CM). We shall assume that the reader is familiar with the positions of the mass centres of uniform solids in the shape of common geometrical figures.* o

* See, for example, (English Universities Press) .

Advanced Level Applied Mathematics, by C.

ss

G. Lambe

284

TT

A C O U R S E I N A P PLI E D MATH EMAT I C S

[CH.

Show that the centre ofgravity of one·half of a uniform right circular cone, cut offby an axial plane, is distant rf from this axial plane, r being the radius of the plane base. (S.U.)

Example 6.

The half cone is symmetrical about the plane OCN, and hence the centre of gravity G lies in this plane. Let ji be the dis­ tance of G from the axial plane of section GAB. Suppose the cone to be divided into semi­ circular elements by planes parallel to the base and let be the thickness of the element distant from the vertex. If is the density of the material of the cone, the mass of the element is where is its radius. The CG of this element lies at G' on N'C', distant 4r'f37T from N'. Hence, assuming gravity to be acting parallel to ON, taking moments of the weights of the elements about an axis through 0 parallel to AB and equating their sum to the moment of the resultant weight force acting at G, we obtain

- - - - - - -

!TTpr''dx,

r'

lh!1rpr'2 4r' 37T

•

T ----j_ :X:

dx p

x

0

0

C

dx = i1rpr2hji.

A

Now, from the similar triangles ONC, ON'C', we deduce that Hence =

11.4.

ji rf7T.

=

r' rxfh.

Inertial Forces. Centrifugal Force

It is occasionally convenient to transform an essentially dynamical problem into a fictitious, but equivalent, statical problem and then to solve the latter by the methods of this chapter. Thus, consider a rigid body in motion under the action of certain external forces. Let m be the mass of a typical particle of the body which, at time t, is at a point whose position vector relative to a fixed origin is r. Then, if F is the total force (internal and external) applied to the particle, we have the equation of motion or

F = mr, F - m'i = 0.

(11.30)

Now consider the statical problem posed if the rigid body occupies the same position at time t, but is stationary and in equilibrium under the action of the same external and internal forces and an additional system of forces, one applied to each particl�, of which -m'i is that applied to the particle m. If we write down the conditions that the

1 1]

EQUILIBRIUM OF FORCE SYSTEMS

285

particles of the body shall be in equilibrium, we obtain a set of equations of which equation (11.30) is a typical member. Mathematically, therefore, the dynamical and statical problems are exactly equivalent. The additional forces which are introduced to compensate for our neglect of the body's motion will be called the inertial forces, since they are proportional t9 each particle's inertia or mass. Thus, each particle is in equilibrium under the action of the internal, external and inertial forces. For example, if the rigid body is held stationary inside a vehicle which is accelerating with an acceleration f of constant direction, each particle of the body will experience this acceleration and the inertial force acting upon the typical particle will be -mf. The inertial forces then form a parallel system, and each is proportional to the mass of the particle upon which it acts. It follows that this system is equivalent to a single force -Mf acting at the mass centre of the body, M being the total mass. It is accordingly justifiable to disregard the body's motion and to look upon it as being in equilibrium under the action of the reactions at its supports, the weight forces and a fictitious inertial force acting at the mass centre. It will be appreciated that this point of view is that most naturally adopted by an observer inside the vehicle. Another particularly simple case is that of a rigid body rotating with constant angular velocity w about a fixed axis. If r is the perpendicular distance of the particle m from the axis of rotation, the acceleration of this particle is w2r along the perpendicular towards the axis of rotation. The inertial force acting upon the particle is therefore mw2r acting away from the axis of rotation. This particular inertial force is called the centrifugal force. Regarding the body as being in equilibrium under the action of the applied forces and the centrifugal forces, correct equations will now be obtained. If the axis of rotation is at a great distance from all the particles of the body, as, for example, in the case of a racing car moving around a circular racing track, the centrifugal force applied to the particle m may be taken to be mw 2r approximately, where r is the distance of the mass centre from the axis of rotation. The centrifugal forces will then all be proportional to the masses of the particles upon which they are supposed to act and will be very nearly parallel. In these circumstances the system of centrifugal forces is equivalent to a single force Mw2r acting at the mass centre. Examples of the use of the concept of centrifugal force will be found in Sections 13.5, 13.9.3 and 19.3. 11.5.

The Principle of Virtual Work

Suppose a particle is acted upon by n forces Fv F2, Fn, whose resultant is R. If the particle suffers a displacement d during which

286

A C O U R S E IN A P P L I E D M A T H E MA T I C S

[CH.

the applied forces remain constant in magnitude and direction, then, as shown in Section 2.2, the work done by the resultant will equal the sum of the works done by the individual forces. It should be clearly understood that additional forces may have to be applied to the particle to cause the displacement to take place, but this in no way affects the validity of the result just stated. In the special case when the forces are in equilibrium the sum of the works done will be R zero. This is the Principle of Virtual Work as it applies to a particle in equilibrium. Thus, suppose a particle of weight W rests in equili­ brium on a rough, inclined plane of slope 6 (Fig. 11.3). Let R and F be the normal and frictional components respectively of the reaction of the plane on the particle. FIG. 1 1 .3.-Particle on a Rough, Inclined Plane To apply the principle of vir­ tual work, we suppose the particle to be displaced a distance x up the plane and we imagine that all the forces applied to it remain constant in magnitude and direction during the displacement. If the displacement were actually caused to take place, F would reverse its sense and become limiting ( =f.LR). Hence the circumstances in which our displacement takes place are purely imaginary and it is for this reason that the work is de­ scribed as " virtual ". In the virtual displacement, by the principle explained above, the total work done is zero. Hence Fx - Wx sin 6 = 0, F = W sin 6. or If the virtual displacement is a distance y along the normal to the sloping plane, the principle of virtual work leads to the equation Ry - Wy cos 6 = 0, R = W cos 6. or If the displacement y were actually carried out, R would vanish and no work would be done by this force. Again, therefore, we must emphasize the imaginary nature of the displacement. We now proceed to generalize the principle so that it can be applied to the problem of a system of bodies in equilibrium. Any particle of this system is in equilibrium under the action of internal and external forces. If, therefore, we imagine that the particles of the system are displaced in any way (not necessarily consistent with the constraints)

1 1]

E Q U I LI B R I U M O F F O R C E S Y S T E M S

287

and that during the displacement both the internal and external forces remain constant in magnitude and direction, the net work done by these forces will be zero for each particle and hence, also, for the whole system. This is the General Principle of Virtual Work.

Again we draw attention to the imaginary nature of the virtual displacement. If one of the bodies concerned is a steel bar it is quite permissible to consider a virtual displacement in which the particles forming one half of the bar move in one direction and the particles comprising the other half move in the opposite sense. Such a dis­ placement could be carried out in practice only by breaking the bar and, during the rupture, the internal forces would fluctuate violently. However, during the purely imaginary or virtual displacement, these internal forces are assumed to remain steady in magnitude and direc­ tion and, provided the bar is in equilibrium, the net work done by all the forces concerned is then zero. It is, however, very often convenient to suppose that the virtual displacement is consistent with the constraints, so that no rupture of a

p/1,-:.:;:_""""'t� I I

I

I I I I

I

II

lr I I I I I

:

/

I

I

I I

I I

I

I

���--�------------------�--�� - - -'N

P

F

M

F

0

FIG. 1 1 .4.-Work Done by Internal Forces

rigid body takes place. The following result explains the reason for this : In any small virtual displacement of a rigid body which does not

lead to rupture the work done by the internal forces is of the second order of small quantities. For let P, Q be any two particles of the rigid body in their equilibrium positions and let P', Q' be their respective displaced positions ( Fig. 1 1 .4) . Suppose these particles act upon one another with equal and opposite forces F when in their equilibrium positions. Then, if P'M, Q'N are perpendicular to PQ, the total work done by

these forces during the virtual displacement, supposing them to remain unaltered in magnitude and direction, is

F . PM - F . QN = F(PM + MQ - MQ - QN) = F(PQ - MN) = F . PQ(l - cos 6), (since PQ = P'Q'), where e is the inclination of P'Q' to PQ and is a small angle if the

288

A C O U R S E I N A P PLI E D M A T H E M A T I C S

[CH.

displacement is small. Now (1 - cos 6) = !62 to the second order in 6 and hence work done = !F . PQ 62• (11.31) Summing over all particle pairs, we prove that the total work done by the internal forces is of the second order as stated. Accordingly, if we restrict ourselves to small virtual displacements which do not rupture the rigid bodies occurring in a problem and work to the first order of small quantities, we may neglect the work done by internal forces. This will be our procedure in the worked examples below. If, on the other hand, it is the internal forces of a body which are to be calculated, the virtual displacement must be so chosen that the body is distorted and the internal forces are permitted to do work, for otherwise they will not appear in our equations. Other forces which do no work in small virtual displacements consistent with the constraints are : l . The reaction at a smooth constraint, for the displacement of the point of contact will be at right angles to the force. 2. The action and reaction at a smooth pivot, since these are equal and opposite and, unless the joint is fractured, the displace­ ments of their points of application are equal and the works they do cancel.

Two smooth rods, inclined at an angle 2oc (oc < !1r), are fixed in a vertical plane so that each rod makes an angle oc with the downward vertical. Along each of these rods a small ring of weight w is free to slide, and the two rings are connected by two equal, uniform 0 bars AB, BC, each of weight W, which are freely jointed to the rings at A and C and are smoothly pivoted together at B. Use the principle of virtual work to show that, in the symmetrical position of equilibrium in which A C is horizontal and either rod makes an angle e with the upward vertical, w = ! tan oc tan e - I. W (Le.U.)

Example 7.

Consider the system formed by the two rods and rings and suppose that it under­ goes a virtual displacement in which e in­ creases by the rings slide on the fixed rods and the bars remain of con­ stant length. In such a displacement we may neglect the work done by the internal w forces in the bars ; the works done by the B action and reaction at the smooth pivot and the reactions of the smooth bars on the rings are all zero. Only the works done by the weight forces will appear in the equation of virtual work. If = = sin (6 + oc) = sin oc = sin e cot + cos 6).

de,

AB, BC

B

AB BC 2a,

OB 2a 2a(

oc

EQUILIBRIUM O F FORCE SYSTEMS

l l]

289

Thus the depths of the points X and A below the point 0 are and y respec­ tively, where sin cot 0< + cos sin cot 0<. When increases by x and y increase by respectively (to the first order), where cos 6 cot 0< sin cos cot 0< The total work done by the four weight forces is and, by the principle of virtual work, this is zero. Hence cos cot 0< - sin cos cot This is equivalent to the relationship stated in the question.

x

6

x = a(2 6 6), y = 2a 6 d6, dx, dy dx = a(2 6)dtl, dy = 2a 6 dtl. 2Wdx + 2wdy 2Wa(2 6 6)d6 + 4wa 6 O
A thin, elastic ring, of mass m and natural diameter a, has modulus il.. It is placed, with its plane horizontal, on a smooth sphere of radius a and it stretches under its own weight. Prove that in its equilibrium position any diameter of the ring subtends an angle 26 at the centre of the sphere, where 21ril.(2 sin e 1) = mg tan 6. (L.U.)

Example 8.

-

Suppose the ring suffers a virtual displacement in which it remains horizontal but slides downwards over the spherical surface, increasing by During the displacement, each element of the ring moves at right angles to the reaction of the smooth surface acting upon it, and therefore these reactions do no work. If is the centre of the ring, its height above 0 is given by cos 6. This height increases by as a result of the displacement, where sin The negative sign indicates that weight of the ring performs work

6

d6.

0

C

OC = a

d(OC)

d(OC) = -a 6d6.

OC suffers a numerical decrease. -mgd(OC) = mga sin 6d6.

The

Since the displacement causes an expansion of the ring, its internal forces also perform work. If is the tension in the ring in its equilibrium position, the total work done by the internal forces will be where is the increase in the length of the ring. But A sin and thus cos Hence the work done by the internal forces is cos The equation of virtual work can now be written down, thus sin cos Also, by Hooke's Law

T

dl = 21ra 6d6.

Tdl, dl l = 21T . C = 21ra 6 -21raT 6d6. mga 6d6 - 21raT 6d6 = 0. -

T = 1Ta � (l - a) = i\(2 sin 6 - 1).

The result required now follows immediately.

L

290

A C O U R S E I N A P P L I E D M A T H EMAT I C S

[CH.

A square ABCD of equal smoothly jointed light rods each of length a lies on a smooth horizontal table; OA, OB, OC, OD are four equal light rigid rods, each of length 2a, smoothly jointed at 0, A, B, C, D, so that all the rods form the edges of a pyramid. A weight W is suspended in equilibrium at 0. By the method of virtual work, or otherwise, find the stress in a rod such as OA (L.U. ) and a rod such as AB.

Example 9.

Consider firstly a virtual dis­ placement in which the rods form­ ing the square remain fixed and the point 0 moves up­ wards along the vertical 00' through the centre of the square. The rods will all increase in length, and hence their internal forces will perform work. = and suppose Let increases by Now

0

ABCD

AO, EO, CO, DO

6 LOAO' 6 d6. AO = AO' sec 6 = .;2 sec 6. a sec 6 tan 6d6. d(AO) = V2 If T is the thrust in the bar A 0,

the work it performs during the A virtual displacement is

B

Td(AO) = V2I aT sec 6 tan 6d6.

By symmetry, the thrusts in the remaining sloping bars each perform the same work. Also

00'

=

.;2 tan 6.

d(OO') = V2a sec• 6d6, and the work done by the weight W is -Wd(OO') VI2 Wa sec2 6d6. The equation of virtual work is 2y2aT sec 6 tan 6d6 - VI 2 Wa sec2 6d6 ""' 0, or T = i W sec 6 cot 6. In the equilibrium position cos 6 = AO'fAO = I/2y2. T = Wfyi4. Hence Now suppose that 6 increases by d6 in such a way that 0 moves vertically, =

-

the sloping bars remaining fixed in length, whereas the bars forming the square all contract. Then = = cos = sin If S is the tension in the work done by the internal forces of will be = sin Similarly for the remaining three horizontal bars.

AB y2AO' 2y2a 6. d(AB) -2y2a 6d6. AB, -Sd(AB) 2y2aS 6d6.

AB

EQUILIBRIUM O F FORCE SYSTEMS

1 1]

00' = AO sin 6 = 2a sin 6. d(OO') = 2a cos 6d9 and the work done by W is - Wd(OO') = -2Wa cos 6d6. The equation of virtual work is 8y'2a5 sin 6d6 - 2Wa cos 6d6 = 1 5= W cot 6 = Wf4y' 14. and hence 4 v' 2

291

Also

0

The principle of virtual work provides us with a very general con­ dition which must be satisfied by all mechanical systems in equilibrium. Its wide scope may be emphasized by showing that the conditions represented by equations (11.1) and (1 1.2) may be deduced as particular cases. Thus, suppose a rigid body, which is in equilibrium under external forces Fi applied at points with position vectors ri, is subjected to a virtual displacement of the simplest type, viz., a small translation specified by the vector da. The particles of the rigid body remain at the same relative distances from one another, and hence the in­ ternal forces do no work. The force Fi does work Fi da, and the equation of virtual work is accordingly ·

da ·

or

n

1 i=

1

Fi

=

(11.32)

0.

This equation is true for all small vectors da, and hence equation (1l.l) follows. Now assume the virtual displacement is a small rotation of the rigid body about some axis through the origin 0. Again the internal forces do no work. If d6 represents this rotation both for magnitude and direction of axis, the displacement of the point of application of Fi is d6 X ri and the work done by the force is d6

X ri · Fi

=

d6 ri •

X Ft.

Hence the equation of virtual work is

or

d6

·

n

1

i=l

ri X Fi = 0.

(11.33)

This equation being true for arbitrary d6, equation (11.2) now follows. The scope of the principle may be further extended to embrace dynamical problems by the introduction of inertial forces (Section 11.4). Then, if m is a typical particle acted upon by a force F and an

292

A C O U R S E I N A P PLI E D MAT H E M AT I C S

inertial force -m'i:, let ar be its virtual displacement. by the forces acting upon the particle is then

(F - m'i:)

•

[CH.

The work done

ar

and, since the system may be regarded as being in equilibrium under the action of all the forces (including inertial forces) , the principle of virtual work requires that

2(F m'i:) ar = 0, 2F · ar = 2mr · ar, -

or

•

(11.34)

the summation extending over all particles of the system. This equation has already been obtained as equation (10.38) and, if we now limit the virtual displacement to be one consistent with the constraints as they exist at the instant under consideration, the development to Lagrange's equations of motion proceeds as described in Chapter 10. The principle of virtual work is clearly one of the most fertile for the theoretical development of mechanics. EXERCISE 1 1

l . A uniform trapdoor of weight W is in the shape of a rectangle ABCD and is smoothly hinged along its edge AB. It is supported at an angle rt. to the horizontal by a uniform rod also of weight W, which is smoothly pivoted to the corner D and rests with its other end on a rough, horizontal plane through A B. The point of contact of the rod with the plane is vertically below C. If the length of the rod is equal to the length of the side AD of the trapdoor, show that the coefficient of friction at the point of contact of the rod with the plane is at least i cot ex. Show also that the reaction at the hinge is a wrench of pitch 2l tan ex 4 + tan2 ex where l = AB. 2. Three equal spheres, each of radius 3a, are placed on a smooth, hori­ zontal plane and fastened together by an inextensible string which surrounds them in the plane of their centres and is just not tight. A fourth sphere, of weight 3W, is placed on top, touching all three of the lower spheres. If the string will break when its tension reaches the value W, show that the radius of the upper sphere cannot be less than a without breaking the string. (L.U.) 3. Three particles A , B, C of equal weights are joined together by light, rigid wires so that A BC is a straight line and AB = BC. The com­ bination is placed on a uniformly rough, horizontal plane and the normal reactions of the plane on the particles are equal. A horizontal force whose magnitude steadily increases whilst its direction remains unaltered is applied at A . Show that equilibrium can never be broken

1 1]

EQUILIBRIUM OF FORCE SYSTEMS

293

by the particles commencing to rotate about B, but that the particles will commence to rotate about C if the acute angle made by the direction of the force with the line of the particles is greater than 60°. 4. Two forces F, G, whose lines of action are at right angles, are com­ bined to form a wrench. Show that the pitch is cFGf(F2 + G2) , where c is the shortest distance between the lines o f action. Find the ratio in which the central axis divides the shortest distance between (Le.U.) the lines of action. OABC, O'A 'B'C' are opposite faces of a cube of side a, 00', AA' being edges. Equal forces act along OA , BE', O'C'. Find the pitch of the wrench equivalent to this system. (Le.U.) 6. Two forces X, act along the straight lines y = x, z = c and y = x z = - c respectively. Find the equations of their central axis and the pitch of the equivalent wrench. Show that as X and vary in magnitude, the central axis always lies on the surface (S.U.) z(x2 + y2) = 2cxy. 7. OABC is a tetrahedron in which A OB, A OC, BOC are right angles.

5.

Y

-

,

Y

-+ -+ -+ -+

8.

9.

10.

11.

12.

13.

Forces "AOA , fLOE, vA C, 6BC act along the edges indicated. If the system reduces to a single force, prove that "A6 VfL. If this condition is satisfied, and if the resultant intersects OA , show that = 0 and that the system of forces must be plane. (Li.U.) The axes of two wrenches coincide with the x and y axes, their forces are P1 and P 2 and their couples G1 and G 2 respectively. Find the pitch of the resultant wrench, and show that its axis is parallel to the x, y plane and distant (P1G 2 - P2G1)/(P12 + P 22) from it. (S.U.) Forces of magnitudes P and Q act along given perpendicular non­ intersecting lines, and are such that P + Q has a given value. Show that the pitch of the equivalent wrench is greatest when P = Q. (N.U.) Forces of magnitude 1, 2, 3, 4, 6 act along the edges OA , OB, OC, AB, BC, CA of unit length of a regular tetrahedron. Determine the total force and the pitch of the equivalent wrench. (Li.U.) The radii of the ends of a frustum of a uniform, solid, right circular cone are a and 2a. Show that the distances of its mass-centre from the ends are in the ratio 17 1 1 . The frustum is placed with its curved surface resting on a horizontal plane. Show that it topples over if tan2 IX > 17/28, where IX is the semi-vertical angle of the cone from which it is cut. (Le.U.) A hollow vessel of thin, uniform material consists of a right circular cylinder of radius a and height h, one end of which is closed by a plane face and the other by a hemispherical shell. Show that it will not lie with a generator of the cylinder in contact with a horizontal table (S.U.) if 2h < - l)a. Four forces are in equilibrium. Show that the invariant r = R G of any two is equal to that of the other two and that the same invariant of any three is zero. =

e

5,

(y5

·

294

A C O U R S E IN A P P L I E D M A T H EMAT I C S

[CH.

I4. AECD is a parallelogram of light, j ointed rods, kept rigid by a diagonal

rod ED. Equal and opposite forces P are applied at A and C in the line A C. Prove that the tension in ED is ED . F AC (Le.U.)

I5. Five equal, uniform rods are freely jointed together at their ends to

form a closed chain AECDEA , and the system hangs freely from A . Prove that, if A E and EC are inclined to the vertical at angles and cfo respectively, then tan c/> = 2 tan Prove that sin is the root between 0 and t of the equation

e

e.

e

( I - 2x) 2

= I +I 6x23xz'

and show that the equation has one, and only one, root in this range. (M.T.) I 6. Four uniform rods AB, EC, CD, DE, each of length a and weight w,

are freely hinged together at E C and D. A light, inelastic thread of length 6af5 joins E and D, the breaking tension for the thread being I3wf8. The ends A and E are held together and slowly moved apart at the same level. By an application of the principle of virtual work, or otherwise, show that the thread tightens when the length A E is approximately I · 685a, and breaks when A E is I4af5. (M.T.) ,

I7. Four uniform rods, each of length a and weight w, are freely jointed to form a rhombus AECD, which is freely suspended from A . A smooth, uniform disc, of radius r and weight w, rests between the rods CE and CD and in the same vertical plane. The joints E and D are connected by a light, elastic string of natural length ja and, in equi­ librium, ED a. Use the principle of virtual work to show that the modulus of elasticity of the string is 3(2r - y3a)wfa if r > y3af2. (L.U.)

=

IS. A framework consists of six equal, uniform rods, each of weight W and length 2a, freely j ointed at their ends to form a regular hexagon

AECDEF, which hangs freely in equilibrium with AE fixed hori­ zontally and DE below A E. The framework is kept in shape by an endless, light, inextensible string passing successively through two small, smooth, fixed rings at A and B and two small, smooth, light rings attached to the rods CD and EF at their mid-points. Show by the principle of virtual work that the tension in the string is 8y2I

+ 9

I4y3 w

(L.U.)

I9. Four equal particles A , E, C, D, each of weight W, are connected by

light strings AE, EC, CD, DA , each of length and then placed at rest symmetrically on a smooth sphere of radius R(l < !1tR) . Prove, by virtl!al work or otherwise, that the tension in each string is (L.U.) tW sin (lfR) v'sec (l/R) .

l,

1 1]

EQUILIBRIUM O F FORCE SYSTEMS

295

20. Six rods are hinged together to form a tetrahedron ABCD ; the rods

BC, CA , AB are light, and the rods A D, BD, CD are equal in length and each of weight W The system stands with A , B, C on a smooth, horizontal table. Prove that the tensions in BC, CA , AB are pro­ portional to sec A , sec B, sec C respectively, where A , B, C are the (acute) angles of the triangle ABC. Prove that the tension in BC is 3aW h cosec2 A cot B cot C, B

where h is the perpendicular distance from D to the plane ABC. (S.U.)

CHAPTER 12

EQUILIBRIUM OF STRINGS AND CHAINS 12.1.

General Equations of Equilibrium

Consider a perfectly flexible, fine, inextensible string in equilibrium in a plane under the action of external forces applied along its length. These forces are in two categories : (i) those which are applied at discrete points, and (ii) those which are distributed continuously over a finite arc of the string. If A (Fig. 12.1 (a)) is a point on the string at which there is a resultant applied force F, there will clearly be a cusp on the string at this point. The particle of the string at A will be in equilibrium under the action of F and the tensions at A, T and T+ _

F

Mds

Nds

T_

/

/t\

(a)

I I I I

/�)

(b)

FIG. 12.1.-Equilibrium of a Plane String

in the two parts of the string meeting at A. These tensions will act along the two tangents to the cusp. By resolving in two directions, we obtain two equations of equilibrium. Let PP' (Fig. 12.1 (b)) be a small arc of length ds of a part of the string subject to a distributed force. Let T be the tension at P and T + dT the tension at P' These forces act upon the element PP' in the directions of the tangents to the string at P and P' respectively. Let these tangents make angles if;, if; + dif; with a fixed line in the plane. If R is the resultant applied force per unit length at P, the external force acting upon P P' will be Rds to the first order. We shall resolve this force along the tangent and normal to the string at P and denote the components by (Mds, Nds) respectively. The figure indicates 296

[cH. 12]

E Q U I L I B R I U M OF S T R I N G S

297

the positive senses for these components. Resolving along the tangent and normal all forces acting upon PP', we obtain as the conditions of equilibrium Mds + (T + dT) cos difi - T = 0, (12.1) Nds + (T + dT) sin difi = 0.

}

These equations are accurate to the first order. Hence, neglecting second-order quantities, cos difi = 1, sin difi = difi and thus Mds + dT = 0,

Nds + Tdifi = 0,

or, proceeding to the limit, dT = -M' KT = -N, ds

(12.2)

where K = difijds is the curvature at P. If the first of equations (12.2) is divided by the second, we obtain d (log T) = MJN = tan cp, difi

(12.3)

where cp is the angle made by R with the normal. When cp is known as a function of ifi, this equation can be integrated for T. For example, in the simple case of a light, taut string in contact with a smooth curve, the reaction of the curve is along the normal and cp = 0. Then log T and hence T is constant along the string. 12.2.

String Hanging in Equilibrium under Gravity

Suppose the end points of a string are fixed and it hangs under gravity in the plane of rectangular axes Oxy, Oy being vertically upwards (Fig. 12.2). R is directed vertically downwards and is the weight per unit length at P. We denote this by w. Taking the fixed direction from which ifi is measured to be the x-axis, by resolving R we obtain M = -w sin ifi, N = -w cos ifi. Hence equations (12.2) take the form . dT = w sm ..,.!., , ds

T

difi ·'· = w cos 'f'• ds

(12.4)

and equation (12.3) becomes

ddf (log T) = tan ifi. .

(12.5)

Integrating, we obtain (12.6) T0 being a constant of integration. When ifi = 0, T = T0 and thus T0 is identified as the tension at the lowest point of the string (provided such a point exists).

298

A C O U R S E I N A P P L I E D M A T H E MA T I CS

[cH.

To proceed further, we must know the manner in which w varies along the string. Firstly, suppose w is constant, i.e., a uniform string. Writing the first of equations (12.4) in the form dT dy w ds = ds ' T = wy + constant

we obtain

by integration. The value of the arbitrary constant will depend upon

FIG.

12.2.-The Catenary

the position of the x-axis from which y is measured. select this position that the constant vanishes. Then

T = wy .

We shall so (12.7)

i.e., the tension is directly proportional to the height above Ox. Eliminating T between equations (12.6) and (12.7) , we obtain an equation of the curve in which the string will hang, viz., y where c = T0jw, or

= c sec !f,

(12.8)

T0 = we.

(12.9)

Thus c is that length of string whose weight is equal to the tension at the lowest point. Differentiating equation (12.8) with respect to x, we find that

��'

tan ¢' = dyjdx = c sec !f tan ¢' or

dxjd!f = c sec !f.

12]

E Q U I L I B R I U M OF S T R I N G S

299

Thus, integrating, x

=

c log (sec ifl + tan ifJ),

(12.10)

it being assumed that the y-axis passes through the lowest point of the string so that x = 0 when ifJ = 0. Equations (12.8) and (12.10) are parametric equations for the curve of the string. Elimination of ifJ will result in the Cartesian equation. Thus sec ifl + tan ifl = e'* · (12.11) But sec2 ifJ - tan2 ifJ = 1 and hence sec ifJ - tan ifJ

=

1

sec ifl + tan ifl

= e -xfc.

(12.12)

Addition of the last two equations yields sec ifJ

=

cosh

(xfc)

y = c cosh

(12.13) c This is the equation of the common catenary. The x-axis is called the directrix of the catenary. If A is the lowest point, OA = c. Other important equations of the catenary will now be found. From the second of equations (12.4) and equation (12.6), we obtain ds 2 d-ifl - c sec ifl and thus s = c tan ifl, (12.14) provided s is measured from A (i.e., s and ifJ vanish together). Then, from equations (12.8) and (12.14), (12.15) y2 = 8 2 + c 2 . Also, from equations (12.11) and (12.12), tan ifJ = sinh (xfc) . Hence, by equation (12.14), s = c sinh �. (12.16) c Example 1. A uniform rod AB, of length 2a, is freely pivoted at one end to a fixed point A . To the other end B of the rod is fastened a heavy, uniform chain which hangs from a fixed point C at the same level as A . The tangent to the chain at the point B is horizontal. The weights per unit length of the chain and the rod are equal. Show that the length of th� chain is 2a( in2 o: + <X)t, where oc = LBA C. (S.U.) and thus

s

X

- ·

cos

300

A COURSE I N APPLIED MATHEMATI C S

[CH.

Since the rod is in equilibrium, taking moments about we obtain cos A sin <X = = cot <X . or It follows from equation that the parameter of the chain is = cot <X. Let be the length of the chain. Then, by equation since = sin <X at sin = + = sin IX sin2 -===--a "= cos sin2 <X , from which the result stated follows.

A,

2wa2 <X, T0 • 2a T0 wa (12.9) c a l (12.15), y c + 2a C, (c + 2a <X)2 l" c2, 12 4ac + 4a2 <X, 4a2( <X + )

c

�

A heavy, uniform, flexible chain rests in limiting equilibrium with its ends on two rough, horizontal tables, for each of which the coefficient of friction between the chain and the table is 1-'· If the equilibrium is limiting at each end, and if the horizontal and vertical p distances between the nearer edges of f - - - - the tables are 2a and h respectively, prove that the length of the hanging portion of the chain is D and the total length of the chain is D ( 1 + � coth �) where D2 = h2 + 4c2 sinh2 c� . (The vertical plane in which the chain lies is perpendicular to the edges of the tables.) (N.U.) Let P, Q have coordinates (x1, y 1), (x 2 , y 2) respectively referred to the usual

Example 2.

h

A

14---- 2 a---__.,

axes. Then, if c is the parameter of the catenary

y1 = c cosh �c . y 2 = c cosh :i.c

But

y1 - y 2 = h and thus h = c cosh ::._c, - c cosh �c = 2c sinh x, 2+c sinh x , 2-c x,, 2c sinh c� sinh x, 2c+ x., (i) since x1 - x 2 = 2a. Also, by equation (12.16) if s 1 and s 2 are the lengths of the arcs AP, A Q respectively s1 = c sinh � . s = c sinh .:l>j, c 2 c Now s1 - s 2 = D (N.B. s 2 is negative) and hence D = 2c sinh x, 2c- x . cosh x, 2c+ x., = 2c sinh � cosh x, + x�. (ii) 2c c =

Xz

1 2]

E Q U I L I B R I U M O F S T R I N GS

301

Subtracting the square of equation (i) from the square of (ii), we now obtain the result D2 - h2 = 4c 2 sinh2

P

�c -

From equation (12.7), the tensions at and Q are Tl = WYv r . = wy •. respectively. These forces must equal the limiting frictions operating on the lengths 1 1 and 1 2 of chain lying on the tables. These frictions are fLWl 1 and JLWl 2 • Hence from which it follows that

) 2c x1 + x 1- x-. 2 1 1 + 1 2 = - (y 1 + y 2 = - cosh ---2 cosh x2c 2c JL JL x x 1 ., = � cosh � cosh c 2c JL = !!_ coth �. c fL by equation (ii) . This completes the solution.

I

+

Two cases of a non-uniform chain hanging under gravity are of practical importance. Firstly, there is the " suspension bridge "

FIG.

wx

12.3.-Parabolic Catenary

problem of a chain whose form is such that when hanging the weight of any element ds is wdx, where dx is the horizontal projection of ds and w is constant. This is equivalent to the problem of a light chain uniformly loaded horizontally, as in a suspension bridge.

302

A C O U R S E I N A P P L I E D M A T H EM A T I C S

[CH.

We take rectangular axes through 0, the lowest point of the chain, Oy being vertically upwards (Fig. 12.3) . Consider the equilibrium of the finite arc OP of the chain, P being any point (x, y) . The forces acting upon this arc are : (i) T0 the tension at 0 ; (ii) T the tension at P, and (iii) wx the weight of the arc OP. Resolving horizontally and vertically, we obtain the equations of equilibrium, viz., T0 = T cos ifi, wx = T sin ifi.

(12. 17)

Division yields the equation tan ifi = dyfdx =

w x T0

and an integration now gives the equation of the curve as (12. 18) Thus the chain hangs in a parabola. This method of attack upon the problem may also be employed to yield equation (12.14) in the case of the uniform catenary. From this equation, the remaining equations of the catenary can be found. Their derivation is left as an exercise for the reader. Secondly, there is the problem of the " catenary of uniform strength ". This is designed to be of variable cross-section so that the stress (Section 13.3) is constant along the whole length of the chain. Thus, those portions of the chain subject to large tensions are corre­ spondingly thicker. Let " be the constant stress, oc the cross-section and T the tension. Then T = oc't'. Also, if p is the density of the material, the weight w per unit length is given by w ocp. Hence =

T = AW, (12.19) where "A = 't'jp. Equations (12.4) are valid and, from the second of these equations and equation (12.19) , we obtain

Integration yields

ds = "A sec ifi. difi

(12.20)

s = "A log (sec ifi + tan ifi),

(12.21)

provided we measure s from the lowest point. Also, from equation (12.20), we have dx difi = dy difi =

dx ds = cos ifi . "A sec ifi = "A, ds . difi dy ds . = sm ifi . "A sec ifi = "A tan ifi. ds • difi

(12.22) (12.23)

EQUILIBRIUM OF STRINGS

12]

Whence,

x

=

"AI/J, y = A log sec !f,

303

(12.24)

the origin being taken at the lowest point. The xy-equation is accordingly y=

:A

X

log sec ).,_ '

(12.25)

We note that as x ---+ ±i"A7t, y ---+ + oo . Thus the chain hangs between the lines x = ±i"A7t and its span cannot exceed "A1t. If the material is strained to breaking point everywhere, "A takes its maximum value Amax. and the span is still less than 7tArnax.· This, then, represents an upper limit for the span which can be achieved with a given ma­ terial. Typical values for steel are -. � 106 lbjsq in, p � 0·3 lbjcu in, and the theoretical upper limit to the span is about 106 in or 16 miles. 12.3.

String in Contact with a Cylinder

If the weight of the string be neglected and the contact is smooth we have already proved that the tension is constant. This justifies the assumption, usually made in elementary work, that the tension is unaltered when a light string passes over a smooth pulley. The mass of the string being negligible, any motion the string may possess does not invalidate this result. Next suppose that the string is light but that the contact is rough and that the string is on the point of slipping in the direction of in­ creasing s. If fL is the coefficient of friction, M fLN (Fig. 12.1 (b)), the senses of M and N being opposite to those shown in the figure. Hence, by equation (12.3), =

d (log T) = dlfi Integration yields

fL.

{12.26) (12.27)

where T = T0 at 1/J = 0. 1/J is the angle of wrap around the cylinder, and the last equation shows that the tension increases exponentially with respect to this angle. The law of tension is observed to be independent of the shape of the cylinder. If a rope is wrapped three times around a cylinder, 1/J = 61t. Then, if fL = 0·5, the ratio of the tensions at the two ends of the rope is e37T = 12,400. It is clear from this result that by the use of this " capstan principle ", a small force can be employed to resist a com­ paratively large one.

304

A C O U R S E I N A P P L I E D MATH EMAT I C S

[CH.

A uniform, circular disc, of weight W, has its plane vertical, and is pressed against a vertical wall perpendicular to its plane by a light string in contact with its rim, one end of which is fixed to a point in the wall above it and in its plane. The other end of the string sustains a weight P, which hangs freely below the disc. The coefficient offriction between the string and the disc is f'; the T wall is sufficiently rough to prevent the disc from slipping on it. Show that in limiting equilibrium, if e is the angle between the upper part of the string and the wall, P(l + cos 6)ef'B = W + 2P.

Example 3.

(Le.U. ) The forces acting upon the disc are all indicated in the diagram. Taking moments about the point of contact with the wall, we obtain cos = where is the radius of the disc. But = Hence = cos +

C, T(a + a 6) P . 2a + Wa, a T Pef'B . P(l + 6)e�'B W 2P.

p

If the weight of the string is to be allowed for, assuming the friction limiting, equations (12.2) take the form

. dT R w sm f, ds = fL +

T

df = R + w cos f, ds

•

(12.28)

Rds being the normal reaction acting upon ds and the direction f = 0 being horizontal. Eliminating R between these equations, we obtain

1s - fLT�t

=

w(sin f - fL cos f),

. ds dT T = w(sm f - fL cos f) . df - fL df

or

(12.29)

Multiplying throughout by an integrating factor e-!L'P, this equation may be written

:f ( Te-""') = we-""' (sin f

- fL cos f)

which, upon integration, yields T

=

J

ef"P we-�'.P(sin f - fL cos f)

��-

;� df

(12.30)

(12.31)

The intrinsic equation s = s(f) for the cross-section of the cylinder being known, equation (12.31) shows how the tension varies with f.

12]

EQUILIBRIUM OF STRINGS

305

A uniform chain, of weight W, is of length equal to half the circum­ ference of a fixed, rough, vertical circle. The chain rests on the upper edge of the circle, extending from one end of the horizontal diameter to the other. A vertically downward force is applied at one end of the chain. Prove that the chain will not move unless the force is greater than 2fLW(1 + ewrr) 7T(1 + fL•) where fL is the coefficient offriction between the chain and circle. (L.U.)

Example 4.

•

Assume that the chain is on the point of slipping. In accordance with the conventions adopted in the general theory, it will be convenient to measure s from the lowest point of the circle. Then the intrinsic equation of the circle is = where is the radius and Y, increases from !1r to tn- along the chain. By equation the tension at any point on the chain is given by

s a.p,

a

(12.31),

T = waei'.P Je-I'.P (sin Y, - fL cos Y,)dofl, = waei'>/J [ (�: � � cos Y, - 2� sin Y, ) e-I'.P + constant} fL• 1 When Y, = !1r, T = 0. Thus constant = � fL. + 1 e-t�'"· Hence, when Y, = f,.,

+ ei'") T - 2fLwa(1 fL. + 1 _

•

which is equivalent to the result stated, since

1rwa = W.

EXERCISE 1 2 l . A uniform string o f length 2 l hangs over two smooth pegs which are level with each other, distant 2a apart. Show that in equilibrium the free ends of the string are both at the same level, and that the para­ meter c of the catenary in which the string hangs is given by l = ce*. (Le.U.) 2. A uniform chain of length l hangs with its end links P and Q threaded on two fixed, smooth rods AB and BC. The rod AB is vertical with B highest, and the angle ABC is 45°. Prove that the distance of Q from AB is l log (1 + y2) . Prove also that the centre of gravity of the chain is at a distance l(1 - y2 + log ( 1 + y2)) (S.U.) from AB. 3. Let T be the tension at any point P of a uniform catenary, T0 that at the lowest point C, and W the total weight of chain from C to P. Show that T2 = T02 + W2• A chain is suspended from two points at the same height. If the length of the chain is 100 ft, its total weight 40 lb and the sag 10 ft, find the greatest tension and the distance apart of the supports. (Li.U.)

306

A C O U R S E I N A P P L I E D MAT H E M AT I C S

(CH.

4. An endless, uniform chain hangs over a smooth pulley of radius a. If the chain is in contact with two-thirds of the circumference of the pulley, show that its total length is 3a 4 .na + (Le.U.) log, (2 + y'3) 5. A string, of weight w per unit length, is tied at one end to a ring which can slide on a smooth, horizontal rail and passes over a smooth peg in the line of the rail with a length hanging vertically. The ring is held at rest by a horizontal force 4aw along the rail and the inclination of the string at the ring is tan-1 i to the horizontal. Prove that c = 4a and that the total length of the string is l la. Show also that the distance between the ring and peg is Sa log 2. (S.U.) 6. A heavy string of length 2l is hung from two points on the same horizontal level, distant 2a apart. Show, by expanding y and s in powers of xfc, that if c is large, the maximum sag of the string, b, is small and that l - a = 2b2f3a approximately. Show also that the tension at the lowest point is wa2(2b, where w is the weight per unit length. (Le.U.) 7. A chain has a small, rough ring attached to each end, and the rings are threaded on a rough, horizontal rod, the coefficient of friction between the rings and the rod being ·l"· Show that, in limiting equilibrium, the ratio of the span of the chain to its length is ,S" log 5, and find the ratio of the dip at the centre to the span. (S.U.) 8. A length of string hangs from two fixed points which are at the same level and at a distance 2a apart. Show that the value of c which makes the tensions at the ends as small as possible is afu, where u tanh u = 1. Show also that this corresponds to the length of the string being 2a sinh u tanh u. (Le.U.) 9. A chain of length 2l hangs between two points A and B on the same level. The tension, both at A and at B, is three times the tension at the lowest point. Prove that the horizontal distance between A and B is l log, (3 + y'S) . (Le.U.) v'2 10. A heavy, uniform chain A B of length l is suspended by its ends A, B. The end A is the lower, and the chain everywhere slopes up from A to B ; at A it makes 30° with the horizontal, and at B 60°. Prove that the vertical and horizontal displacements of B from A are (y'3 - I)l and �y'3l log (!y'3 + I) ; and show that the tension at A equals the weight of the chain. (Le.U.) II. A balloon B is attached to a point A on the ground by a uniform cable AB of length 500 ft and weight 250 lb. The tangent to the cable at A is inclined at an angle tan-1 (f) to the horizontal and the tension of the cable at A is 350 lb wt. Prove that B is at a height 377 ft approxi­ mately above the ground. Find also the horizontal projection of AB. (S.U.)

12]

EQUILIBRIUM O F STRINGS

307

12. A uniform chain of length l is freely suspended between two fixed points A , B. The slopes of the chain at the two points are found to be 30° and 60°. Obtain the possible values of the parameter c, and show that the greatest tension in the chain is equal to that of a length ly3 of the chain, or one-half that value, according as the lowest point of the chain is at one of the ends or occurs between them. (D.U.) 13. Assuming the standard formul<e for the catenary, prove that if a uniform chain hangs in equilibrium over two small pegs at different levels the ends of the chain lie on the directrix of the portion hanging in a catenary. Given that the lengths of the vertical portions of the chain are 11 and l2 and that the vertex of the catenary divides the length of the whole chain in the ratio a1 : a 2, determine the length of the whole chain. (B.U.) 14. A uniform chain, of weight w per unit length, to which a weight W is attached at the middle point 0, hangs symmetrically over two smooth pegs on the same horizontal level and distant 2d apart. The chain is of such a length that, when the system is in equilibrium, a length l of chain hangs vertically over each peg and the tangent at 0 to each half of the chain makes an angle OG with the horizontal. If 0 is at a vertical distance D below the pegs and W = 2wc tan OG, prove that (i) lfc = sec OG cosh dfc + tan OG sinh dfc, (ii) D = l - c sec OG. (M.T.) 15. 0 is a fixed point at a height l/3 above a rough horizontal table. A uniform chain of length l has one end attached to 0 and the free end is drawn along the table until the chain hangs in equilibrium in a vertical plane through 0 with half its length upon the table. Show that the coefficient of friction between the chain and the table must (L.U.) exceed 5/12.

1 6. 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 5l of the chain hangs vertically. Find the parameter of the catenary A C, and show that the distance of the peg C from the vertical wire is 3l log (yl O - 1). (L.U.) 17. A uniform, heavy, flexible string of length l rests in a vertical plane with a length tz in contact with a smooth plane inclined at an angle !1t 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 wly(5(8), and find the horizontal and vertical distances of A from the lower end of the string. (L.U.) 18. A heavy chain, not necessarily uniform, is suspended from two fixed points. The tension T, at any point of the chain where the line density is m, acts at an angle 1 to the horizontal. If s is the length of chain, measured from some fixed point on it, prove that

fs (T sin 1)

= mg.

308

19.

20.

21.

A C O U R S E I N A P PL I E D MATH EMAT I CS

The two ends of a uniform chain of length A are on a rough horizontal rod, one end of the chain being fixed and the other end being free to move. Prove that, if [.L be the coefficient of friction, the chain will rest in equilibrium provided that the distance between the ends is less than A[.L sinh-1 (1/[.L). (L.U.) 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. (M.T.) A uniform, heavy, flexible string lies along the inside of a rough curve in the form of a catenary with its axis of symmetry vertical and con­ cavity upwards. Show that when the string is in limiting equilibrium ew/> cos .p has the same value at the two ends, where [.L is the coefficient of friction and .p is the inclination of thy tangent to the horizontal. (M.T.) A uniform chain of weight w per unit length is in equilibrium in a vertical plane in contact with a smooth surface. If Oy is vertically upwards and Ox is chosen appropriately, show that the tension T at any point (x, y) on the chain is given by T wy. Deduce that if the chain hangs in equilibrium over a smooth cylinder its ends are at the same level. A uniform string has its ends fixed to two points on the x-axis and rotates about this axis with constant angular velocity, the shape of the string remaining unaltered during the motion and plane. If Yo is the greatest distance of the string from the axis of rotation and ex is the angle made by the ends of the string with this axis, show that cos y 2 Y o2 1 cos rf; where if; is the angle made by the tangent to the string with the x-axis at the point (x, y) . An arch of a cycloid is fixed in a vertical plane so that the line of the cusps is horizontal and the vertex downwards. A uniform chain rests on the cycloid, stretching from the vertex to a cusp. If the chain is on the point of slipping, show that the coefficient of friction [.L satisfies the equation ((..1.2 + l)ehr!L 3. Show that, in the case of the catenary of uniform strength, the weight w per unit length varies according to the law w w 0 cosh (s/A), where w0 is the value of w at the lowest point.

=

22.

=

23.

24.

[cH. 12]

(

ex)•

_

=

=

CHAPTER 1 3

DEFORMATION O F ELASTIC BODIES

13.1. Internal Forces and Deformations In Chapter 1 1, when considering the equilibrium of a system of forces acting upon a rigid body, we did not :find it necessary to be able to calculate the internal forces between individual particles. However, when designing a structure such as a bridge or a piece of machinery it is of the utmost practical importance that our knowledge of the internal-force system should be as complete as possible, for if these forces become excessive at some point this part will fracture and the structure will collapse. If a rigid body A comprises part of the structure, let 5 be an imaginary surface which divides A into two parts A 1 and A 2• A 1 is in equilibrium under the action of the external forces applied to it and the internal forces exerted across 5 by the particles of A 2 lying close to 5. This system of internal forces is equivalent to a force F acting through an arbitrarily selected point 0 and a couple M. Assuming the external forces to be specified, by writing down the six conditions (equations (ll.l) and (11.2)) for the equilibrium of A 1, we obtain sufficient equa­ tions to determine the components of F and M. This is known as the Method of Sections. By taking a number of sections of A and applying this method to each, we can often attain to a considerable degree of knowledge concerning the system of internal forces. However, since there are an infinity of possible force systems operating across any section 5, each equivalent to F and M, it is not possible to determine the internal force system in detail by this method. A complete knowledge of the internal forces operating within the body A can be obtained only by taking into account the elastic pro­ perties of the material from which it is formed. No body can be precisely rigid as we have supposed in Chapter I I . A body will always " give " to some extent when acted upon by external forces, i.e., it will suffer deformation. During the process of deformation, the weaker portions of the body will yield more readily than the stronger portions, and will accordingly throw on to them the onus of balancing the applied forces to some extent. The elastically strong portions will therefore tend to occupy regions where the internal forces are relatively intense . From such considerations as these, the dependence of the system of internal forces upon the elastic properties of the body is immediately apparent. In an elastic medium the deformation caused by the system of 309

310

A C O U R S E IN APPLI E D MATHEMAT I C S

[CH.

applied forces is termed the strain. The associated internal forces constitute the stress. From what has been said, it is clear that there is some relationship between stress and strain involving the elastic properties of the medium. The precise nature of this relationship will concern us in Section 13.8. However, before turning our attention to the general problem of the deformation of an elastic body of any shape, there is a simpler one-dimensional case which will serve to introduce the type of problem we shall be considering. This is the problem of the deformation of a straight bar of small cross-section.

13.2. Shearing Force and Bending Moment Any section of the bar being of small area, the stress across it acting upon one portion of the bar will usually be sufficiently represented by a force F acting at the centre of the area and a couple M. The force F s

s

s

cLP T s

(a)

(c)

(b) FIG. 13.1 .-Shearing Force and Bending Moment

is resolved into a component T in the direction of the tangent to the bar at the section P being considered and a component S along the normal (Fig. 13.1 (a)). By Newton's Third law an equal and opposite force and couple must act across the section upon the other portion of the bar. Since each portion of the bar is in equilibrium, the external forces applied to each must just balance the stress at the section. Thus, in the case of the right-hand segment in Fig. 13.1 (a), the sum of the com­ ponents of the external forces acting upon it in the direction PO must equal T. Similarly, the sum of the components in the direction

13]

D E FO RMAT I O N O F E L A S T I C B O D I E S

311

O P of the external forces acting upon the left-hand segment is also T. The position is then that these two parts of the bar are being pulled apart by equal and opposite external forces T and fracture at the section P is being prevented by the action of the components of stress T. T is accordingly termed the tension at P. If the sense of all the forces T are reversed, the bar is in a state of compression at P. We adopt the convention that T is positive when it represents a state of tension. By a similar argument, we deduce that the right-hand portion of the bar is subject to a net component 5 of external force in the downwards direction, whereas the left-hand portion is subject to a component in the upwards direction. These two components of external force are tending to shear and break the bar at P. 5 is therefore termed the shearing force at P. Suppose that the deformation of the bar is small and that the x-axis represents the line of the unstrained bar. If all the external forces lie in the plane of rectangular axes Oxy (usually vertical with Oy downwards) , so will the forces 5, T and the couple M. The shearing force will be taken to be positive if the components of stress 5 at the section are directed as shown in Fig. 13.1 (a).* The bar is then tending to shear in the manner shown in Fig. 13.1 (b). Finally, the external forces acting upon either segment of the bar will have a net moment about F sufficient to balance the couple located there. Thus, in Fig. 13.1 (c) , the external forces acting upon the right­ hand portion have a clockwise moment about P and the forces acting upon the left-hand portion an anti-clockwise moment. These two sets of external forces are tending to bend the bar at P after the manner shown in Fig. 13.1 (c) . M is therefore termed the bending moment at P. In the case of a two-dimensional system of external forces in the plane Oxy, the bending moment is positive if the couples located at P are in the senses indicated in Fig. 13.1 (a), and the bar is tending to bend as in Fig. 13.1 (c) .

A uniform beam AB, of length l and weight w per unit length, is sup­ ported at A and at a point C such that AC a (>!l). Show that the greatest bending moment, without regard to sign, occurs either at C or at a distance l(l - lf2a) from A. Hence show that the value of a for which the beam is (Le.U.) least likely to break is lfy2. By taking moments of the external forces applied to the whole beam about A and C, we find that the reactions X and Y at these supports are given by X = wl(l - lf2a), Y = wl22a.

Example 1.

* P,

=

Mathematical texts usually adopt our sign conventions for the stress forces at but take the positive x-axis in the reverse sense. This is contrary to the normal convention regarding the relative orientation of the x- and y-axes and, moreover, introduces negative signs into the equations and (U!,jl) below. For these reasons, we prefer our set of conventions.

(13.1)

[CH.

A C O U R S E I N A P PLI E D MAT H EM AT I C S

312

A C AP

A,

the If P is a point on the beam between and and distant x from of the beam are : (i) the reaction external forces acting upon the segment Taking moments X, and (ii) the weight wx acting at the mid-point of we calculate the bending moment there to be of these forces about � x � a. M = twx• - wl(l - lf2a)x,

P,

0

y

AP.

X

cB====�ct===�P====�==�A

B

P

w:x.

C, an additional external force Y acts upon AP

and lies between If and the bending moment i s M= = The graph of M against diagram below : M will be where

twx• - Xx - Y(x - a), a � x � l. tw(l - x) 2, x is now easily sketched and is shown in the found to have a negative minimum at x = oc,

oc = l{l - lf2a) . It is clear, therefore, that M is numerically greatest at x = a or x = at these points, M = tw(l - a) 2 and -!wl2{1 - lf2a) 2 respectively.

oc,

and,

M

Assuming that the tendency to break is greatest where the bending moment is greatest, we can reduce the chance of fracture by choosing a so that the largest bending moment is as small as possible. Now, as a increases from !l to l, tw(l - a) • decreases from twl2 to and !wl2 {1 - lf2a) 2 increases from to twl•. Hence the optimum value for a is such that !w(l - a) 2 = !wl2{1 - lf2a) 2, i.e., such that a = lfy'2.

0

0

13]

313

D E F O R M A T I O N O F E LA S T I C B O D I E S

Now consider the case of a beam supported horizontally along the x-axis and loaded, w being the load per unit length at a point P of coordinate x. The element dx of the beam is in equilibrium under the action of the forces shown in Fig. 13.2. Resolving vertically, we obtain or

wdx + 5 = 5 + d5, d5 = dx W.

(13.1)

Taking moments about P', we find that

5dx + M + twdx2 = M + dM,

and, neglecting second-order terms, this yields

dM = 5 dx

(13.2)

·

s + c/ 5

t

M

wd x

5

y FIG. 1 3.2.-Derivatives of Shearing Force and Bending Moment

If at a point Q on the beam there is a concentrated load W, by considering the element with Q at its centre and resolving vertically, we obtain the equation

W + wdx + 5 = 5 + d5, i.e., as dx --+ 0, d5 --+ W This result implies that there is a dis­ continuity in 5 at Q of magnitude W. Taking moments about Q, we obtain M + dM = M + !5dx + !(5 + d5)dx, dM = 5dx + !Wdx, or i.e., as dx --+ 0, dM --+ 0. The bending moment is accordingly continuous across a concentrated load. The vertical reaction R at a support may be regarded as a concen­ trated load -R. We deduce then that there is a discontinuity in the shearing force at a support of magnitude minus the reaction and that the bending moment is continuous at such a point.

314

A C O U R S E IN A P PLI E D M A T H E M A T I C S

[CH.

Two uniform, heavy beams AB, BC, of lengths 4a, 2a respectively, and each of weight w per unit length, are smoothly hinged at B and rest in a horizontal position on three supports at A , D, C, where AD 3a. Find the reactions at the supports and draw the bending moment and shearing force diagrams. (Le.U.) Let X, Y, Z be the reactions at the supports as shown and suppose x is measured from A. To the right of A the shearing force 5 is zero and hence, immediately to the left of A , 5 = -X. Along AD, we have d5 IE: = w (l")

Example 2.

=

and hence, integrating,

5 = wx - X, 0 < x < 3a.

z

X

y

J

],

B 0

(ii)

I

1

It follows from this last equation that immediately to the right of D, and hence, immediately to the left of this point, 5=

3wa - X,

=

5 3wa - X - Y.

Equation (i) is also valid on DC. Integrating and employing the last result to determine the constant of integration, we obtain (iii) = < < From equation (iii), the value of 5 immediately to the right of C is found Its value to the left is accordingly to be Z. But clearly this must be zero. Hence Z= (iv) as is otherwise apparent by resolving vertically for the whole beam. For the bending moment along AD

5 wx - X - Y, 3a x 6a. 6wa - X - Y. X + Y + 6wa,

and thus, integrating,

�� = 5 = wx - X

6wa - Y -

(v)

M= (vi) the constant of integration being zero, since M is continuous at A and is obviously zero .to the right of this point. Similarly, along DC, we find that M= constant. But, from (vi), M = at D and M is continuous at this point. Thus the constant of integration is and M (vii) Now M must be zero at B and C and hence = =

twx• - Xx, 0 � x � 3a,

=

twx• - (X + Y)x + �wa2 - 3Xa 3 Ya twx• - (X + Y)x + 3Ya, 3a � x � 6a. 8wa2 - 4a(X + Y) + 3Ya 0, 18wa2 - 6a(X + Y) + 3Ya 0.

D EFORMATION OF ELASTI C B O D I E S

13]

315

We calculate that = = and then, using (iv) , Z = The shearing force and bending-moment diagrams can now be drawn from equations (ii), (iii), (vi) and (vii) . This is left as an exercise for the reader.

X wa, Y 4wa

13.3.

wa.

Extension of a Bar

Consider a straight bar which is put into a state of stress by external forces applied at points on the bar and directed along it. The shearing force and bending moment will be zero, and the stress across any section of the bar will be represented by a tension T. If a. is the cross­ sectional area of the bar we shall assume that the stress T is distributed uniformly over it and measure the stress by -r = Tfa., i.e., the force per unit area. The bar will be strained and will increase in length. We shall measure the strain at any point on the bar by the length increment per unit length at the point. Thus, if 0 is a fixed point on the bar, P is any point on it and OP = x, when the bar is in its natural state, and OP = x + u when the bar is extended, then u will be a function of x. Let Q have coordinate x + dx when the bar is not extended. When the bar is strained, Q has coordinate

x + dx + Ux

+ dx

= x + dx + u + du dx dx

to the first order and the extended length of the element PQ is seen to be

du dx. dx + dx

Thus, the increase in length experienced by this element is the strain is therefore dufdx.

Hence

e = du -· dx where e is the strain or extension as it is termed in this case.

�: dx and (13.3)

The results of experiments with actual bars formed from elastic materials are summarized by the statement that the stress is pro­ portional to the strain (Hooke's Law). Thus

E = -rfe,

(13.4)

where E is a constant for the material called its Young's Modulus. e is dimensionless, so that E has the same dimensions as -r and is a force per unit area. E is very large for most structural materials, e.g., its value for steel is about 2 X 109 gmfsq em. It follows that the deformations resulting from the application of forces to such materials are very minute.

A C O U R S E IN A P P L I E D MAT H E M A T I C S

316

[cH .

Hooke's Law may be written

du dx

T

£;,

(13.5)

Hence, if the tension T is constant, as for example in the case of a bar stretched horizontally by equal and opposite forces applied at its ends, and the cross-section is uniform, we obtain by integration

u = TxfEIX, point 0. Equation

(13.6)

since u = 0 at the fixed (13.6) shows that all points on the bar suffer displacement proportional to their distances from 0.

A straight bar of material of constant density p and Young's Modulus section at a point P distance x below the point of suspension and u is the vertical displacement of P due to the tension in the bar, show that (1!. + d"u)jr!!!: dxt!_ (log rx) = - E dx2 dx . If the extension is constant along the bar, show that IX decreases exponentially below the point of suspension. Neglecting the small vertical displacements u of the particles of the bar, the element dx at a distance x below the support is in equilibrium under the action of a tension T upwards and a tension T + dT together with a weight prxdx downwards. Hence dT + prxdx = 0, dT dx = -piX. But equation (13.5) is valid for the bar, and hence, eliminating T between this equation and the one just obtained, -piX = dT = !_ ( EIX du ) , dx dx dx drx du - prx = EIX d2u dx2 + E dx ([;/ Hence !� (J! d"u)j IX dx = E + dx2 r!!!:dx, which is equivalent to the result required. If dufdx = constant, d2ufdx2 = 0, and hence l diX "' dx = constant = -k. Integrating with respect to x, we find that log IX = -kg + constant, ex = a:0e-kx, or where "' = at x = 0.

E is suspended from a point on it to hang vertically. If"' is the area of the cross­

Example 3.

_

"'o

13.4. Flexure of a Beam Suppose that a straight beam of small cross-section is subjected to external forces and bending takes place. We shall suppose that,

13]

D E F O RMAT I O N OF E LA S T I C B O D I E S

3 17

after deformation, any two plane right sections of the beam, originally parallel, remain planes and meet in a line l whose direction is inde­ pendent of the particular pair of sections being considered. If p is a plane perpendicular to l, the bending is said to be parallel to p. Consider two adjacent right sections of the beam which, after flexure, are inclined to one another at an angle 6 and meet in the line L (Fig. 13.3) . Since the cross-sections are close together, we shall

r . :§ .· '

I

.�

'

'' ''

'' ''

' ,

·.

'

�

,

R

j

·.' e : '

• L

L

FIG. 13.3.-Flexure of an Element of a Beam

treat them as congruent figures. Let P, P' be corresponding points of the two sections. Then the straight line P P' will be normal to the planes of the sections before flexure. The particles of the beam lying upon this line will form an arc of a curve after deformation, and we shall further assume that the arc P P' intersects the displaced planes of section orthogonally. This is equivalent to the assumption that there is no shearing of adjacent sections. Let 00' be a particular arc PP' which suffers no elongation when the beam is bent. This arc may be external to the beam, but, for reasons given below, is usually internal. We now take rectangular axes Oxy in the plane section containing 0. Oy is taken to be perpendicular to L. It will be clear that our assumptions concerning the state of strain within the beam imply that 0 could be replaced by any point on the x-axis. This axis is termed the neutral line of the section. Let P have coordinates (x, y) . Then, if OL = R, P is distant y + R from L and, after flexure, arc PP'

=

(y + R)6.

(13.7)

But, before the beam is strained, the two plane sections are parallel and P P' = 00' = R6. Thus P P' undergoes an elongation y6 and

318

A C O U RS E IN A P P L I E D MAT H E M A T I C S

[CH.

its extension is accordingly y6/R8 = yjR. Assuming that Hooke's Law is applicable to the filament P P', the stress at its ends will be given by " = Ey/R, ( 1 3.8) ' where E is Young s Modulus for longitudinal extension of the material of the beam. We now know the distribution of stress across the section of the beam and can compute the tension, shearing force and bending moment. Suppose we choose the y-axis to pass through the centroid G of the cross-section. Let (0, y) be the coordinates of G. Let doc be a small element at P of the cross-section rx. The force acting upon doc is 'l'drx = Eydrx/R in a direction normal to oc. Replacing all such forces by parallel forces acting at G together with the appropriate couples, we show that the total stress is equivalent to a force T

=

l

l!_ ydrx R "

(13 .9)

acting at G in a direction normal to oc and couples M=

E

R

l!(Y - y)drx,

i

E N = - j[ "' xydrx,

(13. 10)

with axes parallel to Ox and Oy respectively. By ,definition, T is the tension and the resultant of the couples M and N is the bending moment at oc. According to this approximate theory, the shearing force is zero, a result to be expected since we have assumed that there is no shearing strain. We shall always be concerned with beams for which any section is symmetrical about Oy. Then N = 0 and the bending moment couple lies in a plane parallel to the plane of bending. G being the centroid of oc, rxy = T

and hence Also M

=

Erxy/R.

( 13.1 1) (13.12)

� 1 (y - y)2drx + 11 (y - y)drx,

(y - y) 2drx, i1 EI/R,

= =

=

1"' ydrx

(13.13)

where I is the second moment of area of oc about an axis parallel to Ox through G. If T = 0, or is negligible, as it will be in all examples we shall give

D E F O R MA T I O N OF E L A S T I C B O D I E S

13]

319

below, then y = 0 and G lies on the neutral line of the section. R is then the radius of curvature of the locus of centroids of cross-sections of the beam. Hence, if K is the curvature of this locus, EIK = M. .

(13. 14)

EI is called theflexural rigidity of the beam and, assuming this constant along the beam, the last equation shows that the curvature of the beam is everywhere proportional to the bending moment.

A light, straight spring of length l and flexural rigidity EI is clamped vertically at its lower end. Show that the weight which must be attached to its upper end to bend it through a right angle is approximately

Example 4.

3 ·438EI/12• Let be the weight which must be attached to the upper end 0 of .... the spring. Taking axes hori­ zontally and vertically down­ wards, we know that the curve of the spring is tangential to The bending moment at the point ( , is and hence, by equation (13.14) ,

W

:c

Oy

Ox.

P x y) Wx EIK = Wx.

Let .p be the angle made by the tangent to the spring at with Then, in the usual notation,

P

Ox. dl/J K =
Hence

C

�� ��

,

d.fo

cos "' dX = which, upon integration, yields 0

sm since .p =

--r-:;:-�---====---•0

-

Ox

0 when x = 0.

,/,

'f'

=

w

EI X'

w

2Ei

x2

'

This may be written

x• = k2 sin .p, k2 = 2EifW.

where

At the clamp C, .p = i1r and hence, from (i), Now the length of the spring is given by

x = k. • l = lds = {sec .fodx = {..;:4 : x4 11 y 1dt- 4, where t = xfk, =k t 0 k F(1/y2, 77/2), •

=

V2

1

(i)

320

A C O U R S E IN A P P L I E D MAT H E MAT I CS

[CH.

where F is the elliptic integral of the first kind. From tables, we find that F(1fy'2, 7r/2) = 1·8541 . Hence, substituting for k in the last equation and solving for we obtain

W,

w

= 3·438 72 ·

EI

In the particular case when the flexure of the beam is small we shall take an x-axis along the line occupied by the mean centres of the cross­ sections in the unstrained state. After deformation the line of mean centres will be a curve y = f(x) , whose gradient y' will be a small quantity. Hence, neglecting the square of y', we have for this curve 3 K = y" /(1 + y'2) /2 = y" and thus, by equation (13.14) , Ely"

=

M.

(13.15)

If M is known as a function of x, two integrations will now yield the approximate deformed position of the beam. Alternatively, employing equations (13.1) and (13.2), we have

(

dzy d2 EI dx2 dx2

)

=

d2M w dx2 =

(13.16)

and, if the load w and flexural rigidity EI are known at all points along the beam, by four integrations we can obtain the equation of the dis­ placed beam. In particular, if w and EI are constants, this equation is of the fourth degree.

A uniform rod ABC, of weight w per unit length, is supported at its ends A, C and at a point B of its length, the three points ABC being at the same level. If AB a, BC = b, show that the reaction of the support at A is w(3a2 + ab - b2)f8a. Hence show that the rod can remain in contact with all the supports only if y'1 3 - 1 a y' 1 3 + 1• �--�

Example 5.

=

6

z


2

y

(D.U.) X

Z be the reactions at the supports respectively. Then Let the bending moment at all points of the beam is given by

X, Y,

A, B, C 0 � x � a, M = !wx2 - Xx, = !wx• ...,. Xx - Y(x - a), a '(; x � a + b .

D E FO RMAT I O N O F E L A S T I C B O D I E S

13]

321

Substituting for for in

M in equation (13.15) and integrating twice, we obtain x (0, a) Ely = .\wx4 - tXx" + Ax, since y = 0 when x = 0. Now y = 0 when x = a. Hence A = tXa2 - 214wa3• But Ely' = twx3 - !Xx2 + A, Ely' = twa• - tXa2• and so, a t x = a, Since y' is everywhere continuous, this gives its value immediately to the left of B. Hence integrating equation (13.15) once, we now find that for x in (a, a + b), Ely' = twx• - !Xx2 - !Y(x - a)• - i.;wa8 + tXa2• A further integration yields Ely = -.\wx4 - tXx3 - tY(x - a)" + (tXa• - 214wa3)x, the constant of integration being zero, since y = 0 when x = a. But y = 0 at x = a + b and hence ,'.,w(a + b)' - tX(a + b)3 - tYb3 + (tXa2 - ;\-wa8)(a + b) = 0, X(a + b)(2a + b) + Yb2 = !w(a + b)(3a2 + 3ab + b2). (i)

Taking moments about C for the external forces applied to the whole beam, we obtain (ii) (i) and (ii) can now be solved for and to give

X(a + b) + Yb = !w(a + b)2• X Y X = w(3a2 + ab - b2)f8a, Y = w(a + b)(a2 + 3ab + b2)f8ab. Also, since clearly X + Y + Z = (a + b)w, Z = w(3b2 + ab - a2)f8b. therefore This result is also obtainable from the corresponding result for X by inter­ change of a and b. For the rod to maintain contact with all supports, X, Y, Z must all be positive. Y is clearly positive. X and Z are positive provided 3m2 + m - 1 > 0, 3 + m - m2 > 0, where m = afb. The first inequality requires that m < -v'136 - 1 or m > v'136 - 1 •

The second inequality is satisfied provided

1 - v'13 < m < 1 + v'13• 2 2 Since m must be positive, both inequalities are satisfied provided that v'13 - 1 < m < v'13 + 1 6 2 Example 6. Assuming that the bending moment at a point of a slightly bent, thin beam is ElfR, where El is the flexural rigidity and R the radius of curvature at the point, prove that, if both ends of a thin, uniform strut, of length l, are clamped, the greatest thrust which it can exert without bending is 47r2Elfl•. (Li.U.) •

M

(1)

322

A C O U R S E IN A P PL I E D MAT H E MAT I C S

[CH.

Suppose that the strut bends slightly under the action of inwardly directed forces at the clamps. Let the system of forces at each clamp be equivalent to a force and a couple as shown. We shall neglect the weight of the strut, so that the forces act along the line of the clamps. Taking axes as shown, the bending moment at a point on the strut is given by

P

P

G

Q(x, y)

M = G - Py.

Hence, by equation

(13.15), Ely" = G - Py, y" + n•y = GfEI, or where n2 = PfEI. The general solution of this equation is y = A cos nx + B sin nx + GfP. But y = y' = 0 at x = 0 and x = l. Hence A + GfP = 0, A cos nl + GfP = 0, -A sin nl = 0, B = 0. If A = 0, then G = 0 and the beam is not bent. Thus, for bending, we must have A = -GfP, B = 0, sin nl = 0, cos nl = l. The last two equations imply that nl = 2k1r, where k is an integer. Hence P = 4k2'"2Elfl2• Taking k = 1, we find that the smallest value of P for which bending can take place is given by P = 4'"2EI/l2• It may be remarked that, if P exceeds this value, the bending becomes pronounced and the approximate theory, which is limited to small deflections of the beam, is no longer applicable. The value calculated for P is accord­ ingly the critical load which can be supported by the strut. 13.5. Whirling Shafts

If a straight, thin shaft can rotate in bearings 0 and X (Fig. 13.4) , when its angular speed w reaches a critical value .n the phenomenon of whirling is observed. That is, the shaft exhibits a deflection from its natural straight line and, if w is further increased, this deflection will become pronounced and the shaft will break. The shaft cannot therefore be employed for the transmission of angular velocities in excess of .n. To calculate n, we shall suppose that at this speed the deflection

D E F O RMAT I O N O F E L A S T I C B O D I E S

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323

is everywhere small, that the relative shape of the shaft remains un­ altered during the rotation and that the shaft always lies in a plane. We shall neglect the weight of the shaft. Introducing a centrifugal force which acts upon each element, we reduce the problem to one of statics as explained in Section 11.4. Taking axes Oxy in the plane of the shaft as shown in the figure, the element at P(x, y) describes a circle of radius y. Taking ds to be the length of the element and m the mass per unit length at this point, we calculate that the centrifugal

FIG. 13.4.-Whirling Shaft

force is mdsw2y in a direction parallel to Oy. It follows that we may regard the shaft as at rest and deflected by a load mw2y per unit length. Equation (13.16) is applicable, yielding

d2 ( EI d2y) - mw2y. (13.17) dx2 dx2 In the particular case when EI and m are constant along the shaft,

this equation reduces to

where k4

=

mw2jEI.

d4y dx4 - k4y, .

(13.18)

The general solution of this equation is

y = A cosh kx + B sinh kx + C cos kx + D sin kx, (13.19)

A B, C , D being constants of integration to be determined by the boundary conditions. As shown in the example below, these boundary conditions can be satisfied only if the value of k is chosen appropriately. This circumstance fixes the value of w for whirling. ,

A uniform, solid, circular shaft of length l and diameter d rotates in collinear bearings at its ends, neither of which exerts any directional constraint. Show that the whirling speeds of the shaft are the values of w which satisfy w2 = (TT4Ed2f16pl4) X (14, 24, 34, .), where E is Young's modulus and p is the density of the material of the shaft.

Example 7.

(Li.U. ) Since the bearings impose no directional constraint on the shaft, the bending moment is zero at = and = Hence, by equation = at these points and the boundary conditions are y= = at x = 0 and =

y" 0

x 0 y" 0

x l.

(13.15),

x l.

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A C O U R S E I N A P PL I E D M A T H EMAT I C S

[CH.

Substituting in equation (13. 19), we find that these equations imply

A= B sinh kl + B sinh kl -

If B =

C = 0, D sin kl = 0, D sin kl = 0.

D = 0, the shaft is not whirling.

Hence, for whirling, either sinh kl = 0 or sin kl = 0. Since kl =1= 0, the first alternative cannot be accepted. The only possibility therefore is that kl = n1r, (i) where n is an integer. Now m = !1rpd2 and = 7Td4f64. Hence

I

k4

=

16p w2/Ed2•

Substituting in (i), we obtain w2 = n41T4Ed2f16pl4,

which is the result stated. Taking n = 1, we obtain the maximum speed n at which the shaft can be safely driven, viz.,

13.6. Analysis of Strain

Let OX1XJ(3 be a fixed, rectangular, cartesian frame S in the vicinity of an elastic body (Fig. 13.5 (a)) . When the body is deformed by the application of forces, suppose the particle P, originally at the point (X1 , X2, Xa) is displaced to the point (X1 + Uv x2 + u2, Xa + ua)· The quantities (u1, u2, Ua) will be functions of (Xv x2, Xa) . for otherwise the body has simply suffered a translation and has not been strained. Let Px1x2x3 represent a rectangular frame s with its origin at P and its axes parallel to those of S. We shall suppose that s moves with P as the body is deformed. Let Q be a particle of the body near to P having coordinates (x1 , x2, x3) relative to s before the body is strained. (xv x2, x3) will be assumed to be so small that their squares may be neglected. Prior to the displacement, Q has coordinates (X1 + Xv x2 + x2, Xa + Xa) relative to s and hence, after deformation, its coordinates are : xl + x1 + u1 (X1 + X1 , x2 + X2 Xa + Xa) . x2 + x2 + u2 (X1 + x1 , x2 + x2, Xa + Xa) . Xa + Xa + Ua (X1 + x1 , x2 + x2, Xa + Xa) . ,

or, to the first order,

au1 aul + Ul + aX Xl + aX Xz + 1 2 au2 au2 X X2 + X2 + U2 + ax X1 + aX2 2 + 1 au3 au3 Xa + Xa + Ua + ax xl + ax x2 + 1 2 X1 +

Xl

au1 X ax3 a , au2 X aX3 a , au3 X ax3 a .

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D E FO RMAT I O N O F E L A ST I C B O D I E S

325

where ul> u2, u3 and their partial derivatives are to be calculated at the point (Xl> X2, X3) . At the same time P has coordinates (X1 + ul> x2 + U2, x3 + u3), and therefore the coordinates of Q relative to s are now a aul aul + ul x3, x + 1+ ax1 1 ax2Xz ax3 au2 au2 au2 + 1 + BX ( 1 3.20) 2 + aX3 Xa, X aX1Xl 2 a au3 a + u3 X2 + 1 + u3 X3 . aX1Xl aX2 aX3 An observer situated at P and moving with this particle will ac­

(

) (

) (

)

-+

cordingly detect a change in the components of the vector PQ, joining

}l'_p'---- "-.

q'

X,

x,

0

x,

(b) (a) FIG. 13.5.-Deformation in the Neighbourhood of P

the particles P and Q, from the values (xl> x2, x3) to the values given at -+

(13.20) , caused by the strain. Let PQ' be the displaced position of this vector (Fig. 13.5 (b)) . Then the displacement of the particle Q, -+

as observed from P, is given by the vector QQ'. -+

-+

Since

-+

QQ' = PQ' - PQ, the components (8x1 , 8x2, 8x3) of this displacement will be the difference -+ -+ between those of PQ' and PQ. Thus where a;j = au;jaXi.

8x; =

3

,L l a;jXj, (i = 1 , 2, 3),

j�

(13.21)

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A C O U R S E I N A P P L I E D M AT H EM A T I C S

[CH.

If the a;j are constants, the characteristics of the strain in the vicinity of any pair of particles of the body are the same and the strain is said to be homogeneous. Denoting the column matrices

(xx2x31) , (lllxllxx312) l

by X and

llx respectively

and the 3 X 3 matrix (a;j) by A, the equation (13.21) can be written in the form llx = Ax. (13.22) Let (x1 ', x2', x3 ') be the coordinates of Q' relative to the corresponding column matrix. Then x'

=x+

s

and let x' denote

llx = x + Ax = (A + I)x,

(13.23)

where I is the unit matrix. Equation (13.23) relates the positions of Q relative to P before and after deformation. Now suppose that a second strain, characterized by a matrix B, is given to the body. Let x" determine the new position of the particle Q relative to P. Then, as in the case of equation (13.23) for the first strain, x'' =

(B + I)x'

(13.24)

Combining equations (13.23) and (13.24), we find that x"

= (B + I) (A + I)x = (BA + A + B + I)x, .

(13.25)

i.e., the net effect of the two strains is that of a single strain of matrix BA + A + B. Had the strains been given to the body in the reverse order, the resultant strain would have corresponded to the matrix AB + A + B. Since, in general, AB -=!= BA, the resultant effect of two strains is not independent of the order in which they are applied. If, however, each strain causes displacements of the particles of the body which are very small, then the quantities u; are small, and hence the elements of A and B are small. In these circumstances we can neglect AB and BA, and the resultant strain is determined in each case by the matrix A + B( = B + A) . For small deformation, therefore, the order in which a series of strains is given to an elastic body is of no consequence and the net effect, as determined by the matrix of the strain, is the sum of the effects of the individual strains applied separately. The Principle of Superposition is accordingly applicable, and this possibility greatly simplifies the solution of these problems. From this point onwards, all strains will be assumed infinitesimal. In any strain a certain component of the displacement of Q relative to P can be represented by a simple rotation of the elastic b0dy without deformation. This component is of no interest to us in this section, since it will not be responsible for the stresses engendered within the

13]

D E F O RMAT I O N O F E L A S T I C B O D I E S

327

body by the strain. We shall first eliminate this component therefore. Let us write equation (13.21) in the form 3

where We note that

3

ox; = j!= l C;jXj + j != l e;jXj, c;j = �(a;j - aj;), eij = �(a;J + aj;).

(13.26) (13.27) (13.28) (13.29)

e;j is called the symmetrical component of a;j and C;j the anti-sym­ metrical component. It is clear that c11 = c22 = c33 = 0. Hence, putting 81 = c23, 82 = c3 1, 83 = c1 2 , we can write jL= l C;jXj = (8aX2 - 82X3 , 81X3 - 8ax2, 82X1 - 81x2). (13.30) If the body were given a small rotation a = (81, 8 2 , 83 ) about P, the resulting displacement of Q, viz., a X r where r = (xi> x2 , x3), would 3

•

have the components appearing in the right-hand member of equation (13.30). The contribution to the displacement of Q of the first sum in the right-hand member of equation (13.26) accordingly corresponds to a rigid-body rotation about P and could be eliminated by suitably rotating the frame s as the deformation developed. We shall assume that this is done and then

3 oX; = j L= l e;jXj. (13.31) This equation represents a pure strain, and the coefficients e;j are termed the components of strain at the point P. Since e;j is symmetric, only six of the components can take arbitrary values. Equation (13.31) can be written in the matrix form

ax = Ex. If x' determines the position of Q relative to taken place, then x' = (E + I)x.

(13.32)

s after deformation has (13.33)

This transformation from the coordinates X; to the coordinates x/ is linear, and it follows, therefore, that in the vicinity of P, where the components of strain are appreciably constant, straight lines and planes are transformed into other straight lines and planes respectively. If the strain is homogeneous, this will be the case over the whole solid. E can be decomposed into a sum of six matrices E; (i = 1 , 6), in each of which all but one of the components of strain are zero (e12 , e21 counting as one component, etc.) . The general strain represented by equation (13.32) can then be obtained by the

A C O U RS E IN A P P L I E D MATHEMAT I C S

328

[CH.

superposition of the six strains with matrices Ei. We will consider each of these elementary types of strain separately. Firstly, suppose e11 -=fo 0, but all other components of strain are zero. In this case (13.34) I.e., all displacements are parallel to the xcaxis and proportional to the particles' distances froi:n the x2x3-plane. This is a simple extension of magnitude e1 1 in the direction of the x1-axis. Any straight line parallel to this axis has its direction unchanged and its length altered from l to l(1 + e11) , and a line perpendicular to the axis is simply displaced laterally without any alteration. Similarly, if e22 -=fo 0 (or e33 -=jo 0) and all other components vanish, there is an extension in the direction of the x2-axis (or x3-axis). Secondly, suppose e12 = e21 -=fo 0, but all other components vanish. Then equations (13.31) read : (13.35) We observe that there is no displacement in the direction of the x3-axis. Consider the rectangle PABC (Fig. 13.6 (a)) . Let B be the point ""•

s'

p

A

(a)

.x:,

A

P

FIG. 13.6.-Pure Shearing Strain

(b)

(xv x2, 0). According to equations (13.35), the displacements suffered by the points A, B, C upon deformation are (0, e21Xv 0), (e12x2 , e21xv 0), (e12x2, 0, 0) respectively. It follows that the parallelogram PA 1 B1C1 represents the rectangle after deformation. Now AA 1 = e21x1 • Hence LA 1 PA = e21• Similarly, LC' PC = e12 = LA 1 PA. If, after deformation, the parallelogram is rotated as a rigid body through an angle -e12 about P (a rotation which does not affect the state of strain), the original and strained positions of the rectangle will be as shown in Fig. 13.6 (b). The strain has been supposed small, so that, to the first order, B1C' lies along BC. The type of deformation shown in Fig. 13.6 (b), in which every particle is displaced parallel to a fixed plane, its

13]

D E F O RMAT I O N O F ELAST I C B O D I E S

329

displacement being proportional to its distance from the plane is called a shearing strain. If e is the angle of shear, then e = 2ei2 · We can now state that any strain is equivalent to a succession of three extensions in the directions of three coordinate axes and three shears parallel to the three coordinate planes. However, by proper choice of axes, the shears can be eliminated. For consider the matrix equation (13.32) characterizing a pure strain. Let Px1*x2*x3* be new rectangular axes obtained from the axes Px1x2x3 by a rotation. Let the column matrix x* determine the new coordinates of a point whose original coordinates were given by x. Then the coordinate transformation can be written

x = Jx*,

(13.36)

llx = Jllx*,

(13.37)

where J is an orthogonal matrix. Also

where llx* defines the displacement of the point x* relative to the new axes. Substituting in equation (13.32), we obtain

Jllx* = EJx*, (13.38) or, multiplying on the left by J-1, llx* = J-1EJx*. (13.39) J-IEJ is the matrix specifying the components of strain relative to the new frame of reference. It is a well-known result of algebra that it is always possible to find an orthogonal matrix J such that J-1EJ is a

diagonal matrix. t

We shall accordingly suppose that

(13.40) and equations

(13.39) then read (13.41)

We observe that the components of strain corresponding to shears have vanished, with the result that the strain is now exhibited as a succession of three extensions parallel to the axes. The directions of the axes Px1 *, Px2 *, Px3* are termed the principal directions of strain at the point P, and e11*, e22*, e33* are called the principal strains at P. It is clear that any straight line through P along one of the principal directions has its direction unaltered by the strain. t See, e.g., mann) .

A University Algebra, by D.

E. Littlewood, Chapter III (Heine­

330

A C O U R S E IN A P P L I E D MATHEMAT I C S

[CH.

A body is deformed by internal stress so that the particle which in the strained state is at the point (x, y, z) referred to a set offixed rectangular axes has undergone the displacement whose components are 3 K(2x - y + z), - 3 K(X + y), K(3x + 5z), where K is a small constant. Show that there is no rotation of any small portion of the body in the neighbourhood of (x, y, z), that one principal extension is 3 K, and determine the other two principal extensions. (M.T.)

Example 8.

(

In the notation we have been employing, A= 6 -3 -3 3

-3

0

3

0 5

)

and, this matrix being symmetric, it represents a pure strain without rotation, and may therefore be denoted by E. The latent roots of E are the roots of the equation -3 3 - >.

16

-3 -3 3 o These prove to be >. = -4, 3, 9.

>. 0 I 5 - >. I

=

o

.

The diagonal matrix to which E can be reduced by an orthogonal trans­ formation is, in this case, therefore K -4

(

0 0) 0 3 0 0 0 9

Hence the principal extensions are -4K, 3 K, 9 K. The first, being negative, represents a contraction.

13.7. Analysis of Stress Let P be a point in an elastic medium in a state of stress and let dS be any small element of area containing P. Then the particles of the medium to one side of the element will exert forces upon the particles to the other side. This system of forces is equivalent to a force F acting at P and a couple G. The moment of G will be equal to the sum of the moments of the forces about P, and GjdS will tend to zero as the dimensions of dS are diminished to zero. We shall therefore assume that dS is sufficiently small for GjdS to be neglected. F will not, in general, be normal to dS. The force per unit area acting across dS is F/dS and the limit of this vector as dS -+ 0 defines a vector T called the stress vector at P. Thus T = lim (FjdS).

dS -+ 0

(13.42}

T clearly depends not only upon the point P, but also upon the orienta­

tion of the element dS. Let Ox1x2x3 be a rectangular, cartesian frame of reference in the vicinity of the elastic body and let 't'1, 't'2, 't'3 be the stress vectors at P(x1, x2, x3) corresponding to elements dS which are normal to the x1-, x2- and x3-axes respectively. In each case it is conventional to

D E FO RMAT I O N OF E L A S T I C B O D I E S

13]

331

assume that the stress vector represents the force exerted across the element by the particles on the side opposite to 0 on the particles on the same side as 0. By Newton's third law, the forces exerted across the element in the reverse sense are equal and opposite to these. We again emphasize that these stress vectors are not necessarily in the directions of the axes. Let the components of -t1 in the directions of the axes be "w -r12, -r1 3 respectively. We resolve -t2, -t3 into their com­ ponents according to the same notation. Thus "ii is the xrcomponent of the stress vector -t;. The nine quantities "ii are called the com­ ponents of stress at P. We shall show that, when these are known,

A

---� x,

�0-

:x.,

FIG. 1 3.7.-General Stress Vector and Components of Stress

the stress vector corresponding to any element dS at P can be calculated (equation (13.47)) . They therefore completely determine the state of stress at P. The components -r11, -r22 , -r33 are along the normals to the respective plane elements across which they act and are referred to as the normal components of stress. The reader should verify that if "u is positive, the elastic material is in a state of tension in the direction of the x1-axis. A negative normal component indicates compression. The remaining components act in the planes of the surface elements and are termed the shearing components. Consider a small tetrahedron of material PABC with an apex at P and the faces meeting at P parallel to the coordinate planes (Fig. 13.7). The face ABC can have any orientation. Let (lv l2 , l3) be

332

[cH.

A C O U R S E I N A P P L I E D M A T H E MAT I C S

the direction cosines of the normal to ABC, if it is directed away from P. Regarding ABC as a small plane element, let T be the correspond­ ing stress vector for the forces exerted across the element by the matter outside the tetrahedron upon the matter inside. If the area of ABC is � . the net force exerted across this element is (13.43) where (T1 1 T2, T3) are the components of T in the directions of the axes. Now area BPC

= 11� , area CPA = 12�, area APE = 13� ,

(13.44)

and hence the forces acting across these faces upon the material within the tetrahedron are - 11�-r:1 , - 12�-r:2, - 13�-r:3, and these have com­ ponents as follows : Face BPC : (-11�"w -11�"12, - 11�'"�"d · Face CPA ( -12� '"�"21, -12� '"�"22, -12�'"�"12) . Face APB : ( - 13�"av -13�'"1"32, -13�'"1"33) .

}

(13.45)

Finally, the material within the tetrahedron may be acted upon by a " body force ", e.g., gravity. We shall assume that this is proportional to the mass of the material within the tetrahedron . Thus, if p is the density of the material at P and p is the perpendicular distance of P from the face ABC, this force can be taken to be }pp �g. Assuming that the tetrahedron is in equilibrium, by resolving all forces acting upon it in the direction of the x1-axis, we obtain the equation Tl� - 11�"11 - 1z�"zt - 1a�'"�"31 + PP�gl

= 0, ·

(13 . 46)

where g1 is the x1-component of g . As p -?- 0 and the tetrahedron shrinks to zero in all its dimensions, this equation becomes quite accurate in the form T1

= "n11 + '"�"2112 + '"�"311s .

Similarly, by resolving along the remaining two axes, we obtain

"12l1 + '"�"22l2 + "azla, Ta = '"�"1ali + '"�"2al2 + "aala.

T2 = I.e.,

T; =

3

j

!= l "iili, (i = 1 , 2, 3) .

(13.47)

This equation permits the calculation of all components of T when the components of stress are known. Let P be at the centre of a rectangular parallelepiped of elastic material with its edges parallel to the axes (Fig. 13.8). We shall

13]

D EFORMATION OF ELASTI C BOD I ES

333

suppose the parallelepiped so small that the state of stress within it can be assumed uniform. Then the forces exerted on any face of the parallelepiped by the surrounding material will have a resultant acting at the centre of the face and the forces acting on opposite faces will be equal and opposite. In the figure the components of these forces parallel to the axes have been shown for three orthogonal faces of areas Av A 2 , A s . We shall assume that the " body forces " acting

.x.,

FIG. 13.8.-Symmetry of the Components of Stress

upon the parallelepiped have a resultant acting through P. The parallelepiped is in equilibrium. Hence, taking moments about an axis through P parallel to Ox1, we obtain

a2, as being the lengths of the edges parallel to the x2- and Xs-axes respectively. Now A 2a2 , A sas both equal the volume of the parallele­ piped. It follows from the last equation therefore that (13.48)

Similarly

The number of independent components of stress is accordingly six only (cf. components of strain). If T now denotes the column matrix having elements (Tv T2 , Ts), I the column matrix having elements (lv l2, l3) and U the symmetric matrix (-r;j) , equation (13.47) can be written

T = Ul. This equation is the counterpart for stresses of equation

(13.49) (13.32) for

[cH.

A C O U R S E I N A PPLI E D MATHEMAT I C S

334

strains and, as in the case of the latter equation, it may be simplified by a rotation of axes. t The result is the equation

T* = U*l* ,

(13.50) (13.51)

where

I.e., we can always choose axes such that the shearing components of stress vanish. Across an element normal to the x1*-axis therefore, the stress is itself normal to the element and of magnitude -r11 *. -r11 *, -r22*, -r33* are termed the principal stresses at P, and the new axes are called principal axes of stress. It follows from equation (13.50) that the stress across a plane element whose normal has direction cosines (11*, l2*, l3*) relative to the principal axes has components (T1*, T2*, T3*), where (13.52) To summarize, we have proved that the stress at any point of an elastic medium may be regarded as a combination of three pure tensions (or compressions) in three orthogonal directions. Example 9.

A t a point in an elastic medium the stress is a pure shear represented all other components vanishing. Show, by by the non-zero component rotating the axes through an angle of }1r about the x 3-axis, that a pure shearing stress is equivalent to a tensile and a compressive stress both of magnitude 7' 1 2 in perpendicular directions.

7'1 2,

The matrix U takes the form =

u

(

..�. T � )'

Suppose the axes are rotated about Ox3 through an angle o f }1r. Let Ox1*x 2*x 3* represent the new frame of reference. The displacement vector according to the equations x transforms into the vector

x* (x 1 * + x 2*)/y2, x 3 x 1 = (x 1* - x 2*) /y2, X 2 x* 3 • i.e., the orthogonal matrix J of the transformation is given by J = ( lfy2 -l/y2 0 ) l/v2 lfv2 o 0 0 l =

=

'

For an orthogonal matrix, J-1 Hence

J-1

=

=

i, where j denotes the transpose of J .

( - lfv2 lfv2 0

lfv2 lfv2 0

o o

l

)

.

T and 1 are both vectors (the latter a unit vector), and hence transform in the same manner as the displacement vector x in equation (13.36) .

t

1 3]

D E F O RMAT I O N O F E L A S T I C B O D I E S

(�

335

It now follows from equation (13.51) that 0 U* = J-1UJ = '1" 1 2 -?" 12 0

i.e., the principal stresses are 7"12, -7"1 2 , 0. This is what was to be proved. There is, of course, a corresponding result for pure shearing strain. Thus, a pure shearing strain e 12 is equivalent to an exhmsion e 1 2 in the direction Ox1* and a contraction e 12 in the direction Ox 2*.

13.8. Generalization of Hooke's Law In this section we shall be concerned with the mathematical relation­ ship between the stress within an elastic medium specified by the components "'ii and the resulting strain characterized by the com­ ponents eij · Taking principal axes of strain at a point P, the medium is extended in the directions of these axes by amounts ell, e22, e33 (we omit the asterisks henceforward). There is no shearing strain. Assuming that the medium is isotropic (i.e., its elastic properties are non-directional), there will be no shearing stress across a plane element normal to any one of these axes, for, on account of the symmetry, there is no reason why such a stress should exist in one direction rather than in another. The stress is accordingly normal to such plane elements, and we deduce that the axes are also principal axes of stress. Let "'n• "'22, "'aa be the principal stresses. According to Hooke's Law, stress is proportional to strain. We shall generalize this law by supposing that, provided the strain is not excessive, the principal strains are linearly related to the principal stresses. Thus, we write eu = a"'u + b("'22 + "'aa) ,

ez2 = a"'22 + b( + "'u) ' + b("'n + "'22l · = eaa

a"'aa

"'aa

}

(13.53)

The form assumed for these equations is justified by the assumption of isotropy, for e11 must be the same function of ( "'n • "'22, "'aa) , that e22 is of ("'22, "'aa• "'n) and that e33 is of ("'aa• "'n• "'22). Equations (13.53) indicate that the elastic properties o f an isotropic medium can be specified in terms of two elastic moduli a and b. Consider now the particular case of a material so stressed that the principal axes are in the same directions at every point and the principal stresses are ("'u • 0, 0), where "'n is constant. A cylindrical bar stretched longitudinally will approximate this state. The material is then subjected to a purely tensile stress. The principal strains are given by equations (13.53) to be ell = a"'n• e22 = b"'u e33 = b"'u (13.54) But, by the definition of Young's Modulus (equation (13.4)),

•

Hence

E = "'u/eu. a = 1/E

(13.55)

336

A C O U RS E I N A P P L I E D MAT H E M A T I C S

[cH.

It is found by experiment that an elastic material subject to tensile stress contracts laterally. The ratio of the lateral contraction to the longitudinal extension is denoted by cr and is called Poisson 's Ratio for the material. But, by equations (13.54), cr

Thus Equations

(13.56)

= -e22/e11 = -e33/e11 = - bfa. b = - crfE.

(13.53) can now be written in the form 1 en = E {"u - cr ("22 + "aa)} , 1 e22 = {"22 - cr ('t"aa + "u)},

(13.57)

E

1

eaa = E {"aa - cr ('t"u + '�"22)} . Consider the particular case of an isotropic medium in a state of pure shearing stress of intensity s. As shown in the example at the end of the last section, the principal stresses are then '"n = s, '1"22 = - s, "aa = 0 and equations (13.57) reduce to the form e1 1

=

(1 + cr)sfE, e22 = - ( 1 + cr)sfE,

e33 =

0. . (13.58) (1 + cr)sfE. As

I.e., the strain is also a pure shear of magnitude shown on p. 329, this corresponds to an angle of shear of 6 = 2(1 + cr)sfE. Hence

s

E

e = 2(1 + cr) = !J.·

(13.59)

The ratio !J. of the shearing stress to the resulting angle of shear is called the modulus of rigidity of the medium, and equation (13.59) exhibits its relationship with Young's Modulus and Poisson's Ratio. Another particular case of interest is when '"n = T22 = "aa = "• i.e., when the medium is stressed uniformly in all directions. From equations (13.57) we find that the strain is also uniform in all directions and that

e11 = e22 = e33 = e = (1 - 2cr)"�"fE.

(13.60)

In this case, a unit cube has, after deformation, edges of lengths 1 + e. Since e is small, this implies that its volume has increased to (1 + e) 3 = 1 + 3e. The increase in volume per unit volume of a deformed elastic medium is termed the dilatation and is denoted by � ­ Thus, in this case, or

A

k

= =

3e

=

3(1 - 2cr)'1"/E,

T/A =

E/3(1 - 2cr).

(13.61).

D E FO RMAT I O N O F E L A S T I C B O D I E S

13]

337

If ... is negative, � will be negative and k represents the ratio of the compressive stress to the resulting cubical compression. k is accordingly referred to as the modulus of compression. Solving equations (13.57) for the principal stresses, we obtain

"�"n = A(e11 + e22 + e33) + 2!.1.ew "�"22 = A(eu + e22 + eaa) + 21.1.e22• "�"33 = A (en + C22 + e33) + 2!.i.e33•

}

(13.62)

where 1.1. is defined by equation (13.59) and

A = (1 + cr)Ecr (1

-

(13.63)

2cr)

A and 1.1. are the constants of Lame. Exercises

Obtain the relationships p. (3,\ + 2p.) E= k =

,\ + p.

p.E

9p. - 3E

'

,

k =

.\ + iJL,

a = 2j. - 1 •

E

a=

,\

2(.\ + p.) '

(E - 2p.) . >. = p. 3p. - E

Since, clearly, k is a positive quantity, equation (13.61) indicates that a <

t.

E and 1.1. are the elastic moduli most readily obtained by experiment and their observed values (lbfsq in) for a few materials are given in the table below : Material Steel Cast-iron Copper, cast . Silver Platinum Phosphor-bronze

E 30 13 13 11 24 13

X X X X X X

p. 106 106 106 106 106 106

12 4 4 4 9 5

X

106

X X X X

106 106 106 106

X 106

13.9. Simple Cases of Elastic Deformation In general, the problem of the deformation suffered by an elastic body when it is subjected to external forces is a very complex one. The deformation is governed by three sets of equations : (i) the partial differential equations relating the displacements ui to the components of strain e;j ; (ii) the Hooke's Law equations relating the components of stress and strain, and (iii) the equations of equilibrium of a typical element of the body. For equilibrium of such an element, we have already shown that it is necessary that "�"ij = "�"ii· However, these are not sufficient conditions, and three other partial differential equations must be derived to complete the set of sufficient conditions.

338

A C O U R S E I N A P P L I E D MATH EMAT I C S

[CH.

The complete set of equations has to be solved for the displacements

ui under given boundary conditions, e.g., at a point on the surface of the

body where no forces are applied, all components of stress must vanish. This section will be devoted to a number of problems in which the presence of spherical or axial symmetry so simplifies the statement of the above conditions that a solution is readily found.

13.9.1. Spherical Shell Suppose that a spherical shell of homogeneous isotropic elastic material, internal radius a and external radius b, is subject to uniform normal pressures over its interior and exterior surfaces. Let r be distance of a point P in the medium from the centre 0 and u the dis­ placement of P constituting the strain. Then, on account of the spherical symmetry, u will be a function of r alone and will be in the direction OP. The principal directions of stress and strain at P will be along 0 P and along any two perpendicular directions in the plane normal to OP. From equation (13.3), the radial extension at P is dufdr. Consider an element at P in the plane normal to OP and subtending a small angle e at 0. All displacements being radial, after deformation this element will continue to subtend the same angle at 0, but its distance from 0 will have increased from r to r + u. Hence its elongation is ue and its extension is uejre = ufr. Thus the principal extensions are

e11 = dufdr, e22 = e33 = ufr.

(13.64)

The corresponding principal stresses will be taken to be

(13.65) and then, by equations

(13.57), we have E du dr = '1'1 - 2cr'l'2 , E u-r = (1 - cr)'1'2 - cr'l'1

(13.66) (13.67)

There remain the equilibrium conditions. Consider a thin, hemi­ spherical shell of material having its centre at 0, internal radius r and external radius r + dr (Fig. 13.9). Neglecting its weight, it is in equilibrium under the action of a normal stress '1'1 over its inner surface, a normal stress '1'1 + d'l'1 over its outer surface and a normal stress '1'2 over its rim. The forces acting upon the inner surface clearly possess a resultant along the axis of symmetry. If dS is an element of this surface whose normal makes an angle ex with the axis of symmetry, the con­ tribution to the resultant of the force acting upon the element is

13]

D E FO RMAT I O N O F E L A S T I C B O D I E S

-r1dS cos oc = -r1dS', where dS' is the projection of of the rim. The resultant is therefore

339

dS on to the plane (13.68)

the integral being evaluated over the circular area bounded by the rim. It follows immediately that the normal stress over the outer surface has a resultant along the axis of magnitude

1t1'2-r1 + d(1tr2-r1) = 1tY2-r1 + 1t fr (r2-r1)dr .

I I I I I I I I I

"

I ' I ' '\ ' I ' I I I I 1 I

'

0

dS' c,

(13.69)

FIG. 1 3.9.-Equilibrium of a Hemispherical Element

Finally, the stress over the rim has a resultant along the axis of magnitude

21tY-r2dr.

(13.70)

For equilibrium, we require that 1t

or

fr (r2-r1)dr � (r2-rl)

=

21t1'-r2dr,

(13.71) 2r-r2 . Equations (13.66), (13.67) and (13.71) determine the solution to the

problem.

=

Eliminating u between the first two equations, we obtain

(13.72)

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A C O U R S E I N A P P L I E D MATH EMAT I C S

Using equation

(13.71), we eliminate -r2, thus d2-r1 + � d-r1 - 0 dr2 r dr · ·

[CH.

(13.73)

The general solution of this equation is

-r1 = A + Bfr3•

(13.74)

(13.71) now yields -r2 = A - Bf2r3. . (13.75) and, from equation (13.67), we find that (13.76) Eu = (1 - 2 a)Ar - (1 + cr) 2rB2 • The constants of integration A and B are determined by the bound­ Equation

ary conditions. Suppose that over the inner surface of the shell there is a normal pressure of Po and over the outer surface a normal pressure of p 1 . Then -r1 = -p0 over r = a and -r1 = -P1 over r = b. Then, from equation (13.74), substituting these boundary conditions and solving for A and B, we get

(13.77) and the problem is completely solved. In the case of a solid sphere of radius b subjected to a uniform pressure p1 over its surface, -r1, -r2 must be finite at r = 0, and hence B = 0. Also -r1 = -P1 over r = b, and thus A = -p1. The solu­ tion to the problem is therefore

(13.78) I.e., the displacement is proportional to the distance from

13.9.2. Cylindrical Shell

0.

A similar problem is that of a cylindrical shell of internal radius a and external radius b, with plane ends normal to the axis. Uniform normal pressures are supposed applied over each curved boundary surface and over the two plane ends. We assume that the displace­ ment of any point P is in the plane containing P and the axis of the cylinder. We also suppose that the principal axes at P are : (i) in the direction of the perpendicular from P to the axis ; (ii) parallel to the axis, and (iii) perpendicular to both these directions. Letting r be the length of the perpendicular to the axis and u the component of P's

13]

D E FO RMAT I O N O F E L A S T I C B O D I E S

341

displacement along the perpendicular to the axis, we have for the principal extensions :

e11 = dufdr, e22 = ufr, eaa = e,

where e will be taken to be a constant. Then, if "I • -r81 "a are the principal stresses, equations

E -ur = "2 - cr (-rs + ) Ee = "a - cr (-rl + -r2) .

"1 ,

(13.79) (13.57) yield (13.80) (13.81) (13.82)

Consider an element of the shell bounded by two co-axial cylinders of radii r and r + dr and an axial plane. This is in equilibrium under the action of a normal stress "I over the inner surface, a normal stress "I + d-r1 over the outer surface and a normal stress -r over the straight 2 edges. Over the plane ends there is a normal stress "a• but these clearly cancel. As for the sphere, we show that the resultant force acting upon the inner surface is equal to that which would result from a normal stress "I over the projection of the inner surface on to the axial plane, viz., 2rl-ri, where l is the length of the shell. The resultant force acting on the outer surface is therefore 2rl-r1 + d(2rl-ri). The resultant force on the two straight edges is 2l-r2dr. For equilibrium, it is required that

d(2rl-ri) = 2l-r�r, d dr (ni) = "2·

or

(13.83)

u and "a between equations (13.80)-(13.82), we obtain (13.84) Elimination of -r2 by the use of equation (13.83) now yields d2-r2l + � d-rl - 0 (13.85) dr r dr - • · Eliminating

The general solution of this equation is

-r1 = A + Bfr2. Whence, from equation (13.83), "2 = A - Bfr2, and, from equation (13.82), = Ee + 2crA. -r3

(13.86) (13.87) (13.88)

342

A C O U R S E I N A P P L I E D MAT H E M AT I C S

(13.81), we obtain Eu = (1 + cr) [(1 - 2cr)rA - �] - crEer.

[cH.

Also, from equation

(13.89)

As a particular case, suppose the inner surface of the shell is sub­ jected to a uniform normal pressure of Po and the outer to a uniform normal pressure of p 1 and let the plane ends be free from force. Then -r1 = -p0 over r = a, -r1 = -P1 over r = b and -r3 = 0. This will be so if

e = -2crA JE. (13.90)

13.9.3. Rotating Disc

Suppose that a thin disc rotates with uniform angular velocity w in its own plane about its centre 0. We shall introduce centrifugal forces so that we can neglect the motion. We shall assume radial displacements, u being the displacement at a distance r from 0. At any point P in the disc, the principal axes are : (i) along OP ; (ii) perpendicular to OP and in the plane of the disc, and (iii) parallel to the axis of rotation. We assume that there is no stress in the latter direction. Then, the principal strains are

e11 = dujdr, e22 = ujr, e33 and Hooke's Law leads to the equations

(13.91) Consider a semi-circular element of the disc with its centre at 0, inner radius r and outer radius r + dr. If t is the thickness of the disc, the normal stress over the inner surfaces has a resultant 2rt-r1 and over the outer surface a resultant 2rt-r1 + d(2rt-r1). Forces -r2tdr act at the two ends of the semi-circular arc. The centrifugal forces act along the radii, and their resultant is computed by the method employed for the resultant of the normal stresses over the inner surface. If p is the density of the disc, the resultant centrifugal force proves to be pw2rdr x 2rt. For equilibrium, it is necessary that i.e., Eliminating obtain

d(2rt-r1) + 2pw2r2tdr = 2-r2tdr, d dr (r-rl) + pw 2r2 - -r2 . u and -r2 between equations (13.91)

and

(13.92) (13.92), we (13.93)

13]

D E FO RMAT I O N O F E L A S T I C B O D I E S

343

The general solution of this equation proves to be

(13.94) = A + � - � (3 + cr)p(i)2r2. Whence, from equation (13.92), we find that (13.95} "1'2 = A - � - � (1 + 3cr)p(i)2r2, and, from equations (13.91}, Eu = {1 - cr)Ar - (1 + ) � - � (1 - cr2)p(i)2r3 . . (13.96) If the disc is of radius a, is solid to the centre and is free from applied force at its rim, = 0 over r = a and is finite at r = 0. Hence B = O and '�'I

cr

"�'t

This implies that

= 1 (3 + cr)p(i)2 (a2 - r2 ), = fP(i)2 [ (3 + cr)a2 - (1 + 3cr)r2] , Eu = kP(i)2 (l - cr}r[(3 + cr)a2 - (l + cr)r2]. '�'I "�'2

At r

= a,

u = p (i)2a3 tl - cr)f4E ,

giving the increase in radius of the disc caused by the rotation. EXERCISE 1 3 1 . A light, horizontal beam AB, of length 1 2 ft., is supported at its points of trisection and carries weights of 80 lb, 60 lb at A , B respectively. Obtain algebraic expressions for the bending moment at a point at distance x ft from A in the different parts of the beam, and draw a bending-moment diagram. State where the maximum bending moment occurs and write down its value in lb-ft. (Le.U.) 2. Two equal, uniform rods AB and BC, each of length 6a and weight 6wa, are freely hinged together at B and rest horizontally on three supports at A , C and D, where D is the point in BC distant a from B. Obtain expressions for S and M as functions of the distance x from the end A . Find at what point the bending moment has its greatest numerical value, and show that this greatest value is 9wa2J2. (S.U.) 3. A uniform beam, of length 4 ft and weight 20 lb, is supported on two supports at the same level 1 ft from each end, and carries a load of 10 lb at its mid-point. Draw shearing-force and bending-moment diagrams, and find the points at which a hinge could be inserted without upsetting equilibrium. (Le.U.)

344

A C O U R S E I N A P P L I E D MAT H E M A T I C S

[CH.

4. A uniform plank AB, of weight W and length 16 ft, is horizontal, supported at its ends. It would just break if an extra load 2W were placed at its mid-point. Find at what distance from A this same load should be placed just to break the plank, if the plank were clamped horizontally at A and had B unsupported. (The maximum tendency to break may be assumed to be at the point where the bending moment is greatest.) (Le.U.) 5. A uniform beam, of length 2a and weight 2wa, rests on two supports at the same horizontal level at equal distances z from its ends. Show that if z < ta, the bending moment B will have positive and negative values. Draw the graph of B for the value z = ia and obtain the maximum positive and negative values of B . Show that to ensure that the greatest numerical value of B shall be as small as possible, the bending moments must be numerically the same at the centre and supports. Find the corresponding value of z and show that the greatest bending moment is then !wa2 (3 - 2y2) . (S.U.) 6. A uniform rod AB, of length 2a and mass m, can turn freely about a fixed pivot at A , and is released from rest in a horizontal position. Show that, when the rod is inclined at an angle e to the horizontal, the tension and shearing force at a point distant y from B are respectively (3y - 4a)y ( 16a - 3y)y '-----;: ---;--"--' ;-=- mg sin e and mg cos e. 8a2 1 6a2 Obtain an expression for the bending moment at this point. (N.U.) 7. The ends of a light beam rest on supports at the same level and distant l apart. If the beam carries a load W at its mid-point and its flexural rigidity is EI, show that the mid-point suffers a deflection of Wl3f48EI. Show further, that if the load is redistributed uniformly over the beam the deflection is reduced by a factor �8. A uniform, heavy, elastic beam, of length 2a, rests on two supports at the same horizontal level each at distance b from the mid-point of the beam. Find the relation between a and b if the mid-point of the beam is at the same horizontal level as the ends. (S.U.) 9. A uniform beam AB, of length 2l and weight w, per unit length is clamped horizontally at one end A and is free at the other end B. The beam rests on a support at its middle point C, the height of the support being adjusted so that B is on the same level as A . Find the height of the support at C above the level of A and prove that the pressure on it is 12wlf5. (Li.U.) 10. A uniform beam, of weight W and length l, rests on two supports, each at a distance !a from the middle section of the beam. Show that the middle of the beam is deflected upwards if afl < 6 - V 30. (S.U.) 1 1 . A uniform beam, of weight W and length 2a, is clamped at its ends in such a way that the ends and the middle are on the same level. Show that the small slope of the beam at its ends is Wa2f24EI. (S.U.)

13]

D E F O RMATI O N O F E LA S T I C B O D I E S

345

12. A uniform beam is simply supported at its ends and at its mid-point, the ends being in a horizontal line. What is the depth of the middle support below the level of the other two if all three reactions are equal? Take the length of the beam as 2a, the weight per unit length as w and the flexural rigidity as EI. (Li.U.) 13. A girder AB, of length L, is simply supported at both ends and carries a load varying uniformly from 0 at A to w at B. Find the formula for the deflection of the beam at a distance x from A . (L. U.) 14. A uniform, thin lath, of length l and constant flexural rigidity EI, 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. (L.U.)

15. A uniform beam, of length 21 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 H of the total load and find the deflection at the midpoint. (L.U.)

1 6. A light, uniform pole, of length l 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 ex 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 + n2)y = n2a + n2 (l - x) tan ex, where D = dfdx and Eln2 = T cos ex. Solve this differential equation and show that na = tan ex (tan nl - nl) . (L.U.)

17. Assuming that the bending moment at a point of a thin beam is ElfR, where EI is the flexural rigidity and R the radius of curvature at the point, show that a thin, uniform strut of length l will not bend under a longitudinal thrust less than r.2Elf!2. (Li.U.)

18. A light strut is clamped at one end and freely pinned at the other in line with the clamp. Show that the strut will just buckle under a compressive force P, where p = n2EJfl2, l being the length of the strut and n the least positive root of the equation tan x = x.

19 . A uniform, light, flexible rod of length l is clamped so that when it is

unloaded it is vertical, with the clamp at the lower end. If a weight W, sufficient to bend the rod, is attached to the upper end, prove that,

346

A C O U R S E I N A P P L I E D MAT H E M A T I C S

[CH.

in its position of stable equilibrium, the shape of the rod is given, in a usual notation, by 2El y 2 - a2 = (1 cos ,P) , -

W

-

where y is the deflection from the vertical line through the upper end, and a is the horizontal separation of the two ends. By the use of Euler's approximate theory, show that, if the curvature of the rod in the bent position is small, the critical load is given by W = !rt2Elfl2 • (Li.U.) 20. Derive the equation d4y

21.

22.

= K4y (K4 = mCJl2/El) dx4 for the small lateral deflection of a vertical whirling shaft in steady motion. The mass per unit length of the shaft is m and the other symbols have their usual meanings. If the length of the shaft is l and each of its ends is constrained by a vertical clamp, show that the least critical angular speed of whirling is (z12fl2)y(Elfm), where z1 is the smallest positive root of the equation cosh z cos z = 1 . Show that z1 is approximately 3rt/2, and that a closer approximation is 3rt/2 + sech (3rt/2) . (Li.U.) In a given strain of an isotropic material the particle P, whose co­ ordinates referred to given rectangular Cartestian axes are x"', is dis­ placed during strain through small distances u"' parallel to these axes, where u1 = K(x 3 + x1) , u 2 = K(x1 + x 2), u 3 = K(x2 + x 3), and K is a constant. Determine the principal stresses in terms of (M.T.) K, E and cr. If (in the us·ual notation) en = e 22 = e a s = e, e 2a = 0, where e is constant, prove that the most general expressions for the components u1 are u1 = ex1 + F(x 2 , x 3) , u 2 = ex 2 + x3f(x1) + g(x1) u 3 = ex3 - xd(x1) + h(x1), where the functions F, f, g, and h are arbitrary, and hence find the most general expressions for u1 if i = k, (M.T.) i =I= k. Calculate the principal stresses at the point (x1, x 2, x 3) when the dis­ placement vector has components given by the equations u1 = -kx 2x3 , u 2 = kx3x1, u 3 = 0. An empty spherical cavity in a large mass of metal has radius a. Air is pumped into the cavity to a pressure P. If fl. is the modulus of rigidity of the metal, show that the radius of the cavity increases by Paf4f1.. ,

23.

24.

13]

D E FO RMAT I O N OF E L A S T I C B O D I E S

347

25. A gun barrel is formed from two circular cylindrical shells of the same metal, one shrunk on top of the other. There is no longitudinal stress in either tube. When unstressed, the inner tube has internal and external radii a and b respectively and the outer has radii b - e: and c at the surface of contact is respectively. Show that the pressure given by (c2 - b2) (b2 - a2)Ee: 2b3(c2 - a2) When the gun is fired, the pressure over the inner surface of the inner tube is p0• Show that the principal stresses in the inner tube are given by a2 (c2 - r2) b2 (r2 - a2) - r2 2 a 2) "1 = - 2 2 (b a 2) r (c a2 (c2 + r2) b2 (r2 + a2) - 2 2 "z = r2 (c2 a2) r (b a2 ) Calculate the principal stresses in the outer tube. 26. A disc of radius a is in one piece with a shaft of negligible diameter through its centre. When the disc is rotating with angular velocity w, show that the hoop-tension -r2 takes its maximum value §pw2a2 (3 + cr) at the shaft. If the disc is only keyed to the shaft, show that the maximum hoop-tension is doubled, i.e., this arrangement is half as strong as the former. (Hint : Assume that the disc is an annulus with internal radius b and let b ---+ 0.)

P

P=

Po

_

_

.

Po

_

_

P.

P,

AN S W E R S

2

EXERCISE

I. 1 0·26 h.p. 5. tav (afk). Ft cos wt. 15. 2w

3

EXERCISE

11. (Sag cot 6 cosec 6)!. 19. (.,.13 + 1).,.!(!ag).

3. (2 + .,;2) ( 1 - ;)r. 3e 10. !c log 4---- y e-:r· 24. 5aw2f3.

2. 12·52 m.p.h. 9. 60 m.p.h. 21. 2Tr/3.,.13k. 12. mgcf(h + c). 20. (llagf2)l.

14. 6mg.

EXERCISE 4

5.,.!5.,a•'a -· 1. .,.!(8p,f9c), .,.1(29/45). 6. � 15. �c. 16. r = (2t2 + 2.,.!6t + 4)i. 19. 4aw2(2 sinh wt - 3 cosh wt). 22. �; ( 1 - �t ( 1 + �)r. 24. (i) 6 = a2D.f(a - ut)2, (ii) T = a�D.2f(a - ut)3 • 2Traf(2.,u + aD.). 28. h/3. EXERCISE 5 2 5. �1, 136I, where I = m.,.!(2gh) cos l. V cos 2a./(1 + 2 sin a.). 7. tan-1 .,.!e. 8. mu(1 + e)b.,.!(a• - b2)/a2, u{(1 + e) � - er-l e_ If}!! ___!!!!____ If}!! _ 4/Lhe 12· _ 1 - e '\/ g , 1 - e '\/ g (1 - e)2 - en) 14 ( 1gh 3 . /13• V2.,a(l gh) . en 1 (1 - e) . 15. Energy = !(3E + Q) ; tan-1 {t(j - 1 /} with original motion. EXERCISE 6 8. Vag. 7. 1 : 2, 1 : -l. l 12Mg(1 2 sin 6) 15 · {a( 16M + 3M' cosec� 6)} · 21 a2gt"/2(a• + k") ' 22. 0, !.,, �. 24. .i2{m + Mx2f(a2 + x2)} = 2Mg( .,.!5a - va• + x2). 25. cos-1{(a2 - l2)lf.,.!3l}. 28. Wa(.,.!3 cos !6 - 1)2 - t Wa cos 6; stable. 31. The lower position of G is stable. 30. {ga(3.,.!3 - 4)}!. 34. 27r��· 348 a..

•

•

·

"'•

V

•

'

]fY

ANSWERS

349

EXERCISE 7 S.C. a circle of radius a; B.C. a circle of radius !a. 3.8. Parabola, 6. aw. vertex 0, axis perpendicular to .>.. 15. -!M(b" + 3x2), -!M(a• + 3y2) , Mxy. 20. -!M(b" + c2), -!M(c2 + a•), -!M(a2 + b2), !Mbc, !Mea, !Mab. EXERCISE 8 3. R = 80mgf49, llmgf49. 4. V39mgf28. 1. v (24gfl7a). l' { F =2M2gx(2x + l) t 7· {2M2x + (m•2m2glx M2)l} 2M2x + (m2 - mM + M2)l} · mM + 14. iMg(9 - e-:eJ9a), /8Mg(9 + e-zJ9a) . 16. 1811'/ 19. 15. 4y2af3. 17. 2awfy5. 19. KE = ma2[2B2 + 46� cos•� 6 + (22 cos• 6 + 1) �•], AM = 2ma2[26 cos2 6 + (2 cos 6 + 1)]. 20. 21w { y2 - !v5 + log �� ! i} · . = awr ' angu1ar ve1ocr.ty == a•w • 22. Radra. 1 ve1ocrty 3r• + a (3r• + a•)f � cos (r/> - 6)] , 23. KE = t.mt•[8a• +�9k·�· + 12ka AM = ml2[t6 + k2 + k(B + �) cos (rp - 6)]. k = y2 - 1. 34. v = Mgf3k. 25. awfg. EXERCISE 9 3. (6b - lOa) : (2a - 3b). 1. 1f3m. 2. 51/21, 21/21. 4. a2f3(l + a) from mid-point of bar on the side of A . 9. mV"(a2 + b2)/2(3p2 + a• + b2). 6. tm(2u + v), tm(u + 2v). 11. 6]/7ma, 15]f7ma, 9]/7m. 16. 3uf4a. 13. 3y2 Vf8a. 12. 3 : (2 cos oc + 1). 19. fmy(2gh) sin 2oc. 18. 9f"fml. EXERCISE 10 4. 2ma2w6(i + cos 2 6) , 2ma2w6 sin 6 cos e. 6. v 12gf5a. 12. -7g sin r�.f5a, 3g sin r�.f20a. p1, are positive roots of 14. 21Tfp1, 2, where abp421Tfp - (3a + 2b)gp2 Po+ 2g2 = 0 . 15. KE = tma•a• + 2mx•, PE 2b.>. (x - b - 2a cos II) " - 4mgx. 21. 21r,J¥;• 21r,j�. 17. 21rvMa/(M + m)g. 20. 21r�• 21rJ;g · i --l 22. 21r(a ± b) ; p = !(a + b)i + !(a - b) , q = !(a + b)l - !(a - b)l ; q, = L ·sin qt cos pt. EXERCISE 11 4. G2 : F2• 5. ta. X2 - Y2 2XYc 6. (X - Y)x = (X + Y)y, z = X• + y• c ; P = x• + y2 10. R = 5, p = 29f25-yl2. 8. (P1G1 + P2G2)/(P1" + P:J'). =

A C O U R S E I N A P P L I E D MATHEMAT I C S

350

EXERCISE I2 7. 4/5 Iog 5. I I . 324 ft. 3. 52 lb wt ; 97 · 3 ft. 1 6 . c = 3l. I3. 2(a1l1 - a2l2)/{a1 - a2) . I2. y' 3l/2 or y'3lf4. W(e�"" - I) W(I - e-"'") 3 + y' IO I9. {a) (b) I7. tl y'2 + log , y'8l. I 8 4,.., , I I + v'2

}

{

•

_

EXERCISE I 3

e•�""

_

Mmax. = 320 lb ft at X = 4. 2. x = 3a. 5. z = (y'2 - I)a. Hinges at y'3 ft from each support. 4. 1 ft. 8. 3a3 - I 2ab2 + 4b3 = 0. mgy2 (2a - y) cos 6fi6a2• l l wl4fi20EI. I2. 7wa4f72EI. wx(L" - x2) {7L2 - 3x2)/360EJL. b = a(sec nl - 1 ) , (n2 = WfEI) ; M = Wa sec nl; W = 7T2Eif9l2• 5Wl3f96EI, W = total load. T11 = T22 = KE(4a + I)/(1 + a) {1 - 2a), .,.33 = KE(2 - a)/(I - 2a) {I + a). 22. u1 = ex1 + ex2 - bx3 + p, u2 = - ex1 + ex2 + ax3 + q, r; a, b, e, p, q, r all constants. u3 = bx1 - ax2 + ex3 23 . .,.11 = -p.kr2, .,.22 = T33 = p.kr2, where r2 = x1 2 + x 2• 2 a2 {e2 - r 2) b 2 {e2 - r2) 25. Tt = P o r• (e• - b ") r• (e• - a•) a2 {e2 + r2) b 2 {e2 + r 2) + .,.. = 2 2 r {c - a2) P o r2 {c2 - b 2)

I. 3. 6. 9. I3. 14. I5. 21.

0,

+

P,

P.

•

INDEX (The numbers refer AccELERATION vector, 11

cartesian components of, 12 constant, 28 intrinsic components of, 13 polar components of, 12 Air resistance, 37, 63, 75 Angular velocity, 7 Anomaly, eccentric, 99 real, 99 Aperiodic motion, 44 Apse, 87 Apsidal distance, 88 Bending moment, 3 1 1 Body centrode, 1 7 1 Capstan principle, 303 Catenary, common, 298 directrix of, 299 parabolic, 302 uniform strength, 302 Central axis, 279 Central force, 85 Centre of mass, 283 Centrifugal force, 285 Compression, 3 1 1 Conservative system, 142 Constraints, 142 Couple, 277 Damping, 45 critical, 46 Degrees of freedom, 153 Dilatation, 336 Dyne, 22 Earth, rotation of, 243 Elastic impact, 124, 232 Elliptic harmonic motion, 66, 104 Energy, conservation of, 72 equation of, 72, 140 in orbital motion, 92 kinetic, 70 potential, 71 Equilibrium of rigid body, 271, 291 Equimomental systems, 185 Equivalent-force systems, 276 Erg, 24 Escape velocity, 94 Eulerian angles, 248 Euler's equations, 242 Extension of a bar, 315 External forces, 140, 191

to pages.)

Flexural rigidity, 319 Flexure of a beam, 3 1 6. Forced oscillations, 4 7 Forces, 2 1 generalized components of, 254 moment of, 203, 272 parallelogram of, 277 scalar moment of, 272 transmissibility of, 277 Forcing function, 4 7 Frame of reference, 4 rotating, 240 Friction, 273 angle of, 274 coefficient of, 274 cone of, 274 laws of, 274 limiting, 274 Generalized coordinates, 152, 253 Gramme, 22 Gravitational constant, 102 Hodograph, 10 of orbital motion, 90 Holonomic systems, 254 Hooke's constant, 33 Hooke's law, 33, 315, 335 Horse-power, 27 Impact of spheres, 124 Impulse, 1 18, 226 Independence of forces, 23 Independence of rotary and translatory motions, 210 Inertial forces, 284 Inertial frame, 20 Initial motions, 132, 214 Instantaneous axis, 173 Instantaneous centre, 170 Internal forces, 140, 191, 287, 309 work done by, 140 Inverse square law, 88 Isolated system, 201 Jet velocity, 216 Joule, 24 Kepler's equation, 100 Kepler's laws, 101, 202 Kinetic energy, 70 of particle system, 148 of rigid body, 145, 246

ii

INDEX

Lagrangian, 261 Lame, constants of, 337 Littlewood coordinates, 65 Mass, 2 1 Mass ratio, 217 Modulus of compression, 337 of rigidity, 336 Momenta! ellipse, 180 Momenta! ellipsoid, 183 Moment of inertia, 146 Momentum, angular, 203, 241 conservation of, 205 equation of, 204, 210 of particle system, 208 linear, 22, 1 1 8 conservation of, 22, 124, 197 equation of, 191 of particle system, 191 Neutral line, 317 Newton, 22 Newton's first law, 20 Newton's law of gravitation, 102 Newton's second law, 21 Newton's third law, 21, 121, 1 3 1 Normal mode, 134 Nutation, Eulerian, 244 Orbit, 87 conic, 89 time in, 99 Oscillations, forced, 48 on elastic string, 33 small, 154, 261 Parabola of safety, 59 Parabolic motion, 56 Parallel axes, theorem of, 184 Parallel forces, 282 Particle, 3 Pendulum, compound, 146, 192, 205 simple, 75 simple equivalent, 147 Periodic time, 30, 99 Perpendicular axes, theorem of, 179 Phase lag, 47 Pitch, of force system, 280 of screw motion, 178 Poisson's ratio, 336 Polygon of forces, 24 Position vector, 4 Potential energy, 71, 142, 260 of an elastic string, 71 Pound, 22 Poundal, 22 Pound weight, 22 Power, 26 Precession, 250 pr-equation, 103 Principal axes of inertia, 1 8 1

Principal strains, 329 Principal stresses, 334 Product of inertia, 180 Projectile, 55 range of, 56, 61 time of .flight of, 57 Recoil, 122 Relative acceleration, 1 1 Relative velocity, 5 Resonance, 48 Response characteristic, 4 7 Restitution, coefficient of, 125 law of, 125, 232 Resultant, 24 Rocket motion, 2 1 6 Rocking bodies, 159 Rotating frame, 240 Rotation about pivot, 241 Screw motion, 178 Sections, method of, 309 Shear, angle of, 329 Shearing force, 3 1 1 Simple harmonic motion, 30, 75, 146 amplitude of, 3 1 frequency of, 3 1 period of, 30 phase of, 31 Slug, 22 Space centrode, 1 7 1 Speed, 4 Stability of equilibrium, 155, 160 Steady state, 46 Strain, 310, 324 components of, 327 homogeneous, 326 pure, 327 shearing, 328 small, 326 superposition of, 326 Stress, 310, 330 components of, 331 Surface of revolution, motion on, 1 10 Suspension bridge, 301 Tension, 3 1 1 Terminal velocity, 39, 65 Top, spinning, 247 Transient, 46 Variable mass, 215 Velocity from infinity, 93 Velocity vector, 4 cartesian components of, 6 intrinsic components of, 6 moment of, 86 polar components of, 7 Vertex of trajectory, 38, 57 Vibrations, sympathetic, 137 transverse, 1 33

INDEX Virtual displacement, 253 Virtual work, 286 Viscous friction, 43 Watt, 27 Whirling shaft, 322

Wire, motion on a rotating, 106 Work, 24 equation of, 70 Wrap, angle of, 303 Young's modulus, 315, 336

iii

A COURSE I N A P P L I E D MAT H E M A T I C S

PHYSICAL SCIENCE TEXTS General Editor

SIR GRAHAM SUTTON, C.B.E., D.Sc., F.R.S. Director-General, Meteorological Office, Formerly Dean of the Royal Military College of Science, Shrivenham, and Bashforth Professor of MathematicaJ Physics.

ADVANCED LEVEL APPLIED MATHEMATICS by c. G. LAMBE, B.A., Ph.D. APPLIED MATHEMATICS FOR ENGINEERS AND SCIENTISTS by C. G. LAMBE, B.A., Ph.D. ADVANCED LEVEL PURE MATHEMATICS by C. J. TRANTER, O.B.E., M.A., D.Sc. TECHNIQUES OF MATHEMATICAL ANALYSIS by C. ]. TRANTER, O.B.E., M.A., D.Sc. GENERAL PHYSICS AND SOUND (To Advanced and Scholarship Level) by D. H . FENDER, B.Sc., Ph.D. HEAT (To Advanced and Scholarship Level) by A. J. WooDALL, O.B.E., Ph.D., F.Inst.P. LIGHT (To Advanced and Scholarship Level) by C. B. D AISH, M.Sc. ELECTRICITY AND MAGNETISM (To Advanced and Scholarship Level) by C. G. WILSON, M.Sc., A.Inst.P. EXPERIMENTAL PHYSICS (To Advanced and Scholarship Level) by C. B. DAISH, M.Sc. D. H. FENDER, B.Sc., Ph.D. A COMPENDIUM OF MATHEMATICS AND PHYSICS by DOROTHY MEYLER, M.Sc. SIR GRAHAM SuTTON, C.B.E., D.Sc., F.R.S. ELECTRON PHYSICS AND TECHNOLOGY by J. THOMSON, D.Sc. E. B. CALLICK, B.Sc. PRINCIPLES OF ELECTRONICS by M. R. GAVIN, M.B.E., D.Sc. ]. E. HouLDIN, Ph.D. A COURSE IN APPLIED MATHEMATICS (Covering B .A. and B.Sc. General Degrees) by D. F. LAWDEN, M.A. PHYSICS FOR ELECTRICAL ENGINEERS by W. P. Jo LLY, B.Sc.

In Preparation

REACTOR PHYSICS AND TECHNOLOGY by J. WALKER, Ph.D., F.Inst.P. D. ]AKEMAN, Ph.D., A.Inst.P. ELEMENTS OF MATHEMATICAL PHYSICS

A COURSE IN

APPLIED

MATHEMATICS By DEREK F. LA WDEN, M.A. Professor of Mathematics, University ofCanterbury, New Zealand

VOLUME I I Part III : Field Theory Part IV : Hydromechanics

THE ENGLISH UNIVERSITIES PRESS LTD 102

NEWGATE STREET

LONDON, E.C.l

First printed 1960

© Copyright Derek Lawden, 1960

F.

Printed in Great Britain for the English Universities Press, Limited, by Richard Clay and Company, Ltd., Bungay, Suffolk

GENER A L EDITOR 'S FOR EWOR D

by SIR GRAHAM SUTTON, C.B.E., D.Sc., F.R.S.

Director-General, Meteorological Office, Formerly Dean of the Royal Military College of Science, Shrivenham, and Bashforth Professor of Mathematical Physics

present volume is one of a number planned to extend the Physical Science Texts beyond the Advanced or Scholarship levels of the General Certificate of Education. The earlier volumes in this series were prepared as texts for class teaching or self-study in the upper forms at school, or in the first year at the university or technical college. In this next stage, the treatment necessarily assumes a greater degree of maturity in the student than did the earlier volumes, but the emphasis is still on a strongly realistic approach aimed at giving the sincere reader technical proficiency in his chosen subject. The material has been carefully selected on a broad and reasonably comprehensive basis, with the object of ensuring that the student acquires a proper grasp of the essentials before he begins to read more specialized texts. At the same time, due regard has been paid to modem developments, and each volume is designed to give the reader an integrated account of a subject up to the level of an honours degree of any British or Commonwealth university, or the graduate membership of a pro­ fessional institution. A course of study in science may take one of two shapes. It may spread horizontally rather than vertically, with greater attention to the security of the foundations than to the level attained, or it may be deliberately designed to reach the heights by the quickest possible route. The tradition of scientific education in this country has been in favour of the former method, and despite the need to produce techno­ logists quickly, I am convinced that the traditional policy is still the sounder. Experience shows that the student who has received a thorough unhurried training in the fundamentals reaches the stage of productive or original work very little, if at all, behind the man who has been persuaded to specialize at a much earlier stage, and in later life there is little doubt who is the better educated man. It is my hope that in these texts we have provided materials for a sound general education in the physical sciences, and that the student who works conscientiously through these books will face more specialized studies with complete confidence. THE

v

PREFACE MY intention has been to write a book on applied mathematics which will be of assistance to students reading for science degrees at a level lower than that of Honours. It is assumed that the reader has suc­ cessfully completed a course of study in the subject to the Inter­ mediate standard of his degree course or to the Advanced Level of the General Certificate of Education. It is also assumed that the reader will be attending a coarse of lectures in pure mathematics, during which he will study the theory of linear differential equations, both ordinary and partial, the definition and evaluation of surface and volume integrals and the notation and techniques of vector analysis. All these branches of the subject are now common to the courses arranged for this class of student by the various university bodies in the United Kingdom. Vector methods have accordingly been employed without comment at any point in the argument where it was felt they arise quite naturally. However, no previous knowledge of the properties of the vector operators " grad ", " div " and " curl " is assumed of the reader, these being developed ab initio as required. Since it is expected that the majority of the book's readers will not be mathematical specialists, who could be expected to complete for themselves an argument not made fully explicit, very few links have been omitted from the chains of reasoning, with the result that the reader of average ability should be able to make considerable progress without outside assistance. This should commend the work to those many students who read for External General Degrees of London University by part-time attendance at technical-college classes. However, a serious attempt has been made to present a fairly complete logical structure. I have not therefore avoided diffi­ cult passages in the development of the subject, either by ignoring them, appealing to the reader's intuition or by offering an argument by analogy. The use of any of these stratagems may be justified in a book written for use by engineers or others who are primarily inter­ ested in acquiring the facility to use a tool. However, it is assumed that the readers of this book are motivated, at least partially, by a desire to study applied mathematics for its own sake as an <esthetic­ ally pleasing and elegant structure. Those engineers who appreciate that a proper understanding of the tools they employ will greatly assist them in their efficient use, should also find the book helpful. Another type of student to which the book may appeal is the man who is reading for Honours in mathematics but who, for one reason or vii

viii

P R E FA C E

another, finds that the texts t o which he is referred make too great a demand upon his abilities. The book should carry such a student a long way towards his goal, after which he may feel sufficiently con­ fident to study more comprehensive texts. The subject matter of the book divides into four parts, Part I Dynamics (Chapters 1-10), Part II Statics (Chapters ll-13), Part III Field Theory (Chapters 14-18), Part IV Hydromechanics (Chapters 19-21). Part I includes a detailed discussion of the dynamics of a particle and of a rigid body moving in a plane, and an introduction to the theory of the general three-dimensional motion of a rigid body. This introduction explains how considerations of angular momentum may be employed to derive Euler's equations for the motion of a rigid body pivoted at a fixed point, establishes Lagrange's equations and illustrates their application by a number of easy problems on three­ dimensional motion. In Part II equations of equilibrium for a system of rigid bodies acted upon by a three-dimensional system of forces are derived as particular cases of previously established dynamical equa­ tions. This Part concludes with an introduction to the theory of elasticity. Part III gives an account of the Faraday-Maxwell theory of the electro-magnetic field together with an introduction to the Newtonian theory of gravitational attraction. The equations of the electro-magnetic theory are expressed in terms of an unrationalized Gaussian system of units, since the examination papers of the various university bodies which have been consulted indicate that such a system is normally employed by mathematical departments in their teaching. However, an appendix has been added at the end of this Part, in which an explanation of the advantages to be gained from rationalizing the unit system is given and the Giorgi system is described and related to the system employed in this book. In Part IV the equilibrium of a fluid in a general gravitational field is discussed. A chapter is devoted to the one-dimensional flow of an ideal fluid in a smooth pipe, the theory being illustrated by the problem, now very topical, of the rocket motor. The general equations of motion of an ideal fluid are obtained in a final chapter and the characteristics of a number of simple irrota­ tional flows and flows due to rectilinear vortices are investigated. By analogy with electrostatics, the velocity potential cp is related to the flow velocity by the equation q = -grad cp. Throughout the book important results and principles are illustrated by worked examples, some of which are original but many of which have been taken from examination papers set by the following bodies : Birmingham University (B.U.) Cambridge University (Mathematical Tripos) (M.T.) Durham University (D.U.) Leeds University (Le.U.)

PREFACE

IX

Liverpool University (Li.U.) London University (L.U.) Manchester University (M.U.) Nottingham University (N.U.) Queen's University, Belfast (Q.U.) Sheffield University (S.U.) Other problems, from the same sources, for working by the student, will be found in the sets of exercises at the ends of the chapters. The source of each such example or exercise is indicated according to the abbreviation scheme shown above. The author wishes to express his thanks to the bodies concerned for permission to make use of this material. The author is also indebted to Sir Graham Sutton, F.R.S., for a number of suggestions which have led to improvements in the text and to his brother, Mr. J. G. Lawden, who is responsible for the pre­ paration of the figures and diagrams. D. F. LAWDEN

University of Canterbury

CONTENTS PART III : FIELD THEORY

14.

FIELD INTENSITY AND POTENTIAL

. 14.1. Law of Force Between Point Charges 14.2. Fields and Lines of Force 14.3. Elementary Distributions 14.4. Two-Dimensional Fields 14.5. The Potential Function 14.6. Potential Due to Elementary Distributions 14.7. The Electric Dipole 14.8. The Magnetic Dipole . 14.9. Potential Energy of a Charge Distribution

353 353 354 357 365 368 374 379 383 390

15.

GAuss's THEOREM.

IMAGES . 15.1. Flux and Gauss's Theorem . 15.2. Applications of Gauss's Theorem . 15.3. Divergence. Green's Theorem. Poisson's Equation 15.4. Elementary Solutions of Laplace's Equation 15.5. Force on a Charged Conductor 15.6. Uniqueness Theorem. Harmonic Functions 15.7. Images 15.8. The Complex Potential Appendix .

399 399 402 406 411 416 418 424 432 437

16.

PoLARIZED MEDIA

. 16.1. Dielectrics. Polarization 16.2. Poisson's Equivalent Distribution . 16.3. Harmonic Functions and Image Charges 16.4. Fields in Cavities 16.5. Energy of an Electrostatic Field . 16.6. Fields Due to Magnetized Materials 16.7. Uniformly Magnetized Bodies

443 443 444 451 456 457 459 463

PoissoN ' s E QUATION .

xi

Xll

CONTENTS

17.

STEADY ELECTRIC CURRENTS

. 17.1. Currents in Wires. Kirchhoff 's Laws 17 .2. Electrical Generation of Heat 17 .3. Equation of Continuity 17.4. Differential Form of Ohm's Law 17.5. Heat Generated by a Three-Dimensional Flow

18.

E LECTRO-MAGNETISM

18.1. Magnetic Shells . 18.2. Fields Due to Simple Current Loops 18.3. The Biot-Savart Law 18.4. Force Acting upon a Circuit in a Magnetic Field 18.5. Equations of a Steady Electro-magnetic Field 18.6. The Vector Potential . 18.7. Electro-magnetic Induction. Faraday's Law 18.8. Maxwell's Equations . 18.9. Energy Considerations. Poynting's Vector 18.10. Boundary Conditions . 18.11. Electro-magnetic Waves Appendix

.

4 70 470 474 476 478 484

489 489 490 495 497 503 512 514 524 529 531 533 537 545

APPENDICES A AND B

PART IV : HYDROMECHANICS

19.

HYDROSTATICS 19.1. Pressure in a Fluid 19.2. Equilibrium of a Fluid in a Gravitational Field 19.3. Relative Equilibrium of Rotating Fluids 19.4. Stability of Floating Bodies. Metacentres

555 555 556 561 564

20.

ONE-DIMENSIONAL M OTION OF AN IDEAL FLUID 20.1. Pressure in an Ideal Fluid in Motion 20.2. Equation of Continuity 20.3. Euler's and Bernoulli's Equations 20.4. Flow of a Homogeneous Liquid 20.5. Compressible Flow. The Rocket Motor 20.6. Vibration of Elastic Strings . 20.7 . The Propagation of Sound Waves along Pipes

573 573 573 576 579 583 589 600

.

21.

CONTENTS

xiii

GENERAL MOTION OF A FLUID 21.1. Stream Tubes 21.2. General Equation of Continuity. Irrotational Motion . 21.3. Particular Cases of Irrotational Motion 21.4. Euler's Dynamical Equations 21.5. Kinetic Energy of an Irrotational Flow 21.6. Stream Function. Circulation 21.7. Vortex Motion Appendix

611 611 614 617 621 629 631 637 641

ANSWERS INDEX

648 651

.

•

PART I I I

F IE L D THEOR Y

CHAPTER 14

FIELD INTENSITY AND POTENTIAL 14.1. Law of Force Between Point Charges

According to Newton's Law of Gravitation (Chapter 4) two particles of masses m1 and m2 at a distance r apart attract one another with a force F given by the equation

(14.1) y being the gravitational constant. If the particles are now replaced by small bodies of negligible mass, which have been given electric charges,* it is found that a force of attraction or repulsion between the charges arises which obeys a similar inverse square law. If the charges are of the same sign, the force is one of repulsion, whereas if they bear opposite signs the force is one of attraction. In either case, if ev e2 specify the charges both for magnitude and sign, the force between them is given by

F - k erle22. .

(14.2)

Coulomb being the first to obtain experimental verification of this law, it is referred to as Coulomb's Law. The constant k, unlike y, depends upon the medium in which the charges are immersed. Until Chapter 16, we shall suppose that the charges are placed in a vacuum or in air, since the value of k for the latter is found to differ only very slightly from its value for a vacuum. It will be convenient to suppose the unit of charge to be selected so that for a vacuous medium, k = 1. Thus, the electrostatic unit of charge is defined to be that charge which, when placed one centimetre from a similar charge in a vacuum, experiences a force of one dyne.t In these units, therefore,

(14.3) - er2le2 . . If e1 and e2 have the same sign, F is positive. F having a positive F -

sign accordingly indicates a repulsion between the charges, whereas its having a negative sign indicates attraction. In the gravitational case,

* In this, and the subsequent chapters of this Part, we shall assume that the reader is familiar with the simpler electro-magnetic phenomena and the terms which are employed to describe them. t This unit is too small for practical purposes. Instead we employ the coulomb, which is 3 X 109 e.s.u.

353

354

A COURSE I N APPLIED MATHEMATICS

[cH.

however, F calculated from equation (14.1) will be positive although there is an attraction between the particles. This difference in sign between the two cases will recur throughout the subsequent field theory, a theory which otherwise is largely common to the two classes of phenomena.*

14.2. Fields and Lines of Force Any particle placed in the vicinity of a distribution of matter will experience a force, each particle of the distribution attracting it accord­ ing to Newton's Law. Similarly, an electric charge placed near other charges will also be acted upon by a force. This phenomenon will be observed throughout a region, having more or less indefinite bound­ aries, with the matter or charge distribution in its interior. Such a region is referred to as a gravitational (or electric) field. Suppose that a particle of unit mass is placed at a point in a gravi­ tational field. Assuming that its introduction does not cause any rearrangement of the matter distribution, the force which acts upon it is termed the field intensity due to the matter distribution at the point. In this way we associate with each point of the field a vector G and, when a formula is available by the use of which we may calculate the magnitude and direction of this vector at all points of the field, the latter is completely specified. If the particle of unit mass is replaced by another of mass m, the force which acts upon m due to any particle of the matter distribution will, according to equation (14.1), be different in magnitude from the corresponding force on the unit particle by a factor m. The direction of the force remains unaltered, however. The resultant force F acting on m due to the whole distribution is accordingly given by F = mG . .

(14.4)

In a similar fashion, by introducing a unit positive charge into an electric field, we can find the electric intensity vector E at any point. If, however, part of the charge distribution responsible for the field is to be found on bodies which are electrical conductors, this charge will be free to move through the material of the conductor if attracted or repelled by other charges. It follows that the introduction of the unit charge will result in a redistribution of charge and a consequent distortion of the field we are attempting to investigate. To overcome this difficulty, we can imagine the field to be probed by a charge e of such small magnitude that it has negligible effect upon the other charges present. The force F which acts upon e when it is placed at any point is given by the equation F=

eE, .

and hence, if F is measured, E can then be found.

(14.5)

* The gravitational law of attraction is converted into the electrostatic one by putting y = - I .

1 4]

F I E L D I N TEN S I T Y A N D P O T E N T I A L

355

As a particular case, consider the gravitational field due to a particle of mass m at 0. The force exerted on a particle of unit mass at a point P is of magnitude ymfr2 where r = OP, and acts along PO. Thus, if

,

�

OP = r,

(14.6) Similarly, the field due to a charge e at 0 is given by the equation

E = 3re r .

{14.7)

.

Any field can be mapped by the drawing of lines of force. A line of force is a curve drawn in the field having the property that the tangent to it at any point is in the direction of the intensity at that point. An arrow on a line of force indicates the sense of the intensity vector. The lines of force due to a single charge or particle are clearly straight lines radiating from the charge or particle.

FIG. 14.1 .-Equation of Lines

of Force

To calculate an equation of a line of force in a field due to a set of collinear charges (or particles) , we proceed as follows : Let ev e2 , be the charges at points P1, P2 , P3 , e3 , (Fig. 14.1) and let P be any point on a line of force in the field. Let (r1 , 61) be polar co­ ordinates of P with respect to P1 as pole and the line of the charges as initial line. h, 62), (r3 , 63), are defined similarly relative to •

•

•

•

•

•

•

•

•

356

A COURSE

IN

[ctr.

APPLIED MATBEMAT1CS

P2 , P3 , as poles. Let r/>1 be the angle made by the tangent to the line of force at P with the radius vector Then •

•

•

sin 'I' ,n

s

=

P1P. r1 dfds\

(14.8)

where is the arc length parameter measured along the line of force. The component of the electric intensity due to at P in the direction of the normal to the line of force at this point is

e1

� del r\1 sin rPl r1 ds . .

(14.9)

=

This component of the intensity due to the whole charge distribution is accordingly � de1 � de + r1 ds + r2 ds2

·

·

·

0

=

(14.10)

·

being equal to zero, since the resultant intensity is in the direction of the tangent by the definition of a line of force. If y is the perpendicular distance of from the line of charges, then

P

and hence, by equation

(14.11)

(14.10),

el Sill. e1 deds1 + ez Sill. e2 deds2 + .

.

.

Integrating this latter equation with respect to

e1 cos e1 + e2 cos e2 + .

. . =

=

0

.

(14.12)

.

s, we obtain

constant.

(14.13)

This equation is true along any line of force and, by taking different values for the constant, we obtain all lines of force of the family. Two charges, 4e and -e, are placed at a distance a apart. Show that there is a neutral point at a distance a from the charge -e. Show that the line offorce which leaves the positive charge at an angle !7T to the line of charges goes off to infinity and ultimately makes an angle cos-1 ! with the l-ine of the charges.

Example 1.

(D.U.)

Let N be a point on the line of charges distant a from -e and 2a from 4e. The intensity at N due to the negative charge has magnitude efa2 and that due to the positive charge has magnitude 4e/(2a)2 = efa2• These are in opposite senses. Hence the resultant intensity at N is zero. The equation of any line of force in the field is 4e cos 81 - e cos 8 2 = C, 81 and 8 2 being as shown in the diagram. For the particular line of force we are to consider, when is at A , 8 1 = !7T and 8 2 = 7T. Thus C = e and for this line of force 4 cos 81 - cos e z = 1.

P

F I E LD I N T E N S I T Y A N D P O T E N T I A L

14] When

P has receded to infinity, 6 1 = 6 2 = 6 say.

4 cos e proving the result stated.

- cos e

= a

cos e

=

357

Thus

1,

In general, if (l, m, n) are the direction cosines relative to rectangular axes Oxyz of the electric intensity E at a point P having coordinates (x, y, z), the components (Ex, Ey, Ez) of the intensity are given by

Ex = lE, Ey = mE, E. = nE. (14.14) Hence (Ex, Ey, E.) form a triad of direction ratios of E. If P' is a neighbouring point (x + dx, y + dy, z + dz) on the line of force through P, direction ratios of P P' are (dx, dy, dz). But PP' is in the same direction as E. Hence dx dy dz (14.15) E., = E1 = £; These equations determine the family of lines of force in the field.

14.3. Elementary Distributions In this section we will calculate the field intensity vector due to a number of elementary distributions of matter (or electricity) .

14.3(1). Uniform Bar

Consider firstly the field due to a uniform bar AB (Fig. 14..2) of length l and mass m per unit length. Let P be any point in the field and PN = p the perpendicular from P on to the line of the bar. Let Q be any point on the bar and let QN = x. The intensity at P due to the element QQ' = dx of the bar is ym dxjPQ2 in the direction PQ. But, if LNPQ = e, x = p tan e and hence dx = p sec2e de, where LQPQ' = de. Also PQ = p sec e. Thus the intensity at P due to the element is equal to

(14.16)

A C O U R S E IN A P P L I E D M A T H E M A T I C S

358

[ctt .

Suppose a circle be constructed with centre P and radius p, inter­ secting PA, PB in A 1, B 1 respectively. Consider a uniform bar of line density m occupying the circular arc A 1 B1• Let RR' be the element of this curved bar corresponding to the element QQ1 of the straight bar. Since RR1 = pde, the intensity at P due to this element is ymp dejpz = ym dejp in the direction PR. We conclude that the elements QQ' and RR1 have precisely the same effect at P and consequently that the curved bar A 1 B1 produces the same intensity at P as the straight bar. 8

FIG. 14.2.-Field Due to a Uniform Bar

P due to the curved bar is directed along Let LAPN = ex, LBPN = �. so that LAPB = � - ex, LCPN = t (ex + �) and LCPQ = e - t (ex + �). The component of the intensity due to RR 1 in the direction PC is then By symmetry, the field at the bisector PC of LAPB.

found to be

Y; cos {fJ

-

i(ex + �)} d(J

and the total intensity along this line is found by integration to be

2'j t cos {6 - t (ex + �) } dO = 2;m sin t (� - ex)

. LAPC. (14.17) . = 2ym --p be in the direction PC, this is its Slll

Since the intensity is known to magnitude. The lines of force must be such that the tangent to any one of them at a point P must bisect the angle APE. Any hyperbola with its foci at A and B possesses this property, and we therefore conclude that the family of lines of force in any plane containing the bar is a family of confocal hyperbol::e as shown in Fig. 14.2. A uniformly positively

F I E LD I N T E N S I T Y A N D P O T E N T I A L

14]

359

charged bar will possess the same field with lines of force directed away from the bar. If the bar extends to infinity in both directions, LAPB = 1t and hence LAPC = t7t. Equation (14.17) shows that the field intensity due to such a bar at a point distant p from it has magnitude 2ymjp. The direction of the field will be everywhere perpendicular to the bar, and the lines of force will accordingly be straight lines normal to the bar. B is a point on a semi-infinite, uniform, straight rod of line density a and A is the accessible end. If the rod is bent at B so that the portion BA makes an acute angle 6 with the remainder, show that the couple at B due to the mutual attraction between 1X the two portions is "'1 ,. ya2AB cot !6. (L.U.)

Example 2.

Let be a point on AB such that B x. Let = dx be an element of AB. If is parallel to the infinite portion of the rod, the intensity at due to this portion will act along the bisector of the angle and will be of magnitude

P PP'

P

P

=

PX

BPX

2Ya"

sin H1T - 6) = r! cosec !6. x x sm v The force F acting upon the element of mass adx can now be found by application of equation (14.4) . I t is in the same direction as the intensity and of magnitude

A

PP'

F

ya•

= x dx cosec

!6.

As shown in Chapter l l , this force is equivalent to an equal and parallel force acting at B, together with a couple of B moment equal to the moment of F about B, viz., Fx sin !(1T - 6) = Fx cos !6 = ya2 cot !6dx. Integrating over all elements of AB, we now find that the net couple has moment ya2AB cot !0.

14.3(2). Uniform Circular Disc We shall establish a result of a general nature, applicable to any uniform, flat plate, and then apply it to the case of a circular disc. Let C (Fig. 14.3) be any closed curve, not necessarily plane, and let P be a point in its vicinity. If P be connected to the points of C by straight lines, the conical apex formed at P is called the solid angle subtended at P by C. To obtain a measure of this angle, we imagine a sphere of unit radius constructed with P as centre. The conical surface with apex at P will intersect the sphere in a closed curve C'. The area of the sphere enclosed by this curve C' is taken as the measure of the solid angle at P. The reader should compare this method of

360

A COURSE IN APPLIED MATHEMATlCS

lcH.

measuring a solid angle with the method of measuring a plane angle in radians by means of the length of an arc of a unit circle. Clearly C' divides the spherical surface into two regions, either of which may be thought of as being bounded by C'. The smaller region is regarded as

FIG.

14.3.-Solid Angle

the measure of the solid angle w and the larger as the measure of the reflex solid angle D. It will be seen that w

(14.18) + Q = 4rr, . and that the largest solid angle is 4rr. In the particular case when C is a circle and P lies on the axis of C (Fig. 14.4), C' is also a circle. Let IX be the semi-vertical angle of the

FIG.

14.4.-Solid Angle Subtended by a Circle

right circular cone forming the solid angle. Then the smaller segment of the sphere bounded by C' has area 2rr(1 - cos IX) and this is accordingly the solid angle subtended at P by C. Consider any uniform plane plate of surface density cr bounded by the ' contour C (Fig. 14.5). Let P be any point in the field due to the plate and let PN be perpendicular to the plate. The intensity at P due

14]

F I E LD I N T E N S I T Y A N D P O T E N T I A L

361

to an element dA of the plate is ycrdAJPQ 2 in the direction PQ and the component of this intensity along PN is therefore given by

dG = ycr dAPQcos2 6 .

(14.19)

n

Let d w be the solid angle p subtended at P by the bound­ ary of dA. Let dA 1 be the element of the surface of the sphere of centre P and radius PQ cut off by this solid angle. Then dA 1 is similar to the element of area dw of the unit sphere centre P, and hence dA = PQ 2dw. But dA 1 is nearly a plane area with PQ as its normal and dA is a plane area with PN in the direction of its normal. Since the angle between PQ and PN is 6, this c is the angle between dA and dA 1• Also dA 1 is the ortho- FrG. 14.5.-Field Due to a Uniform Plate gonal projection of dA, provided these elements are small. Hence dA 1 = dA cos 6. It now follows that dA dA cos 6 . 1

I

dw - PQ2 - PQ2

(14.20)

Thus

dGn = ycr dw. Integrating over all elements dA of the plate, we find that the com­ ponent of the total intensity at P along the normal to the plate is Gn = ycr J dw = ycrw,

.

(14.21)

where w is the solid angle subtended at P by C. In the particular case when the plate is circular and P lies on its axis, this component of the intensity is given by the expression

(14.22) 2rcycr(1 - cos ) being the angle shown in Fig. 14.4. But, by symmetry, the total intensity at P must be directed along the axis of the disc, so that the formula (14.22) gives the total intensity in this case. o:

o: , .

A C O U R S E IN A P P L I E D M A T H E M A T I C S

362

[CH.

A portion of a paraboloid of revolution y2 = 4ax (0 � x � a), where is the distance from the axis of symmetry, is of uniform density p. Show that the gravitational force at the focus is of magnitude 41Typa(1 - log 2) . (D.U.)

Example 3. y

Consider the contribution to the intensity at the focus S of the elemental disc between the planes x and x + dx and of radius y. We have PS = a + x, by the focus-directrix property and SN = a - x. Thus cos LPSN =

SN a-x = SP -a + x·

Since p dx is the surface density of the disc, it now follows that the intensity at S due to this element is

(

21Typ dx 1 -

in the direction

SO.

f

=

41Typx dxf(x + a)

Integrating, we calculate that the total intensity at S is 41Typ

14.3(3) Uniform Sphere a

: � ;)

0

a x dx -- = 41Typa(1 - log 2). x + a

Let be the surface density of a spherical shell of radius a and let P be a point, either inside or outside the shell, distant r from the centre 0 (Fig. 14.6). Consider the contribution to the field at P of an ele­ mental ring of material intercepted on the surface of the sphere by two planes perpendicular to OP. Let Q, Q' be points, one on each edge of the ring and lying in a plane through OP. Let LQOP = e and LQOQ' = de. The ring can be opened out into a strip of width a de and length 2rrQN = 2rra sin e. Its area 1s accordingly 2rra2 sin e de and its mass is 2rrcra2 sin e de. The intensity at P due to any particle m of the ring is ymju2, where u = PQ and its component in the direction PO is ym cos cpfu2, where

14]

FIELD INTENSITY AND POTENTIAL

363

� = L OPQ. The component along PO of the intensity due to the whole ring at P is therefore

7tcra2 � y cos � "" 2y "" ym cos � £.m � s e cos 'f' A.. de 2 u -

-

-

£.

m •

.

(14.23)

But, from considerations of symmetry, we can conclude that the intensity at P due to the ring is directed along PO. Thus, equation (14.23) defines the total intensity due to the ring at P.

FIG. 14.6.-Field Due to a Spherical Shell

Now so that

u2 = a2 + r2 - 2ar cos e,

(14.24)

u du = ar sin e de.

(14.25)

Also

(u2 + r2 - a2)j2ur.

(14.26) Substituting from the last two equations into equation (14.23), we cos � =

find that the intensity due to the ring can be written

r:�a ( 1 + r2 � a2 ) du.

(14.27)

Integration with respect to u will now yield the intensity at P due to the whole shell. There are two cases. If P is outside the shell, the extreme values of u corresponding to the elemental rings at A and A' are (r - a) and (r + a) respectively. The intensity is then

y1tcra 1 r+a ( 1 + r2 - a2 ) du = 4y1taa2 L M r2 r2 r2 , (14.28) u2 4M2 cr is the total mass of the shell. Equation (14.28) r-a

-

=

where M indicates that at an exterior point a uniform spherical shell attracts =

A C O U R S E I N A P P L I E D MATHEMAT I C S

364

feR.

like a particle of equal mass at its centre. This remarkable result is due to Newton. If, however, P lies inside the shell, the limits of u will be found to be (a - r) and (a + r) , so that the integral for the intensity is

yrr:a a+r 1 + 2 � a2 ) du = 0. u r Ja-r (

(14.29)

yMr3 = yM r. a3y2 aa-

(14.30)

r

Thus there is no field within the shell. If a charge e be given to an insulated spherical conductor, it will be proved later (Section 15.3) that it resides on the surface. On account of the symmetry, it will be spread uniformly over the surface and its field outside the conductor will be indistinguishable from that of a point charge e at the centre. Inside the conductor, there will be no field. The problem of a solid sphere in which the density depends only upon distance from the centre may now be solved by dividing the body into concentric spherical shells and integrating their contributions. At an external point, each such shell attracts like a particle of equal mass at the centre, and hence the whole sphere is equivalent to a particle of equal mass at its centre. At an internal point P, only those shells not containing P make any contribution to the intensity at P, and these all attract at P like a particle of equal mass at the centre 0. If the sphere is uniform, of mass M, radius a, and if OP = r, the total mass of the elemental shells of radii less than r is Mr3ja3 • Their attraction at P is consequently of magnitude

Thus, inside a uniform sphere, the field intensity is directed towards the centre and is proportional to the distance from the centre. A uniform sphere of radius a and density cr is surrounded by a uniform spherical ocean of outer radius b and density p. Find the attraction of a point in the ocean at distance r from the centre, and show that it will be a minimum at a point within the ocean if

Example 4.

fp <

cr

(

:�.)

·

(L.U.)

It will be convenient to regard the distribution as comprising a sphere of radius b and density p impregnated with matter of density (cr - p) occupying a concentric sphere of radius a. The intensity at P due to the first sphere (for which P is an internal point) is given by equation (14.30) to be 1jyrrpr.

The intensity at P due to the second sphere (for which P is an exterior point) is 4yrra3 (cr - p) 3r2

•

The total attraction at P is therefore given by G

=

[

�yrr pr +

� (cr - p)J

1 4]

FI ELD I NTENS ITY AND POTENTIAL

365

dG = Y7T [p 2a3 a p) dr a - ra ( - J and G is a minimum if dGfdr = 0, i.e., if r3 = 2 (� - 1 ) a3. Now

4

The value of r given by this equation will correspond to a point within the ocean if a3 < 2

(� - l ) a3 < b3,

which conditions are equivalent to the inequalities stated above.

14.4. Two-Dimensional Fields The field due to a number of parallel bars of infinite length has the same characteristics in any plane perpendicular to the bars. Such a field is therefore completely determined by its configuration over any such plane section. The field is said to be two-dimensional. The electric field due to parallel-line charges of infinite length is similarly two-dimensional in this sense. Suppose that ev e2 , are the charges per unit length on a set of parallel-line charges which intersect a particular plane of section in the points Pv P2 , (Fig. 14.7) . Let P be any point on a line of force in the plane and let (r1 , 61), (r2 , 62) , be polar coordinates of P relative to poles Pv P2, respectively and any initial lines through these points. The intensity at P due to the line charge through P1 was shown in Section 14.3(1) to be 2edr1 and to act in the direction •

•

•

•

•

•

•

•

•

•

•

.

366

A C O U R S E I N A P P L fE D M A T H E M A T I C S

[CI-1 .

P1P. If c/>1 is the angle between the tangent to the line of force at P and P1P, the component of this intensity along the normal is . de1 , 2e -1 sin -�.. (14.31) ds r1 't'I = 2e1 -

where s is the arc length parameter on the line of force.

FIG. 14.7.-Lines of Force due to Line Charges

The total normal component of intensity due to all the line charges 1s zero. It follows that

2e1 deds1 + 2e2 deds2 +

·

·

·

=

0

·

·

Integrating this equation with respect to s, we obtain e1e1 + e2e 2 + . = constant, an equation which is valid along any line of force. .

.

(14.32) (14.33)

Show that, if three parallel wires carrying unifo1'm charges -e, 3e, - e per unit length meet a cross-section of the field in collinear points A, 0, B, where A O = OB = a, then there are two equilibrium points in the plane. Prove also that the lines of force which go to infinity are separated from those which do not by the curve r2 = a 2 (4 cos 2 (l - 1), where 0 is the origin and (l is measured from OB. Draw the lines offorce. (M.T.)

Example 5.

Let Q, R be the neutral points and let OQ = OR = x. Since the intensity vanishes at these points, - � + � - � = 0. Thus x = y3a.

x-a

x

.x + a

FIELD INTENSITY AND POTENTIAL

14]

367

After a little thought it will be understood that the configuration of lines of force must be as sketched in the diagram. The curve of separation between the lines which leave 0 and depart to infinity and those which terminate at A or B is that formed from the two lines which pass through Q and the two which pass through R.

I

' ' I I I

I

I

,'

I

I I I

I

. , I

I

,

I

I

/

,

;("

; ,.

.,.

"

"

� �

R

Q

. .

.

I

I . . . I

,

, I

.

. \ . \ \ . \ \ \ \

�

\

\

\

\

Let P be any point on the line of force OR and let Ov a 2 be defined as shown. Then -ea 1 + 3ea - e6 2 = constant, 3a = a 1 + a. + constant. or But this line terminates at R with a 1 = 6 2 = 6 = 0. Hence the constant in the last equation must be zero, and 3o = a 1 + e • . From the geometry of the figure it is clear that r sin a r sin e tan a 1 = tan e2 = r cos e - a, r cos 6 + a Hence tan 6 1 + tan ° • ) tan 36 = tan (6 1 + 0 2 = l - tan 61 tan e . r2 sin 26 = r2 cos 26 - a•' and it follows, upon solving for r2, that sin 36 a• = a2 (4 cos• a - 1). r2 = sin () This equation includes the other three arcs of the curve of separation, since it is unaltered when 6 is replaced by -a or ., ± a. .

368

A C O U R S E IN A P PLI E D M A T H E M A T I C S

[cH.

14.5. The Potential Function

If a particle is moved in any manner from one point to another in a gravitational field, the work done by the force acting upon the particle due to the field is independent of the particle's track between the two points. For suppose this were not so. Let A, B (Fig. 14.8) be the two points and assume, if possible, that there are two paths B rl> r2 j oining them such that the work done by the attraction on the particle as it moves from A to B via ri is WI and via r2 is W2 , where WI > w2. Imagine that two smooth tubes are bent into the shape of the curves r 1, r 2 and are fixed rigidly in position along these paths. Suppose that the tubes are j oined at A and B by small, smooth sections of large curvature for the purpose of A FIG. 14.8.-Two Paths in a Conservative diverting a particle, in motion in Field one tube, into the other. If, now, a particle is introduced into the circuit at A and is projected along the path rI with sufficient momentum to enable it to reach B, work of amount W1 will be performed upon it whilst it moves between these points. It follows from equation (3.51) that the particle's KE increases by WI . It now moves along the second path r 2 from B to A and the work performed upon it is W2• In the second tube it accordingly suffers a KE loss of W2• However, upon arrival back at A its KE has increased by a net positive amount of WI - w2. The particle can now be permitted to perform further circuits indefinitely, and the energy of the system will continually increase in defiance of the law of conservation of energy. Hence WI :t w2. But neither is WI < W2, for, in this case, by allowing the particle to complete circuits in the opposite sense, we should again be able to defy the law of energy conservation. Hence WI = W2. A field having the property we have j ust shown to be possessed by a gravitational field is said to be conservative (cf. conservative system, Section 6.3). We infer that an electrostatic field is also conservative by a similar argument to that which has just been applied to the gravi­ tational field. Let D be a fixed datum point in an electrostatic field and let P be any other point. If a point charge at P is moved to D by any route, assuming that the charge distribution is not altered during the motion, an amount of work is performed by the field force acting upon the -

14]

FIELD INTENSITY AND POTENTIAL

369

charge which, after D has been chosen, depends only upon the position of P in the field. This work is called the potential energy which the charge has when at P. This definition is in accord with that already introduced in Section 3.4, and the argument of that Section is relevant to the motion of the charge in the electric field. The potential at P is defined to be the PE of a unit positive charge at this point.* We shall denote this quantity by V, so that V is a function of the co­ ordinates of P. In calculations involving the potential, it is generally convenient to take D to be at infinity, i.e., at a great distance from the charge distribution responsible for the field, where the intensity is negligible. The potential function in a gravitational field is defined similarly as the PE of a particle of unit mass. Let P be any point in an electrostatic field at which the potential is V. Let P' be a neighbouring point at which the potential is V + dV. If a unit positive charge be moved from P to the datum point via P', the work which will be done by the field will be V. But the work done by the field over the arc of this path from P' to the datum point is V + d V. It follows that the work done by the field as the charge moves from P to P' is -dV. Let PP' = ds and let Es be the com­ ponent of the intensity in the direction P P'. Then the work done over the arc PP' is also Esds. Thus or

Esds = -d V, E

s

av

= --, OS

(14.34)

partial derivative notation being employed, since V is a function of three coordinates. Equation (14.34) states that -E is a vector whose component in any direction is the rate of change of V in that direction. Such a vector is termed the gradient of V and is denoted by grad V or \l V. Thus E = -grad V . . (14.35) Consider the points in the field at which the potential takes an assigned value V0 • Since V is a function of the coordinates (x, y, z) of points in the field, the equation V = V0 (14.36) defines a surface called an equipotential. By taking different values for V0, we generate a family of equipotentials. Consider the equi­ potentials corresponding to the values V, V + dV of the potential. * If the work is measured in ergs and the charge is an electrostatic unit, the potential is in e.s.u. 1/300 of an e.s.u . of potential is the practical unit called the volt.

370

A C O U R S E IN A P P LI E D M A T H E M AT I C S

[cH.

Let P be a point on the first equipotential and P' a neighbouring point on the second (Fig. 14.9). Let PN be normal to the first equipotential, intersecting the second in N. Then, if dV is very small, PN will be normal to the second equipotential and we can assume L PN P' = right angle. Taking P P' = ds, PN = dn, LP'PN = 6, we have

V

V -t- d V

FIG. 14.9.-Gradient of v

oV dV dV dn av Ts = ds = dn . ds = on cos e. (14 · 37) Consider the vector of magnitude oVI on in the direction PN. According to equation (14.37), its component in any direction p P' is oVI OS. It follows that this is the vector grad V. Thus, if n is the unit

normal at P to the equipotential through P, then grad v

= oV on n.

(14.38)

It now follows from equations (14.35) and (14.38) that the lines of force are the orthogonal trajectories of the equipotentials. Example 6.

If r is distance measured from a point 0 and V = f(r) , the equipotentials are spheres concentric on the point 0. -+ If P is any point and OP = r, the unit normal at P to the sphere through P is rjr. Also, a av 8n = i[r f(r) = f' (r) .

Hence

'r grad V = f ( ) r. r

In particular, if V = efr,

E = -grad V = _e. r. r

This implies that efr is the potential function corresponding to a point charge e at 0. This will be proved to be the case below.

The components of grad V in the direction of rectangular axes Ox, Oy, Oz are oVIox, oVIoy, oVIoz. Hence oV ov Ez = Ex = - ' (14.39) ox oz Also, if i, j, k are unit vectors along the axes, we can write oV . ov . oV k . grad V = (14.40) ox 1 + oy J + oz . -- ·

14]

371

F I E LD I N T E N S I T Y A N D P O T E N T I A L

I t is often convenient t o look upon " grad " as an operator which, when applied to a scalar function of position, generates a vector function. Clearly v

= grad =

. 8

. 8

ox + J oy +

1-

k 8

-·

(14.41)

8z

Since V is a scalar, whereas E is a vector, it is more convenient to specify a field in terms of the potential than the intensity. The components of intensity can always be found when the potential is known by the use of equations (14.39) . We shall now obtain the potential due to a particle of mass m. Let P be any point in the field distant r from the particle (Fig. 14.10). Let Q be any point on a path joining P and a point oo at a great

p

0

m

FIG. 14. 10.-Potential Due to a Particle distance from 0, the position of m. Let OQ = p . Then the force which acts upon a unit particle at Q is ym/p 2 and the work done by this force as the unit particle is moved from P to oo is -

f oo 0

1'2 cos � ds,

m

(14.42)

p

where � is the angle shown in the figure between the tangent to the path and OQ, and s is the arc length parameter measured along the path from P. But cos � = dpfds, so that the work done may be written

- Io"' ym ds11:£ ds -1"' ypm2 dp ym "' I I m =

p2

=

This is the potential at

P.

r

_y .

r

=

P

r

(14.43)

372

A C O U R S E IN A P PL I E D M A T H E MAT I C S

[cH.

Putting y = -1 and m = e, we find that the potential at a distance r from a charge e is given by V = r�- . (14.44) A conducting body comprises atoms, the electrons of which are free

to move, under the influence of an electric field, from one part of the conductor to another. If such a body, charged or uncharged, is placed in the vicinity of a charge distribution, the free electrons will flow through the material until the resultant field inside the conductor has been reduced to zero, when the flow will cease. It follows that in a field due to charges on conductors and other bodies, if it is supposed that the charges are stationary, so that the field is electrostatic, there can be no field intensity within the conductors. Hence, at any point in a conductor, the derivative of the potential in any direction is zero. But this can be the case only if the potential takes a constant value throughout each conductor. Thus we are entitled to speak of the potential of a conductor, provided the charge distribution is static. A charge distribution in a conductor caused by an external field is called an induced charge. It will be proved in Section 15.3 that, under static conditions, any charge on a conductor must reside entirely upon its surface. The datum point at infinity employed in the definition of electrostatic potential is usually taken to be at a distant point on the Earth's surface. At this point the potential is zero and, since the Earth is a conductor, we deduce that its potential is everywhere zero. If, therefore, any point is connected to the Earth by a conducting wire, i.e. is " earthed ", its potential is immediately reduced to zero. In an insulating body, on the other hand, charges are not free to move through the material, and the potential can alter from point to point. Since the work done in any displacement by the resultant of a number of forces acting upon a particle is equal to the sum of the works done by the individual forces, the potential at a point due to a distribution of point charges or particles is equal to the sum of the potentials due to each separately. Thus, if ev e2 , are a set of point charges and r1, r2 , are the distances of a point P from them, the potential at P due to the distribution is given by •

•

•

•

•

•

(14.45) Although, according to modern views, both matter and electricity are atomic in character, it is frequently convenient in theoretical work to replace an actual distribution of point charges or particles by an hypothetical continuous distribution, spread over either a three­ dimensional region of space or over the surface of some body. A

1 4]

F I E LD I N T E N S I T Y A N D P O T E N T l A L

373

volume distribution of charge is then specified by stating the density of the charge at every point of the distribution. Then, if p is the charge density at a point, the charge which will be found in any small region of volume dv enclosing the point will be p dv. Similarly, a distribution of charge over a surface will be defined by a surface density function cr such that if dS is a small element of the surface, the charge upon it is cr dS. Suppose that P is a point in the field due to a volume distribution of charge p. Let dv be the volume of an element of the region occupied by the charge. Provided this element is small, the charge p dv it encloses may be regarded as a point charge and the potential due to it at P will be p dvjr, where r is the distance of P from dv. The potential at P due to the whole charge distribution is then

V=

J

P

;v ,

(14.46)

the volume integral being calculated over the whole region in which charge is present. Similarly, the potential due to a surface distribution is given by the equation

V = j cr;s,

(14.47)

the surface integral being taken over all the charged surfaces. It follows that the potential due to a combined volume and surface distribution is given by

(14.48) A particle moving in a plane has polar coordinates (r, 6) and is acted on by radial and transverse forces (R, T) which are functions of (r, 6) . If a potential energy V exists, show that d V = -Rdr - Trd6, and hence that oRfo6 = o(Tr)for. Examine which of the following fields of force (in which k denotes a constant) is conservative, and find the PE when it exists: (a) R = k/r3, T = 0, (b) R = k cos 6, T = -k sin 0, (c) R = 0, T kr2• Li. U.)

Example 7.

=

(

Let V be the PE of the particle at P(r, 0) and let V + dV be its PE at the neighbouring point P'(r + dr, 6 + d6). If the particle moves from P to P', to the first order of small quantities the components of its displace­ ment along and perpendicular to OP will be dr and r d6 respectively. The work done by the field forces will accordingly be R dr + Tr dO. But, as proved on p. 369, this work must also be -dV. Hence d V = -R dr - Tr d6. (i) If, now, d6 = 0, i.e., 6 is constant during the displacement, we can deduce from this equation that av = -R. (ii) Br

374

A C O U R S E IN A P P L I E D M A T H E M A T I C S

[cH.

Similarly, by taking dr = 0 , we can show that av = - Tr. ae

Hence

(iii)

a o2 V oR = = Br (Tr) . iJ8 oroe

(iv)

We have proved that this is a necessary condition that the field should be conservative. It is also a sufficient condition. The field of case (a) clearly satisfies this condition. Its potential must be such that av ar

= rk"'

Hence V = -kj2r2 + constant.

T

R p

0

In case (b) the condition is again satisfied and V must be such that av

er

= -k cos e,

av . = -kr sm e. ae

p

Thus, v = - kr cos 0 = -kx, where X = r cos e is the %-coordinate of measured parallel to the initial line 6 = 0. We note that o VJox = -k, so that the field intensity in the direction of the x-axis is everywhere equal to the constant k. The components of intensity along y- and z-axes perpen­ dicular to Ox are zero. It follows that we are here dealing with a uniform field, whose lines of force are parallel lines in the direction Ox. In case )

(c

oR = 0, iJ(J

a

8r (Tr)

= 3kr2,

and the field is not conservative.

14..6. Potential Due to Elementary Distributions

In this section we will consider the potential due to the distributions considered previously in Section 14.3.

1 4]

:F l E .t D I N T E N S I T Y A N D P O T E N T I A L

375

15.6(1) Uniform Bar Employing the notation of Section 14.3(1) and Fig. 14.2 the potential at P due to the element QQ' of the bar is But PQ = p sec e, be written

y mdx. PQ

dx = p sec2e de and hence - ym sec e de.

this contribution may

/3 1 (sec � ! !an � ) V = -ym sec e de = -ym log sec a an a

The net potential at P due to the whole bar is therefore IX

(14.49)

(x2 + p)t, the potential at P is equal to p2) ! v = - ym 1XXAB (x2 dX+ p2)t - - ")'m log XBXA ++ (x(xBA22 ++ p)t XB + rB (14.50) = - ym log + ' XA rA where rA = A P, rB = BP. Alternatively, since PQ =

This result may be expressed in another form thus : Since, by the Cosine Rule, rA2 = rB2 + l2 - 2lxB, rB2 = rA2 + l2 + 2lxA, it follows that equation (14.50) may be written

l) 2 - rA2 - ym log rB(rB2 + - (rA - l) 2 ' (rB + l - rA) b + l + rA) = - ym 1og (rB + rA - l)(rB - rA + l)' r + 1 A = - ym og r.1 + rBrB -+ (l

V=

(14.51)

From this last form of V, we deduce that the equipotentials are given by r.a + rB = constant . (14.52) and hence that the section of these surfaces by any plane containing the bar is a family of confocal ellipses with their foci at A and B. These ellipses intersect the family of confocal hyperbol<e forming the lines of force orthogonally (a well-known geometrical :figure) as appears in Fig. 14.2.

376

A C O U R S E IN A P P L I E D M A T H EMAT I C S

[CH.

If P is opposite the centre of the bar, the potential at this point may be calculated as double the potential due to either half of the bar, viz.,

V = -2ym f

sec 6

de = -2ym log (sec � + tan �) .

(14.53)

In this case, let us choose a new datum point from which to measure V, in such a position that the potential is altered by a constant 2ym log l. Then V = 2ym log l - 2ym log (sec � + tan �) . = 2ym log {l cot �/(cosec � + 1)}, (14.54) = 2ym log {2pj(cosec � + 1)}. . Now suppose that the length of the bar approaches infinity so that � --+ tn-. By inspection of equation (14.54), we see that in the limit

V = 2ym log p.

(14.55)

This is the potential function due to an infinite bar. It will be observed that as p --+ IX!, V --+ IX!, This implies that the work which will be done by the field force when a particle is brought from infinity to P will be infinite, indicating that a point at infinity is unsuitable for use as a datum point. This accounts for the necessity of changing the datum point during the above calculation. In the case of the infinite bar, therefore, the equipotentials are the coaxial cylinders p = constant, all of which are intersected orthogonally by the lines of force. The intensity in the direction of p increasing, i.e., along a line of force, is

= -� (2ym log p) = - oV ap ap

-

2ym , p

(14.56)

as obtained previously. A uniform bar is of length 2a and mass m. Calculate the grav-itational potential due to the bar at points distant 2a from its centre 0: (i) at P on its perpendicular bisector, and (ii) at Q on the line of the bar produced. The bar is free to rotate about an axis through its centre perpendicular to OP and OQ. A particle of mass M is placed at P and the bar commences to rotate from 1'est under the particle's attraction, until it points directly at the particle. If w is its angular velocity in this latter position, show that

Example 8.

w2

yM Iog 2 y 3 ' ---;za - 6 1 + v5

_

At P, rA = rB = y5a. Also l = 2a. Hence, by equation (14.51), ym 2 y'5a + 2a m = log ' = - ya log t(l + yG) .

Vp

2a

A t Q , rA = 3a, rB = a.

2 y 5a - 2a

Hence

4a + 2a ym log 4a" VQ = - ym = - a log y'3. 2a - 2a

1 4j

377

FIELD INTENSITY AND POTENTIAL

It now follows that the PE of the particle M in its initial and final positions takes the values ymM l + y'5) -a log t(

and

ymM - -- log y'3 a

respectively. The loss in the PE of the system is accordingly the difference between these quantities, viz.,

2y'3 . ymM 1 -- og a l + y'5 By the principle of energy, this loss is compensated by a corresponding increase in KE. But, if w is the angular velocity of the bar in its final position, its KE is then ima2w2• Hence 2 y' 3 ymM !ma2 w2 = -- log --a l + y'5 showing that w2 takes the value stated.

14.6(2) Uniform Circular Disc

Let P be a point on the axis of a uniform circular disc of radius a and centre 0. By considering the contribution to the potential at P of an elemental ring of radius r and thickness dr and then integrating, the reader should be able to show that the potential at P due to the whole disc is given by

V = - 27tycr[ y(z2 + a2) - z],

.

(14.57)

where cr is the mass per unit area of the disc and z = OP. Differentiating this result with respect to z, the expression for the intensity may be found.

(14.22)

14.6(3) Uniform Sphere Dividing. a uniform spherical shell into elemental rings as shown in Fig. 14.6, the reader should show that the contribution of such a ring to the potential at P is - 21t'ycra dujr (employing the notation of Section 14.3(3)). By integration it can now be established that, if P is outside the sphere, the potential there is - yMjr, i.e., is the same as that due to a particle of mass equal to that of the sphere and placed at its centre. If, however, P lies inside the sphere, the potential takes the constant value - yMja, implying that there is no field within the sphere. This confirms the results of Section 14.3(3). The potential at any point due to a stratified sphere can now be found by dividing it into elemental shells, finding the contribution due to each, and then integrating. At an external point any such sphere produces a potential - yMjr, where M is its total mass. At interior points, however, the potential is dependent upon the manner of varia­ tion of the density. Thus, if the sphere is uniform and P is a point distant r from the centre 0, the contribution to the potential at P of all

378

A C O U R S E IN A P P L I E D M A T H E M AT I C S

shells of radius less than mass placed at 0, viz.,

[cH.

r is equal to that of a particle of identical

(14.58) A shell of radius x ( > r) and thickness dx en­

where p is the density. closes P and contributes a potential

y . 4rtx2dx = - 4rtypX dX. X

(14.59)

The total potential due to all shells of this type is

-4rtyp [' x dx = - 2rtyp(a2 - r2),

a being the radius of the sphere. Hence, adding (14.58) and (14.60), we obtain the final result V = -irtyp(3a2 - r2), .

(14.60) the expressions

(14.61)

determining the potential within a uniform sphere. The radial intensity within the sphere is

yM r, --artypr = -7 (14.62) in agreement with equation (14.30). By putting y = - 1, the potentials due to corresponding electric -Tr

iW

=

4

charge distributions are easily found.

Of two concentric conducting shells the inner, a sphere of radius a, is given a charge e and the outer, a sphere of radius b, is earthed. Show that the ratio of the charge on the inner shell to its potential is abf(b - a) and calculate the charge on the outer shell. Let e' be the charge on the outer sphere. There is spherical symmetry so

Example 9.

that the charges will be distributed uniformly over their respective con­ ductors. Let P be a point between the spheres and distant r from their common centre. The potential at P due to the two spherical layers of charge is given by v

=

e

r

e'

+ · 5

(i)

Over the outer sphere, V = 0, since this conductor is earthed. Hence Substitution in equation (i) shows that e' = -e, i.e., the charge on the inner sphere induces an equal charge on the outer. Thus V = 0 when r = b.

(ii) Putting r = a in this latter equation, we find that the potential of the inner conductor is

1 4]

F I E LD INTENSITY A N D POTENTI A L

379

Hence

e ab (iii) V = b - a· The system we have been examining is called a spherical condenser, and the ratio (iii) is its capacitance. Clearly the capacitance measures the charge which will be absorbed by the inner conductor for unit rise in its potential. It is clear from equation (iii) that capacitance has the dimension of a length

and hence the e.s.u. is the centimetre. If, however, the potential difference between the plates of a condenser is measured in volts and the charge on the insulated plate in coulombs, the capacitance is in farads. 1 farad = 9 X 101 1 em. If the earthed sphere is removed, the potential of the remaining sphere is given by

V = a� and hence its capacitance alone is a. However, by making the difference (b - a) very small, the capacitance of the combination of two spheres can clearly be made very large. This accounts for the superiority in practical utility as a reservoir for electric charge of the combination of two spheres over a single sphere. More will be said concerning condensers (or capacitoys) in the next chapter.

14.7. The Electric Dipole

In certain circumstances the equal positive and negative charges resident within an atom become slightly displaced from one another and the atom behaves electrically like a combination of two point charges - e and +e a small distance d apart. Such a combination is referred to as an electric dipole, and the atom is said to be polarized. It is usual to assume that d is small compared with all other distances being considered, so that the dipole can still be regarded as a point entity. The vector whose magnitude is ed and whose direction is

[cR.

A C O U R S E IN A P P L I E D M A T H E M A T I C S

380

that of the displacement from the negative to the positive charge is termed the moment of the dipole. We shall denote it by m. Let 0 (Fig. 14.11) be a point midway between the charges and let P be any point in the field of the dipole. Taking 0 as pole of polar coordinates and the line of the charges as the initial line in the sense

FIG. 14. 1 1 .-Potential Due to a Dipole

shown in the figure, let P be the point (r, 8) . Then, since d is small, we have, to the first order in this quantity,

AP = r + !d cos 8, A 'P = r - td cos 8, and thus, the potential at P is given by V= =

e

e

e

- = A 'P - AP r -

ed cos 8 rz

__ _

=

[( 1 - 2r-d cos 8) - 1 - ( 1 + -2rd cos 8 )-1]

m cos 8 2

__ ' _

r

(14.63)

(14.64)

approximately throughout to the first order in d. Since mr cos 8 = m · r, where r =

�

OP, we may also write

m·r V = -3- · r

(14.65)

Another formula for the potential which we shall find useful in the sequel may be obtained as follows : Regarding P as a fixed point and 0 as a variable point, 1/0P = 1/r is a scalar function of position defined

14]

F I E LD I N T E N S I T Y A N D P O T E N T I A L

for points 0 in the region surrounding on p. 370, it will be noted that grad

=m

Referring to the example

U) = ?· .

a negative sign being omitted, since (14.65) V

P.

•

(14.66)

-+

PO = - r.

grad

381

Hence, by equation

0}

(14.67)

It must always be remembered when using this formula that the gradient is to be calculated on the understanding that 0 is the variable point. If (E,, Eo) are the radial and transverse components of the intensity at P, we may show, as in the Example on p. 373 (equations (ii) and (iii)) , that 8V Eo = --r1 (14.68) 86 · Substituting for V from equation (14.64), we therefore obtain 6 E _ 2m cos r3 ' r -

6. Eo = m sin ra

(14.69)

Alternatively, we may proceed vectorially thus. If E is the intensity -+

at P and P' is a neighbouring point such that OP' = r + dr, the work done by the field as a unit positive charge moves from P to P' is E dr. But, according to the argument on p. 369, this work must be -dV. Hence •

E . dr

Now r dr.

=

-d

(m . r )

=

ra

-

m . dr + 3m . r dr. r4

ra

(14.70)

r2 = r2 and hence, by differentiation, we conclude that r · dr = Equation (14.70) can therefore be written

·

E dr

=

(

·m + 3m · r r) dr . rs ra

--

--

·

This equation being true for all small displacements that

·r m 3m -rs r. . r

E = -a +

( 14. 71)

dr, we may infer (14.72)

This latter result exhibits E as the sum of two vectors, one along the -+

line of the charges and the other along 0 P.

382

(cH.

A C O U R S E IN A P P L I E D M A T H E M AT I C S

To obtain the equation of the family of lines of force, suppose ¢> is the angle made by the direction of the intensity at P with the radius vector from 0. Then, if r = r(6) is the polar equation of the line through P, Er 1 dr (14. 73) r d6 = cot ¢> = Eo = 2 cot 6. Integration with respect to 6 now yields the result log r or

=

2 log sin 6 + constant, r = C sin26 . .

(14.74)

As an exercise the reader should now sketch the assembly of lines of force. Suppose that the dipole is placed in a uniform field of intensity E (constant in magnitude and direction) . Equal and opposite forces eE will act upon the two charges as shown in Fig. 14.12. These forces constitute a couple of moment eEd sin 6. The axis of this couple is

��------------��� eE + E'

eE___/

7

L �E

FIG. 14.12.-Dipole in a Uniform Field

normal to the plane of the vectors m and E. Hence, if the vector specifies the couple both for magnitude and direction of axis, then

G = m X E.

.

G

(14.75)

The reader should check that the sense of the vector product has been selected correctly. If the dipole is placed in any electric field whose potential is V, its PE is the sum of the PEs of its constituent charges, viz., where A, A ' are the positions of the charges -e, +e respectively. But (VA' - VA ) is the increase in the field potential as we proceed from A to A' and, since AA' = d is small, we have very nearly

av

VA' - VA = -- d OS '

(14.76)

1 4]

F I E L D I N T E N S I T Y A N D P O T E NT I A L

383

where o VJ os is the derivative of V at the dipole's position in the direction AA ' Thus, if W is the dipole's PE, .

ov ov. W = ed = mOS OS

But oVI OS is the component of the field intensity E in the direction of m. Accordingly, we have finally -

W = -m · E. (14.77) If the charges - e and + e in Fig. 1 4 .11 are replaced by two parallel, infinite, line charges, one having a charge -e per unit length and the other a charge +e per unit length, distant d apart, the combination so formed is called a line doublet. Equation (1 4 .55) (with y = -1) shows that the potential at P is given by V

= 2e log AP - 2e log A ' P, = 2e [log ( l + t cos e ) - log ( 1 2ed cos e r

2m cos e r

-

i, cos e)J, (14.78)

where m = ed is the moment of the doublet and we have approximated to the first order in d. The reader may now calculate the radial and transverse components of the intensity by differentiation of this potential. 14.8. The Magnetic Dipole

It is a well-known property of a bar magnet that, if it is broken down into smaller pieces, each such piece exhibits all the properties of the original bar magnet. This subdivision may be continued to the stage where individual molecules of the magnetic substance are obtained and each molecule will be found to behave like a small bar magnet . It is found that this behaviour may be accounted for on the hypothesis that the molecule comprises two magnetic charges or poles, equal in magnitude but opposite in sign, separated by a very small distance . It has not proved possible to isolate these magnetic charges ; they invariably occur in pairs as we have described. Thus, the fundamental magnetic entity from which all magnets are supposed constructed is a dipole. Experiment reveals that the behaviour of the dipole is consistent with the hypothesis that magnetic poles attract or repel one another c

A C O U R S E IN A P P L I E D M A T H E M A T I C S

384

LcH.

according to an inverse square law of force, the force being one of re­ pulsion between like poles and of attraction between unlike poles. We shall choose the unit pole to be such that when it is placed one centimetre from a similar pole the force of repulsion is one dyne. Equation (14.3) is then valid for the calculation of the force in dynes between two magnetic poles of strengths e1, e2 a distance r em apart. The moment of a magnetic dipole is a vector quantity, defined in the same manner as the moment of an electric dipole. We can now extend the concepts of field intensity, potential, lines of force, etc. to the case of the field due to a distribution of magnetized material, though, since magnetic charges are never free to flow through the substance of a material, there is no magnetic body corresponding in its properties to an electric conductor. The Earth itself possesses a magnetic field, so that, if a magnetic dipole is freely pivoted in the Earth's vicinity, it is acted upon by a couple m X H, where m is the dipole's moment and H is the intensity of the Earth's field at the position of the dipole. This couple will be zero only when m and H are parallel. It follows that the dipole will rotate into alignment with the direction of the field. We may accordingly employ a small dipole to explore the direction of the Earth's field and it is found in this way that the direction is approximately north-south at most points on the Earth's surface. That pole of a magnet which points in the northerly direction when it is freely pivoted is therefore called the North pole and the other pole is termed the South pole. It is conventional to measure the N pole positively and the S pole negatively, so that the direction of the moment is along the line

SN.

Magnets of moments JL, JL', JL' are pivoted at the vertices A , B, C respectively of a triangle in which the angles B and C are equal. They are in equilibrium when they lie along the internal angle bisectors with their positive poles inside the triangle. Show that JL'f!L = 8 cosec �B cos3B (2 3 cos B) . (M.T.)

Example 10.

-

Let A C = b and BC = a. Then the components of the intensity at C due to the magnet at A are 2JL cos !A along A C, " b

JL sin !A . and --3- perpend1cular to AC. b Hence the component in the direction IX perpendicular to JL' at C is 2JL cos tA . JL sin tA -z;a- cos !C ,...-� sm li1C. •

JL'

-

Similarly, the component in the direction IX of the intensity at C due to at B is 2JL' cos tB . JL' sin !B --a3 - sm 21 C --a-3- cos li�c. -

14]

FIELD INTENS ITY AND POTENTIAL

385

But the net component of the intensity at C in the direction a must be zero, since the field at C is along the axis of the magnet there. Hence, putting A = 7T - 2B and C = B, we obtain the equation

to (2 sin B sin iB - cos B cos �B) = � sin !-B cos !-B. A

Or, since a = 2b cos B, J-L'!J-L = 8 cosec !-B cos3B (4 sin2!-B - cos B), = 8 cosec !-B cos3B - 3 cos B) .

(2

Consider two magnetic dipoles of moments m and m' at the points -+

Let PP' = r. The field at P' due to m is given by equation Referring to equation (14.77), we see that the PE of m' in the field due to m is given by m · m' - 3m · r m' · r. W - --

P, P'. (14.72).

-

rs

ra

.

(14.79)

Knowing the PE of a system of dipoles, its equilibrium configurations can be found by the method of Section 6.5. The stability of these configurations can then be discussed in the usual manner. Two magnetic dipoles, each of moment M, are free to rotate about their centres A , B in a plane through AB; in any position the axes of the dipoles make angles 6, 6' with AB. If there is a uniform external magnetic field of intensity H in the direction AB, prove that the work that must be done to move the dipoles to any position from the position 6 = 6' = 0 is M• MH(2 - cos 6 - cos 6') + + cos (6 - 6') - 3 cos 6 cos 6'},

Example 11.

aa-{2

where a is the length AB. Hence prove that the position 6 = 6' = 0 is stable provided that Ha3 > - M. (M.T.)

A C O U R S E IN A P P L I E D M A T H E M A T I C S

386

[CH.

Suppose the magnet at A is first brought into position. If m is its vector moment, its PE when in the position shown is -m H. Now suppose that the second magnet of moment m' is brought up to B. ·

�

If AB =

r,

the field at B

. IS

_I!!

3m · r

+

a3

a5

r +

H.

The magnet's PE in this field is

3m • r m' • r -u --� - --------as -- - m'

m • m'

Hence, the total PE of the system is given by

W = m as· m'

_

3m · r m ' · r

_

as

m H •

_

·

H

.

m' . H,

M• = - 3 [cos (a - a') - 3 cos a cos 6'] - MH (cos a + cos 6'). a

When a

= a' = 0,

W=

-

2M2

as - 2MH.

Accordingly, the increase in the system's PE as the magnets are moved from the configuration a a' 0 to that shown in the diagram is given by P

=

=

= MH(2 - cos a - cos a') + M• ii3{2 + cos (a - W)

-

3 cos e cos a'}.

P is equal to minus the work done by the forces of the system, and hence is equal to the work done on the system. Regarding P as the PE of the system, we have

�: = MH sin e - �2 [sin (8 - W) - 3 sin a cos a'], �:, MH sin a' + �2 [sin (8 a') + 3 cos a sin 6'], -

=

= 0.

and both these derivatives are zero when e = a' Thus P is stationary in this position and it accordingly represents an equilibrium configuration. Also B2P M2 = MH cos e ii3 [cos (e a') - 3 cos e cos S'J, Ba• -

:a·:a, �3· [cos (e =

B2P = MH cos 8' aw•

and, when 0 = 6' = 0, B2P

862

= MH +

2M2

as •

-

-

-

9') - 3 sin e sin 6'], M2

aa

[cos (a - 6') - 3 cos 0 cos 8'],

B2P M2 = BaBe' aa ·

B2P 2M2 = MH + a;a · Be'•

F I E LD I N T E N S I TY A N D P O T E N T I A L

14]

387

Hence, P is a minimum when e = 6' = 0 and the equilibrium is stable provided

132P 132P 1362 136'2

> ( 13136•13P6' ) •'

2M2 M2 MH + as > as · Ha3 > -M.

i.e., if i.e., if

In the case of two magnetic dipoles at P and P' (Fig. 14.13), the action of that at P upon that at P' can be represented by a force F and a couple G (Chapter l lJ . Since the field intensity due to m will not vary appreciably over the length of m', the couple may be calculated from equation (14.75). It is given by

G=m

,

X

( - m;a + 3m · r r) , -ys

m' Xs m + 3m · r m' X r. r r"

(14.80)

F may be calculated from the PE of the system as given by equation (14.79). Suppose that m' is displaced without rotation to the point whose position vector relative to P is r + dr. The total work done by F m'

---:...op<-----t• m

FIG. 14.13.-Force System Between Two Dipoles

the forces of the system is F · dr, and this must equal the decrease in the PE, viz., - dW. Hence

F . dr = -d ( m y. 3m' 3m · ry5m' . r ) . . dr = 3mr4 m' dr + 3m · drrs m' · r + 3m . rrm' s - 15m · rrsm' · r dr. _

•

·

388

A C O U R S E I N A P P L I E D M AT H E M A T I C S

But r dr = written

r . dr (see

p. 381 ) .

[CH.

Hence this latter equation may be

_ ( 3mr·s m' r + 3m'rs r m + 3mrsr· m , F . dr 15m · r m' . r r ) · dr. . •

-

r

7

Since equation (14.81) is valid for all small displacements have F = ar + bm + em', 3m · m'

a

where

y5

(14.81)

dr, we must ( 1 4 . 82)

15m · r m' . r y7

R m'

FIG. 14.14.-Coplanar Magnets

If the magnets are coplanar and r makes angles 6, 6' with m, m' re­ spectively as shown in Fig. 14.14, the angle between m and m' is (6 - 6'} and hence, from equati
W

' mm = 3 r

=

m

'

': r

[cos (6 - 6') - 3 cos 6 cos 6'], (sin 6 sin 6'

- 2 cos 6 cos 6').

(14.83)

Let (R, T) be the components of the force F acting upon m', along and perpendicular to r. Let G be the couple. Suppose that r, 6, 6' in-

14]

F I E LD I N T E N S I T Y A N D P O T E N T I A L

crease by small quantities dr, the forces of the system is

de, de' respectively.

389

The work done by

R dr + Tr de + G(de - de'). . This must equal the Hence

(14.84) decrease in the PE of the system, viz., - dW.

-dW = R dr + (Tr + G)de - G de ' . By putting de, dO' equal to zero, we show that oW · R = -or

(14.85) (14.86)

Similarly

oW. . oW G = 86' Tr + G = - 88, (14.87) Substituting for W from equation (14.83) into these latter results, we calculate that

R = 3mr:n' (sin e sin e' - 2 cos e cos e') , 3mm' sm. (e + 0') , T- ----;4

(14.88)

G = mr7' (sin e cos e ' + 2 cos e sin e').

As an exercise, the reader may also deduce these results from equations (14.80) and (14.82). A fixed magnetic doublet of moment M has its centre at 0 and its axis along Ox. A uniform field H acts parallel to Ox. A second magnetic doublet of moment M' is free to swing about its centre P in the plane POx, where OP3 = MJH and LxOP = e. Find the periodic time of small oscilla­ tions of the second magnet about its position of stable equilibrium, taking I to be its moment of inertia. Discuss the case e = }1r. (M.T.) Let ¢ be the angle made by the axis of M' with OP in the position of stable equilibrium. The field at P due to M and H is

Example 12.

_

� rs

�

+

3M • r + r H, rs

where OP = r. The PE of M' is accordingly given by · W = M · M' _ 3M r M' · r _ M' . H, =

rs

MM' ra

rs

3MM' cos (e - ¢l - ---;;.--- cos e cos ¢ - M'H cos (e - ¢),

= - 3M'H cos e cos ¢, since rs = MfH. Clearly W is a minimum when ¢ = 0. Hence M is in a position of stable equilibrium when its axis lies along OP.

390

A C O U R S E I N A P P L I E D MATH E M A T I C S

[CH.

When oscillating about this position, its energy equation is F�· 3M'H cos e cos "' = constant, or, provided .p is small, 3M'H I - cos 6(constant - 2) . = -

·

2

-

This i s the energy equation o f SHM o f period 27T

I I \1 3M'H cos 6 .

------. H

In the particular case when e = !1r, W is identically zero. The equi­ librium is now neutral and M' will settle in any position. At the point P the fields due to M and H cancel one another, i.e., P is a neutral point.

14.9. Potential Energy of a Charge Distribution Suppose that charges ev e2 , en are initially in a diffuse state at •

•

•

infinity, i.e., are separated by great distances, and are then assembled at the points P1, P2 , Pn of a region respectively. We will calculate the increase in the PE of the charges resulting from this process. This increase will represent the work which will be done when all the charges return to infinity, e.g., when all the charges are earthed. We shall assume that the charges are brought up from infinity in such a manner that at any stage the charges at Pv P2, • • • Pn are xen respectively (x � l). The final PE increment is, of xe1, xe2 , course, quite independent of the manner in which the distribution is assembled. If vl is the potential at pl due to the final charges Pn, the potential at P1 at the stage when the e2, en at P2 , charges at P2, Pn are xe2 , xen will be x V1 by equation Pn may (14.45) . Similarly, at this stage, the potentials at P2 , x Vn. where V2, Vn are the final potentials be written x V2, at these points. When this stage has been reached, suppose that the process is continued by the addition of charges from infinity so that •

•

•

•

•

•

•

•

•

•

•

•

•

•

•

•

•

•

•

•

•

•

•

•

•

•

•

14]

F I ELD I N T E N S I T Y A N D P O T E N T I A L

391

endx the charges at P1, P2 , Pn are increased by e1dx, e2dx, respectively. The increase in the PE associated with the charge at P1 is then e1 V1x dx. Since V1 is the potential at P1 due to all charges but the one actually at Pv it will be observed that we are neglecting the increase in the PE caused by the combination of the charges xe1 and e1dx. This omission is justified by the fact that these two charges were originally present at infinity as part of the charge e1 and, when this charge was divided, there must have been a loss of PE equal exactly to the gain which results when it is reformed at P 1 Thus, when x increases by dx, the total increment in the PE is •

•

•

•

(14.89) x now yields the PE of the final charge n 1 W = (e1 V1 + e2 V2 + . . . + enVn) 1 x dx = !i!= 1e;V;. 0 . (14.90)

Integration with respect to distribution, viz.,

Similarly, in the case of a gravitational field due to particles of masses mn situated at points where the potential of the dis­ Vn respectively, the increase tribution takes the values Vv V2 , in PE which results when the particles are assembled from infinity is given by

mv m2,

•

•

•

•

•

•

n W = t iL= 1 m;V; .

(14.91)

Since the potential of a matter distribution is everywhere negative, This loss of PE, which arises when matter condenses from an infinitely diffuse state into a compact mass, will be compensated exactly by an increase in the KE of the particles concerned. When the condensation has proceeded to the stage where the particles are in constant collision with one another the KE of their motion appears as heat and the temperature of the condensation rises. This is the mechanism whereby a star is formed by condensation from a gas cloud. When the temperature of the star has risen to a certain critical value libera­ tion of the internal eriergy of constitution of the separate atoms of the star can commence, and the further development of the star is then largely governed by the evolution of these atomic processes. Suppose that, instead of a discrete distribution of charge, we are presented with a continuous distribution over a certain region r of space and over various surfaces S. r may be such that it is possible to move from any point of the region to any other point without passing outside the region, or it may comprise a number of separate regions of this type. If r is divided into a large number of volume

W is also negative and represents an actual loss of PE.

392

[cH.

A C O U R S E I N A P P L I E D MATH EMAT I C S

elements, of which dv is a typical one, each such element may be re­ garded as a point charge p dv, where p is the charge density at dv, and its contribution to the sum _LeV appearing in the formula (14.90) for the PE of a distribution is p Vdv, V being the potential at dv. The net PE associated with the volume distribution of charge is accordingly

k

t pVdv.

(14.92)

.

Similarly, if cr denotes the surface density of charge at any point on any of the surfaces S at which there is an element of area dS, the con­ tribution to the PE of this part of the distribution is !

L crVdS.

(14.93)

The total PE is therefore given by

W = ! jr pVdv + t

L crVdS.

(14.94)

If one of the surfaces 51 is a conductor, then V is constant over this surface and its contribution to the PE is

tV 1 cr dS = teV,

(14.95)

s,

where e is the total charge on the conductor. Thus, if an electro­ en on conductors at static field is entirely due to charges ev e2 , potentials Vv V2 , Vn respectively, the energy of the system is •

•

•

•

•

•

n W = ! i=L1 e,Vi . .

(14.96)

Thus, in the case of the spherical condenser of the Example on p. the total energy is

378,

te2 (� - �) = e2 (b - a)j2ab.

If the inner sphere is conn < ted to the outer by a wire, this energy will appear as heat caused by t e discharge current which will flow through the wire. Three conducting concentric spherical shells, ofradii a, b, c (a > b > c) are insulated from each other and carry charges ev e 2, e 3 respectively. Find their potentials. Calculate the loss of energy which occurs when the two outer shells are connected by a wire. (S.U.)

Example 13.

14]

F I E LD I N T E N S I T Y A N D P O T E N T I A L

393

It follows from the results of Section 14.6(3) that the potentials of the conductors r r r are v., respectively, where

= a, = b, = c v.

=

el

+

V6, V, + e e. a a,

The energy of the distribution is therefore =

+

!e1 V. + !e 2 Vb !e3V, <:i' + e 22 + e_l_ + e2e3 + e1(e2 + e3).

2a 2b

2c

b

a

When the two outer shells are connected by a wire, let a charge e flow from one to the other so that the charges become e 1 + e, e2 - e, e3• The potentials of the conductors are now given by v.

= (e 1 + e) + a(e2 - e) + e3 = e1 + e2a + e3,

Vb = e1 a+ e + (e2 - be) + e3, V' = e1 a+ e + e 2 b e + �.c + But v. = V6• Hence e e 2 e 3 and the potentials are -

=

v.

=

e l + e. + e a = v ' b

a

394

A C O U R S E IN A P P L I E D M A T H E M A T I C S

[cH.

The energy o f the changed distribution i s now found to be

� (e12 .e 22) + (b - � + fc) e32 + � (e2e3 + e3e1 + e 1e 2) +

and the loss of energy is accordingly

! a - D (e. + e,)•. Example 14. If M is the mass of a uniform sphere of radius a, prove that the loss of PE in assembling the sphere is 3yM2f5a. Consider an elemental spherical shell of radius r and thickness dr. If p is the density of the sphere, the mass of the shell is 47Tpr2dr. Also, equation (14.61) gives the potential at all points of the shell. It follows that the shell's contribution to the integral (14.92) is -trr2yp2r2 (3a2 - r2)dr. The PE of the whole sphere is now obtained by integration to be

-1'1r2yp2 f r2(3a2 - r2)dr 0 = -H1r2yp 2a6• a

= Since this is equivalent to the result stated. Alternatively, the result may be obtained by assuming the sphere to be constructed in concentric spherical layers. When the radius of the sphere has grown to the potential at its surface is If a layer of matter of thickness and mass is now brought from infinity and deposited upon this surface, the further loss in PE is X = Integration with respect to now shows that the loss in PE sustained when the whole sphere is assembled is

M 41Ta3pf3, r,

as before.

-47Typr2/3. dr 41Tpr2dr 41Tpr2dr t1rypr2 ¥1r2yp2r4dr. r a J,.l 1T"yp• f r'dr = �:7T•yp•a6 0

EXERCISE 14 1 . Three equal, small spheres of mass m gm and each carrying a charge e electrostatic units are suspended by strings of length l em from the same point. If d em is the distance between a pair of the spheres when the system is in equilibrium, show that 9e4l2 = d2 (m2g2d4 + 3e4) .

Why, for a similar system consisting of charges and strings of length J...l, do the masses have to be m (and not J...m) for the charges to rest at distance J...d apart? (M. T.) 2. Draw a rough sketch of the lines of force in any plane through AB due to point charges 9e, -e at the points A , B, and show that the angle that a line of force that ends at B makes with AB at A qmnot exceed .38° 57'. (M. T.)

J...e

14]

FIELD INTENSITY AND POTENTIAL

395

3. A point charge 4e is at A , and a point charge - e at B, where A B = a. Show that there is a neutral point N on AB produced, such that AN = 2a. Show that, in any plane through A B, there is a critical line of force which leaves A at an angle 7t/3 with AB, and cuts A N at N. If P is a point on this line of force, and the angles PA N and PBN are denoted by e and c{>, show that sin te = t sin tc{>. Show also that, if the bisectors of the angles PAN, PBN meet at Q, the locus of Q is the circle on MN as diameter, where M lies on A B and A M 2af3, and hence find a geometrical construction for determining points on the critical line of force. (M. T.) =

4. Point charges - e, + 3e, -e (e 0) are placed in a straight line at points A , B, C respectively, where A B = BC. Show that the greatest angle that a line of force from B to C can make at B with the line ABC is cos-1 ( 1/3). (S.U.)

>

5. Point charges e and 2e are placed relative to axes Ox, Oy at the points (2a, 0) and ( - a, 0) respectively. Sketch the lines of force in the Oxy plane, and show that the lines of force which start from the charges e and 2e respectively are separated by a line of force which passes through the equilibrium point xja = 5 - 3y'2, y = 0, and which has the asymptotes y ± xy'B (x 0) . (S.U.)

>

=

6. O(xyz) is a rectangular system of axes, and a charge + 2e is placed at

the point (0, - a, 0), whilst charges of - e are placed at each of the points (0, a, 0) and (0, 2a, 0). Show that lines of force which leave charge + 2e at right angles to the axis Oy intersect the xz-plane in a circle, radius y'(d2 - a2) , given by (3a - 2d) 2 ( d2 + 3a2) 4a2d2 • (S.U.) =

7. Three charges e, - 4e, e are placed at collinear points A, B, C, where B is equidistant from A and C. Sketch the lines of force and show that any line of force leaving C will reach B at an inclination to BC less than 7t/3. (Li.U.) 8. A thin, uniform lamina of surface density cr is of infinite length and has breadth a. P is a point on the perpendicular to the plane of the lamina through a point 0 on one edge, and OP k. Show that the attractive force at P has components 2ycroc: and 2ycr log sec oc: parallel and perpendicular to PO respectively, where oc: = tan-1 (afk). An infinitely long bar of uniform density p has a semi-circular cross­ section of radius r. If P and Q are the ends of the straight edge of a cross-section, show that the force at P along PQ is 1tYpr. (L.U.) =

'A 'A'

9. A finite wire L of line density is bent to form an arbitrary portion of the catenary y c cosh (xjc) . An infinite line distribution L' of matter of constant density is placed along the line y = 0. Show that the resultant force on L due to L' is perpendicular to L', of magnitude 2y'AA'lfc, where l is the length of the projection of L on L', and passes through the mid-point of this projection. (L.U . ) =

396

A C O U R S E I N A P P L I E D MAT H EMAT I C S

[CH.

10. AB, A 'B' are two parallel rods of line densities, m, m' and are placed with the line j oining their middle points perpendicular to the rods and of length c. Show that the attraction between the rods is 2ymm' (AB' - A A ') . c 1 1 . Three uniform rods each of length a and line density "A are j oined at their ends to form an equilateral triangular framework. A fourth uniform rod, of length 2af3 and line density "A', is placed along the line which is perpendicular to the plane of the triangle and passes through its centroid. If the ends of the rod are distant af3 and a from the centroid, show that the attraction of the framework on the fourth rod is (L.U.) 3y"AA' log {7(19 - 8y3) /13}. 12. Show that for a uniform cylinder of length 2h, density p, and semi­ circular cross-section of radius a, the attraction at the centre of the rectangular face is of magnitude a + (a2 + h2) l (L.U.) 4pyh log . h 13. An infinite plane lamina is of surface density cr. Show that the attraction at any point outside the lamina is 27tycr. (Hint : Employ equation (14.21) .) 14. A uniform solid, of mass M, occupies the space generated by one com­ plete revolution, about the line y = 0, of the area bounded by the curve ay2 = x3 and the lines y = 0, x = a. Find the attraction at the origin. (L.U.) 15. A uniform disc has radius c and mass M. A uniform, straight wire has length c and mass m. The wire is perpendicular to the disc, one end being at the centre. Prove that the mutual force of attraction is 2yMm(2 y2) /c2• (L.U.) 16. Gravitating material is distributed uniformly, mass m per unit length, along two infinite, skew, perpendicular lines. Find the attraction between the lines. Three parallel, infinite, uniform rods are each of mass m per unit length, and cut a perpendicular plane at points A , B , C . A fourth infinite, uniform rod, of the same mass per unit length, lies in the plane A BC. Show that the resultant gravitational force on the fourth rod is 2ym27t or 6ym21t according as it does or does (D.U.) not cross the triangle ABC. 17. Show that at a point P on the bed of a lake, whose depth is uniform and equal to d for a considerable distance around P, the acceleration due to gravity exceeds its value at a point of the surface above P by the amount 47tyd(jp pw) , where p is the earth's mean density and pw the density of water. (D.U.) 18. There is a spherical cavity inside an otherwise uniform sphere of density p . Show that the field intensity within the cavity is every­

{

}

-

-

---+

where 47typOC, where 0 is the centre of the cavity and C is the centre of the sphere.

1 4]

19.

F I E LD I N T E N S I T Y A N D P O T E N T l A L

397

A sphere, with centre at 0 and radius a, consists of uniform material except for a spherical cavity of radius b ( < !a), with centre at 0', where 00' = ta. Prove that the force vanishes at a point inside the material and on the line through 0 and 0'. If this point is at a distance af6 from 0, show that 2 b3 2a3 (L.U.)

7

=

•

20. A smooth, straight tunnel of negligible cross-section is bored through a uniform sphere of density p. Show that a particle placed in the tunnel will oscillate in SHM of period (3rr/yp) t.

21. Prove that the potential of a circle of uniform surface density cr and radius a at a point on its circumference is 4ycra (L.U.)

.

22. Prove that if a solid sphere of uniform density p is divided into two parts by a plane which cuts the surface in a circle of radius b, the attraction between the two parts is rr2yp2b4/3. (Hint : Calculate the attraction of the whole sphere on one part.) (L.U.)

23. A body consists of a uniform sphere of radius a and density p, with the addition of a surface distribution, of mass m per unit area, over the part of the sphere cut off by a right circular cone with vertex at the centre and semi-vertical angle ex:. Find the gravitational potential at points on the axis of symmetry external to the sphere. (D.U.) 24. Prove that the potential of a uniform, thin, hemispherical shell, of mass M and radius a, at a point on the axis within the hemisphere and distant x from the centre of the sphere is - � + y(a2 + x2) ' X X a and deduce the attractive force there. If a particle is released from rest at the centre of the sphere, show that it will strike the surface with velocity {2(y2 - l )yMfa}t. (L.U.)

� [I

]

25. A particle of mass m is placed on the axis of a solid hemisphere of radius a and mass JYI at a distance a from the plane face on the side opposite the curved surface. Show that the work done in rotating the hemisphere through a right angle about a diameter of its plane face is y m (5 - 4y2) . (L.U.)

�

26. Three concei1tric, thin, spherical shells are of radii a, b, c (a < b < c) ; the first and third are connected by a fine wire through a small hole in the second, and the second is connected to earth through a small hole in the third. Show that the capacity of the condenser so formed is c2 ab + (M.T.) b -a c - b"

27. Prove that the PE of a small magnet of moment fl pivoted at a distance r from another small fixed magnet of moment 11.' and in the direction of the axis of fl ' is 2 _ J:!:_ cos

y3

I

e'

398

A

C O U R S E IN APPL ! E D MATH E M A T I C S [cH. 14J •

when the axis of fl. makes an angle () with the axis o·f !L ' . A uniform field H now acts in addition from the magnet fl. towards the fixed magnet. Prove that the previously stable position of equilibrium becomes unstable when H > 2f1.'/r3• (M.T.) 28. Two doublets, each of moment M, are mounted with their centres at fixed points at a distance r apart. They can rotate in the same plane and are constrained by a frictionless mechanism to be always parallel and in the same sense. Find the positions of equilibrium, discuss their stability and show that the work which must be done to turn the system from a position of stable to one of unstable equilibrium is (M.T.) 3M2fr3• 29. Prove that if there are two magnetic molecules of moments M and M' with their centres fixed at A and B, where AB r, and one of the molecules swings freely, while the other is acted on by a given couple, so that when the system is in equilibrium this molecule makes an angle with AB, then the moment of the couple is 3MM' sin 28/2r3(3 cos2 8 + 1) !-. (M.T.) 30. A system of charges consists of + 2e at the origin and -e at the points ( ±a, 0, 0). Show that at great distances the potential assumes the form v - ea2(3 cos2 e - l)/r3• (Li.U.) 31. A thin, uniform, circular disc of mass m and radius a is assembled from matter originally in a state of infinite dispersion. Show that the loss of PE that occurs is 8ym2f3rra. (Hint : Employ the result of Ex. 2 1 .) 32. A uniform, spherical shell of gravitating matter, in which the volume of the matter is equal to that of the hollow it encloses, contracts under its own gravitation into a uniform solid sphere of the same density. Prove that the work done by the gravitating forces is 3(2! - l )yM2j l Oa, where M is the mass and a the radius of the solid sphere. (L.U.) 33. The density p of a sphere, of mass M and radius a, at a distance r from the centre of the sphere is given by p = p0 (a - r)/a where p0 is a con­ stant. Show that the gravitational attraction at a point P, distant x( < a) from the centre of the sphere, is - yMx(4a - 3x)fa4• Hence, or otherwise, show that the loss of gravitational potential energy in assembling the sphere from a state of diffusion at an infinite distance is 26yM2/35a. (L.U.) =

e

=

CHAPTER I 5

GAUSS'S THEOREM.

POISSON'S EQUATION.

IMAGES

15.1. Flux and Gauss's Theorem

Consider a field at each point of which a vector A is defined, e.g. , A may be E or H, the electric oi magnetic intensity vectors. As we have already seen, the direction of the vector field at all points o£ a region can be exhibited in a diagram by the drawing of curves having the· property that the tangent at any point is in the direction of A at that point. These A-lines can also be constructed in such a manner that the magnitude of A at points of the field can also be read from the diagram. Let et be a small plane area at a point P in the field, oriented so that the A-lines in the neighbourhood of P are all normal to et. Let n be the number of lines intersecting et. Then it is necessary to arrange that the number of A-lines constructed in the neighbourhood of P makes nfet, the number of lines per unit normal area, proportional to the magnitude of A at P. If this rule for the construction of lines is obeyed at a large number of points in the field, the resulting diagram will convey an immediate visual impression of both the direction and strength of the A-field. In those regions where the lines are numerous the field strength is relatively large, whereas in regions where the lines are sparsely distributed the field is weak. In general, of course, it will be necessary to initiate lines at various points of the field if the above rule is to be followed. However, it will be proved in Section 15.2 that, in the particular cases of the fields we have considered in the last chapter, if lines are started at the charges in an appropriate fashion, it is not necessary to initiate further lines in free space. This makes the construction of the diagram particularly easy in these cases. If 5 is any closed surface constructed in the A-field and dS is any surface element, let An be the component of A in the direction of the outward normal to dS, i.e., along the normal in the sense outwards from the region enclosed by 5. Then the surface integral

(15.1) is termed the flux of A out of S. Consider the contribution to the flux of dS. Let dS' (Fig. 15.1) be the orthogonal projection of dS on to a plane in the neighbourhood of dS normal to the direction of the A-lines there. Then dS' = dS cos e, where e is the angle between the 399

400

A COURSE

tN

A P P L I E D MATH E M AT I C S

LcH.

planes of dS and dS' and hence also between their normals. But the normal to dS' is in the direction of A. Hence An = A cos 6 and

(15.2) Since A dS' is proportional to the number of lines intersecting dS' and this number is clearly equal to the number intersecting dS (as­ suming that dS' and the distance between dS' and dS are so small that AndS = A dS cos e = A dS'.

A

I

IL I

dS - - +------f

FIG. 15.1 .-Flux Through a Surface Element

no lines are initiated in the region between the elements), the contri­ bution of dS to the flux out of S is proportional to the number of lines intersecting dS in the outwards direction. We conclude that

N

ex.

number of lines leaving 5,

.

(15.3)

any line entering the region enclosed by S being counted negatively. This argument yields a pictorial representation of N, but equation (15.1) must be regarded as the fundamental definition of the flux. The manner in which the flux of electric intensity out of any closed surface S in an electrostatic field is related to the electric charges re­ sponsible for the field appears from Gauss's Theorem. This states that

the flux of field intensity out of any closed surface in an electrostatic field is equal to 47t times the total charge enclosed by the surface. In other words, the number of lines of force leaving a surface is 47t times the charge enclosed. To prove the theorem, consider first the case of the field due to a single charge e at a point P. Let S be a surface enclosing P (Fig. 15.2). Construct a conical surface with its apex at P and let the solid angle so formed at P be small and equal to dw. This cone will intersect S an odd number of times (three times in the figure) cutting off elements d51, d52 , from S at Q 1, Q2 , respectively. Let 61 be the angle between the outward normal to d51 and the axis of the cone. The intensity at Q1 due to e is e/PQ12 and its component along the outward •

•

•

•

•

•

15]

GAUS S ' S THEOREM

normal at this point is e cos 81/PQ12 • d51 to the flux out of S is

401

Hence, the contribution of

(15.4) the equality being established by an exactly similar argument to that which leads to equation (14.20).

FIG. 15.2.-Gauss's Theorem

The normal to d52 makes an obtuse angle with the conical axis directed from P. It follows that the contribution of d52 to the flux is -e dw. The contribution of d53 is +e dw and so on. It now follows that the net contribution of all surface elements within the cone is e dw - e dw + e dw

-

. . . =

e dw,

.

(15.5)

since the number of terms is odd. All elements of S can now be taken into account by a summation with respect to all the elemental solid angles into which the space surrounding P can be divided. Thus flux out of S

=

J E,.dS e J dw =

=

rr

4 e.

(15.6)

This proves the theorem in this particular case. Now suppose that e is external to S. The above argument can be repeated with the modification that the cone will now intersect S an even number of times. The reader should satisfy himself of the truth of this statement by drawing the appropriate diagram. The net con­ tribution to the flux of all surface elements within the cone is therefore e dw - e dw + e dw -

. .

.

=

0, .

(15.7)

A C O U R S E I N A P P L I E D MAT H EMAT I C S

402

[CH.

since the number of terms is now even. The total flux out of 5 is accordingly zero. eN, of Finally, suppose that the field is due to charges ev e2 , which the first M are inside 5 and the remainder are outside. Let EN be the intensities due to the charges separately. Then Ev E2 , the resultant field intensity E is given by E = El + E2 + . . . + EN. (15.8) If En is the component of E along the outward normal to 5 at any point, it follows that •

.

.

•

•

•

.

{15.9) En = Ein + E2n + + ENn . and hence that the flux out of 5 is given by i EndS = fs E1nd5 + i E2nd5 + . . . + i ENndS, (15.10) = 47t (e1 + e 2 + . . . + eN), . •

•

•

by the two special cases already treated. This completes the proof. The corresponding theorem for a gravitational field G may be ex­ pressed thus :

( GndS = - 41tyM,

(15.11)

ls

where M is the total mass enclosed by 5. 15.2. Applications of Gauss's Theorem

In certain cases of symmetrical charge distributions it is possible to calculate the field intensity immediately by application of Gauss's E

FIG. 15.3.-Charged Spherical Shell

theorem. Thus, consider again the case of a spherical conducting shell of radius a possessing a charge e (Fig. 15.3). Let P be any point outside the shell and let E be the field intensity at this point. Con-

15]

GA U S S ' S T H E O R E M

403

siderations of symmetry allow us to assert that E is everywhere radial and is of constant magnitude over a sphere S through P concentric with the shell. The flux of field intensity out of S is

L E dS

E J dS = 47tr2E. s By Gauss's theorem, this flux is equal to 47te. Hence E 2re

(15.12)

=

(15.13)

=

as before.

The density of an infinite circular cylinder of radius a at a distance r from its axis is proportional to y'(a2 - r2). Show that the attraction at an internal point is 2yM [aa (a2 r2) 3/2] a3r (L.U.) where M is the mass of unit length of the cylinder. The attraction E is clearly radial every­ where and depends only upon the distance r from the axis of the cylinder. Let us choose a co-axial cylinder of radius r and of unit length as the surface S of Gauss's theorem. The attraction has no component in a direction normal to the plane ends of S, and hence the flux from these ends is zero. Over the curved surface of S, E is constant and everywhere

Example 1.

__

_

_

•

normal to the surface. The flux from this surface is accordingly

- 1 EdS s

=

-Ef dS = -21rrE,

s

1

_--.-,.e::.. r

the negative sign being appropriate since E is directed inwards relative to To calculate the mass of the matter enclosed by consider the mass between two co-axial cylinders of radii and + having the same length as The volume between these cylinders is and the density of the matter they contain is where is a constant of proportionality. The mass enclosed between them is therefore

S,

x S. 21rx dx,

E

S.

x dx

ky'(a2 - x2),

k

- - - - -,

21rkxy'(a2 - x2)dx.

Integration over the range (0,

r) now yields the total mass inside S, viz., (i) 21rk��y'(a2 - x2)dx = i1rk[a3 - (a2 - r2) 312], 0

Application of Gauss's theorem (15.11) yields immediately E =

41Tyk 3r [a3 - (a2 - r2) 3/2],

(ii)

404

A C O U R S E I N A P P L I E D M A T H E M AT I C S The mass of unit length of the infinite cylinder is found by putting in the expression (i) . Thus Substitution for

M = f1Tka3,

or

3M3 k = 21Ta

[cH.

r=a

k in equation (ii) leads to the result required. Example 2. A distribution of electric charge is spherically symmetrical about an origin 0. If r is distance measured from 0, and the total charge within a sphere of radius r, centre 0, is qr•(i). (e-.!• - e-2'/") show that the potential is (Li.U.) and that as r ----+ 0, the charge density----+ 3qf47Ta3• Taking the surface S of Gauss's theorem to be a sphere with centre 0 and radius r, if E is the radial field intensity at all points of S, the flux out of S is 4?Tr2E. Hence, by Gauss's theorem, 4?Tr2E = 41Tqr•(i2 (e--.1• - e-•,1•), E = a!12 (e--.1• - e-2'1"). or But E is minus the derivative of rp in the radial direction and is a function of r only. Thus dr� = _!{a• (e-,la - e-•d•) ' q, !f. (e--.1• - !e-••1•) + constant. a As r ----+ q, ----+ 0 (assuming datum point at infinity) . It follows that the constant of integration is zero. Over a small sphere of radius r, the average charge density is � � c�- + O(r2)] . 4?Ta2r (e--.1• - e-••'•) = 4?Ta2r a by expansion of the exponentials. Letting r ----+ 0, we find that the density at r = 0 is 3qf41Ta3• =

oo ,

As an exercise the reader may compute the field due to a uniform, infinite, straight wire employing a Gaussian surface in the form of a circular cylinder of unit length with its axis along the wire (see Example l above). An application of Gauss's theorem of a more general nature concerns tubes of force. If C is any closed contour in an electrostatic field, the lines of force which intersect C form a surface called a tube of force (Fig. 15.4). Let A 11 A 2 be the areas of two right sections of a tube of small cross-section and suppose that the portion of the tube between these sections contains no charge. Let E11 E2 be the magnitudes of the intensity at the two right sections. Applying Gauss's theorem to the closed surface formed by the tube and its two right sections, we deduce that the flux from this surface is zero. But the flux from the

15]

GAUS S ' S THEOREM

405

curved part of the surface is clearly zero, since the intensity is every­ where tangential. It follows that the flux into one right section must equal the flux out of the other, i.e.,

(15.14) Thus, if A is any right section of the tube and E is the intensity there, AE is constant for the tube provided we confine our attention to a stretch of the tube which does not encounter charge. The constant value of AE is referred to as the strength of the tube. It will be noted that the field intensity is inversely proportional to the cross-section, so that in regions where the tube is narrow, the field is comparatively strong, whereas a relatively broad tube indicates a weak field.

FIG. 15.4.-Tube of Force

As explained in Section 15.1, the flux across the sections A 1, A 2 may be measured by the number of lines of force which cross them. Equation (15.14) may therefore be interpreted as stating that the number of lines of force within the tube is the same throughout any length which does not encounter charge. This means that no lines of force can ever be initiated in free space. A charge must always be present at the source of any line of force. This justifies a statement on p. 399. Now consider a tube of force which connects two charged conductors. Suppose the tube is extended a little into the body of each conductor and then closed. The resulting closed surface has zero flux across it, since the field is tangential to the tube and is zero within each conductor. Hence, by Gauss's theorem, the total charge contained is zero. This implies that if +e is the charge within the tube on one conductor, then - e is the charge within the tube on the other. Thus, in a field due to charged conductors, a tube initiated by a charge +e on the surface of one conductor either proceeds to infinity or terminates at a charge -e on the surface of another conductor.

A C O U R S E I N A P P L I E D M A T H EMAT I C S

406

15.3. Divergence.

Green's Theorem.

[CH.

Poisson's Equation

Returning to a consideration of the A-field referred to in Section 15.1, let P be any point in this field and 5 a closed surface surrounding it. If V is the volume bounded by 5, we consider the ratio NJV, where N is the flux .out of 5. As we shall prove immediately, this ratio tends to a definite limit, independent of the shape of 5, as this surface tends to zero in all its dimensions, eventually collapsing upon P. The value of this limit is called the divergence of A at the point P, and we write div A = lim !!. . (15.15) V-+0 V To prove that this equation provides us with a unique definition of div A, we require Green's Theorem, viz. :

Let Oxyz be a system of rectangular axes in some region of space and let P, Q, R be functions of (x, y, z) defined over a region r bounded by a surface 5 and with continuous first partial derivatives. Then l (lP + mQ + nR)d5 = Jr ( �: + �; + ��) dv, (15.16) where (l, m, n) are the direction cosines of the outward normal to d5. z

X.

FIG. 15.5.-Green's Theorem

Suppose that 5 is met in at most two points by any parallel to Oz. Then a contour C can be drawn on 5 dividing it into a lower portion 5 1 and an upper portion 52 , so that any parallel to Oz meeting 5 intersects it in two points, one belonging to 51 and one to 52 (Fig. 15.5). Let

GA U S S ' S T H E O R E M

1 5]

407

Let A be the region of this plane enclosed by the projected curve. Then, if R = R(x, y, z) and if z = cfo1 (x, y), z = cfo2 (x, y) are the equations of 51 and 52 respec­ tively,

C be projected orthogonally on to the xy-plane.

r oR J r Tz dv

= =

r l.a dA

j��'' oRoz dz,

L [R {x, y, cfo2 (x, y)} - R {x, y, cfo1 (x, y)}] dA,

the final integral being a double integral evaluated over A . The con­ tinuity of oRfoz has been assumed when performing the integration with respect to z. If y is the acute angle made by the normal to d5 with Oz, and dA is the projection of d5 on the xy-plane, then dA = d5 cos y. Hence

f nR dS f nR{x, y, cfo1 (x, y)}d5, L nR{x, y, cfo1 (x, y)} sec y dA, - { R{x, y, cfo1 (x, y)}dA, =

s,

s,

=

=

since, over 5v n = cos (1t - y) = - cos y (see Fig. 15.5) . Since, over 52 , n = cos y, it follows similarly that

Hence,

1 nRd5 J R{x, y, cfo2(x, y)}dA. 1 nRd5 l( nRd5 + 1 nRd5 = 1r °0Rz dv. =

S,

s

=

A

s,

s,

(15.17)

If 5 is met in more than two points by some parallels to Oz, by dividing into a number of sub-regions, each bounded by a surface which satisfies our condition, applying the result (15.17) to each and then adding, we can show that this result is true quite generally. The details are to be found in pure mathematical texts. Exactly parallel reasoning can now be employed to prove that r

L lP d5 [ �: dv, L mQ d5 Jr �; dv. =

(15.18)

=

(15.19)

Adding equations (15.17)-(15.19), we obtain Green's theorem. We shall now identify the surface 5 of Green's theorem with the

408

A C O U R S E I N A P P L I E D M A T H E M AT I C S

[CH.

surface S employed in the definition of div A at equation (15.15). If Ax, Ay, A. are the components of the field vector A in the directions of the axes Oxyz, the component in the direction of the outward normal to S is (lAx + mAy + nA.). Hence,

N = I (lAx + mAy + nA z)dS = h ( O�x + 0�Y + 0�•) dv, s (15.20)

1( r OX

employing Green's theorem to establish the second equality. oAx + oAv + oA. dV . 1 . dlV A - 11m -

oy x + OAy + oA. =M ox oy oz V-+O V

OZ ) '

Thus

-

,

(15.21)

since, over a small region r, the integrand is appreciably constant at its value at the point (x, y, z). Equation (15.21) shows that div A has been unambiguously defined and, in addition, provides us with an expression from which it may be calculated.* A special case of Green's theorem, which will be of the greatest importance later, can now be derived by setting P = Ax, Q = Ay, R = A. in equation (15.16). The result of these substitutions may be written in the form

L A ndS = Jr div A dv. .

(15.22)

h p dv .

(15.23)

In the special case of an electrostatic field, suppose this is due to a continuous distribution of charge of density p. If S encloses a region r , the total charge within S is given by the volume integral

Hence, by Gauss's theorem,

.

fs EndS = 4rr Jr p dv.

. . (15.24)

The left-hand member of this equation may be expressed as a volume integral by the use of Green's theorem in the form of equation (15.22). The result is that

or

h div Jr (div

E E

dv = 4rr h p dv, - 4rrp)dv = 0.

(15.25)

* If, formally, we take the scalar product of the vector operator 'V (equation (14.41)) with the vector A, we obtain the right-hand member of equation (15. 21) . Accordingly, div A is often denoted by 'V A. •

15]

409

GAU S S ' S THEOREM

This integral is zero for all possible regions of space be so only if the integrand is identically zero,* i.e., if div E = 4rrp.

r,

and this can

(15.26)

This is one of the fundamental equations of the electrostatic field, the other being equation (14.35). If we eliminate E between these two equations, we obtain 'V2V = div grad V = - 4rrp, . (15.27)

where '72 denotes the operator " div grad ". This notation is suggested by the equivalent expression for div grad V, viz., 'V 'V = 'V2V. The form this latter equation takes in terms of rectangular cartesian coordinates (x, y, z) is most important. It may be found by sub­ stituting the components of E, as given by equation (14.39), in the expression (15.21) for the divergence. We then deduce from equation (15.26) that •

fJ2V + fJ2V + fJ2V "2 = - 4rrp . . <52 <52 uX uy uZ

(15.28)

Thus, relative to rectangular cartesian coordinates,

'V 2

fj2 fj2 fj2 fJx2 + fJy 2 + fJz2'

(15.29) a result which is obtainable by formally squaring equation (14.41). Equation (15.27) is called Poisson 's Equation. In the special case when p = 0, i.e., in free space, this equation takes the form 'V2V = 0, (15.30) and is called Laplace 's Equation.

Problems in electrostatics generally reduce to the determination of solutions of Poisson's or Laplace's equation satisfying certain conditions peculiar to each problem. We will amplify this statement in Sections 15.4 and 15.6. Exercise

A closed surface S embraces other closed surfaces Sv 52 , Sn, no two of which intersect. The region between S and the remaining surfaces is denoted by r. Show that the identity (15.22) remains true provided the surface integral is replaced by a sum of surface integrals taken over the surfaces 5, S1, S 2, 5,., the direction of the normal being outwards from r in each case. •

•

•

•

•

•

It follows from equation (15.26) that, since E vanishes everywhere within a conductor, p = 0 over such a region. Thus, the charge on a

* For, if the integrand were not zero at a point P, assuming it continuous in the neighbourhood of P, a region r surrounding P could be found throughout which either div E - 471p � > 0 or div E - 41Tp � m < 0. In either case, the integral could not be zero.

M

410

A C O U R S E IN A P P L I E D M A T H E M A T I C S

[CH.

conductor always resides upon its surface. This fact leads us to inquire what form equation (15.26) takes at a surface distribution of charge. Let cr be the charge density on the surface S (not necessarily conducting) and let dS be an element of S. Imagine a cylinder constructed with dS as a right section and projecting from S a short distance on either side (Fig. 15.6) . Let the ends of the cylinder be right sections and suppose its length negligible by comparison with dS, i.e., a quantity

s

' /

r

-- -- --- --- - - -

' .._ I

'-- - - - - - - - - - - - - _ _ _

)

E2n

FrG. 15.6.-Field Intensity at a Surface Charge

of the second order. Then, if Ern, E2n are the components of the intensity on the two sides of S in the direction of one of the normals to dS, the flux out of the two ends of the cylinder is (E1ndS - E2ndS) . The flux across the curved surface is negligible by comparison with this quantity. Hence, by Gauss's theorem applied to the whole surface of the cylinder, E 1ndS - E2ndS = 47t adS, E rn - E2n = 47t cr . (15.31) or This equation implies that there is a discontinuity of 47tcr in the value of the normal component of the intensity as we pass across S. In the particular case when S is the surface of a conductor, we can put E 2n = 0, since there is no field in the body of the conductor. Hence E, = 47tcr, (15.32) where En is the component of the field along the outward normal. Although the normal component of the intensity is discontinuous across a surface distribution, the component in the direction of a tangent is continuous. For the potential distribution over one side of the sur­ face will be the same as that over the other side, and it follows, there-

15]

GAUS S ' S THEOREM

411

fore, that the derivative of the potential in any direction tangential to the surface does not depend upon the side being considered. In particular, since the intensity is zero inside a conductor, the tangential component is zero just outside the conductor, i.e., the intensity and the lines of force are both normal to a conducting surface. Equation (15.32) accordingly gives the total intensity at a conducting surface. 15.4. Elementary Solutions of Laplace's Equation

Let us first consider the case of an electrostatic problem in which the field is due to uniform distributions of charge over a number of parallel planes. Clearly the potential will vary only in a direction perpendicular to these planes. Hence, if an x-axis is taken in this direction, V is a function of x alone and Laplace's equation reduces to (15.33) The general solution of this equation is

V = Ax + B,

(15.34)

where A and B are arbitrary constants. Since dV jdx = A , the field intensity is uniform and of magnitude -A . The lines of force are all straight and parallel to Ox. The values of A and B will depend upon circumstances peculiar to each problem of this type. Suppose that the field is due to a single charge distribution of constant surface density a over the plane x = 0. For reasons of symmetry, the intensity on both sides of this plane can be taken to be of magnitude E, and the directions of the intensity will be normal to the plane and in opposite senses. By equation (15.31), we obtain 2E = 47ta or

E = 27ta.

(15.35)

The field is now completely determined. The field due to a distribution of charge over an infinite plane con­ ductor may be calculated similarly. However, since there is no field within the conductor, equation (15.31) yields the result (15.36) determining the magnitude of the field. Now consider a parallel-plate condenser formed from two infinite parallel plane conductors, one of which is earthed (Fig. 15.7). Let the earthed plate be in the plane x = 0 and the other plate in the plane x = t, t being the distance between the plates. Suppose the insulated plate is given a charge which raises its potential to V = .p. The potential between the plates is given by equation (15.34) , where A

412

[cH.

A C O U R S E I N A P P L I E D M A T H E M AT I C S

and B have to be chosen so that V = when x = t. It is easily found that V

=

0

when

x=0

¢>xjt.

and V = ¢>

(15.37)

The field in the direction of x increasing is -dVJdx = -¢>ft. Thus, the normal field at the plate x = t is ¢>jt and, if cr is the surface density of charge, q, . ¢>jt = 4n: or (15.38) =

cr 4n:t"

cr

The normal field out of the earthed plate is -¢>jt, and hence the surface density of the charge induced on this plate is - cr . This result is otherwise obvious, since every tube of force leaving a charge on the insulated plate must end on an equal charge of opposite sign on the earthed plate.

t !

o

I I

'

t

- o-

I '

I

'

I. I

I

I I I ' I

tI

I

y

t

I I y I

I I I I '

V· ¢

V= O

FIG. 15.7.-Para!lel-plate Condenser

Equation

(15.38) may be written cr 1 � = 4n:t'

(15.39)

revealing that the capacitance per unit area of this type of condenser is 1/4n:t. In practice, the plates cannot be of infinite extent. Hence, if A is the area of either plate and we neglect the departure of the field from uniformity near the edges of the plates, the total capacitance will be given by

(15.40) We note that C can be made very large by reducing the distance, between the plates to a small value. Another type of problem in which Laplace's equation can be reduced to an ordinary differential equation, and consequently solved in general terms, is that in which the field is caused by charges distributed uni­ formly over co-axial cylinders of infinite length. Taking the common axis of the cylinders to be the z-axis, if p denotes the perpendicular

GA U S S ' S T H E O R E M

15]

413

distance of the point (x, y , z) from this axis, then V is a function of only. But p = (x2 + y2) i. Hence

p

oV dV op x av (15.41) ·ox = dp ox = (x2 + y2) !- dp ' dV � ( dV ) o2 V = �c y2 dV + ox2 ox (x2 + y2) !- dp J = (x2 + y2) � dp (x2 + y2) i ox dp x d2 V op y2 dV (x2 + y2) t dP + (x2 + y2) ! dp2 ox y2 dV + x2 d2 V · (15.42) + 2 (x y2) ! Tp x2 + y2 dp 2 X

•

X

Similarly

x2 dV + y2 d2 V (15.43) + 2 (x y2) � crp x2 + y2 dp 2 · · Also o2 Vjoz2 = 0, since opjoz = 0. Hence Laplace's equation may be

written in the form

or

(15.44)

Thus A single integration now yields

or

p dV dp = constant, dV = A . dp -p ·

(15.45)

A further integration gives finally

V = A log p + B. (15.46) Since the variation of V is confined to directions perpendicular to Oz, i.e. , along p, the field is everywhere radial. Its intensity is - dVjdp and hence, by equation (15.45) is inversely proportional to the distance

from the z-axis. As a special case, we shall consider the problem presented by a cylindrical condenser. Let the outer of two infinite co-axial conducting cylinders be earthed and let the inner be charged to a potential cp. If a is the radius of the inner cylinder and b is the radius of the outer cylinder, we have to choose A and B in equation (15.46) to satisfy

414

A C O U R S E IN A P P L I E D MAT H E MAT I C S

the conditions found that

V = cfo when p = a and V = ( bl p) V - cP log log (bla)' _

and

E

P

=

cP

p log

0

when p =

b.

[CH. It will be

(15.47)

·

(15.48)

(bla) ' ·

where Ep is the radial field intensity. Putting p = a in equation (15.48), we obtain the normal intensity at the inner cylinder. Thus, if a is the surface density of charge on the cylinder, by equation (15.32)

cP

alog

-

(bla) - 4rr a.

(15.49)

The charge on unit length of this cylinder is accordingly

2TCaa = cfol2 log (bla)

(15.50)

and the capacitance of this length is therefore given by --- - · - 2 log1(bla)

C -

--

(15.51)

·

Finally, we will obtain the general form of the solution to Laplace's equation in the case of spherical symmetry, i.e., when the field is due to charges distributed uniformly over concentric spheres. Taking the common centre of the system at the origin of the axes Oxyz and denoting the distance of the point (x, y, z) from 0 by r, we observe that V is a function of r alone. Taking r = (x2 + y2 + z2) ! , the reader should be able to prove, in a manner similar to that by which equation (15.42) was established, that

with similar expressions for o2 VI oy2 , therefore equivalent to the equation

Writing this in the form

o2 VI oz2 .

Laplace's equation is

d2 V + � dV - 0 dr2 r dr - ·

(15.53)

!_ � ( dV ) 0 r2 dr r2 dr -

(15.54)

_

'

two integrations lead to the general solution for

V

=

Jt

r + B.

V, viz., (15.55)

GAUSS ' S THEO REM

15]

415

This result also follows from the theory of Section 14.6(3), the first term of the right-hand member of equation (15.55) representing the contribution to V of the charged shells which do not enclose the point (x, y , z) and the second term representing the contribution of the shells which do enclose this point. We conclude this section with an example of the use of equation (15.55) . An insulated, thin, spherical shell of radius a, charge Q, potential Vv is sur1'ounded by a concentric, thin, insulated, spherical shell of radius 2a, charge - 2Q, potential V2, which in turn is surrounded by a concentric, thin, insulated, spherical shell of radius 3a, charge 3Q, potential V3• Find V1 , V2, V3 in terms of Qfa and show that 3 V3 = V, + 2 V2 . (S.U.)

Example 3.

Measuring r from 0 (see diagram), let A V = -; + B

in the region I . The normal field intensity over the shell r = a is then given by

( dr ) r = a = a•

V - d--

A

- ·

The surface density of charge over this shell is Q/4na2• Hence Also we must have V = V1 over r = a. Hence v, = a + B. Q

Throughout the region l therefore v=Q

u - �)

+ v, .

From this equation we find that the normal field intensity over the inner surface of r = 2a is

( ��) r = 2a = - 4�2

in the sense towards 0. The density of the charge on this inner surface is therefore -Qfl6na2 and the total charge on the surface must be -Q, as also follows from the fact that all tubes of force which leave charges on r = a must terminate at equal and opposite charges on r = 2a. The remaining charge -Q resides on the outer surface. Since V = V 2 over r = 2a, v = v1 . •

D

-

Sl.

2a

(i)

A C O U R S E IN A P P L I E D MATH EMAT I C S

416 is

[CH.

Since the charge on the outer surface of we can now show that in the region

V2 ,

But

V= V

3

over

Eliminating

r = 3a .

r = 2a is -Q and its potential 2 V = O"" (_!:_2a _ ! ) + V2• Hence

1'

v. _- 6aQ + v

•.

(ii)

Qfa between equations (i) and (ii), we obtain the result 3V3 = V1 + 2V2. The charge on the inner surface of r = 3a must be Q (since -Q resides on the outer surface of r = 2a) and hence a charge 2Q is present on the outside of r = 3a . It follows that the potential in region 3 is given by v = 2Qr , there being no constant term, since V -+ 0 as r -+ . Putting r = 3a, we find that v = 2Q 3a . 3

oo

Equations (i) and (ii) now show that

vl = Qfa, v. = Qf2a.

The reader should also obtain these results by employing the method used in the example on p. 378.

15.5. Force on a Charged Conductor

The charge on any element of the surface of a charged conductor is subjected to a force which is the resultant of the attractions and re­ pulsions of the remaining charges on the conductor and of any other charges contributing to the field. Since the charge on such an element is quite free to move in all directions but that normally outwards from the conducting surface, under equilibrium conditions the force must act in this direction. This is also to be expected from the fact that the field is everywhere normal to a conducting surface. Consider an element of the conducting surface of area dS, sufficiently small to be regarded as plane. Let cr be the surface density of charge on this element. The charge on dS will occupy a very thin layer, across which there will be a very rapid change in the field intensity from a zero value on the side of the layer interior to the conductor, to a value 47tcr at the exposed face. To calculate the force on dS, we must take account of this variation in the field in which the charge is im­ mersed. Let AA , BE' (Fig. 15.8) represent the interior and exposed faces respectively of the region occupied by the charge on dS. Taking an x-axis perpendicular to these faces, all components of the intensity perpendicular to Ox are zero at both faces. We shall assume that such components remain zero between these faces, i.e., that the field is '

1 5]

GAUS S ' S THEOREM

417

everywhere directed along Ox. The intensity E is then a function of x alone and, from equation (15.26), the charge density p is given by

1 dE · p = 4rr (15.56) - dx The charge lying between the planes x, x + dx is p dx dS and the force upon it is Ep dx dS. The net force upon dS is accordingly 4 1"' dE dx = dS 1 "" E dE = 2rro2d5. dS Ep dx = dS E 4rr 4rr 0 dx lx, (15.57) �

�

Thus, the force on the surface of a charged conductor is 2rro2 per unit area. A

.X

I

0

Frc.

A'

� v E� � �

xv

� � � �v

B

x.-tax.

:x:2

:X.

'

B

15.8.-Force on a Charged Conductor

We can now calculate the force on either plate of the parallel-plate condenser considered in the previous section. It is

2 2rro2A = A,P Srrt2 •

(15.58)

Since this force is a simple function of rp, the potential difference between the plates, by measuring it we can obtain rp. This is the principle of the attracted disc electrometer, a description of which instrument will be found in almost any physics textbook.* A spherical balloon of negligible mass has a charge e uniformly dis­ tributed over its surface. The pressure inside is p while that outside is kp, but the balloon maintains its natural unstretched shape on account of electrostatic repulsion. Find the radius of the balloon. (D.U.) If a is the radius of the balloon, the surface density of charge is ef41Ta2• The force acting upon a small element dS of the balloon's surface due to the E.g., General Degree Physics, Part V, by C. J. Smith (Arnold) .

Example 4.

*

A COURSE IN APPLI ED MATHEMATICS

418

[CH.

charge is therefore e2dSj8TTa4• This is balanced by a resultant inwardly directed force due to the two gas pressures of p(k - l)dS. Hence

p(k - l)dS or

a

15.6. Uniqueness Theorem.

=

=

e•ds 8TTa4, e2 •/ \1 8TTp(k - 1)'

Harmonic Functions

The majority of electrostatic problems require the calculation of the field due to charges carried by conductors of known shapes and positions. Either the total charge on each conductor or its potential will be specified. Occasionally, a volume distribution of charge will also contribute to the total field, in which case it is to be expected that the density of such charges will be stated at all points of the regions they occupy. Point charges of given magnitudes and positions may also occur. When all these data are specified in respect of any electro­ static problem it is uniquely soluble. This is the important Uniqueness Theorem, which we shall now prove. 5, be the surfaces of the conductors whose potentials Let 51> 52 , or total charges are known, and let 5 be a sphere of very large radius R, whose centre lies in the region occupied by the conductors. Let r denote the region lying between the surfaces 5; and 5. We shall assume that r is occupied by a volume distribution of charge of known density p ( p will be zero at points where there is no charge) and a number of point charges. If possible, let V, V' be two distinct potential functions, defined throughout r, and each representing a possible solution to our problem. Then, at any point of r not occupied by a point charge, we shall have •

•

•

'7 2 V = '7 2 V' = -4rcp. Let

U = V - V'.

Then

'7 2 U = 0 . .

(15.59)

In the neighbourhood of a point occupied by a point charge e, we must have

v = r-e + VI , V' = �r + vI I ,

where r is distance measured from the charge and Vl> V/ are finite in the neighbourhood of the charge (this is a consequence of equation (14.45)). Hence

u = VI - VI '

and is accordingly finite in such a neighbourhood and hence throughout r.

GAU S S ' S T H E O R EM

15]

419

If the potential of the conductor 5; is specified, V and V' take identical values over 5; and hence U = 0 over this surface. By equation (15.32), the surface density of charge over a conductor . av' m the other, u" / u"n denotmg 5J IS. 1 av m. the one case and 1 Tn -

-

4rr on

.

4rr

differentiation along the outward normal. the conductor is known, we must have _

Hence, if the charge QJ on

_!___!.S �on d5 = __!__4rr l(s; oV'on d5 = Q · 4rr ; _

J

and thus, by subtraction, we know that

lrs; �!! on d5 = o.

p.

(15.60)

Now, by the extension of Green' s theorem set as an exercise on 409, since u is finite throughout r,

n 1 d1v. (U grad U) dv = -•I= I J.s; U "auun d5 + J. U auun d5. 8

r

But

c.-

(15.61)

( oU ( oU ls, U on d5 = U ls, on d5,

since U is constant over a conductor and hence these integrals are all zero, either because U = 0 (when the potential is given) or by virtue of equation (15.60) (when the charge is given) . Also, by the identity proved in the Appendix to this chapter, div (U grad U) = U div grad U + grad U grad U, = (grad U)2, in consequence of equation (15.59). Finally, over a sufficiently large sphere 5, the charge system will behave (so far as the lowest-order terms in 1/R are concerned) as a single charge Q at the centre. To this order of accuracy •

V = � + 0(1/R2),* V ' = � + 0(1/R2), and hence U = V - V' = AfR + 0(1/R 2) over 5. It follows that ?U ( au (d terms in ls U on d5 = 1 U oR d5 = -A 2 ls R-.; + higher-order 1/R 2 = _ '!_� + 0(1fR2), --+ 0 as R --+ s

*

oo .

This represents the condition to be satisfied by the potential function " at infinity ".

420

A C O U R S E I N A P PLI E D MATH EMAT I C S

Letting R ---+ oo, therefore, in the limit equation

[eli.

(15.61) reads (15.62)

where r is the whole of space outside the conductors. But the inte­ grand of this integral is essentially positive, and equation (15.62) therefore implies that ( V U) 2 = 0. 'V U = 0, Hence i.e., 'V V = 'V V'. Thus the field-intensity vectors for the two hypothetical solutions are identical. This proves the theorem. A consequence of the Uniqueness Theorem of great significance is that if, by any means, we are able to determine a solution of Poisson's equation which yields the correct total charges on all conducting surfaces involved in a problem and for which this charge is specified and which also yields the stated potentials over all other conductors and behaves in the appropriate manner in the vicinity of the point charges and at infinity, then the problem has been solved. In order to take advantage of the uniqueness principle, it is useful to have available a set of simple solutions of Laplace's equation. Any solution of this equation is called a harmonic junction. If Oxyz is a rectangular cartesian frame of reference, r = (x2 + y2 + z2) i- is the distance of the point P(x, y, z) from 0 and 6 is the angle made by OP with Oz, then simple harmonics are :

(a) -Ez + B, i.e., the potential in a uniform field of intensity E in the direction of Oz (see equation (15.34)) ; (b) 1jr, i.e., the potential due to unit charge at 0, (c) cos 6jr2, i.e., the potential due to a dipole of unit moment placed at 0 and directed along Oz. The reader should verify by direct substitution that each of these functions satisfies Laplace's equation. Also, since this equation is linear, any multiple or any linear combination of multiples of these solutions will be itself a solution. Thus,

V = -Ez + r� cos 6 .

(15.63)

is harmonic. We will now consider a particular problem, viz., that of an earthed conducting sphere of radius a placed in a field which, before the intro­ duction of the sphere, was uniform and of intensity E (Fig. 15.9). Taking the origin 0 at the centre of the sphere and the z-axis in the

1 5j

THEOREM

GA U S S ' S

421

direction of the field, we know that before the sphere is placed in position the potential is given by

V = -Ez = -Er cos e,

(15.64)

neglecting an arbitrary constant whose value can in no way affect the field, since this depends upon derivatives of V. The distortion in the field caused by the charge induced upon the sphere must vanish at an

FrG. 15.9.-Earthed Conducting Sphere in a Uniform Field

infinite distance from 0 and must be a function of the variables (r, e) alone, i.e., must be axially symmetical about Oz. The solution (15.63) is of this type. Writing it in the form

we require that

V = - (Er + � ) cos e,

V shall be zero over A E + -2 0 . -

Thus

a

V

-

a

=

,

-E (

r = a

1.e., 1

r -

for all

(15.65) e.

3

"f A - E a

� ) cos e

This will be so if

(15.66)

.

(15.67)

is a solution of Laplace's equation satisfying all the prescribed con­ ditions, and it is therefore the unique solution to the problem. The radial component of field intensity is

E,. =

-

�� = E ( 1 2:S3) cos e. +

.

(15.68)

422

A C O U R S E I N A P P L I E D MATH EMAT I C S

[cH.

At the surface of the sphere the field intensity is now found to be 3E cos e by putting r = a in this equation. By equation (15.32) , the surface density of charge is then given by cr

=

3E cos e. . 47t

(15 .69)

For 0 < e < f7t, this charge density is positive, whereas, for f7t < e < TC, it is negative, as indicated in Fig. 15.9. The total charge induced is clearly zero. The charge on the positive hemisphere can be found by considering the charges on ring-shaped elements of the surface having their centres on Oz. The charge is " " 2TCa2 sin e de = !Ea2 sin e cos e de = !Ea2 . (15.70)

l

l

The transverse component of field intensity (see equation ( 14.68)) is given by aa . 1 av (15.71) Eo = = -E 1 - -3 sm e . r ae r - -

(

-

)

The magnitude and direction of the field at any point can now be calculated. The lines of force have been sketched in Fig. 15.9. A variant of the problem which has just been solved is that of an insulated charged conducting sphere placed in a uniform field. V must be chosen so that it : (i) satisfies Laplace's equation ; (ii) is constant over the spherical surface ; (iii) yields a surface distribution of charge whose sum total is equal to the given charge, and (iv) becomes identical with the potential of a uniform field at infinity. If we take

V = re - E ( r ra-2a) cos e -

-

'

.

(15.72)

the second member, as has been shown, makes the potential constant over the sphere (in fact zero) and contributes zero total charge to its surface. The first member is the potential due to an insulated sphere possessing a charge e, and it is accordingly constant over the given sphere and contributes a total charge e (alternatively, this may be verified by direct calculation of the surface density) . All conditions are therefore satisfied by V, and equation (15.72) determines the field due to a conducting sphere with a charge e placed in a uniform field of strength E. A field due to charges on infinite, parallel, cylindrical conductors is two-dimensional in character, i.e., it has the same form in any plane perpendicular to the generators of the cylinders. To calculate such fields, we therefore seek potential functions possessing the same pro­ perty. Assuming that the axes of the cylinders are parallel to the z-axis, if p is the perpendicular distance of the point (x, y, z) from this

15]

GAUS S ' S THEOREM

423

axis and c{> is the angle made by this perpendicular with a fixed direction, the following functions are harmonic :

(a) -E p cos c{> + B, i.e., the potential of a uniform field E in the direction c{> = 0 ; (b) -2 log p, i.e. , the potential due to an infinite line charge of unit density along the z-axis ; (c) 2 cos cfo/p, i.e., the potential due to a line doublet of unit moment placed along the z-axis. These harmonics are particularly useful in the solution of problems involving circular cylinders. For example, the reader should verify that the potential appropriate to an earthed conducting cylinder of radius a in a uniform field E perpendicular to its axis, is given by

V = -E ( p - �) cos c{>,

(15.73)

whereas, if the conductor is insulated and has a charge e per unit length, then

V = -2e log p - E ( p - � ) cos cf>. (15.74) Example 5. An insulated and uncharged conducting shell of radius a is brought into a region where there is a uniform electrostatic field of intensity E. Show that, if the shell is cut into two pieces along a plane at distance d from the centre and perpendicular to the field, a force of 9E2(a• - d4) /l6a2 is necessary to prevent the two pieces from separating. (D.U.) •

FORCE

The appropriate potential function in this case is that given by equation

(15.67), which has been seen to yield zero total charge on the sphere. The surface density of charge is accordingly given by equation (15.69). The force per unit area acting upon the spherical surface is 2?Tu2 = 9£2 cos2 6/87T,

and acts along a radius. Consider the element of the spherical surface between two parallel planes perpendicular to the applied field as indicated in the diagram. The resultant

424

A C O U H. S E l N A P P L I E D M A T H E M A T i C S

lnr.

force on this element acts along the axis of symmetry. Hence, resolving all forces acting on its parts in this direction, we find that the net force is 9 2 cos2 e . cos e . 21ra2 sin e de = E2a2 cos3 e sin e de. 8 Integration now yields the resultant force on one portion of the sphere, viz., "' � E•a• cos3 e sin e d() = l·6 E2a2(1 - cos4 oc)

::

f

�

0

which is equivalent to the result stated. The force on the other portion is found by integrating over the interval (oc, 1r) . A force of the same magnitude but opposite in sense is obtained. The net force acting on the whole sphere is accordingly zero. Thus, a solid conducting sphere shows no tendency to move under the influence of the field.

15.7. Images

In the case of certain problems, it may be shown that the field due to a number of point charges satisfies all the relevant conditions and hence, by the uniqueness theorem, must be the field required. Certain of these charges will be included in the formulation of the problem, but others will have been introduced to generate a resultant field having the required charac­ teristics. These latter, purely hypothetical charges, are called image charges. As an example, consider the problem of calculating the field due to a point charge p' 8 e placed at a point P in front . �--------'1--------1.!!.--::./ -e of an infinite, earthed, con­ ducting, plane slab (Fig. 15. 10) . The field in the region to the right of the face of the slab has to satisfy the following conditions : (i) its potential must be zero over the slab face ; (ii) its form in the vicinity of P must be FIG. 15.10.-Point Charge and Infinite Plane that appropriate to a point Conductor charge there, and (iii) its potential must approach zero as we recede to an infinite distance from P in any direction. All these conditions are satisfied by the field due to the given charge e and an image charge -e at P', where P' is the mirror image of P in the slab face. The potential at any point Q on the slab face due to this system of charges is /

(15.75)

15]

GAU S S ' S T H E O R E M

425

as required. The other conditions are clearly satisfied by the field due to this charge system. It follows that this field in the region to the right of the slab face is identical with that which arises when e is placed in position. The field inside the conductor has to satisfy conditions which are entirely different from those listed above. It follows that this field is not identical with that caused by e and its image over this region. The field within the conductor is, of course, zero. The field intensity at Q in the direction normal to the slab and out­ side the conductor is given by 3

2e cos e - -2e En - - PQ (fi cos 2 _

_

e.

(15.76)

d being the distance of P from the slab. The surface density of the charge induced at Q is accordingly found to be a

_

_!_

47t

_ _ e cos e. En 27td2 3

(15.77)

If the slab is removed and the image charge introduced, there will be no detectable change in the field in the neighbourhood of P. We deduce that the force acting upon e will also be unaltered. But this force due to the image charge is e2/4d2, acting along the line PP'. This, therefore, is the force which is exerted upon the charge e by the charge it induces on the slab. An equal and opposite force acts upon the conductor.

FIG. 15. 1 1 .-Point Charge before an Earthed Conducting Sphere

The reader should sketch the lines of force linking the point charge and the conductor. Another problem which may be solved by the method of images is that of a point charge e placed outside an earthed conducting sphere (Fig. 15. 1 1 ) . Let P be the position o f the charge and let P' be the point inverse to P relative to the sphere. We place an image charge

426

A COURSE I N APPLI ED MATHEMATICS

[CH.

e' at P'. Since the potential at any point Q on the sphere is to be zero, we require that e 0 or P'Q e' e' (15.78) P'Q + PQ = PQ- = - e OP . OP' = OQ 2, Now OP OQ . or (15.79) OQ = OP' But LQOP is common to both the triangles QOP', QOP, and the sides including this angle have just been proved to be in the same ratio. Hence the triangles are similar and we have P'Q OQ a = (15.80) PQ = OP T . where a is the radius of the sphere and f = OP. The requirement (15.78) is accordingly equivalent to the condition ·

ae e , = -T .

(15.81)

If e' is selected to satisfy this equation, the potential over the whole spherical surface due to e and its image charge will be zero. The field over the region external to the sphere due to the two charges therefore satisfies all the prescribed conditions, and must accordingly be the actual field required. The reader should show that the force between the charge e and its image, and hence the attraction of the sphere for e, is

afe2

(15.82) Example 6. Within a spherical hollow of radius a in an earthed conductor, equal point charges e are placed at equal distances f from the centre on the same diameter. Show that each is acted on by a force of magnitude 1 . e [(a44aap (Li.U.) - f4)' + 4j•J 2

A'

- cte

T

8

I

GAU S S ' S THEOREM

15]

427

Let and B be the positions of the two charges e. If image charges -aefj are introduced at the inverse points and B', these, together with the original charges, make the potential over the spherical surface zero, as required. The field within the sphere due to these four charges is accordingly that sought for. The force acting on the charge at due to the remaining three charges is found to be

A

A'

A

and this reduces to the expression stated.

The number of lines of force leaving a point charge e is, by Gauss's theorem, 4rte. Let S be any surface bounded by a closed curve C drawn in the field due to e and let w be the solid angle subtended at e by C. Since the field is spherically symmetrical about the point charge, a fraction w/4.-rr of the total number of lines leaving e will lie inside the solid angle w and will consequently intersect 5. Hence the number of lines leaving S in a direction away from e is ew. In other words, this is the flux out of S. Now consider the field generated by a system of point charges e1, e2, eN. Let C subtend angles wv w 2, wN respectively at these charges. Then the net flux out of S •

•

•

•

•

•

(15.83) where the positive sign before any term is taken if the sense " out­ wards from S " is also the sense " away from the corresponding charge " and the negative sign otherwise. If this system comprises a set of charges together with their images and S is a portion of the surface of a conductor relative to which the image charges have been chosen, let cr be the charge density over 5. The total charge on S is then

18 cr dS = 1 18 EndS = 41rt 4rt

X

flux out of S =

1

4� ! ± ew, (15.84)

by equation (15.83) . We are thus provided with a formula from which the charge on any portion of a conducting surface may be found immediately the image system has been determined. In particular, if S is the whole surface of a conductor (and therefore closed) and if e1, e2 , er are the charges of the system generating the field which are inside S, the total charge on S is e1 + e2 + . . . + er, if the charge is induced on the outside, and is minus this quantity if the charge is induced on the inside. This result follows from the fact that S subtends a solid angle 4rt at each of the charges inside S and zero solid angle at any charge outside 5. Thus the charge on a conductor is the sum of the image charges it contains. In the case of the system •

•

•

428

A COURSE IN APPLIED MATHEMATICS

[CH.

of Fig. 15.11, for example, the charge on the spherical conductor IS e ' = -aejj. A point charge e is placed outside an insulated conducting sphere of radius a carrying a charge Q. Show that, even if e and Q are of like sign, the force between the point charge and the sphere is attractive provided that the distance of the point charge from the centre of the sphere lies within a certain range. Verify that when Q = 7ej18 this range extends from a to 2a. (Li.U.) Let A be the position of e and let A ' be the inverse point. If OA = f, we place an im1>-ge charge -aeff at A ' so that the potential over the sphere is reduced to zero. Finally, placing a second image charge (Q + aejf) at 0, the potential over the sphere is raised to a constant value and the charge upon it is the sum of the image charges it contains, viz., Q, as required. The field outside the sphere due to e and its images satisfies all the conditions of

Example 7.

the problem, and is accordingly the one we are seeking.

A

e

The force acting upon

e due to the two image charges is ae2f (Q + aejf)e !" (f" - a•) • towards 0. This is easily shown to be positive, and the force therefore attractive if Q and e have opposite signs. If, however, Q and e are both positive, the force is attractive only provided ae2f ae• Qe (!• - a•) • - f3 > p' a3e(2f" - a2) > Q, i.e., if f(J• - a•) • 2x2 - 1 Q i.e., if x(x2 - 1)2 > e' where x = ffa. The derivative of the left-hand member of this inequality with respect to x is - (6x• - 3x2 + 1)Jx2(x2 1) 3 • Since f";?:. a, then x ";?:. 1, and this derivative is negative for such values of x. Hence the left­ )lapd member of the above inequality decreases from + oo as x ipcreP.Ses from -

15]

GA U S S ' S T H E O R E M

429

unity, and the inequality will be satisfied for a range of values of x, 1 :::;; x :::;; x0, where x0 is a root of the equation 2x2 - 1 Q. = x (x2 - 1)2 In the case Q = 7ef18 this equation will be found to have a root x = 2. Hence the force acting on is attractive if < 2 .

e e a :::;; f a Example 8. A metal hemispherical dome of internal radius a rests on horizontal ground. A point charge e is placed at a height !a vertically above the centre of the hemisphere. Prove that the induced charge is divided between the dome and (M.T.) the ground in the ratio 1 3 2. Let the point charge e be at the point A and let A' be the point inverse :

to A relative to the hemisphere. The hemisphere is earthed, so that its potential is zero. This can be ensured by introducing an image charge at A '. These two charges do not, however, give zero potential over the plane ground. To satisfy this requirement, we introduce further image charges and at the points B, B' which are the mirror images of A , A ' respectively in the plane of the ground. These two extra image charges also contribute zero potential over the hemisphere. Hence, the field within the dome due to the system of four charges satisfies all the given conditions and is accordingly identical with the actual field.

-4ef3

-e

4e/3

-e 8 �e

a'

To find the charge induced on the inside of the dome, we shall employ equation (15.84) . Con.>ider the term contributed by the charge at A '. cos L OA'C = t, and hence the solid angle subtended at by the dome is 2Trj5. The term required is therefore -2ef15. Also, cos L OA C = ! and the solid angle subtended at A by the dome is accordingly 16Trj5. The term contributed by the charge at A is now found to be -4ef5. Similarly, the solid angles subtended by the dome at B and B' are 4Trf5 and

A'

430

[CH.

A C O U R S E IN A P P L I E D MATHEMAT I C S

The contributions of the charges at these points to the charge induced on the dome are consequently and respectively. The total charge induced on the dome is + = However, the net charge induced on the dome and the ground it covers is Thus, the equal to minus the charge enclosed by these surfaces, viz., charge induced on the ground must be The result required is there­ fore proved.

27Tf5.

ef5 -2efl5 - l·s e - te i e - 1� e - fte.

-e. -2efl5. Example 9. An electric dipole of moment M is placed at a point P at a distance f(>a) from the centre 0 of an earthed conducting sphere of radius a. Find the potential at every point of space if: (i) the dipole points along OP; (ii) the dipole is normal to OP, and show that the image system consists of a dipole and a charge in the first case, but of a dipole only in the second case. (Li.U.)

e

p

-e

Q

In case (i), let the dipole comprise a charge e at P and a charge -e at Q as shown in the diagram. These give rise to image charges at the inverse then OP' = points P' and Q'. If PQ = and OQ' = to the first order in Hence P'Q' = The image charges at P' and Q' are and + respectively. = These charges are equivalent to two charges of equal magnitude but opposite in sign at the points P' and Q', together with a charge at Q'. If is very small and correspondingly large, so that = M, the image system will be seen to comprise a dipole of moment

"8f, a2ff a2f(f + "8f) "8f. a2"8fff2. -aeff aef(f "8/) aeff - ae"8f/P aeff -ae"8f/P e eof =

a2ff - a2"8f/P of

J X P'Q' J a;r = J: M =

X

situated at P' and a charge at the same point. The dipole is directed along P'O. The potential at every point outside the sphere can now be computed. The potential within the sphere is, of course, constant and equal to that at its surface. We note that a charge is induced upon the sphere. If the dipole is normal to OP, the charges are given the positions shown in the diagram. The image charges are at P' and at Q'. By similar triangles, P'Q' OP' PQ = OP = f = yo

-aMf/2

-aMff2

±e -aeff a2ff a2

and hence the moment of the dipole P'Q' is �

ae X P'O' = T

I!!'

f

a2 = pas M,

X yo PQ

aeff

GAUSS ' S THEOREM

15]

431

since . PQ = M. In this case the image system is a dipole of moment a3Mff3 at the inverse point and opposite in direction to the dipole of moment M. No charge is induced upon the sphere.

e

The problem presented by an infinite line charge parallel to the axis of an infinite circular conducting cylinder may be solved by the introduction of appropriate image line charges. Thus, if the cylinder is of radius a (Fig. 15.12) and if the line charge is of magnitude e per unit length and passes through the point A where OA = f, we introduce an image line charge -e at the inverse point A '. If P is any point on

FIG. 15.12.-Infinite Circular Cylinder and Line Charge

the cylinder, the potential at all points on the cylinder is constant, as required, if 2e log A P - 2e log A ' P = C, i.e., if APJA 'P = constant. (15.85) But, as proved for the case of the sphere and point charge, AP OA f = = constant. = A ' p OP a Hence the image charge is appropriate.

(15.86)

432

A C O U R S E IN A P P L I E D M A T H E M A T I C S

[CH.

The charge induced upon unit length of the cylinder is equal to the image charge enclosed in this length, viz., -e. If the cylinder is insulated and uncharged, a second image line charge must be introduced at 0 of magnitude e per unit length. This maintains the potential over the cylinder constant, but, since the charge enclosed by unit length of the cylinder is now zero, this body is uncharged. 15.8. The Complex Potential

A very powerful technique for dealing with two-dimensional electro­ static problems is furnished by the properties of analytic functions of a complex variable. If Ox, Oy are rectangular axes in a plane perpendicular to the generators of the infinite charged cylinders re­ sponsible for the field, let us regard the x-axis as the real axis and the y-axis as the imaginary axis of an Argand plane. Putting z = x + iy, consider any function f(z). Let f(z) = U(x, y) + iV(x, y). It is shown in texts devoted to pure mathematics * that if j(z) is to possess a unique derivative, then it is necessary that

oU oV oU = - oV. (15.87) ox oy ' oy ox the Cauchy-Riemann equations. Eliminating U between

These are these equations, we obtain

or

o2 V o2 U - 8o22V Bx2 - oxoy y• \7 2 V = 0.

(15.88)

Thus V is a possible potential function and, provided it satisfies the boundary conditions for a problem, it will define the unique solution to the problem. If w = f(z) provides the solution to an electrostatic problem, w is called the complex potential of the field. Its real and imaginary parts, viz., U and V, are said to be conjugate functions. Since the derivative of w is unique, it may be calculated by differenti­ ation parallel to the x-axis. Thus

=

,.,

-Ev - iE

(15.89)

(Ex, E11) are the components of the field intensity. Hence (15.90) I �; I = (F:.�2 + Ey2) l = E. E.g., Functions of a Complex Variable, by E. G. Phillips, Chap. I (Oliver an
where

*

Boyd).

GA U S S ' S T H E O R E M

15]

433

It follows from equations (15.87) that

8U &U + 8V 8V ax ay

ax ay

_ -

0

·

(15.91)

This is the condition that the two families of curves U = constant, V = constant should cut orthogonally. But the family V = constant is the family of equipotentials. Hence the family U = constant is the family of lines of force. If one of the cylindrical conductors intersects the xy-plane in a curve C of which AB is an arc (Fig. 15. 13), the charge per unit length on the y

B

FIG. 15.13.-Charge on a Cylinder

portion A B of the cylinder is Q=

Bcr ds = B Ends, JA 47t JA 1

(15.92)

where s is the arc length parameter on the curve A B. Let if; be the angle made by the tangent to AB with the x-axis. Then En = -Ey cos if; + Ex sin if;,

oV

8V .

= - cos if; - - sm if; ay ox au dx au dy _ - ox ds + oy ds ' dU Hence

'

(15.93) (15.94)

A C O U R S E I N APPLI E D MATHEMAT I C S

434

[CH.

Thus, following the curve with the conductor to our left, the increment in U as we proceed from A to B gives the charge per unit length on this segment of the cylinder. To make use of the results just obtained, our procedure will be an indirect one. We shall consider some well-known analytic functions and determine the electrostatic problems for which they provide the solution. w

15.8(1)

= z2

We have

U + iV = (x + iy)2 = x2 - y2 + 2ixy. Thus the equipotentials are the rectangular hyperbola; xy = constant, and the lines of force are the rectangular hyperbola; x2 - y2 = constant. Any two equipotentials can represent conductors, e.g., the two positive axes for the one conductor and the hyperbola xy = c2 for the other. Their potentials will be 0 and 2c2 respectively. The reader should sketch these two curves and the lines of force between them. Along the conductor represented by Ox, dU _ En = dU ds = dx = 2x Hence = Enf4rc = -xf2rc and the charge is negative and its density _

·

cr

is proportional to the distance from 0.

15.8(2)

w

= log z

We have

U + iV = log (x + iy) = log r + ie, where (r, 6) are the polar coordinates of the point (x, y) . Thus the equipotentials are the radial lines () = constant, and the lines of force are the concentric circles r = constant. We can take any pair of radial lines () = ex, () = � to be two conductors at potentials and �· Then, if ex > �, along () = ex we have dU ! En - dU ds - dr - r ' _

_

ex

_

i.e., cr = 1f4rcr and the charge density is inversely proportional to the distance from 0.

15.8(3)

w

We have

= 1/z - iy • U + z. V = XX� + y2

The equipotentials are the curves - y = constant

X2 + y2

V,

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GA U S S ' S T H E O R E M

435

i.e., the system of co-axial circles

x2 + y2 + yjV = 0.

These circles all pass through 0 and are tangential to the x-axis there. The lines of force form the orthogonal co-axial system x2 + y2 - xjU =

0

which are all tangential to the y-axis at 0. The reader should sketch these two systems. In polar coordinates V=

- ! sin e r

=

! cos (B + !7t). r

Referring to equation (14.78) , we observe that this is the potential due to a line doublet of moment ! directed along the negative y-axis. If f(z) is analytic, so is if(z). But

ij(z) = - V + iU

and we can interpret the curves U = constant as equipotentials and the curves V = constant as lines of force. It is easily verified from equations (15.87) that "V 2 U = 0, as we expect. Thus, in each of the three cases considered above, any pair of lines of force may be regarded as two charged conductors, in which case the old equipotentials become the new lines of force. For example, the complex potential w = i log z is appropriate to the field between two co-axial circular cylinders, the radial lines e = constant then represent­ ing the lines of force. Also, w = - 2ie log z has imaginary part - 2e log r ; this being the potential for an infinite line charge of strength e, - 2ie log z is the complex potential for such a charge.

Two parallel, infinite, line charges of strengths e and -e intersect the Argand plane in the points (a, 0), (-a, 0) respectively . Show that the complex potential is given by w = -2ie log zz -+ aa. Deduce that the capacitance per unit length of an infinite circular cylinder of radius r, whose axis is distant d from an infinite earthed plate is {d + (d: - r')} · l/2 log If P is any point z in the Argand plane and A is the point (a, 0), the vector AP represents the complex number (z - a). Hence the complex potential of the line charge through A is -2ie log (z - a). Similarly, the complex potential of the line charge through A'( -a, 0) is 2ie log (z + a). The com­ plex potential for the two charges is therefore w = 2ie log (z + a) - 2ie log (z - a) z-a = - 2ie log z + a·

Example 10.

�

A COURSE I N APPLIED MATHEMATICS

436

Hence

[cH.

V = -2e log !; � :1 = -2e log ��

and the equipotentials are the curves for which = constant. The loci defined by this equation are well known to form a system of co-axial circles with as limiting points and the imaginary axis as radical axis.* Also zU= arg = 2e arg z arg z z = = The lines of force are accordingly the loci for which = constant. These form a co-axial system of circles with as common chord. This system is, of course, orthogonal to the system of equipotentials.

APfA'P

A, A'

2e + aa { ( - a) - ( + a)} 2e(LPAx - LPA'x) 2e LAPA'. LAPA' AA' y

Let us choose the imaginary axis and a circle of radius from among the equipotentials and treat these as two conductors. Then our complex potential provides the solution to the problem of the field between these conductors. Considering the two possible positions of see diagram), since these points lie on the same equipotential = = .\. Thus, if is the centre of the circle and =

r

B, B' P ( ABfA'B AB'/A'B' C OC d, d + r - a = �-- d + r = .\ d+r+a a+d-r and hence a = v'(d2 - r2), .\ = rf{d + y(d2 - r2)}. E.g., Analytical Geometry, by A. Robson, Vol. I, Chap. 5 (C.U.P.) . *

437

l l>J The potential of the circular conductor is -2e log ,\

=

2e log

d + y (d" - r2) r

·

Since AP = A 'P when P lies on the imaginary axis, its potential is zero. If we proceed around the circular conductor from keeping the conductor to our left, A'PA increases from 0 to 2TT. The increment in is accordingly 4TTe and, by equation (15.94), the charge per unit length of this cylinder is e. Hence, the capacitance per unit length of a cylinder of radius r whose axis is a distance d from an earthed plane proves to be as stated. Clearly the two line charges provide an image system for the circular cylinder and the infinite plane.

B,

U

APPENDIX Let be a system of rectangular axes and let u (x, z) , v (x, z) be a scalar and a vector function of position respectively. Denoting the components of v in the directions of the axes by v", Vu, Vz, the corresponding components of the vector uv are uvx, uvy, uv. and hence

Oxyz

y,

div (uv)

=

=

=

i( ( U

u

)

uvx

+

�(

)

uvv

+

OVz OVy OVx + + OZ OX iJy

div v

+

v

·

)

grad u.

i( +

)

uv. ,

Vx

y,

OU

8);

Vy

+ OUoy +

Vz

OU ()z '

EXERCISE 15 l. By applying Gauss's theorem, determine the attraction at a point inside an infinitely long, uniform, solid cylinder. An infinitely long, uni­ form, solid cylinder of radius a has a cavity in the form of a sphere of radius a with its centre on the axis of the cylinder. Show that inside the cavity the equipotential surfaces form a system of hyperboloids of revolution. (L.U. )

2. A plane slab of matter, infinite in extent in all directions, is of thick­ ness d. The density at any point P in the slab distant x from one plane face is p, where p is a function of x. If G is the intensity of attraction at P, by applying Gauss's theorem to a cylinder whose generators are perpendicular to the faces of the slab and whose ends are in the planes parallel to these faces and distant x and x + dx from one of them, show - 4rryp. Deduce that, if p is proportional to x that dGjdx =

G

=

(

2rryM 1

-

2;2) ,

where M is the mass per unit area of the slab.

3. Using Gauss's theorem, find the potential at all points due to a uni­ form volume distribution of total charge 2e contained inside a sphere of radius a. Two point charges each equal to - e can move freely inside this uniform distribution. Find the configuration of equilibrium . (Li.U. )

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A C O U R S E I N A P P L I E D MATH EMAT I C S

[cH.

4. A distribution of charge, none of which lies outside a sphere of radius a and centre 0, produces a potential given by V

5.

6.

7.

8.

=

ga {exp (1 - rfa) + 1}, r :::;; a,

2Q V = -, r � a, r where r is the distance from the point 0 and Q is a constant. Find how the charge is distributed. (S.U.) A spherical conductor of radius a is kept by batteries at a constant potential if>. It is surrounded by a concentric, spherical conductor which is kept at zero potential. Under the internal forces this latter conductor contracts from radius b to radius b1 . Show that the work done by the electrical forces is if> 2a2(b - b1) /{2(b 1 - a) (b - a) }. Find the energy supplied by the battery. (Li.U.) Three concentric, hollow, conducting spheres have radii a, b, and c (a < b < c) and carry total charges q1 , q2 , q3 • Find their potentials. The inner and outer spheres are now connected by a fine wire, and form one plate of a condenser ; the middle sphere is the other plate. (Li.U.) Show that the capacity is b2 (c - a)f(b - a) (c - b) . A conducting sphere of radius a is surrounded by a concentric, con­ ducting, spherical shell of inner and outer radii b, c respectively (a < b < c) . Find the electrostatic energy of the system in the following two cases : (i) sphere at potential if>1, shell earthed ; (ii) sphere earthed, shell at potential if>2 • (Li.U.) Obtain expressions for the energy of a parallel-plate condenser, neglecting edge effects. Show that the energy is an increasing function of the distance d between the plates if the charge is kept constant, but a decreasing function of d if the potentials are kept constant. (Li.U.)

9. ·An insulated, spherical conductor, formed of two hemispherical shells in contact, whose inner and outer radii are b and b', has within it a concentric, spherical conductor of radius a and without it another concentric, spherical conductor, of which the internal radius is c. These two conductors are earth connected and the middle one receives a charge. Show that the two shells will not separate if 2ac > be + b'a. (M.T.)

10. A parallel-plate condenser is formed by three conducting plates, each of area A ; the two outer plates are earthed and are at a distance 2d apart, while the inner plate has a charge E and is at distances d ± x from the outer plates. Neglecting edge effects, find the potential difference between the inner and outer plates, and the mechanical (M.T.) force on the inner plate. l l . Three infinitely long, co-axial, circular cylinders have radii a, b, c (a < b < c) , and are made of conducting material. The cylinders of

15]

12.

13.

14.

GA U S S ' S T H E O R E M

4 39

radii a and c are earthed and the cylinder of radius b is charged. Show that the condition that this charge should be equally distributed between the inner and outer surfaces of the cylinder is that b2 = ac. Two semi-infinite, earthed, conducting planes meet at right angles at a common edge. A charge e is placed on a bisector of the angle between the planes at a distance d from the common edge. Find the image system and show that the charge is attracted towards the common edge with a force (2y2 - 1 ) e2f4d2 • A point charge e is placed outside an insulated, uncharged sphere of radius a at a distance f from the centre. Determine the image system, and prove that the surface density cr at a point of the sphere distant r from the point charge e is given by 4rrcrajr3 = e[r3 - f(f2 - a2) ] . (Li.U.) A point charge -e is situated at the centre C of a hollow, earthed, conducting sphere of internal radius a . A second point charge + ke is at a distance f(< a) from C. Find the potential at every point within the sphere. If k is positive, show that the resultant force on the charge ke vanishes if f is the root between 0 and a of the equation f4 - kaj3 - 2a2f2 + a4 0. (Li.U.) Two semi-infinite, conducting planes A and B intersect at right angles, and are earthed. A point charge e is placed in the corner formed by the planes, at distances a and b from A and B respectively. State the conditions satisfied by the potential cf>, and use the method of images to determine cf>. By choosing the planes A , B, and the plane through the charge perpendicular to them as the coordinate planes of a rect­ angular system of axes, derive an expression for the surface density of induced charge at any point on A , and hence show that the total charge on this plane is =

1 5.

-� tan-1 (�) .

(Li.U.)

16. A point charge e is situated at P at a distance c from the centre 0 of a conducting sphere of radius a at zero potential. Find the force acting on the point charge. Another point charge e', situated at Q so that POQ is a straight line and PO OQ, is such that the net force on e is zero. Show that 4ac3(c2 + a2) 2 e' (Li.U.) e- = (c2 a2) 2 { (c2 + a2) 2 - 4aca} ' =

-

17. A conducting hemisphere of radius 3a is placed with its base on an infinite conducting plane maintained at zero potential. If a point charge is placed on the axis of the hemisphere at a distance 4a from its centre, on the same side of the plane as the hemisphere, show that the ratio of the total induced charge on the plane to that on the hemisphere is 7 : 13. (Li.U.) 18. Show that the surface density of the charge induced on a spherical conductor at zero potential by an electric charge q varies inversely as (S.U.) the cube of the distance from the charge q.

440

A C O U R S E I N A P P L I E D M A T H E M A 'fl C S

[cH.

19. A conductor, in the form of an infinite plane with a hemispherical boss of radius a, is at zero potential in the presence of a point charge e at a point P. The line OP makes an angle 6 with the normal to the plane, where 0 is the centre of the circular base of the hemisphere and OP f Obtain the image system. Show that, if a;j is small, the component of attraction on the point charge normal to the plane is greater than it would be in the absence of the boss by an amount e2a3 cos 6 (1 + 3 cos 26) j• approximately. (S. U .) =

.

20. A positive point charge at A is fixed in front of a large conducting plane which is earthed. Show that the tubes of force, which start on the point charge and make an angle if> or less with the normal from A to the plane (not with the normal produced) , end on a circular region of the plane whose radius subtends an angle 6 at A , where sin !if> y2 sin !6. (M.T.) =

2 1 . The centre of an insulated, uncharged conducting sphere of radius a is at the middle point of the straight line joining two equal point charges of electricity, which are a distance 2c apart. If ajc is small, show that the force on either charge would be increased in the approxi­ mate ratio

if the sphere were removed.

(M.T.)

,

22. A point-charge e is placed inside a hemispherical hollow of radius a in an uninsulated conductor. The point-charge is on the axis of symmetry at a distance f from the plane boundary. Show that it is in equilibrium if ffa is equal to a root of the equation xs - 8x7 - 2x4 - 8x3 + 1 0. Show also that this equation has one and only one root between 0 and 1 . (M.T.) =

23. An insulated, uncharged, conducting sphere of radius a is placed with its centre at a distance a cot c;:, where c;: < i7t, from a point charge e . Prove that the charge induced on the hemisphere nearest to the point charge is . 3Cl . Cl -e sec c;: sm -2 sm 2. (M.T.) 24. A conducting surface consists of two infinite planes which meet at right angles and a quarter of a sphere of radius a fitted into the right angle. If the conductor is at zero potential and a point charge e is symmetrically placed, with regard to the planes and the spherical surface, at a great distance f from the centre, show that the charge induced on the spherical portion is approximately - 5ea3j7tj3. (M.T.)

1 5]

GA U S S ' S T H E O R E M

441

25. A long, cylindrical, conductor of radius a, at zero potential, is placed in an electrostatic field perpendicular to its axis and originally uniform and of magnitude E If the cylinder consists of two halves in contact on the diametral plane perpendicular to the original direction of the field, show that two equal and opposite forces of magnitude 2aE 02/3rt, per unit length of the cylinder, are necessary to hold the halves together. End effects are neglected. (S.U.) 26. If r is the position vector of any point on a closed surface S and n is the unit outward normal to S, show that the volume enclosed by S is

0

•

�

L

r

·

n dS.

27. If V is a scalar function of position defined throughout a region r bounded by a surface S and a is any constant vector, prove that div( Va) = a · grad V. Deduce, using Green's theorem, the result

jr

grad V dv =

fs Vn dS,

where n is the unit outward normal to S. 28. r is the region between non-intersecting closed surfaces S v S 2 , SN and a surface 50 which encloses them. V, V' are finite differentiable scalar functions of position throughout r. Applying Green's theorem (equation (15.22)) to the functions V grad V' and V' grad V, show that •

1r ( V\7 2 V' - V'\7 2 V)dv = .� ols;( ( V ,_

:

0 • 0

V'

•

•

�:) dS,

where 8/on denotes differentiation along the normal to S; in the sense outwards from r. Deduce that if V 1, V 2 , VN are the potentials of N conductors when their charges are ev e 2 , en respectively and if V1', V 2 ', Vn' are their potentials when their charges are e/, e 2 ', en'· then �e V' = �e'V. (This is Green's Reciprocal Theorem.) 29. Show that the complex potential •

.

•

•

•

•

•

•

w =iE

.

•

•

•

(� - z)

solves the problem of an earthed conducting cylinder of radius a placed in a uniform field E. Deduce that the equation of the family of lines of force is

(� + r) sin e = constant,

where (r, 6) are polar coordinates. Show also that the charge induced on unit length of the cylinder comprises a positive charge aE(rt spread over one half of the cylinder and an equal negative charge over the other half. w = i cosh-1 (zfa), (a real) 30. If x = a cos U cosh V, show that y - a sin U sinh V. =

442

A C O U R S E I N A P P L I E D MAT H E M A T I C S [cH. 1 5]

Deduce that the curves U = constant are confocal hyperbolce and the curves V constant are confocal ellipses. What electrostatic problem is solved by the complex potential w? The two plates of an infinite cylindrical condenser intersect any plane normal to its generators in an ellipse of semi-axes a, b and the straight line joining the foci. Show that its capacitance is a+b l jlog a b. 31. Show that the complex potential w = zi solves the problem of the field between a semi-infinite plane and a parabolic cylinder. Show that the lines of force form a system of confocal parabolce. 32. Show that the complex potential =

_

z- a

- 2ie log + z a solves the problem of the field between a line charge of strength e and a parallel, infinite, earthed plane a distance a away. Deduce that one half of the charge induced on the plane lies within y2a from the line charge. w

=

--

CHAPTER 1 6

POLARIZED MEDIA 16.1. Dielectrics.

Polarization

It is found experimentally that when certain insulating materials, such as mica or glass, are placed in an electrostatic field, they become polarized, i.e., any small portion of the material exhibits the properties of an electric dipole. Such materials are called dielectrics. This phenomenon can be explained in either of two ways. The atoms of any material comprise positive and negative charges and, if the material is a conductor, the negative charges are free to move through the body of the substance under the action of any applied electric field. If, how­ ever, the material is an insulator, this flow of charge cannot take place and an applied field can only shift very slightly the charges in an atom. The positive charges shift in the direction of the field and the negative charges in the opposite direction. The small separation between the charges produced in this manner creates a dipole whose axis is in the direction of the field. Alternatively, we may suppose that the molecules of the material are naturally dipoles, but that, in the absence of an electric field, these dipoles arrange themselves in a random fashion. Any small volume of the material containing a large number of molecules will not then appear to be polarized, since the constituent dipoles will cancel one another out. If, however, an electric field is applied, a couple given by equation (14.75) will be applied to each dipole, tending to align it with the field. Any small volume of the dielectric will now contain a preponderance of dipoles with axes point­ ing approximately in the direction of the field, and accordingly will appear to be polarized. The first explanation is considered adequate to explain the polarization of such dielectrics as glass and mica, but some liquids, such as water, and gases, such as ammonia, are inherently polarized, and for these both effects contribute to the polarization observed. Let dv be any small element of a polarized dielectric. This element will behave as a dipole whose moment will be assumed to have mag­ nitude proportional to its volume. Thus, we shall take the moment of the element to be P dv. P is then the moment per unit volume in the neighbourhood of dv and is termed the polarization. We shall assume further that the polarization at any point in a dielectric is proportional to the electric field intensity E at the point and is in the same direction. This is true for media that are isotropic (i.e., have the same properties in all directions from a point) but is not true for certain crystals. The 443

444

A C O U R S E IN A P P L I E D M A T H E M A T I C S

[CH.

assumption we have made is suggested by the two explanations of the phenomenon of polarization that we have already advanced. Thus, we have the equation P = kE, (16.1) where k is called the considered.

susceptibility of the dielectric at the point being

16.2. Poisson's Equivalent Distribution

Suppose, then, that we require to calculate the contribution made to an electric field by some polarized dielectric occupying a region of space r bounded by a surface S. Let dv be an element of dielectric at the point Q (Fig. 16. 1). This element behaves as a dipole of moment P dv. A

FIG. 16.1 .-Potential Due to a Polarized Medium

The field intensity E or potential V at any point outside the dielectric can be found by integrating the effects due to all such elementary dipoles comprising the dielectric. To compute the potential at a point within the dielectric, we suppose that a small cavity is cut in the di­ electric at the point being considered. V is then computed in the cavity and its limit found as the dimensions of the cavity all approach zero. E cannot be defined in this way, for it is found that the limit to which the intensity tends is not independent of the shape of the cavity. This remark will be amplified in Section 16.4. The limit approached by V is unique, however. We shall accordingly proceed to find a formula for the potential which will be valid at all points both inside and outside the dielectric and leave the precise definition of E until a later stage in the argument. It will be appreciated that until this stage has been reached, equation (16. 1) has not been accorded a precise meaning. Let A be a point at which we wish to compute the potential V due to the dielectric. The contribution of the element dv at the point Q is, by equation (14.67), (16.2) where r is the distance AQ (Fig. 16.1). As has been explained already on p. 381, the differentiations implicit in the gradient have to be

16]

P O L A R I Z ED M E D I A

445

carried out on the assumption that A is a fixed point and Q a variable one. The net potential at A due to the whole dielectric is therefore (16.3) Now, as was proved in the Appendix to the last chapter, if u is a scalar function of position and v a vector function of position, then Putting u ment,

div uv = u div v + v · grad u. . (16.4) = 1/r and v = P in this identity, we obtain after rearrange­ P · grad

U) = div (P/r) � div P.

(16.5)

dv - lr( divr P dv.

(16.6)

-

Hence, equation (16.3) may be written in the form V = ( div (P/r)

lr

The first integral of the right-hand member may be transformed by Green's Theorem in the form of equation (15.22). Thus PndS - ( div P dv. (16.7) V=

1

s

r

lr

r

where Pn is the component of P along the outward normal to S. By equation (14.47), the first term of this last expression for V is the potential at A due to a surface distribution of charge over S having density Pn at any point. By equation (14.46), the second term of this expression is the potential at A due to a volume distribution of charge over r having density -div P at any point. When computing V we may consequently replace the polarized dielectric by these two charge distributions which together form what is known as Poisson's

Equivalent Charge Distribution.

We can now define the field intensity E at any point in the usual manner as the force exerted upon unit positive charge placed at the point. For, if the point P is within the dielectric and this material is not fluid, we can imagine it replaced in the neighbourhood of P by a liquid having electrical characteristics identical with those of the solid medium. This permits the transport of a unit positive charge from P to a neighbouring point P', and hence the argument culminating in equation (14.35) is valid, i.e., we have (16.8) E = -grad V, . defining E uniquely.* * A more penetrating analysis reveals that and E change very rapidly in the interstices between the polarized molecules of a dielectric. Our definitions yield a " smoothed-out " field.

V

446

A C O U R S E I N A P PL I E D MATHEMAT I C S

[cH.

Alternatively, E can be defined as the field within a small cavity in the Poisson distribution, for the field within a cavity in a volume distribution of charge approaches a unique limit as the cavity collapses upon a point. V being the potential function for this distribution, equation (16.8) follows immediately, indicating that the second defini­ tion is equivalent to the first. Consider the field due to a volume distribution of charge of density p in a dielectric medium. We replace the polarized dielectric by its Poisson distribution so that the charge density is altered to the value p - div P. Then, from equation (15.26), or

div E = 47t(p - div P), div (E + 47tP) = 47tp.

(16.9) We now define a new vector D, called the electric displacement, by

means of the equation

D

= E + 47tP,

(16.10)

(16.9) takes the form div D = 47tp. (16.11) This is the form of equation (15.26) which is appropriate in a dielectric medium. In empty space, clearly P = 0, D = E and equations (15.26) and (16.11) are identical. Equations (16.1) and (16.10) show that D = E + 41tkE = (1 + 47tk)E, or D = KE, . (16.12) where K = 1 + 41tk . . (16.13) K is termed the dielectric constant of the medium. In free space K = l. Equations (16.8), (16.11) and (16.12) are sufficient to enable us to solve any electrostatic problem in which dielectrics are involved. Eliminating E and D between these equations, we obtain an equation for V, viz., div (K grad V) = - 47tp, (16.14) so that equation

a ( av) o ( oV) o ( oV) ox K ox + oy K oy + oz K oz = - 41tp. (16 · 15)

or, in cartesians,

If K is constant throughout the medium, this latter equation may be written

(16.16)

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POLARI ZED MEDIA

447

which should be compared with equation (15 .27) . If also there is no free charge, p = 0 and V satisfies Laplace's equation (15.30), i.e., V is harmonic. Let 5 be a surface of discontinuity between two dielectric media 1 and 2, the susceptibility on one side of 5 being k1 and that on the other side k2 • A subscript n will indicate a component in the direction of the normal to 5 drawn from 5 into the medium l . If P1 , P2 are the polarization vectors in the two dielectrics, when these are replaced by their equivalent charge distributions, surface charges of densities - P1n, P2n appear on 5. For the purpose of achieving absolute generality, we will suppose that there is also a free surface charge of density a over 5. Then, by equation (15.31),

E1n - E2n = 47t (cr - P1n + P2n), (E1n + 41tP1n) - (E2n + 41tP2n) = 47tcr,

or i.e.,

(16.17)

This equation implies that there is a discontinuity in the normal component of the displacement of 47tcr across 5. In the absence of free charge, Dn is continuous. Since D = 0 within a conductor, at the surface Dn = 47tO". (16.18) There is no discontinuity in the potential across 5, and hence, as proved on p. 410, any tangential component of E is continuous. In particular, E, and hence D also, is normal to the surface of any con­ ductor. Suppose that a field is due to a volume distribution of charge of density p in a region occupied by dielectric. Let 5 be any closed surface in the region enclosing a volume r. From equation (16.11), we obtain

47t lr p dv = Jr div D dv = l

Dnd5,

having employed Green's Theorem to effect the transformation from a volume to a surface integral. The last equation is equivalent to the statement 47t x charge enclosed by S = flux of D out of 5. (16.19) This is the generalized form of Gauss's Theorem. The theorem may be applied to the calculation of the D-field due to various charge distribu­ tions possessing spherical or cylindrical symmetry, etc. , in the manner already described in Section 15.2. Thus, in the case of a conducting sphere carrying a charge e and immersed in a dielectric of constant K extending to infinity, at distance r from its centre D = efr2, and hence E = efKr2 • The presence of the dielectric accordingly reduces the magnitude of the field by a factor 1/K. E

448

A COURSE IN APPLI ED MATHEMATICS

[CH.

If we imagine the sphere to shrink to a point charge e, the field intensity at a distance r from the charge is ejKr2, and hence the force exerted upon another point charge e' placed at this distance from e is given by

ee' Kr2

(16.20)

F = --- ·

This is the general form of equation (14.3) (Coulomb's Law) . D-lines and tubes may be defined in the obvious manner, and will possess all the properties found to be possessed on pp. 404-5 by the lines of force in free space.

A condenser comprises two conducting surfaces, one enclosing the other and separated by a vacuum. The outer conductor is earthed. If C is the capacitance, prove that when the space between the conductors is filled by dielectric of constant K, the capacitance is increased to KG. Let q, be the potential of the inner conductor, a the surface density of the charge upon it and e the total charge. Then

Example 1.

The potential V in the vacuum between the conductors satisfies Laplace's equation and the boundary conditions V q, over the inner conductor and V = 0 over the outer. By the uniqueness theorem, there is only one potential function having these properties. The density a is given by the equation

=

a = 411T En =

-

1 av

47T an '

where 8Vfon denotes the derivative of V along the normal outwards from the inner conductor. Now suppose that the space between the conductors is filled with dielectric and the potential of the inner conductor is kept constant at by addition or subtraction of charge. V still satisfies Laplace's equation and the same boundary conditions. It follows by the uniqueness theorem that the distribution of V is unaltered. However, if is the new surface density of charge 1 av , = 477 477 o =

a'

a = Dn

K

n Ka. Thus the charge density is increased everywhere by a factor K. The total charge on the inner conductor is accordingly Ke and the new capacitance is C' = Ke = KG. -

P O L A R I Z ED M E D I A

1 6]

449

A condenser consists of two similar parallel plates at a distance d apart, one of the plates being earthed. The space between the plates is filled with a substance whose dielectric constant increases uniformly from one of the plates to the other. Neglecting edge effects, show that the capacitance per unit area is 4wd log (K2JK1) ' where Kv K2 are the values of the dielectric constant at the two plates.

Example 2.

(Li.U.) If x is measured from the insulated plate in a direction perpendicular to this plate, V is a function of x alone and equation (16. 15) takes the form V !:_ = 0. Integrating, we obtain

where

A is a constant.

Now

d dx (K dx ) K dV dx = A '

and hence

Ad dV = dx K1d + (K2 - K1)x' Integration yields the result V = B log {K1d + (K 2 - K1)x} + C, where B = Adj(K2 - K1) and C is another constant. ·when x = d, V = 0 and when x = 0, let V = tf>. Then 0 = B log (K d) + C, = B log (K21d) + C.

Thus If

a

is the charge density on the insulated plate, 1 a - 4,;- Dn - 4,;- n -

_ K1 E - K1 ( dV) 4w dx B(K 2 - K1) (K2 - K1) " 47Td log (K2/K1} 4wd The capacitance per unit area is therefore K - K1 "4> = 47Td log2 (K2JK1) ' Example 3. Three concentric thin spherical conductors A, B and C have radii a, b and c (a < b c). The space between A and B is filled with dielectric of constant K, and that between B and C is filled with dielectric of constant K'. The conductors A and C are earthed, and B carries a charge of total amount Q. Show that Q is divided between the inner and outer surfaces of B in the ratio Ka(c - b)JK'c(b - a). (Li.U.) Let V be the potential function in the region 1 between the spheres A and B and let V 2 be this function in the region 2 between the spheres B _

_

a

<

1

x-o

A COURSE IN APPLIED MATHEMATI C S

450

I

C.

[CH.

and In both regions and the potential satisfies Laplace's equation and possesses the property of being spherically symmetrical about 0. Thus, from equation (15.55),

Over r

= a, V 1 = 0.

2

v1 = A; + B . Hence B = -A fa and thus V 1 = A U - !) ·

Similarly, in region 2, since

V2 = 0 over r =

c,

we have

The potential is continuous across the conductor

__

B and hence

a1 is the surface density of charge over the inner surface of B, !!_ ( 8 V 1) KA 2_ D.,. _- 47T a 1 = 47T - 47Tb2. or It follows that the total charge over this surface is Q1 = 47Tb 2a 1 = -KA. I f a2 i s the surface density o f charge over the outer surface of B, a• = -!: (o;"),�b = !:�· The total charge on this surface is therefore Q 2 = 47Tb 2a2 = K'C. If

r�b

We can now deduce that

Q1 Q�

use having been made of equation (i) .

(i)

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P O L AR I Z E D M E D I A

451

16.3. Harmonic Functions and Image Charges

The uniqueness theorem introduced in Section 15.6 can be extended to the case of an electrostatic field in the presence of dielectric media. Suppose that, in any particular problem, a field can be found whose potential satisfies equation (16. 14) and which leads to the specified value of the potential or of the total charge on each conductor and which also satisfies the following conditions at any boundary between two dielectrics : (i) The potential is continuous across the boundary, and (ii) Dn is continuous across the boundary, unless a surface charge exists there, in which case there is a discontinuity of an amount given by equation (16.17) .

Suppose also that, in the neighbourhood of any point charge e, the potential function approaches infinity after the manner of ejKr, where r is distance measured from the charge. Then the extended Uniqueness Theorem states that this field is the unique solution to the problem. We can accordingly employ harmonic functions and image charges to solve problems of this nature in the manner already explained in the previous chapter. p

FIG. 16.2.-Dielectric Sphere in

a

Uniform Field

Thus, suppose that a sphere of dielectric constant K2 is introduced into an infinite medium of dielectric constant K1 (Fig. 16.2). Before the sphere was embedded in the medium, let there have been a uniform field of magnitude E. Our problem is to calculate the field in the medium and in the sphere after the introduction of the latter. The presence of the sphere will distort the uniform field in the infinite medium, but this distortion will be negligible at great distances from the sphere. Accordingly, if V1 is the potential function in the medium it is reasonable to assume that V1 Er cos 6 + A2 cos 6, (16.21) =

-

r

452

A C O U R S E I N A P P L I E D MATHEMAT I C S

[CH.

where (r, 6) are spherical polar coordinates as indicated in the figure. The first term of this expression for V1 corresponds to the uniform field and the second to the distortion caused by the sphere. The second term approaches zero as r -+ CY:J, as required. Within the sphere, the potential function v2 cannot involve a harmonic function of the type cos 6jr2, since this would become infinite at the centre 0, whereas there is no point charge at this point. We shall therefore assume that V2 = Br cos 6 (16.22) We shall now show that the two boundary conditions at the surface of the sphere can be satisfied by appropriate choice of A and B. First, the potential must be continuous across the spherical surface and hence, if a is the radius of the sphere, we must have •

.

- Ea cos 6 + Aa2 cos 6 = Ba cos 6. This is true for all 6 if

Secondly, the normal displacement must be continuous. sense of the outward normal to the sphere, we have

(16.23) In the

Din = KlEin = -K1 ( 8ur�1 )r � a = K1 ( E + 2a1 ) cos 6, D2n = K2E2n = - K2 ( 8u�r-2 )r � a = - K2 B cos 6. Thus D 1n = D 2n for all 6 if K1 ( E + 2a1 ) = - K2B, 2K1A + K2a3B = - K1a3E. (16. 24) or if Solving equations (16.23) and (16.24) for A and B, we obtain KI - K2 a3E B = - 3KI E A - - 2K K2 + 2K1 I + K2 ' Thus, K1 - K2 a3 ) E cos 0 , V1 = - ( r + 2K (16.25) 1 + K2 r 2 1 Er cos 0 . V2 = - K 3K (16.26) 2 + 2K1 .

Equation (16.25) indicates that the sphere has the same distorting effect as would a dipole of moment

KI - K _ a 2K1 + K22 a E

placed at 0 with its axis in a direction opposed to the field.

1 6]

POLARI ZED MEDIA

453

Equation (16.26) implies that the field within the sphere is uniform, in the direction of the original field E, and of magnitude 3K1Ej(K2 + 2K1) . The problem of the field within a spherical cavity cut in a dielectric in which there was originally a uniform field E is a particular case of the problem we have just solved. Taking K2 = I , we find that the field is 3K1Ej(1 + 2K1). It will be noted that this field is not of magnitude E, showing, as was stated in the last section, that we cannot define the field intensity at a point in a dielectric as the field in any small cavity at the point. It will be proved in Section 16.4, however, that by choosing the cavity to have a certain particular shape, such a definition of the field intensity can be devised.

FrG. 1 6. 3.-Image Systems for Point Charge and Infinite Dielectric

As an example of the use of image charges, we will now solve the problem of a point charge e placed in front of an infinite plane face of dielectric of constant K (Fig. 16.3) . We will suppose that the field in the vacuum to the right of the plane face is generated by the charge e at P together with an image charge e ' at P', P' being the optical image of P in the plane face (Fig. 16.3 (a)). The field in the dielectric, on the other hand, cannot be generated by such a system of charges, for the potential at P', a point of this region, is not infinite. We will assume that this field can be generated by a charge e " at P. P is not a point of the region, and hence the infinity in the potential at this point is not significant. The fields in the two regions must satisfy the two boundary con­ ditions stated at the beginning of this section, over the plane face.

454

[cR.

A C O U R S E I N A P P L I E D MATH EMAT I C S

Let Q be any point of this face. Then the potentials at Q in both fields are identical if

e + e' = e".

i.e., if

(16.27)

(It should here be noted that the distorting effect of the dielectric upon the field in the vacuum surrounding P is represented by the image charge at P', i.e., for convenience of calculation we imagine that the dielectric is removed and e' substituted. This implies that the con­ tribution of the image charge to the potential at any point in the vacuum is to be calculated as if e ' were itself placed in a vacuum, not immersed in a dielectric.)

e

FIG. 16.4.-Field due to Point Charge and Infinite Dielectric

The field intensity at Q in the vacuum is the vector sum of ef PQ2 along PQ and e'jP'Q 2 along P'Q. Resolving in the direction of the normal n, we obtain En =

e'

P'Q 2

cos 6

e - PQe 2 cos 6 = (e, - e) cos PQ 2

•

(16.28)

Similarly, this same component of the intensity at Q in the dielectric is En =

e" cos e. - PQ 2

(16.29)

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P O LA R I Z E D M E D I A

455

Dn is accordingly continuous at Q if cos 6 Ke" (e 1 - e) P = cos 6, Q2 PQ 2

e - e1 = Ke". . (16.30) It will be found that conditions (16.27) and (16.30) are both satisfied if we choose e 1 and e " so that

i.e., provided

K-1 e = - - --- e (16.31) K+1 ' With these values for the image charges, the fields due to the two image systems are identical with the fields in the two regions separated by the plane face. Since the field in the dielectric can be generated by a single charge at P, the lines of force in this region are straight rays emanating from P. The complete field is represented in Fig. 16.4. The attraction of the image charge e1 for the charge e is K - 1 e2 eel . . = (16.32) - 4J2 K + 1 4d2 ' I

where d is the distance of e from the plane face. This force therefore represents the attraction of the dielectric for e.

An electric dipole of moment M. is situated at the centre of a spherical cavity of radius a in an infinite dielectric of dielectric constant K. Show that the field in the dielectric is equal to that of a dipole of moment 3Mf(2K + 1) situated in free space. Find the potential in the cavity . (S.U.) Taking spherical polar coordi­ nates (r, 6) as indicated in the p

Example 4.

diagram, let V 1, V 2 be the potential functions in the spherical cavity and in the dielectric respectively. v l must involve a term M cos 6fr2 corresponding to the infinity in the potential at 0 due to the dipole there. The distortion caused by the dielectric we will represent by the harmonic function A r cos 6. Thus M cos 6 V1 = + Ar cos 6. V

--r2-

cannot involve a term in

r cos2 6, since the potential in the dielectric does not become infinite with r. We accordingly try B cos e v2 = r2 . The boundary conditions over r = a are : (i) V1 = V2, . o--ar-V1 = K o V2 {11.. ) Dnt = Dn 2t I.e., ar·

2

I

456

A C O U R S E IN A P P L I E D M A T H E M A T I C S These lead to the equations

Whence

� + A a = �,

-

[CH.

2:: + A = - 2�B.

3M - 2M A = - 2K 2K + I aa' B = 2K + ! "

I n the dielectric therefore

3M cos a 2 V2 (2K + l)r =

and this is the potential function due to a dipole at 0 of moment 1) . In the cavity 1 = ;o cos a . + 1 aa vl

3M/(2K +

(

16.4. Fields in Cavities

2K - 2 r ) M 2K

Suppose that a small cavity exists in a dielectric. Any point within the cavity is in free space, and the field there can be found by replacing the dielectric by the equivalent charge distribution. This will com­ prise : (i) a distribution over the external surface of the dielectric of density Pn ; (ii) a distribution over the interior of the dielectric of

--P

FrG. 16.5.-Cylindrical Cavity in a Dielectric

density -div P, and (iii) a second surface distribution over the walls of the cavity of density Pn. If the cavity is very small, the con­ tribution to the field intensity at a point within it of the distributions (i) and (ii) will be the same as the field at this point in the Poisson distribution in the absence of a cavity, i.e., the field E as defined in Section 16.2. The distribution (iii) will, however, introduce a further component of field intensity, and it is this component which depends upon the shape of the cavity. In particular, suppose that the cavity is cylindrical with its axis in the direction of the polarization P (Fig. 16.5) . The equivalent charge distribution over the walls of the cavity comprises a layer of density P over one circular end and a layer of density - P over the other end. P being everywhere tangential to the curved surface of the cylinder, Pn = 0 over this surface and there is no charge there. Let 0 be the

16]

POLAR I ZED M E D I A

457

centre of the cavity and let a be the angle subtended at 0 by a radius of one of the circular ends. By equation (14.22) , the intensity at 0 due to the charge on either circular end is 2rrP(1 - cos a) in the direction A O. Thus the total intensity at 0 due to the two layers of charge is 4rrP(1 - cos a) . This is in the direction of P. It follows that the total intensity at 0 is E + 4rrP(1 - cos o:) . . (16.33) It will be observed that this intensity depends upon o: , and hence upon the shape of the cavity. There are two particular cases of especial significance. First, suppose o: is zero, so that the cavity is " needle-shaped ". Then the expression (16.33) reduces to E. Thus E is the field in a needle-shaped cavity, and we are now provided with an alternative definition of E. Secondly, suppose o: = !rr, so that the cavity is " disc-shaped " . The expression (16.33) now becomes E + 4rrP = D. Thus D is the field in a disc-shaped cavity. This is sometimes taken to be the definition of the displacement vector.

Exercise

By neglecting the Poisson distribution over the curved walls of any disc-shaped cavity, show that the component of the field in the direction of the cavity's axis is equal to the component of the displacement in this direction.

16.5. Energy of an Electrostatic Field

Suppose that a charge distribution of density p exists in a region r of a dielectric or vacuous medium and gives rise to a potential V. By equation (14.94) (the argument leading to this equation is valid when dielectrics are present) , the PE of the distribution is given by

W = ! [ pV dv.

(16.34)

W = 8� [ V div D dv.

(16.35)

Outside r, p = 0, and hence the volume integral in this equation may be evaluated over any region enclosing r. We shall choose to integrate over the interior I of a large sphere 5, whose centre lies in r and which has a radius R. Then, since div D = 4rrp,

But it follows from the identity (16.4) that

V div D = div VD - D · grad V, = div VD + D . E, 8 equation (16. ) having also been employed. Thus

W = 8� [

div VD dv +

;rr [ D

. E dv.

(16.36) (16.37)

458

[cH.

A COURSE I N APPLIED MATHEMATICS

The first integral of the right-hand member of this equation may be transformed into a surface integral by application of Green's Theorem in the form of equation (15.22) . As a result we obtain

W = 8� l VDndS + 8� 1

D·E

dv. .

(16.38)

At a great distance R from the charge distribution its dimensions become negligible by comparison with R, so that it will behave like a single charge Q, where Q= Thus, upon S, we have

[ p dv.

( 16.39)

V = QjR, Dn = QjR 2

approximately.

(16.40)

Hence (16.41)

If we now suppose that R -+ oo , this surface integral tends to zero. In the limit therefore, equation (16.38) yields

W = 8� j D · E dv,

(16.42)

the region of integration extending over the whole of space. This result shows that we may regard the energy of an electrostatic field as being distributed over the whole of space with a density at any point of D E/8rr: In an isotropic medium D = KE and this energy density is KE2j8rr:. Maxwell believed that this interpretation of the result (16.42) corresponded to a physical fact, viz., that the presence of electric charges was responsible for a state of strain in an all-pervasive medium called the aether and that the energy associated with the strain in any neighbourhood had density D E/8rr:. Today, however, equation (16.42) is not accorded any such fundamental significance, but is simply accepted as a mathematical . device by which the energy may be calculated when the field has been determined. •

.

,

·

A sphere, ofradius a, of homogeneous dielectric material of constant K, is placed in a uniform field of electric intensity E. Find the potential at all points of the field. Assuming electric energy to reside in the field, show that the ratio of the amount of energy in the sphere to the amount of energy in the same volume before the sphere was introduced is 9KJ (K + 2) 2• (S.U.)

Example 5.

The potential at all points of the field can be found by putting K1 = 1 and K 2 = K in the equations (16.25) and (1 6.26). The field within the dielectric sphere is found to be uniform and of magnitude 3EJ(K 2). The energy density is accordingly constant and equal to 9KE 2j871'(K + 2) 2 ,

+

l6j

P O :L A R l Z E D M ED I A

459

The tota1 energy within the sphere is therefore 3KE 2a3j2(K + 2)�. Before the sphere is introduced, the •energy density is E2j87T and the total energy within the spherical region is E2a3J6. The ratio of these two total energies is now found to be as stated.

16.6. Fields due to Magnetized Materials

Any piece of magnetized material comprises a vast number of mag­ netic dipoles. The magnetic field it causes may therefore be cal­ culated by a method similar to that by which we found the electric field due to a block of polarized dielectric. If dv is any small element of the material, it will behave like a magnetic dipole of moment I dv. I is termed the intensity of magnetization at the point occupied by the element. If r is the region occupied by the material, we may prove, as in Section 16.2, that the material's contribution to the potential n at any point may be found by replacing it by a volume distribution of magnetic poles of density -div I over r, together with a surface distribution of poles over the boundary 5 of density In, viz., the com­ ponent of I in the direction of the outward normal to 5. Since iso­ lated poles cannot exist, this will be recognized as a mathematical device having no basis in reality. These distributions of magnetic poles are referred to as Poisson's distribution of imaginary magnetic matter. The field intensity H at a point is now defined as the force acting upon unit positive pole and hence H = - grad n

{16.43)

H may now be shown to be the field due to the Poisson distribution and the field in a needle-shaped cavity. The proofs proceed exactly as in the case of a polarized dielectric. The differential expression of Gauss's Theorem at a point within the material is (16.44) div H = -4rt div I, or div B = 0, . (16.45) where B = H + 4rtl. (16.46) The vector B (the counterpart of D) is termed the magnetic induction. Before the above equations are soluble in any particular case, we must be provided with a relationship between the vectors I and H. The magnetization within most materials is governed by the simple law I = KH, .

(16.47 )

where K is called the magnetic susceptibility of the material and is a very small quantity. For the large majority of materials, K is negative, and such materials are said to be diamagnetic. Thus, for water, K = -0 · 77 x I0-6• However, in the case of a small class of materials called paramagnetic materials, K is positive. For example, K = 3 x I0-4 in the case of manganese.

460

A C O U R S E I N A P PL I E D M A T H E M A T I C S

[CH.

The simple law of equation (16.47) breaks down completely if we attempt to apply it to a few materials, having very pronounced mag­ netic properties, called ferromagnetic materials. This class comprises iron, cobalt, nickel and a number of alloys containing these metals. The behaviour of these metals when they are subjected to a varying magnetic field is exceedingly complex, and will be found described in physics textbooks. However, provided that the variations in the field from a mean value are relatively small, I is approximately linearly dependent upon H, and thus (16.48)

I = I0 + KH. .

I0 is the intensity of magnetization which would remain within the material if the field were reduced to zero and the law were valid for this value of H ; it is termed the intensity of permanent magnetization. The law (16.47) may be looked upon as the particular case of equation (16.48) when the material has no permanent magnetization. Combining equations (16.46) and (16.48), we obtain or

B = (1 + 4rrK)H + 4rri0, B = {LH 4rri0,

+

(16.49) where fl. = 1 + 4rrK and is termed the permeability of the material. fl. is less than unity for diamagnetic materials and greater than unity for paramagnetic and ferromagnetic materials. In the latter case, fl. can be very large indeed, values of 20,000 being obtainable. Equations (16.43), (16.45) and (16.49) are sufficient to determine the field when the distribution of permanent magnetism is known. Eliminating B and H, we obtain div ( fL grad n) = 4rr div I0. (16.50) In regions where the permanent magnetization is constant in mag­ nitude and direction or is non-existent this equation reduces to div ( fL grad n) 0, (16.51) and, if fL is constant over such a region, n satisfies Laplace's equation. If r is any region bounded by a surface S, then =

fs

dS = £ div B dv = 0,

En

.

(16.52)

i.e., the flux of induction out of any surface is always zero. This is the form of Gauss's Theorem appropriate to a magnetic field. Lines of induction can be defined in the usual way, and it follows from Gauss's Theorem that these can only be initiated by magnetic poles. How­ ever, since isolated poles cannot exist, lines of induction must either be closed curves or proceed to infinity in both directions. By applying Gauss's Theorem to a small disc at a surface of dis­ continuity between two magnetic media, it may be proved that En is

J? O L A R I Z ED

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MEDIA

461

continuous across such a surface (d. continuity o£ Dn in the absence of a surface charge) . The magnetic potential n is also continuous. B, like D, is the field in a disc-shaped cavity in the magnetized material, the axis of the disc being parallel to the intensity of magnetization. By analogy with the electrostatic case, we conclude that the energy associated with a given magnetic distribution may be regarded as being spread over the whole of space with a density B Hj8rr. The methods which proved successful in solving particular electro­ static problems may be employed in similar circumstances when the fields are magnetic. We conclude this section with a selection of such problems.

·

A magnetic dipole of moment M is placed at a distance d from the infinite plane face of a region occupied by soft iron ofpermeability!-'· If the axis of the dipole is perpendicular to the face, show that it is attracted to the iron by a force 3M2 1 8d4 + 1 .

Example 6.

•

,_. -

!-£

p

o'

Since soft iron can retain no permanent magnetism, the magnetic potential 0 is a harmonic function both inside and outside the iron. In the empty space, let 0 = 0 1 and suppose this function is generated by the dipole M at 0 and an image dipole of moment M' at 0', where 0' is the mirror image of 0 in the plane face. Then, if the axes of the dipoles are parallel, M cos 8 + M' cos e: .

O1 =

y2

r'2

Within the iron let 0 = 02 • If this function can be generated by an image dipole of moment M" at 0 and in the same direction as M, then M" cos e

n. =

--r-.- ·

462

A C O U R S E IN A P P L I E D MAT H E M A T I C S

r r'

Over the plane face !1 is continuous. Since = and e = this face, this boundary condition leads to the equation

rr

[CH.

- 6' over

(i) M - M' = M". Equations ( 14 . 69) give the radial and transverse components of the intensity due to a dipole. Resolving these further in the direction of the axis of the dipole, we obtain the component in a direction perpendicular to the face of the soft iron contributed by the dipole M, viz.,

M

r• (3 cos2 e -

1).

I t now follows that o n the vacuum side of the plane face,

B,. = =

M yo

(3 cos2 e - 1) +

r'• (3 cos• 0' - 1)

M'

M + M'

--r-3- (3 cos2 0 - 1).

Similarly, o n the opposite side o f this face,

M" B,. = � (3 cos2 0 - 1).

r

But B,. is continuous across this face and hence

M + M' = I-'M".

(ii)

From equations (i) and (ii); we now deduce that

M' =

� � � M.

The force acting upon M can be calculated as the force of attraction between M and M'. This may be found from the first of equations ( 14.88) by setting e = 0' 0. The result is that there is an attraction of 6MM' 3M2 1-' - 1 = · �-' + 1' =

8d' Example 7. An infinite mass of iron bounded by two parallel planes is uniformly magnetized to intensity I in a direction perpendicular to these planes. In it there is a right circular, cylindrical cavity with generators parallel to the direction of I and equal in length to the diameter of an end. Assuming that the cavity does not alter the uniformity of the intensity of magnetization, show that the magnetic force at the centre of the cavity is 2y2rri in the direction opposite to that of I . (S. U.) (2d) •

--• 1

EJ 0

1

-1

Since div I = 0, the Poisson distribution comprises two layers of magnetic poles of densities over the infinite planes, and I two layers of the same densities over the circular ends of the cavity. The field at 0 due to each infinite plane charge is obtained by putting y = 1, a = and et. = trr in the expression (14.22) for a field due to a disc. The result is 2Trl. Each plane contributes a field of this magnitude at 0 in a direction opposite to that of I. The remaining surface distributions over the two discs each contribute a field

±I

I

POLARI ZED MEDIA

l 6J

in the same direction as I .

463

The net field at 0 is accordingly

4rrl - 4rrl(l

-

cos

in the direction opposed to I.

45°) = 2yl2rrl

A spherical shell of magnetic material, permeability f', is bounded by concentric spherical surfaces, the external and internal radii being a and b 1·espectively. A small magnet of moment M is placed at the centre C of the spherical cavity. Show that the external field is the same as that which would be produced by a small magnet of moment 9MP-[ (21-' + l)(f' + 2) - 2(1-' - 1)2(bfa) 3J-l at C, in the absence of the shell. (S.U.)

Example 8.

Let 01, 0 2 , 0 3 be the potentials in the three regions shown in the diagram. Taking spherical polar coordinates we assume that

3

p

(r, 6), M cos e 111 = -�2- + A r cos 6, r B cos e 11 2 = -- + Cr cos 6, r2 n. D cos a = r• . A r cos 6 represents the distortion

introduced by the presence of the shell into the field in the neighbour­ hood of The potential being continuous across the boundaries of the shell, A, D must satisfy the conditions

M.

B, C,

M B b2 + A b = j}2 + Cb, B(i2 + Ca = D(ili'

B., across these boundaries leads to the conditions ( - b3 2B + C) , 2M - [)3 + A 2Jl. ( �-' - �13_ + c) = a3 a3

The continuity of

= I-'

Solving in these four conditions for D, we find that

D [(21-' + 1 ) (!-' + 2) = But the potential function 0 can be generated by a dipole of moment D at the point C. We have therefore obtained the result required. -

3

2(1-' - l)2(bfa)3] 9Mf'.

16.7. Uniformly Magnetized Bodies

In this section we will consider permanent magnets for which the intensity of magnetization I is constant in magnitude and direction throughout the body. Thus, consider a straight bar of uniform, small cross-section IX,

. A C O U R S E I N A P P L I E D MAT H EMA'TI C S

[cR.

uniformly magnetized so that I is in the direction of the axis of the bar (Fig. 16.6). Since div I = 0, Poisson's equivalent distribution com­ prises surface distributions of magnetic poles over the two ends of the

FIG. 1 6.6.-Field Due to a Bar Magnet

magnet, having densities ± I. Since rx is small, each distribution is approximately equivalent to a single pole. The bar magnet accordingly behaves as though it possessed two poles of strengths ± J at each end. The lines of force are now easily sketched in as shown in the figure. It will be observed that, within the mag­ net, the field H is in the direction opposite to that of I. The couples exerted on the element­ ary dipoles of which the magnet is composed are therefore in a sense tending to demagnetize the material. We accordingly refer to the field within the mag­ net as a demagnetizing field. The lines of magnetic induction coincide with the lines of force FIG. 16.7.-Field Due to a Uniformly outside the magnet, but, since Magnetized Sphere they must form closed loops, they run in a contrary direction to the lines of force inside the magnet. Now consider a uniformly magnetized sphere of radius a (Fig. 16.7).

P O LA R l Z E :b M E D I A

I 6J

Outside the sphere, the potential shall assume that

n1

must vanish at infinity, and we

(16.53) Inside the sphere, the potential

n2

n2 =

Over the spherical surface,

must be finite at 0, so we shall try

Br cos e .

n1 = 02

.

(16.54)

and hence

Ba3• B can be found from equation (16.46).

(16.55)

A =

Since En is continuous across

the surface of the sphere,

- B + 4rd. From equations (16.55) and (16.56) we now find that A = irra3I, B = !rrl. 2A 3 =

a

(16.56)

(16.57) Thus, the field inside the sphere is uniform, of magnitude !rri and

in the opposite direction to I,

H

i.e.,

=

-!rrl.

(16.58)

This is the demagnetizing field. The induction B within the sphere is given by B = H + 4rrl = !rrl. . (16.59) The lines of magnetic induction are in the same direction as the mag­ netization, and are shown in Fig. 16.7 as closed loops. The field outside the sphere is identical with that of a dipole of moment !rra3I placed at 0 with its axis in the direction of I. Thus, the radial and transverse components of the field outside the sphere can be found from equations (14.69) to be

Hr _

8rra3I cos e 3r3

'

Ho -

_

4rra3I sin e. 3ra

At a point on the spherical surface, these reduce to

Hr = !rri cos e,

H0

=

!rrl sin e.

(16.60)

Hence, if a is the angle made by the intensity with the tangential plane at such a point, tan a = Hr/Ho = 2 cot e. (16.61)

Now, to a first approximation, ti;e Earth behaves magnetically like a uniformly magnetized sphere. Thus, if A. is the magnetic latitude

A C O U R S E i N A P P L I E D MATHEMAT I C S

[CH.

of a point on the Earth's surface, A = t7t - e and equation (16.61) may be written (16.66) tan a = 2 tan A •

•

a is the angle made by the magnetic field with the horizontal at the point in question and is termed the angle of dip.

EXERCl:SE

16

spherical conductor o f radius a i s coated with a uniform thickness d of shellac (constant K) . Show that the capacity of the conductor is increased in the ratio (Ka + Kd) : (Ka + d) . (Li.U.) 2. If the surface of separation between two dielectric media of specific inductive capacities K1 and K2 carries a charge cr per unit area, and the intensity just inside the first medium is E1 at an angle 0( 1 to the common normal, derive equations which will determine the appropriate magnitudes E2 and 0(2 for the second medium. In particular, show that l. A

K2 cot 0(2

=

K1 cot 0( 1

( 1 + K-E1 cos0( 1 1

47tcr

)·

(Li.U.)

3. A point charge e is situated at the centre of a spherical cavity of radius a in a dielectric medium of constant K filling all space. Find the electric field at all points and also any surface or volume density of charge due to polarization. (Li.U.) 4. A spherical conductor of radius a carries a charge and is surrounded by a spherical shell of dielectric of constant K whose inner radius is a. Find the radius of the outer boundary of the shell so that half of the electrostatic energy may reside in the part of the field occupied by the (Li. U.) dielectric. 5. Three concentric, hollow, conducting spheres A , B and C have radii a, b and c (a < b < c) . The spaces between A and B, between B and C and outside C are filled respectively with uniform dielectrics of con­ stants Kv K2 and K1 . A is earthed, B is uncharged and C carries a charge e. Find the charge on A . (M.T.) 6. A condenser consists of two concentric metal spheres of radii a, b(a < b) separated by a dielectric of constant K. Find the electrostatic energy when the inner and outer spheres have charges e1, e2 • If the outer sphere is connected by a fine wire with the outer surface of an equal, uncharged condenser whose inner sphere is kept at zero potential, the condensers being so far apart that their mutual induction is negligible, prove that the resulting loss of electrostatic energy is (e1 + e2) 2 (b + Ka - a)f2b(2b + Ka - 2a) . (M.T.) 7. The plates of each of two condensers of equal capacity are thin, con­ centric, conducting spheres of radii a and 2a. In one condenser half the region between the spheres is filled with homogeneous dielectric, the boundary being part of a diametral plane. Between the plates of the other condenser there is a spherical shell of the same dielectric

1 6]

8.

9.

P O L A R I Z ED M E D I A

467

fitting closely on to the inner plate. Show that the outer radius of this shell is (2K + 2)af(K + 2), where K is the dielectric constant. (M.T.) A very long, earthed cylinder of radius a, surrounded by a uniform thickness b - a of dielectric of constant K, is placed with its axis perpendicular to a uniform electric field. Find the electrostatic potential distribution in the dielectric and in the space outside. (Li.D.) A conducting sphere of radius a is surrounded by an insulating shell of dielectric constant K. The shell is concentric with the sphere and has internal and external radii a and b respectively. The system is situated in a uniform electric field of intensity F. Show that the potential at points outside the sphere of radius b is e b3F cos 8[a3 (2K + I) + b3 (K - I)] r - Fr cos e + r2 [2aa (K - I) + b3 (K + 2) ] e where is the total charge on the sphere and r, 8 are coordinates with the centre of the sphere as origin and the direction of the field as axis. (S.D.) A circular right cylinder of radius a and length 2a is uniformly mag­ netized with intensity I parallel to the axis. Determine B and H at any point on the axis, inside and outside the magnetic material. (S.D.) A spherical shell of magnetic material of constant permeability fl. is bounded by concentric spheres of radii a, b(a < b) . At the centre of these spheres is situated a small magnet of moment M, and the magnetic field at an infinite distance from the shell is of magnitude F and has the same direction as that of the moment of the small magnet. Show that the magnetic potential outside the shell is ( - Fr + N/ r2) cos 8, where N{b3 ( fl. + 2) (2f1. + 1) - 2a3 (fl. - 1) 2} b3{ 9Mf1. + (b3 - a3) (2f1. + 1) (fl. - 1)F} and r, 8 are radius and polar angle with respect to the position of M as origin and its direction as polar axis. (Li.D.) A small magnet of moment M is placed at a point at distance h from the plane face of a semi-infinite mass of soft iron of permeability fl.; the axis o f the magnet makes an angle 8 with the normal t o the plane face. Show that the moment of the couple acting on the magnet has magnitude (fl. - I)M2 sin 28 I 6(f1. + I)h3 A circular cylinder of soft iron of permeability fl. and infinite length is introduced into a uniform magnetic field of strength H so that its axis is perpendicular to the direction of the field. Show that the intensity of magnetization induced in the iron is constant and is given by [L - I H . I = fl. + 1 21t '

10.

I I.

=

12.

13.

468

A C O U R S E I N A P P L I E D M A T H E MA T I C S

[CH.

14. An infinitely long, circular cylinder of hard magnetic material is mag­ netized so that the intensity of magnetization I is uniform and per­ pendicular to the axis. Show that a demagnetizing field of strength 2rr:I exists in the material. 15. A magnetic field of intensity H exists in a medium of permeability f.l. · A spherical cavity of radius a is cut in the material and a small magnet of moment M placed at its centre, the axis of the magnet making an angle ex: with the field H. Show that the field in the cavity comprises two components : (i) the field due to the magnet, and (ii) a uniform field making an angle � with the axis of the magnet, where 3[J.Ha3 cot � = cot + . ex: 2 ( [J. - 1) M sm . cx: (Hint : Resolve M along and perpendicular to H and superpose the solutions to two problems.) 16. A very long, cylindrical rod of radius a and specific inductive capacity K is placed in a uniform electric field £0 with its axis at right angles to the lines of force. Show that the field intensity inside the rod is uniform and of magnitude 2E0((K + 1), and derive an expression for the potential of the field outside the rod. (Li.U.) 17. A piece of magnetic material is uniformly magnetized to intensity I. Express the magnetic potential due to the material as a volume integral and show that this is the same integral which would be ob­ tained for the component of the electric intensity in the direction of the magnetization due to charge of density I occupying the same region as the material. Hence calculate the internal and external fields due to a uniformly magnetized sphere. 18. Employing the first result of the previous exercise, solve Ex. 14. 19. An infinite line charge, e per unit length, is placed parallel to the plane face of a semi-infinite slab of dielectric of constant K. Show that it is attracted to the slab with a force K - I e2 K+ per unit length, a being the distance of the charge from the plane face. 20. Two particles with charges e and e' are placed in dielectrics having constants K and K' respectively. The surface of separation is plane and bisects the line joining the particles at right angles. If d is the distance between the particles, show that the force acting upon e is e {2e' + e ( I - K'(K)} (K + K')d2 2 1 . A conducting sphere of radius a is surrounded by a material of dielec­ tric constant K whose outer boundary is a concentric sphere of radius 2a. If the system is placed in a uniform field E, show that equal amounts of positive and negative electricity are separated on the surface of the sphere, that of positive sign being 9Ea2K 5K + 7"

� · a-

16]

POLARI ZED MEDIA

469

22. A small, spherical cavity is cut in a permanent magnet. Show that the magnetic field within it is H + where H is the magnetic field and I the intensity of magnetization in the magnet. A soft iron sphere with a small, spherical cavity at the centre is placed in a uniform magnetic field. Show that the field within the iron approximates to uniformity when the radius of the cavity tends to zero, and find the limit of the ratio of the fields in the cavity and the iron. (M.T.) 23. A sphere with centre at the origin and radius a is magnetized so that the components of intensity of magnetization are A(x, y, z + a) . A small magnet of moment fl. is held at a point on the axis of x at a distance c(c > a) from the centre of the sphere and with its axis parallel to the axis of z. Find the mechanical interaction between sphere and magnet. (M.T.) 24. A sphere, of uniform intensity of magnetization I, has a concentric, spherical hollow. Find the magnetic force in the hollow. If the hollow were cubical and concentric, with two faces perpendicular to the direction of magnetization, find what would be the value of the magnetic force at the centre of the sphere. (M.T.)

!n:I,

CHAPTER 17

STEADY ELECTRIC CURRENTS 17.1. Currents in Wires.

Kirchhoff's Laws

Employing a battery or dynamo, it is possible to maintain two different points in a conducting medium at different potentials. In such circumstances a static distribution of charge is impossible, for an electric field will be present within the conductor and the free electrons will move under its influence. Such a motion of charge within a conductor constitutes an electric current. If the conductor is a fine wire the flow of charge is one-dimensional. This is the simplest case of the type of problem we shall be considering in this chapter and will be disposed of immediately. Suppose that the two ends of the wire A and B are maintained at potentials VA and VB, where VA > VB · As we proceed along the wire, the potential falls steadily from the value VA to the value VB, and, if s is distance measured along the wire from A, fJVjfJs will accordingly be negative. But - fJVjfJs is the component of the electric field along the wire in the sense of s increasing. We conclude that at all points of the wire there is an electric field tending to cause a flow of positive charge in the direction A -+ B. Since, however, it is the electrons within the atoms of the conductor which are free to move and these possess negative charges, the actual flow of charge which takes place is from B to A . The strength of the electric current at any point of the wire at any instant is measured by the rate at which charge is being transported across the section of the wire through the point. Thus, if a charge de crosses the section in time dt, the current is i = dejdt in the sense in which the charge is moving.* In the case of a metallic conductor, de is negative, and hence the current in the direction of electron flow is negative. We conclude that, in such a case, the current in the direc­ tion opposite to that of the electron flow is a positive quantity. Thus, when no negative sign is given to the measure of the current, its sense must be taken to be in the direction leading from points at high potential to points at a lower potential. If the potentials VA and VB are maintained constant, as, for example, when A and B are connected to the two electrodes of a battery, the flow of current along the wire will be steady, i.e., the current at any point on the wire will not vary with the time. The charge density at any point of the wire will also be constant and hence, if IX, � are two * If e is in coulombs, i is measured in amperes, 470

Thus 1 amp.

=3

x

109 e.s.u.

[cH. 1 7]

STEADY ELECTRIC CURRENTS

471

cross-sections of the wire the quantity of charge between ct and � will be invariant with respect to t. This implies that, over any period of time, equal quantities of charge will cross ct and � i.e., the currents at ct ' and � will be equal. Thus, under steady conditions, the current strength is the same at all points of a wire. It is found by experiment that, when the current flow in a wire is steady, the current strength i is proportional to the potential difference between the ends of the wire, viz., VA - Vn (Ohm' s Law). We may therefore write (17.1) where R is a constant of the wire called its resistance. If the potential is measured in volts and the current in amps., R is in ohms.

E c

D

�

(b)

-�R:..-�

p .... .

e

-

l

j

,.._ �2---· Q

---l �-__...

FIG. 17.1.-Electrical Network

W

Consider now an electrical network formed from wires meeting at /unctions such as A, B, C, . . . (Fig. 17.1 (a)). Some of the arms of the network (e.g., AB) will contain batteries responsible for the currents flowing through different parts of the network. Assuming steady conditions, the rate at which positive charge is arriving at the junction A (in reality, the rate at which negative charge is leaving this point) is given by the sum of the currents leading into A, viz., (i1 + i2 + i3 + i4) (Fig. 17. 1 (b)). Since, under steady conditions, charge does not accumulate at A, (17.2)

This is a particular expression of Kirchhoff's First Law, which states that the algebraic sum of the currents feeding into a /unction of a network

is zero.

Let PQ (Fig. 17.1

(c)) be any arm of the network containing a battery

472

A C O U R S E I N A P P L I ED M A T H E M AT I C S

[CH.

which maintains a steady potential difference of C between its plates. C is termed the electromotive force (e.m.f.) of the battery. Let i be the current flowing into the battery terminal at the lower potential. Then, if R1 is the resistance of that part of the arm along which the current flows into the battery and R2 is the resistance of that part of the arm along which the current flows awp.y from the battery, the net drop in potential VPQ as we proceed from P to Q is given by VPQ =

Rli - c + R2i = Roi - c;

(17.3)

where R0 is the total resistance of the arm. The potential difference between the terminals of any actual battery is found to fall below C as the current taken from it increases. This decrease is directly proportional to the current flowing, so that the p.d. is always given by p.d. = C - ri, (17.4) where r is a constant of the battery called its internal resistance. The e.m.f. C is observed to be the p.d. between the plates on open circuit (i = 0) . Replacing C in equation (17.3) by the p.d. given in equation (17.4), we obtain

VpQ = R0i - C + ri = Ri - C,

(17.5)

where R is the total resistance of the arm including the internal resistance of the battery. If now ABCDA is a closed circuit in the network, the net potential drop around it must be zero, i.e., VAB

+ VBc + Vcn + VnA = 0.

(17.6)

But each term of the left-hand member of this equation is given by an equation of the type (17.5). Thus

2,(Ri - C) = 0, .

(17.7)

where the summation extends over all the arms of the closed circuit. Either the clockwise or anti-clockwise sense of description of the circuit must be designated as the positive sense, and then all currents in this sense will take positive signs, whereas currents in the opposite sense will have negative signs. C will have a positive sign if we pass from the low-potential plate to the high-potential plate as we proceed across the battery in the positive sense. Equation (17. 7) can also be written in the form 2,Ri = 2,C, . (17.8) i.e., the algebraic sum of the resistance drops around a closed circuit in a network is equal to the algebraic sum of the e.m.f.s in the circuit. This is Kirchhoff's Second Law.

STEADY ELECTRIC CURRENTS

17]

47 3

A network consists of a hexagon of equal wires, each of resistance r, together with wires each of resistance 3r joining two pairs of opposite corners. If a current C is sent in and taken out at the two remaining corners, find the current in each of the wires, and show that the equivalent resistance of the network is llrf8. (N.U.) Taking account of the symmetry and Kirchhoff's first law, the currents in the various arms of the network may be expressed in terms of C and i as shown. Applying Kirchhoff's second law to the circuit ABEFA, we obtain the equation irC + 3ri - r(!C - i) - irC = 0, i = iC. or

Example 1.

The currents in the various branches of the network are now determined.

c

A

The p.d. between and CD, viz.,

D

C

A and C is equal to the sum of the drops over AB, BC trC + r(iC - i) + trC = :!.81.rC.

It follows that the whole network behaves as a single resistor of resistance

llrf8. A telegraph wire has n faults F; each with resistance R1; the resistance of the wire between F;_1 and F1 is r;. After fault F,. the wire is earthed, the resistance of the wire between Fn and the earth being rn+ 1• One terminal of a battery is attached to the other end of the wire, and the second terminal is earthed. Find a difference equation for the potential at a fault: and in the case when aU R; are equal to R and aU r; are equal to r, show that the potential at F,, is Vs = A cosh sc;; + B sinh sc;;, where A = -B tanh (n + l)c;;, B is a constant, c;; = cosh-1 ( 1 + rf2R) . (Li.U.) Let C; be the current in the wire between the faults F;_ 1 and F;. Then, applying Ohm's Law to the wire F8F8+, we obtain the equation

Example 2.

Vs - vs+ I = r,+ , cs+ l' . (i) The current flowing to earth from the fault Fs is clearly C8 - Cs+ t· Applying Ohm's Law to this leak to earth, we find that V, = R,.(C_. - C-<+ 1). (ii)

A COURSE IN APPLIED MATHEMATICS

474

[CH.

From equation (i),

I

�

I

Cs + 1 = - ( Vs - V8 u) · r�1 Substituting in equation (ii) and rearranging, we obtain a difference equation for V., viz., Cs = - ( V8_ 1 - Vs),

F,

fn Cs-Cstt

I If rs

= r, Rs = R for all s , this equation simplifies to

V8 + 1 - 2 ( 1 + rj2R) Vs + V8 _ 1 = vs+ 1 - 2 cosh IX Vs + VS - 1 = or where IX = cosh-1 (l + rj2R). Let V8 = x' be a solution of this difference equation. Then x2 - 2x cosh IX + 1 = (x - eiX) (x - e-IX) = or Thus x = eiX or e-IX and the general solution for V8 is Vs = PesiX + Qe-siX = A cosh SIX + B sinh SIX,

0, 0,

0, 0.

where A = P + Q, B = P - Q and are arbitrary constants. But the point Fn+ 1 is at earth potential. Hence V8 = when s = n + 1 .

I t follows that or

0 A cosh (n + 1)1X + B sinh (n + l)IX = 0, A = -B tanh (n + l)IX.

This completes the solution to the problem.

17 2 Electrical Generation of Heat .

.

Consider a wire PQ of resistance R whose ends are maintained at constant potentials VP and V Q· Let i be the steady current flowing in the wire. The motion of the free electrons which participate in the flow of charge along the wire comprises two components : (a) a purely random motion which persists even in the absence of a p.d. between the ends of the wire, and (b) a drift in the direction Q --* P. The component (b) is very small (about 1 cmjsec) compared with component (a), but it is this component which constitutes the physical basis of the electric current. The currents associated with the random motion have an overall mean value of zero in any direction, and hence may be disregarded when we are taking the macroscopic point of view. The rate of drift of the charge entering the wire at P is very little different from its rate of drift as it leaves the wire at Q. It follows that the KE

STEADY ELECTRIC CURRENTS

17]

475

associated with this drift is very nearly unaltered as we pass from P to Q. Consider the charge in motion between P and Q at time t (Fig. 17.2 (a)) . After a further time ?lt has elapsed, the distribution of this charge will be as represented by Fig. 17.2 (b). The change in the distribution may be represented as the removal of a charge i?lt from the neighbourhood of P to the neighbourhood of Q. This involves a loss in PE of i?lt(VP - VQ) . As has been remarked, this loss of PE is not compensated by a corresponding increase in KE, but appears as heat in the wire. The physical mechanism by which this heat is generated P.-----, Q I I

(a)

(b)

:I

""

iH

I

•l

FrG. 17 .2.-Loss of PE in Current Flow

is as follows : Under the influence of the electric field charge is con­ tinually being accelerated, but any KE gained is almost immediately lost by impact between the moving charge and the fixed atomic nuclei. Under the influence of these impacts, the energy of vibration of the nuclei is augmented and an increase in temperature of the wire becomes apparent. The rate of loss of PE is seen to be i(Vp - VQ) = i2R (by equation (17. 1)). This, then, is the rate at which heat energy is generated in the wire. Denoting the rate of heat production by h, we have (17.9)

Four points A, B, C, D are connected by five wires: DA, BC, each of resistance R,· DC, AB each of resistance 2R; and DB of resistance !R. The points A, C are connected by wires, oftotal resistance !R, to a battery ofe.m.f. E. Show that the current in AB is Ef5R and that the rate at which energy is expended by the battery is 8£2f15R. (D.U.) Let x, y, z be the currents in AB, AD and ED respectively. Then, by

Example 3.

application of Kirchhoff's first law, we find that the currents in the remaining branches are as indicated in the diagram. Applying Kirchhoff's second law to the circuits we obtain the equations = kRz R (x - z) - 2R ( z) - !Rz !R(x + = E . 2R + - z)

2Rx + y+ x R(x +

ABDA, BCDB, ABCEA, Ry 0, 0, y) =

A C O U R S E I N A P P L I E D MA T H E M A T I C S

476

Solving for x, y,

X

z,

we find that

= Ef5R,

Thus, the current in

= Ef3R,

z

[CH.

= -2Efl5R.

A B has been shown to be Ef5R. y

8

X

R

2R

z

A

y

x-z

Y, R

c

J.R

R

y+

Z

D l, R

E

The total current taken from the battery is x + y = 8Efl5R. The potential drop across the network connected to the battery terminals is E. The rate at which heat is being generated in this network is therefore 8£2f15R, and this must be the rate at which energy is being supplied by the battery.

1'i 3 Equation of Continuity .

.

In this section we will commence the study of electric currents flowing in three-dimensional media. Let dS be an element of area at a point within the conducting medium, oriented in such a manner that the direction of the drift of charge at the point is normal to dS. At any time t, let the rate at which charge is crossing dS be di. Then dijdS is the rate of charge flow per unit area at the point. The vector of magnitude dijdS and in the direction of the charge drift is denoted by j and is called the current density at the point. If the charge in motion is of negative polarity (as in the case of metallic media), dijdS will be negative and j must be taken to have magnitude - dijdS and direction opposite to that of the drift. The pattern of the flow within a conductor is completely specified when j is given at all points at each instant t. If the flow is steady, j will be independent of t. If, at any instant, curves are drawn within the conductor having the property that the tangent at any point is in the direction of the current

17]

STEADY ELECTRIC CU RRENTS

477

density at the point, such curves are called lines of flow. All the lines of flow intersecting a closed contour within the conductor generate a tube offlow. Now let d5 be an element of area to which j is not normal. Let d5' (Fig. 17.3) be the orthogonal projection of d5 upon a neighbouring plane which is perpendicular to the direction of flow. Provided d5 and d5' are very close together, there will be a negligible rate of accumulation of charge inside the cylinder of which d5, d5' are the J

ds'

ds

'

\

j Frc. 1 7 .3.-Flow Across an Oblique Element

plane ends. Hence the rate of flow across d5 will equal the rate of flow across d5', viz., j d5'. But d5' = d5 cos e, where e is the angle between the planes of d5 and d5 ' . Thus, the rate of flow across d5 is j d5 cos e = j cos e . d5 = jnd5,

(17.10)

jn being the component of j in the direction of the normal to d5. We can now obtain an expression for the rate of flow i across any surface 5 (not necessarily closed) by integration of the rates of flow across its elements.

The result is

i = l jnd5,

(17.1 1)

i.e., i is the flux of current density across 5. The sense in which the normal to 5 is to be taken will depend upon the direction across 5 in which we wish to compute the total rate of flow. At points within a conducting medium which are connected to batteries or dynamos, charge enters or leaves the medium. Such a point is called an electrode. An electrode must always be of finite dimensions, and we shall suppose it bounded by a surface 5. The rate of flow across 5 at any instant is given by equation (17. 1 1) and is called the strength of the electrode. Now suppose that 5 denotes a · closed surface within the medium

478

A C O U R S E IN A P P L I E D M A T H E M A T I C S

[CH.

enclosing a region r devoid of electrodes. Then, at any instant, the rate of flow out of S must equal the rate at which the total charge within S is decreasing. Let p be the charge density at any point of r and let dv be a volume element at this point. The rate at which the charge within dv is decreasing is -o(p dv)jot = - dvopjot. It follows that the rate of decrease of the total charge within r is

- 1 �at dv. r

The rate of flow of charge from the region is equal to

£ j,dS £ div j dv, =

having employed Green's Theorem. Thus

£ div j dv - jr �� dv, [ ( �� + div j ) dv 0 . . =

or

=

(17. 12)

Equation (17.12) is true for all regions r of the type being considered and this can be the case only if the integrand is everywhere zero, i.e., if

op . · (17.13) at + dlv J = o. . This is known as the equation of continuity. If the flow is steady, p is independent of the time, and the equation of continuity reduces to div j = 0.

(17.14)

1'1.4. Differential Form of Ohm's Law

Suppose that a steady electric flow is taking place in a homogeneous isotropic conducting medium. Consider the portion of a tube of flow of infinitesimal cross-section between two right sections dS, dS' (Fig. 17.4). Let this portion be of small length ds and let V, V + dV be the potentials at its ends dS, dS' respectively, the direction of flow being from dS to dS'. If j is the current density at dS, the current flowing in the tube is of strength j dS. It is reasonable to assume that the current flo� within a tube is governed by the same law as that within a wire, i.e., Ohm's Law and hence that i.e.,

V - (V + dV) = Rj dS, - dV = Rj dS,

(17. 15)

where R is the resistance of that portion of the tube being considered. It is found by experiment that the resistance of a uniform wire is

STEADY ELECTRIC CURRENTS

17]

479

directly proportional to its length and inversely proportional to its cross-section. Thus, we shall take

R = " dsfdS, . (17.16) where the constant " is termed the specific resistance of the medium. Substituting in equation (17.15), we obtain

- dV = "J ds, oVfos = "i·

or

(17.17)

-

But - avIOS is the component of the electric field in the direction of the current flow at dS. Now, as has already been stated, the rate of drift of charge in a conducting medium is always quite small, the �------ ds

--------�

ds'

V+dV

FIG. 17.4.-Portion of a Tube of Flow

momentum gained by the units of charge being continually destroyed by impact with the atoms of the medium. It follows that the normal component of the acceleration v2fp of a charged particle as it drifts under the influence of the field is negligible, and hence that the force acting upon the particle is in the direction of its motion. This implies that the field intensity E and the current density j are in the same direction. Thus, -oVfos = E and equation (17.17) may be written in the vector form

E = 't"j,

or

j

=

(17.18)

crE,

where cr = 1 /" and is called the specific conductivity of the medium. This is the differential form of Ohm's Law. To summarize, the equations governing the steady flow of electric charge through a homogeneous isotropic conducting medium are

. �=

-

grad

0, j = crE.

d1v J

=

V

'}

.

(17.19)

If the medium is not homogeneous or is not isotropic, the last of equations (17. 19) must be replaced by a more complex equation. F

480

A C O U R S E I N APPLI E D MAT H E M A T I C S

}

[CH.

The equations (17. 19) should be compared with the equations of an electric field in an uncharged dielectric, viz., E = -grad div D = 0, D = KE.

V,

(17.20)

It will be observed that by replacing D by j and K by a, the electro­ static field equations are converted into the current :flow equations. We are accordingly led to expect that the solutions to many problems of electrostatics may be interpreted as solutions to problems of current :flow. This is indeed the case. Thus the solution to the problem of a uniform dielectric sphere introduced into an infinite dielectric medium in which there is a uniform field (p. 451) has its counterpart in the problem of a uniform conducting sphere introduced into an infinite conducting medium through which there :flowed originally a uniform current of density j. The reader should show that, if a1 is the con­ ductivity of the infinite medium and a2 that of the sphere, then the current in the latter is uniform and of density 3a2jf(2cr1 + a2). Suppose that a steady :flow in a conducting medium i s fed from electrodes of strengths ii> i2 , in. Let 5 be any closed surface constructed in the medium. Then, since there is no accumulation of charge within 5, the rate of :flow across 5 must equal the sum of the strengths of the electrodes within 5. Thus, the flux of j out of a closed •

•

•

surface 5 is equal to the sum of the strengths of the electrodes inside 5.

This is Gauss's Theorem for steady :flows. Consider a small spherical electrode of strength i within an infinite medium of conductivity a. Let 5 be a sphere of radius r concentric with the electrode. There being spherical symmetry, the :flow will be everywhere normal to 5 and of constant density j over 5. By Gauss's Theorem 4rrr2j = i.

Hence j = i/4rrr2 . It follows that E is radial and of magnitude if4rrcrr2. We accordingly have the equations . � -�- · · - 4rrar = � 2' E = (17.21) J 4rrr 4rrar-2' V -

Now consider a small electrode in the form of a circular disc set in an infinite plane plate of conductivity a and thickness t. Taking 5 to be the curved surface of a concentric circular disc of radius r, assuming radial :flow and applying Gauss's Theorem, we find that in this case i 2 E2rratr' V = - 2rt crt log r. (17 .22) A similar analysis, which we shall leave for the reader to perform for himself, shows that the two-dimensional :flow from an infinite

STEADY ELECTRIC CURRENTS

17]

line electrode of strength equations

i

481

per unit length is characterised by the

. = � E = -�- V = - � log r. 2 21tcrr' 21tr' 7tcr

J

.

(17 .23)

Problems of steady current flow can, of course, be solved directly without conversion into the corresponding electrostatic problem. It follows from equations (17.19) that V is a harmonic function, and consequently all the methods which are available for the solution of electrostatic problems (method of images, etc.) can be employed in this new field. The boundary conditions which, by the uniqueness theorem, completely determine the solution, are : (a) At a boundary between a conductor and an insulator jn = 0, since there is no current flow into the latter material. (b) At a boundary between two different conducting media jn is continuous, since as much charge must leave one conductor across any element of the boundary as enters the second conductor across the same element. Across such a boundary, the potential is either continuous or jumps by a fixed amount, depending upon the two materials in contact. Any discontinuity is generally small and will be neglected in our subsequent work. (c) We shall assume that the material of any electrode has infinite conductivity. Thus, in an electrode E = 0 and V = constant. Either the potential of the electrode must be stated or the flux of j across its surface (i.e., the current entering or leaving by the electrode) if a problem is to be uniquely soluble.

The space between two perfectly conducting, co-axial, cylindrical shells ofradii a and b (a < b) is filled with uniform material of specific conductivity u. Prove that the conductance per unit length between the shells is 2Truf1og(bfa) .

Example 4.

(Li.U.) Let the outer conducting shell be earthed and the inner be maintained at potential rf>. Then, just as in the problem of the cylindrical condenser, we prove that the radial field E between the shells is given by "' - · E = p-log

(bfa)

where p is distance measured from the common axis (see equation ( 15.48) ) . Thus the current density i n the radial direction at the inner shell i s given by

J. = a logurp(bfa) "

The total current leaving unit length of the inner cylinder is . . 27rurp 2 � · log Hence, the resistance per unit length of the cylinders is 1 t� log 2Tru and the reciprocal of this quantity is the conductance.

= TraJ = =

(bfa)

(�)a ,

i, where

482

A C O U R S E I N A P P L I E D MATH EMAT I C S

[CH.

Four equal, circular, perfectly conducting electrodes of small radius 3 are placed with their centres at the corners of a square of side a in an infinite sheet of metal of thickness t and uniform conductivity a. One pair of opposite corners are at one potential and the other pair at a different potential; show that the resistance between the pairs is

Example 5.

1 a log 2Mt oy2'

(M.T.)

Let currents i enter at the electrodes A , C and leave by the electrodes B, D. Then, if r1, r 2, r3, r4 are the distances of any point P in the plane of the conductor from the four electrodes A , B, C, D, respectively, by equations (17.22) we deduce that the potential at P is given by

r,

"

r,

If P is on the circumference of the electrode A , r1 = o, r 2 = r4 = a, r3 Thus, if V1 is the potential at the electrode A (or at C), i

a

v l = 2rrut log 3y2' Similarly, if V 2 is the potential at either of the electrodes B or D,

Hence

i ov2 log v 2 = 2rrut a.

vl - v . = rrut log oy2 ' i

and the resistance between A C and BD is

a

1 a - v-. -vl---= 2rrut log -- · 2i oy2

=

y2a

.

S T E A D Y E LE C T R I C C U R R E N T S

1 7]

483

Two equal infinitely long perfectly conducting cylinders of radius c, with their axes parallel at a distance 3c apart, are embedded in a uniform medium of resistivity filling all the rest of space. Show that the equivalent resistance, per unit length parallel to the axes, for current flow from one cylinder to the other, is :: log v'5 + I . (S.U.) v'5 1

Example 6.

-r,

1T

-

Take any right section of the two cylinders and let 0, 0' be the centres of the two circles obtained. Choose points on 00' such that they are inverse with respect to both circles. Then, if = = = 3 - x and hence = Solving for we find that = !(3

Q, Q'

OQ'

c

OQ O'Q' x,

x(3c - x) c•. x + y'5)c.

x

p

Introduce image line electrodes of strengths If P is a point distant the potential at P is given by (equation (1 7.23))

i, -i per unit length at r1, r 2 from Q, Q' respectively, V = - 2iT- log r + 2-i-r log r2 = 2i-r- log rr2l Over the cylindrical electrode with its axis through 0', r /r1 O'Q'fc xfc = !(3 + y'5) (see p. 426) . Hence, over this surface the2 potential due to the image electrodes assumes the constant value V1 = 2i-r" log !(3 + y'5), Q, Q' respectively.

7T

1

TT

7T

-

·

=

=

and one boundary condition has been satisfied. Over the other cylinder the potential takes the constant value 2 = z;;. log 3 Thus both boundary conditions are satisfied and the image system is appropriate to a flow of charge between the cylindrical electrodes. The flux of current density from unit length of the cylindrical electrode with axis through 0' is equal to the flux from unit length of the image electrode at viz., Thus, a total current enters by one cylindrical electrode and leaves by the other. The effective resistance between the electrodes is accordingly T = - log !(3 = - log · �

V2 i-r

Q,

i.

+ v'5.

i

VI - v. T

. -

1T

+ y'5)

1T

y5 + l v'5 - l

-

A COURSE I N APPLIED MATHEMATICS

484

[CH.

17.5. Heat Generated by a Three-Dimensional Flow

Consider again the element of a tube of flow illustrated in Fig. 17.4. The current flowing in this element is f dS, and its resistance is given by equation (17.16). Hence, from equation (17.9), we deduce that the rate of production of heat in the element is

(j d5) 2

x

" � -.pds ds = -.pdv, d =

(17.24)

where dv = dS ds in the volume of the element. This result may be described by saying that the rate of heat production within the medium is h per unit volume, where

" h -.p =}_2 = crE 2 = j =

(J

•

E.

(17.25)

Suppose that current is flowing in a medium bounded by a surface S which is insulated. Let currents i1, i2 , . in enter the medium at perfectly conducting electrodes bounded by surfaces Sv 5 2 , Sn Vn respectively. We and maintained at potentials V1 , V2 , shall denote by r the region between the surface S and the electrodes. The rate of heat generation per unit volume of the medium is •

•

•

•

•

•

•

•

j · grad V = V div j div Vj = - div Vj (17.26) having employed equation (16.4). The total rate of heat production is j ·E

=

-

-

therefore equal to

-1 div Vj dv -1 Vfn dS - _i 1 V}n dS . =

r

s

r = l s,

(17.27)

But }n = 0 over S, since this surface is insulated, and hence the first integral of the right-hand member of the last equation is zero. Also, over 5,, V = V, (a constant) and thus

1 V}n dS Sr

= Vr

1 }n dS Sr

= -

Vrir, .

(17.28)

it being necessary to introduce a negative sign since }n is to be taken in the sense outwards from r whereas ir is the flux of current density across Sr into r. We have now calculated that the total rate of heat production is

n

z v,.i,.

r=l

.

(17 .29)

This result is easily explained by supposing that each electrode is maintained at its potential by a battery, one of whose plates is con­ nected to the electrode and the other to earth. In each unit of time,

1 7]

STEADY ELECTRIC CURRENTS

485

the rth battery will raise a charge ir through a potential gradient of Vr and hence will perform work of amount V,.ir. The total rate at which work is being performed by the batteries is then given by the expression (17.29) . The energy of the system is accordingly being increased at this rate and appears in the form of heat. EXERCISE 1 7 1 . Five resistances AB, A C, BC, BD, CD of 15, 75, 30, 1 50 and 1 2 ohms respectively are connected as indicated by the lettering, and the points A and D are connected to a battery. Find the equivalent resistance of the network. (D.U.)

2. A length of uniform wire of resistance 5r is bent to form the sides of a closed, regular pentagon ABCDE, the centre 0 of which is connected to A , B, C, D and E by equal lengths of straight wire, each of resist­ ance r. Find the equivalent resistance of the network to a current which enters at 0 and leaves at A , and show that if the wire OA is removed the equivalent resistance is increased in the ratio 1 1/6. (Li.U.) 3. Six equal, uniform, conducting wires form a tetrahedron ABCD. A given current is introduced at A and removed at a point P of BC such that BPfPC = 'AffL . Prove that the current in AD is independent of the position of P, and that the ratio of the current in BP to that in CP is ('A + 3 fL) /(3 'A + fL) . (Li.U.) 4. Current is supplied by a motor working at rate W to two circuits in parallel, one of which contains accumulators of voltage E0 and resist­ ance r on charge, and the other lamps of total resistance R. Prove that the current through the lamps is E0 + {E02 + 4rW( 1 + r/R)}t(M.T.) 2(R + r) 5. A system of conducting wires, all of equal resistance, form the edges of a cube. Find the currents carried by the wires if a battery is con­ nected without extra resistance to two adj acent vertices of the cube. (M.T.) 6. An electric current passes along a uniform cable from A to B. The resistance from A to B is R, the potential at A is V, and the cable is earthed at B. If, owing to a leak at some intermediate point X, cur­ rent of amount C' reaches B when current of amount C is sent out from A , find the ratio of the length AX to the length AB. Show that the resistance of the leak is C' (CR - V) . (M.T.) (C - C') 2 7. Two long, straight, parallel wires are joined by cross wires of the same material at equal distances, forming an infinite ladder of equal squares, the resistance of a side of a square being r. A current enters and

4 86

A C O U R S E lN A P P L I E D MATii E M A T i C S

[CH.

leaves the network at the ends A , B of one of the cross wires. If the currents in successive segments of one of the long wires, measured from A , are u1, u2, u3, • • • show that Un - 4un + l + Un + 2 0 (n > 1}. Show also that Un = u 1 ( 2 - y'3)n - I, and that the equivalent resistance of the network is r/y'3 . (M.T.) 8. A uniform wire A 0An has resistance nr and is connected to earth by An_1 , wires each of resistance 2r at the (n - 1) points A v A 2 , where A 0A l A lA 2 . . . A n _ 1An. A 0 is maintained at a potential V0 and An is earthed. If V8 is the potential at A8, show that V8 _ 1 - fr V8 + Vs+ r = 0, (M.T.) where 1 � s � n - 1 . Find V8 and the current in A 0A 1 . 9. A simple circuit of resistance R is connected to the terminals of a battery of e.m.f. E and negligible internal resistance. Show that the time-rate of production of heat is E2fR. When two points of the circuit are connected by a wire of resistance 3R/8 it is observed that the rate of heat production is doubled. Find the current passing (M.T.) through this wire. =

•

=

=

•

•

=

10. A cable of infinite length has one end maintained at a potential v.. The cable leaks to earth along its whole length, the resistance per unit length of the insulating sheath being R1 . If R is the resistance per unit length of the cable, show that the potential V at distance x from the " live " end is given by v

= v.e-"'"',

where m2 = RfR 1 • (Hint : Consider a length dx of the cable and note that the resistance of the insulator surrounding this element is R 1fdx.) 1 1 . Two uniform, conducting media (T 1, T 2) have a plane interface 0) , and extend throughout space on the negative (T 1} and positive (T 2} sides of the plane. A point source of current i. is situated in the medium T 1 at a distance a from the interface ; determine the current :flow. Show that the total current :flow into the second medium is T 1i./(T 1 + '1' 2} , and sketch the current lines in the case when '1'1 > '1' 2 • (S.U.)

(z =

12. A perfectly conducting sphere of radius a is surrounded by a concen­ tric, spherical shell of inner radius a and outer radius fa, and of uniform resistivity T ; the rest of the space is filled by a uniform medium of resistivity T '. Show that the rate of outflow of current from the con­ ducting sphere, when it is maintained at potential v. relative to points at an infinite distance, is (S.U.) 13. A conducting sphere of radius a and uniform resistivity Tz is embedded in a conducting medium of uniform resistivity T 1 filling the rest of space, and modifies a steady current :flow which would otherwise have a uniform current intensity C 1 in the direction. The value of the

z

1 7]

STEADY ELECTRIC CURRENTS

487

specific inductive capacity K is assumed to be unity everywhere. Show that the potentials inside and outside the sphere are given by 3-r1-r2 C 1 V1 r cos av + A , (2 -r2 + -r 1) aa + ( "�" + "1" 2) C 1 2 1 + V0 -r 1C 1r cos va cos av A r (2 "�" 2 "�"1) respectively. Find the charge density at any point on the spherical interface. (S.U.)

=-

=

-

14. A circular hole of radius a is drilled in a fiat sheet of metal of constant thickness and conductivity. Show that if the flow of current in the sheet is uniform at great distances from the hole, the expression for the potential is cp

= A (r + �) cos 8,

the axis of the hole being r = 0. (Li.U.) 15. Two perfectly conducting concentric spherical shells of radii a and b(a < b) are separated by a medium of specific resistance -r. Show that the equivalent resistance between the shells is _2_

(� - I)

4rc a b · 16. Two perfectly conducting, parallel, infinite, plane plates are separated by a medium whose specific resistance increases linearly with the distance from one plate, from a value -r 0 to another -rl" If d is the distance between the plates, show that the equivalent resistance between two corresponding portions of the plates of areas A is d(-r0 + -r 1}/2A . Prove also that if I is the current leaving unit area of one plate, the charge density between the plates is I ( -r1 - -r0) f4rcd. 17. Three perfectly conducting, spherical electrodes of small radius a are at the vertices of an equilateral triangle of side a in an infinite medium of conductivity a. If a current i enters by one electrode and a current ii leaves by the other two, show that the equivalent resistance is

! G - �)

· 8 a 18. A current enters an infinite plane plate equally by three perfectly conducting, circular electrodes of equal radii a and placed at the vertices of an equilateral triangle of side a. It leaves by a similar electrode at the centre of the triangle. Show that the equivalent resistance of the combination is 2-r I a log 3rct \ a:a ' where -r is the specific resistance of the material of the plate and t is its thickness.

)

19. A semi-infinite block of metal of conductivity a has an insulated plane face. A small, perfectly conducting, spherical electrode of radius a,

488

A C O U R S E I N A P P L I E D MATHEMAT I C S

[cH.

17]

distant a from the plane face, is maintained at a potential V, the potential at infinity being zero. Show that the maximum current density on the insulated face is 4craV 3y3a2• 20. Two small, spherical, perfectly conducting electrodes of radii a and a' are embedded in an infinite medium of conductivity cr, their centres being at a distance 2a apart. Show that the resistance between them is approximately R =

1 (I + .!._ I) . _ a 4n-cr a a' -

If the medium is bounded by an infinite plane surface from which each of the centres of the electrodes is at a distance b, show that the resistance is

1 (1

1

)

(M.T.) + 4n-cr b - y(a2 + b2) • 21. Current enters an infinite, plane plate of thickness t and conductivity cr by a perfectly conducting, circular electrode of radius a. Current leaves by an electrode of radius a. If a is small and f is the distance between the centres of the electrodes, show that the equivalent resist­ ance between the electrodes is f2 - az log ( �) · 2n-crt 22. A circular disc of thickness t, radius a and conductivity cr has set into it two perfectly conducting, circular electrodes of small radius a, one being at the centre of the disc and the other at a distance f from this point. Show that the equivalent resistance between the electrodes is af log n-at ay(az - j2) " (Hint : If the offset electrode is of strength i, introduce an image electrode, also of strength i, at the inverse point.) 23. An infinite medium of conductivity cr contains two perfectly conducting, spherical electrodes, one of radius a and the other of small radius a. Show that, if a current i enters by one electrode and leaves by the other, the equivalent resistance between the electrodes is (f - a) 2 a + --ap p - az ' 47tcr La where f( > a) is the distance between the centres of the two spheres. (Hint : Introduce image electrodes at the inverse point and at the centre of the larger sphere.) R

1

1

_ _

1 r1

J

CHAPTER 1 8

ELECTRO-MAGNETISM 18.1. Magnetic Shells

As a result of the researches of Oersted and Ampere in the early nineteenth century, it became known that an electric current gives rise to a magnetic field which is steady when the current is steady. To be precise, it was discovered that a small loop of wire carrying a current behaved like a magnetic dipole whose moment M was in the direction of the normal to the plane of the loop and had magnitude proportional to the area dS of the loop and to the current i flowing in it. The sense of M was found to be that of the longitudinal motion of a right-hand screw rotated in the sense of the current flow. Thus, if n is the unit normal to dS whose sense is related to that of the current flow in the manner just described, M

=

ki dS n,

(18.1)

where k is a constant of proportionality. If i is measured in electro­ static units, dS in sq em and M in the units employed in Section 14.8, it is found by measurement that 1/k = 3 X 1010 = c, c being a constant which will later be identified with the velocity of light in cmjsec. We may, however, choose to measure the current i in new units so chosen that k = 1. Then

(18.2) i dS n. The new units of current are called electromagnetic units and clearly (18.3) 1 e.m.u. = c e.s.u . . Until Section 18.8 we shall assume all currents to be measured in e.m.u. Equation (18.2) establishes a link between electric and magnetic phenomena. The current i constitutes a flow of electric charge under M

=

the influence of an electric field, whereas M is a magnetic moment. Prior to the discoveries of Oersted and Ampere, there was no reason to believe that electric and magnetic effects were in any way related, and they have hitherto received separate, though similar, mathematical treatments in this book. In this chapter, however, we shall treat electric and magnetic effects as two facets of the same fundamental phenomenon of electro-magnetism and shall construct a theory of the

electro-magnetic field.

489

490

A C O U R S E I N A P P L I E D MATH EMAT I C S

[CH.

Consider any surface S bounded by a closed loop C as shown in Fig. Imagine two families of curves drawn on this surface to form a network. If a current i is supposed to flow around each mesh such as UVWXU, XWYZX, etc., the resultant current in any side of a mesh such as WX will be zero, for there will be equal and opposite contributions from the currents in the two meshes having WX for a side. This will not apply to sides such as QR which belong to one mesh alone. Thus, the re­ sultant effect of all the cur­ rents i will be that of a current P i flowing around the loop C. Conversely, we may calculate the magnetic field due to a current flowing around C by integrating the effects of the currents flowing in the in­ dividual meshes. Each mesh and its associated (b) (a) FrG. 18.1 .-Magnetic Field Due to a Large current behaves like a dipole of Current Loop moment i dS, dS being the area of the mesh. The axis of this dipole is along the normal to dS and in the sense determined by the right-hand screw rule. The aggregate of such dipoles constitutes a shell of magnetic material in the form of the surface S, magnetized so that a layer of north poles exists on one side of the shell and a layer of south poles on the other side (Fig. 18.1 (b)) . This is termed a magnetic shell. i is the moment per unit area of the shell, and is referred to as its strength. The magnetic potential at the point P (Fig. 18.1 (b)) due to an ele­ ment dS of S is given by equation (14.64) to be i dS cos fJjr2• But, as proved on p. 361, dS cos fJ/r2 = dw, the solid angle subtended at P by dS. Hence, i dw is the potential at P due to dS and the net potential at P due to the whole shell is given by

18.1 (a).

n

=

L i dw

=

iw, .

(18.4)

where w is the solid angle snbtended at P by the loop C. The field due to the current loop can now be found from the equation H = - grad 18.2. Fields Due to Simple Current Loops

n.

(18.5)

We will first calculate the magnetic field caused by a steady current i flowing in a long, straight wire. The loop will be completed by leads from a battery to each end A, B of the wire, all effectively at infinity

18]

E L E C T R O -MA G N E T I S M

491

(Fig. 18 . 2 ) . We shall assume that the wire and leads lie in a plane p. Let P be any point in proximity to the wire. Consider the unit sphere with centre at P. The solid angle subtended by the current loop at P intersects the surface of the sphere in two great circle arcs NXS and NYS, where NS is the diameter parallel to the wire. The great circle NXS lies in the plane containing P and the wire and the great A �-------.�--� p

N

y

8 �------�---J FIG. 18.2.-Magnetic Potential Due to a Long, Straight Current

circle NYS lies in the plane through P parallel to p. The area of the lune bounded by these two great-circle arcs measures the solid angle subtended at P. Let 0 be the angle between the perpendicular OP (=r) from P to the wire and the plane p. Then the angle between the great-circle arcs bounding the lune is 1t - e and the area of the lune is accordingly 2(1t - e). Hence the potential at P is given by

n = 2i(7t - e) .

(18.6)

Let us calculate the field in any plane perpendicular to the wire. Let the wire intersect this plane in 0 (Fig. 18.3) and let P be any point in the plane having polar coordinates (r, e), e being measured from the plane of the current loop. If (H, H0) are the radial and transverse components of the magnetic field at P (there is no component of the field in the direction parallel to the wire, since n does not vary in this direction) , then, as at equations (14.68) , H, = -

an = 0, or

1 an = 2i ae -:y·

Ho = - r

.

(18.7)

These results indicate that the field is of magnitude 2ijr and in a trans­ verse direction everywhere. Consequently, the lines of force are

492

A C O U R S E IN A P P L I E D M A T H E M A T I C S

[cR.

concentric circles about the wire. In Fig. 18.3 the sense of the lines of force is anti-clockwise, since H0 is positive and the current flows out from the paper. We conclude that the directions of the field and cur­ rent flow are related by a right­ hand screw rule. The solid angle subtended by a circle at a point P on its axis has been proved to be 27t(1 - cos ) in Section 14.3(2), ex being the angle subtended at P by any radius of the circle. Thus, if a current i flows around the circle the magnetic potential of its field at P is given by

ex

FIG. 18.3.-Field Due to a Current in a Straight Wire

Q

= 27ti(1 - COS ex) .

If z is the distance of centre 0 of the circle and a is the circle's radius

(18.8)

P from the (18.9)

The field at magnitude

P will clearly be along the axis of symmetry OP and is of (18. 10)

At the centre of the circle, the field is therefore 21tija . A solenoid is obtained practically by winding a large number of turns of wire around a circular, cylindrical former. For the purposes

0

FIG. 18.4.-Field on the Axis of a Solenoid

of calculation, a current i flowing through such a solenoid may be replaced by a large number of separate circular currents arranged side by side with their centres on an axis Oz (Fig. 18.4). Let n be the number of loops per unit length of the solenoid and let a be its radius. Con­ �ider the loops between two planes perpendicular to the axis and distant z il,.nd z + t/z from 0, Their n11mb�r is n dz, and eg.ch contributes a

18]

ELECTRO-MAGNETISM

field given by equation is accordingly

493

(18.10) at 0. The field at 0 due to these loops 2rcnia2dz (z2 + a2F'

Integrating, we obtain for the field at 0 due to the whole solenoid

- 2rcma. 21z,(z2 +dz a2)�· z, a cot e, where e is the angle indicated in Fig. 18.4, this H-

Putting z = equation becomes

H=

2rcni �a sin e de = 2rcni(cos [3 - cos

rx ,

) .

(18.11)

where rx , [3 are the angles subtended at 0 by the radii of the ends of the solenoid nearer and further from 0 respectively. If the solenoid is very long and 0 is a point on its axis in the vicinity of the centre, then rx = rc, [3 = 0 and the field is given by H = 4rcni. (18.12) Draw the equipotential curves and lines of force in a cross-section of the magnetic field due to flow and return currents j in two long, parallel wires. Show that, if a small magnet of moment is free to turn about its centre P, and the plane through P at right angles to the wires cuts them in A , B, then

Example 1.

M

in equilibrium the axis of the magnet passes through the circumcenlre of the triangle A PE, and the magnet will oscillate about this position in time AP . BP 1T \/ jM . AB ' where is its moment of inertia. (M.T.) Suppose that the current flows out from the paper at A and into the paper at B. If P is any point in the plane perpendicular to the wire and 6v 6 2

f2I .

I

A COURSE I N APPLIED MATHEMATICS

494

[cH .

are the angles shown in the diagram, the potential at P due to the two wires is given by

n

and hence LAPB are constant. If P moves along an equipotential, Thus P will move along an arc of a circle passing through A and B. The family of equipotentials is accordingly the set of co-axial circles having AB for common chord. The lines of force must cut the equipotentials orthogon­ ally. It follows that the system of co-axial circles orthogonal to the equi­ potential system and having A, B as limiting points constitutes the family of lines of force. It will be noted that the family of equipotentials is identical with the family of lines of force for the field due to two similar parallel, infinite line charges of opposite signs (see Example, p. 435) . The equi­ potentials in the electric field r� are the lines of force for the magnetic field. The equipotential through P is the circumcircle of the tri­ angle APE . If C is the centre of this circle, CP is a radius and hence, also, a tangent to the orthogonal circle through P. But this orthogonal circle is the line of force through P. Hence, a small magnet pivoted at P will lie in equilibrium with r, its axis along CP. If LAPC = [3, LBPC = <X, AP = r 1, BP = r 2 , the strength of the field at P is B A 2" 2" H = ..2 sin f3 + ..2.. sin <X. But

<X

r1

= goo

-

A,

r2 f3 = goo

-

B

and hence this equation is equivalent to

2" 2" H = .J cos B + .1 cos A, r2 ri 2" = _.1_ (r1 cos A + r2 cos B), rtro 2j . AB

=

' AP . BP

H the magnet at P is rotated from its equilibrium position through an ·angle 6, the restoring couple is MH sin 6 (see equation (14.75)) . The equation governing its oscillation about equilibrium is accordingly

Ili + MH sin 6 = If the oscillations are of small amplitude, they are simple harmonic .of p.eri.o.d

0.

21T

I I

'V MH

=

7T

/ 21 . AP . BP_

'V jM . A B

18J

495

E L E C T R O - M AGN E T I S M

18.3. The Biot-Savart Law

Let a closed loop C subtend a solid angle w at a point P (Fig. 18.5). If H is the magnetic field at P due to a current i flowing around the loop, the work which will be done by the field on a unit pole when it is

FrG. 18.5.-Change in Solid Angle Due to Circuit Displacement

moved from P through a small displacement da is H da. This work will also be equal to the decrease in the potential. Thus •

H · da =

-

d O. = - i dw . .

(18.13)

To calculate the change in w caused by the displacement from P, it is convenient to regard P as a fixed point and to suppose that each element ds of the circuit undergoes a displacement - da, the circuit moving into the position of the loop C'. The contribution to dw of the element ds will then be equal to the solid angle subtended at P by the parallelogram QRR'Q'. Let n be the unit normal to this parallelogram in the outwards sense from the region generated by the displacement of C. Then, if dS is the area of the parallelogram,

�

�

QP = r and 1> is the angle between n and QP, the solid angle subtended

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•

[cR.

at P by the parallelogram is dS cos rf>fr2 = dSn . rjr3 But dSn = X so that the solid angle contributed by the parallelogram is

ds da,

I r

3

ds da X

r

·

I r

= -3

(18.14)

·

ds of C, we obtain 1 r xr3 ds. dw = da ·

By integrating over all the elements

(18.15)

.

o

(18.13) and (18.15), we now deduce that

. H . da

From equations

da r x ds. .

da . . j r xra ds = da . . j ds rxa r. (18.16) This last equation is true for all small displacements da. Thus j dsr� r. (18.I7) H i If e is the angle between r and ds, equation (18.17) may be interpreted as stating that each element of the circuit C contributes to the field =

2

-

c

2

--

=

c

--

c

at P a component i ds sin ejr2 in a direction perpendicular to the plane containing both the element and P. If P were to rotate about an axis along in a right-handed sense, its initial direction of motion would be that of the field due to This is the Biot-Savart Law.

ds

ds.

A wire carrying a current i is in the shape of a plane curve r = f(6), (r, 6) being polar coordinates. Show that the magnetic field at the pole is of magnitude · de t r the integration being extended over the length of the wire. Deduce that the field at a focus of an elliptical wire carrying a current i is 2mfl, where l is the semi­ latus rectum. Show also that the field at the centre is mLJA , where L is the perimeter and A is the area. If q, is the angle between the tangent to the curve r = f(6) and the radius vector, the contribution to the field at the pole 0 of an element ds of the circuit at P is i ds sin 6fr2 in a direction perpendicular to the plane of the curve. But sin q, = rd6fds and hence the net field at 0 is

Example 2.

J

- ·

j

H = i i � ds = r ds

if�r

The wire is in the shape of an ellipse with focus at 0 if

l = 1 + e cos f!,

r

l being the semi-latus rectum and e the eccentricity.

magnetic intensity at 0 is

ff"(l + e cos 6)d6 = 27Tifl. 0

In this case the

497

ELECTRO-MAGN E T I SM

18]

If 0 is at the centre of the ellipse, its polar equation is _.!: = �os2 e + �n· a , H=i

!2" 0

a2

r2

b2

Thus

a, b being the semi-axes.

/cos• a sin2 e i '\/ ---;2 + � da = ab

!2"

v a• sin2 a + b2 cos2 6 da.

0

0

X

But, if x = a cos t, y = b sin t are parametric equations of the ellipse, its perimeter is given by L

! 2"

=

Va2 sin2 t + b2 cos• t dt.

0

Thus, H

=

§ = 1riLjA .

18.4. Force Acting upon a Circuit in a Magnetic Field Suppose that a closed loop carrying a current i is introduced into

a magnetic field H.* To calculate the PE of this system, we shall suppose the loop replaced by an equivalent magnetic shell whose boundary is the loop. Let dS be an element of this shell and let n be the unit normal to dS in the sense from the negative to the positive side of the shell (see Fig. 18.1 (b)) . Then the moment of the element is i dSn and, by equation (14.77), its PE in the field H is - i dSH n = -iHndS. The net PE of the shell, and hence of the current circuit, is therefore ·

W=

-

i

j HndS

= - iN,

* I .e., H does not include the field of the current itself.

(18.18)

498

A COURSE

IN

A P P L I E D MATH EMAT I C S

[CH.

where N is the flux of H across the surface of the shell in the positive sense (this is the sense of progression of a right-hand screw rotated in the direction of the current) . If 511 52 are the surfaces o f two shells having the circuit for boundary, and S is the closed surface formed from 51 and 52 , since div H = 0, by Green's Theorem

L HndS j div H dv = 0, 1 HndS + 1 HndS = 0 . =

or

s,

s,

(18.19)

If the fluxes across 51 and 52 are taken in the same sense, this latter equation implies that (18.20) Thus the flux is independent of the shell selected. We accordingly refer to N as the flux through the circuit. If the circuit is immersed in a magnetic medium, the element dS of the magnetic shell will lie in a disc-shaped cavity. The component of the field in the cavity, in a direction perpendicular to its plane faces, is equal to the component of the magnetic induction B in this direction (see Exercise, p. 457). But this is. the direction of the normal to dS and hence the component of the field along the normal is Bn. The PE of the elementary dipole dS is therefore - iBndS and the total PE is given by W= -i

j EndS

= - iN , .

(18 .21)

where N is the flux of magnetic induction through the circuit. Since div B = 0, this flux is independent of the surface over which it is calculated. If the circuit is displaced slightly, the current being maintained constant, its PE will change to W + dW. Then -dW is the work done by the forces acting on the circuit. Sup­ pose this displacement con­ sists in a distortion of the circuit whereby an element ds is translated from the position ds PQ to the position P'Q', the FIG. 18.6.-Force on a Current Element current loop being maintained unbroken by wires PP', QQ' -+ -+ (Fig. 18.6) . Let PP' = QQ' = dr. The flux through the circuit increases by the flux through the parallelogram PP'Q'Q. But dr X ds

499

E L E C T R O - MA G N E T I S M

18]

is a vector equal in magnitude to the area of the parallelogram and having direction normal to its plane. The flux through the parallelo­ gram is accordingly B · dr X ds. Hence

ds = iB X ds · dr. (18.22) If F is the force acting upon the element ds, the work done is F · dr. dW

Thus

=

-i dN = - iB

F · dr

·

dr X

=

-iB X ds · dr.

(18.23)

This equation is valid for all dr and it follows that

F = ids X B.

(18.24)

Equation (18.24) shows that each element of the circuit is subjected to a force iB ds sin 6, where 6 is the angle between the element and the magnetic induction. The direction of this force is perpendicular to the element and the induction and is in the sense of progression of a right-hand screw being rotated from the direction of the current to that of the induction. This is Ampere's Law. In the above analysis, B is the induction of the applied field, i.e., it does not include the induction due to the current i itself. Thus the force F does not include the action upon the element ds caused by the magnetic field of the charge flowing in the remainder of the circuit. If B is the total induction at ds due to the applied field and the current i, F is the total force upon ds. If, however, the circuit is regarded as a rigid body and the resultant action upon it is required, the field of the current may be neglected, for the action of the current upon itself must be represented by a system of forces in equilibrium. This action will accordingly contribute to the internal forces in the wire only. i' Show that two parallel infinite straight wires carrying currents i, i' at a distance r apart, attract or repel one another with a force 2ii'jr per unit length.

Example 3 .

If the currents are in the same senses as shown in the diagram, the field at a point P on the current i' due to the other current is of magnitude 2ifr and in a direction normal to the plane of the paper and into it. The force F acting on an element at P is accordingly 2ii'dsfr in the direction shown. It follows that the force per unit length of the conductor is 2ii'fr, tending to attract the wires together. If the currents are in opposite senses, the force is one of repulsion.

- - .!'-

-

...,F,__,.J.. _ -r; p

500

A C O U R S E IN A P P L I E D MAT H E M A T I C S

[cH.

Show that the force on a circular current of radius a due to a magnetic doublet of moment m, lying on and along the axis of the circle, at a distance f from the centre, is

Example 4.

(S.U.)

If P is any point on the circular current, in the notation of the diagram the field at P due to the dipole at 0 possesses components (H.. Ho) given by 2m c�s e H, = , r By Ampere's Rule, the force acting upon an element ds of the circuit at P due to the component H, is iH,ds in a direction opposite to that of Ho and the force due to the component H0 is iH0ds in the direction of H,. Resolving these forces in directions (i) along the axis of the coil and (ii) per­ pendicular to this axis, it is clear that the net component in the second 0

m

direction is balanced by an equal and opposite component acting upon the element at P', a point on the circuit diametrically opposed to P. The net component along the axis of the circle is S i iH,ds sin 6 + iHods cos e = � sin e cos e ds. r

Integrating over all the elements of the circle, we find that the resultant force acting upon the circuit is 3mi .

ra

sm e cos e

f

ds

=

6.,mia . --rs-

sm e cos a,

along the axis. This is equivalent to the result stated.

A

current i flows round a circular wire and a current i' along an infinite straight wire in the same plane. Find the field H due to the current i', and hence show that the mutual force between the two circuits is 4.,ii' (sec <X 1), where 2<X is the angle subtended by the circle at the nearest point of the straight wire. (S.U.)

Example 5.

-

1 8]

E L E C T R O -MAGN E T I S M

501

Let PP' be an element of the circular wire subtending an angle d6 at the centre 0. Let OA = f be the perpendicular from 0 to the straight wire and let LPOA = e. The distance of the element from the straight wire is f - a cos 6 and thus the magnetic field at this point due to the current i' is

2i'f (f - a cos 6) in a direction perpendicular to the plane containing the two wires. The force exerted on the element PP' is accordingly 2aii'd6 PP' = j - a cos e j - a cos 6 along PO. Only the component in the direction A O will contribute to the resultant force on the circle. The resultant force is therefore a repulsion of 2ii'

!.,

-rr

2aii' cos e d6

j"' ..,!"'

f - a cos e = 4aii' = 4n

0

0

cos e d6 f - a cos e

cos 6 d0 cosec Ct - cos a'

since cosec c:< = ffa. This integral may be evaluated by changing the variable to t, where t = tan !9, the result being as stated in the question.

Consider an element ds of a wire carrying a steady current i. Let dS be the cross-section of the wire at the element and let p be the density of the charge which is flowing there. If v is the velocity of the charge through the element, in time dt the distance moved by the charge is v dt, and hence the volume of charge passing through the element is v dt dS. This volume corresponds to a quantity pv dt dS of charge and thus the current is given by i = pv dS. Since v and ds are in the same direction, ids = pV dS ds = ev, where e is the total charge flowing in the element at any instant. It now follows from equation (18.24) that the force acting upon the element due to the action of a magnetic field is given by F = ev X B. (18.25)

502

A C O U R S E IN A P P L I E D M A T H E M A T I C S

[CH.

If we assume that the force acting upon an element of a circuit actually acts upon the charges moving around the circuit, equation (18.25) will give the force exerted upon a charge e moving with velocity v. We further assume that the same force will act upon the charge when it moves across a magnetic field in the absence of a conductor, e.g., as when an electron moves in the space between the cathode and anode of a wireless valve. It is assumed in equation (18.25) that the point charge e is expressed in e.m.u. If, however, e is measured in e.s.u. the force due to the magnetic field is ev X Bfc. If, in addition, an electrostatic field of intensity E is present, this will contribute a second force eE. Thus, the resultant force acting upon the point charge will then be given by

F = eE +

e

c

-

v X

B.

(18.26)

The space between two circular, co-axial cylinders is evacuated and electrons are set free with negligible initial velocity at the surface of the inner cylinder. They are accelerated from the inner cylinder, of radius a, towards the outer, of radius b, by means of a constant potential difference V between the cylinders. If a uniform magnetic field H is applied parallel to the axis of the cylinders, show that electrons will just reach the outer cylinder if 8mc2b2 V = eH2(b2 - a2) 2, where e is the charge and m the mass of an electron, the variation of m with (S.U.) velocity being neglected.

Example 6.

0

Taking polar coordinates (r, 6) as indicated, the electric intensity at the point P is given by equation (15.48) to be E=

v �, r log (bfa) �-

in a radial direction. This field contributes a force eE acting as shown on an electron at P. If the magnetic field is directed into the plane of the diagram, by equation . (18.25), each of the velocity components (r, rli) contributes a force acting upon

1 8]

ELECTRO-MAGNETISM

503

the electron. The magnitudes and directions of these forces are shown in the diagram. We can now write down radial and transverse equations of motion for the electron, viz., (i)

m(r - ril•) = r lo;�bfa) - �Hril, � � !:_ (r2il) = Hr. (ii) r dt c Equation (ii) may be written d eH d � " dt (r•e) - mc dt ('lir ) · Whence, upon integration, we obtain eH (r2 + constant) . r•e = 2mc But, when r = a, e = 0 and hence eH (r - �)• . e = 2mc r Substituting for e in equation (i), we find that .r = d ( u•) = eV + e•H• ( a4 - r) . mr log (bfa) 4m2c2 rs Iii . Since r = 0 when r = a or b, integrating this latter equation with respect to r over the interval (a, b), we show that 0=� Sm2c2b2 (b2 - a2)2' m -� _

...

which is equivalent to the result stated in the question.

18.5. Equations of a Steady Electro-magnetic Field Suppose that a steady current i flowing in a closed loop C is responsible for a magnetic field H. We will calculate the work done by the field forces which act upon a unit magnetic pole as it is moved around a closed path C'. Over any element of the path the work done is -d!J., the decrement in the potential along this element, and hence the total work done, is equal to the net decrease in the magnetic potential as the path is described. Now = iw, where w is the solid angle subtended by C at a point in the field. If the path C' does not link the circuit C (Fig. by considering successive positions P1, P2, on the path it will be found that, after the completion of a circuit of C', w returns to its original value at P1• It follows that the net work done by the field forces is zero for this type of path. If, however, the path C' links the circuit C (Fig. and, commencing at P1, we move around the path via the successive positions P2 , P3 , • • • back to P1, a continuous change in the solid angle causes it to vary in the manner indicated. Commencing with the value wl> it takes the value zero at some point on the path between the points P2 , P3 and thereafter is negative. Upon arrival back at PI > the solid angle assumes the value - w5• The net decrease in w is accordingly

n

•

•

•

18.7 (a)),

18.7 (b))

(18.27)

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A C O U R S E I N A P P L I E D MATHEMAT I C S

[CH.

4rri .

and the work which has been done by the field forces must be It will be observed that the sense in which the path has been de­ scribed is related to the direction of the current by a right-hand screw

C'

(a)

(b)

FIG. 18.7.-Ampere's Circuital Rule

rule, i.e., if a right-hand screw is rotated in the sense of the motion around it will advance in the direction in which the current threads This is taken to be the positive sense of description of a closed path threading a circuit. We have therefore proved viz. ,

C'.

C'

Ampere 's Circuital Rule, when a unit magnetic pole is taken around a path threading an electric circuit carrying a current i, in the positive sense, the work done by the field forces is 4rri. In the case of a path threading the p circuit n times, the work done will be 4rrni. c,

A consequence of the rule just proved is that the magnetic potential due to a steady cur­ rent is a multi-valued function. For the work done by the field forces as a unit pole is removed from a point P in the field to infinity will depend upon the FIG. 18.8.-Multi-valued Potential number of times the path to Function infinity threads the circuit. Thus, in Fig. if the work done along the path is 01, and that along the path c2 is n2, the work which will be done when the pole is taken from infinity to the point p via the path c2 and thence to infinity

18.8,

C1

1 8]

505

ELECTRO-MAGNETISM

via the path C1 will be n1 - n 2• But this is a closed path threading the circuit three times, and hence

(18.28) In general, the difference between the potential corresponding to any two paths to infinity from the point being considered is Thus, if n0 is the potential corresponding to any one path, the general value of the potential is given by the equation

47tni.

n = n0 +

47tni.

(18.29)

As a particular case, consider the potential due to an infinite straight current. An expression for this has been obtained as equation It will be observed that the angle 6 is arbitrary to the extent of an integral multiple of thus making the potential arbitrary as regards the addition of an integral multiple of Another consequence of Ampere's Rule is that a steady electro­ magnetic field is not conservative. This does not imply any violation of the law of conservation of energy, since such a field can be main­ tained only by the expenditure of energy in an external source (battery or dynamo) . If a unit pole is taken around a closed path threading the circuit, as will be shown later, an e . m.f. is induced in the circuit opposing the flow of current so that extra energy must be expended in the external source, and this exactly accounts for the work per­ formed by the forces of the field. Now consider an electro-magnetic field caused by a steady flow of charge through a region of space occupied by a conducting but non­ magnetic medium. If C is any closed path drawn in the medium (Fig. and is the current density, the rate of flow of charge across any surface bounded by C is

(18.6).

27t,

18.9)

47ti.

5j

i=

Ljnd5,

and this is independent of since div = If r is the position vector of any point on C relative to some origin 0, the work done by the field as a unit pole is moved over the element dr of the path at C is H · dr. The net work done by the field forces as the pole describes C can now be found by integration over C. We shall assume that Ampere's Rule may be applied to the path C, so that we have

5,

Ja H

·

dr

=

j 0.

4m = 47t L jnd5.

.

(18 30) .

If the conducting medium is magnetic, the field H at any point will be the resultant of two components : (i) a component due directly to the transport of charge, and (ii) a component due to the mag­ netism induced in the medium. The work contributed by the second

506

A C O U R S E I N A P P L I E D MATH EMAT I C S

[CH.

component when a unit pole is taken around a closed path is zero, for the field of the induced magnetism is caused by a distribution of dipoles and is accordingly conservative. Equation remains valid in this case also therefore.

{18.30)

s

0

FIG. 1 8.9.-Relationship Between H and j

{18.30)

To convert equation into a differential relationship, it is convenient first to introduce a new vector function called the Let A be any vector function of position (see Section and let be any closed contour drawn in the A-field. If r is the position vector of any point of so that dr is a vector element of (as in Fig. the scalar quantity

15.1)

C,

C

c = fa A · dr,

.

curl. C 18.9),

(18.31)

where the line integral is taken around the contour C, is called the circulation of A around C. If rectangular axes Oxyz are defined in the field so that the points at the ends of the element dr have coordinates (x, y, z), (x + dx, y + dy, z + dz), then dr has components (dx, dy, dz). Let A have components (Az, A y, Az). Thus A · dr = Axdx + A ydy + A.dz and = fa (Axdx + A ydy + Azdz) . . (18.32) Suppose that dS is any small plane element of area in the A-field containing within it the point P(x, y, z) and bounded by a contour C (Fig. 18.10). Let (l, m, n) be the direction cosines of the normal to dS, taken in the sense which is related to the direction of description of C by a right-hand screw rule. Let P(�, 'YJ, �) form a set of rectangular axes parallel to the set Oxyz and let (�, 'YJ, q be the coordinates of any C

18]

E L E CT R O -M A G N E T I S M

C

point of relative to these new axes. can then be written c

=

507

The circulation of A around

l (A�;d� + A11dYJ + A,d�)

C

(18.33)

.

z

FIG. 18.10.-Curl of a Vector

(�, YJ , q are small, A�; = Ax (x + �' y + YJ , z + �) oA x OAx = A x + OAx ax � + ay_ YJ + az � ' A11 = A11 (x + � ' y + YJ , z + �) (18.34) OA y = A + OaxA y � + BAy ay YJ + az � A, = A z (x + � ' y + YJ, z + �) oA z OAZ )" = A z + OAZ ax � + ay YJ + az "' where Ax, A11, Az and all their derivatives in the right-hand members of these equations are to be calculated at the point P (x, y, z). Thus Ax ( YJ d� + OAx ( � d� = Ax l( d� + &oxAx lc( � d� + &oy oz lc lc c o o o + A v l dYJ + �y l � dYJ + �y l YJ dYJ + �y l � dYJ But, by Taylor's Theorem, since

11

'

'

c

I

+ A z lc( d� + OoxA z (a � d� + OoyA z l( YJ d� + OozA :l � d� l c c (18.35) .

508

[CH.

A C O U R S E IN A P P L I E D M A T H E M A T I C S

Now

f d�

a circuit of

C

= [ � [a =

is

0, since � returns to its initial value when a completed. Also fa � d� = [ !�2 [ c = 0 for the same

reason. Similarly, four other integrals present in the right-hand member of equation may be equated to zero. On the other

(18.35)

hand, the value of the integral

C' C

fa

'Y)

d� is unaffected if it is computed

around the projection of on the �'YJ- or xy-planes, for the coordinates (�, 'YJ) take the same values around as they do around Thus

C'

C.

d� = 1 'Y) d� = - (projection of dS on the xy-plane) , = -n dS, (18.36) having regard to the direction of description of C'. Similarly, L � d'YJ = n dS,

1

a

1J

c·

f 'YJ d� - fa � d'Y) a fa � d� - fa � d� =

=

=

= m

l dS,

(18.37)

dS.

(18.35) can now be written in the form oAy ) + ( oA x oA . ) + ( oA y oAx ) � = ( oA. as oy oz l az ox m ox oy (18.38)

Equation

_

_

_

n.

Suppose we define a vector curl A such that its components in the directions of the axes at any point (x, y, z) are given by

oA oA oA x oA . oA oA x - ( oy. ozy '. oz ox '. oxy oy ) · (18.39) Then the component of the vector in the direction (l, m, n) is given by the right-hand member of equation (18.38). It follows that the com­ curl A

_

_

_

_

ponent of curl A in any direction is

dS

lim �. dS as_,.o

(18.40)

where is at the point (x, y, z) and has its plane normal to the direction in which the component is being calculated. Thus curl A is not de­ pendent upon the particular axes Oxyz selected.* * Alternatively, equation (18.38) shows that the inner product of curl A and the vector (l, m, n) is invariant with respect to the axes and hence curl A is a vector. If, formally, we take the vector product of the vector operator \7 (equation (14.41)) with the vector A, we obtain the right-hand member of equation (18.39). Another notation for curl A is therefore v X A.

1 8]

509

E L E CT R O - MAG N E T I S M

We can now prove Stokes' Theorem, which states that if S is any surface bounded by a closed contour C and lying in the A-field, then L n · curl A dS = Jo A · dr, (18.41) where n is the unit normal to S whose sense is related to the sense of de­ scription of C by a right-hand screw rule. Let S be divided into a large number of small elements by two families of curves drawn upon it (Fig. 18.11). Consider the contribu­ tion to the surface integral appearing in equation (18.41) of a typical element dS. n · curl A is the component of curl A in the direction of the normal to dS and is therefore equal to cfdS, where c is the cir­ culation of A around the boundary of dS in the sense n

indicated in the figure. Thus

n·

curl

A dS = c.

The surface integral is ac­ cordingly equal to the sum FIG. 1 8. 1 1 .-Stokes' Theorem of the circulations around But the elements of the common boundary of two such elements contributes to the cir­ culations around each element, and these contributions will be in opposite senses and hence will cancel. However, an arc such as which is part of the boundary of one element only, will yield a con­ tribution to the net circulation which will not be cancelled. It follows that the sum of the circulations around the elements of is equal to i.e., is equal the sum of the contributions of all elements of like This proves the to the right-hand member of equation theorem. Applying Stokes' Theorem to the left-hand member of equation we obtain

S.

PP',

C PP', (18.41).

(18.30),

ln Ln.

· curl H

dS = 47t L j

•

S

n dS,

(18.42) 47tj)d5 = 0. . This latter equation is true for all surfaces S drawn in the electro­ or

(curl H -

magnetic field so that we may conclude that

or

curl H curl H

=

47tj 0, 47tj.

=

(18.43)

510

A C O U R S E IN A P P L I E D M A T H E M A T I C S

[CH .

It follows at once from this equation that H can no longer be generated from a scalar potential. For, if H = -grad n, curl H = 0 (see of Appendix to this chapter) . This contradicts equation

§1 (18.43).

+

H2 , Suppose that at any point in an electro-magnetic field H = H1 where H1 is the intensity directly attributable to the currents flowing and H2 is the intensity due to the magnetism induced in the medium and to any permanent magnetism it may possess. The field due to a current flowing in a wire is identical with that of a certain shell situated But, at all vacuo, except at points within the material of the shell. points outside the shell the field satisfies the equation div H = 0. It follows that this equation is valid at all points of the field due to the current. Hence, since tubes of flow of small cross-section in any steady current distribution are equivalent to linear conductors, we shall assume that the field H1 satisfies the equation div H1 = 0. Also, by equation the field H2 due to the magnetized material satisfies div I. Thus the equation div H2 = H2) = div H1 div H = div (H1 div H2 = div I, div(H = 0. or B=H+ Putting

in

(16.44),

-41t +

-41t

+

+ 41tl) 41tl,

(18.44) (18.45)

we accordingly obtain the equation

(18.46)

div B = 0.

B is, of course, the magnetic induction at, the point in the field. To summarise, the equations of a steady electromagnetic field are as follows : cu�l H = d1v B = 0, B=H+

41tj,

41tl,

}.

(18.47)

The intensity of magnetization I will be related to the total field in­ tensity H by an equation such as At points of the field at which there is no current flow, a potential function n exists such that

(16.48).

H = - grad

n. .

(18.48)

For the closed tubes of current flow may be replaced by magnetic shells, none of which pass through the point under consideration, and the field due to this distribution of magnetic matter can be generated from a potential function. A steady current I flows in a straight wire whose section is a circle of radius a, and the current density inside the wire is uniform. A ssuming . A mpere's Circuital Rule, find the magnetic field H at a distance r from the axis of the wire. Deduce that curl H is: (1) zero when r > a, and (2) parallel to the (Li.U.) current and of constant magnitude 4Ifa2 when r < a.

Example 7.

1 8]

E L E CT R O - M A G N E T I S M

511

We shall assume that the lines of force are circles with their centres on the axis of the wire and with their planes perpendicular to this axis. We shall also suppose that the intensity H depends upon the distance r from the axis of the wire alone. A line of force of radius r lying inside the wire embraces a current r2Ifa2• Applying Ampere's Rule to this line of force, we obtain the equation 2rrrH = 4rrr2Ifa2, or H = 2Irfa• . . (i) If the line of force lies outside the wire, it embraces the total current I, and hence Ampere's Rule yields 2rrrH = 4rri, or H = 2lfr. (ii)

H

Taking axes Oxyz with Oz along the axis of the wire and in the direction of the current, at a point P(x, y) inside the wire the components of H in the directions of the axes are Hx = -H sin 6 = -Hyfr = - 2Iyfa•, Hu = H cos e = Hxfr = 2Ixfa2, H, = 0, having employed equation (i) . The components of curl H can now be derived. They are aH. aH. (cur1 H)x = 8)1 - & = 0,

( curI H)u = 8Z

aH,.

-

--ai" = 0, aH.

(curl H). = aH" - aH"' = 41fa•. ax

ay

Thus curl H is a vector in the direction of Oz and of magnitude 4Ifa2 • This verifies equation (18.43) . If P is outside the wire H,. = 2lyf(x2 + y2), Hu = 2Ixf(x2 + y2), H, = 0 and it may now be checked that all components of curl H are zero. -

G

A C O U R S E I N A P P L I E D M A T H E M AT I C S

512

[CH.

An infinite, straight wire carrying a current I is situated at a distance a from the plane face of a semi-infinite block of soft iron of permeability 1-'· Show that the wire is attracted to the iron with a force J2 (tL 1) / (/k + 1) a per unit (Li.U.) length. Let 01, 02 be the magnetic

Example 8.

-

potentials outside and inside p the magnetic medium re­ spectively. Let 0 1 be gene­ rated by the straight current I and a similar image current I' occupying the position of the optical image of I in the plane face. Let 0 2 be gene­ rated by an image current I" I" coincident with I. All these currents will be assumed to [' flow from the diagram to­ wards the reader. At a point P on the plane face nl = 2I6 - 21'6 + constant, 0 2 = 2I"6 + constant, the constants making allowance for the fact that1the reference line from which the angle e in equation (18.6) is measured is arbitrary. Now the potential is continuous across the plane face and thus (i) I - I' = I". Resolving the intensities at P due to the three currents in the direction of the normal to the plane face and applying the condition that En is con­ tinuous across this face, we obtain 2I 2J' . 2I" r sm e + r sm e = �"r sm e. .

or

.

I + I' = From equations (i) and (ii), we find that

J.ki".

(ii)

I ' = 1!:_=-J:. I /L

+

1.

The force of attraction per unit length between the parallel currents I, I' a distance 2a apart is II' !L - 1 I = a �-' + 1 · a•'

which is accordingly the attraction of the block for I.

18.6. The Vector Potential It is shown in the Appendix (§4) that a solenoidal vector (i.e . , one whose divergence vanishes identically) is expressible as the curl of another vector function. B is such a vector, and hence we may write B = curl where

A is called the vector potential.

A,

(18.49)

E L E C T R O - M A GN E T I S M

1 8]

51 3

The curl of a gradient being identically zero, A is arbitrary to the extent of a gradient. Thus, if A' is a vector potential, the vector A defined by A = A' + grad rfo, (18.50) We have

rfo being arbitrary, is also a vector potential for B. 'V 2rfo. div A = div A'

+

(18.51)

We can now choose rfo such that 'V 2rp = - div A',

(18.52)

for rfo will be the potential of a charge distribution of density p = div A' /471: and hence is uniquely determined if A' is defined at all points. With such a choice of rfo, div A = 0. (18. 53) Equations (18.49) and (18. 5 3) accordingly define A uniquely. Thus, consider a magnetic dipole of moment m and let r be a position vector whose origin is at the dipole. Then it is easily verified that A=

m X r.

y3

(18.54)

For, taking the z-axis along the line of the dipole, we have m = (0, 0, m) , r = y, z) and hence

(x,

Ax = Thus

-myjr3, A y = mxjr3, A z = 0.

curl A =

(

3mxz 3myz 3mz2

_

y5 ' y5 ' y5

3m ·-r m = --,s r - 1-3 '

y3 )

�

(18.55)

(18.56)

which is the magnetic intensity due to the dipole (equation (14.72)) and hence the induction B. Also div A = -

! (';?) � (7:) +

= 0. .

(18.57)

Now suppose that the field is due to a steady current flow in a medium of constant permeability fL, j being the current density specifying the flow. Then, from equations (18.47), (18.49), we deduce that curl curl A = curl B = fL curl H = 471:fLj .

(18.58)

Expanding the left-hand member of this equation (see §7 of Appen­ dix), we find that (18. 5 9) grad div A - 'V 2A = 471:{1-j, or, by appeal to equation (18. 5 3) , 'V 2A = - 471:fLj.

(18.60)

514

A C O U R S E IN A P PLI E D M A T H E M A T I C S

Thus, the x-component of

Ax

[CH.

A satisfies the equation 'V 2A x = -41ttJ.fx,

indicating that is the potential in the electric field due to a charge distribution of density t-tix· But this potential is unique and is given by

A x = f t-tfrx dv'

where r is distance measured from the element dv to the point at which Ax is to be calculated. The volume integral is evaluated over the whole of space. Similar expressions may be found for the remaining components of We deduce that

A.

(18.61) In the special case where the current flows in a fine wire the element dv can be taken to be a small segment ds of the wire. Then, if IX is its cross-section and i is the current flowing

f dv = � . IX dS = i ds. IX Also, if ds is the vector element of the wire, since the current flow is in the direction of this element, we can write

j dv = i ds

and therefore

(18.62) (18.63)

the integral being a line integral evaluated over the whole length of the w1re. Thus t-tidsjr is the contribution to of the element ds of the wire. Substituting in equation (18.49) , we obtain the contribution of this element to B, viz.

A

'ds r

curl � =

t-t�. r curl ds t-tids X r r3 ,

.

ds X grad

'

(r)

� ,

(18.64)

where we have employed the expansion of curl (uv) obtained in the This is the Biot-Savart Law. Appendix

(§5) .

18.7. Electro-magnetic Induction. Faraday's Law From what has been said in the previous sections of this chapter, it will be appreciated that the electric and magnetic components of an electro-magnetic field are intimately related to one another. When a

ELECTRO-MAGNETISM

1 8]

515

charge moves relative to an observer, the latter will be able to detect a magnetic field. Thus a moving electric field gives rise to magnetic effects. Also, as will be proved immediately, a moving magnetic field produces electric effects. Neither the electric nor the magnetic aspect of the electro-magnetic field can therefore be treated in isolation from the other. Let us now consider the electric field generated by the motion of a magnetic field having induction Suppose that the field is being translated with velocity and let e be a charge at rest. Then, re­ garded from the point of view of an observer moving with the field , the charge e has velocity -v relative to the field and, by equation X if e is measured in e.s.u., it will experience a force It follows that an observer at rest with the charge will detect an electric field of intensity X at the point occupied by the charge, for, by definition, the electric intensity is the force which acts upon unit stationary charge. Thus,

B.

v

-ev Bfc.

(18.25),

-v Bfc

E = -c1 B x v .

(18.65)

is the electric intensity generated by the motion of a magnetic field

B with velocity v.

FIG. 18.12.-Circuit Moving in a Steady Magnetic Field

B

v

Now suppose that the field moves with velocity relative to a closed conducting loop The free charge in the wire will be acted upon by forces causing it to circulate and a current will flow. Thus an observer will detect an e.m.f. operating around the loop. The electric field at any point on the wire is given by equation Let denote the potential distribution along the wire which would account for this field. Then, if are the potentials at the ends of an

C.

(18.65).

-+

element PQ

V, V + dV

( =dr) of the wire (Fig. 18.12), by equation (14.34) dV = -E dr = - -c1 B X v · dr. ·

V

[CH.

A C O U R S E IN A P PL I E D M A T H E M A T I C S

516

The decrease in the potential around the whole circuit in the sense of dr is therefore

C=

- 1 dV = �c 1 a

a

B X v·

dr =

_!c 1 · dr X a

C is the e.m . f. operating around the circuit.

B

v.

(18.66) C

Let us now calculate the change in magnetic flux through during the short time interval Regarding the magnetic field as stationary, has velocity -v relative to it and moves from to in the time

dt.

C

C C'

18.12).

dt PP' = QQ' = - vdt. --+

--+

(Fig. Let PQ move to P'Q'. Then The flux of induction through the parallelogram PQQ' P' can now be calculated as on p. to be B dr X v . If PQQ' P' is de­ scribed in the sense indicated by the order of the letters, the flux just calculated is to be taken as threading it in the positive or right-handed sense. Taking the flux of induction through and in the positive sense in relation to the sense of description of (viz., that of dr) , the flux through is now seen to be less than that threading by an amount

·

499,

( dt) -

C C

C'

C'

C

- dt f B · dr X v. (18.67) Thus, if N is the flux through C and N + dN is that through C', - dN = dt l B dr X v, fff = l dN or (18.68) c dt I.e., the e.m.j. induced in a circuit is proportional to the rate of increase of the flux of magnetic induction through the circuit and is in the negative sense in relation to the sense in which the flux is taken. It is found by experiment that this law is obeyed when the change in the flux is produced by any other mean�, e.g., by fixing the circuit C �

.

a

-

--

·

-·

and varying the current in a solenoid in its neighbourhood. The existence of this effect was first demonstrated experimentally by Faraday, who described the phenomenon as that of

electro-magnetic

induction.

In the remainder of this section we shall assume C to be measured in e.m.u . , in which case the factor occurring in equation is superfluous. The chief practical application of the principle just explained is the If a loop of wire is rotated in a magnetic field about some axis the flux through the loop will vary and an e.m.f. will be induced in the wire. By breaking the loop at some point and connecting the two ends to an external apparatus this e.m.f. may be employed to pass

1jc

alternator.

(18.68)

ELECTRO-MAGNETISM

18]

517

an electric current through the apparatus. The details will be found described in textbooks of electrical engineering. If the field is uniform and of strength H and if the loop is plane, of area A and is rotated about an axis perpendicular to the direction of the field, when the plane of the loop makes an angle e with the field direction, the component of the intensity perpendicular to the plane of the loop will be H sin e and the flux through the loop will be AH sin e. It follows from equation that the e.m.f. induced in the loop at this instant is given by

(18.68)

rff

= - � (AH sin 6) = - A Hw cos e, .

(18.69)

where e is the rate of rotation. If the loop is replaced by a plane coil of turns, an e.m.f. is induced in each turn and the resultant induced e.m.f. is - A Hnw cos e. In this case, if is constant, we may write e = and then

w= n wt

w

rff =

n

- AH w

cos

wt = -rff0 cos wt. .

(18.70)

Thus the e.m.f. generated across the terminals of this type of alternator fluctuates with the time in a simple harmonic manner and is called an Devices are available for transforming this e.m.f. into a constant e.m.f., either within the machine (the commutator) or externally (the rectifier) . If the transformation is carried out within the machine the latter is then termed a Now suppose that the induction B is due entirely to a current flowing in a loop Provided the intensity of magnetization produced in the magnetic material (if any) surrounding the loop is always directly proportional to the field intensity H, the induction B at any point will be fixed in direction and directly proportional to Thus, the flux of induction through will be given by an equation of the form

alternating e.m.f.

dynamo.

i

C.

i.

C

N = Li,

(18.71) where L is a constant depending only upon the geometry of C and is called the self-inductance of the loop. The reader should verify that the flux N is positive with respect to the sense of the current flow, and hence that L is positive. If the externally applied e.m.f. in the circuit is increased, causing the current i to increase, an e.m.f. rff is induced in the loop given by the equation rff

di = - dN dt = - L dt

-·

(18.72)

It will be noted that the sense of this e.m.f. is such as to oppose the tendency of the current to increase. It is accordingly referred to as a If, on the other hand, decreases, equation shows that the induced e.m.f. operates in a sense tending to oppose the decrease.

back e.m.J.

i

(18.72)

A C O U R S E IN A P P L I E D MATHEMAT I C S

518

[cH.

In most practical circuits the flux through the circuit is concentrated in one or two elements termed An inductor generally comprises a large number of turns of wire wound upon a soft iron to form a solenoid. The magnetic field in the soft iron is re­ latively intense compared with the field in the vicinity of other parts of the circuit, and the contribution of the latter field to the total flux through the circuit may therefore generally be neglected. The flux through an inductor will be given by an equation such as so that its self-inductance may be defined as the flux per unit current. Alternatively, it follows from equation that L is the back e.m.f. induced in the coil when the current increases at unit rate. In par­ ticular, the back e.m.f. in volts induced when the current increases at a rate of ampjsec is taken to be the self-inductance of the element in The henry is the practical unit of inductance.

inductors.

former

(18. 71),

L

henries.

�(05 c.;t

(18.72)

1

"->

L

L

R

R

(a)

FrG. 18. 13.-Simple A.C. Circuits

Consider the circuit of (a) , comprising an inductor of and resistance connected to the terminals of an self-inductance If is the current flowing alternator generating an e.m.f. 6'0 cos by application of Kirchhoff's second law and equation at time we derive the equation

Fig. 18.13 R

L

(18.72),

t,

D = djdt,

i

di = R ' wt - L dt z,

(18.73)

(LD + R)i = 6'0 cos wt.

(18.74)

o 0

0 cos

or, if

wt.

Solving this linear equation with constant coefficients by a well­ known method, we obtain + cos sin A e -Rt/L + 0 ' 2 co o + 2

·

_

wt wL wt R 2 w2L the first As t ---+

R

(18.75)

oo, term in this being an arbitrary constant. expression for i tends to zero, and hence can be neglected after some time

A

E L E CT R O - M A G N E T I S M

18]

519

has elapsed (generally of the order of a fraction of a second) . This component of is accordingly referred to as the The re­ maining component of the current persists and is called the in the In the steady state therefore

i

transient.

forced steady state. R cos (J)t + (J)L sin CJlt, z - Co R2 + (J)2L2 _ (CJlt - oc) Co (R2 + CJlR2L2 2+)l (cos J)2L 2 ' (18.76) (R2 ::2L2)l cos (CJlt - ) where tan a = CJlLjR. We observe from equation (18.76) that, in the steady state, i oscillates in a sinusoidal manner with the same frequency as the applied e . m.f. but lags behind it by a phase angle Also, if i 0 oscillation

.

_

oc ,

is the amplitude of the current, then

io = Cof(R2 + CJl2L2)l, C0 = i 0 (R2 + CJl2L 2)l. .

or

(18.77)

oc.

(18.77)

Equation indicates that the relationship between the ampli­ tudes of the e . m.f. and the current is similar to that obtaining between the e . m.f. and the current in Ohm's Law, the quantity + playing the role o f the resistance. This quantity i s called the of the circuit. If a condenser of capacitance is included in the circuit as in Fig. and if is the charge which has been removed from one plate and added to the other at time the potential drop across the plates is Thus, by equating the total p.d. around the circuit to the total e.m.f., we obtain

(R2 (J)2L2)l impedance

18.13 (b) qfC.

C t,

q

o cos (J)t - L didi = Rz. + cq ·

(0 .Q>

But the rate of accumulation of charge on the positive plate of the condenser is equal to the current flowing into this plate. Thus = Hence

i dqjdt.

(18. 78) This equation may be analysed by methods similar to those employed in Section to determine the motion of the dynamical system governed by equation ( . The reader should prove that the forced oscillation is given by C0 cos((J)t + �) . = z + where tan � =

2.4

2 86).

{R2 ((J)L - 1/(J)C) 2} t' (1/(J)C - (J)L)fR.

(18.79)

A COURSE IN APPLI ED MATHEMATICS

520

[CH.

i0 is the amplitude of the current, then (18.80) tff0 Zi0 , where Z {R2 + (6lL - 1/6lC)2}i = the impedance. Clearly i0 is a maximum when Z is a minimum. For variable 6l, this condition obtains when ((,)L - 1/(,)C) = 0, i.e., when 1 (18.81) (,) = 6l = VLC . is termed the resonant angular frequency of the circuit. If

=

=

R

c.JR

If there is no applied e . m . f. in the circuit the equation governing the current flow is

( LD2 + RD + f) i

=

0.

(18.82)

This equation is similar to that determining the motion of the dynamical system shown in Fig. viz., equation As in the case of the dynamical system, if the condenser is initially charged, it will discharge itself according to one of two modes : (i) a damped oscillatory mode, or (ii) an aperiodic mode. The condition for an oscillatory discharge will be found to be that The reverse inequality is the < condition for an aperiodic discharge.

2.9,

(2.72).

R 2(LfC)i.

An alternating e.m.f. of amplitude E and frequency w/2rr is supplied to a coil of inductance L and resistance R. If the coil is shunted by a condenser of capacitance C and resistance S, show that the circuit can be replaced (as far as permanent current is concerned) by a non-inductive resistance provided that CR2 - L = w 2LC(CS2 - L). (L.U.)

Example 9.

Let i = i1 + i 2 be the total current taken from the alternator and let ±q be the charges on the condenser plates at time t as shown in the diagram.

Then

(i) Applying Kirchhoff's second law to the circuits A BEFA , A CDFA , we obtain the equations di . E cos wt - L dtl = Rlv E cos wt = Si2 +

t·

Differentiating the second of these equations and employing equation (i), we may write them in the form (LD + R)i1 = E cos wt,

( SD + �) i 2 = - Ew si

n

wt.

The steady-state solution of the first equation has already been found at equation (18.76) . Thus R cos wt + wL sin wt .

_

11 - E

R2 + w2L2

•

1 8]

E L E CT R O - M A G N E T I S M

521

The corresponding solution of the second equation is ECw(wCS cos wt sin wt) ' . -

z. =

Thus i = i1 + i 2 = E

1 + w2C252

[ ( R• +Rw•L2 + 1 .:·�:g•s•) cos wt + L - 1 + ':�c• .s•• ) sin wt] . + ( R• � w•L2 A

B i,

fees wt

L

('..)

R

c D

E

F

If, however, the circuit behaves like an alternator connected across a pure resistance, i will be in phase with the alternating e.m.f. and the term in­ volving sin wt in the expression for i must be absent. Hence wL wC = R2 + w2L2 1 + w2C252'

which is equivalent to the condition stated.

Suppose that two circuits CI > C2, are in the vicinity of one another. The flux through C2 due to a current flowing around C1 will be pro­ portional to Let it be Then the e.m . f.
i1•

i1

Mi1 •

0 2 - -M didtl , D

If

(18.83)

-

i1 is measured in amps,
only upon their geometrical properties and relative positions. If the two circuits lie in a medium of constant permeability vector potential due to the current l in cl is given by

i

A

= 1-'-�·1 1 dsr 1 a,

_,

•

flo,

the

(18.84)

522

A C O U R S E I N APPLI E D MAT H E M A T I C S

ds1

C1 r A C Js n · B dS = fa n curl A dS = L A

[CH.

where is a vector element of and is distance measured from this element to the point at which is to be calculated. The flux of in­ duction through any closed curve in the field is ·

.

dr, .

(18.85)

S being any surface bounded by C. In particular, the flux through c2 is · ds2 1 fJ.Zlds2 1c,yds1 = fL21 1c, 1c, ds1--r-' (18.86) ' c, ds2 being a vector element of C2 and r the distance between ds1 and ds2. It now follows that (18.87) This is Neumann 's Formula for the mutual inductance between the two circuits. It shows that M is symmetrical with respect to the circuits, and hence the flux through cl due to a current i2 in c2 must be Mi2 • Suppose that a coil of n1 turns and self-inductance L1 is closely wound on a soft iron former. When unit current flows through a turn of the coil, let B be the flux through it. All turns will be assumed to generate the same field so that the total flux passing through the coil will be n1B. This flux threads each turn of the coil so that the total flux linking the coil is n12B. Thus L1 = n12B. Now suppose that a second coil of n2 turns is wound on top of the first. The flux n1B due to unit current in the first coil will be assumed to link each turn of the second coil so that the mutual inductance is M = n1n2 B. The self-inductance L2 of this second coil is given by L 2 = n22B. Thus L1 = n12B, L2 = n22B, M = n1n2B. (18.88) .

We observe that

.

'

M = VL1L2.

(18.89)

In practice, part of the flux due to the first coil will fail to link all turns of the second coil so that Thus = < v where is called the between the coils and is less than unity. Now let the first coil be connected to the terminals of an alternator, whilst the second coil is left on open circuit (Fig. Neglecting the resistance of the first coil, if is the current through it and is the applied e.m.f., then

k

M L1L2• coefficient of coupling i1

M kv' L1L 2,

18.14).


(18.90)

E L E C TRO - M A G N E T I S M

1 8]

The e.m.f. induced in the second coil is then (18.83),

tff2 = - MDi1

=

-

523

1!2, where, by equation

� 1!1'

(18.91)

1

If we assume equation (18.89) to be satisfied, then

,ff2 = - Jf: {fl

(18.92)

having employed equations (18.88).

FrG. 18.14.-Simple Transformer Circuit

Thus the applied and induced e.m.f.s are in the same ratio as the turns in the two coils. This is the principle of the By choosing the turns ratio appropriately, an alternating e.m.f. applied in a may be converted into an alternating e.m.f. in the of any desired amplitude.

transformer.

primary circuit secondary circuit

A coil of negligible resistance and self-inductance L is inductively coupled to a coil of negligible resistance and self-inductance N, the mutual inductance of the two coils being M. The ends of the coil N are connected to the plates of a condenser of capacity C, and an electromotive force E sin nt is applied to the ends of the coil L. At time t = 0 the currents are zero and the condenser is uncharged. Prove that at a subsequent time t the current in the coil N is nME (cos nt - cos n0t), (LN M")(n• n o") 1 where n02 L{C(LN - M2)}- • (Li.U.)

Example 10.

_

_

=

Let i1, i 2 be the currents in the primary and secondary circuits respec­ tively, and let q be the charge on the condenser. Equating the potential drops around each circuit to the e.m.f.s, we obtain the equations di di . E sm nt - L 1 - M dt2 = 0, dt di1 di N 2 = !l . . M dt dt c •

_

_

(ii)

A C O U R S E I N A P P L I E D M A T H E MA T I CS

524

Also dqfdt = i 2 • Hence

LDi1 + MDi 2 = E sin nt, . MD2i1 + (ND2 + lfC)i 2 = 0. .

[CH. (iii) (iv)

Elimination of i1 between equations (iii) and (iv) yields

{(LN - M2)D2 + L{C}i 2 = -nME cos nt, nME . (D 2 + n0 ") 2 2 - cos nt, LN M• _

or

_

where n02 is defined in the question. L, M c

Esin nt ("'..)

The general solution for i 2 now proves to be . . 2 2 = A cos n0t + B sm n0t +

(n•

_

nME n o•) (LN

_

cos nt. M•)

At t = 0, q = 0. Hence, from equations (i) and (ii), di 2{dt = 0 initially. Also i 2 = 0 at t = 0. Thus A

=

-

nME (n2 - n02) (LN - M2) '

B

=

O.

This proves the result stated.

18.8. Maxwell's Equations In this section we shall obtain the equations governing a non-steady electro-magnetic field. Let C be any closed curve drawn in the field. If the flux of magnetic induction through C is changing and C is a wire, we know that an e.m.f. will be induced in the wire of magnitude and sign given by equa­ tion (18.68). If E is the electric field responsible for this e.m.f., the latter is given by

S

tff = l E · dr.

If is any surface with C as boundary, the flux of induction through C is

N = fa B

·

n

dS,

(18.93)

1 8]

E L E C T R O - M AGN E T I S M

525

n being the unit normal to S in the sense positive relative to the direc­ tion of description of Then, by Faraday's Law,

C.

f f

or

dt 1s B · n dS' dr = -1 ! � . n dS. s c at

E · dr =

c

c

E

·

_! � c

(18.94) C is not oc­ (18.94) may be

We shall assume this equation to be valid even when cupied by a wire. Then, by Stokes' Theorem, equation transformed into B. ! n · curl E dS = n

1

or

l(

-1 � ��)

a dS, at . n dS = 0.

s c

s

curl E +

(18.95)

This equation will be true for all surfaces S drawn in the field and this can be the case only if

1a

B curl E = - ;; -ar ·

(18.96)

In this equation E is measured in e . s.u. and B in e . m . u. Now consider equation valid in a steady electro-magnetic field. As is proved in the Appendix to this chapter, div curl H = Thus, equation implies that div = and this contradicts equation unless the flow is steady. Hence equation cannot be valid for a varying field. It is accordingly necessary to modify equation so as to remove this discrepancy and in such a way that the modified equation reduces to equation under steady conditions. Following Maxwell, we carry out this modification as follows : We first assume that the equation

(18.43),

(17.13)

(18.43)

(§ 3)

(18.43)

(18.43)

div n = is valid in all circumstances.

0. (18.43)

j 0

4rrp

Then

. 1 an ap 1 a (18.97) at = � at d.IV n = dIV 4rr at ' I t is here assumed that p and n are measured in e . s . u. We now conelude from equation (17.13) that 1 an . dIV ( (18.98) 4rr at + J ) = 0' . where j is now being measured in e.s.u. These units for j will bP ·

assumed throughout the remainder of this chapter.

526

A C O U R S E I N A P P L I E D M A T H E M AT I C S

[CH.

Let us define a new quantity J by the equation

1 8D 47t 8t ' . div J = 0. •

"T = J +

so that J is called the

(18.99) (18.100)

total current and comprises two components : (i) j, 8 the conduction current, and (ii) ; �, the displacement current. Since 7t D = E + 47tP, . 1 8E 8P displacement current = (18.101) 47t Tt + 8t · The second term of this expression arises from the currents which must flow within the molecules of a dielectric consequent upon a change in the polarization. However, the first term may be non-zero in a vacuum and cannot be identified with the motion of any actual charge. Nonetheless, it is an essential constituent of J if this quantity is to be solenoidal. In e.m.u., the total current is and, since it is solenoidal, we can write

J'fc

47t c

curl H = - J without contradiction. obtain

Substituting for curl H =

J in the last equation,

} (47tj �) 8 + 8

, .

we

(18.102)

(18.43)

which is the modification of equation we have been seeking. However, equation cannot be established as a logical con­ sequence of more elementary principles, and for its complete justifica­ tion it is necessary to examine its consequences and to check these experimentally. This has been done, and the results confirm the expectations. We shall now collect together all the equations governing a variable electromagnetic field. They are :

(18.102)

curl H = curl E =

� (47tj �) --1 -8B8t , 8 + 8

c

,

(18.103)

0, 47tp. J In these equations (Maxwell's Equations) it is assumed that p, j, D, E div B = div D =

are measured in e.s.u. and B, H in e . m . u.

In addition, we shall make

18]

527

E L E C T R O - M AG N E T I S M

use of the following relationships which are valid in an isotropic homo­ geneous medium

D = KE,

j = o-E.

B = tLH + 4rtl0,

.

(18.104)

Since B is solenoidal, we may express it as the curl of a vector potential Then aA �� curl E = curl A = curl - I c at c at · 1 aA Thus curl E + = 0, . (18. 105)

A.

(

c at

(

)

)

and, by a theorem proved in the Appendix (§2), this implies that E+

1 aA

c

at = - grad V,

where V is a scalar potential function.

Hence

E = - grad V -

1 aA

c ar ·

(18. 106)

If the conditions are steady, this equation reduces to E = -grad V in agreement with equations (17. 19). Assuming the dielectric constant K and the permeability lL to be con­ stant and there to be no permanent magnetism present over the region of space being considered, we have a 4rt[Lj + tLK . curl curl A = curl B = lL curl H = a

�)

�(

Expanding the left-hand member and substituting for E from equation (18. 106) in the right-hand member, we obtain

{

}

. 4rt[L . tLK av 1 a2A grad drv A - \7 2A = c J - c grad at + c 7fi2 .

(18. 107)

But A is still arbitrary to the extent of a gradient, and we take this opportunity of determining it completely, by requiring that it should satisfy the equation (18. 108) (cf. equation (18. 5 3)) .

Equation (18. 107) then reduces to

" 2A - tLK �2A = - 4rttL . J v c2 at2 c . •

This is the non-steady state form of equation (18.60) .

(18. 109)

528

A C O U R S E I N A P P L I E D MATHEMAT I C S

(18.108), we obtain . ( 1 8A )

Also, from equation fLK o2 V = -d1v C2 8[2

c Tt ,

= div(E + grad V) , (equation = div E + V 2 V, =

�fl

+ V 2 V, (equations

(18.106)),

(18.103)),

since K is constant over the region being considered. scalar potential satisfies the equation fLK o2 V 'V 2 V _ = _ ot2 K fl ·

4rr

c2

[CH.

Thus the

(18.110)

This is the generalised form of Poisson's equation. Having found A and V to satisfy and E, are all determined. = and both V and satisfy In an uncharged dielectric p = an equation of the form

(18.108), (18.109) (18.110), B, 0, j 0

H, D A

(18.111) wave equation,

This is the and it may be shown that any solution of the equation represents a wave being propagated with velocity c f VfLK. * The general solution of the one-dimensional wave equation is discussed in Section Also, in an uncharged dielectric, both E and can b e shown to satisfy the wave equation thus : From equations fLK a2E ! = _1:: � curl = _ curl curl E = _ curl .

18.11.

c

But

oB ot

H (18.103),

c ot

H

c2 ot2

curl curl E = grad div E - V 2E = - V 2E,

since div E =

0.

Hence

(18.112) Similarly in the case of H. Thus each component of E and H is propagated through space as a wave (an electro-magnetic wave), the velocity of propagation being cfv'fLK. In vacuo this velocity of propagation is c = 3 X 1010 cmfsec. Now light waves, wireless

waves and X-rays are all found to be propagated with precisely this velocity, a fact which strongly suggests that they are of an electro­ magnetic nature and constitutes strong evidence for the validity of Maxwell's equations. * E.g., Mathematical Theory ofElectricity and Magnetism, by Sir J. Jeans, 5th ed., et seq. (C.U.P.).

pp.

520

1 8]

ELECTRO-MAGNETiSM

529

Finally, in this section we will obtain a first integral of Maxwell's equations for the case of a homogeneous isotropic conductor. Since = crE =

j

crDJK,

471"0" p. dw J = R0" dw D = K Hence, by the equation of continuity (17.13) op + 47t cr (18.113) at K P = 0 Integration of this equation with respect to t at a fixed point (x, y, z) •

·

yields

(18.114)

K/47t cr

(x,

where -r = and Po depends upon y, z) . It follows from equation that, whatever the original charge distribution, the charge at any point in the conductor will decay exponentially. The rate of decay is governed by the constant -r, which has the dimension of time and is called the When = -r, = and about two-thirds of the initial charge has dispersed from each point. For copper, = x and hence -r = sec. The dispersion of charge in a con­ x ductor is therefore exceedingly rapid, and for most purposes it may be assumed that any charge always resides on the insulated surfaces of conductors or upon surfaces of discontinuity between them.

t

(18.114)

time of relaxation. K 1, cr = 5 ·3 1017

p 0·37p0 1 · 5 I0-19

18.9. Energy Considerations.

Poynting's Vector

Suppose that a varying electro-magnetic field exists in a region r bounded by a closed surface Then the rate at which the field energy within r is decreasing must equal the rate W at which work is being done upon the charges within r together with the rate at which energy is escaping from r across This latter rate can be expressed as the flux across of a vector called the Let e be any charge within r moving with velocity v. It is acted upon by a force

5.

5.

5

( �

e E+

Poynting Vector.

v x B

)

and hence the rate at which work is being done upon it is

( �

e E+

p

v X B

) = ·

v

eE v. ·

(18.115)

If is the charge density of flowing charge, the charge flowing within an element of volume is and the rate at which work is being done

dv p dv

530

A

C O U R S E IN A PP L I E D M A T H E M AT I C S

[CH.

upon it is • Hence the total rate at which work is being done upon charges within r is

pE v dv.

W

=

£ pE · v dv.

(18.116)

j =

(18.117)

The current density vector j is given by Hence the integral

pV.

(18.116) can be written in the form £ E · j dv.

(18.118)

This integral gives the rate of heat production within r (Section provided none of the work done increases the KE of the charges appreciably. This will not be the case, for example, in the space between the cathode and anode in a wireless valve. From Maxwell's equations, we have

17 5) .

,

.

c

1 oD 47t curl H - 47t m· Substituting for j in the integral (18.118), we obtain W = 4__!__7t l(r ( cE · curl H E . oD ) dv. ot J=

-

(§6), div (E X H) H curl E - E

(18.119)

But, from a result proved in the Appendix =

Hence

·

•

curl

H.

oD dv. W 47tC l(r H · curl E dv - 47t lrr d1v (E X H) dv- 417t l(r E • 7ft (18.120) 1 8B But curl E - c; ot . cE X H/47t, equation (18.120) is equivalent to Thus, writing S ( ( oB + oD W __!__ 47t lr H . ot E · ot ) dv fr div S dv, ft ( f1H2 tKE2 ) dv - Is S . n dS, . (18.121) = -£ c

=

.

=

=

=

_

-

having employed Green's Theorem.

Now

2 2 87t (� + KE )

U = __!__

(18.122)

18]

E L E C T R O -MAGN E T I S M

531

is the total electric and magnetic energy density in the field. equation (18.121) in the form

-� [ U dv

=

W + £ S n d5, . ·

Writing

(18. 123)

we see that the rate of decrease of the field energy within r is equal to the rate at which work is being done on the charges within r, together with the flux of S across S is the Poynting vector and its flux across must represent the rate of energy flow across this surface. If the field is steady, there is clearly no flow of energy across and it will be found that the flux of S across the closed surface is zero. However, S is not necessarily zero, and hence its flux across an element need not be zero. Thus S does not, in general, determine the rate of energy flow at a point, i.e., only the flux of S across a surface is significant.

5

5.

5,

d5

closed

18.10. Boundary Conditions In this section we will determine the form taken by Maxwell's equations at a surface of discontinuity. Let be a surface separating two regions 1 and 2 having different electric and magnetic characteristics. In order that our results shall be as general as possible, we shall suppose that a current flow of infinitesimal thickness S (a is taking place in and also Y that is the site of a surface charge distribution cr . If is a line element drawn normal to the current and is the on current across is the surface current density at the element. The vector i :2. having magnitude and direction that of the flow is the

5

current sheet) 5 5 ds 5 di ds, dijds dijds surface current density �� 0 Let ABC D be a small rectangle R with its edges BC, AD each in one face of 5 and its plane normal to the current flow (Fig. 18. 15) . Taking x- andy-axes parallel to the FIG. 18.15.-Conditions Across a Current Sheet sides AB, AD of the rectangle respectively, let x = xv x = x2 , y =YI> y = y2 be the edges. Then, if j is the current density in the region occupied by the K

rectangle, the current across it is the flux of j across the figure, i.e., is (18. 124)

.iii

[CH.

A C O U R S E IN A P PL I ED M A T H E M AT I C S

532

y

since, provided the rectangle is small, i will be independent of but rapidly varying with However, the current across the rectangle Hence is also =

x. BC . i (y2 - y1)i.

i=

j

Jx, j dx, . x,

(18.125)

the directions of i and being identical. is not necessarily normal to Now suppose that the rectangle the current flow. Then, over the rectangle

ABCD

K oi )

= c1 (

an 4 . + = -c 7tJ m

curl H

dS = � L ( 47tj + � � ) . n dS,

)

1 (4

.

(18.126) Thus, if n is the unit normal to R in the outwards sense from Fig. 18.15 cur1 H

towards the reader,

Ln

·

7tJ + -cr m

.

or, employing Stokes' Theorem, this may be written

,

( H · dr

)R

= c1 n ['dy lx, ( 1 n · (y2 - Y1) lx, ( x, ·

=c

1h

x,

K oj ) dx,

47tj + a at

0j 47tJ + --;;- 8t

. K

) dx, (18.127)

In calculating the circulation of H around R, we shall neglect the contributions of the edges since H is finite everywhere, whereas the thickness of the current sheet will be supposed small by comparison with and Thus, if are the components of H along in the regions and respectively, then

BC BC, AD

AB, CD AD. Htv Ht2 1 2 L H . dr = Htl (Y2 - Yl) - Ht2 (y2 - Yl), . (18.128)

(y2 - y1) is sufficiently small. (18.128}, we obtain

provided and

Comparing equations

(18.127)

{18.129) in being the component of i in the direction of n.

This is the boundary condition sought. The second term in the right-hand member of equation is frequently negligible by comparison with the first term, unless the current fluctuations are very rapid.

(18.129)

18]

E L E C T R O - MA G N E T I S M

533

In the absence of a current sheet we note that all tangential com­ ponents H1 of H are continuous across Since B (unlike remains finite as the thickness of the sheet ap­ proaches zero, a similar argument applied to the Maxwell equation

5.

j)

curi E = -

1 8B

-

c ­at

5.

shows that all tangential components of E are continuous across By consideration of the flux out of a disc with its plane faces in the two faces of it may be proved that the equations

5,

div D = 4rrp,

div B = 0,

imply that the normal component of D is discontinuous by 4rr a (equa­ tion (16 . 17)), whereas the normal component of B is continuous. These complete the set of boundary conditions. 18.11. Electro-magnetic Waves Let Oxyz be rectangular axes. We shall first obtain solutions of Maxwell's equations which are independent of the variables (x, y) , i.e., we shall be supposing that, at any instant, the electric intensity is the same at all points of any plane parallel to the xy-plane . Similarly, the magnetic intensity does not vary over such a plane. This is the case of Suppose that the electro-magnetic disturbance is being propagated through a dielectric medium having constant permeability fL and which is uncharged . Then, since = 0 and the partial derivatives of the components of E and H with respect to x and y vanish, Maxwell's equations take the form,

plane electro-magnetic waves.

K,

j

oHy

K 8Ex

8Hx

K oEy

O

- �c oEz at '

fL 8Hx

oEy

- Tz = --;; Tt ' - az = - c Tt ' Tz

c -at '

_

oE.c

Tz

=

O=

fL

oHy

- c; Tt ' _ f!- 8Hz c

at

(18. 130)

,

The final four of these equations imply that both Ez and Hz are constant. We shall take Ez = Hz = 0, thereby ignoring possible uniform electric and magnetic fields in the direction of the z-axis. Such fields have no significance in relation to the wave phenomena we shall be discussing.

A C O U R S E I N A P P L I E D MAT H E M A T I C S

534

[CH .

Considering the first four equations (18.130), we obtain

a2Ex - fl. a2Hy - fl. a ( aHy) fl.K a2Ex az2 - -c az8{ - c at & C2 8t2 · Thus, Ex satisfies the one-dimensional wave equation 82¢. 1 82¢. (18. 131) 8z2 = tl2 8t2 ' where (18.132) a = cfvfLK. . Similarly, it may be shown that Ey, Hx, Hy satisfy this equation. To find the general solution of equation (18. 131), we change the in­ dependent variables from (z, t) to (u, v) by means of the transformation u = z - at, v = z + at. _

_

_

-

Thus

and

8¢. = 8¢. 8u + 8¢. 8v = 8¢. + 8¢. 8z 8u 8z 8v az au 8v 82 - � ( 8 + 8¢. ) - � ( a¢. + 8¢. ) 8u + � ( 8¢. + 8¢. ) av 8z2 8z 8u 8v au 8u 8v 8z av 8u 8v az 82¢. + 2 a2¢. + 82¢. = au 2 8u8v 8v2 ' _

Similarly

_

a2¢. - a2 ( 82¢. 2 82¢. + 82¢. ) 8u2 - 8u8v 8v2 · 8t2

Equation (18. 131) is therefore equivalent to

82¢. au8v = 0 · Integrating with respect to u, we obtain 8¢. = G(v), 8v where G(v) is an arbitrary function of v. A further integration with respect to v now yields = f(u) + g(v), where j(u) is an arbitrary function of u and g(v) is the integral of G(v)

and therefore arbitrary. Thus, we have proved that



= f(z - at) + g(z + at), (18.133) wheref and g are arbitrary functions. Nowf(z - at) takes the constant value j(<X) at all points and times satisfying the equa.tion

z = at +

<X,

E L E C T R O -MA G N E T I S M

18]

a.

535

Such points advance along the positive z-axis with velocity At determining the distribution of /-values along the z-axis at this instant. These values are thereafter propagated represents a plane wave with velocity a along Oz. Thus (z The profile advancing in the positive sense along Oz with velocity of the wave is determined by the form of the function f. Similarly, g(z+ represents a plane wave proceeding with velocity in the reverse sense. Since E. H. = both the and H-waves are of the type, i.e., neither possesses a component in the direction of propagation. Let us suppose that Ex, Ey are being propagated as waves in the direction Oz. Then

t = 0, j(z - at) = j(z)

f

at) transverse

at)

0,

=

a.

E-

a

f(z - at), Ey = F(z - at). (18.134) Hence, from equations (18.130), we deduce that oHy _ aK J' (z _ at) • oHx = _ aK F' (z _ at) oz oz =

Ex

_

and thus

=

Hy

=

c

c

J�j(z - at),

Hx =

-

•

J�� (z - at), (18.135)

any further arbitrary functions being incompatible with the remaining equations We now observe that

(18.130).

(18.136) Thus the electric and magnetic intensities are always orthogonal to one another and to the direction of propagation. If we choose == then Ey = Hx = and the electric intensity is everywhere parallel to Ox and the magnetic intensity to Oy. An electro-magnetic wave for which the directions of and H are everywhere the same is said to be and the plane containing the direction of propaga­ tion and the vector H is the In the case just considered, this is the plane Oy . Consider the equation

F 0,

0

E plane of polarization. z

polarized

q,

=

f(x cos e + y sin e - at).

(18.137)

f(rx) at all points and times satisfying X COS 6 + J sin 6 = at + rx. . (18.138) For given time t, this represents a plane parallel to Oz, distant (at + rx) from this axis and whose normal is inclined at an angle e to Ox. As t increases, this plane moves in the direction of its normal with velocity a. Equation (18.137) accordingly represents a plane wave being propagated in a direction parallel to the xy-plane and making an angle e with Ox. � will take the value

A C O U R S E IN A P P L I ED MATH EMAT I C S

536

The equation 'f' = A..

. A s1n cu

(t

-

X cos e + y sin e ) ' a

[CH.

(18.139)

therefore represents a particular plane wave of this type. At a fixed point (x, y, z) , cf> oscillates sinusoidally with angular frequency cu. Thus, the wave corresponds to a beam of Suppose this beam is being propagated through a dielectric medium characterized by constants fl. and occupying the region x < 0. ' IL Let the region x > 0 be occupied by a dielectric with constants be the velocity of wave propagation through this second and let medium. Let e be the angle of incidence of the incident beam (Fig. 18 . 16) upon the plane of discontinuity x = 0. Maxwell's equa­ tions and the boundary conditions over = 0 can be satisfied only by assuming that the incident beam gives rise to a reflected beam and a transmitted beam Let e' be the angle of reflection and cf> the angle of refraction. of the various beams. They are Consider the contributions to

monochromatic light.

K,

K',

a'

I

x

R

T.

Ey . . _ A Sln cu ( t (1) I) Ey . _ A , Sln cu

(ii)

R) E

(iii)

T) Ey = B s1n. cu

11 -

x

-

(t (t

-

_

-x

x

y -

. - A sm cu

(t - X

cos e + y sin e

a

+ A ' sin cu and, in the region x > 0

Ey _- B s1n. cu ( t _

x

cos e' + y sin e'

a

cos cf> + y sin cfo

Thus, in the region x < 0

E

e + y sin e ) , a

cos

a

) (t

_

'

-x

cos cf> + y sin a

'

cos

1>)

)

),

.

e' + y sin e' ) a •

Ey is continuous across = 0 if y sin e ' ) _ sin e ) ) y a'sin "''f' . . . cu ( t - y A sm -- + A , sm cu ( t - -- - B sm cu ( t a a for all y and t. This can only be so if x

__

sin e sin e' sin cf> ---a = = -----a' ' .

-a-

and

A + A ' = B.

(18. 140)

E L E C T R O - MA G N E T I S M

1 8]

537

E

H

The boundary conditions on the other components of and lead to equations and determine the amplitudes of reflected and trans­ mitted waves. From equations we deduce that

(18.140)

(18.140),

e

(18.141) (18.142)

= e',

a

sin e sin 4> - a: ·

(18.141) is the well-known law of reflection, (18.142) Snell's Law of Refraction. The ratio

Equation is

? = J��

�

and equation

(18.143)

= v, .

coefficient of refraction

between the two media. Unless, when is the assigning values to the constants K, K', an allowance is made for the y

R

I

FIG. 18. 16.-Refl.ection and Refraction of Light

inertia of the atomic charges responsible for the polarization of the dielectrics, the value obtained from equation for v will differ greatly from the experimental value at the high frequencies correspond­ ing to visible light waves. However, for waves of much lower fre­ quency (wireless waves) there is good agreement between theory and experiment.

(18.143)

APPENDIX 1. curl grad V = 0 The x-component of curl grad V is o• v o• v o o 8)i (grad V), - az (grad V). = oyoz - ozoy

= 0.

538

A COURSE I N APPLIED MATHEMATICS

A =A

A=

[cH.

Similarly for the remaining components. 2. If curl 0, then exists such that grad Suppose curl is identically zero over some region of space r. Then, if is any closed contour in r, we shall suppose that it is always possible to find a surface S bounded by and lying wholly in r. Such a region is said to be By Stokes' Theorem

V

C

nected.*

V

fA dr = ls( AdS = V = ! PA · dr a

n

•

curl

·

C simply con­

0.

Let F be a fixed point in r and P a variable point. Then .F

does not depend upon the path connecting F and P. Hence, if P has coordinates

(x, y, z),

V = V(x, y, z)

and is a single-valued scalar function of position defined in Let P' be a point in the neighbourhood of P and let PP'

f A · dr = V, f ' dV = J PA · dr = A,ds, A p

and hence where Thus

P'

.F

.F

A, is the component of

Then

p

in the direction of PP'.

A , = av as ·

A = V. 8Av) + ( 8Ax 8A,) + ( 8Av az ay az ax oz ax A A = Ax = fBydz, Av = -fBxdz, A, =

This being true for all small displacements from P, grad 0 3. div curl For the left-hand member is � �

A= � ( 8A , ax ay

r.

= ds. A · dr = V + dV,

_

_

_

and this is identically zero. 4. If div B = 0, then exists such that B curl A We shall show that a suitable is defined by the equations

8Ax) !Jy

0.

Both are still arbitrary to the extent of any function of We have

Ax, Ay

_

x and y.

8A, 8Av Bx 8Ax 8A, - By ay az . az ox 8Ay _ 8Ax = - ! ( 8Bx + oBu) dz = JoB,dz (since div B = 0) ox 8y oz ox 8y = B, + C(x, y), _

_

_

-

and

*

By taking C to be a curve embracing the inner cylinder, the region between two parallel, infinite cylinders is seen not to be simply connected.

1 8]

E L E C T R O -M A G N E T I S M

539

where C is an unknown function of and y. By proper choice of the arbitrary functions associated with and C can be eliminated. Then B = curl A. 5. curl = curl v v X grad The x-component of curl is 0 0 + ay ( ) - BZ ( ) = ay - Bz

A,

uv u

x Av,

-

u uv OVy) au au UVz UVy U ( av. oy - ()z V, Vy and this is the x-component of u curl v grad u. Similarly we can identify the other components of the two members. 6. div (u X v) = v curl u - u · curl a a a div (U X V) = BX (uuVz - UzVy) + ay (u,v, - u,v.) + (u,vy - UyVx) · oz -

·

v

v curl u - u · curl v ·

= _

v x

( au. - OUy) v• ( au. - au.) v. (OUy - au.) ay az az ax ax ay ) ov 0 ov. av. ) u. ( ay _ • u• ( az ax!'• u• ( OVy _ ov•)

v

+

•

+

_

_

_

az

ax ay

= div (u X v) . 7. curl curl A = grad div A - v •A The x-component of curl curl A is

a a oy (curl A). - Fz (curl A)., oA•) � ( aA. oA,) , � ( oAv = ay ax ay az az ax o•A. a•A, a•A. a•A. = axay + axaz - ay• - ---ai"' oA __?__ ( aA. aA, ) a•A. o"Av a•A . • = ax ax + ay + az - ax• - ay• ---ai"' 0 d'lV A - v 2A • . - ax _

_

_

-

But this is the x-component of the vector grad div A - v •A. Similarly, the y- and z-components of curl curl A will be found to equal the corresponding components of this last vector. Thus curl curl A = grad div A - v •A.

EXERCISE 18 1 . A pair of circular coils of radii a, b and of m and n turns respectively carry the same current i in the same sense and are placed parallel on a common axis at a distance !(a + b) apart. Find the magnetic force they produce at a point on the axis distant !a, !b from the two coils and show that on the axis near this point the field is so uniform that its first two differential coefficients vanish at the point if mfa2 = nfb2 and its third in addition if a = b, m n. (M.T.) =

540

[CH.

A C O U R S E I N A P P L I E D M A T H E MAT I C S

2. The significant part of an electric circuit consists of a current C flow­

ing in the sense Oz in a wire along the line x = 0, y = a, and in the sense zO in a wire along the line x = 0, y - a, the coordinate axes being right-handed. Show that the magnetic potential at a point (x, y, z) due to the circuit is =

A -

_1

2c tan

x2

_2_yaoc-2x---,a

+

-

2'

where A is a constant, and that, if there is in addition a uniform field of strength H, parallel to Oz, a magnetic needle at (k, 0, 0) will rest in equilibrium making an angle with Ox, where + k 2) H(a2actan e = -· (M.T.) 4

e

-

3 . Thin, conducting wire forming a square circuit of side 2a conveys a steady current prove that the magnetic intensity at the centre of the square is and state its direction in relation to the sense of the current. (M.T.)

j; 4y2jja,

4

. Three long, parallel wires conveying steady currents

are cut by a perpendicular plane in the vertices A , B, C of a triangle in which the angles B and C are equal. Prove that on the line through A per­ pendicular to BC the only point at which the magnetic intensity vanishes lies on the circumcircle of the triangle ABC. If the triangle ABC is equilateral, each side being of length a, find the magnitude and direction of the mechanical force per unit length on the wire through B. (M.T.) 5. Wire is wound closely in circles of latitude on a hemispherical surface of radius a. The total current carried by the windings between any two circles of latitude is k times the difference of latitude in radians between the two circles. Show that the magnetic field at the point where the radius of symmetry cuts the other half of the completed sphere is (7tk/a){log (3 + (Li.U.)

-2j, j, j

2y2) - y2}.

6. A current i (e.m.u.) flows in a wire in the form of an ellipse of axes 2a,

2b(a > b) and a current i' (e.m.u.) flows in an infinite, straight wire parallel to the minor axis of the elliptic wire, in the same plane with it, and at a distance d(> a) from its centre. The currents in the two wires are in the same sense at the points nearest together. Show that the mutual force between the two circuits is 47tii' (bja)

{ ( 1 - �)

-!

-

1} .

(S.U. )

7. A wire follows the circumference of a circle of radius a except for an

arc of angular length across which it follows the chord. The loop is suspended from a point opposite the mid-point of the chord so that its plane is perpendicular to a long, straight wire which passes through its centre. When the currents are I and I', show that the torque on the loop is 4II'a (sin 01: - 01: cos 01:) . (S.U.)

201:

18]

E L E CT R O -M A G N E T I S M

541

8. A circuit, carrying a current i, is in the form of a plane curve s sym­ metrical with respect to an axis OA in its plane. An infinitely long, straight wire carrying a current i', passes through 0 and is normal to the plane of the curve s. Show that the mechanical forces acting on the circuit reduce to a couple about OA of magnitude

4ii'

J

sin 0 cos cf> ds,

where 0 is the angle which the line drawn from 0 to a point P of s makes with OA , cf> is the angle the tangent at P makes with OP and the integral is taken along half the curve on one side of OA . Hence show that, in the case of a circular circuit of radius a when 0 is on its circumference, the couple is 2rtii'a. (S.U.) 9. Calculate the field due to a long, straight wire carrying a current by

application of the Biot-Savart Law. 10. By resolving along the normal to a line of force and integrating with respect to the arc length, prove that the equation of a line of force in a field due to currents i 1 i 2 , i,. flowing along n infinite parallel wires may be written in the form rli, r2i, rni,. = constant, where r 1 r2, rn are distances measured from the n wires. What is the mathematical relationship between this field and the electrostatic field due to line charges iv i 2, in per unit length on the wires? •

0

,

•

•

0

•

•

0

•

•

•

•

1 1 . A current circulates around a wire bent into the shape of a semi­ circle with its diameter. Show that at a point on a line through the centre perpendicular to the plane of the semi-circle the intensity makes an angle

cf>

with the line given by tan cf>

=

� tan <X, where 2<X is the

7t

angle subtended by the diameter at the point.

12. A particle, whose mass is m, and charge e, is projected in the plane z = 0 with speed v and is subject to a magnetic field whose intensity, H, is constant throughout space, and whose direction is that of the z-axis. Show that the particle remains in the plane z = 0, that its speed is constant, and that it describes a circle of radius mvcfeH. (S.U.) 13. An electron of mass m and charge -e is initially at rest at the origin of coordinates (x, y, z) . A constant electric field acts along the x-axis and a constant magnetic field along the z-axis. Show that the path of the electron is a cycloid. (D.U.)

14. Between two infinite plane conductors x = 0, x a there is a uniform electric field E parallel to the x-axis, and a uniform magnetic field H parallel to the z-axis, the axes being rectangular. Particles of mass m and charge e are emitted with negligible initial velocity from the con­ ductor. Assuming that space-charge effects may be neglected, show that none of the particles will reach the plane x = a if eaH2 > 2mc2E. (Li.U.) =

542

A COURSE IN APPLIED MATHEMATICS

[CH.

15. A particle of mass m and charge e e.m.u. is under the action of gravity and a uniform horizontal magnetic field of strength H e.m.u. It is released from rest at a height 2m2gfe 2H2 above a horizontal plane ; determine the position of the point where the particle first reaches this plane. (M.T.) 16. An iron ring in the shape of a toroid whose axis is a circle, has wound upon it turns of wire per unit length. If a current flows in the wire, by applying Ampere's Rule show that the magnetic intensity along the axis is

n

i

47tni.

17. An infinitely long, circular, cylindrical cavity is cut in an infinite medium of permeability fl.· A current flows along a straight wire in the cavity parallel to its axis. Show that the magnetic field in the cavity can be regarded as due to this current and a parallel image current flowing in the medium and that the field in the medium can be regarded as due to a current i" flowing along the given wire and a parallel current flowing in a wire along the axis of the cavity. Deduce that the wire in the cavity is attracted towards the nearest part of the cavity with a force 2 ([1. 1)i2 (fl. + 1)d per unit length, where d is the distance between i and its image in the medium. 18. A current i flowing in an infinite, straight wire is at a distance f from and parallel to the axis of an infinite circular cylinder of soft iron of permeability fl.· The radius of the cylinder is a (
i

i'

i'

-

i'

-

19. A circular coil of wire of radius a, self-inductance L and resistance R is rotated about a diameter in a uniform external magnetic field of strength H. The angular velocity of the coil is c:u , constant, and the axis of rotation is perpendicular to the direction of the magnetic field. Prove that the periodic part of the current induced in the coil is 1ta2Hw . (R sm wt - wL cos wt), R 2 + w 2L2

if at time t = 0 the plane of the coil is perpendicular to the direction of the field. Find the mean rate (j2R) at which heat is generated in the coil, and hence the mean couple that must be applied to the coil to maintain its motion. (M.T.) 20. A small magnet of moment m cos pt is placed along the axis of a circular wire of radius a, inductance L and resistance R, at a distance c from its centre.

1 8]

21.

22.

23.

24.

ELECTRO-MAGNETISM

543

Find the flux of magnetic induction through the wire and the current induced in it. Show that the mean rate of heat production in the wire is (M.T.) A condenser, of capacitance C and leakage conductance G, is con­ nected, while charged, to a coil, of inductance L and series resistance R. Show that the discharge is oscillatory if (RC + GL) 2 < 4LC(I + RG) . (D.U.) Two terminals A, B are connected by a wire of resistance 2R in series with a condenser of capacity If(Rw), and also by a second wire of resistance R and self-inductance 2Rfw. An alternating e.m.f. E cos wt is applied between A and B . Show that, in the steady state, the total current flowing between A and B is E(3 cos wt + sin wt)f5R. (D.U.) An alternating voltage E cos nt is applied at t = 0 to a circuit of negligible resistance and consisting of an inductance L and capacity C in series. If n2 = I fLC, and if the initial current and initial charge are zero, show that the current for t > 0 is (Ef2nL) ( sin nt + nt cos nt) . (Li.U.) Two circuits, each containing a condenser and a self-inductance, are placed near together so as to have a mutual inductance M . Their resistances are negligible. Show that the system has two real periods of free electric oscillation, and that if the circuits are moved so as to increase M, the longer period increases and the shorter period decreases. (M.T.)

25. Two square circuits, each of side a, are placed in parallel planes at distance a apart, so as to form the boundaries of opposite faces of a cube. Prove that the coefficient of mutual induction is 2 + y2 (M.T.) Sa log 1 3 + I + y3 - 2y2 . + V 26. Two equal, circular loops of radius a lie opposite each other, a distance c apart. Show that the coefficient of mutual induction is cos e M �2 21ta2 2a2 - 2a2 cos O ) t dO . (c2 + 0

(

=

)

12,.

If c is large, deduce that M12 27t2a4fc3 ; and find the force of attraction for unit currents flowing in the same direction round the coils. (Hint : The P.E. of one loop in the field of the other is - M 1 2i 1i 2, (Li.U.) where i l> i 2 are the currents.) 27. State Maxwell's field equations, and give the constitutive relations valid for a non-ferromagnetic homogeneous isotropic medium. If the charge density p is zero, show that the conduction current j satisfies the equation =

H

li44

A C O U R S E I N A P PL I E D M A T H E M A T I C S [en. 1 8] The region z � 0 is occupied by metal, in which a current flows parallel to the x-axis. The strength of the current on the surface z = 0 is IX cos wt. Neglecting the last term in the above equation, show that the solution satisfying the boundary conditions is

j = (f, 0, 0) , where and

f = IXe-pz cos (wt - pz) ,

(Li.U.)

p2 = 2m;f.1.wf c2•

28. By putting A

Va in Stokes' Theorem ( 1 8 . 4 1 ) , where a is an arbitrary constant vector, show that =

l

n

X grad V dS =

fa

V dr.

29. Show that for a plane electro-magnetic wave in a dielectric medium,

the :field energy is everywhere equally divided between the electric and magnetic :fields. Show that a possible

30. The plane z = 0 is occupied by a metal sheet. solution of Maxwell's equations is given by

E = A [cos (nt - =fc), 0, (IX/ �) sin (nt - rxxfc)]e-fJzfc, = A [O, - (nf�) sin (nt - IXXjc) , O] e-fJzfc,

H

for z > 0, and E = A [cos (nt - =fc) , 0, - (IX/�) sin (nt - IXXfc)]efizfc, H = A [O, (nf�) sin (nt - rxxfc) , O]efizfc,

for z < 0, where n, IX and � are real and 1X2 - �2 n2• Determine the current in the sheet that is necessary to maintain the oscillation. (M.T.) =

31. A plane wave is being propagated in the direction of the z-axis in a uniform isotropic conductor. Assuming E = (0, E11, 0) , H = (Hx, 0, 0) and that these quantities are independent of x and y, show that

� 1tcrE11 az c oEy f.l. oH,. = oz c Tt "

oHx =

(4

+

)

K oE• at '

Assuming a solution o f the form where

rx,

E11 = Ee-t:J.Z cos w (t - nzjc) , n, w are constants, show that n2 = !f.I. [(K2 + l 67t2cr2fw2) t + K], IX2

�

�;22 [(K2 + l 61t2cr2jw2) i - K] .

APPENDIX A TO PART III

UNITS IN a first course of electro-magnetic theory such as this book provides, since the experimental bases for the laws of attraction between point charges and poles are very familiar and easily comprehended, and since a similar law has already arisen in the theory of gravitation, these laws form the natural starting-point for the mathematical argument. By measuring electrical quantities in e.s.u. and magnetic quantities in e.m.u., these fundamental laws assume their simplest form, and for this reason such units have been employed throughout the preceding chapters. This mixed system of units is called the Gaussian system. If, however, it is desired to exhibit the mathematical theory as a logical structure arising from certain postulates, a greater degree of elegance is attained and a smoother and more rapid development achieved if, in the case of electrostatics, these postulates are taken to be : (i) that the circulation of E around any closed curve is zero (Conservation Law) ; (ii) that the flux of D out of a closed surface is 47t times the charge enclosed (Gauss's Theorem), and (iii) that D = KE. From this point of view, the occurrence of the factor 47t in Gauss's Theorem is artificial, and it can be eliminated by choosing as a new unit for D the old electrostatic unit multiplied by 47t. The differential form of Gauss's Theorem then becomes div D =

p

and the relationship between D and E is

E. D=K 47t Putting Kj47t

= K, we have

D = KE,

where K is the dielectric constant in our new system of units. In a vacuum, K = l and K = l/47t = 0 ·0796, i.e., 0 ·0796 is the dielectric constant of free space. Similarly, fundamental postulates for the theory of the magnetism due to steady currents may be taken to be : (i) that the circulation of H around a closed curve is 47t times the current enclosed (Ampere's Law) ; (ii) that the flux of B from a closed surface is always zero (Gauss's Theorem) , and (iii) that B = fLH. Again, it is natural to suppress the 47t by adopting a unit for H which is 47t times the old 545

546

A C O U R S E I N A P P L I E D MAT H E M A T I C S

electro-magnetic unit. becomes

Then the differential form of Ampere's Law

curl H = j. = 4rq.�.H = vH, where v = 4rtfJ.. In a vacuum, v = 4rt, i.e., the permeability of free space is now 4rt. In order that H shall still measure the force acting upon unit pole, the unit of pole strength must be decreased by a factor 1/4rt. The new system of units so obtained is called the rationalized Gaussian B

Also

system and, in terms of these units, Maxwell's equations take the form

= � (j + 88�) . 1 oB curi E = - - - ' at div D = p div B = 0.

curl H �

c

We also have the relationships D

= KE,

B

= vH,

j = O'E.

On the other hand, the laws of attraction between point charges and poles are more complicated. Thus, since K = 4rtK, equation (16.20) becomes

ee' . F = 4rtKr 2 Magnetic poles of strengths p, p' are of strengths pj4rt, p' /4rt in the old

units, and the force between them is accordingly given by P,-n ' _P'"'' -_ F16rt2fJ.r2 -- 4rtvr2 ' .r _

.r _

Gaussian units, rationalized or otherwise, are unsuitable for ordinary practical use on account of their magnitudes. Some improvement could be effected by basing a Gaussian system upon the m.k.s. system (see p. 22) instead of upon the c.g.s. system. None of the formul;:e derived in Chapters 14-18 would then require any modification, though c would have the value 3 X 108 m.sec.-1• However, a Gaussian system also suffers from the disadvantage that its units are not con­ sistent with one another, with the result that the constant c appears in Maxwell's equations. c is the mathematical expression of the in­ consistency between the electrostatic and electro-magnetic systems of units employed together in a Gaussian system. In a consistent system, c will be replaced by unity and Maxwell's Equations corre-

APPEND I X A

TO P A R T I I I

547

spondingly simplified. However, the adoption of such a system must further complicate one, at least, of the two laws of attraction. By international agreement, the practical system of units preferred by engineers and physicists is a rationalized and consistent m.k.s. system. Such a system is uniquely defined when the magnitude of any one unit has been specified. The unit of current is taken to be one­ tenth of an electro-magnetic unit and is called the ampere. The system so defined is known as the Giorgi system. We proceed to develop it in detail. The charge which flows past a point in a wire carrying a current of 1 amp in 1 sec is called the coulomb. An electrostatic field has unit intensity if a charge of 1 coulomb experiences a force of l newton. If 1 joule of work is done by the field forces when a charge of 1 coulomb is moved between two points, the P.D. between them is l volt. We prove, as before, that E = - grad V, and hence the unit of field in­ tensity is the volt m-1 . The farad is the capacitance of a condenser whose insulated plate receives a charge of 1 coulomb when its potential is raised by l volt above that of the earthed plate. Since the system is rationalized, the flux of D from a closed surface is equal to the charge enclosed. Thus the unit of flux of displacement is the coulomb and the unit of D is the coulomb m-2 • The dielectric constant K is determined by the equation D = KE. The unit of K is therefore coulomb m-2 volt m 1 Now Also

_

farad m_1 .

l coulomb = I0-1 e.m.u. = I0-1 c e.s.u. (c = 3 x 1010 approx.) 1 volt = 1 joule/coulomb = 107 erg/10- 1 c e.s.u. = l08c-1 e.s.u.

Thus the practical unit of field intensity is

l voltjm = l06c-1 e.s.u. The flux of displacement from a charge of l coulomb is also 1 coulomb in Giorgi units. This charge is I0-1c e.s.u., and hence the flux from it is 47tcl0-1 expressed in e.s.u. Thus 1 coulomb = 47tc10-1 e.s.u. of flux of D. For the Giorgi unit of D we therefore have 1 coulomb m-2 =

47tcl0-5 e.s.u. of D.

548

A C O U R S E I N A P P L I ED M A T H EMAT I C S

Since K

=

DjE, the practical unit of dielectric constant is 47tcl0-5 - 41tC21o-u e.s.u. 106c-1

In e.s.u. the dielectric constant of a vacuum is unity. Hence, in Giorgi units it is 1011/47tc2 = 8·854 X I0-1 2 farad m-1 . This quantity is generally denoted by K0• Consider two charges e, e' coulombs a distance r metres apart in a medium of dielectric constant K farads m-1 • Expressed in e.s.u., these quantities are I0-1 ce, I0-1 ce', l02r, 47tc2 10-11K. Hence, by equa­ tion (16.20) , the force between them in dynes is

10-2c2ee'

47t210-7Kr2· Expressed in newtons, the force is given by

F=

ee'

-

47tKr2

.

This is the form taken by Coulomb's Law in the Giorgi system. The ohm is the resistance of a wire through which a current of 1 amp passes when the p.d. between its ends is 1 volt. If the flux of magnetic induction through a circuit is changing and the e.m.f. induced is 1 volt, the rate of change of flux is taken to be 1 weberjsec. Thus the weber is the unit of magnetic flux and the weber m-2 is the unit of induction B. If a circuit carrying a current of 1 amp is threaded by a flux of 1 weber, its self-inductance is 1 henry. In a rationalized system the unit of H is chosen so that its circulation around a closed curve embracing a current is equal to the current multiplied by the number of turns the curve makes around the current. In the Giorgi system therefore, H is measured in amp-turn m-1 . Unit pole is now defined as that pole which, when placed in unit field, is acted upon by a force of 1 newton. This unit is therefore newton j oule j oule sec = volt sec = weber = = ' amp m-1 amp coulomb i.e., pole strength is measured in webers. The unit of permeability is so chosen that Thus it is

B = vH. weber _1 weber m-2 m = henry m-1 . = amp m-1 amp -o-

--

APPEN D I X A TO PART I I I

549

If the flux through a circuit is changing at the rate of l weber sec-1 , Hence the rate of change of flux is lOB e.m.u. sec-1 . We conclude that

l volt (= lOB e.m.u.) of e.m.f. are induced. l

weber

= lOB

e.m.u. of flux.

Hence the Giorgi unit of B is l

weber m-2

= 104

e.m.u.

Suppose that a current of l amp is flowing in a circuit. The circula­ tion of H around any closed curve linking the circuit once is l amp in Giorgi units. Now l amp = I0-1 e.m.u. of current, and hence this circulation is 4rd0-1 in e.m.u. Thus amp = 47tl0-1 e.m.u. of circulation of H. The Giorgi unit for H is accordingly l

3

l amp m-1 = 4d0- e.m.u. of H. Unit pole in the Giorgi system is

l newton 105 _ l _3 e.m.u. - - 10 B e.m.u. amp m 1 - 41t 10 41t The new unit of permeability is l 104 l weber m-2 7 _3 e.m.u. - -10 e.m.u. 1 1 amp m 41t 10 41t The permeability of a vacuum is l e.m.u. = 47tl0-7 henry m-1 • This is the permeability of free space, often denoted by f-lo· Let poles of strengths p, p' webers be r metres apart in a medium of permeability v henries m-1. Expressed in e.m.u. these quantities are IOBpj47t, IOBp' j47t, l02r, l07vj47t. The force between the poles is accordingly _

1

_

_

-

Or, in newtons, F =

PP '

47tVY 2'

-

This is the Giorgi form of the law of force between magnetic poles. Since the Giorgi 5ystem is rationalized and consistent, expressed in units from this system Maxwell's equations take the form curl H

=

.

an

J + at '

8B curl E = - at ' div D = p, div B = 0.

550

A C O U R S E I N A P P L I E D M A T H E M AT I C S

We also have the equations D = KE,

B = vH,

j = o-E.

In free space it will be found that E and H satisfy the wave equation ()2 \7 2 - Ko!lo ()t2" The velocity of wave propagation is accordingly 1 - c m sec-1 .v1Ko!lo - 100 as found employing the Gaussian system. We conclude this appendix by presenting a table listing the more common electro-mag'letic quantities, the names of their Giorgi units and the magnitudes of these units in e.s.u. and e.m.u. Giorgi unit

=e.s.u. X m

Capacitance Charge Current Dielectric constant Displacement Electric intensity Electric potential Flux of induction Inductance Magnetic induction Magnetic intensity Magnetic moment Magnetic potential Permeability

Farad Coulomb Ampere Farad m-1 Coulomb m-• Volt m-1 Volt Weber Henry Weber m-• Amp-turn m-1 Weber-m Amp-turn Henry m-1

l O-•c• l 0-1c l 0-1c 47T l 0-11 c2 47T lo- •c l O•c-1 l OBc-1 l OBc-1

Pole strength Resistance

Weber Ohm

Quantity

m

= e.m.u. x n n

IQ-9 l0-1 l 0-1

47Tl 0- 11 47T 10-•

10 6 lOB

!_ l 01 0c-1

lO B 109 104 47T l o-a 101 0 . __!_ 47T

_!_ l 07c-•

__!_ 107

l 0 9c2

l O•c-1

47T l 0-3c

47T 47T l0-1c 47T

lOBc-1 __!_ 47T l09c-•

47Tl 0-1

47T

l OB _!_ 47T 109

APPENDIX B TO PART Ill

PHYSICAL SIGNIFICANCE OF THE FUNDAMENTAL VECTORS IT has been shown in Chapter 16 that E and D are the electric intensity vectors at the centres of needle-shaped and disc-shaped cavities respectively. This will be true provided the dimensions of these cavities, though small by comparison with other dimensions occurring in a problem, are nevertheless large by comparison with the distances between adjacent molecules of dielectric. For, if this latter condition is not satisfied, the effect within the cavity of the dielectric cannot be approximated by a continuous Poisson distribution as we supposed, and the discrete nature of the charges within the dielectric must be taken into account. There will then be very rapid variations in the electric intensity even at the centre of the cavity, and the value of this quantity at any point can only be found when the precise positions and mag­ nitudes of the neighbouring point charges are known. In a larger cavity these rapid variations will be confined to a film, a few molecular diameters in thickness, extending over the walls of the cavity ; the field will be sensibly uniform at the centre. Similarly, the potential V, as defined in Section 16.2, is that at the centre of a cavity whose di­ mensions are large by comparison with the average intermolecular distance. Consider, then, a small point charge e, such as an electron, which is moved from a point A to another B in a dielectric. As it approaches and recedes from the discrete charges constituting the dielectric, the force acting upon it will vary rapidly, both in magnitude and direction. However, if it is moved from the centre of a small cavity at A to the centre of a small cavity at B and these cavities have dimensions which are large by comparison with the average intermolecular distance, since the actual field due to the discrete charges must still be conserva­ tive, the work done by the electric field is e(VA - VB) · Since E = -grad V (equation (16.8) ) , this is precisely the work which would be done if E were the actual field intensity within the dielectric. Thus E is a " smoothed-out " model of the actual field, giving accurate results in any problem where the discrepancy between the assumed continuous charge distribution and the actual discrete distribution can have little effect on the result. A single electron passing through a dielectric is not such a case. The electron may suffer a large deflection, not pre­ dicted by the Maxwell theory, if its path passes close to an atom within the dielectric. 551

552

A C O U R S E I N A P P L I E D MA1' H E M A T 1 C S

Similar remarks apply to the vectors H and B and to cavities within magnetic materials. There is an additional point to be made in this case, however. If a unit magnetic pole is taken once around a closed contour in a region containing permanent magnets and magnetic materials, but no currents, the work done by the field is generally taken to be

T H·

dr =

0,

since H is the gradient of a single-valued potential function. If, however, the permanent magnetism is assumed to be caused by circulating charges within the atoms of the magnetic material and the contour threads these elemental current circuits, the work done is, by Ampere's Law, not zero but 47t times the sum of the currents linked. It may be proved that this sum is

and hence the work done is

47t T I · dr = T H

·

dr

+ 47t T I · dr = T B . dr.

In these circumstances it will be correct to regard B, rather than H, as the " smoothed-out " field. If, therefore, the magnetic pole which is employed to explore a magnetic field is assumed to be capable of penetrating the atoms of a magnetic material, the vector B is correctly designated the magnetic intensity. If, on the other hand, we suppose its path to lie between these atoms but never through them, then H must be taken as the magnetic intensity. The second point of view seems the more reasonable for practical purposes, and we have adopted it in this book. If the former standpoint is adopted, by permitting a pole to circulate around a closed line of induction, energy can be abstracted from the field due to permanent magnets. Each time the pole pene­ trates a polarized atom, induction effects will reduce the internal energy of this atom by an amount equal to that by which the pole increases its KE. Such a method of taking energy from permanent magnets is not practicable, and consequently B can only be regarded as the field intensity in a very academic sense. The reader should be warned, however, that certain authors,* guided by the special theory of relativity, where E and B are comparable quantities, interpret B as the force per unit pole. Since the modern view is that a magnetic pole is an entirely fictitious entity, this disagreement between teachers is of no serious consequence. p.

* E.g., Electrodynamics, by A. Sommerfeld (Academic Press, New York, 1 952), 10.

PART I V HYD R O M EC H ANICS

CHAPTER 19

HYDROSTATICS 19.1. Pressure in a Fluid

The class of fluids is subdivided into liquids and gases. A liquid placed in a container occupies a definite volume and possesses an upper boundary surface. A gas, on the other hand, adjusts its volume to that of the container. A mixture of gases, such as the atmosphere, becomes more tenuous with height but has no precise upper boundary. Liquids are virtually incompressible, whereas gases yield quite appreciably to any change in the applied pressure. However, although all fluids are able to resist compressive forces to some extent, and hence can remain indefinitely in a state of compression, it is a fundamental property of both liquids and gases that they cannot support any shearing stress and remain in a state of equilibrium. If, at any instant, shearing forces operate across a small plane element drawn in the fluid, the fluid particles to one side of the element will slip in the plane of the element relative to the particles on the other side. This slipping will continue until all shearing stress has been eliminated, when the fluid can rest in equilibrium. Thus the action across any small plane element drawn in a fluid in equilibrium is always in the direction of the normal to the element. It is also a characteristic property of a fluid that this action can never be in the nature of a tensile stress, i.e., it is always compressive. Let us now adapt the analysis of stress carried out in Section 13.7 to the particular case of a fluid. The shearing components of stress at a point are all zero. Let "n • T22, "aa be the normal components of stress relative to rectangular axes Oxyz (these must all be negative). Let T be the stress across an element whose normal has direction cosines (l1, l2, l3) . Then, since T is normal to the element, its com­ ponents are (l1T, l2 T, l3T) . Equations (13.47) accordingly simplify to

or

T=

"u

=

"22 = "aa·

(19.1) (19.2)

Equations (19.2) imply that the stress across a small plane element at a point in a fluid is independent of its orientation, and hence we may allude to the stress at a point in a fluid without reference to any direction. This stress T will be negative. I T I = p is called the pressure at the point. 555

fili(J

A C O U R S E IN A P P L I E D M AT H E M A T I C S

[l'H.

19.2. Equilibrium of a Fluid in a Gravitational Field

Let Ox, Oy, Oz be rectangular cartesian axes constructed in a region occupied by a fluid and in which there exists a gravitational field. Suppose the fluid is in equilibrium under the attraction of the field. Let P be the point (x, y, z) (Fig. 19.1). Let p be the pressure and V the gravitational potential at P. Then p = p(x, y, z), V = V(x, y, z) . Let aA be a small plane area containing P and parallel to the yz-plane. z

(· )P

{)p'

- s"'-

FIG. 19.1 .-Fluid in Equilibrium under Gravity

Construct a right cylinder with aA as base and length ax. Consider the particles of fluid inside this cylinder. They are in equilibrium under the compressive forces exerted by the fluid outside the cylinder over its surface and the weight of the fluid inside the cylinder. The resolved parts of these forces in the direction of Ox must cancel. The forces exerted on the curved surface of the cylinder are normal to Ox and hence can be disregarded. The force acting on the plane end containing P will be paA , if aA is small. The force acting on the other plane end is similarly p 'M , where p' is the pressure at P' (x + ax, y, z) in this end. The volume of the cylinder is aA ax, and hence the mass of fluid inside it is p?>A ax approximately, where p is the density of the fluid at P. p also may be a function of (x, y, z) . The intensity of gravitational attraction at P in the direction Ox is - 8Vj8x. The component of the weight force in the x-direction is accordingly - p?>A ax8Vj8x. For equilibrium, we must have paA - p 'M - p?>A ax

��

=

0,

] 9]

H Y D R O STATICS

P

or But

-

av P, = p ox ax.

P ' = p(x + ax, y, z) = p to the first order, by Taylor's Theorem.

5{)7

.

+ �� ax,

(19.4)

Hence,

av ap - ox ax = p ox ax, av 1 ap = - ox p ax·

or

(19 . 3)

(19.5)

Similarly, we may prove that av - 8y

1 ap

p 8y'

av 1 ap - 8z = p az ·

(19 · 6)

Since - 8 Vj8x, - 8Vj8y, - 8 Vf8z are the three components of the vector grad V and, similarly, - 8pj8x, - 8pj8y, - 8pj8z are the com­ ponents of grad p, equations (19.5) and (19.6) may be summarized in the form 1 - grad V = grad p. (19.7) p -

It follows from equation (19.7) that the vectors grad p, grad V are parallel and hence that the normals at a point P to the equipotential and surface of constant pressure through the point are identical. This being true for all points P in the field, the family of equipotentials must be identical with the family of surfaces of constant pressure (isobars). Thus, in the case of a fluid in equilibrium in a uniform field, the isobars must be horizontal planes and, in the case of a liquid, its free surface (over which p = constant) is also horizontal. From equation (19.7) we also deduce that or i.e.,

curl(p grad V) = -curl grad p = 0, p curl grad V + grad p X grad V = 0, grad p X grad V = 0.

Hence the vectors grad p and grad V are parallel, and this implies, as before, that the surfaces of constant density coincide with the equipotentials. In the case of a uniform field therefore, the density must be constant over any horizontal plane, i.e., the fluid will be horizontally stratified. If p is a function of p alone, equation (19.7) can be integrated. In these circumstances we say that the fluid is barotropic. If, how­ ever, p depends upon other quantities (e.g., temperature) in addition

558

A C O U R S E I N A P PL I E D M A T H E M A T I C S

[CH.

baroclinic. In the case of a barotropic fluid, we consider the function J dpjp. It will be a function of p alone and hence of (x, y, z). We have _?_J dp !:___ ( dp ) op ! oP (19.8) ox p - dp J p • ox - p ox . _ o f dp Hence - oV ox - ox p' to the pressure, the fluid is

_

and similarly

_

oV o dp - oy - oy J -;;- · _

These three equations imply that

- oVoz - ozo J dp-;;- · _

_ oG - oG - oG ox oy oz 0, _

where

_

G = V + j dpjp. It follows that V + j d: = constant C.

(19.9)

(19. 10)

The simplest case is that of a homogeneous liquid, for which p is constant throughout. Equation (19.10) then takes the form

or

V + Ep = C, p = C - pV.

(19. 1 1 )

If the gravitational field is uniform, as it is, for example, over small regions close to the Earth's surface, taking the z-axis in the direction of the upward vertical we have V = gz, where g is the field intensity. Hence p = C - pgz. (19.12) If we take the datum plane z = 0 to be the surface of the liquid,

p = Po over this plane, where Po is the atmospheric pressure. Thus C = Po and (19. 13) P = Po - pgz. · It is clear that pressure increase is, in this case, proportional to depth in the liquid. (See Advanced Level Applied Mathematics, by C. G. Lambe, E.U.P., Chapters 10, 1 1 for a detailed discussion of this particular case.) Now consider the Earth's atmosphere. Upward currents cause the air to rise and its pressure to diminish. It expands therefore and its temperature falls. The rate at which the temperature falls

HYDROSTAT I C S

19]

559

with increasing height above the Earth's surface is approximately constant in the denser lower levels of the atmosphere (up to a height of about 10 km) and is called the temperature lapse rate. We shall denote its value by �(� = 6 ·5° C per km). Thus, if T0 is the tempera­ ture at the Earth's surface and T is the temperature at a height z, T = T0 - �z . .

(19.14)

The general equation of state of a homogeneous gas, which is in a condition far removed from liquefaction, is p = BpT,

(19. 15)

where T is the absolute temperature and B is a constant depending on the composition of the gas. We shall assume that this equation is valid for the gaseous mixture constituting the lower levels of the atmosphere. Air currents within this gas ensure that it is thoroughly mixed and therefore of uniform composition. We shall, however, suppose that these currents are not sufficiently intense to prevent our proceeding as though the atmosphere were in equilibrium. As a first approximation we shall suppose the air temperature to be independent of the height, i.e., we take T = Tm (a mean constant value) . Then p = pjBTm and hence

f dpp = BTm log p.

The gravitational potential at a height z is V = gz and consequently equation (19. 10) becomes in this case gz + BTm log p = C . .

(19.16)

Or, if p = P o when z = 0 (i.e., at the surface), then C = BTm log Po and thus (19. 17) where ). = gjBTm· In these circumstances therefore, the pressure is reduced exponentially with increasing height. More exactly, we may assume that the temperature varies with the height according to equation (19.14) . Then

p

p- B ( T0 - �z) and equation (19. 10) becomes gz + B

j (T0 - �z) 1 = constant. .

(19. 18)

GflO

A CO U R S E I N A P P LI E D

M A T H E M A T I C' S

[en.

Differentiating with respect to z, we obtain

p ( To - �z) B

dp = dz

- g,

gdz B (T0 - �z)'

or which, upon integration, yields log p =

When z = 0,

J� log (T0 - �z) + constant . .

p = Po and thus

constant = log p0 -

(19.19)

J� log T0•

Substituting for the constant, we find that

p = Po ( 1 - �) g/Bf3.

(19.20)

The standard atmosphere of the International Commission for Air Navigation assumes T0 = 288° A, � = 6 · 5 x w-s oc per em, g = 981 em. sec- 2 and B = 2·87 X 106, up to a height of 11 km. In this zone therefore p = Po (1 - 0·0226z) 5·2s, (19.21) where z is in km. At a height of 11 km, p = 0 · 222p 0 . Above this height it is assumed that the temperature remains constant at 216·5° A. Equation (19.17) is valid in this zone therefore and we have p = 0·222Poe-O·l58(z-ll) = 1 ·26Poe-0·15Bz, .

(19.22)

where, again, z is height above the Earth's surface in km. A quantity of liquid of uniform density p is in equilibrium in the form of a sphere of radius a under the mutual attractions of its particles. Show that the pressure at the centre is i1ryp2a2.

Example 1.

If r is the distance measured from the centre, we have already proved (equation (14.61)) that V = f-rryp(r2 - 3a2)

defines the potential distribution inside the sphere. From equation (19.11), we obtain = C f7Typ2(r2 - 3a2) = C' f7Typ2r2. But, when

Putting

p r = a, p = 0.

-

ff7Typ2a2. Hence i1ryp2(a2 - r2) .

Hence C'

=

p= r = 0, we obtain the pressure at the centre.

H l]

H Y D R O S TA T I C S

!)Ill

If the absolute temperature T at a height z is a given function f(z) of the height, show that the ratio of the pressures at two heights z1 and z2 is given by z log P = g f 'fdz ' P: -k , (z) z k being the constant in the equation p = kp T. As an aeroplane ascends the temperature and pressure are simultaneously recorded and a curve is drawn plotting the absolute temperature against the logarithm of the pressure. Prove that the height ascended between two readings is k X' - gf T dx, where x = log p. (M.T.) Assuming the gravitational field uniform, with the z-axis vertically up­ wards, we may take V = gz. Then, from equations ( 19.6), l dp = -g. p dz Since p = kpT, this may be written g 1 dp g (i) p dz = - k T = - kj(z) ·

Example 2.

x,

Integrating between the limits (z1, z2), we now obtain log If

�

=

- i{!1:)" z,

x = log p, equation (i) is equivalent to dx = - g ' dz kT � or dx -- _g'!_ y· Integration between the limits (xv x2) now yields the result stated. 19.3.

Relative Equilibrium of Rotating Fluids

If a fluid rotates around a central axis after the manner of a rigid body, so that there is no motion of its particles relative to one another, it is said to be in relative equilibrium. For example, if no air currents existed in the Earth's atmosphere, it would be rotating with this body and would be in a state of relative equilibrium. If we introduce a centrifugal force, which is supposed to act upon each particle of the fluid, the rotational motion may be neglected (see Section 1 1 .4) and the problem treated as a statical one. The centrifugal force acting upon a fluid particle P of mass m distant r from the axis of rotation is mw 2r, where w is the angular velocity. If PN (Fig. 19.2) is the perpendicular on to the axis of rotation, this force acts along NP. In addition to the gravitational field, therefore, we have introduced a centrifugal field whose intensity is w2r at P in the direction NP. Consider the field having a potential

562

A C O U R S E IN A P P LI E D M AT H EMA T I C S

function V = - foo2r2• The component of field intensity direction NP is

m

[CR.

the

(19.23) V has zero derivative in any direction perpendicular to NP, and hence the component of field intensity in such a direction vanishes. Ac­ cordingly, the centrifugal field may be regarded as having been derived from this potential function.

FIG. 19.2.-Centrifugal Force on a Fluid Particle

If Vg is the potential function of the gravitational field, the equili­ brium condition (1 9. 10) yields the equation Vg - lw2r2 +

J d:

=

C. .

(19 . 24)

Thus consider a liquid of uniform density p rotating with angular velocity oo about the z-axis (assumed vertically upwards) in a uniform field g. Equation (19.24) yields gz - fw2(x2 + y2) + P. p

At the liquid surface, p

=

=

C

.

.

(19.25)

p0, the atmospheric pressure. Thus

x2 + y2 •

=

2g z + constant . . (l) 2

(19.26)

This is the equation of the liquid surface, which is seen to be a para­ boloid of revolution. Any other surface of equal pressure is of the same type. Suppose that a spherical attracting body, such as the Earth, is surrounded by an ocean of density p, the whole rotating in relative equilibrium with angular velocity oo. The gravitational potential within the ocean, if we neglect its own attraction, is -yM/R, where

1 9]

563

HYDROSTATICS

M is the mass of the spherical core and R is distance measured from its Equation (19.24) is therefore

centre.

At the free surface, p

=

� - Y:: - !ch2

=

C .

(19.27)

.

0 and hence this has equation (19.28)

This surface is approximately an oblate spheroid. Let a be its equa­ torial and b its polar radius. At the poles r = 0, R = b. Hence C = yMjb. At the equator r = R = a. Hence -

y� + iw2a2

or where

=

Y�,

(19.29)

is the intensity of the field at the surface of the solid core. is termed the ellipticity of the free surface. Equation (19.29) shows that this quantity is very nearly equal to aw2j2g. For the Earth, this result gives a value 1 · 73 X 10-3 for the ellipticity, whereas, in fact, it is about 3 ·3 X 10-3 . The discrepancy is explained by the fact that the solid core is not a sphere in this case, being itself distorted from a spherical form by the Earth's rotation. g

(a - b)ja

A container, in the form of a right circular cylinder of radius a, is filled with gas. It is set rotating about its axis with angular velocity w. When the gas is in equilibrium relative to the container and its temperature has returned to its original value, show that the resultant force on either cir­ cular end has been increased by a fraction 1 ;\ (e 1 - A) 2 2 where A = w a f2k and k is the con­ stant in the equation of state p = kp. (Gravity should be neglected.) Since p = kp, Jd: = k log p,

Example 3.

>.

-

and equation (19.24) takes the form k log p - iw2r2 = C. Before rotation commences, let the pressure be w = 0 and hence C = k log P. Thus =

p Pe"''r'/2k

.

P.

Then

p = P when (i)

564

A COURSE I N APPLIED MATHEMATICS

r dr

[cH.

r

Consider a circular element of one end of the container of inner radius and outer radius + (see diagram) . The pressure is uniform over this element and is given by equation (i) . The force on the element is accordingly

2rrPrew'r'f2kdr. 2rrPJarew'r'f2k dr = rra;P (e.\

The total force on the plane end is now seen to be 0

-

1),

where ,\ = w2a2f2k. The original force on this end was rra2P. The fractional increase is therefore as stated.

19.4.

Stability of Floating Bodies. Metacentres

Consider a body floating in equilibrium in a homogeneous liquid at rest in a uniform gravitational field (Fig. 19.3) . Two forces only act upon the body, its weight W acting vertically through the body's

----,r

- --: --- -- - - - :

t

- - - - - - - -

�.---

H T

Frc.

19.3.-Body Floating in Equilibrium

mass centre G and the upward thrust T of the liquid acting through the mass centre H of the displaced liquid (Archimedes' Principle, see Advanced Level Applied Mathematics, by C. G. Lambe, E.U.P., Section l l . l ) . For equilibrium i t i s necessary that T = W and that they should act along a common vertical line. But T is equal to the weight of the liquid displaced. It follows that a floating body displaces its own weight of liquid. Also G and H must lie on the same vertical line. H is called the centre of buoyancy. To assess the stability of the equilibrium, consider first the effect of a small translation of the body without rotation. Such a displacement can be resolved into a horizontal and a vertical component. Clearly the force system is in no way affected by a horizontal displacement and the body remains in equilibrium, showing no tendency to return to its original position. The equilibrium is therefore neutral relative to such a displacement. If, however, the body is free to move in a

19j

HYDROSTATICS

565

vertical direction, any increase in its depth of submersion will increase the weight of liquid displaced, and hence T will assume a value greater than W. There will accordingly be a resultant upward force tending to return the body to its equilibrium condition. Similarly, if the body is displaced in the opposite sense, a resultant downwards force will be generated, and this also will drive the body back towards equilibrium. Relative to vertical displacements, :floating equilibrium is therefore stable. The effect of rotation of the body upon the force system is more complex. We shall assume that the body is symmetrical about a vertical plane and only consider the effect of a small rotation about an axis perpendicular to this plane. Such a rotation is equivalent to an equal rotation about any other parallel axis, followed by a translation. That the translation does not cause instability, has already been shown, and we shall consequently neglect it. The position of the axis of rota­ tion may therefore be chosen arbitrarily, and we shall take __::o+---�....;-----rA'' x it to be in the liquid surface _cA� and such that the volume of liquid displaced is unaltered by the small rotation. Let Ox, Oy be rectangular axes in the liquid surface, Ox being in the plane of symmetry and Oy the axis of rotation (Fig. 19.4). When the body is tilted through an angle e about Oy, let BB' be the liquid surface and suppose that the centre of buoyancy moves from H to H'. In the displaced position, the upward z thrust of the liquid will act FrG. 19.4.-The Metacentre through H' in a direction perpendicular to the new horizontal BB'. The line of action of this thrust will lie in the plane of symmetry and will intersect the line GH in a point M. The magnitude of the thrust will be equal to W, the body weight, since the quantity of liquid displaced is un­ changed. If M is above G as shown in Fig. 19.4, this thrust, together with the body weight W acting through the mass centre G, constitute a couple which will tend to rotate the body so that AA' moves into coinci­ dence with BB', thus restoring the equilibrium. If, however, M is below _ _ _ _ _

566

A C O U R S E IN A P P L I E D M A T H E M A T I C S

[CH.

G, the sense of the couple will be that tending to increase e. The posi­ tion of M relative to G accordingly determines the stability or otherwise of the equilibrium relative to the type of displacement we have been considering. The limiting position of M as 6 ---+ 0 is called the metacentre of the floating body and the height of M above G is the metacentric height. For stability, the metacentric height must be positive. Let drx be an element of the region rx of the xy-plane bounded by the water line in the equilibrium position. Let (x, y) be a point in drx. The increase in the volume of liquid displaced due to the rotation can be regarded as the sum of the volumes of a number of cylinders based on elements such as drx, a typical one being shaded in the figure. The height of this cylinder is xe, to the first order in e, and hence its volume is xe drx. The total increase in volume is therefore

1 X6 drx 6 1 X drx. =

IX

But this is to be zero, and hence

i x drx

IX

=

0.

(19.30)

Thus the sum of the moments of the elements of rx about Oy is zero, and it follows that the centroid of rx lies on Oy. In view of the sym­ metry, this centroid must be at 0. The axis of rotation must there­ fore be selected to pass though the centroid of rx. Let Oz be a third axis of coordinates perpendicular to AA '. Relative to the three axes, let H have coordinates (p, 0, q) and H' coordinates (p', 0, q'). Let V be the volume of liquid displaced and p its density. H is the mass centre of the liquid displaced by the body when in the equilibrium position. Hence, if gravity is supposed to act in the direc­ tion Oz, the sum of the moments about Oy of the particles of this displaced liquid is V pgp. When the body is in the displaced position, the liquid displaced has been altered by the addition of a wedge A 'OB' and subtraction of a wedge A OB. The effect is to increase the sum of moments about Oy by an amount pg

1 x26 drx. IX

Since H' is the mass centre of the new volume of displaced liquid, the new total moment must be Vpgp'. Hence Vpgp'

or

=

Vpgp + pg

J x26 drx, IX

(19.31)

1 9]

H YD ROSTATICS

567

Now suppose that gravity acts in the direction A A '. The per­ pendicular distance from the xy-plane of the mass centre of the elemental cylinder based on drx is txe approximately. The contribution to the sum of moments about Oy of this element is accordingly pgxe drx x txe !pgx262drx. The change in the sum of moments due to the rotation is therefore =

ipg62

i x2 drx,

•

(19.32)

and the equation corresponding to equation (19.31) is

62 q - q = -2V I

1 X2 drx. ex

(19. 33)

But the right-hand member of this equation is zero to the first order in e. Hence q ' = q to this order of approximation and the displace­ ment of H to H' is parallel to A A ' . It now follows from equation (19.31) that

(19.34) where K is the radius of gyration of the area rx about the axis of rotation. Thus, since HH' = HM . e to the firs(order,

HM

=

;

rx 2.

( 19.35 )

Equation (19.35) determines the position of the metacentre and hence the stability of the equilibrium. For stability, it is necessary that

(19.36) The couple generated by the displacement has moment

W.GM sin e

.

=

W.GM.e .

(19.37)

approximately, since W equals the weight of liquid displaced. In the particular case of a solid of revolution floating with its axis vertical, if a is the radius of the section of the solid in the liquid surface, then rxK2 = !1ta4 and the condition for stability is 7ta4

4V

> GH.

(19.38)

A floating body consists of a light, circular cylinder of radius a and height 2a, with two equal masses fastened at points P, Q on the axis of the cylinder, where P is equidistant from the plane faces of the cylinder. The cylinder floats with its axis vertical and with a vertical height of 3af2 immersed. Show that this is stable if and only if Q is below P and QP > af6. (Le.U.)

Example 4.

568

A C O U R S E I N APPLIED MATHEMAT I C S

[cH .

Let Q be a distance below P. Then G, the mass centre of the body, is distant from P. Hence OG H, the CM of the liquid displaced is midway between the liquid surface and the base of the cylinder. Thus, if 0 is the centre of the base, OH Thus

x

!x

GH

=

= a - tx. = !-a. OG - OH = (a - tx) - !-a = !a - tx. '- Cl I I I

-1-�=-- �p -r- -f =� _

t_

.:X:

In this case, V (19.3S), we obtain

= 37Ta3J2.

},Q

- -tQ

]�

a 1

I I

Hence, for stability, employing condition

ia > !a - !x, x > aj6. or Example 5. A solid is in the form of a uniform right circular cone whose height is equal to the radius of its base. Show that it can float in stable equilibrium in water with its axis vertical and vertex downwards if 1 < Sa < S, where a is its specific gravity. A particle of weight >.Wja is attached to the vertex of the cone, whose weight is W. .Tf Sa < 1, show that the system can float in stable equilibrium with the vertex of the cone downwards provided that (!a)! - a < >. < 1 - a. (S.U.) Let r be the radius of the section of the cone in the water surface. Then r is the height of the conical volume of displaced water and its mass is !J7TY3w, where w is the density of water. The mass of the cone is !t"a3 aw. Since the cone displaces its own weight of water, }1rr3wg = }1ra3awg, r = aat. i.e., (i) If G is the CM of the cone, OG = !-a. If H is the CM of the displaced water, OH = !-r. Hence GH = !-(a - r). The volume of displaced water being }1rr3, condition (19.3S) for stability takes the form 7TY4 arrr > 4(a - r), i.e 2r > a. .•

48

Jl

.

19J

H YDROSTATICS

569

I n view of equation (i), this inequality i s equivalent to

2al > 1, Sa > l.

or

But a < 1, for otherwise the cone would sink. Hence

1 < Sa < S .

0

If

W is the weight of the cone, W = }1ra3awg. If a particle of weight J..Wfa is attached to 0, the total weight becomes }1ra3wg(,\ + a) . This is equal to the weight of water displaced. Hence, if R is the pew radius of the surface section of the cone, }1rR3wg = }1ra3wg(J.. + a), or R = a(,\ + a)!. (ii) Taking moments about 0 to find the new position of the mass centre G, we obtain the equation or

( 1 + �) W. OG = W. £a, =

3a a. + a) But OH = £-R, and hence 3a a £R. GH = ,\ 4( + a) OG

4 (,\

We conclude that the condition for stability is

1rR4 3a tn-R" > 4(,\ + a) a - £R, 2R > >. +a "a.

or

Referring to equation (ii), we see that this implies that or

(,\ + a)� > !a, ,\ > (ia)t - a.

Since is positive, this inequality effectively limits ,\ only when its right­ hand member is positive, i.e., when Sa < l . I t is clearly necessary that < for otherwise the cone i s submerged. Hence, by equation (ii), (,\ + a)t < 1, <1 or The stated inequalities have now been obtained.

,\

R a, ,\

a.

570

A C O U R S E IN A P P L I E D M A T H E M AT I C S

[CH.

EXERCISE 1 9

1 . A certain atmosphere i s assumed t o b e divided into

horizontal layers of uniform temperature, the thickness and temperature of the ith layer being h;, T; (i = 1, 2, . . . The atmosphere obeys the perfect gas law p RpT. Prove that the pressure p. at the top is related to the pressure p6 at the bottom by the formula

n

n).

=

Pa = Pb exp ( - - Ln --'-h· ) · g Ri=

2.

(M.U.)

1 Tl

Prove that at a height z in the atmosphere (assumed to be of constant temperature throughout) the pressure is given by

p

=

p.e-gzflc,

where p. is the pressure at the surface of the Earth and k is a certain constant. A large, spherical balloon of radius r and total weight W floats with its centre at a height h above the surface of the Earth. Show that h is given by =

4rtP.k (r cosh IJ!.

� sinh

)

!J!. k g k 3. A volume V of liquid of density p occupies the region between two concentric spheres and encloses a volume v of gas. The pressure p in the gas is uniform and related to its volume by the equation pv k. Neglecting the mass of the gas, show that in the equilibrium state, v

eghflc

gW

-

·

(M.T.)

=

is given by

� aN = v(7v + V) (v + YP \j 7r

V)-! - 3v3.

4. A gravitating mass of gas is in an equilibrium configuration possessing spherical symmetry, and the pressure and density are everywhere related by p Kp1 + 1/n where K and n are constants. Prove that the =

gravitational potential, if adjusted to vanish at the free surface of the distribution, satisfies the equation

d2 + � d + v2A-n 0, 'I' dr2 r dr + 1) K}-n, G is the constant of gravitation, and r is =

where v2 the central distance. If = 1, show that the radius of the distribution is of its central density to its mean density is rt2/3. =

4rtG{(n

n

rt/v and the ratio (M.T.)

5. Fluid of total mass M and constant density p rotates with angular velocity w about a fixed axis Oz under the action only of a body force per unit mass which is w 2fe2 times the distance from 0 and directed towards 0, where e < 1 . Prove that the surfaces of constant pressure are oblate spheroids whose meridian sections are of eccentricity e. If the pressure vanishes at the boundary, show that the pressure at the centre is

{

}

t w2 9 pM2 ( 1 - e2) 2 2rt2

4e 2

·

(L.U.)

1 9]

HYDROSTATICS

fi7 l

6. A tank in the form of a cube of side 2a is completely filled with homo­ geneous liquid and is set rotating with angular velocity w about a vertical axis through its centre perpendicular to two of its faces. Show that the rotation causes the depth of the C.P. of a vertical side of the tank to decrease by a fraction k/(1 + 4k), where k = aw2f6g.

7. A right circular cylindrical container with plane ends of radius a is just filled with liquid of density p. It is then set rotating with angular velocity w about its axis, which is horizontal. Show that when the liquid is in relative equilibrium, the surfaces of equal pressure are co-axial cylinders with their common axis parallel to the axis of rotation and distant gfw2 from it. Show also that the total thrust on one of the plane ends of the container is

a3 (g + !-aw2) .

8. A uniform cylinder, of weight W, height h, radius (hy6)/4 and S.G. cr, floats in water with its axis vertical. Show that the equilibrium is unstable if 1 < 4cr < 3. Investigate whether, if cr = t, the equi­ librium would be made stable by the addition of a particle of weight iW to the lowest point of the axis. (S.U.) 9. A uniform, thin, hollow, circular cylinder without ends made of material of S.G. s floats in water with its axis vertical. Show that the equilibrium will be stable if the ratio of height to radius of the cylinder (S.U.) is less than 1/y(s - s2) .

1 0. A uniform solid figure of density p is formed by fitting together the plane bases, each of radius a, of a hemisphere and circular cone of height a. The figure floats freely in water with its axis of symmetry vertical and the conical portion downwards. If the equilibrium is stable and no part of the hemisphere is immersed, prove that

(S.U.) t > P > tma. 1 1 . The curved surface of a uniform solid of revolution is formed by the rotation of a parabola of latus rectum 4a about its axis, and it has a plane base perpendicular to the axis. If the solid floats in homo­ geneous liquid with its axis vertical and plane base uppermost, prove that the distance between the centre of buoyancy and the metacentre is 2a. Prove also that, if the height of the solid is 6a, it floats in stable equilibrium in water if its S.G. s is such that 1 > s > ,t. (L.U . )

12. A solid, homogeneous, circular cylinder of radius r, height h and mass M floats in stable equilibrium with its axis vertical in a fluid of density p. When a particle of mass m is placed at the centre of the upper face the equilibrium just becomes unstable. Prove that

where k

=

k2r2 k(M + 2m) 2h2 M+m 1tr2h pj(M + m) .

-

1'

(L.U.)

13. A uniform cube of S.G. s floats in water with two of its faces vertical and one specified edge above the water, the other three horizontal edges being immersed ; show that, if s lies between if and ·!, there are

572

A C O U R S E IN A P P LI E D M A T H E M A T I C S

[en. 1 9]

three positions of equilibrium. Prove that, if s = fie. the cube can float with two of its faces inclined to the vertical at an angle tan-1 1 · 4. (M.T.) 14. A uniform solid of revolution floats with its axis vertical in liquid of uniform density. Define the metacentre and find its distance from the centre of buoyancy. Determine the shape of the solid if this distance is independent of the relative density, assuming that the surface of the solid has a unique tangent plane at each point. (M.T.) 15. A slice of a uniform, solid sphere is bounded by parallel planes. Prove that it will float in stable equilibrium with its plane surfaces horizontal in any liquid of greater density. Prove also that, if the density of the liquid is twice that of the solid, the metacentric height is the same whichever plane surface is uppermost. (M . T.)

CHAPTER 20

ONE-DIMENSIONAL MOTION OF AN IDEAL FLUID 20.1.

Pressure in an Ideal Fluid in Motion

In the previous chapter we have shown that, under the supposition that shearing stress is absent from a fluid in equilibrium, a scalar quantity called the pressure can be defined at each point. When a real fluid is in motion, however, shearing stresses are present to a greater or lesser degree, depending upon the viscosity of the fluid. No real fluid is completely inviscid, but the viscosity of gases is compara­ tively small and, since the shearing forces are roughly proportional to the relative velocities between the fluid particles, the effects of viscosity in both liquids and gases can often be neglected provided the rate of flow is sufficiently small. A hypothetical, completely in­ viscid, fluid is called an ideal fluid. In this, and the succeeding chapter, we shall consider problems of fluid flow where the conditions are such that approximation by an ideal fluid yields satisfactory results. Equations (13.47) were obtained by consideration of the state of equilibrium of the matter inside a small tetrahedron drawn within the strained material. If this material is in motion, additional " mass X acceleration " terms figure in the original equations. However, the mass of the tetrahedron is a third-order quantity and hence, like the terms associated with the body forces, these additional terms vanish upon proceeding to the limit in which the dimensions of the tetra­ hedron all approach zero. As a consequence, equations (13.47) are valid for a strained medium in motion. It now follows, as in Section 19.1, that the stress across a plane element in an ideal fluid in motion is always normal to the element and independent of its orientation. Thus a scalar quantity, called the pressure, is defined at each point of such a fluid. This is no longer true if the fluid possesses viscosity, a fact which accounts for the comparative simplicity of the treatment of the flow of an ideal fluid. 20.2.

Equation of Continuity

Consider an ideal fluid flowing along a narrow pipe of varying cross­ section. If the rate of flow of fluid across any right section of the pipe is uniform over the section, the velocity of flow q at any point in the pipe can be expressed as a function of two variables, viz., the distance 573

574

A C O U R S E IN A P P L I E D M A T H E M A T I C S

[cH.

s, measured along the pipe, of the point in question from some fixed origin 0 taken in the pipe and the time t. Thus q

=

q(s, t) .

(20.1)

.

Similarly, if the density p is uniform over any right section, p

=

p (s, t)

.

(20.2)

.

Since q and p are dependentup on a single length s, the flow is said to be one-dimensional. Let et. be the cross-sectional area of the pipe at a, distance s from 0. Then et. = et.(s) (20.3) •

.

In time dt, the particles of fluid in the vicinity of et. move a distance q dt along the pipe. It follows that a volume et.q dt of fluid crosses et. in time dt, provided dt is small. The mass of fluid crossing et. is accordingly pet.q dt and the mass flow rate across et. is therefore pet.q. Now consider the region of the pipe between the two right sections distant s and s + as from 0 (Fig. 20.1). The rate of mass flow into

y.-

--

I I I

lis ------'1•�: I I

I I


FrG. 20. 1 .-Equation of Continuity

the region is pet.q. The rate of mass flow out of the region at the same instant t is the value taken by this quantity a distance as farther along the pipe, viz., a

pet.q + Fs (pet.q) Bs

to the first order in Bs. to be

(20.4)

The net rate of flow into the region is now seen a (pet.q) Bs. - as

(20.5)

20]

M O T I O N O F AN I D EA L F L U I D

575

The volume of the region being exas, to the first order in as, the mass of fluid it contains at time t is pexas. The rate at which this mass is in­ creasing is a op m (pexas) = exas 8t ' . (20.6) since ex and as do not vary with t. Expressions (20.5) and (20.6) denote the same quantity and hence op at

+ �1 Fsa (pexq) = o . .

(20.7)

Equation (20.7) expresses mathematically that there is no cavitation within the fluid, i.e., that the fluid distribution is everywhere continuous without cavities. It is therefore called the equation of continuity. In the particular case of a pipe of uniform cross-section, ex is a constant, and equation (20.7) reduces to op at

+ Fsa (pq) = 0 · ·

(20.8)

Again, if the flow is steady, i.e., q and p are independent of t, opjot = 0 and the equation of continuity is simply a

as (pexq)

pexq

or

=

=

0,

.

constant,

(20. 9) (20.10)

i.e. the rates of mass flow across all sections of the pipe are the same. If the fluid is a homogeneous liquid, p may be assumed constant. Thus opjot = 0 and the continuity equation is again 0

os

or or

(pexq) = 0, pexq = function t alone, exq = j(t) . .

(20. 11) (20.12)

If the liquid flow is steady, f(t) = constant, and equation (20.12) may be written (20. 13) exq = constant, . i.e., the rate of flow is inversely proportional to the cross-section. A pipe of circular cross-section and length l has a diameter which decreases uniformly with the distance from either end, being a minimum at its mid-point. Homogeneous liquid flows steadily through the pipe, v being its velocity of influx into one end. If k is the ratio of the smallest to the largest cross-sectional diameter respectively, show that the time taken for a liquid particle to traverse the pipe is

Example 1.

I

::v (1 + k + k2) .

576

A C O U R S E I N APPLI E D M A T H E M AT I C S

[CH.

the smallest diameter. Let be the distance Let be the largest and travelled along the pipe by a particular liquid particle at time t. The diameter of the pipe at distance s from the end at which the liquid enters is 2

d

kd

s

d - Ts (d - kd),

provided s < ![. The velocity of flow in this region of the pipe Hence, employing equation (20.13), we obtain

rrd2 [1 - T2s (1 - k)]' dtds = rrd2v,

or

r

d

1

1..----

s -----

is

dsjdt.

kci

Integrating over the range (0, ![) o! s, we find for the time the particle to traverse half the pipe's length 1 T = .!_ . = .!_ (1 + +

- k" 6v 1 - k 6v

T taken by

k k').

Doubling this result, we obtain the result stated.

20.3.

Euler's and Bernoulli's Equations

A fluid particle, distant s from a fixed point 0 in the pipe, has speed q(s, t) at time t. If a time dt now elapses, the distance moved by the particle along the pipe is q dt to the first order of small quantities. Its distance from 0 is accordingly s + q dt at time t + dt. At this point and time the speed of flow is q(s + q dt, t + dt) and, by Taylor' s theorem, 8q 8q q(s + q dt, t + dt) = q(s, t) + q dt + Ft dt, . (20.14) os again neglecting quantities of the second order. The speed of the particle has therefore increased by an amount

( q 8qos + 8t8q) dt,

(20.15)

20]

M O T I O N O F AN I D E A L F L U I D

577

in the time dt. Hence, the rate of increase of the particle's speed at the point s and time t is given by (20.16) But the component of acceleration in the direction of a particle' s motion is equal t o the rate of increase of its speed (Section 1 .4) . It follows that f is the component of the particle's acceleration in a direction along the pipe's axis. It will be observed from equation (20. 16) that although the flow may be steady so that oqjot 0, the fluid particles may still experience an acceleration. This must, of course, occur as they move from regions of low flow velocity to regions where this quantity is large. Consider the fluid particles which lie instantaneously, at time t, in a right cylinder whose plane ends, both of area A , lie one in each of the cross-sections of the pipe distant s and s + as from 0 (Fig. 20.2) . The external forces acting upon this system of particles comprise : =

�----- Ss ------��

FIG. 20.2.-Equation of Motion

(i) the thrust of the fluid surrounding the cylinder, and (ii) the weight of the particles. Let p = p (s, t) be the fluid pressure in the pipe at times t distant s from 0. The fluid pressure over the cross-section distant s + as from 0 is then op (20. 17) as, . p (s + as, t) = p(s, t) + os to the first order. Since the fluid thrust over the curved surface of the cylinder is everywhere normal to the longitudinal axis of the pipe, the resultant fluid thrust in the direction of this axis is

(

pA - p +

)

op op as A = - A as . . os os

(20.18)

578

A C O U R S E IN A P P L I E D M A T H E M AT I C S

[CH.

Let g., be the component of gravitational attraction in the direction of the axis of the pipe. The mass of fluid inside the cylinder is Hence, the component of the weight force in this direction is and the resultant external force along the axis of the pipe is

pA as. pAg.,as

( pg., - �f ) Aas. .

(20.19)

But, neglecting first-order terms, all particles within the cylinder have the same acceleration given by equation (20. 16) . To this order of approximation therefore, this will also be the component of the ac­ celeration of the CM of the fluid particles in the direction of the flow. The product of this acceleration component and the total mass of the system is therefore aq aq (20.20) q at + OS

(

) pA as' .

correct to the first order of small quantities. We can now obtain an equation of motion by equating the expressions (20.19) and (20.20), viz., aq . (20.21) + q osaq = g" p1 ap at os

-

This is Euler's Equation of Motion for the fluid flow. In certain circumstances, Euler's equation may be integrated to yield an energy equation. Let V be the potential function of the gravi­ tational field in which the flow is taking place. Then the component of attraction along the pipe is given by (20.22) (See equation (14.34) .) Also, let rp be defined by the equation cP or

J q ds,

= -

aq, q= - · as ·

(20.23)

cfo is clearly arbitrary to the extent of a function of t. Any function rp satisfying equation (20.23) is called a velocity potential for the flow, since the flow velocity in the direction of s increasing may be derived from it in the same manner that we derive the component of gravita­ tional attraction in this direction from V. We shall investigate the properties and use of the velocity potential in greater detail in Chapter 21.

20]

M O T I O N O F AN I D E A L F L U I D

579

Finally, as at equation (19.8) , we may prove that, provided p and p are functionally related, !!__ dp = ! op (20.24) os p os ' p

J

and then Euler's equation may be written

- oV - _!_ J �P,p � ( !q2 - �� + f d:) 0.

o2cf> _ (.lq2) - osot + _!!_ OS 2

=

OS

OS

v+

or

We now conclude that tq2

- �� +

v+

J d:

=

= C(t) ' .

(20.25)

(20.26)

where C(t) is an arbitrary function of the time. This is Bernoulli's Equation . If the flow is steady, q is independent of t, and hence a velocity potential having the same property can be found. Equation (20.26) then reduces to (20.27)

where C is independent of t, since the left-hand member of this equation is clearly so. In the case of a homogeneous liquid in steady flow along a pipe, p is constant and Bernoulli's Equation becomes fq2 + v + p_ = c, p p = K - fpq2 - pV,

or

(20.28)

where K pC is a new constant. It will be observed that the pressure decreases as the velocity of flow increases. In the case of a liquid, the velocity of flow may be so great that p is theoretically negative. Since a negative pressure cannot be maintained in a fluid, this would imply that cavitation was taking place, and the assumption of continuity within the liquid medium would no longer be justified. =

20.4.

Flow of a Homogeneous Liquid

In this section we will solve a number of problems relating to the flow of a liquid, assumed homogeneous and incompressible. The fact that the pressure in such a liquid and its velocity of flow are related provides us with a means of measuring flow velocities in a pipe by observation of the liquid pressure. A convenient method is to introduce a constriction into the pipe through which the liquid is

[CH.

A COURSE I N APPLIE D MATHEMATICS

580

flowing, as shown in Fig. 20.3. The velocity of flow increases as the liquid passes through the throat, and there is a consequent fall in the pressure . . The difference f),.p between the pressure in the pipe at P and the pressure in the throat at T is measured by means of a differential manometer containing a quantity of mercury and into which the liquid can flow until the mercury is in a state of equilibrium. The complete apparatus is called a Venturi Meter. -

-

-

-----

p

-

-

-

-

-

-

-

-

T

-+-

- -

-

-

-

-

- -

-

-

- - - -f h

_ _ _

1

FIG. 20.3.-Venturi Meter

If h is the difference between the levels of the two ends of the mercury filament and d is the density of mercury, f),.p = dhg, provided we can neglect the weight of liquid in the two arms of the U-tube (a correction is easily made if greater accuracy is desired). Let p 0, q0 be the pressure and velocity of flow at P and p 1, q1 the values taken by these quantities at T. Neglecting the difference in level between P and T, and assuming the flow to be steady, from equation (20.28) we obtain the results Po = K - ! Pq02 - p V, PI = K - t Pq12 - pV, V being the gravitational potential at P or T. Subtraction yields (20.29) f),.p = f p(q12 - q02) . But the continuity equation (20. 13) shows that if rx0 , rx1 are the pipe cross-sections at P and T, then rx0q0 = rx1qp Equation (20.29) is accordingly equivalent to the result qo =

J---,2p(r2 1) -,-f),.p-' -

(20.30)

where r = rx0frx1 is the ratio of the cross-sectional areas of the pipe and throat. Equation (20.30) gives q0 when f),.p has been measured and the constant r of the meter is known.

581

M O T I O N O F AN I D E A L F L U I D

20]

The rate of mass flow through the pipe is p rxoqo = rxorxl

2 p l2 Joco2p!1 oc

(20.31)

-

Two problems in which the flow is not steady will now be solved. A straight pipe, of uniform cross-section, is bent to form a right angle and fixed with one arm vertically upwards. A valve is provided in the horizontal section distant l from the bend. The valve is closed and liquid of density p is poured into the vertical portion of the tube until it fills the pipe from the valve 0 to a height l in the vertical portion. The valve is now opened, and after a time t has elapsed the level of the liquid in the vertical tube has fallen a distance x. Show that x = l(l - cos wt), where w2 = gf2l. Calculate the pressure at aU points in the liquid at this instant.

Example 2.

The cross-section of the pipe being constant, the equation of continuity (20. 12) shows { ----t� that q is a function of t alone. We deduce from equation (20.23) that a velocity potential is given by .p = -qs. But clearly, q = x, and hence application of equation (20.26) yields !x2 + sx + V + 1!. p

=

(i)

C(t),

where s will be measured along the pipe from 0, the initial level of the liquid in the vertical section. In this section, V = -gs, whereas in the horizontal section, V -gl. Let Po be the atmospheric pressure. Then at time t, the boundary conditions are p = Po at s = x and s = x + 21. Thus, substituting these conditions in equation (i), we find that =

tx• + xi - gx + f'! p

=

tx• + (x + 2l)i - gl + 'b. p

Subtracting, we obtain or

(ii)

C(t), =

C (t) .

2li = g(l - x),

i + w2x = w2l.

The solution to this equation, satisfying the initial conditions x at t = 0, is easily shown to be X = l(l - cos wt), as stated.

(iii) =

x

=

0

A C O U R S E I N A P P L I E D M A T H E MA T I C S

582

[CH.

Eliminating learn that

C(t) from equation (i) by substitution from equation (ii), we (s - x)x + V + gx + p� (p - Po) = 0. (iv) But x is given in terms of x by equation (iii) and accordingly equation (iv) is equivalent to P = Po - p[w2(l - x)(s - x) + V + gx]. In the vertical section, V = -gs and the pressure at any point is therefore given by P = Po + pw2(s - x)(l + x). In the horizontal section, V = - gl and the pressure equation is P = Po + pw2(l - x)(2l + x - s). Example 3. A horizontal tube is in the shape of a cone of small apex angle. Liquid of density p extends from the vertex for a distance c along the tube. A small hole is bored in the tube at the vertex and compressed air is pumped into the tube, the pressure at time t being P. The pressure at the other liquid surface is zero. If x is the distance of the nearer liquid surface S from the vertex at this instant, show that x x• 2x ( 1 - -:_) x' x + ( a - �x' + x'4 ) x• = ��. p where x'3 = x3 + c3• Calculate the manner in which P must vary with x if S is to be uniformly accelerated and deduce that when x is large, P = pfc3f3x2, where f is the acceleration. Let cx0 be the cross-sectional area of the tube a unit distance from the vertex, and let be this area at a distance s from the vertex. Then cx(s) = cx 0s2• (i) At time t, let x, x' be the distances of the nearer and further liquid surfaces from the vertex respectively. Then the volume of liquid between these surfaces is !cx0(x'3 - x3). Since the initial volume is !cx 0c3, this leads to the ex

equation or When

s = x, q = x.

by equation (i) .

(ii) Hence the continuity equation •

(x) x. = x x. q = cxcx(s)· so ·

(20.12) shows that

.P = - Jqds = �x.

It now follows that a velocity potential is

We can now write down Bernoulli's Equation in the form

x• 2x 2s4x• - s x• - �s x + Pp = C(t). The boundary conditions are (i) when s = x, p = P, and (ii) when s = x', p = 0. Thus - � %2 - xx + � = C(t), x• - 2x ) . x2 ( 2x'4 ? X 2 - x' - C( t) . .•

X

_

Subtraction yields the equation given in the question.

20]

MOTION O F AN I D EAL FLUID

%2 = 2fx and this latter equation becomes

If the surface nearer to the vertex has uniform acceleration

p

2P

(

= 2'J•"x 4

-

�

x'

+

x• ) .

f,

583 x

=

f,

.x'4

x' from equation (ii), we can now show that p 4pf{1 - ( 1 + :;:) ( 1 + fat�} This equation determines the manner of variation of P with x. If x is so large that powers of cfx beyond the third can be neglected, expansion of the expression for P in terms of such powers indicates that P pfc3f3x2, approximately. Substituting for

=

=

20.5.

Compressible Flow. The Rocket Motor

Suppose that a gas is flowing along a pipe of varying cross-section and that its expansion is adiabatic, i.e., no heat is communicated from the gas to its surroundings, or vice versa. The relationship between p and is then p = k "�, and hence

p

p

f d:

=

y:1

�·

Bernoulli's equation for the flow is accordingly tq2 -

a� + v + _y_ p = C(t). at y- l p

(20 .32)

We will consider in detail the problem of the steady flow of a gas from a chamber, in which the pressure and density are maintained constant,

FIG. 20.4.-The Rocket Motor

through a Venturi tube into an atmosphere where the pressure is p0• When in steady operation, these conditions are approximated in a rocket motor. Fuel and oxidant are pumped into a combustion chamber C (Fig. 20 . 4) , where they combine with the liberation of heat. The products of combustion are therefore formed at a high pressure and escape into the atmosphere along the Venturi. The latter is normally a convergent-divergent passage with a throat T at which the cross-section is a minimum.

584

A C O U R S E IN A P P L I E D M A T H E M A T I C S

[cR.

Since the flow i s steady, equation (20.32) may be written y 2 �q 2 + y -_1 p_p _

=

C,

(20.33)

where we have omitted V, since the effect of gravity is negligible by comparison with the pressure generated in the combustion chamber. In this chamber q = 0. Let the state of the gas there be specified by p = P I · The constant C is now easily found, the equations p = and hence

Pv

q2

=

�

y-1

( PI

_

t . . p

PI

)

(20.34)

The velocity of sound in the gas in the combustion chamber is ci, where (20.35) (see Section 20.7, equation (20.89)). Also, if c is the local velocity of sound at a point in the Venturi, p c2 = y , (20.36)

p

.

Equation (20.34) can now be conveniently written (20.37) showing how the local velocity of sound varies with the velocity of flow. Equation (20.10) is the equation of continuity. Let s be measured along the Venturi from the combustion chamber. Differentiating the equation of continuity with respect to s, we obtain

! dp + ! � + !

p ds

q

dcx O ds ex ds - · _

(20.38)

Differentiation of equation (20.33) also yields

(20.39) and, since p

=

kp"�,

dp - (Y - 1 ) P.. 2 ds p p ds ·· !!:_ (P.) -

Thus

(20.40)

(20.41)

20]

M O T I O N O F AN I D E A L F L U I D

585

Eliminating dpfpds between equations (20.38) and (20.41), we find that

'!!!: - '!:_

(� 1 ) ds�-

q c2 -

ds

·

(20.42)

It follows from this equation that if, at some point, q < c, i.e., the flow velocity is subsonic, drxfds and dqfds are of opposite sign. This implies that to cause q to increase as we move in the direction of flow, rx must decrease. A convergent passage will accordingly cause an increase in the flow velocity when this is subsonic. On the other hand, if the flow is supersonic, i.e., q > c, a divergent passage must be employed to cause an increase in the flow velocity. If q = c, drxfds = 0. This condition is achieved in the throat of the Venturi. We conclude that the velocity of flow can equal the local velocity of sound only in the throat. It must not be supposed, however, that the flow velocity is always equal to c in such a region, for, if drxfds = 0, there is an alterna­ tive to q = c, viz., dqjds = 0. In the throat, therefore, the flow velocity is either a maximum or equal to that of sound. Employing equation (20.37) to eliminate c2 from equation (20.42), we obtain the equation (20.43) where (20.44) If the outflowing gas is permitted to expand into a vacuum to zero pressure, p = 0 at the end of the Venturi, and hence c = 0 there. It follows from equation (20.37) that q then attains its maximum possible value for the specified conditions in the combustion chamber and that this is a as defined by equation (20.44). Integrating both sides of equation (20.43) with respect to s, we calculate the following relationship between q and rx, viz., q(a2

_

]

q2)y - 1 =

A being the constant of integration.

Ajrx,

.

(20.45)

Equations (20.37) and (20.45) completely determine the flow. In certain circumstances A will be determined by the conditions at the Venturi exit. Suppose that at this point rx = rx0, p = Po (atmospheric pressure) . Then the value of p at this point can be found from the equation p = kp"'�, and hence the value of c can be derived. Equation (20.37) enables us to find the appropriate value of q and, if this is set into equation (20.45) with rx = rx0, the value of A appears to follow. However, a more detailed investigation reveals that such a procedure is not always satisfactory.

586

A C O U R S E IN A P P L I E D M A T H E M A T I C S

[CH.

If q is plotted against a for various values of A, a family of curves such as is shown in Fig. 20.5 (a) results. For all values of A , a takes its minimum value when

- '\/fFl Y+l _ - c1 '\J/2 Y+I' .

q - q* - a

-

(20.46)

As we proceed down the Venturi, let a first decrease from some large value to a minimum a* in the throat, increasing thereafter to a a �-------r--, I

'T

J

_ _

(a)

a r------, Y

(b)

FIG. 20.5.-Flow Velocity Characteristics

value a0 at the exit. If A = A 1 , the variation in q down the Venturi may be deduced from the appropriate graph by following the arc from P to Q and then back to R. It will be noted that the conditions corresponding to the arc QST of this curve cannot be realised in practice.

20]

M O T I O N O F AN I D E A L F L U I D

587

q takes its maximum value in the throat and, if q is plotted against s, a curve of the form shown in Fig. 20.5 (b) is obtained. If A = A 2 , we arrive at similar conclusions. If, however, A = A 3, conditions in the throat are represented by the point U and, as IX now begins to increase again, two alternatives present themselves. Either we may follow the upper arc U V, or we may retrace our steps along the lower arc UW. For this value of A , the curve in the qs-plane accordingly splits into two branches at the value of s corresponding to the throat. If A = A4 or As, IX cannot take the values it has in the vicinity of the throat, and hence these curves cannot represent conditions in this region. They do, however, lead to the curves in the qs-plane shown in dis­ continuous line. The exit velocities corresponding to the different values of A can be read from Fig. 20.5 (b) along the line XY. It will be noted that as A increases towards A 3 , the exit velocity also increases, and hence the value of c at the exit as given by equation (20.37) decreases. But �

1-�

c2 = ypfp = ykY p Y, and we can therefore deduce that the exit pressure must also decrease. If A = A 3 , suppose P o = Pc is the exit pressure appropriate to the lower of the two curves shown in Fig. 20.5 (b). For values of the exit pressure greater than Pc , conditions in the Venturi will be represented by a curve such as A = A 1 or A • 2 If, however, the exist pressure is reduced below the critical value Pc, there will, in general, be no curve which can represent the conditions throughout the Venturi and, at the same time, yield the correct pressure at the exit. It is found, in practice, that, in these circum­ stances, conditions between the Venturi entrance and the throat are given by the curve A = A 3 and that this curve will also be followed beyond the throat as far as some point I where the flow velocity q will change in a discontinuous manner to a value corresponding to the point ], after which the solution curve A = As will be followed to the exit. That curve A = As will be chosen which yields the specified pressure at the exit. The discontinuity in q is referred to as a shock wave.

When P o < Pc, therefore, a shock wave will usually exist in the mouth of the Venturi and conditions between the shock wave and the com­ bustion chamber will be quite independent of the exit pressure, being given in all such circumstances by the curve A = A 3 • In this region dqfds is positive, and it therefore follows from equation (20.42) that the velocity of flow is subsonic between the chamber and the throat and is supersonic between the throat and the shock wave. In the throat q = q* (equation (20.46)) and the velocity of flow is equal to the local velocity of sound. If P o > Pc, there is no shock wave. Also dqfds and d�Xjds are of opposite signs at all points in the Venturi. This implies that the ·

A C O U R S E IN A P P L I E D M AT H E M AT I CS

588

[CH.

flow is everywhere subsonic, with q taking its maximum value in the throat. By proper design of the Venturi, it can be arranged that the exit pressure on the curve A = A 3 is equal to the atmospheric pressure In this particular case there will be no shock wave. Equation (20.37), or its equivalent equation (20.34), is then valid over the whole region of flow. Thus the exit velocity, is given by the equation

Pc'

p0•

q0,

(P o),� 1], [1 P1 P1

�b

=

y- 1

(20.47)

P o/PI p0Yjp1Y. P o Pc' ,

= since This equation may, of course, also be employed when the whole flow is subsonic. If < the curve A = A 3 will be followed to the exit, the exhaust gases expanding to atmospheric pressure after leaving the Venturi. In these conditions also there is no shock wave.

A long pipe is in the form of a thin, hollow, circular cylinder of radius a fitted at one end with a conical diverging pipe of circular section. Gas obeying Boyle's law, pfp = k, flows steadily with speed v0 from the uniform pipe into the divergence. A ssuming the flow to be everywhere one-dimensional, show that, where the radius of the conical section is r, the speed of flow is equal to xv0, where

Example 4.

- 1 = A log (p.x),

x2

A = 2kfv0 2,

and

- - -

-

�

-

-

p. =

��

(rfa) 2 •

- - - - -

The equation of continuity (20.1 1) yields pxv01rr2 = p0v01ra2,

(Li.U.)

r

!

_ _ _

or pxr• = p0a2, where p0 is the density of the gas in the cylindrical pipe and p is its density in the conical pipe. If p0, p are the corresponding values of the pressure, Po/P = p 0 f p, and hence Since p

=

kp,

r:

=

k log p .

M O T I O N O F AN I D E AL F L U I D

20]

589

Thus Bernoulli's Equation for the fiow is + k log P0, + k log and hence 2k 1 2 log p� = ,\ log fLX.

!x2v02 x2

20.6.

p !v02 p =

-

=

Vo

Vibration oi Elastic Strings

Before proceeding to an investigation of the problem of the propa­ gation of sound waves along a narrow tube, we shall find it helpful to study the analogous phenomenon of the transmission of a disturbance along a stretched string. Suppose that the string is uniform, of mass m per unit length. We shall neglect gravity, so that when the string is stretched between two fixed points it lies in equilibrium along a straight line, the tension throughout its length having the constant value T. We shall confine T

FIG.

20.6.-Motion of an Element of a Stretched String

our attention to motions of the string in which each particle makes oscillations of small amplitude along a line perpendicular to the equilibrium position of the string. During such small transverse vibrations, the tension of the string will not change appreciably at any point, and the forces acting upon an element Ss will be as shown in Fig. 20.6. Taking an x-axis along the equilibrium line of the string and the y-axis in the direction of vibration, let a point P fixed on the string have coordinates (x, y) at time t. Then y will be a function of x and t. Let P, Q be the two ends of Ss and let the tangents to the string at these points make angles lj;, !f; + Slf; respectively with the x-axis. The acceleration of P is o2yfot2 in the direction Oy and, if we neglect quantities of the first order, this is also the acceleration of all particles of the element PQ. Hence, since mSx is the mass of this element (Sx being its unstretched length) , to the first order its equation of motion is

T sin (lj; + alj;)

-

T sin !f; max �� · . =

(20.48)

590

A COURSE I N APPLIED MATHEMAT I C S

[CH.

But, to the same order of approximation, T sin(if; + 3if;) = T sin if; cos alj; + T cos if; sin alj; = T sin if; + Taif; cos if;, and thus

(20.49)

or, in the limit,

(20.50)

Now, provided that if; is always small, as we shall further assume, if; � tan if; = ayjax. Also, cos if; � l . Equation (20.50) can ac­ cordingly be approximated by

where

a2y 1 o2y = ' ax2 C2 ot2

(20.51)

c2 = Tfm.

(20.52)

Equation (20.51) is the one-dimensional wave equation already encountered in Section 18. 1 1 (equation (18.131)). In that section it was shown that the general solution represented two waves proceeding in opposite directions along the x-axis with speed c. As at equation (18. 133), we shall write the general solution y = f(x - ct) + g(x + ct), .

(20.53)

where f and g are arbitrary functions determining the wave profiles. A particular case of the greatest importance is when the wave profile is a sine wave. By taking f(x) = a sin(�x + {>), .

(20.54)

we obtain a wave of this type proceeding in the positive sense along the x-axis. The maximum lateral displacement of any particle of the string is a, and this is termed the amplitude of the wave. When x increases by 2-rr/ �, the sine function varies through one complete cycle of its values. A complete wave (crest and trough) accordingly occupies a length 2rr:/� of the string, and this distance is referred to as the wavelength :\. Thus (20.55) As the wave proceeds along the string, the displacements experienced by the particles of the string are given by the equation y = f(x - ct) = a sin(�x - �ct + {>), (20.56)

20]

M O T I O N O F AN I D E A L F L U I D

591

It is clear from this equation that the particle at any particular point on the string, i.e., for a given x, executes SHM of angular frequency 2rtcj"A and amplitude a (Section 2.3(2)) . Thus, if f is the frequency of vibration of any particle of the string, c f = ). or ft. = c, .

(20.57)

a very fundamental equation in the theory of wave propagation. If w is the angular frequency of the vibration caused by the passage of the wave along the string, equation (20.56) may be written y = a sin [w(t - xfc) +

ct ,

]

(20.58)

where ct = rt - cp. It will be convenient in many calculations con­ cerning sinusoidal waves to regard the right-hand member of equation (20.58) as the imaginary part of the function i ae [wCt - x!c) + cxJ = A eiw (t- x!c),

where A

=

aei"'.

We shall then speak of the wave y=

A eiw (t- xic),

(20.59)

when only the imaginary part of this wave is intended. A sinusoidal wave proceeding in the opposite direction along the string will similarly be represented by the equation y=

B eiw (t + xlc)1

where, in general, B will be complex. I

A

I

=

aI

eirx

I =a

=

•

(20.60)

It should be noted that

amplitude of real wave.

(20.61)

The form of the functions f and g appearing in equation (20.53) in any particular case will be determined by the initial and boundary conditions of the problem. The initial conditions specify the state of every particle of the string at time t = 0, i.e., its displacement and velocity. The boundary conditions describe the circumstances pre­ vailing at each end of the string throughout the motion. It may be proved that, when the boundary and initial conditions of a problem have been specified, its solution is unique. It follows that if we are able, by any means, to obtain a solution of the form shown in equation ( 20.53) which satisfies these conditions, it must be the solution sought. We shall refer to this important conclusion as the Uniqueness Principle (cf. Section 15.6). Thus, suppose a wave is being propagated along a string and the disturbance arrives at an end where certain boundary conditions have LO be satisfied. It will be found that these conditions can always be satisfied by assuming that a second wave, proceeding in the opposite direction to that of the first wave, arises at the boundary. The first

592

A C O U R S E IN A P P L l E b M A T H E M A T I C S

[CH.

wave is called the incident wave and the second wave, to which it gives rise, is called the reflected wave. The displacement of any particle of the string will then be given by an equation of the form (20.53), in which the two terms of the right-hand member may be interpreted as the incident and reflected waves. If the profile of the reflected wave is so chosen that the boundary conditions are satisfied, it will follow from the principle of uniqueness that the problem has been solved. Two semi-infinite strings are joined at x = 0 and stretched along the x-axis under tension T. The string along the negative x-axis has mass p per unit length, while that along the positive x-axis has mass p' per unit length. A mass m is attached to the strings at x = 0. If simple harmonic waves of angular frequency w arrive from the negative x-direction, show that the reflection coefficient is (c' - c) 2 + w2m2c2c'2 /T2 (c' + c) 2 + w2m2c 2c'2/T2 ' Tfp'. (S.U.) where c• = Tfp, c'2

Example 5.

=

T

X

The velocity of wave propagation along the string of density p is given by c2 = Tfp. Let y = A eiw(t - xjc) represent the incident wave travelling along the negative x-axis. This will give rise at the junction to a reflected wave travelling along the negative x-axis in the negative sense. We shall assume that this waveform is sinusoidal and has the same angular frequency w as the incident wave. The justification for this assumption is simply that we are then able to satisfy the boundary conditions at x 0. Let the reflected wave be denoted by =

y

=

B eiw(l + xjc).

It is to be expected that a disturbance will also be generated in the part of the string x > 0, and we shall assume this is propagated in the positive sense and may be represented by y

=

Ceiw(t- xjc'),

Tfp' determines the velocity of propagation in this part of the where c'2 string. This is the transmitted wave. =

M O T I O N O F AN I D E A L FLU i D

20]

Thus, for x < 0,

593

= Aeiw(t - zjc) + Beiw(t + z/c), (i) y = Ceiw(t- xfc') , (ii) The boundary conditions to be satisfied at = 0 arise from (a) the continuity of y, (b) the equation of motion of m. Clearly, at x = 0, equations (i) and (ii) must yield the same value for y at all times t. Thus Aeiwt + Beiwt = Ceiwt, (iii) or A + B = C. (iv) m moves under the action of the tensions T of the two portions of the For x > 0,

y

x

string (see diagram) and its equation of motion is

m ( �;n = T sin rfr+ - T sin rfr_, 0

where the subscript 0 indicates that the quantity concerned is to be calculated at x 0. Assuming that rfr+ and rfr_ are small,

=

= (�) +; sin .p_ � tan .p_ = ( ��) sin rfr+ � tan rfr+

-o'

the subscripts and - 0 implying that the quantities to which they are attached are to be calculated for a value of x just greater than zero and just less than zero respectively. The equation of motion of is accordingly

+O

m

(v)

Upon substitution from equations (i) and (ii), this reduces to

or

-mw•Ceiwt = Teiwt ( - �c' c + 0c!A - �c B ) ' .mwc ) c. . A - B = ( c' + z7 c

(vi)

The boundary conditions (iii) and (v) should, of course, be applied, not to the complex waveforms (i) and (ii), but to the imaginary parts of these expressions. However, if the two complex members of equations (iii) and (v) are equal, so also are their imaginary parts as is required. Solving for from equations (iv) and (vi), we obtain

B

B

!A - c'c' +- cc +- imwcc'f imwcc'fT-· _

But, from equation (20.61), it follows that the ratio of the amplitudes of the reflected and incident waves is The ratio of the energies inherent in the two waves is equal to the ratio of the squares of their ampli­ tudes (see below) . Thus,

[B[/[A [ = [B/A[.

. = BA\2 = (c' - c)2 + m2w2c2c'2jT•' I (c' + c)• + m•w•c•c'•jT• This ratio is termed the reflection coefficient. .

ratw energ1es

594

A COURSE I N APPLIED MATHEMAT I C S

[CH.

20.6.

Consider again the string of which an element is shown in Fig. The KE of the element is and hence the KE of the length of string lying between the points and at time is

!m8xy2, x = x1 x = x2 , ) 2 dx. . K = -!m l ( &y &t

t

z

(20.62)

z,

The length of this portion of the string is

x2 - x1 + ! lz, ( oxoy) 2 dx, . (20.63) since &yj ox is a small quantity. But the equilibrium length of this part of the string is (x2 - x1), and hence the increase in length is 12 l"'' (a&xy ) 2 dx. The work which must be done against the tension T to cause this increase is given by z, &y 2 (20.64) V = !T l ( ) dx. . ox =

z,

z,

z,

Taking the PE of the string in its equilibrium position to be zero, V is the PE of the displaced string. The total energy associated with this section of the string is therefore given by

= tm l , ( at&y) 2 dx + ! T 1"'' ( ax&y ) 2 dx. . (20.65) In the case of a single progressive wave, i.e., y = f(x - ct) or y = g(x + ct), &y = 1 &y ox ± c &t' x

x,

E

x,

&y 2 T &y 2 &y 2 21 T ( ax ) - 2c2 ( at ) - 21m ( &t ) .

and hence

_

(20.66)

_

•

It follows that in this case the kinetic and potential energies are always equal. In particular, if the wave is sinusoidal so that is given by equation the total energy over a wavelength of the string is

y (20.56), 4rr2Ta2 [>. sm. 2 ( 2rr x 2rrc t + ) dx - 2rr-2 T a2. E � -T � (20.67) _

_

0

.J.. 'I'

_

A-

20]

M O T I O N O F AN I D E A L F L U I D

595

Thus the mean energy per unit length is 2rt2T ):2 a2 = 2rt2mf2a2,

(20.68)

having made use of equations (20.52) and (20.57). It will be noted that the energy is proportional to the square of the amplitude. Returning to the wave equation (20.51), instead of obtaining its general solution, we may seek particular solutions of the form y = XT,

(20.69)

where X is a function of x alone and T is a function of t alone. This manner of solution is said to be by separation of variables. Substitu­ tion for y from equation (20.69) into the wave equation yields, after division by XT, 1 d2X 1 d2T = (20.70) c2T di2' X dx2 The left-hand member of this equation is a function of x only, and the right-hand member is a function of t only. But the equation is to be valid for all values of x and t. This can be so only if each member of the equation is, in fact, the same constant. Denoting this constant by - n2, we derive the two equations d2X + n2X = 0, dx2

d2T c 2n2T - 0' dt2 +

of which the general solutions are

}

(20.72)

D sin net), .

(20.73)

X = A cos nx + B sin nx, T = C cos net + sin net.

D

Thus y = (A cos nx + B sin nx) (C cos net +

(20.71)

is a solution of the wave equation for all values of the constants n, A , B, C, D. This solution can also be written in the form y = (A cos nx + B sin nx) sin (net + Q:) ,

•

(20.74)

where Q: is also arbitrary. Suppose that the string is of length l and is anchored at each of the end points x = 0, x = l. The boundary conditions are then y = 0 for x = 0, x = l and for all t. The solution (20.74) satisfies these conditions if 0, B sin nl = 0 . . A (20.75) =

The second of these equations implies that either B is also zero, or sin nl = 0. The first alternative leads to the trivial solution where the

596

A C O U R S E IN A P P L I E D M A T H E M A T I C S

[CH.

string is stationary. sin nl = 0 if nl = krr, where k is a positive or negative integer. Thus n = krrfl, and a non-trivial solution satisfying the boundary conditions is krrct . krrx l + a . y = B SID -- SID -(20.76) l

.(

) .

When the motion of the string is in accordance with this equation all particles oscillate laterally with SHM at the same frequency and are either in phase with one another or in anti-phase. The amplitude of a particle's vibration depends upon its position in the string, being B sin krrxfl at a distance x from one end. This variation in the amplitude of vibration is shown in Fig. 20.7. These two diagrams

Frc.

20.7.-Normal Modes of Vibration of a String

show the appearance of a rapidly vibrating string in the cases k = l and 3. It will be observed that if k = 2, 3, etc., there are points on the string at which the amplitude of vibration is zero, i.e., the particles at these points are stationary. Such points are called nodes. The nodes occur at equal intervals along the string, being given by the equation krrx p -l- = prr, or x = k l . (20.77) where p = 0, l, 2, . . . k. The points on the string where the ampli­ tude of vibration takes its maximum value are midway between the nodes. Such points are called anti-nodes. A string which is set vibrating in accordance with equation (20.76) is said to be oscillating in its kth normal mode. The characteristic of

20]

M O T I O N O F AN I D E A L F L U I D

597

such a mode, that all particles of the oscillating system move in phase or in anti-phase with one another, has been observed already in relation to systems having but two degrees of freedom (Chapters 6 and 10) . The string, having an infinity of degrees of freedom, possesses also an infinity of normal modes. The appearance of the string oscillating in a normal mode (Fig. 20.7) is that of a wave profile which remains stationary relative to the string. This type of wave is termed a standing wave to distinguish it from the progressive wave. The form (20.53) of the solution of the wave equation is convenient when the problem involves progressive waves being propagated along a string of infinite length, whereas equation (20.76) is more suitable when the vibrations of a finite string are being considered. It should, however, be noted that the solution (20.76) can be expressed in the form (20.53) by employing a trigono­ metric identity relating the product of two sines to the difference of two cosines. The reader should carry out this transformation for himself.

If a string is loaded at its mid-point with a particle twice as heavy as the string, show that the particle can vibrate with a frequency f given by f = Trl�· where l is the length of the string and e is any root of the equation cot e = 26. (D.U.) Let rn be the mass per unit length of the string, so that the particle's mass is 2rnl. Let x = ± !l be the points at which the ends of the string are

Example 6.

anchored.

lm/

Suppose that the vibrations of the two portions of the string for which

x < 0 and x > 0 are determined by the equations y = (A 1 cos nx + B 1 sin nx) sin (net + ) y = (A 2 cos nx + B 2 sin nx) sin (net + ) respectively. The boundary conditions at x = ± !l are both y = 0 for all t. A , cos tnl - B 1 sin tnl = 0, A 2 cos tnl + B 2 sin tnl = 0. y is continuous at = 0. Hence a , a ,

;.·

Thus (i) (ii) (iii)

598

[CH.

A COURSE I N APPLIED MATHEMATICS Finally, since the equation of motion of the particle is

T being the string's tension, the following condition must also be satisfied, viz., (iv)

Addition of equations (i) and (ii) and use of equation (iv) show that A 1 cos = 1 sin

2

!nl 2nlA

!nl.

This implies that either A 1 = 0 or (v) cot = If A 1 = 0, the particle remains stationary and the string oscillates in a normal mode of the type previously discussed, with a node at = 0, i.e., the presence of the particle has no effect upon the motion. The alternative is that should satisfy equation (v) . Putting = this equation becomes cot = The frequency of the vibration is then

!nl nl.

x

n

6 26.

!nl 6,

nc = :;;;: 6c

2,;:

as required. In these circumstances, equations (i) and (ii) reveal that B 1 = A 1 cot = B2 = A 2 cot = The motion of the system is therefore given by the equations y = A (cos sin sin a) , < 0, sin sin a) , > 0, = A (cos where must satisfy equation (v) .

-

!nl nlA1, !nl -nlA1•

nx + nl nx) (net + x nx - nl nx) (net + x

n

Since the wave equation is linear, the sum of any number of its particular solutions is itself a solution. By addition of all solutions of the type (20. 76) , we therefore obtain a solution y =

(

)

� . krcx . krcct .f- Bk sm -- sm -- + a.k , k l l l

•

(20.78)

in which the Bk and rxk are arbitrary constants. Since each term of the right-hand member of this equation satisfies the boundary conditions for a string anchored at x = 0 and x = l, so also does this new solution. It may, in fact, be proved that this solution represents the most general motion of a string anchored in this fashion. Equation (20. 78) shows that any motion of a vibrating string can be represented as the superposition of a number of motions in the normal modes. The constituent of lowest frequency (i.e., k = l) is termed the fundamental, and all other constituents are called the harmonics. The frequency of the fundamental is c(2t, and the frequency of the kth harmonic is k times this value.

20]

M O T I O N O F AN I D E A L F L U I D

599

The KE of a string vibrating in accordance with equation (20.78) is given by k x + rxk Bk sin cos K=

k fml [ k� 1 7c ;

)J dx.

(k�ct

When we take the square of the series, the " cross-products " which arise will make no contribution to K, since

provided k' -=1= k.

11 sm krr:x . 0

•

-1-

sm

11 sin2 krr:x dx

But

0

and hence

-1-

k'rr:x d 1- x �

=

=

0,

tt

(20.79) If the kth harmonic of the vibration were alone excited in the string, the KE associated with it would equal the kth term in this latter series. Equation (20.79) therefore shows that the KE in any vibration of the string is the sum of the KEs associated with the various harmonics calculated as though these existed independently. We may arrive at a similar conclusion in respect of the PE of the string at any instant. Thus, we may prove that

rr:2c2m � k2Bk2 2 ( krr:ct + rxk) - -----;rr- k -=-1 •

sm

V-

-1-

•

(20.80)

The total energy is now seen to be

E = rr:24c2!m ki:= l k2Bk2

(20.81)

and is independent of t, as is to be expected with a conservative system. A uniform string of length l and density p has its ends fixed at the points (0, 0), (l, 0) . The string is drawn aside through a small distance {3 at a point distant b from the origin, and then released from rest. Show that at any subsequent time t the form of the string is given by

Example 'i.

2{312 � Ji• sm . ktrb . ktrx ktrct cos --' sm y = T l tr"b(l - b) k-=-1 T where c2 = Tfp, T being the tension in the string. Find the total energy of this vibrating string. (N.U.) At time t = 0, y is given by the equations y = {3xfb, 0 � x � b, (i) = {3(1 - x)J(l - b), b :::;: x � l, and oyjat = 0 for all x. We have to choose the constants Bk and (f.k in equation (20. 78) to suit these initial conditions.

I

}

600

[CH.

A C O U R S E IN APPL I E D M A T H E M AT I C S Now

(k c

� 1r t oy ..:., sm - cos - + ock ot = k � 1 Bk z z t

k1rc . k1rX

--

and hence, when t =

0,

)

� sm -1 cos IXk· Bk Qt = k2., �1 T

oy

u

Thus oyfot = 0 for all

k7TC . k1rX

X if

Bk cos OCk = 0 for all We shall reserve the values of the Bk as a means of satisfying the other initial condition and shall here take ock = Then, when t = � (ii) y = 2., Bk sm l . k�1 and the Bk must be so chosen that this equation is equivalent to the equations (i). This suggests that we expand the function (i) in a Fourier sine series of period which will be of the form (ii) . The Fourier coefficients Bk will then be given by

k.

!1r. . k7TX

0,

2t

Bk =

2t)o(1 y sm -k1rX -dx, •

l b . k1rxdx + l211 � {3(l - x) sm. k1rx dx, sm - -l21 {3x T b T b 2{3l2 I . k1rb = 1r2b(l - b) Ji2 sm T ' _

0

The solution to the problem is accordingly as stated in the question. in the form The total energy is obtained from equation "' sm E T' k 1 Ji2 _

20.7.

(20 .8I)

pc2f32ZS I . 2 k1rb .,2b2(l - b)2 �

The Propagation of Sound Waves along Pipes

In this section we shall discuss the propagation of a small disturbance in a uniform gas contained in a narrow tube. Employing the notation introduced in Section 20.2, consider the particles of gas which, in equilibrium, occupy the section of the tube distant s along the pipe from the origin. Let � = � (s t) be the small displacement from the equilibrium position experienced by these particles at time t. Then the velocity of the particles in the cross-section a (s) is given approxi­ mately by

,

(20.82) (this is actually the velocity in the section a (s + �). but the difference is a small quantity of the second order in �) . Neglecting quantities of the second order and gravity, Euler's equation of motion (20.21) reduces to (20.83)

20]

M O T I O N O F AN I D E A L F L U I D

601

In practice, the pressure variations are so rapid that the expansion of the gas during the passage of the disturbance is adiabatic. Thus

1 ap 1 dp op

p op

YPo op

y = p os = p dp os = i)2 Bs p02 os' .

(20.84)

where we have replaced p and p by the values taken by these quantities when there is no disturbance, viz., Po and p0, since opjos will be a small quantity of the first order. Thus, the equation of motion may be written (20.85) The equation of continuity is (20.7) . order, we shall write this equation op

or

at

Since q = of,jot is of the first

+ �1 8sa ( Pocx. otac, ) = 0,

since ex. is independent of t. Thus

implying that

:t [p + � :s (cx.f,)J = p

O,

+ E!1ex. uS� (cx.f,) = F(s),

.

(20.86)

where F(s) is an arbitrary function of s. Let us suppose that at some instant t = t0, before the disturbance has been initiated, the gas is in equilibrium. Then p = p0 and f, = 0 identically. Substituting in equation (20.86) , we find that F(s) = p0• Hence P = Po

[1 - � i (cx.f,)J

•

(20.87)

Substitution for p in equation (20.85) now yields

{1

where

}

azc, c2 asa � asa (cx.f,) ' atz c2 = yPo. . Po

(20.88)

(20.89)

602

A COURSE I N APPLIED MATHEMATICS

[CH.

In the case of a pipe of uniform cross-section, equation (20.88) reduces to ()2� ()2� - c2 (20.90) ot2 os2' i.e., the wave equation. The constant c, given by equation (20.89), is accordingly identified as the velocity of propagation of sound in a uniform tube. All the phenomena associated with the propagation of waves along a stretched string have their counterparts in relation to the vibration of a column of gas in a uniform pipe. For example, if the gas occupies an organ pipe of length l, stopped at both ends, equation (20.90) has to be solved under the boundary conditions � = 0 for s = 0 and s = l. This problem has already been encountered in the previous section, and the result indicates that the gas has an infinity of normal modes of vibration, with nodes equally spaced along the tube. Any vibration of the gas can be analysed into a fundamental component of frequency cj2l and harmonics having frequencies which are integral multiples of the fundamental frequency. Two notes are separated by an interval of an octave if the frequency of one is double that of the other. Thus, the second harmonic or the second normal mode has a pitch which is an octave above that of the fundamental. If the organ pipe is open at one end, it is usual to assume that the disturbance within the tube is not communicated appreciably to the atmosphere outside the tube so that p remains steady at the atmos­ pheric value of P o at this end. Thus p = p0 at this end and, from equation (20.87) , we deduce that 0 ( �) = 0 as C(

(20.91)

If C( = constant, this condition becomes

at this section.

0� - = 0. .

(20.92)

OS

Consider an organ pipe of length l stopped at one end and open at the other. The boundary conditions are : (i) at s = 0 , � = 0, and (ii) at s = l, o�jos = 0. A solution of the wave equation is � = (A cos ns + B sin ns) sin (net + e) .

(20.93)

The boundary conditions are satisfied by this solution if A = 0,

En cos nl = 0.

(20.94)

Thus nl = (k + -!)rt, k being integraL The appropriate kth normal mode is therefore given by ;

=

{

B sin (k + -!)

T} sin {(k + -!) -q! + } e

.

. (20.95)

20]

603

MOTION O F AN IDEAL FLUID

The general solution obeying the boundary conditions is obtained by summing over the normal modes, i.e., with respect to k (=0, I, 2, etc.). The frequency of the fundamental (k = 0) is observed to be cf4l, i.e., one-half of the frequency of the fundamental when both ends are stopped. It follows that if one end of a closed organ pipe be opened, the pitch is lowered by an octave. If it is required to calculate the instantaneous pressure at a point in the gas, we may proceed as follows. Since and are small,

(p - p 0)

very nearly.

p

(p - p0)

( ��t (P - Po)

P - Po =

Now

and, by equation (20.87) ,

(20.96)

(ddpp ) o = YPoPo = c2

(20.97) Hence

P - Po =

-c

2

a

;· Po OS

(20.98)

A column of gas of length l is trapped between two pistons, both of which are free to move in a pipe of uniform cross-section "' without friction. The external faces of the pistons are exposed to the atmosphere and the gas is at atmospheric pressure Po and of density P o · One piston is now forced to vibrate according to the equation g = a sin wt. If m is the mass of the other piston and c is the velocity of sound in the gas, show that, when steady conditions prevail, the amplitude of vibration of the second piston is

Example 8.

a

m� j(cos �c - p0c"' sin 0) . c

Let the first piston be at s solution of the wave equation g

=

=

0 and the second at s

A E =

Taking as a

(A cos ns + B sin ns) sin (net + E),

we have satisfied the boundary condition g Thus

= l.

sin (net + E)

=

=

a sin wt at s

a sin wt.

= 0, if

0, n = wfc and A = a. The solution now takes the form �

5 =

( a cos ws + B sm. ws ) sm. wt. c

c

At s = l, we have the equation of motion of the piston of mass m as the boundary condition, viz.,

604

A C O U R S E I N A P P L I E D MAT H E M AT I C S This yields the condition wl . wl p0C<:x a cos c + B Sill c = ;n;;;

whence

(

-a

•

[CH.

)

wl wl Sill �;- + B cos c ,

mw . -wlwl S ill + --- COS c c p 0ca � � _ B = ----� mw � a. . wl wl COS - - -- Sill -C c p0ca __ __

But, at s = l,

<:

s =

( a COS wl + B Sill. wl) Sill. wt,. -C

·C

a sin wt wl mw . wl COS - - - Sill p 0ca c c

proving the result. It should be noted that if the frequency of the impressed vibration is such that w satisfies the equation mw cot "!! = _ (i) p0ca,

c

the amplitude of vibration of the free piston is theoretically infinite. In such circumstances our assumption that the disturbance is everywhere small is no longer valid and the amplitude will be large but not infinite. The column of gas is then said to be in a state of resonance. If m is infinite, i.e., the second piston acts as a rigid stop, equation (i) possesses solutions "!!. = k7T' c

where k is an integer. The resonant frequencies are accordingly kcf2l, i.e., are equal to the frequencies of the normal modes of a pipe of length l stopped at both ends.

The problem of the transmission of sound waves along a pipe of variable section is of importance when designing horns and loud­ speakers. For reasons which will be explained presently, a horn of circular section whose radius increases exponentially with the distance along the axis of the horn transmits sound waves with only small accompanying distortion. Let r be the radius of the neck of the horn at which the source of sound is placed. The radius of the section at a distance s from this source will then be taken to be re•s. Thus a = rrr2e2•s and the equation of wave propagation (20.88) takes the form

l

82� 821', 81', + 2a = 8s2 os £:2 ot2 ' '

(20.99)

Solving this equation by the method of separation of variables, we put 1', = ST, where S and T are functions of s and t respectively. Thus I d25 + 2a dS = 1 d2T c2T dt2 ' ds S ds2

(

)

20]

M O T I O N O F AN I D E A L F L U I D

605

an equation which can be valid for all s and t only if both of its members are equal to the same constant -n2• Hence

d2S + 2a dS + n2S _ - 0' ds2 ds d2T dt2 + c2n2T = 0.

The general solutions of these equations can be written

S = e-as (A eims + Be-ims) ,}

(20. 100) + De-icne, where m2 = n2 - a2• In practice, we find that n is always much greater than a and hence m = n approximately. Thus, taking A = 0, C = I, D = 0 in equations (20.100), we obtain an approxi­ mate solution of equation (20.99) in the form T

= Ceicne

1;

=

Be - as +in(ce-s).

(20.101) Both the real and imaginary parts of this solution must satisfy equation (20.99) separately. Thus, Be-as sin n(ct - s) , 1; (20.102) =

is a possible solution and represents sinusoidal waves proceeding down the horn with velocity c. As these waves advance, their amplitude is attenuated by a factor e-as. The frequency of the note being emitted from the horn is ncj2-rr. It will be noted that neither the velocity of propagation nor the law of attenuation is dependent upon the pitch of the note being propagated. Since any note can be analysed into sinusoidal components of various frequencies, and each of these com­ ponents is propagated along the horn at the same speed and reduced in amplitude by the same factor, they will arrive at the horn' s exit in the correct relationship for recombination into an attenuated form of the same note. Thus, no distortion is introduced by this device. EXERCISE 20 1. A straight tube of uniform bore is bent into an angle 20((0 < ()( < !7':) and fixed with the internal bisector of the angle vertical and the vertex downwards. A column of liquid of density p and length l oscillates in the tube under gravity. Show that the motion is simple harmonic of period , -2l Ttv i cos ()(. If Po is the atmospheric pressure at the free surfaces, show that the pressure in the liquid at the bend is

where

x

Po + 2pgx l -

(

7) cos ()(,

is the distance of either free surface from the bend.

606

A C O U R S E I N A P P L I E D M AT H EMAT I C S

[CH,

2. A tube is in the form of a cone of small vertical angle. It is fixed, vertex downwards, with its axis vertical and filled with liquid. An orifice is now made by a cut at right angles to the axis, the vertex being removed. When the distance of the liquid surface from the original position of the vertex is x, show that the velocity of discharge from the orifice v satisfies the equation a3x

!/x (v2) - (x + a) (x2 + a2)v2 + 2gx4

where a is the distance of the cut from the vertex. that the pressure at the orifice is atmospheric.)

=

0,

(It may be assumed

3. A pump discharges incompressible ideal liquid of density p through a

narrow, horizontal pipe of uniform cross-section, and of length l. The end of the pipe is open to atmosphere (pressure p0) , and the velocity with which the liquid emerges is u = A + B cos wt, where A , B, w are constants (I B I < A) and t is the time. Find the greatest value of l if the pressure in the pipe is to remain positive everywhere, and at all times. If l exceeds this value, show that the pressure can still remain positive if, instead of using a uniform pipe the cross-sectional area of the pipe is arranged to increase linearly with the distance from the free end and find the required rate of increase. (B.U.)

4. Liquid is flowing steadily through a pipe whose two ends have cross­

sectional areas oci and oc 2 and are at depths d i and d 2 respectively below a certain horizontal plane. If the pressure remains constant along the whole length of the pipe, show that the cross-sectional area oc at a point in the pipe at a depth d below the reference plane is given by oc i 2oc 22 (di - d 2) oc2 2 ()( 22d2 + (oc 2 2 - oci2)d' ()(I d i =

5. A gas is in steady flow through a Venturi meter. The cross-sectional areas of the meter at the entrance and at the throat are oci, oc2 re­ spectively. The gas pressure in the entrance and the throat are found to be P I and p2 respectively. If the expansion is adiabatic according to the law p k pY and if the density of the gas in the entrance is P I• show that the velocity of flow into the meter is given by =

2 qi

�- f.J

y - 1 PI =

[1 _ (P 2) Y � 1] PI

(� )�(� r -

.

1

6. Show that the ratio of the pressure in the throat to the pressure in the combustion chamber for a rocket motor operating under supersonic conditions is

(.y-+-i ) 2

_)'

y - 1.

This is termed the critical pressure ratio.

Find its value when y

=

1 ·4.

M O T I O N O F AN I D E A L F L U I D

20]

607

7. Show that the rate of mass :flow from a rocket motor operating under supersonic conditions is

r�.* �[YP1P1 (y ! 1 )��iJ

8. In the case of a rocket motor operating in a vacuum show that, pro­ vided the Venturi is designed to allow expansion to zero pressure, the exhaust velocity is

2 'VIy -y 1

P

t . P1 The temperature in the combustion chamber of a V.2 rocket is 2000° K. and that the exhaust gas obeys the law p = Assuming that = RpT with R = 3 X 0 6, show that the maximum exhaust velocity is about 2 km sec-1.

y

1·4 1

9. A stretched, uniform, elastic string of length 7t has its ends fixed ; the

string is initially in the equilibrium position and is then struck so that its particles are given initial transverse velocities �(x), where �(x) = 0 for 0 � < i7t - a and for i7t + a < � 7t, = for i7t - a < < i7t + a, and where and a are positive constants. If the transverse displace­ ment y(x, t) at time t of a particle distant x from one end satisfies the equation

h

X

h

X

X

(J2y 82y = a2 2 2 8x ' 8t

where a is a positive constant, show that

y(x, t) =

4h I

��

(- ) sin (2n + 1) a sin (2n + 1)x sin (2n + 1)at. )2 1ta n = o (2n

(Le.U.)

1 0. I f I is the impulse of the blow referred to in the previous exercise, show that in the limit as a � 0, 2J ( - 1) n . . sm (2n + 1)x sm (2n + 1) at, y = 1ty ( Tm) n� 0 2n + 1 00

where T is the tension and

m

the mass per unit length of the string.

1 1 . A string of mass p per unit length is stretched with tension T between two fixed points distant l apart. Prove that if, when the string is at rest in its equilibrium position, it is struck at a distance of lfn from one end, where n is an integer, the nth harmonic is not excited. (Hint : Employ a method suggested by Exercises 9 and 1 0.) (S.U.)

12. An infinite string is in a state of tension. At t = 0 its displacement from the equilibrium line of the x-axis is given by y = �(x) and the instantaneous velocities of the points on it are given by y = ifJ (x) . Choose the general solution of the wave equation to suit these initial conditions, and hence show that the subsequent motion of the string is determined by the equation

y = H�(x - ct) + �(x K

+

1 ct) ] + 2c

fx

+ ct

:t -

ifJ (z)dz.

ct

608

A C O U R S E I N A :l? :P L i E b MA't H E M A 'r i C S

LCH.

1 3 . Two semi-infinite strings of masses p 1 and p2 per unit length respectively are joined at x = 0. A train of harmonic waves travelling along the string from the region x < 0 suffers partial reflection and partial transmission at the join of the strings. Prove that the sum of the squares of the amplitudes of the reflected and transmitted waves is equal to the square of the amplitude of the incident wave. (What is the significance of this result in respect of energy?) (N.U.) 14. A long string consists of two portions of different constant line densi­ and its end is anchored to a fixed ties. One portion is of length point. The string is stretched so that the velocities of propagation of transverse waves in the two portions are c (for that of length and c'. Harmonic waves of angular frequency w are being propagated along the portion of indefinite length towards the junction. Show that the amplitude of the waves reflected from this junction is equal to that of the incident waves and that the transmission coefficient at this junction is

a,

cc

a)

'/ ( c'2 cos2 �a + c2 sin2 �a) .

Show also that the amplitude of the vibration at the junction is

(

2 1 + �2 cot2 c2 c times the amplitude of the incident wave. 15 . A uniform string of line density p is stretched in equilibrium along the x-axis with one end anchored at the origin and the other end attached to a ring of mass m free to move on a smooth, straight wire fixed at right angles to the x-axis through the point x = l. An oscillatory force F sin wt is applied to the ring in the direction of the wire. Show that the steady-state vibration of the string is given by wx F sin sin wt c = wl . wl' pwc cos c mw2 sm c

y

wa)--t

-

where c has its usual significance. Discuss the case when w = cffl, where cot f = mflpl. Show that if the ring is at a node, the largest amplitude of vibration of the particles of the string is Ffpwc. 16. A string of length l is stretched between two fixed points with a tension T. An oscillatory force F sin wt is applied at its mid-point in a direc­ tion perpendicular to the string. Show that the steady-state vibrations of this point have amplitude Fe tan wl 2c ' 2Tw

17 . One end of a string of length l is fixed at A and the other end is fastened to the end B of a rod BC of length b which can turn freely about C. Show that the period of a principal transverse oscillation is 2rt/r�, where 1', is a root of the equation M1',2/3p - 1 /b = 1', cot l�,

M O T I O N O F AN I D E A L F L U I D

20]

609

p being the line density of the string, M the mass of the rod, and c (M.T.) the wave velocity for the given tension. 18. Find the periods of the normal modes of vibration of a tense string fixed at the ends. Prove that the period of the gravest mode is almost exactly nine-tenths of that of a simple pendulum whose length is equal to the sag in the middle (due to gravity) if the string be horizontal. If the string consist of two portions of lengths a1, a 2 , and different densities P v p 2, prove that the periods (2rcjp) are determined by the equation

provided

k l2 = P 2 PlfT, k 2 2 = P 2 P /T, 2 Examine the case of p 2 =

0, and explain how T being the tension. the resulting period-equation may be solved graphically. (M.T.) 19. A tube of uniform cross-section contains gas of density p. A diaphragm of mass M per unit area, divides the tube into two sections and is free to move longitudinally. Sound waves of frequency f are propagated along the tube with velocity c and are incident upon the diaphragm, where they give rise to transmitted and reflected waves. Show that the transmission and reflection coefficients are c2 p 2 �����= c 2 p2 + rc2f2M2

rc2f2M 2

and � ����= c 2 p 2 + rc2f2M2

respectively. 20. An organ pipe of uniform cross-sectional area (I( is stopped at one end. An airtight piston of mass m moves smoothly in the tube. Show that the frequency of a normal mode of oscillation of the gas trapped in the tube is f

=

ce

2rcl' where l is the length of the column of gas of density p and e satisfies the equation e tan e = lp(l(fm . 2 1 . A straight pipe P of uniform cross-section divides into two similar pipes P1 and P2 • The gas in P1 and P2 is at the same pressure as the gas in P but has double its density. Simple harmonic sound waves are travelling down P towards the junction, where they are partly transmitted and partly reflected. Show that the ratio of the ampli­ tudes of the reflected and transmitted trains is (\/2 - 1) /2. (Assume that the displacement � and pressure p are continuous at the junction.) 22. A uniform tube of length l contains gas of density p at atmospheric pressure. The tube is closed at one end and a piston of mass M, con­ trolled by a spring of stiffness fL, occupies the other end. Show that w is the angular frequency of a normal mode if tan

wl __ P� = c Mw 2 - [L .

23. A tube of length l is closed at one end and open at the other, and is filled with a gas of mean density p0• A pressure disturbance A p0 sin crt

610

A C O U R S E IN A P P L I E D M A T H E M A T I C S

[cH. 20]

is maintained at the open end by waves passing outside the tube. Prove that the velocity at any point within the tube is cos crt sin {cr(l - x)fc} u = A ' c cos (crlfc) where c2 = dpfdp evaluated for p = p0 and the origin is at the open end. Find how the pressure varies at the closed end. Explain the physical significance of the vanishing of the denominator for certain values of cr . (M.T.) 24. An infinite long, straight tube of unit cross-section contains gases of densities p , p' separated by a smoothly fitting piston of mass M. An harmonic train of sound waves of length 2rt/m is incident on the piston in the gas of density p. Show that in transmission beyond the piston the amplitude of the waves is changed in the ratio 2c : y{(cp + c'p') 2 + M2mzcz}, and the phase by tan-1 Mmcf(cp + c'p') , where c, c' are the velocities of sound in the two gases. (M.T.) 25. A long, straight tube of unit sectional area extends to infinity in one direction (taken as that of the z axis) and is closed at z = - l; at z = 0 there is a freely movable piston of mass m, on both sides of which there is air of density p. Waves in which the displacement is given by a sin K(z + ct) impinge on the positive side of the piston. Show that if the reflected waves are expressed by a1 sin { K(z - ct) + then a12 = a2 ; also show that if a1 is taken equal to a cot le: = cot Kl - Kmfp . (M.T.) 26. Sound waves are being propagated along a tube in which the cross­ sectional area ex depends upon the distance s along the tube according to the equation ex = ex0sn. Employing the method of separation of variables, show that the wave equation (20.88) has a solution of the form � = sl
e:},

(

)

CHAPTER 21

GENERAL MOTION OF A FLUID

21.1. Stream Tubes In this chapter we shall remove the restriction that the fluid motion is to be one-dimensional and will consider more general motions of an ideal fluid in a three-dimensional region. It is usual to describe the flow pattern by means of equations which specify the velocity offlow at a general point of the region and at any instant t. Thus, if rect­ angular axes are erected in the region, the components of the velocity of flow (u, v, w) in the directions of these axes are given in terms of the coordinates (x, y, z) of a point of the region and the time t by equations u = u(x, y, z, t), v = v(x, y, z, t), w = w(x, y , z, t) . (21.1) This is Euler's method of describing the motion. A pictorial representation of the flow is obtained by drawing stream­ lines through the region in which the flow is taking place. A stream­ line is defined to be a curve whose tangent at any point is in the direc­ tion of the flow at that point. If (x, y, z), (x + dx, y + dy, z + dz) are two neighbouring points on a streamline, direction ratios of the chord joining them are dx : dy : dz. But this chord lies along the tangent at (x, y, z), which, since it is in the direction of flow, has direc­ tion ratios (u, v, w). Accordingly dx dy dz (21.2) - = - = -· v w u equations which determine the family of streamlines at any given instant t. As t changes, the family of streamlines will, in general, alter. If, however, the motion is steady, (u, v, w) are independent of t and the streamline configuration does not change. In these circumstances the particles of fluid follow the streamlines, which are therefore also the particle trajectories. This identity between the streamlines and the particle trajectories does not hold if the flow pattern varies with the time. At a fixed rigid boundary surface the velocity of flow can have no component along the normal. The streamlines in this region of the fluid will all lie in the boundary surface therefore. If C is any closed contour drawn in the fluid, the streamlines inter­ secting C will form a surface called a stream tube (Fig. 2 1 . 1). The stream tube will enclose all the streamlines embraced by C. If the flow is steady, any stream tube, and the streamlines it contains, will 611

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A C O U R S E IN A P P L I E D MAT H E M A T I C S

[CH.

remain fixed in space and a fluid particle which is inside the tube at any instant moves along a streamline within the tube, and hence remains inside the tube throughout the motion. If, therefore, the stream tube were replaced by a smooth pipe, the forces acting upon the particles within the tube being unaltered, the flow within the tube would not be changed and the pressure distribution also would remain as before. It follows that, provided the stream tube is of small cross­ section, the flow within it can be analysed in the same manner that the flow of fluid in a narrow pipe was analysed in Chapter 20. In particular, since the flow is steady, the equation of continuity (20.10) is valid, a. being the cross-section of the stream tube at any point along it. - - - -

STREAM L I NE.

!>T R E A M TUBE

FIG. 21.1 .-Stream Tube

Again, by applying Bernoulli's Equation (20.27) to the flow within a stream tube of such small cross-section that it is effectively a stream­ line, we deduce that tq2 + V +

j d: = constant C,

.

(21 .3)

along any streamline, provided the flow is steady. The constant C will not be the same for all streamlines, however. This extension of the range of application of Bernoulli's Equation permits the solution of a number of problems concerning three­ dimensional flows. Thus consider a vessel in which the level of a liquid of density p is maintained constant as it issues from a horizontal orifice near the base. If h is the height of the liquid surface above the orifice, by applying Bernoulli's Equation at the two ends of a stream­ line extending from the surface to a point just outside the orifice, we obtain the result gh +

P._o p

=

C

=

tu2 + b, p

where u is the velocity of flow from the orifice, and it has been as­ sumed that the velocity of flow at the surface is negligible and that the pressure just outside the orifice is equal to the atmospheric pressure Thus u = v2gh, (21.4) which is Torricelli's Theorem.

Po·

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G E N E R A L M O T I O N O F A FLU I D

613

Suppose that a blunt-nosed body is fixed in a uniform stream of liquid of density p. The streamline configuration will assume the pattern shown in Fig. 21.2. The velocity of flow at the points 5 and 5 ' will b e zero. Such points are called stagnation points. Suppose that before the body was introduced the P =:=�;;;:._-4 stream was uniform, the velocity of flow being of constant magnitude u and FIG. 2 1 .2.-Body in a Uniform Stream steady in direction. Then the pressure also had a constant value P o at all points. At a point such as P, some distance away from the body, neither the pressure nor the velocity of flow will be disturbed from their values P o and u when the body is set in position. Let p, be the pressure at the stagna­ tion point 5. Applying Bernoulli's Equation at the two points P, 5 on the streamline P5, we find that tuz

= �!.!., + Po p p

m

where we have neglected any difference between the two points. Thus u ---• U

=

J� (p_,

-

gravitational potential

P0),

(21 .5)

:======�A

----- /.(

----____;;' D

..------� 1----- - - - -- - -

_

_

L

LEV E L

D I F F E RE N C E

FIG. 21.3.-Pitot-static Tube

and the stagnation pressure can be employed as a measure of the velocity of flow in the undisturbed stream. This is the principle of the Pitot Tube. A hollow tube is placed with its nose pointing into the stream whose velocity of flow is to be measured (Fig. 21.3). Liquid

614

A C O U R S E IN APPLI E D MATHEMA T I C S i

[CH.

flows into the tube and exerts a pressure Ps on one end of a column of mercury or other liquid in a vertical manometer tube. The other end of this column is subjected to a pressure P o via a channel which leads from the side of the pipe through which the liquid stream is flowing. The difference (Ps - Po) can be calculated from the difference between the levels of the two ends of the mercury column, and from this the velocity of flow may be deduced. An allowance will have to be made, of course, for the difference in the pressures exerted on the mercury column by the liquid heads A B and CD.

General Equation of Continuity. Irrotational Motion Let q be the velocity of flow at any point within an ideal fluid and

21.2.

let S be a fixed, closed surface in the fluid enclosing a region r. Con­ sider any element dS of S (Fig. 21 .4). The fluid crossing dS in time dt

dS

r

FIG. 21.4.-Fluid Flow across a Closed Surface

will occupy a cylinder of length q dt with its axis in the direction of q. Let n be the unit outward normal to dS and 6 the angle between n and q. The right section of the cylinder has area dS cos 6, and hence its volume is q dt dS cos 6 = q ndS dt. It follows that the mass of fluid crossing dS in time dt is pq • ndS dt, where p is the fluid density at dS, i.e., the rate of mass flow outwards across dS is pq ndS. We now deduce that the rate at which fluid mass is leaving the region r is ·

•

1 pq s

•

n dS .

(21.6)

But, if dv is any element of r, the mass of fluid occupying this element

2 1]

G E N E R AL M O T I O N O F A F L U I D

is p dv and the rate at which its mass is increasing is

615

� (p dv) = � dv.

Thus the net rate at which the mass inside r is increasing is

1 otop dv . .

(21. 7)

r

Now the rate at which the mass inside r is increasing must equal the rate of mass inflow across S. Hence

1 opot dv r

or, by Green's Theorem,

i.e.,

L �� dv L [��

= -

L div (pq) dv, div (pq)J dv 0.

=

+

( pq . n dS , )8

(21.8)

-

=

(21 .9)

This latter equation is true for all regions r, and this can be the case only if

�� + div (pq)

=

0.

(21. 10)

Equation (21.10) is the general equation of continuity, of which equation (20.7) is a special case. In the case of an incompressible liquid of uniform density p = constant and the equation of continuity reduces to div q

=

0.

.

(21. 11)

Consider the special case of uniform incompressible liquid streaming from a point source 0 into a region extending to infinity in all directions and occupied completely by liquid. Let I be the rate at which liquid mass is being introduced at 0. We shall assume that the lines of flow are straight lines directed radially outwards from 0. If a sphere of radius r and centre 0 be drawn in the liquid, the velocity of flow at all points of its surface takes a constant value q and the rate of mass flow across it is 4npr2q. But, since the fluid is incompressible, this must equal I. Thus 4npr2q = I, or

q=

I =m 1-2'

4npr2

(21.12)

where m = I/4np is termed the strength of the source. It will be noted that the velocity of flow varies with the distance from the source in a manner which is identical with that in which the electric field due to a

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A C O U R S E I N A P P L I E D MAT H EMAT I C S

[CH.

charge m varies. The lines of flow are also identical with the lines of force due to the point charge. It follows that the velocity of flow may be derived from a potential function rfo = mfr. We then have q = -grad rfo. . (21 . 13} rfo, of course, no longer has the physical significance it possessed in the theory of the electric field. It is termed the velocity potential. If m is negative, liquid streams into the point 0, which is accordingly referred to as a sink. The combination of a S
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G E N E RA L M O T I O N O F A F L U I D

617

i.e., cp i s a harmonic function. I t follows that the methods used for calculating the potential distribution in electric and magnetic fields are also available for determining the irrotational flow of a homo­ geneous liquid. Such a flow will take place in a region bounded by certain rigid surfaces which may or may not be in motion. At such a boundary the component of the velocity of flow normal to the surface, viz., - ocpfon, where of on denotes differentiation along the direction of the normal, must equal the velocity of the surface in this direction. In the next section we will illustrate the various methods of solution by means of specific problems.

21.3. Particular Cases of Irrotational Motion 21.3 (1). Fixed Sphere in a Uniform Stream Consider a uniform liquid extending to infinity in all directions and suppose all particles of the liquid have a velocity U in the same direction. This is a uniform stream. If Oz is an axis drawn in the direction of flow from an arbitrary point 0, the velocity potential is given by cp = - Uz, (21.18) since then - ocpfox = - ocpjoy = 0, - ocpfoz = U, give correctly the components of the flow velocity along rectangular axes Oxyz. Intro­ ducing spherical polar coordinates (r, e) relative to 0 as pole and Oz as reference line, cp = - Ur cos e . . (21.19) We now introduce a sphere of radius a with its centre fixed at 0 (Fig. 2 1.5) . Assuming that the motion remains irrotational, the new p

FIG. 21.5.-Sphere in a Uniform Stream

velocity potential must be a harmonic function which reduces to the right-hand member of equation (21.19) at infinity. We shall try A (2 1 . 20) cp = - Ur cos e + 2 cos e. r

618

A

C O U R S E I N A P P L I E D MATHEMAT I CS

[CH.

The radial component of the flow velocity is then a.p -" = u cos e ur

2A cos e. r

(21.21)

+ 3

This must be zero at all points on the spherical surface, and this will be the case if A is chosen so that

Hence A = -tUa3 and .p

= - U cos e ( r + ;:2 )

(21.22)

•

The radial and transverse components of the velocity of flow may now be computed. We find that q,

= u cos e ( 1 - �S ) = -� ar r

qo

=

I

-� �t = - U sin e ( ;;3 ) ·I ,

,

l +

(21 .23)

The differential equation specifying the lines of flow is then found in the same way that the corresponding equation of the lines of force due to a dipole was found on p. 382. Thus 1 dr fl!_ 2r3 - 2a3 cot e, r de qo 2rs + as 1 dr 2r3 + a3 dr -2 cot e = 3 - 3 = y 3r2 or r(r a ) de - a3 r de' Integrating with respect to e, we obtain

(3

Cr sm2 e _ - rs as' . -

•

)

(21.24)

where C is a constant. This is the polar equation of the lines of flow which are shown sketched in Fig. 21 .5. The problem of a sphere moving with velocity U through an infinite, homogeneous liquid may now be solved as a corollary to the result just found. We superpose a velocity U in a direction opposite to that of the stream upon the sphere and liquid. This brings the stream to rest at infinity and gives the sphere a velocity - U along Oz. The velocity of flow at all points is changed by the vector addition of a velocity U along Oz, and hence rp is modified by the addition of a term Ur cos e. Thus Ua3 cos e. .p = 2r2 -

2 1]

G E N E RA L M O T I O N O F A F L U I D

619

Or, if the sphere moves with velocity + U along Oz,

1>

= Ua3 cos 2r2 -

(2 1.25}

e.

21.3 (2} . Source Before an Infinite Plane Let m be the strength of a source at a point 0 distant a from an infinite plane barrier (Fig. 21.6) . The flow must be such that the flow velocity at the barrier is parallel to the plane. To ensure this, we introduce an image source of the same strength m at the point 0', where O'N = ON and 0'0 is perpendicular to the infinite plane.

m

o'

a

Cl

m

0

FIG. 21.6.-Source and Infinite Plane with Image

At any point P on the plane, the velocity of flow will be the resultant of the velocities due to the two sources. These velocities are equal in magnitude and are in the directions OP, 0' P respectively. The re­ sultant is accordingly in the direction PN bisecting the angle OPO'. The boundary condition being satisfied, the system of source and image provides us with the unique irrotational solution to our problem. The velocity of flow at any point can now be found as the resultant of the velocities due to the source and its image. The system of lines of flow is identical with the family of lines of force between two equal like charges. 21.3 (3). Line Source and Cylindrical Boundary Let the line source be of strength m and let it intersect a plane perpendicular to its length in A (Fig. 21.7). Suppose that the

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[CH.

axis of a parallel, infinite, circular cylinder meets the plane in a circle centre 0. Let A ' be the inverse point to A relative to the circle. We place an image line source at A ' of strength m and an image line sink at 0 of the same strength. If P is any point in the region out­ side the cylinder occupied by liquid, let the angles PAX, PA 'X, POX be e1, e2, e3 respectively.

FIG. 21.7.-Image System

Then, by an argument similar to that employed to establish equa­ tion (14.33), we can show that for all points P on a streamline mel + me2 - me3 = constant, el + e2 - e3

or

=

constant.

(21.26)

If P lies on the circular section of the cylinder, triangles POA ' , AOP are similar, and hence LOPA ' = L OA P, e2 - ea = 7t - el, el + e2 - ea = 7t.

i.e., or

•

(21.27)

It follows that the circle is a streamline and that the velocity of flow at the cylinder due to the source and its images is tangential to this surface. The boundary condition being satisfied, the flow so deter­ mined must be the unique irrotational solution. A two-dimensional source is parallel to the axis of a rigid cylinder of circular section. In a plane perpendicular to the axis, the source and axis of the cylinder are represented by the points A and 0 respectively; P is a point in this plane and is on the cylinder. Show that the speed of the liquid at P is proportional to !:>.OPA (AP2• (Li.U.)

Example 1.

2 1]

GENERAL MOTION O F A FLUID

621

Referring to the image system shown in Fig. 21 .7, the component of the velocity of flow at P in the direction perpendicular to OP is

�; sin LOPA + ��P sin LOPA '.

If P lies on the cylinder, LOPA ' = LOAP and OP. A 'P = AP ' OA Hence the velocity perpendicular to OP, i.e., the speed of the liquid at P OA . . 7fP Sill LOpA + A . Sill LOA P, P OP =

2m

�:

2m

p• OP Sm6 0PA

(OP . AP sin LOPA + OA . AP sin LOAP) ,

21.4. Euler's Dynamical Equations In Section 21.2 we calculated the equation of continuity from which possible patterns of flow of a fluid between specified boundaries can be derived. However, the forces which must be exerted upon the fluid particles in order that such motions shall in fact take place have not so far been alluded to. Thus the equation of continuity represents a kinematical condition. In this section we shall derive a dynamical equation involving the forces present in the system and from which the internal pressure distribution in the fluid at any point may be determined. Let x be any scalar or vector quantity which can be attributed to any point P within the fluid, e.g., flow velocity, pressure, temperature, etc. Let (x, y, z) be rectangular cartesian coordinates of P and let t be the instant at which x is measured. Then x = x (x. y, z, t) . When a further time dt elapses, the particle of fluid which was at P at time t, moves to a neighbouring point P'(x + u dt, y + v dt, z + w dt), where (u, v, w) are the components of the velocity of flow at P at time t. The value of x at P' is x (x + u dt, y + v dt, z + w dt, t + dt) ox ox ox ox u dt + o v dt + = x(x, y, z, t) + w dt + a dt. y t oz ox (21.28)

Thus, the change in x in time dt as the point at which it is measured moves with the fluid is given by dx

= =

( 8x + ox + oozx ( · grad + oOtx \) dt, ax

q

u

ay

x

v

o

x w + 8t

) dt,

(21.29)

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[CH.

where, if x is a vector, reference may be made to the Appendix to this chapter for an explanation of the significance of the term q · grad X · The rate of change of x in these circumstances (i.e., following the motion of the fluid) is denoted by Dx/Dt and is called the total rate of change of X· From equation (21.29) ox Dx + q grad X · = Dt at •

.

(21.30)

The first term of the right-hand member of this equation is called the local rate of change of x, and the second term of this member is called the convective rate of change of X · The former is the rate of change of x at the fixed point (x, y , z) . Consider now the fluid inside any closed surface S moving with the fluid. Let dv be a fluid element of the region of fluid r enclosed by S. If q is the flow velocity of dv, its momentum is pqdv and its rate of change of momentum is D Dt (pqdv)

= P

Dq dv, . Dt

(21 .31)

since the mass of the element p dv remains constant as the fluid moves. The net rate of change of linear momentum of the fluid inside S is accordingly Dq (21 .32) P Dt dv, r

1

and, by equation (8.4), this must equal the vector sum of the external forces. These external forces will comprise : (i) the pressure over the boundary S due to the surrounding fluid, and (ii) the body forces or forces applied to the individual particles of the fluid due to the presence of gravitational and other fields. If n is the unit outward normal to an element dS of 5, the force exerted on dS by the adjacent fluid will be -pn dS. The vector sum of these forces is accordingly

- k pn dS.

(21.33)

We shall assume that the body force acting upon the element dv is proportional to its mass, and hence may be taken to be Fp dv. In the case of a gravitational field, F will be the intensity. Then the vector sum of the body forces is

1 Fp d11, I'

.

( 21.34)

G E N E RA L M O T I O N O F A F L U I D

2 1]

623

We can now write down the equation of motion, viz.,

£ p �� dv = £ Fp dv - Is pn dS.

.

(21.35)

The surface integral may be transformed into a volume integral by the form of Green's Theorem obtained in the Appendix to this chapter. Thus we obtain

£ p �� dv = l (Fp - grad p) dv, £ ( p �� Fp + grad p ) dv = 0. -

or

Since equation (21.36) is valid for every region integrand must be identically zero, and hence

r

(21.36) of the fluid, the

1 Dq = F - P grad p. Dt

(21 .37)

This is Euler's Dynamical Equation. Employing equation (21.30), Euler's Equation may be written in the form aq 1 (21 .38) + q · grad q = F - grad p. ot P

Another form is obtained by making use of the result found in the Appendix to this chapter, viz., q grad q = grad (tq2) q X curl q. . (21 .39)

.

-

Thus, taking w = curl q, equation (21.38) is equivalent to aq

at + grad (iq2) - q

X

w

=F

-

1

P

grad p. .

(21.40)

This is Lamb's form of the equation of motion and has received practical application to the problem of the " numerical forecasting " of weather. The vector w is termed the vorticity. If the force F is the intensity of a gravitational field, we can put F = -grad V. Also,

as

at equation (19.8) , provided p is a function of p alone,

! grad p = grad p

Hence, equation (21.40) can be written

L

j dpp.

�� + grad ( !q2 + V + j d:) = q X

w.

(21 .41)

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A C O U R S E I N A P P L I E D MAT H E M A T I C S

There are now two special cases of importance. steady, 8qj ot = 0 and hence grad

[CH.

If the fluid motion is

(tq2 + V + j d:) = q X w.

{21.42)

.

The vector q X w is perpendicular to q and hence its component in the direction of flow is zero. This implies that the component of the gradient in this direction is zero. Thus, if ojos denotes differentiation along a streamline,

:s ( iq2 + + f d:) = 0, . iq2 V + j d: = constant, . v

i.e.,

{21.43) (21.44)

+

along a streamline. This result has already been derived as equation {21.3) . A second special case of equation (21.41} is when the motion is known to be irrotational. Then q = -grad cp and w = curl q = 0 .

and hence

{21.45)

�� + grad ( fq2 + V + J d:) = 0, ocfo p grad ( - + 2 q2 + + dp at J ) -- . .

-grad i.e.,

l.

v

o

(21 .46)

This latter equation implies that all space derivatives of the quantity in the bracket are identically zero, and consequently that {21.47) where C (t) is an arbitrary function of t alone. If the motion is also steady, C is an absolute constant. Equation {21 .47) is a generalised form of Bernoulli's Equation. As an example of the use to which equation (21.47) can be put, consider the problem of Section 21.3 (1), viz. , a sphere fixed in a uni­ form stream. The velocity potential is given by equation (21.22). The motion being steady, ocfojot = 0. Also, since the fluid is incompressible and homogeneous,

p

is constant, and thus

j dpfp = pjp.

We shall neglect the effect of gravity. Then equation (21.47) yields -!q = c. " 2 + p_ p

{21.48)

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G E N E R A L M O T I O N O F A FL U I D

An infinity hence

q = U and p = Po say. Thus C = !U2 P = P o + !P(U2 - q2) .

625

+ P o/

P and

(21.49) This equation determines the pressure at all points in the liquid. It is clear that the pressure is least at the point where the speed of flow is a maximum. Employing equations (21.23) , we obtain

(21 .50)

Since r � a, for variable El the maximum value of El = trr. But, for El = rr,

!

and, for variable r, viz., a. Thus

q2 = U2 (

l

as

+ 2r3

)2

q2 will occur where (21.51)

q2 takes its largest value when r takes its smallest, - �P U2•

and hence

(21 .52)

(21 .53) Pmiu. = Po If there is to be no cavitation of the liquid, the pressure must be positive everywhere. This condition is ensured if Pmin. � 0, i.e., if Po �

�P U2•

An infinite mass of in­ compressible liquid contains a spherical bubble within which the pressure is zero. Initially the radius of the bubble is a and is increasing at a rate V. The pres­ sure of the liquid at infinity is P. Show that the bubble will expand to a maximum radius R given by

Example 2.

(

3

V")

R 3 - a3 1 + P 2P Find the pressure at a distance r (>R) from the centre of the bubble when its radius is greatest.

p

·

(S.U.) Let x be the radius of the bubble at time t and let P be any fixed point in the liquid distant r from the centre 0 of the bubble. If q is the velocity of flow at P (in a radial direction) the volume rate at which liquid is moving across a sphere of centre 0 and radius r is 47Tr2q. But, since the liquid is incompressible, this must equal the rate at which liquid is qossing the

A COURSE I N APPLIED MATHEMATICS

626

[cH.

sphere of fixed radius momentarily coincident with the boundary of the bubble. Hence

47r1'2q = 47Tx•x. q = x•x -;o· If q, is the velocity potential, orf> x•x - Fr = roq, = x•xfr. and thus We can now calculate orf>f ot at P. It is given by ?!f> = ! d__ (x2x) ! (2xx2 + x•x). r dt r ot Equation (21. 4 7) now yields 4 '2 p + -p = C(t). (i) - -1r (2xx2 + x•x) + � 2r4 0. But, as r -+ p -+ P. Hence C(t) = Pfp . Also, when r = x, p Thus -2%2 - XX + JP Pfp, (ii) � � � or ) ( x x• + x• = -Pfp. This is a first-order linear differential equation for !x2• Its general solution is tx• = � - !:__, x3 3p where A is a constant of integration. Clearly, as x increases x decreases and eventually becomes zero when the bubble achieves its maximum radius. Thus, when x = R, x = 0. Also, when x = a, x = V. Hence 0 = R3 .:i_ - !:_, 3p A p !V• = a-3 - 3p- · Eliminating A and solving for R, we obtain the result stated. When x = R, x = 0 and hence, from equation (ii), x = - PfpR. Substi­ tuting these values in equation (i) with C(t) = Pfp, we find that or

=

oo ,

=

p =

=

p ( 1 - ¥) '

giving the pressure distribution at this instant.

A

solid hemisphere of radius a rests with its base on the plane level bottom of a stream of great depth, having uniform velocity U and pressure at great distance from the hemisphere. Show that the thrust between the hemisphere, than it would be if the stream were at rest, and the bottom is less by ��1r a with the same pressure at infinity. (Li.U.)

Example 3.

p 2 u•

P

The velocity potential for the flow is identical with that for flow past a sphere, viz., q, = -.: r + cos a.

u ( ;;.)

for the flow velocity derived from this potential is tangential at all points of the plane on which the hemisphere rests, and hence all the boundary con­ ditions are satisfied.

G E N E RA L M O T I O N OF A F L U I D

21]

From Bernoulli's Equation, !q•

627

+ p_p = w• + !:.,p

where we have neglected the gravitational field, since the effect of the stream's motion on the total thrust is required, and the contribution of this field to the thrust is independent of the motion. Putting r = a in equation (21 .50), we find that the speed of flow at the exposed surface of the hemisphere is given by q = t U sin 6 and hence the pressure dis­ tribution over this surface is P = P + tpU2 - tpU2 sin2 a.

This equation indicates that, on account of the flow, the pressure is increased by an amount p U2 (t - t sin2 e) . At the point Q on the hemi­ sphere (see diagram), the resolute of this additional pressure in a direction normal to the plane is p U2 (t - t sin2 e) sin e cos rp. The other two resolutes parallel to the plane face of the hemisphere contribute nothing to the total thrust upon it. An element of surface at Q has area a2 sin e dfl drp, and hence the net increase in thrust perpendicular to the plane is

l'TT 0

de (l'" pa2 U2 (t - t sin2 e) sin2 e cos rp drp

}_f'TT

=

-H-n-pa2 U2,

i.e., the thrust decreases by H-n-pa2 U2 by virtue of the liquid flow. A doublet of strength p. is fixed in a liquid of density p so that the axis of the doublet is perpendicular to an infi·nite rigid plane boundary and at a distance a from it. The liquid fills the space on the doublet side of the boundary. Show that the resultant of the pressure on the boundary due to the presence of the doublet is

Example 4.

3p7rp.2 2a4

(Li.U.)

•

The constituent source and sink of the doublet at A have equal images at A ' (see diagram), which together form a doublet of strength p., whose axis is directed as shown. If P is any point on the plane boundary, the components of the velocity of flow due to the doublet at A along and perpendicular to AP are

qg = _r_s_ P-

sin e

and hence the component parallel to the boundary is 3p. sin a cos e . q. sm e + qo cos a =

rs

.

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A C O U R S E IN A P P L I E D MAT H E MAT I C S

[CH.

Adding the contribution of the image doublet, we find that the total speed of flow at P is

�

�

sin a cos4 fl. q = r sin a cos a = a

Bernoulli's Equation now yields 18�'2 sin2 a cos8 a + P. = f>_r,, a• p P

where Po

is

the pressure at infinity on the plane boundary (i.e., at

e

= !1r) .

"" P

8

Q

a

A

Thus the pressure on the boundary is reduced as a result of the doublet's presence by an amount 18pJL2 P' = --as sin• a cos• a .

Let PN = y and consider the thrust on a ring drawn on the boundary with centre N, inner radius y and outer radius y + dy. This thrust is altered by 21rp'ydy as a result of the flow caused by the doublet. However, since y = a tan a. this can be written 36p7TfL2 21Ta2p' tan a sec• e d6 = � sin3 a cos• a da.

Integrating with respect to 6 over the range 0 � 6 � !1r, we find that the total reduction in thrust is 36p7TfL2 a•

fh'

2 - 3p7TfL . 3 6 cos • 6 d6 sm 2a•

.

The centre of a rigid sphere of radius a is moving with speed V in a fixed direction in an infinite liquid of density p, which is at rest at infinity. Show that at a point on the axis of motion distant r in advance of the centre of the sphere, the pressure intensity exceeds that at infinity by (Li.U.) !pa3{r-2d Vfdt + V2 (2r3 - a3r6)}.

Example 5.

The velocity potential for the liquid motion at any instant has been given as equation (21 .25), viz., va• = • cos 6. q, 2r

G E N E RA L M O T I O N O F A F L U I D

21]

629

At the point at which the pressure is to be calculated the flow velocity is accordingly q=

- ( 8Or"')

B=O

=

Va3fr3 •

The term orf>fot in equation (21 .47) represents the rate of increase in q, at a point fixed in space and not fixed relative to the sphere. At the point P, distance r in advance of the sphere's centre, q, = Va3f2r2

and, if P is fixed,

r

Thus

= - V.

orf> a3 = ot 2r2 a3 = 2r2

•

dV Va3 r, dt ra dV + V2a3 · dt rs

-

From Bernoulli's Equation we now obtain dV - ta•r-• dt

- v•a•r-• + ! V2a 6y-6 + p-p = P-,p

P being the pressure at infinity.

Thus

{ �;. + V'(2r-• - a•r-6)} ,

p = P + !pa• r-•

proving the result stated.

21.5.

Kinetic Energy of an Irrotational Flow

Suppose that a uniform liquid of density p is contained within a rigid surface S and flows irrotationally around a number of rigid bodies bounded by surfaces Sv 52, Sn. Any of these surfaces may be at rest or in motion. Let r be the region between the surface S and the rigid bodies at any instant and let dv be an element of volume in r. The KE of the liquid occupying this element is irq2dv, and hence the net KE of the liquid is given by •

T

•

•

=

tP

£ q2dv.

.

(21.54)

Let 4> be the velocity potential for the flow. From equation (16.4) we conclude that div (cf>q) = 4> div q + q · grad cf>, - q2, = 4> div q - q2 (21.55) =

•

since div q = 0 by the equation of continuity. Thus equation (21.54) may be written T

=

- !P

1 div (cf>q) dv r

=

- tp

f cf>qn dS, :Es

.

(21.56)

by the extended form of Green's Theorem referred to on p. 409. The surface integral is to be calculated over all the boundary surfaces and qn is the component of q in the direction of the normal to any surface

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A C O U R S E I N A P P L I E D M A T H E MAT I C S

[CH.

drawn in the sense outwards from the liquid. Since qn = ocf>Jon , of on denoting differentiation along the normal inward to the liquid,

1

T = -tp

ocf>

dS c/> <> un . .

r.s

(21.57)

A solid sphere of radius a moves along a straight line in an ideal liquid of density p, which is moving irrotationally and is at rest at infinity. Show that the magnitude of the resultant thrust of the liquid on the sphere at any instant is 1;.,a3pf, where f is the instantaneous value of the acceleration of the sphere. (Li.U.)

Example 6.

If U is the speed of the sphere at any instant, the velocity potential of the flow around the sphere is Ua3 .p = '2r2 cos a.

The liquid may be regarded as moving between the solid sphere and a concentric sphere of infinite radius. The contribution of the infinite sphere to the surface integral for T given at equation (21 .57) is zero, for we shall regard this sphere as fixed, and hence qn is zero at all points upon it. Over the surface 5 of the solid sphere

if. = ! Ua cos a,

on �

or (�)

=

Hence the KE of the liquid is given by T = !pU2a

L

r=a

= - u cos a .

cos26 dS.

Dividing the surface of 5 into ring-shaped elements with their centres on the diameter of the sphere which is in the direction of motion, this surface integral may be put into the form T = lrrpU•a•

fo

"

cos2 6 sin a da

= }-rrp U2a3•

If M is the mass of the sphere, its KE is !MU2, and thus the total KE of the system is (!M

+ !1rpa3) U2•

The rate of increase of this quantity must equal the rate at which external forces are doing work on the system. Let F be the force applied to the sphere to cause its motion. Then d FU = dt (!M + }1rpa3) U2

where f = U. Thus

= 2(!-M + }-rrpa3) Uf, F - 1;1rpa"f = Mf.

But, if P is the resultant thrust of the liquid on the sphere opposing its motion, its equation of motion is Hence

F - P = Mf.

G E N E RA L M O T I O N O F A F L U I D

21]

631

It will be observed that P is zero unless the sphere is accelerating. Thus, an ideal liquid in irrotational motion does not impede the uniform motion of a sphere through it. This result may be generalised to refer to the uniform motion of any solid body.

21.6.

Stream Function. Circulation

In this section we shall consider the special case of the two-dimen­ sional motion of a uniform liquid. The liquid can be supposed to flow between a pair of parallel planes and around obstacles in the form of cylinders whose generators are perpendicular to these planes. The flow patterns in all planes parallel to the boundary planes will be identical, so that it is only necessary to calculate the flow in one such

A

0

FIG. 21.8.-The Stream Function

plane. This plane will be taken to be that of rectangular axes Oxy (Fig. 21.8). Let A be any fixed point in this plane and let P be a variable point having coordinates (x, y) . Let Cv C 2 be any two curves joining the points A , P. Construct cylindrical surfaces on C 1 and C 2 having generators perpendicular to the xy-plane and of unit lengths. Since the liquid is to be assumed incompressible, the volume rates at which liquid is transported across these cylindrical surfaces in the senses indicated must be the same. It follows that this volume rate is in­ dependent of the curve joining A and P and may be denoted by if;(x, y, t) . If the flow is steady, if; is independent of t. if; is called the stream function for the flow. In future we shall speak of the rate of flow across a curve such as Cv meaning thereby the rate of flow across the unit cylinder based on cl. Suppose that P moves along a streamline to a point Q. The rate of

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[CH.

flow across the arc PQ of the streamline is clearly zero. It follows that the rate of flow across the curve comprising the arc PQ and the curve C 1 is the same as the rate of flow across C 1 . We deduce that if; takes the same values at P and Q, and hence that the stream function is constant along a streamline. If P, P' are any two points in the xy-plane at which the stream function takes values if;, if;' respectively, by constructing a curve from A through P to P' it becomes clear that the rate of flow across a curve connecting P and P' is if;' - if;. For an observer at P facing in the direction of P' this flow will be in the sense from right to left. If now P' is a point neighbouring on P, we can take P P' = ds and if;' = if; + dif;. The rate of flow across ds is then dif;. But, if qn is the com­ ponent of the flow velocity at P in a direction perpendicular to ds, the rate of flow across this element is qnds. Hence dif; = qnds, oif; = qn · OS •

or

(21.58)

Thus the derivative of the stream function in any direction is the component of the velocity of flow in the perpendicular direction. In particular, if (u, v) are the components of q in the directions of the axes oif; =v ox '

oif; - -u· oy -

·

(21 .59)

If the motion is irrotational with velocity potential cfo , we also have the equations ocfo ocfo - = -U = -v . {21 .60) ' oy . ox Thus, ocfo oif; + ocfo oif; _ 0 {21.61) ox ox oy oy -

' ·

which is the condition that the curves 4> = constant, if; = constant should be orthogonal. We have therefore proved that the system of equipotentials is orthogonal to the system of streamlines (cf. the systems of equipotentials and lines of force in electrostatics). Now, in this case of irrotational motion,

Hence o2if; ox2

o2if; + oy 2

i.e., if; is a harmonic function like cfo.

=

0,

{21.62)

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G E N E RAL M O T I O N O F A F L U I D

633

We have also the relations (21.63) These are the Cauchy-Riemann conditions that �� if; should be the real and imaginary parts of a regular function of the complex variable The theory of functions of a complex variable z = x + iy (i = v' - 1) . may therefore be employed to solve problems concerning the two­ dimensional irrotational flow of an incompressible fluid (cf. Section 15.8) . However, it would carry us beyond the scope of this book to give further details of this very important method, and we shall accordingly refer the interested reader to more comprehensive works.* Consider the special case of a uniform stream of liquid moving parallel to the x-axis with velocity U. If P is the point (x, y) and PN is the perpendicular from P to Ox, consider the rate of flow across the curve formed from the segments PN and NO. It will be Uy in the sense of the positive x-axis. Hence if; = - Uy since the rate of flow de­ fining if; must, by definition, be taken in the opposite sense. If (r, e) are polar coordinates in the xy-plane,

,

if; =

- Uy = - Ur sin e.

(21.64)

Now suppose that a circular cylinder of radius a and of infinite length is placed in the uniform stream with its axis passing through the origin perpendicular to the xy-plane. The stream function for the disturbed flow must be constant over the circular section of the cylinder, since this will be a streamline, and must reduce to the form of equation (21.64) at great distances. Also, if; must be harmonic. We shall try if; = - Ur sin e + :! sin e. r

.

(21.65)

A sin efr approaches zero for large r and represents the distorting effect of the cylinder on the uniform flow. It is harmonic because it is identical with A cos(e - lrr)fr, which is the velocity potential of a line doublet at the origin with its axis in the direction Oy. if; is constant over r = a if

for all e.

( � - Ua) sin e

constant .

This can only be so if A = Ua2 if; = -

*

=

•

(21.66)

Hence

( r - �) u sin e. .

(21 .67)

E.g., Theoretical Hydrodynamics, by L. M. Milne-Thomson (Macmillan).

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A C O U R S E I N A P P L I E D MATH EMAT I C S

[cH.

The streamlines !f; = constant can now be drawn and are similar to those shown in Fig. 21.5. It is left to the reader to verify that the velocity potential is given by cp

=

( �) U cos e.

- r+

.

(21.68)

It follows from the uniqueness theorem that, since this expression for cp is harmonic and satisfies the boundary conditions, it represents the unique solution to our problem. However, this form of the unique­ ness theorem assumes that the velocity potential is a single-valued function. If we permit cp to be multi-valued, other solutions can be found which prove to be of great practical importance. Before cal­ culating these solutions, we will introduce the concept of circulation around a closed contour. The circulation of a vector around a closed loop has been defined in Section 18.5 (equation (18.31)). In hydrodynamics the circulation is always taken to be that of the velocity of flow. Thus, the circulation K around a closed loop C is given by K =

lq

•

dr =

l q1ds, .

(21 .69)

where ds is an element of C and q1 is the component of q tangential to C. Now q1 = - ocpjos if the flow is irrotational and ojos denotes differentiation in the direction of the element ds. Thus (21.70) [cp] being the increase in the value of cp as we make one complete circuit of C. Obviously, if cp is single-valued, the circulation around any closed contour is zero. If, however, cp is multi-valued the circulation around a closed loop can be non-zero even though the motion is irrotational. In calculating the magnetic potential due to an infinite, straight wire carrying a current we have already encountered a multi-valued potential (see equation (18.6) ) . Consider therefore the irrotational flow corresponding to this multi-valued potential, viz., cp

=

_

K6,

21t

(21 .71)

where 6 is a polar coordinate in the xy-plane. The increment in cp as we proceed around any closed curve embracing the origin is -K, and thus the circulation around such a curve is K. On the other hand, the circulation around any contour not embracing the origin is zero. The

2 1]

G E N E RA L M O T I O N O F A F L U I D

635

radial and transverse components of the velocity of flow at the point (r, 6) are given by

q,

=

-�t

=

o,

q9

=

-� :

=

2:,-

(21. 72)

We see that the direction of flow is everywhere at right angles to the radius vector, i.e., the streamlines are circles about the origin (d. the electro-magnetic case). Any circle with its centre at 0 can represent a fixed boundary about which the liquid executes a purely circulatory motion. To compute the value of the stream function at any point P (r, 6), we shall take the point A to be on the unit circle with centre 0 and then join P to A by part of the radius vector through P and an arc of the unit circle. The rate of flow across the arc of the unit circle is zero, since it is a streamline. The rate of flow across the radius vector is accordingly lfi, and hence

1jJ =

[ q9dr ; log r. =

rt

(21. 73)

cfo and ljJ will now be recognised as the real and imaginary parts of the complex function

;: log

z.

It may be verified that ljJ is a harmonic

function. Suppose now that we add terms representing this purely circulatory motion to ljJ and cfo as given by equations (21.67) and (21.68), to obtain

ljJ = �

=

-- ( r - � ) U sin 6 + ;rt log r, l . - ( r + � ) U cos 6 - ;!. j

(21. 74)

It is clear from the nature of the circulatory motion that the additional terms satisfy the boundary conditions on ljJ and cfo over the cylinder r = a (i.e., ljJ = constant and 8cfoj8r = 0). Also, it is apparent from equations (21.72) that the additional terms in the components of q are zero at infinity, and hence that the motion at a great distance is that of a uniform stream. Equations (21.74) accordingly represent a possible flow of a uniform stream about a circular cylinder with circulation K. The streamlines ljJ = constant have been drawn in Fig. 21.9. Let us calculate the resultant force exerted by the stream on unit length of the cylinder. We have l 8cfo a2 � u cos 6, q, = - = 1 or 1 8cfo a2 . K 1 (21 .75) - = 1 + U sm 6 + -q = rt · 9 r o6 r2 2 r J

(- ) -- - ( - )

. . .l

636

A COURSE

iN

[cH.

A P P L I E D M A T H E M AT I C S

Thus, on the cylindrical boundary r = a, q = q9 =

-

2 U sin e + __15_ . 2n:a

(21 . 76)

Assuming steady flow, Bernoulli's Equation now yields p = c - tpq2, . and hence, on r = a, p

=

[

(21 .77)

�]·

K C - p 2 U2 sin2 e - "f!_ sin e + 8n:2a2 n:a

.

(21 .78)

By considering the thrust on that part of unit length of the cylinder corresponding to the element P P' of its cross-section (Fig. 21. 9), we

FIG. 21 .9.-Flow around a Cylinder with Circulation

find that the components of the resultant thrust in the direction of flow and perpendicular to this direction are X = -a

12rr p cos e de, 0

Y = -a

121T p sin e de, 0

(21 .79)

respectively. Substituting for p from equation (21 . 78) , we obtain X = 0,

y

= -KpU . .

(21.80)

We observe that in the absence of circulation (i.e., K = 0) the resultant thrust on the cylinder is zero. If, however, a circulatory motion can be induced to take place a resultant thrust at right angles to the direc­ tion of flow will be applied to the cylinder. By distorting the cross­ section of the cylinder from its circular shape to a non-symmetrical form, circulation about it can be encouraged and a resultant thrust obtained. A cylinder designed to promote circulation with the object of developing a thrust at right angles to the direction of flow of an incident stream is called an aerojoil. The cross-section of a typical

2 1]

GENERAL MOTION O F A FLU I D

637

aerofoil is shown in Fig. 2 1 . 10. The component of the thrust at right angles to the direction of flow is called the lift, and any component in the direction of flow is called the drag. The chief application of aero­ foil theory is to the design of aircraft wings, and the reader interested LI F T

I NC I DE N T STREAM

FIG. 21.10.-The Aerofoil

in the theory of aircraft flight is referred, for further information, to textbooks of aerodynamics.* 21.7.

Vortex Motion

In previous sections we have assumed the fluid motion to be irrota­ tional, i.e., q = -grad cfo, and hence w = curl q = 0. In this section we will discuss a few simple cases of motion with vorticity. Assuming the flow is two-dimensional, if (u, v, w) are the com­ ponents of q along the x- and y-axes and perpendicular to their plane respectively ow ou ov ou - ow ov w = curl q = oy oz' oz ox ' ox oy ' ov ou = o, o, (21 .81) ox - oy .

( (

)

)

since w = 0 and (u, v) are functions of (x, y) only. It follows that the vorticity is everywhere perpendicular to the xy-plane. Consider first the case in which the two-dimensional flow is irrotational outside the region of the xy-plane bounded by the circle x2 + y2 = a2, but for which the vorticity takes a constant value over the interior of the circle (Fig. 2 1 . 1 1 ) . We shall suppose that the liquid extends to infinity in all directions and that no sources or sinks are present. From considerations of symmetry it is clear that the flow velocity at any point P distant r from 0 will be a function of r alone. Also q can have no radial component, for if it had the rate of liquid flow across a circle C with centre 0 and radius r would be non-zero, and this would imply the existence of a source or sink within the circle. The circulation around C is accordingly 2-rrrq. But, by Stokes' Theorem, this must * E.g., Theoretical A erodynamics, by L.

M.

Milne-Thornson (Macmillan) .

638

A C O U R S E I N A P P L I E D MAT H E M AT I C S

[CH.

equal the :flux of curl q = w across the disc whose boundary is C . If, now, P lies in the region of irrotational motion, this :flux will be 1ta2w, since the vorticity is zero throughout the irrotational region. Hence (21.82) If, however, P lies within the circular vortex, the :flux of vorticity will be 1tr2w and then (21.83) Equations (21 .82) and (21.83) indicate that the liquid revolves around 0, the angular velocity diminishing towards zero as we approach infinity. Within the cir­ cular vortex the liquid y rotates like a rigid body with angular velocity tw . It will be noted that the �/ situation is completely / I ,P analogous to that which \ I \ I arose in connexion with \ I I the problem of the electroI x. magnetic field due to a I straight wire carrying a I I I I current. q corresponds to \ I \ \ the magnetic intensity H \ and the quantity ia2w to the strength of the current. We have shown that a circular vortex is rotating FrG. 2 1 . 1 1 .-Motion due to a Circular Vortex like a rigid body with angular velocity tw. In general, it may be proved that, at a point in a :fluid where the vorticity is w, if a small element of the :fluid is instantaneously solidified, it will commence to rotate with angular velocity tw. If the radius a of a circular vortex is so small as to be negligible, it is called a rectilinear vortex. The circulation K around any curve em­ bracing the vortex is equal to the flux of vorticity through the curve, viz., 1ta2w. This quantity is termed the strength of the rectilinear vortex. Equation (21 .82) may now be written ,.

'

' ,

� - -

K

q = 1t ' · 2 r

(21.84)

specifying the velocity distribution due to the vortex. This velocity distribution is identical with that given by equations (21.72), and hence

2 1]

G E N E RAL M O T I O N O F A F L U I D

639

equation (21.71) gives the velocity potential for the flow. Equation {21.73) defines the stream function. At the centre of a rectilinear vortex in a liquid stationary at infinity the velocity of flow is zero. If the liquid possesses a motion in addition to that corresponding to the vortex, it may be proved that the vortex will be carried along by the liquid so that the velocity of its centre will be equal to the velocity of flow of the liquid at that point in the absence of the vortex. The strength of the vortex remains constant during the motion. The typhoon is a natural example of this phenomenon. At its centre the wind velocity is equal to the speed of the air stream supporting it and conditions are comparatively calm. On the periphery of the vortex, however, the rotary motion is dominant, and wind speeds of several hundred miles per hour may be recorded. The velocity at any point of a stream supporting a rectilinear vortex is K/2n c

/(

/(

I
FIG. 21.12.-A Vortex Pair

compounded of a component which would be present in the absence of the vortex and a component due to the vortex. The latter com­ ponent is termed the velocity induced by the vortex. We will now consider the case of two equal, parallel rectilinear vortices separated by a distance c in an infinite liquid (Fig. 21.12). Let K be the strength of the vortices, the circulations about them being in the same sense. The velocity of liquid flow at the centre of either vortex due to the other is Kj2rcc in a direction perpendicular to the line j oining the vortices. By the principle explained in the previous paragraph, it follows that the two vortices have the velocities shown in the figure, and hence rotate steadily about the mid-point of the line joining them with angular velocity Kjrcc2• If the motion outside a rectilinear vortex is irrotational, the methods of problem solution associated with harmonic functions may be em­ ployed in the calculation of the flow. For example, the following problems yield satisfactorily to the method of images.

640

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[CH.

Example 7.

Find the motion of a straight vortex filament in an infinite region bounded by an infinite, plane wall to which the filament is parallel, and prove that the pressure defect at any point of the wall due to the filament is proportional where is the inclination of the plane through the filament and to cos • cos the point to the plane through the filament perpendicular to the wall. (M.T.) Let be the strength of the filament at A . Introduce a parallel image filament at the image point A ' also of strength but with its circulation in the opposite sense. The velocity components at any point P on the wall

a 2a, K

a

K

due to the filaments are of equal magnitude Kj21J'a sec and are equally inclined to the wall. Their resultant accordingly lies in the plane of the wall and the boundary condition is satisfied.

a

p

a

The magnitude of the velocity of flow at P is given by

K K q -- 2 a sec a · 2 cos a -- 'TJ'a cos• a . 1J'

Each vortex moves with the velocity induced by the other, i.e., Kj41J'a parallel to the wall, and thus, as the motion proceeds, the configuration remains that shown in the diagram. The velocity potential at P is given by and thus But

_,_ 'I'

.!:__ a = .!:__ (2a - 71') = 271' .!:__ (11' a) + 271' 271' � = ':_ h at tan a = PNJa -

-

7J'

and hence, differentiating,

sec•a_()

o.

= � � (PN) = 471':•'

P being a fixed point and N a moving point with velo�ity Orp K2 COS2 a at 47T2a2 •

=

Kj471'a.

Hence

G E N E RA L M O T I O N O F A F L U I D

2 1]

641

If p is the pressure at P, Bernoulli's Equation now yields -

. K. p cos2 e + K• • cos4 e + ·-p 41r•a• 21r a

-- P ,

where Po is the pressure at infinity on the wall where e Thus p = Po

and this is the result required.

pK2 cos• -1T2a24

e

Po

=

i1r.

cos 26

A rectilinear vortex, of strength is at a distance f from, and is parallel to, the axis of a solid cylinder of radius a and infinite length : the circulation about the cylinder is such that the vortex remains stationary; show that the magnitude of this circulation is - a2). (Li.U.)

K,

Example 8.

Kf"f(f2

Let K be the circulation around the cylinder. Introducing an image vortex of strength parallel to the given vortex but having circulation in the opposite sense and through the inverse point A ', we find that the stream function at any point P assumes the value

K

.p

=

K log AP - K2,;- log A'P + 2,;-K log OP. 21T

If P lies on the cylinder, APfA'P = OA OP =

f ffa and thus K K = 2 log (ffa) + 2,;- log a = constant.

.P 1T Any right section of the cylinder is therefore a streamline and the boundary condition is satisfied. The velocity of the given vortex is now calculated to be

K

K 21rOA - 21rA A '

and this is zero if K=

K . OA

AA'

=

Kf

f - a2ff

=

Kf"

j2 - a2

APPENDIX be any vector with components (u, v, w) parallel to rectangular axes Oxyz. Let (l, m, n) be the direction cosines of the direction of q. The operator " grad " has components (8fox, of8y, 8foz) parallel to the axes, and hence we have formally 0 0 8 0 0 8 q · grad = u - + v - + w- = q l- + m- + w- · Let

q

ox

oy

8z

(

ox

This equation is taken as the definition of the operator

oy

q

•

grad.

)

8z

A C O U R S E I N A P P L I E D M A T H E M AT I C S

642

[CH .

Let a be any vector function of position and let of os denote differentiation with respect to distance in any specified direction from the point (x, y, z) . Then � = oa � os ox ds

+

oa r!2'. 8y ds

+

oa �. oz ds

� + n �. = l� + m ox oz oy 0 0 0 = (l ax + m + na;z) a. ay

where are the direction cosines of the direction of differentiation. It now follows from what has been established that, if the direction of differentiation is that of the vector q, then

(l, m, n)

q

· grad a = q oa' Fs

i.e., q · grad a is equal to the derivative of a in the direction of q multiplied by the magnitude of this vector. Now q

· grad q

=

u� ox

+ v �oy + w oz�

and the x-component of this vector is accordingly ou ou u- + vox oy The same component of the vector q

(

�)

X

(

+ w -oz8u· curl q is

)

v '!!! -w � - �� ox oy ox · oz

Hence, the x-component of the vector

+ q X curl q is 8u ov ow - 0 1 ( 2 2 2 u ax +_v ax + w ax0 ( 1"2"q2) ax u + v + w ) ox -

q

·

grad q

"2"

-

•

But this is the x-component of grad {tq2) . Similarly for the y- and z-com­ ponents. Hence grad ( iq2) = q · grad q q X curl q.

+

Let be any scalar function of position. Putting P Green's Theorem (equation (15.16)), we obtain

p

1 lp

dS

=

£ ��

=

p,

Q

=

R = 0 in

dv.

But opfox is the x-component of grad p and is the same component of pn, where n is the unit outward normal to the surface S. Thus

fs

pn dS =

fr

lp

grad

p dv

is a valid equation for x-components of the vectors involved. Similarly, it is valid for y- and z-components, and hence valid without restriction.

21]

G E N E R AL M O T I O N O F A F L U I D

64 3

EXERCISE 2 1 1 . Employing Torricelli's Theorem, find the form o f a vessel o f revolution with a small aperture at its lowest point which is such that the level of any liquid it contains descends uniformly. 2. A pitot-static tube, used for measuring the velocity of air flowing in a pipe, gives a pressure difference of h in. of water on the manometer. Show that the velocity of the air at the point in question is given closely by 66yh ftjsec. (Take the density of air as 0·0765 lbjft3. Atmospheric pressure 14·7 lbfin. 2 or 34 ft of water. Acceleration due to gravity g = 32·2 ftjsec2 .) (M.U.) 3. Doublets of strengths l.l. and l.l.' are situated at the points whose cartesian coordinates are (0, 0, c) and (0, 0, a2fc) , their axes being directed towards and away from the origin respectively. Derive an expression for the potential. Show that the flow at all points of the sphere x2 + y2 + z2 = a2 is tangential if [L'/[L (Li.U.) (afc) 3• 4. Prove that liquid motion is possible when the velocity (u, v, w) at (x, y, z) is given by 3x2 - r2, vr5 = 3xy, wr5 = 3zx, ur5 2 where r = x2 + y2 + z2. Show that the streamlines lie in planes passing through the axis of x. (Li.U.) 5. If the motion in an incompressible, homogeneous, irrotational fluid be in two dimensions, prove that if at any instant the velocity be every­ where the same in magnitude, then the direction is also the same. Prove also that if the direction of the velocity be the same everywhere then its magnitude is the same throughout the fluid. (Li.U.) =

=

=

6. A two-dimensional, liquid motion arises from a source at A and a sink, of half the strength of the source, at B. Determine the position of the point C at which the velocity of the liquid is zero. Show that on the circle with centre C and radius ABy2 the velocity is parallel to AB. (Li.U.) 7. Line doublets of equal strength are situated at the points x = a, y 0 and x - a, y 0, their axes making angles ct and 1t ct respectively with the x-axis. Verify that the line x = 0 is a streamline, and find the coordinates of the two points at which the velocity is zero. (Li.U.) 8. In a two-dimensional, irrotational motion there are sources, each of strength m, at the vertices of an equilateral triangle ABC. Prove that the speed of the fluid at a point P is 6m (PG) 2fPA .PB.PC, where G is the centroid of the triangle. (Li.U.) =

=

=

-

9. A source of fluid situated in space of two dimensions is of such strength that 27tpfL represents the mass of fluid of density p emitted per unit of time. Show that the force necessary to hold a circular disc at rest in

644

A COURSE I N APPLIED MATHEMATICS

[cH.

the plane of the source is 2rtpfL2a2fr(r2 - a2), where a is the radius of the disc and r the distance of the source from its centre. In what direction is the disc urged by the pressure? (M.T.) 10. The centre C of a sphere of radius a moves along the z-axis of a fixed rectangular cartesian coordinate system Oxyz. The coordinates of the centre at time t are (0, 0, �). If the sphere is surrounded by liquid in irrotational motion and extending to infinity in all directions, show that the velocity potential at the point P (x, y, z) is given by a3t (z - �)
Deduce that, if OP = r and L POz = 0, then ta3 o y(2aj9A.) . What is the maximum pressure at time t? (Hint : Employ results (M.T.) from Ex. 10.) A sphere of radius a is divided into two halves by a diametral plane and is situated in an infinite liquid of density p which is flowing with uniform velocity u perpendicular to the dividing plane. Show that the pressure between the two halves of the sphere is less by rtpa2u2/16 than it would be if the fluid were at rest under the same forces and with the same pressure at infinity. (M.T.) The space on one side of an infinite plane wall, y = 0, is filled with inviscid, incompressible fluid, moving at infinity with velocity U in the direction of the axis of x. The motion of the fluid is wholly two­ dimensional, in the (x, y) plane. A doublet of strength fL is at a distance a from the wall, and points in the negative direction of the axis 'of x. Show that if fL is less than 4a2U, the pressure of the fluid on the wall is a maximum at points distant y3a from 0, the foot of the perpendicular from the doublet on the wall, and is a minimum at 0. If fL is equal to 4a2U, find the points where the velocity of the fluid is zero, and show that the streamlines include the circle x2 + (y - a)2 = 4a2, where the origin is taken at 0. (M.T.) A point source of strength Q sin nt is placed at the point (a, 0 ,0) inside an incompressible fluid which is at rest except in so far as it is dis­ turbed by the source ; the liquid has aJigid plane boundary coincident =

P P

11.

12.

13.

14.

o

=

21]

G E N E RAL M O T I O N OF A FLU I D

645

with x = 0 and otherwise extends to infinity. Find the resultant force on that part of the boundary inside the circle x = 0, y2 + z2 = b2, and prove that the effect of the source is to produce on the circular area an average normal pull towards the source equal to b4 pQ2 (Li.U.) 32rca2 ( a2 + b2) 2 Derive the equations of motion and continuity in Cartesian coordinates for the motion of perfect incompressible fluid under the influence of external forces. If the density is constant, the flow is steady and two-dimensional, and the external forces are conservative, show that, with the usual notation, (Li.U.) A small, hemispherical solid, of radius a, lies with its flat face in con­ tact with the plane bed of a deep stream. Show that the hemisphere will not be lifted unless U2 > 64ga(cr - p)/33p, where U is the stream velocity at an infinitely large distance from the hemisphere, and p and cr denote the densities of the liquid and hemispherical solid respectively. (Li. U.) A homogeneous liquid of density p is contained in the region between a spherical shell of radius a and a concentric sphere of radius b, where a > b. Find the initial value of the velocity potential when the two boundaries are given velocities U, V respectively in the same direction, and prove that the initial kinetic energy of the liquid is rep {a3b3 ( U - V) 2 + 2 (a3 U - b3 V)2} 3(a3 - b3) (Li.U.) (Hint : Take rp = Ar cos 6 + B cos 6fr2.) Show that the distribution of velocities given by the equations u = a - yf(r) , v = (x - at)f(r), where y2 = (x - at)2 + yz, satisfies the two-dimensional equations of motion of a liquid under no external forces, and find the pressure intensity at any point. •

15.

1 6.

17.

•

18.

19. In an experiment it was noted that the fluid which streamed past a cylinder of circular section possessed only one stagnation point, on the surface of the cylinder. If a is the radius of the cylinder, U the velocity of the stream at infinity, show that the thrust of the fluid on the cylinder was 4rcpaU2 per unit length. (S.U.) 20. A rigid sphere of radius a starts from rest at time t = 0 and moves with a constant acceleration f through an infinite, homogeneous, per­ fect liquid. If P is the pressure at infinity, show that at time t the thrusts on the two hemispheres into which the sphere is divided by a plane perpendicular to the direction of motion are na2P ± i;Mf - 634 Mj2t2ja, where M is the mass of the liquid displaced by the sphere. (Li.U.)

646

A C O U R S E I N A P P L I E D MATH EMAT I C S

[CH.

2 1 . An inviscid, incompressible fluid of density p is in equilibrium between a rigid, spherical surface of radius a and a concentric, spherical, free surface of indefinitely great radius, over which a constant pressure P is maintained. The fluid is attracted towards 0, the common centre, by a force A.fr3 per unit mass, where ). is constant and r is distance from 0. If the inner surface is suddenly removed, prove that the pressure at a distance r from 0 is suddenly diminished by Pa �E_ . + (N.U.) r 2ar 22. A spherical bubble of gas exists in an infinite, incompressible liquid. At an instant when the liquid is everywhere at rest the radius of the bubble is a, the pressure at infinity P, and the pressure of the gas within the bubble is kP (k > 1 ) . Assuming that the pressure-density relation for the gas is pfp� = constant, show that in the ensuing motion the maximum radius of the bubble is given by the real root of the equation (Li.U.) x3 + ax2 + a2x - 3ka3 0. =

23. A sphere whose radius at time t is a(1 + e: sin nt), where a, n, e: are constants and e: is small, is surrounded by liquid of density p which extends to infinity. If there are no external forces, prove that the liquid pressure at any point on the sphere is less than that at an infinite distance by e:pn2a2 sin nt, where the second and higher powers of e: are neglected. (Li.U.) 24. A spherical envelope is situated in an unlimited mass of fluid of con­ stant density p. If the radius of the envelope is increasing at the con­ stant rate q0, show that when its radius is Y0 the pressure intensity at distance r( > r0) from its centre is r p + 2pqo2 --:- ( 1 yo3f4r3) ,

-

where P is the pressure intensity at infinity.

(Li.U.)

25. A spherical bubble of gas is surrounded by liquid, of density p, which extends to infinity. The bubble performs small radial oscillations. Neglecting the inertia of the gas and assuming it to obey Boyle's Law, show that the period of the oscillations is approximately 27ty(pa2j 3P) , where a is the mean radius and P is the pressure at infinity. (Li.U.) 26. A spherical bubble of gas obeying Boyle's Law is initially at rest in an infinite, incompressible liquid, the pressure at infinity being P. The pressure of the gas is initially 2P and the radius of the bubble is a. Show that if R is the radius of the bubble during the subsequent motion, then

pRa

(��)

2

{

= 2P 2aa log

-

� - � (Ra - a3) } ,

and deduce that the maximum radius of the bubble is xa where x is a root of the equation x3 6 log x - 1 = 0. (S.U.)

2 1]

G E N E RA L M O T I O N O F A F L U I D

647

27. A rectilinear vortex, parallel to the z-axis, through the point (0, a, 0) is fixed in liquid on one side of the plane y = 0. Show that the lines of equal pressure in the (x, y) -plane are members of the family of curves (Li.U.) 28. A rectilinear vortex moves parallel to two rigid planes which intersect at right angles. Prove that on the line of intersection of the planes the excess of pressure due to the vortex varies inversely as the square of the distance of the vortex from the line of intersection. (L.U.)

A N SWE R S EXERCISE 1 4 1 6. 2ym21r.

14. 8yM(3 - 2v2)Ja•.

23. - � - -- [ va• + r2 - 2ar cos r

4y?Ta3p

24.

2y1rma

centre.

� [1 - a(a• + x•)-!J .

y

oc

r

- r + a],

=

distance

from

X

28. II = 0, 1r (stable), !1r, �1T (unstable), 6 = angle made by dipoles with r.

EXERCISE 15

Both charges !a from centre on same diameter. ( - .C.) a• Q over r = a. 4. p = Q (� !) exp (1 - rja) inside r = a. u = 4=:2 41ra2 r a 1ra

3. V = � 3 a

·

-

5. Energy from battery = work done by internal forces X 2. 6. q1ja + q 2jb + q3 jc , ( q1 + q2 )jb + q3 jc , ( q1 + q2 + q3 )fc . 8. A ¢> 2j81rd = 21rde 2jA . 7. (i) ab if> 2f2(b - a) ; (ii) t 22 c . b a 1 0. 21rE(d2 - x2)JAd. 21r£2xjA d.

(�

-f-

)

EXERCISE 1 6

3 . Field radial ; efr2 in cavity; eJKr2 in dielectric.

u = - (K - 1)ej4?TKa2

over wall of cavity, p = 0.

5. K2abej[K1 - K2) (b - c)a - K2bc]. (e1 + e 2) 2 • + 2b a" 3Eb3 cos 6 -r In d1electnc = (K + 2)b" + 2 (K _ 1)a• ;2 · (K 1)b3 + (2K + 1)a3 b3 Elsewhere V = E cos 6 -r (K + 2)b" + 2(K 1)a• ro · z is measured along axis from centre in direction of I. a -z a+z a a, z < -a. v (z - a) 2 + a2 v(z + a) 2 + a2 K - 1 a2 3/L 22 £ 0 K + l r - r _cos 6. . 21'- + 1 . . Repulsion along Ox of 4?Ta3>.JLfc4 24. 0 ; -!V21TI(3 + v2) .

4. (K + 1)a. e 12 _ 6 b · 2K a . .

8.

10.

1 6. 23.

( ! !)

(

{

V

{

{

)

)

(

-

_

}

}

}

-

•

EXERCISE 1 7 1. 3 5 ohms.

5. 6. 8. 13.

2. 5rf 1 1 . I/24, I/12, IJ6, 5IJ24 , 7IJ24 (I = total current) . ( V - C'R)JR(C - C') . Vo(2 2n-s - 2•) / (22n - 1) ; Vo(2 2n- 1 + 1 ) /r(2 2n - 1). 3,-1 ('r1 - T2) C 1 cos 6j41T(2,-2 + T1) . 648

9.

4Ef3R.

ANSWERS

l.

�7s (� + g) ·

EXERCISE 1 8

4. 2v'3j2/a perpendicular to BC.

15. Horizontal distance m2gtrfe2H2 from point of release. 19. Couple = TT2a4H2wRf2(R2 + w2L 2). 20. Current = 2TTmpa2(R sin pt - pL cos pt)f(R2 p2V) (a2 + c2)�. 26. 6tr2a4fc4.

30.

+ ;:{3 sin (nt - rx.xfc) along Ox.

EXERCISE 1 9 8. Equilibrium is made stable.

14. Paraboloid of revolution. EXERCISE 20 6. 0·528.

3. lm&x p0/pwB. 23. p = Po + Ap0 sin at sec (alfc). =

649

EXERCISE 21

1 . y = Cx• (Oy along axis of revolution) . 6. C is ( v'2 + 1)AB from B on AB produced. 9 . Towards the source. 7. (0, -a tan !e<) , (0, a cot !<X) . 1 1 . Po - fp;ws - pa2f18t2 where cos e = - 2af9At2. 13. ( ± v'3a, 0) . 17. = {(b3 V - a3 U) r + !a3b3 ( V - U)fr2} cos 6f(a3 - b3) . 18. p = pJrf' (r) dr.

INDEX ( The numbers refer to pages.)

636 Aether, 458 Alternator, 516 Ampere, 489 Ampere's circuital rule, 504 Ampere's law, 499 Ampere, unit, 470, 547 Amplitude of wave, 590 Angle of dip, 466 Anti-nodes, 596 Aperiodic discharge, 520 Atmosphere, Earth's, 558 AEROFOIL,

Bar, field of, 357 potential of, 375 Bar magnet, 464 Baroclinic fluid, 558 Barotropic fluid, 557 Bernoulli's equation, 579, 612, 624 Biot-Savart law, 496, 514 Boundary conditions, 481, 531 Buoyancy, centre of, 564 Capacitance, 379 Cauchy-Riemann equations, 432, 633 Cavity in dielectric, 456 cylindrical, 456 disc-shaped, 457 in magnetic material, 461 needle-shaped, 457 spherical, 453 Centrifugal field, 561 Circular current, field of, 492 Circulation, 506, 634 Co-axial circles, 435, 494 Collinear charges, 355 Condenser, cylindrical, 413 spherical, 379 with dielectric, 448 Conductivity, specific, 479 Conductor, induced charge on, 427 potential of, 372 Conjugate functions, 432 Conservative field, 368 Continuity, equation of, 478, 573, 615 Convective rate of change, 622 Coulomb, 353, 547 Coulomb's law, 353, 448, 548 Coupling, coefficient of, 522 Critical pressure ratio of rocket, 606 Curl, 506 Current density, electric, 4 76 Current density vector, surface, 531

Datum point, potential, 369 Demagnetizing field, 464 Diamagnetism, 459 Dielectric constant, 446 Dielectrics, 443 Dipole, couple acting on, 382 electric, 379 lines of force of, 382 magnetic, 383 moment of, 380 PE of, 383 potential of, 380 Dipoles, mutual PE of, 385 forces between, 387 Disc, field of, 361 potential of, 377 Displacement, electric, 446 Displacement current, 526 Divergence, 406 Doublet, electric, 383, 435 liquid, 616 Drag, 637 Dynamo, 517 Earth's magnetic field, 384, 465 Elastic string, energy of, 594, 599 vibration of, 589 Electric current, 470 Electrode, 477 line, 481 point, 480 strength of, 4 77 Electro-magnetic wave, 528 plane, 533 polarized, 535 Electrometer, attracted disc, 417 Electromotive force, 472 back, 517 Ellipticity of Earth, 563 Energy density in electric field, 458 in magnetic field, 461 Energy of charge distribution, 390 of condenser, 392 of irrotational flow, 629 rate of flow of electro-magnetic, 531 Equilibrium of rotating fluid, 561 Equipotential, 369, 433 Euler's dynamical equations, 578, 623 Farad, unit, 379, 547 Faraday's law, 5 16, 525 651

652

INDEX

Ferromagnetism, 460 Field, 354 two-dimensional, 365, 432 Floating body, 564 stability of, 567 Flow, velocity of, 6 1 1 Fluid, ideal, 573 Flux, 399 through a circuit, 498 Force on charge in magnetic field, 502 on conductor, 416 on electric circuit, 497 Fundamental vibration, 598 Gauss's theorem, for current flows, 480 for electrostatics, 400, 447 for magnetostatics, 460 Gradient, 369 Green's reciprocal theorem, 441 Green's theorem, 406 Harmonic function, 420, 451, 481 Harmonics, 598 Heat generated by a current, 475, 484 Henry, unit, 518, 548 Horn, design of, 604 Image charge, 424, 451 Impedance, 519 Induced charge, 372 on cylinder, 433 Inductance, mutual, 521 self, 51 'i' Induction, electromagnetic, 516 magnetic, 459 Inductor, 5 1 8 Insulator, 372 Intensity, electric, 354, 445 gravitational, 355 Irrotationa1 motion, 616 Isobars, 557 Kirchhoff 's first law, 471 Kirchhoff 's second law, 472 Laplace's equation, 409, 447, 460 Lift, 637 Light, velocity of, 489, 528 Line charge, 365, 435 Line of electric displacement, 448 of force, 355, 433 equations of, 357 of magnetic induction, 460 Magnetic shell, 490 Magnetism, permanent, 460 Magnetization, intensity of, 459 uniform, 463 Maxwell, 458, 525

Maxwell's equations, 526, 549 Metacentre, 566 Moment of current circuit, 489 Monochromatic light, 536 Neumann's formula, 522 Newton's law of gravitation, 353 Nodes, 596 Normal mode, 596 Octave, 602 Oersted, 489 Ohm, unit, 471, 548 Ohm's law, 471, 479 Organ pipe, 602 Paramagnetism, 459 PE in electric field, 369 of electric current, 497 Permeability, 460 Pipe, flow in, 573 Pitot tube, 6 1 3 Plane charge, field of, 4 1 1 Plate, field o f uniform, 361 Poisson's equation, 409, 528 Poisson's equivalent distribution, elec­ tric, 445 magnetic, 459 Polarization, electric, 443 plane of, 535 Polarized atom, 379 Poles, magnetic, 383 Potential, complex, 432 Earth's, 372 of electric field, 369 of particle or point charge, 371 of surface distribution, 373 of volume distribution, 373 scalar, 527 vector, 512, 527 Poynting vector, 529 Pressure in a liquid, 555, 573 Reflection, law of, 537 Reflection coefficient, 593 Refraction, coefficient of, 537 Region, simply connected, 538 Relaxation time, 529 Resistance, 4 71 internal, of a battery, 472 specific, 4 79 Resonant frequency, 520 Rocket motor, 583 Separation of variables, method of, 595 Shock wave, 587 Sink, 616 Snell's law, 537 Solenoid, 492 Solid angle, 359

INDEX Sound waves, 600 Source, line, 616 point, 615 strength of, 615, 616 Sphere, field of uniform, 364 potential of uniform, 377 uniformly magnetized, 464 Spherical shell, field of, 362, 4.02 potential of, 377 Stagnation point, 613 Standing wave, 597 Steady state, 519 Stokes' theorem, 509 Straight current, field of, 490 Stream function, 631 Streamline, 6 1 1 Stream tube, 611 Surface charge, 410, 447 Susceptibility, electric, 444 magnetic, 459 Temperature lapse rate, 559 Torricelli's theorem, 612 Total rate of change, 622 Transformer, 523 Transient, 519

653

Tube o f electric displacement, 448 of force, 404 strength of, 405 Typhoon, 639 Uniform stream, 617 Uniqueness principle, 591 Uniqueness theorem, for electrostatics, 418, 451 for flow of current, 481 Units, electromagnetic, 489 electrostatic, 353 Gaussian, 545 Giorgi system of, 547 rationalized Gaussian, 546 Velocity potential, 578, 616 Venturi meter, 580 Volt, unit, 369, 547 Vortex, rectilinear, 638 velocity induced by, 639 Vorticity, 623, 637 Wave equation, 528 one-dimensional, 534, 590 Weber, unit, 548