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Solved Problems in Classical Mechanics

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Solved Problems in Classical

Mechanics

Analytical and numerical solutions

with comments

O.L. de Lange and J. Pierrus

School of Physics, University of KwaZulu-Natal,

Pietermaritzburg, South Africa

1 3

Great Clarendon Street, Oxfordox26dp

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Preface

It is in the study of classical mechanics that we first encounter many of the basic ingredients that are essential to our understanding of the physical universe. The concepts include statements concerning space and time, velocity, acceleration, mass, momentum and force, and then an equation of motion and the indispensable law of action and reaction - all set (initially) in the background of an inertial frame of reference. Units for length, time and mass are introduced and the sanctity of the balance of units in any physical equation (dimensional analysis) is stressed. Reference is also made to the task of measuring these units - metrology, which has become such an astonishing science/art. The rewards of this study are considerable. For example, one comes to appreciate Newton"s great achievement - that the dynamics of the classical universe can be understood via the solutions of di1erential equations - and this leads on to questions regarding determinism and the e1ects of even small uncertainties or disturbances. One learns further that even when Newton"s dynamics fails, many of the concepts remain indispensable and some of its conclusions retain their validity - such as the conservation laws for momentum, angular momentum and energy, and the connection between conservation and symmetry - and one discusses the domain of applicability of the theory. Along the way, a student encounters techniques - such as the use of vector calculus - that permeate much of physics from electromagnetism to quantum mechanics. All this is familiar to lecturers who teach physics at universities; hence the emphasis on undergraduate and graduate courses in classical mechanics, and the variety of excellent textbooks on the subject. It has, furthermore, been recognized that training in this and related branches of physics is useful also to students whose careers will take them outside physics. It seems that here the problem-solving abilities that physics students develop stand them in good stead and make them desirable employees. Our book is intended to assist students in acquiring such analytical and computational skills. It should be useful for self-study and also to lecturers and students in mechanics courses where the emphasis is on problem solving, and formal lectures are kept to a minimum. In our experience, students respond well to this approach. After all, the rudiments of the subject can be presented quite succinctly (as we have endeavoured to do in Chapter 1) and, where necessary, details can be filled in using a suitable text. With regard to the format of this book: apart from the introductory chapter, it consists entirely of questions and solutions on various topics in classical mechanics that are usually encountered during the first few years of university study. It is

01Solved Problems in Classical Mechanics

suggested that a student first attempt a question with the solution covered, and only consult the solution for help where necessary. Both analytical and numerical (computer) techniques are used, as appropriate, in obtaining and analyzing solutions. Some of the numerical questions are suitable for project work in computational physics (see the Appendix). Most solutions are followed by a set of comments that are intended to stimulate inductive reasoning (additional analysis of the problem, its possible ex- tensions and further significance), and sometimes to mention literature we have found helpful and interesting. We have included questions on bits of 'theory" for topics where students initially encounter di2culty - such as the harmonic oscillator and the theory of mechanical energy - because this can be useful, both in revising and cementing ideas and in building confidence. The mathematical ability that the reader should have consists mainly of the following: an elementary knowledge of functions - their roots, turning points, asymp- totic values and graphs - including the 'standard" functions of physics (polynomial, trigonometric, exponential, logarithmic, and rational); the di1erential and integral calculus (including partial di1erentiation); and elementary vector analysis. Also, some knowledge of elementary mechanics and general physics is desirable, although the extent to which this is necessary will depend on the proclivities of the reader.

For our computer calculations we useMathematica

R1 , version 7.0. In each instance the necessary code (referred to as a notebook) is provided in a shadebox in the text. Notebooks that include the interactiveManipulatefunction are given in Chapters

6, 10, 11 and 13 (and are listed in the Appendix). They enable the reader to observe

motion on a computer screen, and to study the e1ects of changing relevant parameters. A reader without prior knowledge ofMathematicashould consult the tutorial ('First Five Minutes withMathematica") and the on-line Help. Also, various useful tutorials can be downloaded from the website www.Wolfram.com. All graphs of numerical results have been drawn to scale using Gnuplot. In our analytical solutions we have tried to strike a balance between burdening the reader with too much detail and not heeding Littlewood"s dictum that "two trivialities omitted can add up to an impasse". In this regard it is probably not possible to satisfy all readers, but we hope that even tentative ones will soon be able to discern footprints in the mist. After all, it is well worth the e1ort to learn that (on some level) the rules of the universe are simple, and to begin to enjoy "the unreasonable e1ectiveness of mathematics in the natural sciences" (Wigner). Finally, we thank Robert Lindebaum and Allard Welter for their assistance with our computer queries and also Roger Raab for helpful discussions.

Pietermaritzburg, South AfricaO. L. de Lange

January 2010J. Pierrus

Contents

1 Introduction1

2 Miscellanea11

3 One-dimensional motion30

4 Linear oscillations60

5 Energy and potentials92

6 Momentum and angular momentum127

7 Motion in two and three dimensions157

8 Spherically symmetric potentials216

9 The Coulomb and oscillator problems263

10 Two-body problems286

11 Multi-particle systems325

12 Rigid bodies399

13 Non-linear oscillations454

14 Translation and rotation of the reference frame518

15 The relativity principle and some of its consequences557

Appendix588

Index590

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1

Introduction

The following outline of the rudiments of classical mechanics provides the background that is necessary in order to use this book. For the reader who “nds our presentation too brief, there are several excellent books that expound on these basics, such as those listed below. [1Š4]

1.1 Kinematics and dynamics of a single particle

The goal of classical mechanics is to provide a quantitative description of the motion of physical objects. Like any physical theory, mechanics is a blend of de“nitions and postulates. In describing this theory it is convenient to “rst introduce the concept of a point object (a particle) and to start by considering the motion of a single particle. To this end one must make an assumption concerning the geometry of space. In Newtonian dynamics it is assumed that space is three-dimensional and Euclidean. That is, space is spanned by the three coordinates of a Cartesian system; the distance between any two points is given in terms of their coordinates by Pythagorass theorem, and the familiar geometric and algebraic rules of vector analysis apply. It is also assumed ... at least in non-relativistic physics ... that time is independent of space. Furthermore, it is supposed that space and time are su2ciently continuous that the di1erential and integral calculus can be applied. A helpful discussion of these topicsisgiveninGri2thssbook. [2] With this background, one selects a coordinate system. Often, this is a rectangular or Cartesian system consisting of an arbitrarily chosen coordinate originOand three orthogonal axes, but in practice any convenient system can be used (spherical, cylin- drical, etc.). The position of a particle relative to this coordinate system is speci“ed by a vector function of time ... the position vectorr(t). An equation forr(t)is known as the trajectory of the particle, and “nding the trajectory is the goal mentioned above. In terms ofr(t)we de“ne two indispensable kinematic quantities for the particle: the velocityv(t), which is the time rate of change of the position vector, [1] L. D. Landau, A. I. Akhiezer, and E. M. Lifshitz,General physics: mechanics and molecular physics. Oxford: Pergamon, 1967. [2] J. B. Gri1ths,The theory of classical dynamics. Cambridge: Cambridge University Press, 1985.
[3] T. W. B. Kibble and F. H. Berkshire,Classical mechanics. London: Imperial College Press,

5th edn, 2004.

[4] R. Baierlein,Newtonian dynamics. New York: McGraw-Hill,1983.

0Solved Problems in Classical Mechanics

v(t)=dr(t) dt,(1) and the accelerationa(t), which is the time rate of change of the velocity, a(t)=dv(t) dt.(2) It follows from (1) and (2) that the acceleration is also the second derivative a=d 2 r dt 2 .(3) Sometimes use is made of Newtons notation, where a dot denotes di1erentiation with respect to time, so that (1)...(3) can be abbreviated v=33r,a=33v=¨r.(4) The stage for mechanics ... the frame of reference ... consists of a coordinate system together with clocks for measuring time. Initially, we restrict ourselves to an inertial frame. This is a frame in which an isolated particle (one that is free of any applied forces) moves with constant velocityv... meaning thatvis constant in both magnitude and direction (uniform rectilinear motion). This statement is the essence of Newtons “rst law of motion. In Newtons mechanics (and also in relativity) an inertial frame is not a unique construct: any frame moving with constant velocity with respect to it is also inertial (see Chapters 14 and 15). Consequently, if one inertial frame exists, then in“nitely many exist. Sometimes mention is made of a primary inertial frame, which is at rest with respect to the “xed stars. Now comes a central postulate of the entire theory: in an inertial frame, if a particle of massmis acted on by a forceF,then F=dp dt,(5) where p=mv(6) is the momentum of the particle relative to the given inertial frame. Equation (5) is the content of Newtons second law of motion: it provides the means for determining the trajectoryr(t), and is known as the equation of motion. If the mass of the particle is constant then (5) can also be written as m dv dt=F,(7) or, equivalently, m d 2 r dt 2 =F.(8) The theory is completed by postulating a restriction on the interaction between any two particles (Newtons third law of motion): ifF 12 is the force that particle 1 exerts on particle 2, and ifF 21
is the force that particle 2 exerts on particle 1, then

Introduction0

F 21
=ŠF 12 .(9) That is, the mutual actions between particles are always equal in magnitude and opposite in direction. (See also Question 10.5.) The realization that the dynamics of the physical world can be studied by solving di1erential equations is one of Newtons great achievements, and many of the problems discussed in this book deal with this topic. His theory shows that (on some level) it is possible to predict the future and to unravel the past. The reader may be concerned that, from a logical point of view, two new quantities (mass and force) are introduced in the single statement (5). However, by using both the second and third laws, (5) and (9), one can obtain an operational de“nition of relative mass (see Question 2.6). Then (5) can be regarded as de“ning force. Three ways in which the equation of motion can be applied are:

1Use a trajectory to determine the force. For example, elliptical planetary orbits ...with the Sun at a focus ... imply an attractive inverse-square force (see Question

8.13).

1Use a force to determine the trajectory. For example, parabolic motion in a

uniform “eld (see Question 7.1).

1Use a force and a trajectory to determine particle properties. For example,

the electric charge from rectilinear motion in a combined gravitational and electrostatic “eld, and the electric charge-to-mass ratio from motion in uniform electrostatic and magnetostatic “elds (see Questions 3.11, 7.19 and 7.20).

1.2 Multi-particle systems

The above formulation is readily extended to multi-particle systems. We follow stan- dard notation and letm i andr i denote the mass and position vector of theithparticle, wherei=1,2,···,Nfor a system ofNparticles. The velocity and acceleration of the ithparticle are denotedv i anda i , respectively. The equations of motion are F i =dp i dt(i=1,2,···,N),(10) wherep i =m i v i is the momentum of theithparticle relative to a given inertial frame, andF i is the total force on this particle.

In writing down theF

i it is useful to distinguish between interparticle forces, due to interactions among the particles of the system, and external forces associated with sources outside the system. The total force on particleiis the vector sum of all interparticle and external forces. Thus, one writes F i =1 j2=i F ji +F (e i (i=1,2,···,N),(11) whereF ji is the force that particlejexerts on particlei,andF (e i is the external force on particlei. In (11) the sum overjruns from 1 toNbut excludesj=i.The interparticle forces are all assumed to obey the third law

0Solved Problems in Classical Mechanics

F ji =ŠF ij (i,j=1,2,···,N).(12) From (10) and (11) we have the equations of motion of a system of particles in terms of interparticle forces and external forces: dp i dt=1 j2=i F ji +F (e i (i=1,2,···,N).(13)

If the massesm

i are all constant then (13) can be written as m i d 2 r i dt 2 =1 j2=i F ji +F (e i (i=1,2,···,N).(14) These are the equations of motion for the classicalN-particle problem. In general, they are a set ofNcoupled di1erential equations, and they are usually intractable. Two of the four presently known fundamental interactions are applicable in classical mechanics, namely the gravitational and electromagnetic forces. For the former, Newtons law of gravitation is usually a satisfactory approximation. For electromagnetic forces there are Coulombs law of electrostatics, the Lorentz force, and multipole interactions. Often, it is impractical to deduce macroscopic forces (such as friction and viscous drag) from the electromagnetic interactions of particles, and instead one uses phenomenological expressions. Another method of approximating forces is through the simple expedient of a spatial Taylor-seriesexpansion, which opens the way to large areas of physics. Here, the “rst (constant) term represents a uniform “eld; the second (linear) term encompasses a Hookes-law-type force associated with linear (harmonic) oscillations; the higher-order (quadratic, cubic, ... ) terms are non-linear (anharmonic) forces that produce a host of non-linear e1ects (see Chapter 13). Also, there are many approximate representations of forces in terms of various potentials (Lennard-Jones, Morse, Yukawa, Pöschl...Teller, Hulthén, etc.), which are useful in molecular, solid-state and nuclear physics. The Newtonian concepts of force and potential have turned out to be widely applicable ... even to the statics and dynamics of such esoteric yet important systems as "ux quanta (Abrikosov vortices) in superconductors and line defects (dislocations) in crystals. Some of the most impressive successes of classical mechanics have been in the “eld of astronomy. And so it seems ironic that one of the major unanswered questions in physics concerns observed dynamics ... ranging from galactic motion to accelerating expansion of the universe ... for which the source and nature of the force are uncertain (dark matter and dark energy, see Question 11.20).

1.3 Newton and Maxwell

The above outline of Newtonian dynamics relies on the notion of a particle. The theory can also be formulated in terms of an extended object (a body). This is the form

Introduction0

used originally by Newton, and subsequently by Maxwell and others. In his fascinating study of thePrincipia Mathematica, Chandrasekhar remarks that Maxwell"s "is a rarely sensitive presentation of the basic concepts of Newtonian dynamics" and "is so completely in the spirit of thePrincipiaand illuminating by itself ... ." [5] Maxwell emphasized "that by the velocity of a body is meant the velocity of its centre of mass. The body may be rotating, or it may consist of parts, and be capable of changes of configuration, so that the motions of di1erent parts may be di1erent, but we can still assert the laws of motion in the following form: Law I. - The centre of mass of the system perseveres in its state of rest, or of uniform motion in a straight line, except in so far as it is made to change that state by forces acting on the system from without. Law II. - The change of momentum during any interval of time is measured by the sum of the impulses of the external forces during that interval." [5] In Newtonian dynamics, the position of the centre of mass of any object is a unique point in space whose motion is governed by the two laws stated above. The concept of the centre of mass occurs in a straightforward manner [5] (see also Chapter 11) and it plays an important role in the theory and its applications.

Often, the trajectory of the centre of mass

relative to an inertial frame is a simple curve, even though other parts of the body may move in a more complicated manner. This is nicely illustrated by the motion of a uniform rod thrown through the air: to a good approximation, the centre of mass describes a simple parabolic curve such as P in the figure, while other points in the rod may follow a more complicated three-dimensional trajectory, like Q. If the rod is PQ thrown in free space then its centre of mass will move with constant velocity (that is, in a straight line and with constant speed) while other parts of the rod may have more intricate trajectories. In general, the motion of a free rigid body in an inertial frame is more complicated than that of a free particle (see Question 12.22).

1.4 Newton and Lagrange

The first edition of thePrincipia Mathematicawas published in July 1687, when Newton was 44years old. Much of it was worked out and written between about August

1684 and May 1686, although he first obtained some of the results about twenty years

earlier, especially during the plague years 1665 and 1666 "for in those days I was in the prime of my age for invention and minded Mathematicks and Philosophy more than at any time since." [5] After Newton had laid the foundations of classical mechanics, the scene for many subsequent developments shifted to the Continent, and especially France, where [5] S. Chandrasekhar,Newton"s Principia for the common reader, Chaps. 1 and 2. Oxford: Claren- don Press, 1995.

0Solved Problems in Classical Mechanics

important works were published by d"Alembert (1717-1783), Lagrange (1736-1813), de Laplace (1749-1827), Legendre (1725-1833), Fourier (1768-1830), Poisson (1781-

1840), and others. In particular, an alternative formulation of classical particle

dynamics was presented by Lagrange in hisMécanique Analytique(1788). To describe this theory it is helpful to consider first a single particle of constant mass mmoving in an inertial frame. We suppose that all the forces acting are conservative: then the particle possesses potential energyV(r)in addition to its kinetic energy K= 1 2 m33r 2 , and the force is related toV(r)byF=-1V(see Chapter 5). So, Newton"s equation of motion in Cartesian coordinatesx 1 ,x 2 ,x 3 has components m¨x i =F i =-1V21x i (i=1,2,3).(15)

Also,1K21x

i =0,1K2133x i =m33x i ,and1V2133x i =0. Therefore (15) can be recast in the formd dt1L133x i -1L 1x i =0 (i=1,2,3),(16) whereL=K-V.ThequantityL(r,33r)is known as the Lagrangian of the particle. The Lagrange equations (16) imply that the action integral I=3 t 2 t 1

Ldt(17)

is stationary (has an extremum - usually a minimum) for any small variation of the coordinatesx i :

2I=0.(18)

Equations (16) hold even ifVis a function oft,aslongasF=-1V.

This account can be generalized:

1It applies to systems containing an arbitrary number of particlesN.

1The coordinates used need not be Cartesian; they are customarily denotedq

1 ,q 2 ,

···,q

f (f=3N) and are known as generalized coordinates. (In practice, the choice of these coordinates is largely a matter of convenience.) The corresponding time derivatives are the generalized velocities, and the Lagrangian is a function of these6Ncoordinates and velocities: L=L(q 1 ,q 2 ,···,q f ;33q 1 ,33q 2 ,···,33q f ).(19)

Often, we will abbreviate this toL=L(q

i ,33q i ).

1The Lagrangian is required to satisfy the action principle (18), and this impliesthe Lagrange equations

d dt1L133q i -1L 1q i =0 (i=1,2,···,3N),(20) whereL=K-V,andKandVare the total kinetic and potential energies of the system. [2]

Introduction0

1The Lagrangian formulation applies also to non-conservative systems such as

charged particles in time-dependent electromagnetic “elds and damped harmonic oscillators (see Question 4.16). Lagrangians can also be constructed for systems with variable mass. In these instancesLis not of the formKŠV.

1The Lagrange equations (20) can be expressed as

dp i 2dt=F i ,(21) where p i =1L2133q i andF i =1L21q i (22) are known as the generalized momenta and generalized forces. In Cartesian coordinates,pis equal to mass×velocity.

1The action principle (18) is valid in any frame of reference, even a non-inertial

frame (one that is accelerating relative to an inertial frame). However, in a non- inertial frame the Lagrangian is modi“ed by the acceleration, and Lagranges equations (16) yield the equation of motion (24) below ... see Question 14.22. Although the Newtonian formulation (based on force) and the Lagrangian formulation (based on a scalarLthat often derives from kinetic and potential energies) look very di1erent, they are completely equivalent and must yield the same results in practice. There are several reasons for the importance of the Lagrange approach, such as:

1It may be simpler to obtain the equation of motion by working with energy ratherthan by taking account of all the forces.

1Constrained motion is more easily treated.

1Conserved quantities can be readily identi“ed.

1The action principle is a fundamental part of physics, and it provides a

powerful formulation of classical mechanics. For example, the theory can be extended to continuous systems by introducing a Lagrangian density whose volume integral is the Lagrangian. In this version the Lagrangian formulation has important applications to “eld theory and quantum mechanics.

1.5 Non-inertial frames of reference

This section outlines a topic that is considered in more detail in Chapter 14 and is used occasionally in earlier chapters. Often, the frame of reference that one uses is not inertial, either by circumstance (for example, a frame “xed on the Earth is non-inertial) or by choice (it may be convenient to solve a particular problem in a non-inertial frame). And so the question arises: what is the form of the equation of motion in a non-inertial frame (that is, a frame that is accelerating with respect to an inertial frame)?

This leads one to consider a frame S

3 that is translating and rotating with respect to an inertial frame S. These frames are depicted in the “gure below, whereris the position vector of a particle of massmrelative to S andr 3 is its position vector relative to S 3 .TheframeS 3 has originO 3 and coordinate axesx 3 y 3 z 3 .

0Solved Problems in Classical Mechanics

The motion of S

3 is described by two vectors: the position vectorD(t)of the originO 3 relative to S, and the angular velocity1(t)of S 3 relative to a third frame S 33
that has origin atO 3 and axesx 33
y 33
z 33
, which are parallel to the corresponding axesxyzof S. This angular velocity is given in terms of a unit vector

32n(that specifies the

axis of rotation relative to S 33
) and the angled3rotated through in a timedtby 1=d3 dt32n,(23) where the sense of rotation and the direction of

32nare connected by the right-hand

rule illustrated in the figure. Starting from the equation of motion (8) for a single particle of constant massm in an inertial frame S, it can be shown that the equation of motion in the translating and rotating frame S 3 can be expressed in the form (see Chapter 14) m d 2 r 3 dt 2 =F e .(24) Here F e =F+F tr +F Cor +F cf +F az ,(25) where F tr =Šmd 2 D dt 2 ,(26) F Cor =Š2m1×dr 3 dt,(27) F cf =Šm1×(1×r 3 ),(28) F az =Šmd1 dt×r 3 .(29)

Introduction0

We mention that (24) is not a separate postulate, but is a consequence of (8) and the assumptions that space is absolute (meaningr=r 3 +Din the first of the above figures), time is absolute (meaningt 3 =t), and mass is absolute (meaningm 3 =m).

Note that the relationr=r

3 +Dis not simply a consequence of the triangle law for addition of vectors, becauserandr 3 are measured by observers who are moving relative to each other - see Chapter 15. We can interpret the equation of motion (24) in the following way: if we wish to write Newton"s second law in a non-inertial frame S 3 in the same way as in an inertial frame S (i.e. as force=mass×acceleration), then the forceFdue to physical interactions (such as electromagnetic interactions) must be replaced by an e1ective forceF e that includes the four additional contributionsF tr ,F Cor ,F cf ,and F az . Collectively, these contributions are variously referred to in the literature as:

1'inertial forces" (because each involves the particle"s inertial massm);

1'non-inertial forces" (because each is present only in a non-inertial frame);

1'fictitious forces" (to emphasize that they are not due to physical interactions but

to the acceleration of the frame S 3 relative to S). Each of the forces (26)-(29) also has its own name:F tr is known as the translational force (it occurs whenever the origin of the non-inertial system accelerates relative to an inertial frame);F Cor is the Coriolis force (it acts on a moving particle unless the motion in S 3 is parallel or anti-parallel to1);F cf is the centrifugal force, and it acts even on a particle at rest in S 3 ;F az is the azimuthal force, and it occurs only if the non-inertial frame has an angular accelerationd12dtrelative to S.

1.6 Homogeneity and isotropy of space and time

In addition to the fact that the laws of motion assume their simplest forms in inertial frames, these frames also possess unique properties with respect to space and time. For a free particle in an inertial frame these are: First, all positions in inertial space are equivalent with regard to mechanics. This is known as the homogeneity of space in inertial frames. Secondly, all directions in space are equivalent. This is the isotropy of space. Thirdly, all instants of time are equivalent (homogeneity of time). Fourthly, there is invariance with respect to reversal of motion - the replacementt1-t(isotropy of time). These symmetries of space and time in inertial frames play a fundamental role in physics. For example, in the conservation laws for energy, momentum and angular momentum, and in the space-time transformation between inertial frames (see Chapters 14 and 15). In a non-inertial frame these properties do not hold. For example, if one stands on a rotating platform it is noticeable that positions on and o1 the axis of rotation are not equivalent: space is not homogeneous in such a frame. Notwithstanding the fact that, in general, Newtonian dynamics is most simply formulated in inertial space, one should keep in mind the following proviso. Namely, that the solution to certain problems is facilitated by choosing a suitable non-inertial frame. Thus the trajectory of a particle at rest on a rotating turntable is simplest in the frame of the turntable, where the particle is in static equilibrium under the

01Solved Problems in Classical Mechanics

action of four forces (weight, normal reaction, friction and centrifugal force). Similarly, for a charged particle in a uniform magnetostatic field, one can transform away the magnetic force: relative to a specific rotating and translating frame the particle is in static equilibrium, whereas relative to inertial space the trajectory is a helix of constant pitch (see Question 14.25).

1.7 The importance of being irrelevant

There are several obvious questions one can ask concerning Newtonian dynamics, which can all be formulated: 'Does it matter if···?" All are answered in the negative and have deep consequences for physics. The first concerns the units in which mass, length, and time are measured. Humans (and probably also other life in the universe) have devised an abundance of di1erent physical units. In principle, there are infinitely many and one can ask whether the validity of Newton"s second law is a1ected by an arbitrary choice of units. The answer is 'no": the law is valid in any system of units because each side of the equationF=ma must have the same units (see also Question 2.9). Thus, the unit of force in the MKS system (the newton) is, by definition, 1kgms Š2 . This seemingly simple property is required of all physical laws: they do not depend on an arbitrary choice of units because each side of an equation expressing the law is required to have the same physical dimensions. The consequences of this are dimensional analysis (see Chapter 2), similarity and scaling. [6]

The fact that physical

laws are equally valid in all systems of units is an example of a 'relativity principle". Similarly, one can ask whether the mechanical properties of an isolated (closed) system depend in any way on its position or orientation in inertial space. The statement that they do not implies, respectively, the conservation of momentum and angular momentum of the system (see Questions 14.7, 14.18 and 14.19). Furthermore, in Newtonian dynamics any choice of inertial frame (from among an infinite set of frames in uniform, rectilinear relative motion) is acceptable because the laws of motion are equally valid in all such frames. The extension of this property to all the laws of physics constitutes Einstein"s relativity principle. A remarkable consequence of this principle is that there are just two possibilities for the space-time transformation between inertial frames: relative space-time (in a universe in which there is a finite universal speed) or Newton"s absolute space-time (if this speed is infinite) - see Chapter 15. Further extensions of this type of reasoning have led to a theory of elementary particles and their interactions. [7] So, this concept of irrelevance (or invariance, as it is known in physics) which emerged from Newton"s mechanics, and was later emphasized particularly by Einstein, has turned out to be extremely fruitful. The reader may wonder what physics would be like if these invariances did not hold. [6] G.I.Barenblatt,Scaling, self-similarity, and intermediate asymptotics. Cambridge: Cambridge

University Press, 1996.

[7] See, for example, G. t Hooft, Gauge theories of the forces between elementary particles,Ž

Scienti“c American, vol. 242, pp. 90...116, June 1980. 2

Miscellanea

This chapter contains questions dealing with three disparate topics, namely sensitivity of trajectories to small changes in initial conditions; the reasons why we consider just one, rather than three types of mass; and the use of dimensional reasoning in the analysis of physical problems. The reader may wish to omit this chapter at “rst, and return to it at a later stage.

Question 2.1

A particle moves in one dimension along thex-axis, bouncing between two perfectly re"ecting walls atx=0andx=4. In between collisions with the walls no forces act on the particle. Suppose there is an uncertainty1v 0 in the initial velocityv 0 . Determine the corresponding uncertainty1xin the position of the particle after a timet.

Solution

In between the instants of re"ection, the particle moves with constant velocity equal to the initial value. Thus, if the initial velocity isv 0 then the distance moved by the particle in a timetisv 0 t , whereas if the initial velocity isv 0 +1v 0 the distance moved is(v 0 +1v 0 )t. Therefore, the uncertainty in position after a timetis 1x=(v 0 +1v 0 )tŠv 0 t =(1v 0 )t.(1)

Comments

(iA ccordingto (1 ),a ftera time t c =4/1v 0 has elapsed,1x=4, meaning that the position at timet c is completely undetermined. (iiF ort imest2t c , (1) shows that the uncertainty1x24, and one can still regard the motion as deterministic (in the sense mentioned in Question 3.1). However, if we wait long enough the particle can be found anywhere between the walls: determinism has changed into complete indeterminism. (iiiIt is o nlyin the idea l(a nduna ttainable)c ase1v 0 =0(i.e. the initial velocity is known exactly) that deterministic motion persists inde“nitely. (ivIn no n-linears ystemsthe uncer taintyca nincr easem uchf asterw itht ime (exponentially rather than linearly) due to chaotic motion (see Chapter 13).

01Solved Problems in Classical Mechanics

Question 2.2

A ball moves freely on the surface of a round billiard table, and undergoes elastic re"ections at the boundary of the table. The motion is frictionless, and once started it continues inde“nitely. The initial conditions are that the ball starts at a point A on the boundary and that the chord AB drawn in the direction of the initial velocity subtends an angle5at the centre O of the table. Discuss the dependence of the trajectory of the ball on5.

Solution

Because the collisions with the wall are elastic, the an- gles of incidence and re"ection are equal (cf. the angles

6in the “gure). Thus, the angular positions of succes-

sive points of impact with the boundary are each rotated through5(the chords AB, BC,...in the “gure all sub- tend an angle5at O). We may therefore distinguish between two types of trajectory:

15is equal to27times a rational number, that is

5=27p q,(1) wherepandqare integers. Then, afterqre"ections at the wall the point of impact will have rotated through an angle q5=27p(2) from A. That is, the ball will have returned to A. The trajectory is a closed path of “nite length, and the motion is periodic.

15is equal to27times an irrational number. The angle of rotation of the point

of impact with the wall (q5afterqimpacts) is not equal to27times an integer; the ball will never return to the starting position A ... the trajectory is open and non-periodic.

Comments

(iT hisq uestion,lik et hepr eviouso ne,sho wst hats mallca usesca nh aveb ig consequences. Here, the slightest change in the initial velocity can change a closed trajectory into an open one. Consequently, determinism over inde“nitely long periods of time can be achieved only in the unphysical limit where the uncertainty in the initial velocity is precisely zero. (iiO thers ystemssho winge xtremesensitivit yt oinitia lco nditionsc anr eadilyb e constructed (see Questions 3.3 and 4.2).

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(iiiO nthe b asiso fthese, questio nsw erer aisedb yB orna ndo thersco ncerning the deterministic nature of classical mechanics. [1,2]

These examples show that

"determinism is an idealization rather than a statement of fact, valid only under the assumption that unlimited accuracy is within our reach, an assumption which in view of the atomic structure of our measuring instruments is anything but realistic." [2] The examples depict "a curious half-way house, showing not so much the fall as the decline of causality - the point, that is, where the principle begins to lose its applicability." [2] (See also Chapter 13.) At the atomic level uncertain- ties of a more drastic sort were encountered that required the abandonment of deterministic laws in favour of the statistical approach of quantum mechanics.

Question 2.3

The active gravitational mass(m

A )of a particle is an attribute that enables it to establish a gravitational field in space, whereas the passive gravitational mass(m P )is an attribute that enables the particle to respond to this field. (a) Write Newton"s law of universal gravitation in terms of the relevant active and passive gravitational masses. (bSho wt hatt hethir dla wo fmo tionma kesit unnecessa ryto disting uishb etween active and passive gravitational mass.

Solution

(a) The gravitational forceF 12 that particle 1 exerts on particle 2 is proportional to the product of the active gravitational massm A 1 of particle 1 and the passive gravitational massm P 2 of particle 2. Thus, the inverse-square law of gravitation is F 12 =ŠGm A 1 m P 2 r 2

32r,(1)

whereGis the universal constant of gravitation,ris the distance between the particles and

32ris a unit vector directed from particle 1 to particle 2. By the same

token, the forceF 21
which particle 2 exerts on particle 1 is F 21
=Gm A 2 m P 1 r 2

32r.(2)

(bA ccordingt oN ewton"st hirdla w,F 12 =ŠF 21
. It therefore follows from (1) and (2) that m A 2 m P 2 =m A 1 m P 1 .(3) We conclude from (3) that the ratio of the active to the passive gravitational mass of a particle is a universal constant. Furthermore, this constant can be [1] M. Born,Physics in my generation, pp. 78-82. New York: Springer, 1969. [2] F. Waismann, inTurning points in physics. Amsterdam: North-Holland, 1959. Chap. 5.

01Solved Problems in Classical Mechanics

incorporated in the universal constantG, which is already present in (1) and (2).

That is, we can setm

P =m A . There is no need to distinguish between active and passive gravitational masses; it is su2cient to work with just gravitational mass m G and to write (1) as F 12 =-Gm G 1 m G 2 r 2

32r.(4)

Comment

Evidently, the same reasoning applies to the notions of active and passive electric charge. Thus, if one were to write Coulomb"s law for the electrostatic force between two charges in vacuum as F 12 =kq A 1 q P 2 r 2

32r,(5)

wherekis a universal constant, a discussion similar to the above would lead to q A 2 q P 2 =q A 1 q P 1 .(6) Consequently, the ratio of active to passive charge is a universal constant that can be included inkin (5); it is su2cient to consider just electric chargeq.

Question 2.4

The inertial mass of a particle is, by definition, the mass that appears in Newton"s second law. Consider free fall of a particle with gravitational massm G and inertial massm I near the surface of a homogeneous planet having gravitational massM G and radiusR. Express the gravitational accelerationaof the particle in terms of these quantities. (Neglect any frictional forces.)

Solution

The equation of motion is

m I a=F,(1) whereFis the gravitational force exerted by the planet F=GM G m G R 2 (2) (see Question 11.17). Thus a=m G m I GM G R 2 .(3)

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Comments

(i) In many treatments of this topic the factorm G /m I in (3) is absent because it is tacitly assumed that the gravitational and inertial masses are equal. (iiEqua tion(3 )is a pproximateinso fara sit neg lectsa tmosphericd rag( seeQ uestion

3.13) and motion of the planet toward the falling object (see Question 11.23).

Nevertheless, it is an important idealization. The “rst signi“cant work in this connection was by Galileo, who enunciated an empirically based result that, in the absence of drag, all bodies fall with the same gravitational acceleration. This is sometimes referred to as Galileos law of free fall. (iiiGa lileosla w,t ogetherwith (3 ),enco uragedt heh ypothesistha tg ravitationala nd inertial masses can be taken to be the same,m G =m I , and one need consider only mass. This is the weak equivalence principle, which plays an important role in the formulation of the general theory of relativity. (ivBeca useo fi tsimp ortance,n umerousexp erimentsh aveb eenp erformedto test Galileos law, and hence the weak equivalence principle. Modern experiments show [3] that bodies fall with the same acceleration to a few parts in10 13 .Ž See also Question 2.5.

Question 2.5

In Question 4.3 an expression is derived for the periodTof a simple pendulum, tacitly assuming equality of the inertial and gravitational massesm I andm G of the bob. Study this calculation and then adapt it to apply whenm I andm G are allowed to be di1erent, thereby obtaining the dependence ofTon these masses.

Solution

In terms ofm

I andm G , the equation of motion (2) of Question 4.3 is m I d 2 s dt 2 n=Šm G gsin3n,(1) whereg=GM G 2R 2

4see (2) of Question 2.45and other symbols have the same

meaning as in Question 4.3. Then, for small oscillations (|3|21)we see from (1), that (4) of Question 4.3 is replaced by d 2 3 dt 2 +m G m I g

43=0,(2)

where4is the length of the pendulum. Thus, we obtain the desired expression for the period T=276 m I m G 4 g.(3) Whenm I =m G this reduces to the result in Question 4.3.

[3] C. M. Will, Relativity at the centenary,ŽPhysics World, vol. 18, pp. 27...32, January 2005.

01Solved Problems in Classical Mechanics

Comments

(iN ewtonu sedt her esult(3 )in c onjunctionwith exp erimentso np endulumst ot est the equality, in modern terminology, of inertial and gravitational mass. [4]

He was

aware that this test could be performed more accurately with pendulums than by using Galileos free-fall experiment and (3) of Question 2.4. Newton evidently attached importance to these pendulum experiments and often referred to them. He used two identical pendulums with bobs consisting of hollow wooden spheres suspended by threads 11 feet in length. By placing equal weights of various sub- stances in the bobs, Newton observed that the pendulums always swung together overlongperiodsoftime.Heconcludedthat...bytheseexperiments,inbodies of the same weight, I could manifestly have discovered a di1erence of matter less than the thousandth part of the whole, had any such been.Ž [4]

The accuracy of

pendulum experiments was later improved to one part in10 5 by Bessel. (ii) Newton also showed how astronomical data could be used to test the equality of inertial and gravitational mass. [4]

Modern lunar laser-ranging measurements pro-

vide an accuracy of a few parts in10 13 , while planned satellite-based experiments (where an object is in perpetual free fall) may improve this to one part in10 15 , and perhaps even a thousand-fold beyond that. [3] (iiiThe e qualitym P =m A of passive and active gravitational masses in Question 2.3 is based on a theoretical condition (Newtons third law) that is presumably exact.

By contrast, the accuracy of the equalitym

I =m G of inertial and gravitational masses is limited by the accuracy of the experiments that test it.

Question 2.6

By applying the second and third laws of motion to the interaction between two particles in the absence of any third object, show how one can obtain an operational de“nition of relative mass.

Solution

LetF 21
be the magnitude of the force exerted by particle 2 on particle 1, and similarly forF 12 . The equations of motion of the two particles are F 21
=m 1 a 1 ,F 12 =m 2 a 2 ,(1) where them i are the masses and thea i are the magnitudes of the accelerations.

According to the third law

F 21
=F 12 .(2)

From (1) and (2) we havem

2 m 1 =a 1 a 2 .(3) [4] S. Chandrasekhar,Newton"s Principia for the common reader. Oxford: Clarendon Press, 1995.

Sections 10 and 103.

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Comments

(i) Equation (3) provides an operational de“nition of relative mass: one can, in principle, determine the massm 2 of a particle relative to an arbitrarily selected massm 1 by measuring the magnitudes of their accelerations at some instant, in the absence of any external disturbance. (ii) It is clear that the LagrangianLof a system can always be multiplied by an arbitrary constant without a1ecting the Lagrange equations ... see (20) in Chapter

1. For a system of non-interacting particles, whereL=7

1 2 mv 2 , this re"ects the fact that the unit of mass is arbitrary and only relative masses have signi“cance.

Question 2.7

Use a three-particle interaction to show that mass is an additive quantity.

Solution

The equations of motion for three particles interacting in the absence of any other objects are m 1 a 1 =F 21
+F 31
,m 2 a 2 =F 12 +F 32
,m 3 a 3 =F 13 +F 23
.(1)

According to the third law,F

21
=ŠF 12 , etc., and so by adding equations (1) we have m 1 a 1 +m 2 a 2 +m 3 a 3 =0.(2) Suppose particles 1 and 2 are stuck together rigidly to form a single particle. Then a 1 =a 2 =a c , the acceleration of the composite particle due to its interaction with particle 3, and (2) yields (m 1 +m 2 )a c =Šm 3 a 3 .(3) Letm c denote the mass of the composite particle. According to the previous question, for the two-particle interaction of massesm c andm 3 , m c a c =Šm 3 a 3 .(4)

It follows from (3) and (4) that

m c =m 1 +m 2 .(5)

Comment

In thermodynamics a distinction is made between two types of variable. First, there are quantities that are additive when two systems are combined. For example, their volumes, the number of particles, etc. Such variables are referred to as extensive. Secondly, there are quantities such as temperature and pressure that are unchanged when two identical systems are combined ... these are intensive variables. According to (5), the mass of a system is an extensive variable.

01Solved Problems in Classical Mechanics

Question 2.8

Consider a mass dipole consisting of two particles having opposite masses ‚ m(>0) and-m. Describe its motion in the following cases: (a) The dipole is initially at rest in empty inertial space. (bT heco nstituentso fthe dip olein ( a)ha veelectr icc hargeq 1 andq 2 . (c) The charged mass dipole of (b) is placed vertically (with the negative mass above the positive mass) in the Earths gravitational “eld. Assume that the distanced between the particles is negligible in comparison with the distancerto the centre of the Earth.

Solution

Opposite-mass particles repel each other

4this follows from the law of gravitation, see

(4) of Question 2.35. Also, for a negative-mass particle the forceFand the acceleration ainF=mapoint in opposite directions. (a) In empty inertial space the only force acting on neutral particles adistancedapart is the gravitational repulsionF=Gm 2 2d 2 .In response, each particle accelerates at the same ratea=Gm2d 2 in the direction shown: the negative mass pursues the positive mass anddremains constant. The motion eventually becomes relativistic ... see Question 15.13. (b) The net force is the sum of the gravitationaland electrostatic forces: F=Gm 2 +kq 1 q 2 d 2 (k=124710 0 ).(1)

For like charges (q

1 q 2 >0), or for unlike charges (q 1 q 2 <0)with q 1 q 2 >-Gm 2

2k,theforceFis repulsive and the motion is the same

as in (a) with acceleration a=Gm 2 +kq 1 q 2 md 2 .(2)

But, for unlike charges withq

1 q 2 <-Gm 2

2kthe force is attractive.

The directions ofFandaare reversed: the positive mass pursues the negative mass. (cS inced2rthe total force on each mass has the same magnitude, and the resulting acceleration of a vertical dipole is a=F m=GMr 2 +Gm 2 +kq 1 q 2 md 2 ,(3) ‚ Negative-mass particles have never been observed. It is, nevertheless, interesting and instructive to consider the dynamics of such objects.

Miscellanea01

whereMis the Earth"s mass. Again, for like charges or for unlike charges with q 1 q 2 >Šq 2 c where q 2 c =Gm 2 k10 1+Md 2 mr 2 11 ,(4) the forcesFare directed as shown in the first diagram, and the dipole accelerates toward the Earth at a ratea. But for unlike charges withq 1 q 2 <Šq 2 c , the forces are reversed, as shown in the second diagram, and the dipole accelerates away from the Earth. Each particle accelerates at the same rate (3), and sodremains constant. The acceleration increases to the asymptotic value (2) asrincreases.

For unlike charges withq

1 q 2 =Šq 2 c , the accelerationa=0and the dipole remains at rest relative to the Earth.

Comments

(iD espiteits s trangedyna micalp roperties,a m assd ipolew ouldno tv iolatea ny of the laws of physics. [5] For example, despite the acceleration in empty space, energy is conserved because the total kinetic energy 1 2 mv 2 + 1 2 (Šm)v 2 is always zero. (iiThe a ccelerationain (2) and (3) can be controlled (in both magnitude and di- rection) by altering the chargesq 1 andq 2 . The dipole is an 'anti-gravity glider" [5] that can fall, hover, or rise in a gravitational field. (iiiIn a f rametha tis a cceleratinga ta r atea, the total force on each particle is zero because the respective translational forces,ŠmaandŠ(Šm)a, cancel the forces F=maandŠmaon each particle. Thus, the mass dipole is at rest in this frame. It follows that the dipole is unstable with respect to any relative motion of the particles toward or away from each other. It would be necessary to have some feedback mechanism to counter any such drift. (ivO nec anco nsiderv ariationso fthe a bove,s ucha sa ma ssdip olein whic hb oth inertial massesm I are positive, and the gravitational massesm G andŠm G have opposite signs. Or one can consider interactions that point in the same direction, as in a predator-prey problem. (vT hepr ecedingq uestionsjust to ucho nt her atherm ysteriousco ncepto fm ass. Access to the extensive literature on this subject is provided in an article by

Roche.

[6] In the theory of special relativity, mass has the property that it can vary in space and time if so-called 'impure" forces are present (see Question 15.11). Perhaps future, richer theories will reveal further properties of mass.

Question 2.9

Discuss the following statement in relation to Lagrange"s equations: 'In the equation of motionF=mathe units must be the same on both sides".

[5] R. H. Price, Negative mass can be positively amusing,Ž American Journal of Physics, vol. 61,

pp. 216...217, 1993. [6] J. Roche, What is mass?,Ž European Journal of Physics, vol. 26, pp. 225...242, 2005.

00Solved Problems in Classical Mechanics

Solution

In the Lagrange equations for a system of particles (see Chapter 1) d dt1L133q i =1L 1q i ,(1) the units on each side are clearly the same. The right-hand side of (1) gives the (generalized) forcesF i and the left-hand side the rates of change33p i of the (generalized) momenta (see Chapter 1). Therefore, the above statement follows.

Comments

(iT heg eneralizationo ft hiss tatementis: All equations in physics (including all physical laws) have the same units on both sides.12 (2) That is, one has an example of a relativity principle: the laws of physics are equally valid in all systems of units. (iiThe s tatement(2 )is t heba sisf ordimensio nala nalysis,whic hh asfa r-reaching consequences in physics. [7]

Some simple examples follow.

Question 2.10

Solutions to physical problems often involve functions like cosu,sinu, e u ,lnu,···,(1) where the argumentuis a scalar that depends on physical quantities such as time, frequency, mass, etc. Explain whyumust be dimensionless (that is, independent of the system of units used for mass, length and time).

Solution

The result follows by inspection of the Taylor expansions of the functions in (1). (We can, if we wish, take these expansions to be de“ning relations of the functions. [8] )For example, e u =1+u 1!+u 2

2!+···(for allu). (2)

It follows that 1,u,u

2 ,···must have the same physical dimensions, and thereforeu is dimensionless. [7] G.I.Barenblatt,Scaling, self-similarity, and intermediate asymptotics. Cambridge: Cambridge

University Press, 1996.

[8] J.M.Hyslop,Real variable. London: Oliver and Boyd, 1960.

Miscellanea00

Comments

(iI tis a g oodidea to c heckw hetherthe r esultso fa c alculations atisfyt hea bove condition. Thus, an expression likee t/m (wheretis time andmis mass) is clearly unacceptable. (ii) The earliest standards for space, time and mass were related to the human body and human activities. With the introduction of the SI system of units in the nineteenth century, the metre was de“ned by the length of a platinum-iridium bar, the kilogram by the mass of a platinum-iridium cylinder (both preserved under carefully controlled conditions), and the second was related to the rotation of the Earth. In the twentieth century the metre and second were rede“ned in terms of physical and atomic constants. The kilogram is therefore an anachronism in that it is still based on a physical object, and it seems likely that the kilogram will be rede“ned in a more convenient and accurate way, possibly by relating it to Plancks constant. An absorbing account of this topic has been given in Ref. [9]. (Plancks constant is already used in a system of units ... see Question 2.17.)

Question 2.11

Use dimensional analysis to determine the dependence of the periodTof a simple pendulum on its massm,weightwand length4.

Solution

Here, we neglect any dependence ofTon the amplitude of oscillation; this is dis- cussed in Question 2.12. We also assume that the desired function of three variables

T=T(m,w,4)is a power-law relation

T=km 1 w 2 4 3 ,(1) wherek,5,11,12are dimensionless constants. We require that the physical dimensions of each side of (1) be the same, that is [T]=[m] 1 [w] 2 [4] 3 .(2) Here,[Q]denotes the dimensions of the quantityQ(Maxwells notation). In terms of the fundamental units of mass(M),length(L)and time(T)we have ‚ [T]=T, [m]=M,[w]=MLT Š2 (wbeing a force=mass×acceleration), and[4]=L.Thus, (2) can be written M 0 L 0 T=M 1 (MLT Š2 ) 2 L 3 ,(3) which provides three equations in the unknowns5,11and12: ‚ We useTin two senses (a period and also a fundamental unit); which meaning is intended is clear from the context. [9] I. Robinson, Rede“ning the kilogram,Ž Physics World, vol. 17, pp. 31-35, May 2004.

00Solved Problems in Classical Mechanics

5+11=0,11+12=0,Š211=1.(4)

Hence5=Š11=12=

1 2 , and (1) becomes T=k13 m4 w.(5)

Comments

(i) The requirement (2), of equality of dimensions in a physical equation, is the essence of the method of dimensional analysis. It is a consequence of the necessity for physical laws and results to be independent of our arbitrary choice of units for mass, length, time, etc. The numerical values of physical quantities such as velocity, momentum and force do depend on the choice of units but physical laws expressing the relations between these quantities do not. Thus, for example, the lawF=mais valid in any system of units. (iiThe a ssumptionma dein (1 )t hatt hedesir edf ormis a p ower-lawm onomialin the independent variables, is typical of dimensional analysis (see also the following examples). This use of power-law relations should not be regarded as a weakness of the method. In fact, power-law (or scaling) relationships give evidence of a very deep property of the phenomena under consideration ... theirself-similarity:such phenomena reproduce themselves, so to speak, in time and space.Ž [7]

Further,

it can be proved that the dimension of any physical quantityQis given by a power-law monomial: for example, in theM,L,Tclass of units [Q]=M a L b T c ,(6) wherea,b,andcare dimensionless constants. [7] (iiiIn the a boveexa mple,dimensio nala nalysisp rovidese noughindep endent equations to solve for the three unknown quantities5,11and12.Often,thisis not the case (see the following questions). (iv) Withw=mg, (5) becomes T=k6 4 g.(7)

4Strictly,w=m

G gandm=m I ,wherem G andm I are the gravitational and inertial masses, so that (5) is T=k6 m I m G 4 g.(8)

According to the weak equivalence principle,m

I =m G and (8) reduces to (7); see

Question 2.5.5

(vT heco nstantkin (7) has to be determined from a detailed dynamical analysis. This shows thatkis, in fact, a function of the amplitude of oscillation (the maxi- mum arc-lengths) with the simple limitk127ass10. In the next question we examine what happens if we try to use dimensional analysis to obtain also the dependence ofTon the amplitude of oscillation.

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Question 2.12

Use dimensional analysis to determine the dependence of the periodTof a simple pendulum on its massm,weightw,length4and arc-length of swings.

Solution

Instead of (1) of Question 2.11 we now have a power-law relation in four variables: T=km 1 w 2 4 3 s 4 .(1) Hence M 0 L 0 T 1 =M 1 (MLT -2 ) 2 L 3 L 4 ,(2) and so

5+11=0,11+12+2=0,Š211=1.(3)

These yield5=Š11=

1 2 and12= 1 2

Š2. Consequently, (1) becomes

T=k13 m4 w14 s415 4 ,(4) where2is an undetermined number.

Comments

(iB ecauses/4is a dimensionless quantity, we cannot determine the dependence of Ton it by using dimensional analysis. In fact, it is clear that we can replace the factor(s/4) 4 by a power series in(s/4)without disturbing the dimensional balance of (4). Thus, the most general form allowed on dimensional grounds is T=13 m4 w614s415 =6 4 g614s415 ,(5) where6is an undetermined function of the amplitudes/4of the oscillations. A numerical calculation of6is given in Question 5.18. In the limits/410,6127, and it is only in this limit that it is reasonable to assume thatTis independent ofs(as was done in Question 2.11). (iiTh us,i nt hepr esentq uestiond imensionala nalysisha sr educeda nunkno wn function of four variablesT=T(m,w,4,s)to an unknown function of one variable6(s/4). Despite this inability of the method to reduce a result beyond a function of one (or more) dimensionless quantities in most cases, dimensional analysis is a powerful and useful technique, particularly in its application to more complex phenomena (such as the next question). Often, the forms provided by dimensional analysis provide clues on how to perform a more detailed theoreti- cal analysis or how to analyze experimental results. In fact, using dimensional analysis, researchers have been able to obtain remarkably deep results that have sometimes changed entire branches of science... . The list of great names involved runs from Newton and Fourier, to Maxwell, Rayleigh and Kolmogorov.Ž [7]

00Solved Problems in Classical Mechanics

Question 2.13

A liquid having density13and surface tension14drips slowly from a vertical tube of external radiusr. Use dimensional arguments to analyze the dependence of the mass mof a drop on13,14,randg(the gravitational acceleration).

Solution

Assume that the mass of a drop is given by the power-law relation m=k13 1 14 2 r 3 g 4 ,(1) wherek,5,11,12and2are dimensionless constants. Recall that surface tension is a force per unit length:[14]=MLT Š2 2L=MT Š2 . Thus, dimensional balance in (1) requires ML 0 T 0 =(ML Š3 ) 1 (MT Š2 ) 2 L 3 (LT Š2 ) 4 .(2)

Therefore

5+11=1,Š35+12+2=0,Š211Š22=0,(3)

which yield for5,11and12in terms of2

5=1+2, 11=Š2, 12=3+22.(4)

From (1) and (4) we have

m=k13r 3 (13r 2 g214) 4 .(5)

Comments

(i) We see again that the existence of a dimensionless combination ... in this case 13r 2 g214... means that the power-law dependence on this quantity4the number2 in (5)5cannot be determined by dimensional arguments. In fact, we can generalize (5) to m=13r 3 6(13r 2 g214),(6) where6is an unknown function, without disturbing the dimensional balance. Equation (6) is the most general form allowed by dimensional requirements. (iiBeca use13=m2V,whereVis the volume of a drop, (6) can be inverted to read

14r/mg=F(V2r

3 ),(7) whereFis an unknown function. Measurements [10] show thatF(u)decreases slowly from0.2647atu=2to 0.2303 atu=18. The above results are the basis for Harkins and Browns drop-weight method for measuring the surface tension [10] See, for example, A. W. Porter,The method of dimensions. London: Methuen, 3rd edn, 1946.

Chap. 3.

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of a liquid. [11]

Note thatF3=(27)

-1 and therefore it is not correct to make the approximationmg=27r14, as would follow if the surface tension acted vertically around the outer radius of the tube at the instant that a drop breaks away: the phenomenon is more complicated than that.

Question 2.14

A sphere of radiusRmoves with constant velocityvthrough a fluid of density13 and viscosity15. The fluid exerts a frictional forceFon the sphere. Use dimensional arguments to study the dependence ofFon13,R,vand15.

Solution

Assume that

F=k13 1 R 2 v 3 15 4 ,(1) wherek,5,11,12and2are dimensionless constants. Recall that viscosity is the proportionality between a tangential force per unit area and a velocity gradient. So [15]=(MLT -2 ÷L 2 )2(LT -1

÷L)=ML

-1 T -1 .(2)

Then dimensional balance in (1) requires

MLT -2 =(ML -3 ) 1 L 2 (LT -1 ) 3 (ML -1 T -1 ) 4 .(3) Hence

5+2=1,Š35+11+12Š2=1,12+2=2,(4)

and we can express5,11and12in terms of one unknown2:

5=1Š2, 11=12=2Š2.(5)

Thus, (1) becomes

F=k13R

2 v 2 (15213Rv) 4 .(6)

Comments

(iT heexistence o ft hedimensio nlessq uantity15213Rvmeans that the power-law dependence on this number in (6) cannot be determined by dimensional reasoning.

Clearly, we can generalize (6) to

F=13R 2 v 2

6413Rv2155,(7)

where6is an unknown function. Equation (7) is the most general form allowed by dimensional requirements. Thus, dimensional analysis has enabled us to reduce [11] See,forexample,F.C.ChampionandN.Davy,Properties of matter. London: Blackie, 3rd edn,

1959. Chap. 7.

00Solved Problems in Classical Mechanics

an unknown function of four variables to a function of one variable. The dimen- sionless quantity213Rv215is known as the Reynolds number, and it is an essential parameter that governs this phenomenon. The function6is rather complicated in general, although a reasonable approximation can be gi

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