Problems and solutionson mechanics docx

PREFACE This series of physics problems and solutions which consists of seven volumes - Mechanics, Electromagnetism, Optics, Atomic, Nuclear and Particle Physics, Thermodynamics and Stat

Pro~lems and Solutions

Major American Universities Ph.D

C o ~ ~ i l ~ d by:

University of Science and

Edited by:

World Scientific

~ e ~ ~ e ~ s ~ y e ~ o f f d o ~ 'Singapore e ~ ~ n g ~ o n g

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British Library Catafoguing-in-Publi~tion Data

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First published 1994

Reprinted 2001,2002

Major American Zlniversities Ph.D Qualifying Questions and Solutions

PROBLEMS AND SOLUTIONS ON MECHANICS

Copyright 0 1994 by World Scientific Publishing Co Pte Ltd

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PREFACE

This series of physics problems and solutions which consists of seven volumes - Mechanics, Electromagnetism, Optics, Atomic, Nuclear and Particle Physics, Thermodynamics and Statistical Physics, Quantum Me- chanics, Solid State Physics - contains a selection of 2550 problems from the graduate-school entrance and qualifying examination papers of seven

US universities - California University Berkeley Campus, Columbia Uni- versity, Chicago University, Massachusetts Institute of Technology, New

York State University Buffalo Campus, Princeton University, Wisconsin University - as well as the CUSPEA and C.C Ting’s papers for selection

of Chinese students for further studies in U.S.A and their solutions which represent the effort of more than 70 Chinese physicists plus some 20 more who checked the solutions

The series is remarkable for its comprehensive coverage In each area the problems span a wide spectrum of topics while many problems overlap several areas The problems themselves are remarkable for their versati- lity in applying the physical laws and principles, their up-to-date realistic situations, and their scanty demand on mathematical skills Many of the problems involve order-of-magnitude calculations which one often requires

in an experimental situation for estimating a quantity from a simple model

In short, the exercises blend together the objectives of enhancement of one’s understanding of the physical principles and ability of practical application

The solutions as presented generally just provide a guidance to solving

the problems, rather than step by step manipulation, and leave much to

the students to work out for themselves, of whom much is demanded of the basic knowledge in physics Thus the series would provide an invaluable complement to the textbooks

The present volume for Mechanics which consists of three parts -

Newtonian Mechanics, Analytical Mechanics, and Special Relativity -

contains 410 problems 27 Chinese physicists were involved in the task

of preparing and checking the solutions

vi Preface

In editing, no attempt has been made to unify the physical terms, units, and symbols Rather , they are left to the setters’ and solvers’ own prefer- ence so as to reflect the realistic situation of the usage today Great pains has been taken to trace the logical steps from the first principles to the final solutions, frequently even to the extent of rewritting the entire solution

In addition, a subject index has been included to facilitate the location of topics These editorial efforts hopefully will enhance the value of the volume

to the students and teachers alike

Yung-Kuo Lim Editor

INTRODUCTION

Solving problems in course work is an exercise of the mental faculties, and examination problems are usually chosen from, or set similar to, such problems Working out problems is thus an essential and important aspect

of the study of physics

The series on Problems and Solutions in Physics comprises seven vol- umes and is the result of months of work of a number of Chinese physicists

The subjects of the volumes and the respective coordinators are as follows:

1 Mechanics (Qiang Yuan-qi, Gu En-pu, Cheng Jiefu, Li Ze-hua, Yang

2 EZectromagnetism (Zhao Sh-ping, You Jun-han, Zhu Jun-jie)

3 Optics (Bai Gui-ru, Guo Guang-can)

4 Atomic, Nuclear and Particle Physics (Jin Huai-cheng, Yang Baezhong,

5 Thermodynamics and Statistical Physics (Zheng Jiu-ren)

6 Quantum Mechanics (Zhang Yong-de, Zhu Dong-pei, Fan Hong-yi)

7 Solid State Physics and Miscellaneous Topics (Zhang Jia-lu, Zhou You-

De-tian)

Fm Yang-mei)

yuan, Zhang Shi-ling)

These volumes, which cover almost all aspects of university physics, contain some 2550 problems solved in detail

The problems have been carefully chosen from a total of 3100 problems

collected from the China-U.S.A Physics Examination and Application Programme, the Ph.D Qualifying Examination on Experimental High Energy Physics sponsored by Chao-chong Ting, and the graduate qualifying examinations of seven world-renowned American universities: Columbia University, the University of California at Berkeley, Massachusetts Institute

of Technology, the University of Wisconsin, the University of Chicago,

Princeton University, and the State University of New York at Buffalo

Generally speaking, examination problems in physics in American uni- versities do not require too much mathematics They can be characterized

vii

viii Introduction

to a large extent as follows Many problems are concerned with the

various frontier subjects and overlapping domains of topics, having been selected from the setters’ own research encounters These problems show a

“modern” flavor Some problems involve a wide field and require a sharp mind for their analysis, while others require simple and practical methods

demanding a fine “touch of physics.” We believe that these problems, as

a whole, reflect t o some extent the characteristics of American science and

culture, as well as give a glimpse of the philosophy underlying American

education

That being so, we consider it worthwhile to collect and solve these problems and introduce them to physics students and teachers everywhere, even though the work is both tedious and strenuous About a hundred teachers and graduate students took part in this time-consuming task This volume on Mechanics which contains 410 problems is divided into three parts: Part I consists of 272 problems on Newtonian Mechanics;

Part 11, 84 problems on Analytical Mechanics; Part 111, 54 problems on Special Relativity

A small fraction of the problems is of the nature of mechanics as in

general physics, while the majority properly belongs to theoretical me- chanics, with some on relativity A wide range of knowledge is required

for solving some of the problems which demand a good understanding

of electromagnetism, optics, particle physics, mathematical physics, etc

We consider such problems particularly beneficial t o the student as they

show the interrelationship of different areas of physics which one is likely

t o encounter in later life Twenty seven physicists contributed to this volume, notably Ma Qian-cheng, Deng You-ping, Yang Zhong-xia, J i Shu, Yang De-tian, Wang Ping, Li Xiao-ping, Qiang Yuan-qi, Chen Wei-zu, Hou Bi-hui, and C h m Ze-xian

7 August 1991

CONTENTS

Preface

Introduction

Part I Newtonian Mechanics

1 Dynamics of a Point Mass (1001-1108)

2 Dynamics of a System of Point Masses (1109-1144)

3 Dynamics of Rigid Bodies (1145-1223)

4 Dynamics of Deformable Bodies (1224-1272)

1 Lagrange’s Equations (2001-2027)

2 Small Oscillations (2028-2067)

3 Hamilton’s Canonical Equations 12068-2084)

Part I11 Special Relativity

PART I NEWTONIAN MECHANICS

1 DYNAMICS OF A POINT MASS (1001-1108)

1001

A man of weight w is in an elevator of weight w The elevator accelerates (a) What is the apparent weight of the man?

(b) The man climbs a vertical ladder within the elevator at a speed v

relative to the elevator What is the man’s rate of expenditure of energy (power output)?

( Wisconsin)

Solution:

vertically up at a rate a and at a certain instant has a speed V

(a) The apparent weight of the man is

20

F = w f - a = w

9

g being the acceleration of gravity

(b) The man’s rate of expenditure of energy is

and that of the earth be &

*For a more accurate calculation, the orbiting period should be taken as 23 hours

56 minutes and 4 seconds

3

4 Problems €4 Solutions on Mechanics

We have

m u 2 G M m

R R2 ’

where v is the speed of the space station, G is the universal constant of

gravitation, m and M are the masses of the space station and the earth

2 rad/s What is the maximum radius R where he can sit and still remain

on the disk?

( WZsconsZn )

6 Problems d Solutions on Mechanics

Hence the tension of the cord and the acceleration are respectively

A brick is given an initial speed of 5 ft/s up an inclined plane at an angle

of 30" from the horizontal The coefficient of (sliding or static) friction is

p = a / 1 2 After 0.5 s, how far is the brick from its original position? You

may take g = 32 ft/s2

( Wisconsin )

Solution:

Choose Cartesian coordinates as shown in Fig 1.3 For x > 0, the

equation of the motion of the brick is

mx = -mgsintl - pmgcostl ,

giving

and the displacement of the brick is

For t > t l , x < 0 and the equation of motion becomes

so that the displacement of the brick at t = 0.5 s is

5’ 51 + A X = 518 - 318 = 0.25 ft

1006

A person of mass 80 kg jumps from a height of 1 meter and foolishly

forgets to buckle his knees as he lands His body decelerates over a distance

of only one cm Calculate the total force on his legs during deceleration

( Wisconsin)

8 Problems & Solutions on Mechanics

Solution:

The person has mechanical energy El = mg(h + s) just before he lands The work done by him during deceleration is E2 = fs, where f is the total force on his legs As El = E2,

h for which this happens

mu2

for which sin 0 = & , or 8 = 30"

112 of g at the pole What is the escape velocity for a polar particle on the

planet expressed as a multiple of V?

(Wisconsin) Solution:

Let g and g' be the gravitational accelerations at the pole and at the equator respectively and consider a body of mass m on the surface of the planet, which has a mass M At the pole,

10 Problems €4 Solutions on Mechanics

Hence g = 2V2/R

infinity from the planet, the body will have potential energy

If we define gravitational potential energy with respect to a point at

GMm

dr =

GMm

Note that the negative sign in front of the gravitational force takes account

of its attractiveness The body at the pole then has total energy

For it to escape from the planet, its total energy must be at least equal

to the minimum energy of a body at infinity, i.e zero Hence the escape

rotated about its axis at an angular velocity such that the mass slides off

the disk and lands on the floor h meters below What was its horizontal distance of travel from the point that it left the disk?

( Wisconsin)

Solution:

The maximum static friction between the mass and the disk is f = pmg

When the small mass slides off the disk, its horizontal velocity 21 is given

A marble bounces down stairs in a regular manner, hitting each step at

the same place and bouncing the same height above each step (Fig 1.5) The stair height equals its depth (tread=rise) and the coefficient of resti- tution e is given Find the necessary horizontal velocity and bounce height (the coefficient of restitution is defined as e = -vf/vi, where vf and vi are the vertical velocities just after and before the bounce respectively)

( Wisconsin)

Fig 1.5

Solution:

Use unit vectors i, j as shown in Fig 1.5 and let the horizontal velocity

of the marble be Vh The velocities just before and after a bounce are respectively

Problems d Solutions o n Mechanics

which is the necessary horizontal velocity The bouncing height H is given

by the conservation of mechanical energy

1011

Assume all surfaces to be frictionless and the inertia of pulley and cord negligible (Fig 1.6) Find the horizontal force necessary to prevent any relative motion of m l , m2 and M

( Wisconsin )

Newtonian Mechanics

Fig 1.6

Solution:

The forces f i , F and mg are shown in Fig 1.7 The accelerations of

ml, m2 and M are the same when there is no relative motion among them The equations of motion along the z-axis are

14 Problems €4 Solutions on Mechanics

approximate gravitational mass of the galaxy in units of the sun's mass You may assume that the gravitational force on the sun may be approximated

by assuming that all the mass of the galaxy is at its center

where T is the distance from the earth to the sun, v is the velocity of the

earth, m and m, are the maSses of the earth and the sun respectively For the motion of the sun around the center of the galaxy,

where R is the distance from the sun to the center of the galaxy, V is the

velocity of the sun and M is the mass of the galaxy

Hence

M = -

Using V = 2rR/T, v = 2 r r / t , where T and t are the periods of revolution

of the sun and the earth respectively, we have

With the data given, we obtain

M = 1.53 x 101lm,

1013

An Olympic diver of mass m begins his descent from a 10 meter high

diving board with zero initial velocity

(a) Calculate the velocity Vo on impact with the water and the a p p r e

ximate elapsed time from dive until impact (use any method you choose) Assume that the buoyant force of the water balances the gravitational force on the diver and that the viscous force on the diver is h2

The time elapsed from dive to impact is

(b) AS the gravitational force on the diver is balanced by the buoyancy, the equation of motion of the diver through the water is

16 Problems d Solutioru o n Mechanics

1014

The combined frictional and air resistance on a bicyclist has the force

F = aV, where V is his velocity and a = 4 newton-sec/m At maximum effort, the cyclist can generate 600 watts propulsive power What is his maximum speed on level ground with no wind?

in terms of I such that the mass will swing completely round in the circle shown in Fig 1.8

( Wisconsin)

Fig

Newtonian Mechanics 17

Solution:

Take the mass m as a point mass At the instant when the pendulum

collides with the nail, m has a velocity 2) = m The angular momentum

of the mass with respect to the point at which the nail locates is conserved during the collision Then the velocity of the mass is still II at the instant after the collision and the motion thereafter is such that the mass is constrained to rotate around the nail Under the critical condition that the mass can just swing completely round in a circle, the gravitational

force is equal to the centripetal force when the mass is at the top of the

circle Let the velocity of the mass at this instant be v1, and we have

A mass m moves in a circle on a smooth horizontal plane with velocity

vo at a radius & The mass is attached to a string which passes through

a smooth hole in the plane as shown in Fig 1.9 (“Smooth” means

frictionless.)

(a) What is the tension in the string?

(b) What is the angular momentum of m?

(c) What is the kinetic energy of m?

(d) The tension in the string is increased gradually and finally m moves

in a circle of radius & / 2 What is the final value of the kinetic energy?

18 Problems d Solutions on Mechanics

Fig 1.9

(e) Why is it important that the string be pulled gradually?

( Wisconsin)

Solution:

the circular motion, hence F = mug/&

(a) The tension in the string provides the centripetal force needed for (b) The angular momentum of the mass m is J = mvol&,

(c) The kinetic energy of the mass m is T = mvi/2

(d) The radius of the circular motion of the mass m decreases when the tension in the string is increased gradually The angular momentum of the

mass m is conserved since it moves under a central force Thus

or

211 = 2v0

The final kinetic energy is then

(e) The reason why the pulling of the string should be gradual is that

the radial velocity of the mass can be kept small so that the velocity of the

mass cam be considered tangential This tangential velocity as a function of

R can be calculated readily from the conservation of angular momentum

Newtonian Mechanics 19

1017

When a 5000 Ib car driven at 60 mph on a level road is suddenly put into

neutral gear (i.e allowed to coast), the velocity decreases in the following manner:

where t is the time in sec Find the horsepower required to drive this car

at 30 mph on the same road

Useful constants: g = 22 mph/sec, 1 H.P = 550 ft.lb/sec, 60 mph =

20 Problems Solutions on Mechanics

1018

A child of mass m sits in a swing of negligible mass suspended by a

rope of length 1 Assume that the dimensions of the child are negligible compared with 1 His father pulls him back until the rope makes an angle

of one radian with the vertical, then pushes with a force F = mg along the arc of a circle until the rope is vertical, and releases the swing For what

length of time did the father push the swing? You may assume that it is

sufficiently accurate for this problem to write sine M 6 for 0 < 1

The solution of this equation is 8 = Acos(wt) + Bsin(wt) - 1, where the

constants A and B are found from the initial conditions 8 = 1, b = 0 at

t = 0 to be A = 2, B = 0 Hence

A particle of mass m is subjected to two forces: a central force fi and

a frictional force f2, with

22 Probkms d Solutions on Mechanics

(b) If the mass of the pulsar is about 1 solar mass (- 2 x 1030 kg or

N 3 x 105Mearth ), what is the maximum possible radius of the pulsar? (c) In fact the density is closer to that of nuclear matter What then is the radius?

( CUSPEA )

Solution:

(a) Consider the limiting case that the Crab pulsar is just about to disintegrate Then the centripetal force on a test body at the equator of the Crab pulsar is just smaller than the gravitational force:

Newtonian Mechanics 23

where mp is the mass of a proton and is approximately equal to the mass

mH of a hydrogen atom This can be estimated as follows:

is equilibrium when the rings are at points 30" distant from the highest point of the circle as shown in Fig 1.11, find the relation between the three weights

( UC, Berkeley)

Fig 1.11

24 Problems d Solutions on Mechanics

Solution:

Assume the string is also weightless As no friction is involved, the tensions in the segments AC and A E of the string must be the same Let the magnitude be T For the ring A to be at rest on the smooth loop, the resultant force on it must be along AO, 0 being the center of the loop; otherwise there would be a component tangential to the loop Hence

LOAE = LOAC = LAOE = 30"

The same argument applies to the segments B D and B E Then by

symmetry the point E at which the string carries the third weight must be

on the radius H O , H being the highest point of the loop, and the tensions

in the segments B D and B E are also T

Consider the point E Each of the three forces acting on it, which are in equilibrium, is at an angle of 120" t o the adjacent one As two of the forces

have magnitude T , the third force must also have magnitude T Therefore the three weights carried by the string are equal

1022

Calculate the ratio of the mean densities of the earth and the sun from

6 = angular diameter of the sun seen from the earth = +"

1 = length of 1" of latitude on the earth's surface = 100 km

t = one year = 3 x lo7 s

g = 10 ms-2

the following approximate data:

( UC, Berkeley)

Solution:

Let r be the distance between the sun and the earth, Me and Ma be the

masses and Re and R, be the radii of the earth and the sun respectively,

and G be the gravitational constant We then have

A parachutist jumps at an altitude of 3000 meters Before the par&

(a) Assuming that air resistance is proportional to speed, about how (b) How far has she traveled in reaching this speed?

chute opens she reaches a terminal speed of 30 m/sec

long does it take her to reach this speed?

After her parachute opens, her speed is slowed to 3 m/sec As she hits the ground, she flexes her knees to absorb the shock

(c) How far must she bend her knees in order to experience a deceleration

no greater then log? Assume that her knees are like a spring with a resisting force proportional to displacement

(d) Is the assumption that air resistance is proportional to speed a reasonable one? Show that this is or is not the case using qualitative arguments

( UC, Berkeley)

Solution:

(a) Choose the downward direction as the positive direction of the

x-axis Integrating the differential equation of motion

26 Problems €4 Solutions Mechanics

where c is the distance of knee bending and v is the speed with which she hits the ground, considering the knee as a spring of constant k Taking the

deceleration -lOg as the maximum allowed, we have

Newtonian Mechanics 27

1024

A satellite in stationary orbit above a point on the equator is intended to send energy to ground stations by a coherent microwave beam of wavelength one meter from a one-km mirror

(a) What is the height of such a stationary orbit?

(b) Estimate the required size of a ground receptor station

(Columbia)

Solution:

to the spin angular velocity of the earth and is given by

(a) The revolving angular velocity w of the synchronous satellite is equal

with the horizontal

(a) Solve for the accelerations of ml, m2 and the tension in the string

(b) Find the smallest coefficient of friction for which the inclined plane

(Columbia) when p is very large

will remain at rest

28 Problems d Solutions on Mechanics

Solution:

equations of motion of m l and m2 are (see Fig 1.12)

(a) When p is large enough, the inclined plane remains at rest The

ml +m2 (b) The inclined plane is subjected to horizontal and vertical forces (see

Newtonian Mechanics 29

1026

A particle of mass m is constrained to move on the frictionless inner

(a) Find the restrictions on the initial conditions such that the particle (b) Determine whether this kind of orbit is stable

surface of a cone of half-angle a, as shown in Fig 1.14

moves in a circular orbit about the vertical axis

(Princeton)

Solution:

particle are

(a) In spherical coordinates ( r , O , v ) , the equations of motion of the

m(i: - re2 - T + ~ sin2 0 ) = F, ,

where 1 is its distance from the vertex 0 (see Fig 1.15) For motion in

a circular orbit about the vertical axis, i = 1 = 0 With 1 = lo , Eq (2) becomes

30 Problems €4 Solutions on Mechanics

The right-hand side of Eq (3) is constant so that @ = constant = $0, say The particle has velocity vo tangential to the orbit given by vo = lo@o sin a

Equation (3) then gives

v; = 910 cos a ,

which is the initial condition that must be satisfied by YO and lo

that lo becomes 10 + Al, $0 becomes $0 + A@ Equation (2) is now

(b) Suppose there is a small perturbation acting on the particle such

-9 cos a ,

or

A1 - 21&A@ sin2 a - A@: sin2 a = lo+’ sin2 a - g cos a ,

where A1 is shorthand for d2(Al)/dt2, by neglecting terms of orders higher

than the first order quantities A1 and A@ As the right-hand side of this equation vanishes on account of Eq (3), we have

A1 - 2lo$oA$ sin2 a - A@: sin2 a = 0 (4)

There is no force tangential to the orbit acting on the particle, so there is

no torque about the vertical axis and the angular momentum of the particle about the axis is constant:

Substituting 1 = lo + Al, + = $0 + A$ into Eq (5) and neglecting terms of

the second order or higher, we have

Eliminating A@ from Eqs (4) and (6), we obtain

A[+ ( 3 ~ : sin2 a ) 01 = o

As the factor in brackets is real and positive, this is the equation of a

“simple harmonic oscillator” Hence the orbit is stable