Guide To Mechanics PDF

 

GUIDE T

MECH

NICS

 

David A. Towers Senior Lecturer Lect urer in Mathematics University Consultant Editor

o

Lancaster

Titles available

Linear Algebra D. Towers Abstract Algebra C. Whitehead Numerical Analysis J. Turner Mathematical Mathema tical Modellin Modelling g D. Edwards Mathematical Methods J. Gilbert Analysis F. Hart

M. Harrison Harriso n

 

  uide to Mechanics Mechanics Philip Dyke

Head o the Department o Mathematics and Statistics Polytechnic South West

and Roger Whitworth Head o Math Mathemat ematics ics Droitwich High Scho School ol

M

CMILL

N

 

© Phi Philip lip Dyke

Roger Roge r Whitworth

1992

All rights reserved. No reproduction, copy or transmission of this publication may be made without written permission. No paragraph of this publication may be reproduced, copied or transmitted save written permission accordance with the provisions of with the Copyright, Designs or andinPatents Act 1988, or under the terms of any licence permitting limited copying issued by the Copyright Licensing Agency, 9 Tottenham Court Road, London W1P 9HE. Any person who does any unauthorised act in relation to this publication may be liable to criminal prosecution and civil claims for damages. First published 1992 by TH MACMILLAN PRESS LTD Houndmills, Basingstoke, Hampshire RG21 2XS and London Companies and representatives throughout the th e world ISBN 978-1-349-11116-9 978-1-349-11116-9 ISBN 978-1-349-11114978-1-349-11114-5 5 eBook) DOII 10.1007/978-1-349-11114 DO 10.1007/978-1-349-11114-5 -5

A catalogue record for this book from the British Library.

is

available

 

  o

ttilie and

om

 

  ONTENTS

Editor s foreword Preface

xi

1 KINEMATICS 11

Introduction

1.2

Definition o f kinematic quantities One-dimensional models

13 14 15

1.6 17 18 19 1 10

2

Graphical representation

Calculus and rates o f change Constant acceleration Conclusions from experimental data Two- and three-dimensional models Resolution o f vectors Two-dimensional parametric motion

FORCES AND NEWTON S LAWS IN ONE DIMENSION

1

1 1 4 4 9 12 15

19 23

24 31

21

The nature o f force

31

22 23

Newton s laws Resistance and the particle model

34 42

3 FORCE AS A VECTOR

49

Modelling force forcess

49

Resolution

54

33

Resultant force

56 57

35

Equilibrium Friction Newton s laws in vector notation

31

3.2 3.4 3.6 4

x

USING ENERGY

59 64 71

41

Introduction

71

42

Work

71 vii

 

CONTENTS

43 44

Energy Connected particles

76 89

5 COLLISIONS 51 Introduction 52 53 54 55

56 57

58

93

93

Impulse and from Newton s laws Collisions in momentum the real world Conservation of momentum via impulse for colliding bodies Newton s experimental experime ntal law Direct collision between a particle and a fixed barrier Oblique collision between a particle and a fixed barrier Oblique collision between two particles

6 MOTION UNDER GRAVITY IN ONE DIMENSION Introduction 61 Motion with no resistance 62 Motion with resistance proportional to speed 63 Motion with resistance proportional to the square 64 speed Terminal velocity 65

73 74 75

76 77

7.8 79 7 10 7 11 7 12 7 13

7 14 viii

100 100 103

107 110 114 9

119 119 124 of

PROJECTILES Projectiles, 71

7.2

93

motion in the real world Independence of horizontal and vertical motions Velocity as a vector Assumptions for modelling projectile motion Magnitude and direction o f the velocity o f a projectile a givenrelative instance Discussingatmotion to horizontal and vertical directions he path of a projectile: projectile: the trajectory Direction of travel and magnitude of velocity Two trajectories Envelope o f trajectories The motion of a projectile relative to an inclined plane Motion on an inclined plane referred to axes along and perpendicular to the plane he direction of travel at the point of impact o f a projectile with an inclined plane Real problems with projectiles and inclined planes

130 134

42 142 142 144 144 147 148 152 155

156 159 161 163

168 170

 

CONTENTS

8 CIRCULAR MOTION 8.1 Introduction 8.2 Polar coordinates and angular displ displacement acement 8.3 Angular velocity and angular acceleration 8.4 Some observations of circular motion 8.5

Acceleration towards the centre of ahorizontal circle o f motion The analysis of problems involving motion in a circle 8.7 The conical pendulum 8.8 Modelling problems of motion in a vertical circle 8.9 Motion in a circle and connected particles 8.10 Vector methods and circular motion 8.11 Vector formulation for constant angular velocity 8.12 Vector formulation for non constant angular velocity 8.13 Circular orbits 8.14 Angular velocity as a vector

8.6

9 VIBRATIONS Introduction 9.1 9.2 Simple harmonic motion 9.3 amped motion 9.4 Forced Forc ed oscil oscillatio lations ns 1

176

176 176 177 178 180 181

184 188 192 196 198 199 199 200 203 203 204 215

219

VARIABLE MASS PROBLEMS 10.1 Introduction 10.2 Deriving the equations 10.3 Problems of a practical nature involving varying mass

244

ROTATING AXES

256

11.1 11.2

11.3 11.4

Introduction Preliminary notions of rotating frames Rotating coordinate sys system temss The rotating Earth

233 233 234

256 256 259 272

Solutions to exercises

287

Index

291

ix

 

EDITOR S FOREWORD

Wide concern has been expressed in tertiary education about the difficul ties experienced by students during their first year of an undergraduate course containing a substantial component of mathematics. These difficul tiess have a numbe tie numberr of underlying causes includi including ng the change of emphasis from an algorithmic approach at school to a more rigorous and abstract approach in in undergrad undergraduate uate st studies udies the greater expectation of inde independ pendent ent study and the increased pace at whi which ch materia materiall s presented. The books in this series are intended to be sensitive to these problems. Each book s a caref carefull ully y selected short introductory text on a key are areaa of the first-year syllabus; the areas are complementary and largely self contained.. Throughout the pac contained pacee of development s gentle sympathetic and carefu carefully lly motivat motivated. ed. Clear and detai detailed led explanat explanations ions are provided and important concepts and results are stressed. As mathematics s a practical subject which s best learned by doing it ratherr than wat rathe watchi ching ng or reading about someone el else se doi doing ng it a particular effort has been made to include a plentiful supply o f worked examples together toge ther with appropr appropriate iate exercises ranging in di diffi fficu cult lty y from the straight forward to the challenging. When one goes goes ffell ellwal walking king the most brea breathtak thtaking ing v view iewss requi require re some expenditure of effor effortt iin n orde r to gai gain n ac acce cess ss to them them:: nevertheless the peak s more likely likely to be reached if a gen gentle tle and interesting rout routee s chosen. The mathemati mathematical cal peaks attainable in these books are every bit aass exhilar exhilar ating the paths are aass gentle aass we could fi find nd and the interes interestt and expectation are maintained through throughout out to prevent the spi spirit ritss from fl flag aggi ging ng on the journey. Lancaster 1987

x

David A Towers Consultant Editor

 

PREF

CE

Many students in higher educa education tion wi will ll be experiencing courses in mechan ics for the first time or may have found the option o f mechanics difficult when it was studied as part of an A-level mathematics course. In recogni tion o f these factors this text progresses at a gentle pace and requires no previous knowledge o f mechanics. The approach adopted throughout is to motivate particular areas of mechanics through the reflection o f real-life problems and practical activities. The amount of algebraic manipulation required at any stage is ke kept pt to a minimum which w wee feel wi will ll allow students studen ts to wor work k with the mechani mechanics cs concepts w with ith greater grea ter confidence. This type o f approach will go some way to accommodating the changes that have occurred in the teaching of mathematics in schools in GCSE courses and at A-level. The number of students who have studied mechanics at school or in further education has decreased in the past ffeew years with the introduction o f other options at A-level. For those who have studied studi ed mecha mechanics nics before we feel confident tha thatt the appro approach ach we hav havee aadop dopted ted wi will ll o offer ffer a refreshing contrast to their experience of older texts. In writing writing this boo book k we have had to take the t he decisio decision n to omit any study of rigid rigid body me mechanics chanics so th that at th thee text would be shor shortt but sti still ll ac accommo commo date a range of courses in higher education. The course of study here is based on o n particle mechanics whic which h we feel w wil illl establish a soun sound d founda tion for the later study o f rigid body mechanics. As with with oth other er guid guides es iin n thi thiss series this text is suitable for self-study. Answers to exercises are provided s well as hints on how to solve the more demanding problems. P.D.

R.W.

x

 

1 KINEMATICS

1 1 INTRODUCTION The study of the motion of bodies requires a structured understanding of the fundamental quantities of displacement and time. This study is called kinem tics and it will provide a basis for later modelling in other branches of mechan mechanics. ics. Fro From m time and displacement we derive the quantities velocity veloc ity and acceleration. All of these with the exception of time are vector quantities and can be expressed in an algebraic vector form. Not surpris sur prising ingly ly theref therefore ore the sstudy tudy of vecto vectors rs is crucial to the study of kinematics and all mechanics. We shall start our study by considering some kinematic quantities with which you may already be familiar. Everyday language provides us with an intuitive compreh comprehension ension of these quantities but in some cas cases es can lead to serious misunderstan misunderstanding ding particularly when con considering sidering vectors. When a car is trav travelli elling ng along along a road and the speed speedomete ometerr reads an unchanging 3 km per hour the drive driverr natural naturally ly aassum ssumes es that the speed is constant. The fact is that if the car is cornerin cornering g or goi going ng do down wn or cclimb limbing ing up a hill it s accelerating despite the constant speed shown on the speed ometer. speedomet er. IIn n the ffoll ollowi owing ng section we begin to establish the concepts o f displacement velocity velocity and acceleration. In parti particular cular we cla clarify rify the distinction between speed spee d and velocity velocity ofte often n used aass syn synonyms onyms by nonmathematicians and the cause of the appa apparent rent contradiction of the acce accelerlerating car with its constant speedometer reading.

1 2 DEFINITION OF KINEMATIC QUANTITIES The foll followi owing ng formal definit definitions ions of displacement distance velocit velocity y speed and acceleration should help us to make a start in overcoming the aforementioned Consider misconceptions. the fixe fixed d points

and Q

illustrated in Figure 1.1. The 1

 

GUIDE T O MECHANICS

p

Q

Fig 1 1 displacement from P to Q represented by the vector Q = s s the translation that s needed to move the point P to the point Q. The inverse of thi thiss operation the displacement displacement from Q to P is represented by the vector QP Thus: Q

PQ

=

The magnitude of the displacement from P to Q s the distance Q and the orientation of the line segment Q s its direction. t follows therefore that the displacement from Q to P has equal magnitude but that its direction differs by Jt radians or 180 Displacement clearly satisfies the requirements requir ements o off a vector in h having aving bot both h magnitude aand nd direction. Displacementt vectors must of course obey the la men law w of vector ad addition. dition. Th us as illustrated i n Figure 1.2 the displacement PQ QR s equivalent to the displacement PR and so we can write: 0

PQ

•

QR

=

PR

R Q

p

Fig 1 2 s

The velocity of a the bodynot the nrate of change ofhave its displacement wi with th respect to time. vUsing notatio ation of calculu calculus s we that:

 

KINEMATICS

v

dx dt

As veloc velocity ity depends depe nds on displacement the then n velocit velocity y is a derived vector and has a dependent magnitude and direction. The magnitude of velocity is speed. Speed a scalar is thus not dependent on direction. The acceleration a, of a body is the rate of change of its velocity with respect to time. Again calc calculus ulus nota notation tion gi give ves: s:

Accelera tion iiss derived from a vector sso Acceleration o it must be a vector its itself elf poss possess ess ing both magnitude and direction. Acceleration is the most dif diffi ficu cult lt concept to t o ap appreci preciate ate intuitively. t can be non-zero when the speed of a body is unchanged but the direction of motion varies. There are numerous examples of motion with constant speed but non-zero acceleration and some will be discussed in later chapters. In Figure 1.3 whi which ch illustrates the process of cornering with constant speed the vel velocit ocities ies of motion at points and Q , which occur at a one-second interval are giv given en as V and V2 • Note that V F V2 •

v

B p

c Fig.

13

BC VI The triangle Figure 1.3 also shows the vector triangle for V2 formed is isosceles as V = IvzI. The acceleration over the one-second VI> the direction BC represents the direction of this interval is V 2 acceleration and the length BC is its magn magnitud itudee which is non-zero. Note that the direction of the acceleration is not the same direction as V or V2 •

will

be isthe study of from one-dimensional models of motion.first Theconsideration understanding that developed this study can then be Our

3

 

GUIDE

TO MECHANICS

easily rendered to two and three dimensions, using vector notation, with the algebra unchanged. This is an important advantage of using vector notation in mechanics.

1.3 ONE DI DIMENSION MENSIONAL AL MO MODELS DELS

The special case of motion in a straight straigh t line is a usual starting point for th thee study of kinematics. Examples are the motion of a body falling vertically under gravity o r that of a particle attached to a spring lying on a smooth horizontal table. Although we still maintain our vector approach, we can see that all quantities can be expressed as negative and positive values along the direction of motion, the x-axis say, represented by the unit vector i. The position vector r with respect to the origin, at time t, is: r

=

x t)i

The velocity vector is then: r

dx

= 1 dt dt

.

XI

The acceleration acceleration vector is similarly represented as:

t

is usual for the vector formulation in terms of i to or

be

omitted and for the

The

directio direction n to bethat repr represented esented by a - asibody gn. from following defi at definitions nitions will ll apply: Given the displacement of the origin time wi t is x t) in a given direction, then velocity, v is x t) and the th e accel acceleratio eration, n, a, is x t) or v. Note that, in one dimension, it is not neces necessary sary to use bold face for vectors as there is no ambiguity.)

1.4 GRAPHICAL REPRESENTATION

common practice to express one-dimensional motion in graphical form. Consider the following simple example. Two cars A and B are moving at constant speeds in the same direction t is

along parallel straight tra fficc lanes. Car A has speed a speed of 12 ms ms-1 andtraffi passes an observer 2 s aafter A 4

of

10

At

1•

ms-- time ms Car B dhas what tim e an and at

 

KINEMATICS

x m)

o

B

~

~

~

~

~

5

~

~

~

A

15

t 5)

Fig 1 4

what distance from the observer will car B overtake car A The solution of problems of this type shows the value of representing journeys in a graphical form. For example, Figure 1.4 shows the journeys for both A and B. The origin 0 t = 0) is taken to be the point when A passes the observer. The point of intersection of the two straight lines represents the time and place when A and B meet. Accurate drawing of the figu figure re wi will ll lead to the corre correct ct solution; the time is 12 s after car A passes the observer, and the distance is 120 m away from the observer. This type of diagram, called a displacement-time graph can be used to determine determi ne timing and scheduli scheduling ng of events, and is thus valuable in creating timetables. I t provides direct pictorial representation of journeys and events. I t should be noted that the gradient of each curve, represented by i is the velocity, where negative gradients represent reverse motion. n cases like the one shown in Figure 1.4, where the speed is constant, the graphs are straight lines. Mathematically, the velocity-time curve is a more rewarding graphical representation of the journey of a body in a straight line. Consider the following example of a train journey from Leicester to Nottingham, which for our purposes has been divided into five parts A B C, D and E. The motion can be assumed to be a straight line. line. Note tha thatt the gradient aatt any point on these curves is a measure of the acceleration of the train at that instant, and the area below the curve is the distance travelled. The journey is represented by the velocity-time curve in Figure 1.5: A:

B:

Starting from rest, the train travels with constant acceleration for the first 1 minutes of the journey. Iation t thenismoves with4 constant velocity of 70 kmhzero) for minutes.

1

that is, its acceler 5

 

GUIDE TO

MECHANICS

O

~

~

~

Fig c

~

;

~

r

15

The train undergoes non-uniform non-uniform acceleration for 5 minutes reach ing a maximum speed o f 100 kmh-1 •

D: Braking results results in a constant retardation from its speed E:

___

t m in)

of

100 kmh-1

but this results in the train overshooting the station. Having fi firs rstt come to rest it then the n reverses into the station achieving a maximum reversing velocity o f 10 kmh-1 • This time it comes comes to rest at the platform.

The interpretation o f this type o f graph is not as easy as it was for displacement-time graphs. You may find it helpful to consider other journeys that you have experienced in the same way. A journey across a busy town is clearly a highly complex version ofthis example. You Yo u will will fi find nd it constructive to try to develop a velocity-time graph for such a journey. I t should be noted that in velocity-time graphs the acceleration is represented by the gradient o f the curve. Negative gradients represent retardations. Constant accelerations are represented by straight lines. Example

1 4

A body moves along a straight line with an a n initial velocity o f 5 ms-1 • I t then accelerates at 7 m s 2 for a certain period. For the next 10 s it has a

retardation of 1 m s 2 • The total distance travelled during the motion is which the body bod y has an acceleration o f 450.0 m . Find the length o f time for which ms--  • 7 ms Solution

Figure 1.6 shows a sketch For

of

the velocity-time graph for the

our

the

been

completed of veloanalysis journey divided divi ded intojourney. three parts Athe purpose city ty at th thee en end d o f ishas an d C. The veloci u and at 6

 

KINEMATICS

Fig 1 6

the end of B s v. The times o f the fi firs rstt and last periods o off accele acceleration ration are t and T respectively. Using the fact that acceleration s represented by the gradient of the curve, we can write: u 7 =

5

for

t

v =

u

for

10

v

1

B

for C

T

Solution Soluti on of these equations for u, v and T in terms u

=

5

7t

v

=

T

=

o

t

gives:

7t - 35

Using the fact fact tha thatt total distan distance ce travelled eq equals uals the are areaa unde underr the graph for the completed journey, we obtain: 450.5

=

1

1

5)t

l u

l u

v)10

340t

24

1

lvT

Substituting for u and v gives:

o

=

56t 2

-

and by factorising we see that:

o

=

4 14t - 1) t - 6 7

 

GUIDE

TO MECHANICS

The solutio solution n = 1114 gives < 0 so this can the initial acceleration is therefore 6 s

EXER

be

disregarded.

The

time of

ISES 1 4

1 De Describe scribe in words the motion illustrated in the displacem displacement-time ent-time graph shown in Figure 1.7. 5

o ~

~ ~ ~ ~

Fig.

17

2 Figure 1.8 shows the velocity-time graphs for the motion of four different bodies. Describe what might be happening in each case and for each each one sketch the corresponding displac displacementement-time time curve. a)

b)

v

 t

o ~ - - - - - - - - - - - - - - - - - - - e)

o

d)

v

O

r

~

7

~

v

~

Fig. 8

v

18

~

 

KINEMATICS

3 A hovercraft crosses the English Channel Chann el a distance o f 52 km in a time cruising sing speed of 80 kmh- 1 • Choose o f 40 minutes. t is capable of a top crui the velocity-time curve from those shown in Figure 1.9 that best illustrates the journ ey explaining your choice and completing the scales scales o n both axes. b)

a)

Fig.

19

4 A tube train travels travels a distance of 43 432 2 m starting star ting and finishing a t rest in 1 minute. t first accelerates a t 3 ms-2 , then travels with constant velocity and finally retards a t 1 ms-2 • Find the time taken in each of the three stages stages of the journey. 5 Two cars start sta rt t o move from a point on a road. ar A star starts ts first first from rest and moves with a constant acceleration of 3 ms-2 • Two seconds later car B starts and maintains a uniform velocity velocity of 16 ms-1 • Show that the cars will be level twice and find the time during which car B leads car A

1.5, C LCULUS

ND R

TES OF CH

NGE

Let us consider the relationship:

dx =

dt

v

where v is written as a function of t. T o find x in terms o f t we simply integrate both sides to give: x =

f

v

t

Clearly the indefinite indefinite integration here results in the introduction of an arbitrar y constant. This constant is is evaluated by knowing the value of x for t

somewe require the th e distance travelled in in the time interval a f

t

b we can 9

 

G U ID E T O MECH

NICS

find this, its value being that of the definite integral:

f t

b

a

v dt

should be noted that if v chang changes es sign, sign, the integral over the whole range

does not represent the actual distance travelled. The value of the area is the distance, but the process of integration means that areas below the t-axis are negative negative aand nd w wiill be subt subtracte racted d fr from om those areas above the t-axis. A similar approach for: dv =

dt

where a the acceleration, v in terms of t given by:

is

a function function of t leads to a solution for velocity,

v

fadt

Alternatively, considering a as a function dv

of

x we can write:

dv

v

dx

dt

This gives a solution for the velocity, this time in terms of x as: ~ V 2

=

f a dx

Examples 1.5 1

The velocity, v of a body moving along a straight line at time t by: v

=

3t 2

-

2t

a) acceleration When t

10

=

=

d dt

= 6t

- 2

0 the acceleration

is

given

3

Find a) the iinit nitial ial acceleration and b) the disp displacement lacement when the displacement is 5 m when t = 1 Solution

is

- 2 ms-

2•

t =

2 if

 

KINEMATICS

(b) displacement

=

x

=

v dt

=

t3

-

t2

+

3t

+c

(c = constant)

When t = 1, x = 5 and 5 = 1)3 -  1)2 + 3(1) + c. This gives c = 3; so when t = 2 we have:

+

2)3 -  2)2

x =

3(2)

+

3 = 11

The displacement displacement is thus 11 m. acceleration of a block on a table when attached to a spring is given 2. The acceleration by: a=5

lOx

where x is the block s distance from th e spring s fixed fixed end x = 0). If, at the start of the motion, x = 0 and the velocity velocity is then \/2 0 ms-1 , find the distance o f the block from t he spring s fixe fixed d end when the block is first at rest.

Solution

As

suggested, we express the acceleration as: v

dv

=5

dx

- lOx

1)

Integration then gives:

1 v2 = 5x - 5x 2 + c When x

=

2 0, we are given 10

that

=0

The

solution x =

1

(2)

v = \/20 whence:

- 0 + c

When the block is at rest, v =

o= o=

(c = constant)

O

or

c

= 10

Thus, equation (2) becomes:

5x - 5x2

+

5 x - 2) x

10

+ 1)

feasible as the block must mu st stay on one side of is not feasible

the origin; hence, the block first comes to rest at x = 2 m. 11

 

GUIDE

EXER

TO MECHANICS

ISES 1 5

1 The acceleration, a ms ms--2 , o f a particle moving in a straight line is a = 7 - 2t where t is the time in seconds. I f the velocity, v is 12 ms ms-- I when t = 2, then calculate: (a) (b) (c (c)) (d)

the time when a = 3 ms-2 ; v in terms of t; the maximum velocity; the distance travelled in the fir first st second.

2 A body

metres from a point after t second secondss wher wheree x speed and acceleration of the body after 2 s is x

=

r2. Find the

3 A body starts sta rts from A and its displacement from A after a time t seconds is given by x = 2t 3 - 5t 2 20t 4 Find the acceleration when the velocity of the body is 24 ms ms-- I . What will be the body s displacement from when the velo velocity city is 24 ms ms-- I ?

- X2 , where x represents the body s displacement from its starting position O. If at the start s tart of the body s motion, its ve veloci locity ty is 2 ms ms-- I , find:

4 The acceleration of a body is given by a = xV 4

(a) v in terms of x (b) the distance of the body from 0 when at rest; (c (c)) the maximum veloc velocity ity of the body.

1 6 CONSTANT CONSTANT ACCELERATI ACCELERATION ON

Constant acceleration is a special case of motion. t rarely occurs in real problems, as we will find when we look at resistance models in later chapters. As a guide, it is best to consider acceleration to be non-constant, unless we have evidence or justification to the contrary. t is usual to derive three equations to model motion with constant acceleration. A simple example of such motion with constant acceleration is illustrated in the velocity-time graph in Figure 1.10. This allows us to derive three constant acceleration formulae. f the acceleration, a has a constant value ao  then we have: dv

=a

dt 12

°

 

KINEMATICS

v

u

O ~ - - - - - - - - - - ~ - -

 

Fig 1 10

Solution of this equati equ ation on gives gives v = aot c, where c is a constant. t is usual to define the initial velocity as u Thus, when t = 0, v = u and this gives: 3)

The displacement, displacement, s is then given by a second integration as: s t

=

1

ut

constant

ot 2

is usual to take an origin such that the displacement, s at time t

zero to give:

s

=

ut

1 2

-aol 2

=

°s 4)

Note that s replaces the usual notation for displacement, x in the constant acceleration formulae.) A third equation giving the velocity, v as a function of displacement, s can be obtained by eliminating t from equations equatio ns 3) and 4). The same equation can also be derived using integration, by considering ao to be a function o f s 1

2

=

aos

Together with the condition that v

constant

= u when s = 0, this gives: 5)

Equations 3), 4) and 5) are usually usually termed ter med the th e constant acceleration formulae. The notation is standard, except that is often used for acceleration. All three equations als also ol.1O, b bee obtained obtaine d easil easily y from the geometry of the velocity-time curve in can Figure as follows. 13

 

GUIDE TO

MECHANICS

Using the fact that acceleration is the gradient of the velocityvelocity-time time graph gives:

ao

v

u

t

This is equation 3). Using the fact that displacement is the area under the velocity-time graph gives: s

=

1 2

- u

v)t

This is also obtained by eliminating ao betwee between n equatio equations ns 3) and 5).

Example 1.6

The driver of a car is approaching a set of traffic lights. 1When he is 50 m away from them and travelling with a speed of 72 kmh- , he notices they are red. He immediately applies the brakes. f the maximum retardation that his brakes can create is 1.5 ms-Z can the car come to rest before it arrives at the lights? The car comes to rest if its fi fina nall velocity, v, is zero, and we shall assume that the retardation is constant at 1 . 5 ms-1 . Given that the acceleration is constant constant,, we can apply equa equations tions 3), 4) and 5) with v = 0, ao = 1 . 5 and u = 72 kmh- or 20 ms- to find s. The equation that links these quantities is equa equation tion 4) 4).. Thus, inserting the values for u v and ao gives: Solution

I

I

=

Hence, s

=

20 2

2 -1.5)s

133.3 m. Clearly, the car does not come

to

rest in time.

EXERCISES 1 6

1 The brakes of a train are able to produce a retar retardati dation on of 1.5 ms ms--2 • The train is approaching a station and is scheduled to stop at a platform there. How far awa away y fro from m th thee station must the train apply iits ts brakes if it is travelling at 100 kmh- 1 ? f the brakes are applied 50 m beyond this point, at what speed will the train enter the station? 2 A ball is projected vertically downwards and describes 100 m in the 14

 

KINEMATICS

tenth second of its motion. Calculate its velocity of projection if its acceleration can be assumed to be 10 ms-2 • 3 A trai train n P sets off from a station A and travels directly towards a station B accelerating uniformly at 2 ms-2 • A t the same time, a second train is pas passing sing through station B travelling towards station A with uniform speed 30 ms-1 • After what time will the trains meet if the stations are 4 km apart? The trains meet at C Determ Determine ine the acceleratio acceleration n required by at C in order for it to arrive at station A at the same time that P arrives at station B.

4 Two trains A and B are standing in in a station on adjacent tracks ready to leave in opposite directions. A man is sitting in train A opposite the engine of train B. Both trains start star t to move: A accelerates uniformly to a speed of 72 kmh- 1 in 200 m; B accelerates uniformly to a speed of 54 kmh- 1 in 50 s. f the man notes that it takes 15 s before the end of train B passes him, how long is train B? 5 A re relay lay runner running at a speed u begins to slow down at the constant rate of a when approaching her team mate, mat e, who is at rest. Her team mate sets off with accelera acceleration tion b What is the greatest distan distance ce tha t can separ ate them at the time the te team am mate starts if they are to exchan exchange ge the bat baton? on? Assume that the two runners meet in order to exchange the baton.)

1 7 CONCLUSIONS FROM EXPERIMENTAL DAT DATA A

Experimentation allows us to test the mathematical models we use in mechanics against what can be expected in practice. This process is called validation Involvement in experimental work is an integral part of the

study of kinematics. t allows models to assume greater purpose. The following example is an experimental test for uniform acceleration.

Example

1 7

The following table gives the results obtained from an experiment. Here s represents the vertical distance upwards from the observer, in metres, of a body after a time t in seconds.

s

o

1

2

2.0

3.5

4.0

3 2.5

4 22.0 15

 

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TO MECHANICS

Is this data consistent with uniform acceleration? we look at the results graphically, we obtain the displace ment-time graph shown in Figure 1 l1 a). Estimating gradients at the times given leads to the velocity-time graph shown in Figure 1.11 b). Solution

a)

5

5

f

b)

v

10

Fig 1 11

appears clear from from ou ourr estim estimates ates that th thee da data ta is the result o f uniform acceleration. Estimation o f the acceleration from the velocity-time graph confirms this. n alternative approach to this same problem is to assume that the data is a result of uniform acceleration and to seek a contradiction. We write: t

where So u and a are constants constants that must be dete determined rmined from the data data.. The fact that So = 2.0 follows immediately from the value of s at t = 0 and we are then required to find unique solutions for u and a from the other values. This results in our solving simultaneous equations in unknowns. Note that at least four sets o f values of s a n d t are required to confirm constant acceleration for this particular time interval. t is possible to adopt a similar approach when the experimental data consists of pairs of values of the velocity, v and the distance, s from the observer. n

of

table, wedetermine have usedthe ouracceleration estimate using the gradien gradients ts of the graphthe in following Figure 1.12 a) to the relation: 16

 

KINEMATICS

a)

v

3

2

OL

3

4

3

4

5

b) v

5

0

Fig.

5

S

1 12

v ds

s

1

2

3

4

5

v

3.61

3.16

2.65

2.00

1.00

0.42

0.47

0.57

0.75

1 .5 0

1.5

1.5

1.5

1.5

1.5

v t

of

To degree of accuracy one decimal decimal place this appea appears rs to conf confirm irm that the adata is consistent with constant acceleration. 7

 

GUIDE TO

MECHANICS

Alternatively, as before, if we are able to assume uniform acceleration so tha thatt equa equation tion 5) 5)::

is vali valid, d, tthen hen tabulating va values lues of give a straight line.

s

1

13.0

v2

2

and s

v

as

in the follo followin wing g ta table ble should

2

3

4

5

9.99

7.02

4.00

1.00

The resulting graph of v2 against s is shown in Figure 1 12 b ). We see th that at the graph indeed indee d show showss a linear relationship of v2 against s which confirms our assumption that the motion is due to uniform acceleration. t cuts the vertical axis at u and its gradient is 2a This allows allows the values o f a and u to be determined directly. 2

EXER ISES 1 7

1 The motion of ttwo wo parti particles cles is described in the following table. Each particle starts from the origin, 0 at time t = O The symbol s represents the displacemen displacementt in metres from 0 v the velocity in ms- 1 and t the time in seconds. Particle 1

v

-2.5

-2

1

Particle 2

-1

1

4

8

s

7

15

1

15

7

5 7

Confirm that each particle is moving with constant acceleration and determine the acceleration and initial velocity of both particles. 2 A car starts from rest and covers s metres in t seconds. The following

table represents the motion of the car for the first 8 s. t

1

2

3

4

5

s

4

11

21

34

50

78

69

91

116

Plot the displacement-time graph and from it plot the velocity-time t as

graph foracceleration? values of described in in the table. Is the da data ta consistent wi with th constant 18

 

KINEMATICS

1.8 TWO AND THREE DIMENSIONAL MODELS Many of the comments made here apply to all vectors but we will use velocity as our example of a vector. From the th e definition of a vector quantit quantity y the velocit velocity y of a body can be represented simply as a directed direct ed line segment its magnitude being pro portional to its length and its orientation representing the direction. This allows velocities of 30 ms ms--1 north 20 ms ms-- 1 on a bearing of 120° 40 m s 1 SW and 25 ms ms--1 on a bearing of 320° to be represented as shown in Figure 1.13. Note that the north direction here has been defined by the direction allocated to the first case; all other directions have then been measured relative to it. t is just as easy to relate the velocit velocities ies of for example football players on a field of play to the direction of play.

30

Fig. 1.13

Similarly an acceleration of Similarly o f 2 ms ms--2 NW o r a displacement of 10 m on a bearing of 80° can be represented as line segments as shown in Figure 1.14.

Fig. 1.14

Any vector quantity as well as having magni magnitude tude and dire directio ction n wi will ll also satisfy the triangle law o f addition whi which ch wa wass illustra illustrated ted earlier. Using velocity we can consider th thee follow following ing examples of the th e triangle law law..

Examples 1 B 1

1.

n

airplane wishes to fly north in a wind whose speed is 50 kmh- from  

19

 

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c 200

50

A

Fig 1 15

the SW. I f the plane s engines create a forward velocity o f 200 kmh- in still air, what happens to the plane if it steers a course due north? 1

Figure 1.15 shows the velocities of the plane and wind and also the addition of these velocities using the triangle law o f vector addition. The velocity that results, from the addition of the plane and wind velocities, is represented in magnitude and direction as a line segmentt parallel to and equal in length to A C I t is possi segmen possible ble to dete determine rmine the magnitude of the resulting velocity, v = A C), and its direction, e approximately by scale scale dr drawing. awing. However, How ever, it is best done by calculation, because greater accuracy is obtained. Simple use o f the sine and cosine rules gives: Solution

v  = 50 2

v =

2

Z -

2 x 50 x 200 x cos 45 0

238.0 kmh-1

and: 50 sin

e e=

238.0 sin 135 0 8.50

The result is that the plane will follow a course 238.0 kmh-l.

of

8.5 0 with speed

2. The pilot in our example example needs to fi find nd the direction that tha t she should stee steerr

to travel north.

20

 

KINEMATICS

E

o

Fig 1 16

The velocities are shown in Figure 1.16. n this case, the combined effects effects of the wind and plane velocities must b bee V as shown in Figure 1.16. The magnitude of V = D E can be found, but more important for the pilot, the angle p will tell her the direction that she must travel. The calculation involves the repeated use of the sine rule: Solution

sin

sin

p

5

45°

2 p

10.2°

and: V

2

sin 124.8°

sin 45° V = 232.3 kmh-1

The pilot must steer a course of 349.8° and completes her journey at a speed of 232.3 kmh-l. n this case, the wind has assisted her on her journey and she can predict an earlier than expected arrival time.

Observers at an airport may reflect on the time discrepancies between outward and return journeys on flights (not just due to crossing time zones ). This is a consequence of the effect just considered. They may also notice that, during take-off in a high wind, an aircraft appears not to be steering in the direction of travel. To an observer on the ground, it will 21

 

GUIDE TO

MECHANICS

usually travel in a crab-like motion across the sky. This is als also o predicte pred icted d by our model example. These problems are called relative velocity problems. In the example, the velocity of the plane relative t o the air in which it travels was used to find its true velocity for a given wind speed. Other sources for relative velocity problems are: swimmers in fast flowing rivers; the direction of steam from a moving train; wind assistance in athletics events; the direction assumed by the sail o f a yacht.

EXERCISES 1 8

1 A swimmer swimme r who can swim swim at a speed o f 5 ms-1 in the still water o f a swimming pool needs to cross a river whose width is 20 m. The river flows a t 3 ms-1 and she sets off directly across across the river. Find the time it takes her to cross and the distance she drifts down the river while crossing. What direction would she need to set off in if she is to cross the river directly? Why is it not possible for her to cross the river directly if it flows at a speed greater than 5 ms- 1 ? 2 The time taken for an airplane to fly betwee n two cities cities and B a distance of 600 km, is about 2.5 h when the plane steers a course o f 3 ° to AB. f the plane s speed in still air is 250 kmh-t, find the direction it must steer and the time taken to do the return journey.

Fig.

1 17

3 O n a sailing boat, the direction set by the sail is that of the velocity o f the wind relative to the boat, as illustrated in Figure 1.17. The sail is is set at an angle of 45° to the boat s motion and an observer o n the shore 1

measures the boat s s peed to be 6 ms- • Find the wind speed if its direction, shown from flags on the shore, is at 30° to the boat s motion. 22

 

KINEMATICS

1 9 RESOLUTION OF VECTORS

triangle law o f addition that t hat we have already discusse discussed d can als also o lead to a procedure for the t he resolution of vec vectors tors.. In the previous section we combined the component velocities of the wind and the plane in still air to

The

find the resultant true velocity. n alternative viewpoint is to consider velocities in velocities in thei theirr compone component nt parts whi which ch wi will ll be our aim here. In later problems invo involvin lving g aany ny vector quantity you may fin find d tha thatt m much uch of the algebraic manipulation is removed by adopting this process. t is common practice in all branches o f mechanics to consider the resolution of the vector quantities using vector addition in a right-angled triangle. In I n Figure 1.18 we consider a veloc velocity ity of 20 ms-1 in a direction of 60° represented by OP so that O A is in an eastwards direction and A P a northwards direction. This gives the velocities represented by O A and A P as compon components ents of the original vel velocit ocity. y. In this case we ssay ay tha thatt they are resolved compon components ents the triangle OAP being right-angled.

a

A

Fig.

1 18

The original velocity is equivalent to the velocity represented by O A together with the velocity represented by AP. These resolved components are: 20 cos 30° = 17.32 ms- 1 east

and

20 sin 30° = 10.00 ms- 1 north

is convenient and standar standard d notation to introduce unit vecto vectors rs i and j defined in the east and nor north th directions respectively respectively.. These all allow ow a v veloci elocity ty of 20 ms-1 in the direction of 60° to be written as: t

17.32i

1O 00j 23

 

GUIDE TO MECHANICS

EXERCISES 1 9

1 Using i and j component unit vectors in the east and north directions, represent the following velocities in vector form: (a) 45°; (c) 24 kmh-t, ms-t, east; 2

(b) (d) 125 ms-t, kmh-t,300°; SW.

i and j are defined as in exercise 1.9.1, find the magnitude and direction of the velocities represented by: f

(a) 50i;

b) 12i - 9j;

(c)

lOi

+

lOj;

d)

12j.

1.10 TWO DIMENSI DIMENSIONAL ONAL PARAME PARAMETRIC TRIC MOTION I t is possibl possiblee to examine many of the ttwo-dimensional wo-dimensional (2D) v vecto ectorr concepts of kinematics easily easily using using a micro s graphics mod modee aand nd tthe he moti motion on o f its cursor. Position on a micro s screen requires the use of a Cartesian coordinate system and, as a result, it is not surprising to find i and j

component vectors being easily adapted for use. I f for example, we consider the motion of a cursor whose position vector for time t ~ 0 is:

then simulation of the cursor s motion describes the path as illustrated in Figure 1.19. The velocity vector of the cursor is given by: v =

dr

dt

= 2i - 4tj

Note that th at different differentiation iation has been carried out tterm erm by term for each o f the i and j components. A second differentiation gives the acceleration as a constant vecto vector: r: dv a = - = t

4j

In general, for a body whose position vector is described by the vector: ret 24

= x t)i + y t)j

 

KINEMATICS

j

Fig 1 19

the velocity vector

is: r

v = dt =

and the acceleration vector

x

dy

t

t

J

is:

The direction of the velocity vector relative to the x-axis tan-1

:

I

= tan-1

is

given

as:

:)

This is just the direction of the tangent at that point of the path. We conclude concl ude that tha t the t he velocity velocity of a body at any instant is along the tangent ta ngent to its path at that point. The magnitude of its velocity is given by:

Consider a ship at a point A whose vector position is a at time t = O f the constant velocity vector of the ship is u then the n Figure 1.20 shows shows the position P of the ship at subsequent times when t > O The position vector of the ship ship at time t is as a resu result lt given by: by:

r

=a

u

5

 

GUIDE TO MECHANICS

p

o Fig 1 20

This is the vector equatio equation n of a straight line passi passing ng through in a direction u. Expressing the position positio n iin n terms of the parameter t can help in the solution of many problems. For example example,, the clo closest sest approach the position when they are nearest) of two bodies, travelling with a constant velocity, can easily be studied using this vector formulation.

Examples 1.10 1. Two ships P and Q sai saill at the same time from ports and B respec tively. Port B is 25 km due north of A f the ship P sails with a velocity of 1 kmh- l on a bearing o f 3 ° and ship Q sails with a velocity of 1 kmh- l due east, find:

a) after what time th they ey are clo closest sest to each other; b) their distance of closest closest approach; c) the bearin bearing g of Q from P at this time. Solution f we take port A as the origin o f coordinates, then the positions of P and Q, after a further t hours, will be given by: r

=

10 sin 300 i

cos 300 j t

1

and

r

=

lOti

25j

Figure 1.21). This gives the position vector of Q relative to P at time t as: r

26

-

r

=

ti

25 - 5V3t j

 

KINEMATICS

r

Q

:------

p

A

Fig 1 21

The distance, D between P and Q at time t is given by the magnitude of this vector and its direction represents the bearing of Q from P: D2 =

25 - 5V3tY

5ty = 625 - 250V3t

oot 2

a) This quadratic function has a minimum value when t = 5V3/4 , which gives the time of closest approach after 2.165 h o r 2 h 9 minutes 54 s. b) The distance of closest approach is then given by substituting this value of t in the equation for D2 to give: D2 = 625

4

D = 12.5 km

The distance o f closest approach is thus 12.5 km. c) Th Thee vector position of Q relative to is then: r

- rp = 6.25V3i

The bearing of Q from

6.25j

at close closest st appro approach ach is then:

tan- 1 V3) = 60° W of S An alternative solutio solution n of this problem is achieved by drawing a relative velocity diagram. We proceed as in Figure 1.22 for both the ships

27

 

GUIDE TO MECHANICS

a

- vp

Fig. 1.22 motion relative to the ship Vp

P

The velo velocitie citiess o f the ship shipss are: and

= Wi

v   = 5i

5v 3j

so that: V 

-

Vp

= 5i -

5v 3j

This velocity has a bearing of 150° and its magnitude is 10 kmh-l. The closest approach is represented by the distance: N

= 25 cos 60° = 12.5 km

2. A ship sa sails ils from a po port rt with a velocity of 10 kmh-1 due north out to sea. A t the same time, a customs patrol boat, at sea 50 km due east o f the port, is radioed and instructed to intercept in tercept it. f the patrol boat travels at its top speed of 15 kmh- 1 , what course must it steer to intercept the ship as soon as possible? Can the patrol boat intercept the ship before it enters international waters, 50 km from port? Solution f we define the direction of travel of the patrol boat to be 8° W of north, then the velocities of the ship and patrol boat can be illustrated as in Figure 1.23. Using i and j unit vectors, the veloc velocities ities o f the ship and patrol boat can be written as: Vs

= 10j

and

Vp

=

- 1 5 sin 8 i

15 cos 8j

The velocity of the patrol boat relative to the ship is given by: Vp -

28

Vs

= - 1 5 sin 8i

15 cos 8 - 1O j

 

KINEMATICS

I s

1

patrol boat

50

port

Fig. 1.23 For interception, this velocity relative to the ship must be eastwards, towards the shi ship. p. This can only happen if the j component of o f the rela relative tive velocity is zero, which requires that cos e = 2/3, that is e = 48.2°. The patrol boat thus steers a course N 48.2° W to intercept the ship. The distance to sea of the ship at the time of interception is: 5

tan

e

=

44.72km

The ship will, as a result, be intercepted before it reaches international waters.

EXERCISES 1 10

1 With distances measured measu red in nautical m mile iless and velociti velocities es measur measured ed iin n knots, three ships A Band C are observed from a coastguard station. A t noon, they have the following position and velocity vectors relative to the station: rA =

a) b)

j

i

r 

=

rc

= 9i

4j

3i

j

VA

= i

V

= 2i c

=

j

6i

j

Find the position vector of the three shi ships ps after an hou hour. r. Prove that, if the shi ships ps continue wi with th the same v veloci elocities, ties, two of them will collide, and find the time when this happens.

2 A ship P is travelling travelling due east at 12 kmh- 1 and at a certa certain in instant a shi ship p Q is 4 km due south of P f the velocity of Q is 16.5 kmh- 1 on a bearing of 7 75 5°, fi find nd the time ttake aken n until the shi ships ps are closest. Find tthe he bea bearing ring of Q from P a further hour later. 29

 

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3 Two roads intersect at 9 ° at a point P. A maJ;l A is cycling at 1 km along one of the roads toward towardss and at a certain instant is 400 m fn P. A then observes a second man B 300 m from the junction a running towards it at a t 6 kmh-1 • Find the time when the men are nearest each other and the distance between them. 4 A ship sail sailss due n nort orth h at 20 kmh-l. t observes ano anothe therr ship sai sailin ling g 1 bearing of 45° at 15 kmh-1 and at a distance of 5 km due west. Find t distance of closest approach between the ships. Radio contact can only be maintained between the ships when t distance between them does not exceed 5 km. How long after the i sighting will radio contact be maintained?

30

 

2 FORCES AND NEWTON S LAWS IN ONE DIl\1ENSION

2 1 THE NATURE OF FORCE We now introduce introduce the concept offor offorce. Whenin for forces ces are studied the is called statics; study when they arece. studied conjunction withalone, kinematics tics,, then the area of stu study dy is called dynamics. f a body changes its velocity, we conclude that a force acts on it. Consider the motion of parachutists falling from an airplane: (a)

first, they fall vertically downwards as a result o f the force acting first, acting on them in that direction (Figure 2.1(a)). Their speed increases as they move downwards. The vertical fforce orce involved is principall principally y the weight which is the force of the Earth Ear th s attraction act acting ing on the parachutist. n addition, there are resistance forces. Resistance forces will always oppose motion when they occur. t

a)

resist nce

w ight

w ight

Fig 2 1 3

 

GUIDE TO MECHANICS

(b) Afte Afterr the parachute opens (Fig (Figure ure 2.1(b)), 2.1(b )), tthe he parachut parachutist ist s spee speed d wi will ll eventually reach a stage when it stops increasing. In this case, the velocity is no longer changing and all forces acting on the body must cancel out. In fact, the magnitude of the resistance for force ce is then equal to the magnitude of the weight (see Chapter 6). In the cas casee of a body iin n a state of equilibrium that is, at rest, the total forc forcee acting on a body must also be zero. Consider the following cases of a body P in equilibrium: (a) When the body is suspended by a string to hang freel freely y (Figure 2.2 a)), the weight is supported by an upward force in the string, the tension. a)

b)

tension

1

hrust

~ W e i g h t

c)

d)

reaction

normal reaction

friction

push w ight

w ight

e)

reaction

w ight

w ight

Fig. 2.2 (b) When the body is supported o n a spring from below (Figure 2.2(b)), the weight is supported by an upward force in the spring, the thrust. (c) When the body is resting on a horizontal surface, the weight is supported by an upward force supplied by the surface, the reaction or normal reaction (Figure 2.2(c ). (d) Whe When n the body is resting o n a horizontal surface while while being pushed push ed by a horizontal force, the th e weight is agai again n balanced by the normal rea reaction. ction. 32

 

FORCES AND NEWTON S LAWS IN

ONE

DIMENSION

additional resistance force, friction acts to balance the pushing force. The friction force acts tangentially between the surface and and opposes the motion that the pushing force is trying to create (Figure 2.2(d) ). (e) When the body is resting on a rough rou gh inclined plane, plan e, the weight w wil illl pull n

the particle down the plane, if unopposed by a friction force acting up the plane. The normal reaction again opposes the pulling effect of the weight towards the plane (Figure 2.2(e) ). alternative tive way of vie viewin wing g this problem pro blem (Figure 2.2(f) is to con n alterna sider one reaction force applied by the plane. This supports the body on which the weight acts and must act vertically. Clearly, this single force combines the effect of the normal reaction and friction forces. When forces do not balance, motion will result. In the last example, if the friction force is removed, motion occurs down the plane; that is, in the opposite direction to which the force of friction acts. Example 2.1

Figures 2.3(a) and (b) show two cases of forces not in equilibrium. Draw diagrams to indicate the direction o f motion o f a body when (a) on a string and (b) being dragged along a rough inclined plane. a)

b)

norm l reaction

friction

weight

d)

Fig. 2.3 Solution

Figures 2.3(c) and (d) show the resultant forces, shown as

double-headed arrows.

 

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We know that, in reality, bodies have finite rather than infinitesimal dimensions. n order to appreciate the important consequences that arise from this, consider the rectangular packing case, illustrated in Figure 2.4, resting on a rough horizontal floor. norm l reaction

a)

b)

normal reaction

....----

friction

pull

----4_ pull

friction

weight

weight

Fig. 2.4

Figure 2.4(a) , the force o f reaction balances the weight and the pulling force, P is exactl exactly y balanced by tthe he force of friction, F The result is that the case remains in equilibrium. n Figure 2.4(b) 2.4(b) , al alll the for forces ces are maintained in th thee same magni magnitude tude b but ut we find that, in this situation, the case topples. What in fact happens is that the case rotates about the bottom right-hand corner. This cannot be predicted by resolving forces. Clearly, the final motion of any body, whether we are concerned with statics o r dyn dynamic amics, s, depends upon both translation t ranslation and rotat rotation. ion. The finite dimensions of a body make it necessary to take into account the effects o f its its rotation. One method of rest restrictin ricting g equilibrium to transla translation tion only iiss to eliminate the dimensions of a body. For these purposes, we usually reduce the size o f a body so so tha thatt it does not po posse ssess ss any ffin init itee dimension - it has only position. This assumption is fundamental t o the particle model as opposed to the rigid body model when both rotation and translation must be taken into account. This text is based on particle models. n

2.2 NEWTON S LAWS Newton s thr three ee law lawss of motion cannot b bee proved. They were arrived at by painstaking observation and measurement, and a great deal of inspiration by Sir Isaac Newton in 1687. We will consider them one by one through illustrative examples. 4

 

FORCES

ND NEWTON'S LAWS IN ONE DIMENSION

Ice hockey players use a 'puck', a small disc that slides virtually without friction on the ice. Once it is hit, it travels in a straight line at a constant speed until another player stops it or deflects it, or the puck hits a wall o r the goal net. This is an example of Newton's first law. NEWTON'S FIRST L W Every body wi will ll remai remain n a t rest or continue to move with uniform velocity unless an external force is applied to it. Another illustration of this law is the spacecraft Pioneer 1 which is leaving the solar system and will soon be so far from all other planets and stars that no forces will act upon it. I t will then carryon in a straight line until acted upon by an external force, possibly another civilisation that can read the message messa ge on iitt Altho Although ugh the fi firs rstt section of this chapte chapterr dis discusse cussed d the nature of forces, 'force' has not been formally defined. In fact, Newton's first law is the nearest we get to a definition. A force is that which moves a previously stationary mass o r changes the velocity of a moving one. Newton's second law tells us someth something ing more about the n nature ature o f forc force. e. A ball falling freely inwith air (without airvalue, resistance) subject to the force of gravity. t accelerates a constant called isthe gravitational acceleration given the letter g. The force of gravity is a constant, so is g the acceleration it causes on a falling body. This illustrates Newton's second law. NEWTON'S SECOND L W When an external forc forcee is applied to a body, the force produces an acceleration. This acceleration is directly proportional to the t he forc force. e. The constant of proportionality is the (constant) mass of the body, that is force = mass x acceleration Mathematically, this law can be expressed as follows:

Until Chapter 10, the mass, m will always be taken as a constant. Newton's third law is easi easily ly illustrated by you, the t he read reader. er. Unless you are reading this while travelling, you are assumed to be stationary. You do not fall, so the chair o r floor must be exerting a force equal, but in the opposite direction, to your weight. f you push against a wall, the wall must be exerting a resistive force on your push. A glance back at Section 2.1 will provide further illustrations of Newton's third law, which can be expressed as follows. 35

 

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GUI

NEWTON S THIR LAW When a body B exerts an equal and opposite force on A .

exerts a force force on a body

B

A body resting in equilibrium on a table has zero acceleration. From Newton s second law, the net force on the body must be zero. Hence, the weight of the body, acting weight acting do downward wnwardss (Figure 2.2 2 .2(c (c», », must exactly exactly match the normal reaction of the table, acti acting ng upwards. upwards. Similarly Similarly,, a train moving moving on a straight track at a constant speed has no net force acting on it. The tractive force of the engine is just enough to overcome all the different resistive forces acting upon it. These remarkable rema rkable laws, laws, recently celebrating their 300th anniversary, stil stilll adequately describe all mechanics, except the innermost workings of the atom and the motion of objects whose speed approaches that of light itself 3 x 1 8 ms ms-- I   . Before considering an example, a brief word needs to be said about units. Most of us, these days, use SI units (Systeme International d Unites). Distance is measured in metres, and the units of velocity and acceleration is

follow straightforwardly from and kinematics. rate ofper change of distance with respect to time, hence isVelocity measured the in metres second (ms-   . Acceleration is the rate of change of velocity with respect to time, and hence is measured in metres per second (ms-2 , read as metres per second squared). Newton s secon second d la law w of motion tells us that tha t force is equal to the th e product of mas masss and acceleration. The SI unit of mass mass is the kilogram (kg), hence the unit of force is kilogram metre per second squared (kgms-2  . This is considered rather clumsy, hence a new name, the newton (N), is used instead. instead. One O ne newton is the force required to give a mass of 1 kg an acceleration of 1 ms ms--2 • Units do not tell us about direction; however, al alll units can be expressed in terms of mass, length and time, and an equation is certainly wrong if the units of both sides, calculated in terms of mass, length and time, do not balance. f there is a balance, however, the I

equation mayor may not be correct. Only in some imprecise sense is the likelihood likelihoo d of correctness increased. The momentum of a particle is a vector vector quantity equal to the product pr oduct of its mass mass and its velocity. velocity. Th Thee units of o f momen momentum tum are a re kgms-I • f th thee velocity of a particle is constant, then so is its momentum, for constant mass. Newton s second law is: dv

F=m-

 t

or for a constant mass: F =

d(mv) t

36

 

FORCES

ND

NEWTON S LAWS IN

ONE

DIMENSION

This relationship, in words, is that force equals the rate of change of momentum. Integrating both sides of this equation with respect to time between the limits = t and = t2 gives: 1)

The right-hand si side de is the change in momentum brought br ought aabout bout by the for force ce F. The integral of the force with respect to time is called impulse and the integration integr ation of Ne Newton wton s second law law shows that impulse is equal to the change in momentum. The SI units of impulse are newton seconds (Ns). From the fore foregoin going g equa equation, tion, these are also also the units of momentum. Sin Since ce the newton, when expressed in more fundamental units, is kgms-\ this is consistent with momentum being expressed in these same units as kgms- 1 •

Example 2.2.1

Figure 2.5 2.5 sh shows ows the executive toy known as New Newton ton s cradle. Analyse and describe what happens when ball 1 is pulled aside and released.

2

3

4

5

Fig. 2.5

Solution Each ball has equal mass, m When ball 1 is pulled aside and released, it swings down and hits ball 2 which is of course stationary. Ball 1 stops, but what has happened to its momentum? Usually, the balls are made of metal, which can be assumed to be perfectly elastic. As ball 1 swin sw ings gs down, its its spee speed d increases, and so does the ma magnitude gnitude of o f its momen tum as a result. (This increase in speed is due to an increase in kinetic energy, a term explained in Chapter 4.) Ball 1 collides with ball 2. The impact produces an impulse which must be equal to the momentum lost by

ball 1 on impact, from the integral of Newton s second law, equation (1). This impulse is transmitted across all the balls in turn, but balls 2, 3 and 4 7

 

G U ID E T O MECHANICS

cannot move. The impulse is thus given to ball 5 which moves off with the same momentum momen tum as w was as lost by ba ball ll 1 on impact. Ball 5 then swi swings ngs up an and d back down, and the process is reversed and repeated indefinitely. In reality, the balls eventually come to rest due to the air resistance on balls 1 and 5, and the balls not being perfectly elastic.

Momentum is a very important concept, especially when considering the impact of masses (see Chapter 5). We will meet other concepts such as work and energy. However, all of these mechanical quantities can be expressed expres sed in terms of the fundamental units, mass, length a nd time. Let us break here to conside considerr a purely computational example based on Newton s second law.

Example 2 2 2

A mass of 4 kg law:

is

displaced a distance x from the origin, x = 3 sin 2t

What

is

(m)

0; :n: 4 s;

=

:n: 2 s

Newton s second law gives: force = mass x acceleration

Differentiating x

= 3 sin

2t gives:

dx

-

dt

= 6 cos

(ms- i  

t

A second differentiation gives:

~ 38

according accor ding to the

the force acting on the mass that produces this displacement:

(a) at time t (b) after = (c) after = Solution

0

d2 dt 2

.

=

12

sin

t

(ms-2 )

 

FORCES AND NEWTON S LAWS IN ONE DIMENSION

From Newton s second law: d x

force

1 2 sin 2t)

4x

m dt  

4 8 sin 2t

(newtons)

We can now substitute any value of t to obtain the force at that time: (a) (b) (c)

t t t

0 gives force nl4 gives force nl2 gives force

0 N; 4 8 sin 0 4 8 sin nI2) 4 8 N; 4 8 sin nil) 0 N.

This answers the question fully. Note here that x 3 sin 2t is an oscillato oscillatory ry function of the type that is studied extensively in Chapter 9 (vibrations).

Apart from enabling us to compute the force, as in the foregoing example, Newton s second law provides an explana explanation tion for many everyday occur rences. A car travelling at high speed hits a brick wall. Its (negative) acceleration is very large. Some data might be: velocity before impact 9 kmh- (250 ms-   , velocity after impact = 0 ms-I, mass of car = 1500 kg, duration of impact = 1 s. We calculate acceleration from the change in velocity per unit time: I

I

0 250

acceleration accelerat ion

1

-250 -25 0 ms ms--2

Newton s second seco nd la law w the then n gi give ves: s: force = 1500 x (-250) = -3.75 x

1

5

N

This very large force suggests the kind of damage that might be expected by driving into a brick wall. Here is another example. When a lorry is on a slope, making angle 8 with the horizontal (Figure 2.6),, the n there is a component of weight, mg sin 8, in the direction down 2.6) the slope. f the lorry weighs 1 tonnes (or 1 000 kg) and the slope is 1 ° (or 0.174 radians), then the net force down the slope is: mg

sin 8

1

000 x 9.81 x 0.174

17 3 f

the lorry

is

N

parked, park ed, by Newton s third llaw aw,, th thee brakes must exert

17 3

N 39

 

GUIDE T O MECHANICS

Fig. 2.6 of force in order to stop the lorry from rolling or sliding down the slope.

(Friction of course is an aid here since it prevents motion, so if force of the brakes, then strictly: 1703

=

F

is

the

friction

F

and 17 3 is a safe value for F in that it is the most F has to be. When you stand s tand in a lift, how much do you weigh? f th thee lift is stationary or moving with constant speed, then your weight is the same as it is now (assuming (assum ing that you are ar e no nott reading this in an accelerating lift ). f the lift lift is acceleratin accel erating, g, then t hen the situation is as shown in Figure 2.7. is the reaction, m your weight and a is the acceleration of the lift. You may b e forgiven

0

m

Fig 2 7 4

 

FORCES AND NEWTON'S LAWS IN ONE DIMENSION

for thinking that R = mg and that your weight weight is the same, no matter what the value of a However, you would be wr wrong ong Wha t we have to do is to relate a the acceleration, to a fixed origin. This is shown as O. The floor of the lift is then travelling with acceleration a relative to 0 and Newton's second law applied to the occupant is: mg R

=

ma

Rearranging this equation gives: meg

a

=

R

and we have a statement of Newton's third law (as far as you, the occupant occu pant ofthe lif lift, t, are concerned). concerne d). f a = 0, mg = R as expected. f a > 0, th e lift lift is accelerating downwards, mg > R so the reaction of the floor o f the lift, which is after all what you actually feel, is less than your weight. f the lift cable snaps and a = g, you become 'weightless', not because you actually f

<

a are but because R = 0 and you feel as if you are. 0, the lift accelerating accelerati ng upwards, u pwards, mg < R and you feel heavier because R > mg

EXERCISES

1

is

2.2

winter and a stone is thrown across the surface of a frozen pond. Neglecting friction, what will be the path of the stone? t is

2 A ball is falling through the air towards the Earth. I t is travelling with a constant velocity. Explain this in terms of Newton's first law. 3 A geostationary satellite is one that remains above the same point on the Earth s surface. t is thus not moving from the point of view of an earth dweller. How can this be explained from Newton's first law? 4 Newton's cradle was was introduced introduc ed in example 2.2.1 2.2.1 (Figure 2.5). 2.5). Analyse what happens if: (a) balls 1 and 2 are drawn aside and released relea sed simultaneously; simultaneously; (b) balls 1 and 5 are pulled aside to the same height and released simultaneously; (c (c)) balls 1 and an d 2 are ar e drawn aside, as is ball 5, 5, al alll to the same height and an d released simultaneously. (All motion may be assumed to be one dimensional. ) 4

 

GUI

E TO

MECHANICS

5 Sho Show w tha that, t, if the displacem displacement ent of a mass is a quadratic quadra tic function it is being subjected to a constant force.

of

time,

6 The Th e displacement o off a mass of 5 k kg g iiss gi given ven b by y the formula: x t) = t 3

-

6t 2 + 4t + 7

(m)

Find the force in newtons that gives rise to this displacement and deduce at what value of time this force is momentarily zero. 7 The wal walll of death is a fairground attract attraction ion where the victim stands on the inside of o f a large spinning cylinder, facing the axis axis.. As tthe he spin increases, the victim is pressed against the wall. The floor then falls away, awa y, but the victi victim m remains pinned to the wa wall ll.. Explain this in terms of Newton Newt on s third law. law. Hint: You need the definition of friction.) 8 A window clean cleaner er sometime sometimess u uses ses a cradle with p pulleys ulleys to hau haull himsel himselff f

to different floors of aistall block. the the inexten window cleaner and the cradle W office and this is suppor is supported tedweight by fouroflight sible ropes of equal tension, find this tension in terms o f W if the cradle plus window cleaner: (a) (b) (c) (d)

is stationary;

moves u upwards pwards at a cons constant tant speed spe ed of 0.5 m mss-l ; moves downwards at a consta constant nt speed of 0.5 ms ms--l ; moves upwards at 0.5 ms-l after starting from rest a second earlier.

(Take g

=

9.81 ms-  • The effects of friction may be neglected.)

9 Discuss how the answer to exercise 2.2.8 will inclusion of friction.

be

modified by the

2 3 RESIS RESISTANCE TANCE AN AND D THE PARTICLE MODEL

Only in a highly idealised world do balls slide o r roll on a horizontal table and never stop. In reality, resistance acts so as to slow down motion. For solid objects, we can distinguish two kinds of resistance. There is friction between solid surfaces, which is treated fully in Chapter 3, and there is frictio fri ction n betwee between n an object and iits ts flui fluid d surroundings - this frict friction ion is usually called drag Almost all of the time, we will be concerned with bodies moving through air o r water. As a sol solid id objec objectt does this, it needs to push fluid aside in order to make progress. The more streamlined an 42

 

FORCES AND NEWTON S LAWS IN ONE DIMENSION

t ngent to

urve

o pproxim tion

qu dr tic

5. Applying the law of conservation of momentum to the man and trolley gives:

o=

10 x 5 - v

Solving this for v gives

0.625 msm s- . Hence, the

70 x

-v)

the trolley moves with a velocity velocity o f the ball is 4.375 ms- . that

of

EXERCISES 5.4

1 A woman wom an climbs on to a light trolley standing a t one end. She then moves quickly to the other end, stopping when she arrives there. She then jumps j umps off off the trolley. Describe the th e motion motio n o f the trolley during the woman s movements, assuming no friction between the trolley and the ground. 2 Two putty-like masses m and 2m are travelling in the same straight line but in opposite directions, with speed u, when they collide and unite. Find the magnitude and direction of the velocity velocity o f the combined masses masses and determine the loss in energy. 3 A cannon of mass is free to move o n a smooth horizontal track. f the cannon fires a shell, with muzzle speed v, horizon horizontally tally along along th e t rack,

102

 

COLLISIONS

find the resulting recoil velocity generated in the firing. 4 A bullet bulle t

of

mass

1

of

the ca nnon and calculat calculatee the energy

g is fired horizontally into a block f

on

of

wood

of

mass is

1 kg, mswhich reststhe velocity a smooth horizontal plane. the bullet s veloc velocity ity I , find 1000 of the block, if the bullet emerges from the block with a speed of 500 ms- I .

5. 5.5 5 NEWTON S EXPERIMENTAL EXPERIMENTAL LAW Consider the example of two bodies of masses 2 kg and 3 kg moving in a collision course on a smooth surface: they hit each other and move independently after the collision. n Figure 5.9, the motion is illustrated. b efore the collisio collision n are 4 ms-I and 1 ms- I , their direction being he speeds before as indicated. t is our aim t o calculate their respective speeds, v and V after the collision.

v

v Fig. 5.9

Conservation

of

momentum at the collision gives: 2

which simplifies

to

X

4

3

X

1=

v

3V

give: 11

=

v

3V

(5)

This equation gives a relationship between the two velocities but not a solution. We seek a second relationship between v and V so that values for v and V can be obtained. here is a second law governing the collision of such bodies, Newton s experimental law of restitution. t takes account of energy losses a t collision. t can be stated as: f two bodies collide, their relative velocity before impact is e times the ir relative velocity velocity after impact. he constant e is called the coefficient o f restitution for the two bodies concerned. he case e = 1, which is unattainable in practice, is called a perfectly elastic

1 3

 

GUI DE T O MECHANICS

collision, and there is no energy loss. I f e = 0, the collision is termed inelastic and the bodies unite o r coalesce. This was the case in the bullet and block probl em example 5.4.1). For the velocities as shown in Figure 5.6, the restitution law would give: -e U - u

V - v

=

Returning to the problem in this section, if we know have:

that

e = 112

then

we

1 2

-- 4-1)=V -v

=

1.5

Equations Equat ions 5) and 6) can v

=

be

6)

v

solved by simple algebra to give:

1.6 ms-1

and

V

=

0.1 ms- 1

Clearly, the energy loss and its dependence on e does warrant further study. In Figure 5.10, the graph of energy loss loss for the th e two bodies bodie s we have

5

4

 n n

3

.Q > ~ D

I D

2

o

0.2

0.6

0.4

restitution

Fig.

5 10

0.8

1.0

104

 

COLLISIONS

just considered co nsidered is shown for values of e from 0 to 1. Note that when e = 0 an inelastic collision the energy loss loss meas ured as the difference in the kinetic energy before and after the collisi collision on is greatest. greatest . When e = 1 the elastic case it is zero.

Example 5.5 small spheres A a n d C lie in a straight line o n a smooth table. Their masses are m m and 4m respectively. Sphere A is projected towards sphere B with a speed of 8 ms- I . f the coefficient of restitution is 114 find the velocities o f the three spheres after three collisions and show that there can be no more collisions. Three

c0

a 2m

  m

v

Fig. 5.11 In Figure 5.11 the speeds o f the spheres before and after each collision are shown. f the motion of the spheres is modelled as particles then for the first collision we have: Solution

conservation

of

momentum: 8

= VI

2 V 

Restitution Resti tution give gives: s:

This gives VI = 1.33 ms- I and V  For the second collision: conservation

of

= 3.33

ms- I •

momentum: 6.67

=

2 V3

4V4

1 5

 

GUIDE TO

MECHANICS

Restitution gives:

V

=

l

This 0.56 ms- and Forgives the third collision:

V4

=

l

1.39 ms-

.

conservation conservatio n of mom momentum: entum: 2.4 2.44 4

2 V6

= Vs

Restitution gives:

This gives Vs = 0.69 ms- l and no more collisions.

V6 =

0.88 ms- l • As

Vs

<

V6

<

V 4 there can be

EXER ISES 5 5

1 A ball of mass m is moving with speed 2u and collides with a second ball of mass 5m moving with speed u in the opposite direction. f the coeffici coef ficient ent of re restitution stitution is 112 find find the velocities of the balls aft after er impact impa ct and the loss in energy due to the collision. 2 A ball fal falls ls from a heigh heightt of 20 m and at the same time tim e a ball of the same mass is projected vertically upwards from the ground with speed ms \ f the coefficient o f that they in sais directfi restitutionsobetween betwee n the meet two ball balls nimpact. d th thee time tha thatt elapses before 112 fin the two balls reach the ground.

40

3 Thre Threee sp spheres heres of mass masses es m 3m and 6m whose speeds are 6u, 2u and u respectively are moving in the sa same me dire direction ction in the same str straight aight line in the order given. f the collisions that take place are perfectly elastic show that the first two spheres will be brought to rest. 4 Three particles of masses m l , m 2 and m are at rest on a smooth horizontal table. The first sphere is given a speed u towards the second. The coefficients of restitution are e and e between m l and m 2 , and m 2 and m respectively. Determine how the values o f m u m 2 , m 3, u, e and e may be allotted so that the total number o f collisions is a maximum.

106

 

COLLISIONS 5 6 DIRECT COLLISI COLLISION ON BETWEEN A PARTICLE PARTICLE AND A FIXED BARRIER n

Figure 5.12, a particle

of

mass m

shown in collision with a fixed

s

s

s V

barrier. collision after reversed The barrier. U and supplies Its an velocity impulse Ibefore to the the particle, which has its motion by the collision. This s not a case in which conservation of momentum can apply, as the momentum supplied to the barrier through the impact s lost this system resembles that in Figure 5.3 b)).

Fig. 5.12 The impulse momentum relationship applied to the particle only gives: I

=

mV - m -U)

=

m V

U

loss in energy can still be analysed using the restitution law. Applying the law to the impact gives:

The

-e -U-O)

=

V-O

to give the reduced relationship in this case of: eU= V

Example 5.6 A ball m s dropped from a height h on to a horizontal floor floor and it rebounds rebound s to a height l/2)h. Calculate a) the value of the coefficien coefficientt of restitution, e, betwe en the ball and the floor, floor, b) the impulse impulse applied by by the floo floorr to the ball and c) the time until bouncing ce ceases. ases.

107

 

GUI

E TO MECHANICS

---

 0

h

our Y h

v

0

v Fig. 5.13

Solution

Figure 5.13, the ball is shown at the point of impact. The velocities immediately before and after the impact are also shown. f the velocity just before the impact is v, we have: n

v2 = 0

+ gh

v = V 2gh) f

the velocity after impact is V we can also say:

0= V 2

-

1

2g x - h 2

V = v gh) a) Applying the law of restitution to the ball at impa ct gives: gives:

eV 2gh) = v g h ) 1

e= v

b) The impulse momentum relationship gives for the impulse I: 1 = mV gh) - m -V 2gh» = m v g h ) V 2

+1

c) When the vel veloci ocity ty immediately aft er imp act with the floor is V then the time t o the next bounce is given by:

0= Vt- gt 2 2

1 8

 

COLLISIONS

which gives the time as 2V/g. In theory, there will be an infinite number of bounces. As the velocity at each bounce is multiplied by a factor e we have that the total time of bouncing is: 2v

2ev

g

g

2e 2 v

-

--

g

2 ~ v

--

g

where v = v (2gh). Note that the time to the first impact is, in effect, half of a complete bounce. This series can be rewritten as: 2v ( g

-1

e

e

2

+ e3

...)

The geometric progressio progression n can be summed to infinit infinity, y, as e < 1, to give: e 2v 8(1-

1

=

(1

I h) g V

4 \ V2) 3

EXERCISES EXERC ISES 5 6

1 A small body slides over a smooth hori horizont zontal al flo floor or hitti hitting ng a wall along its normal. I f the time taken for the body to return from the wall to its point of projection is twi twice ce the jou journe rney y to the wall wall,, fin find d tthe he coeffic coefficien ientt of restitution between the body and the wall. 2 A ball is projected vertically upwards with a speed U from the th e flo floor or of a room of height h. I t hits the ceiling and then returns to the floor, from which it rebounds, managing to just hit the ceiling a second time. Calculate the coefficient of restitution between the floor and the ball, and the ceiling and the ball, if they are equal. 3 A ball is dropped on to a horizontal plane. I f the coefficient of restitution between the ball and the plane is e, show show that the averag averagee speed for each bounce decreases at a rate of e per boun bounce. ce. 4 Two smooth spheres A and B of equal radii and masses 2m and 3m respectively lie on a smooth horizontal table. A is given a velocity so that it hits B at rest, directly, which then goes o to strike the wall normally. I f the co coef effic ficien ientt of restitution between the th e spheres is 112 and

109

 

GUIDE TO MECHANICS MECHANICS

that between B and the wall is 114 show that there is a total of three collisions only. 5 Two particles

and

B of

masses m and

M

are placed so that they are in

aand line to a B wall. coefficient between B perpendicular is e and between andThe the wall is e . of f restitution is projected towards the wall with with speed u, so that it strikes particle B first, find how the total number of colli collisions sions can be maximised by suitable choice of m M , e, e and u.

5.7 OBLIQUE COLLISION BETWEEN A PARTICLE AND A FIXED BARRIER In Figure 5.14, a particle of mass M is shown at the point of impact with a

smooth barrier. Although this type of problem appears three dimensional, the motion moti on can alwa always ys be considered to take t ake place in a plane containing th e normal to the barrier and the particle s initial velocity, provided that the barrier is smooth.

L Fig. 5.14 itss impact In Figure 5.14, the particle is shown to have a speed U before it with wit h the barrier, at an angle a wi with th the normal, that is, velocity vector U. Its speed after impact is V at an angle ~ with the normal, that is, velocity vector V. As the plane is smooth, the impulse I applied to the particle will be normal to the plane. f the impulse vector is I , then the vector equation for the impulse momentum relationship gives: I= M V

M U

unit vectors i and j are taken as shown in Figure 5.14, then the relationship becomes: f

I j = M V sin

V

cos

~ j )

- M U sin a i -

U

cos a j

110

 

COLLISIONS

Solving gives a decoupling of the components as: V

=

cos

U

~

cos a

is

which the impulse momentum equation applied along the normal to the plane, and V sin ~ = U sin a which is the conservation of momentum along the plane. The restitution law supplies the second governing equation for the motion, applied normal to the plane where energy loss occurs, to give: e

cos a

=

V cos

~

Given the direction a and the magnitude U of the incoming velocity, it possible to find the direction ~ from: tan

~

=

is

tan a

e

and magnitude:

of the outgoing velocity.

Examples 5.7 1

A ball of mass m is droppe dropped d from a height h on to a plane inclined at a to the horizon horizontal. tal. f the ball rebounds so that its motion is initially horizontal, calculate a) the ccoef oeffic ficien ientt of restit restitution, ution, e, between the ball and the plane, plane , and b) the lo loss ss in k kinetic inetic energy due to the impact. Solution Figure 5.15 shows the motion o f the ball just before and just after impact. impact. The speeds just before and after impact are v and V The information given allows the angles before and after impact to be determined in terms of a . The ball falls vertically a height h to give:

v2

=0

v

=

2gh

y (2gh)

111

 

GUIDE TO MECHANICS

o \

\

\

\

v

Fig 5 15

Conservation of momentum along the plane at impact gives: m v sin a

m V cos a

=

and the law o f restitution gives: ev cos a

V sin a

=

a) Dividing Dividing these two expressio expressions ns leads to: e

=

ta tan n2 a

b) Loss Loss in kin kinetic etic energy at impact is:

1 m v2 2

_1.

2

m V2

which can be rewritten to give: ~

v 2 1

-  sin2 a

e2

cos a»

which simplifies to give: 1 2 mv

2

coss 2 a l - e2   co

=

mgh l - t a n2 a )

2. A large smooth horizontal circular table has a vertical rim around its edge. Show that, if a small body is projected from a point P at the edge of the table, in a direction making an angle a with the radius to P so that after two impacts the body returns to P then: coe a =

e

e2

112

 

COLLISIONS

Fig 5 16

In Figure 5.16, the possible path of the particle from P is shown. The angles made by the direction of motion after the first and second impacts, at Q and R are shown as ~ and . The speeds u v and w represent the speeds between the impacts shown. The geometry of the circle allows the incident angle at Q to be identified as a and at R as while a + + = 1/ 1/2) 2)Jt Jt.. For the th e point poin t Q: Solution

e tan a

=

tan

e tan

=

tan

7)

~

For the point R

Removing

~

8)

from equa eq uatio tio n 8) gives gives::

i

e tan ~ = tan

Jt

-  a +

The expansion formula for tan a +

e tan Eliminating tan result.

~

~

=

= cot a + then gives:

1 - t a n a tan tan a + tan

~

~

usin using g equation 7) and rearranging gi give vess the require d

3

 

GUIDE TO

MECHANICS

EXERCISES 5.7 1 A particle of mass m moves on a smooth plane so that its speed before impact with a fixe fixed d barri ba rrier er is u at an angle angle a and afterwards its spee speed d is v 13

at an angle withbarrier with the normal. coeffi fficie cient nt of restitution betwe between en the particle and the is e The coe a) b)

f f

u u

= 2 ms- a = 60° and e = 114 find v and 13 = 5 ms-I, a = 30° and 3 = 45° find e and the loss in kinetic l ,

energy.

2 A smooth rectangular table B CD of width d has a vertical rim along each of the th e edges B and CD of length 21d A small sphere is projected from the point with speed u at an angle of tan- l 3/4) with the side B down the table. The coefficient of restitution between the sphere and the rim is 112. Show that the particle makes four collisions with the rim before leaving the table. 3 A ball is projected with speed u at an inclination a to the horizontal from a point P a distance d from a smooth vertical wall. After striking the wall, it again strikes the point P for a second time. f e is the coefficient of restitution between the wall and the ball, show that: eu 2 sin 2 a = ga l

e

4 A ball is projec projected ted fr from om the bottom of a smooth plane of angle a up the line of greatest slope at an angle 8 to the horizontal. The coefficient of restitution between the plane and the ball is e Show that, if e > cot 8 - a - 2 t an a , the ball will bounce down the plane after its first collision.

5.8 OBLIQUE COLLISION BETWEEN TWO PARTICLES Oblique collisions between two particles represent the most demanding analysis that we shall carry out in the case of collisions. I t is here that the advantages of vector notation become aapparen pparent. t. In fact, using the impulse momentum equation equat ion fo forr each particle separately, and the th e la law w of restitution to consider energy losses losses along the no normal rmal aatt impa impact, ct, allo allows ws us to solve an any y such problem. The meth method od of ana analys lysis is can be summarised as follows, with reference to Figure 5.17, where two smooth bodies of masses m and M are shown at impact.. The veloc impact velocity ity v vectors ectors before impact are u aand nd U, aand nd aft after er impact v and V respectively. The mutually opposite impulse vectors are I and I

4

 

COLLISIONS

Fig. 5.17 The governing equati equations ons for the motion are aass fol follow lows. s. Note that the motion of the two particles particles need not be sol solely ely in the same plane.) From the impulse momentum relationship for M I =

and for

MV - MU

m

=

mv

- mu

From the law of restitution: -e(u - U) . n

where n

is

=

v - V) . n

the unit vector in the direction of I.

Clearly, the compactness of vector notation is a great advantage in expressing the general equations. The same equations, when written in component form, are extremely lengthy. As an example, if we wish to model the motion of snooker balls as particles on a smooth table, we will only need to consider motion in two dimensions. t is also the case that the image ball is usually, but not always, stationary. The example that follows demonstrates some of the simplifications.

Example 5 8 A snooker ball A travels on a smooth table and impinges on a second snooker ball B which is at rest. f ball A is travelling with speed U in a direction e o the line of centres at impact, find the velocity velocity of B after impact and the value of e n order that A s path suffers the greatest deflection.

5

 

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this special case, the motion of s such that it is a t rest before impact and during impact the ball, which we assume to be smooth, s subject to an impulse impulse o f magnitude I along its line o f centres. This means that momentum can only change along the line of centres and, as a result, must start to move in this direction. Solution

In

v I>

e

- -Q_.;;:;a. v_

L, Fig. 5.18

Figure 5.18, the motion of the balls before and after the impact s shown. The unit vectors i and j are also also defined. The gover governing ning equation s for the motion, mo tion, assumi assuming ng that the spheres can be modelled as particles, are as follows. From the impulse momentum relationship for A In

Ii

For

= M V cos

ct i

+

V

sin ct>j - M U cos 8i +

U

sin 8 j

B: i

= mvi

These give:

The

U

sin 8 =

V

sin

ct>

U

cos 8 =

V

cos

ct>

9) v

10)

restitution law gives: - e U cos 8

=

V

cos

ct>

- v

Subtracting equation equa tion 11) 11) from equ ation ati on 10) 10) give gives: s:

11)

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COLLISIONS

v

=

1 2 u cos a 1

e

the velocity o f as required. We now need to maximise the deflection Adding Addi ng equation equa tionss 10) an and d 11 11)) gi give ves: s: 1 2 u cos a 1 - e

= V

eI

-

cos

eI

a for

variation in

a.

12)

and divid dividing ing equa equatio tion n 9) by equatio equa tion n 11) 11) give gives: s: tan

eI

=

2 tan

a

- e

This allows the deflection to be analysed using the expansion formula for tan eI> - a) to give: tan

eI>

- a)

=

eI

tan 2 tan2 a

t n

a

1- e

As 0 ~ a l/2)3t, l/2) 3t, then ta tan n a s an increasing function for all values of a n this range. We shall look here for the maximum value by differentiating with respect to tan a. Differentiation Differen tiation gi give ves: s: d tan eI> - a)) d tan

1

a)

e

1 - e - 2 tan2 a)

2 tan ta n2

a-

e2

The stationary value occurs for: ta tan n2

a = -21   1

- e

and by considering values greater than and less than tan a in the given range, this can be seen to be a maximum. So, the angle a is given by: tan

a= V ( ~ ( 1 - e

EXERCISES 5.8 1 A small sph sphere ere o f mass m is travelling with with speed u along a straight straigh t line. t is met by a second similar sphere travelling at n angle of 30° to the

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fi firs rstt sphere spher e s mo motion tion wit with h sspeed peed u They collide. f the coefficient of restitution is 112 determine the new direction of travel o f each of the spheres. of mass m travelling with speed u is in collision with a 2 A sphere similar stationary sphere B of mass 2m. The direction of motion makes an angle of e with the line connecting their centres. f after the collision sphere is travell travelling ing in a direction at right angles to its original motio motion, n, find the value of the coefficient of restitution between the two spheres. 3 The assumptions that have been made regarding the motion of a snooker ball are that it should be modelled as a particle travelling on a smooth table. Discuss the effects of removing the assumptions.

118

 

6 MOTION UNDER GRAVITY IN ONE DIMENSION

6 1 INTRODUC INTRODUCTION TION

In this chapter, we shall examine one-dimensional motion under gravity. By their very nature, the problems are restricted to those involving bodies moving mov ing vertic vertically ally un under der the influ influence ence of the Ea Eart rth h s gravity, with o orr without resistance. Nevertheless, falling raindrops, parachutists, balls and so on, and upwardly projected bullets, missiles and balls comprise a very important class of problems. Because o their limited scope, we can solve virtually all such problems by examining just five types. First, there is motion without resistance. There are then upwards and downwards mo tions, each under the action o two different types of resistance. This chapter is very much a prelude to two others. Chapter 1 takes into account that some masses either lose material (for example, rockets) or gain material (for example, raindrops) as they move under gravity, while Chapter 7 generalises the analysis to apply it to two dimensions by considering projectiles. We will continue to emphasise the importance of the set procedure for solving mechanics problems: drawing a diagram, deciding on an origin, deciding on a direction for the x-axis, drawing in the forces and then, and only then, writing down Newton s second law o motion.

6 2 MOTION WITH NO NO RESISTANCE

Tradition has it that Newton first had the idea for his laws o motion by observing a falling apple. We have already discussed the appropriateness o

as

masses (see is Chapter 2); valid, therefore, we know thatmodelling to modellarge a falling apple particles as a particle perfectly provided it is not spinning (and, even then, it models linear motion).

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r

Fig. 6.1

Figure 6.1 shows a particle in motion under gravity. For now, we neglect air resistance, so that the only force acting is gravity itself. Gravity is assumed to be constant, hence this is a special case of motion under constant acceleration, which which wa wass dealt with in Chapt Ch apter er 1 Let us derive the equation of motion for the simp simple le system system sh shown own in Figure 6.1. Note that we have chosen chosen the x-ax x-axis is to point in the th e opposite direct direction ion to the t he force due to gravity; therefore: m g = m x acceleration

1)

We recall that acceleration can be expressed in a number of different forms, so we need to select one that is convenient for our particular problem. You should use dvldt if you are interested in obtaining velocity as a function of time, and v dvldx if you you are interested inter ested in obtaining velocity velocity as a function of distance. Finally, you should use d xldt   and integrate twic twice if you want displacement distance above as a function of time.e Ou Our r equation of motion, after cancellin cancelling g theground) mass m from bot h sides, both sideo s,f is thus one of the following: dv

dt

=

g

dv

= g dx

2)

In order ord er to become adept at a t choosing choosing whi which ch of these equations equati ons to apply, we

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MOTION UNDER GRAVITY IN ONE DIMENSION

need to solve particular problems. Here are some examples that will help to focus these ideas.

Example 6 2 1

apple o f mass 0.3 k kg g fa fall llss from a tree tree.. Find a) its velo velocity city after 2 s, ignoring ignori ng ai airr resistance and as assumi suming ng it has not hit the grou ground, nd, and b) the distance it travels. An

Solution

a ) We wish to fi find nd v as a function of time, hence we use the first of the equations 2): dv

-

=

g

dt

Integrating this with respect to time gives: v

A

gt

=

where is an arbitra arbitrary ry constant. We determi determine ne the value of by usi using ng the condition that at time t = 0, v = O Hence, = O Taking g = 9.81 ms-Z we obtain: v

gt

=

=

9.81t

3)

Hence, for a particular time, v is compl completely etely determined. In particular, when t = 2, v = 1 9 . 6 2 ms ms-- The velocity is negative since the apple is travelling downwards, that is, in the negative x direction. Also note that the mass does not enter into this problem. This fact was foreseen by Galileo when he drop dropped ped stones from the leaning tower of Pi Piss a in the early seventeent seven teenth h century. They al alll hit the gro ground und at the same time despite being of different mas masses. ses. Rat he herr mor moree spectacularly spectacularly,, a feather and a rock were, more recently, dropped on the Moon with similar results. The latter experiment would not work on Earth owing to air resistance, which affects the velocity of falling bodies, especially that o f the feather.) b) We now need information connecting distance and time. Therefore, we integr integrate ate equa equation tion 3) by substituting dx / dt for v The equation: I .

dx

-

dt

=

gt

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GUIDE T O

MECHANICS

thus becomes: x

=

1 2 gt 2

-

B

-

(4)

where B is an arbitrary constant. The evaluation o f B requires the us usee o f a boundary condition. However, to solve the problem we do not actually need to know B. To see this, this, we note tha thatt from equ equati ation on (4 (4)) at time t = 0 (when the apple was still on the tree): x O) = B

(5)

Here, x O) denotes the value o f x at t = 0, the standard functional notation. We now put t = 2 into equation (4), with g = 9.81 ms - as before, to obtain: 2

x 2)

=

19.62

B

(6)

The total distance travelled is the difference between x O) and x 2), and performing the subtraction eliminates B:

x 2)

-

x O)

=

19.62

m

This value is negative because the apple is heading towards the origin, so that x 2) is smaller than x O). The distance travelled, in absolute terms, is 19.62 m. That Tha t an answe swers rs the problem, b but ut 19.62 m is very large - far larg larger er th than an the hei height ght of aan n aver average age app apple le tr tree ee In normal circumstances, the apple hits the ground in far less than 2 s The foregoing rather silly example applies onlyt tois apples that always fall down holes o r from apple trees adjacent to cliffs. important to question your results. In a more serious example, such questioning could result in modifying some o f the underlying assumptions, after, o f course, checking all calculations very carefully.

Example 6 2 2

A boy throws an apple with a speed high does it rise? Solution

of

10 ms- vertically upwards. How

Since here we wish to find velocity

1

as

a function

of

distance, this

122

 

MOTION

UNDER

GRAVITY IN

ONE

DIMENSION

time we choose v dv/dx to represent acceleration. As in the previous example, Figure 6.1 will apply. So will equation (2), namely: dv v -

=

g

dx

Integrating this equation with respect to x gives: v2

-

2

=

C

gx

(7)

where C is a constant of integration. integration. This time, time, at the start st art of the motion, x = 0 and v = 1 ms- I • We note that v is now positive since it is travelling in the direction of Ox Substituting these values into equation (7) with g = 9.81 ms- 2 , we obtain: C=

5

The equation for v is thus: v=

- 2gx

1

(8)

The maximum maximum height is attai attained ned when the velo velocity city is instantaneously zero. This, from equation (8), occurs at the height x given by: x

5

5

= - =g

9 81

=

5 lOm

We are now ready to include air resistance in our modelling. First, however, try these exercises.

EXERCISES 6 2

1 A boy drops a stone down the th e bottomless pit . He hears a splash 5 s later. How far has the stone fallen before hitting the water? (You may assume that the sound arrives instantaneously.) 2 A ball is dropped from a tower 2 m high. impact, how high does it bounce?

f

its speed

is

halved on

123

 

GUIDE TO

MECHANICS

3 A stone is thrown vertically upwards from the ground with speed 2 ms- t . A second stone is thrown upwards from exactly the same spot with wit h exact exactly ly the same spee speed d but after a time of 3 s has elapsed. How long does it it take for the stones to colli collide? de? How far above the grou ground nd does the collision take place? 4 A gir girll lean leanss out of an upstairs win window dow and drops a bal balll to her br broth other er who is standin standing g on the ground a distance m verti vertically cally below her. t precisely the same moment as the girl releases the ball, the boy throws another anot her ball wit with h speed ms- t vertically up at her. When and where do they collide? 5 A trai train n travelling with a const constant ant spee speed d of 1 100 00 ms ms-- t is approaching a bridge which is 3 m high high.. Some naughty child children ren want to drop a ttomat omato o on to the cab. How far must the train be from the bridge when they release the tomato if they are to succeed?

6 3 MOTION WITH WITH RE RESISTANCE SISTANCE PROPORTIONAL TO SPEED

all times, it is important to distinguish between velocity and its magni tude, the speed. This is never more so than when relating force force to spe speed, ed, as is done in this section and in Section 9.3. In both of these sections, we are concerned with one-dimensional motion, so the use o f vectors is more confusin conf using g than helpf helpful. ul. How However, ever, tto o emphasise that we me mean an the absolute a bsolute value of velocity, the speed, we use the notation Ivl In Chapt Ch apter er 2, we we discu discusse ssed d the modell modelling ing of rresistance. esistance. We showed tha t, t

on dimensional dimensional grounds, we migh mightt expect resistance to mo motion tion throu through gh air, called drag, to be proportional to the square of the speed. Observations confirmed conf irmed thi thiss for spheres, aatt least over a certain siz sizee ran range, ge, and th e same observationss showed a linear relationsh observation relationship ip betwe between en resistance aand nd spee speed d for small masses. Therefore, although it is possible mathematically to propose other laws, for example, resistance proportional to the inverse of speed or to displacement, such law lawss have no experimental e xperimental back-up. We thus restrict attention to res resist istanc ancee proportional to speed and resistan resistance ce propo proportional rtional to the square of speed. In this section, we discuss resistance proportional to speed only. Resistance, as the name implies, always opposes motion. Thus, the upwardss and do upward downward wnwardss mot motions ions are best tre treate ated d separate separately. ly. This is not essential here, as m k v changes sign as v changes sign. However, t is essential for more complex resistance laws.) These motions are shown in Figure 6.2. The equations of motion are:

124

 

M O T IO N UNDE R GR

a)

VI T Y IN ONE D IM E N SIO N

b)

Fig 6 2

(going up)

(going down)

(9)

(10)

and in each case R = mklvl where m is the mass and k is a constant dependent depen dent onl only y on the size size and shape o off the particle . The fo foll llow owin ing g examples will help to establish these ideas.

Examples 6 3 1

n apple is thrown vertically upwards with a velocity v f resistance proportional to speed, calculate the maximum height reached.

is

Solution Once again, we use v dv / dx for acceleration. Equation (9) applies, as does Figure 6.2; hence:

-mklvl -

m

= m v dv

dx

As the apple is going up, v is in the same direction as x and so is positive, thatt is, vl = v k is the drag constant, tha c onstant, whi which ch wa wass explained in Chapt Chapter er 2 Cancelling m and rearranging gives: v dv kv

=dx

+g

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GUIDE TO MECHANICS

Some manipulation of the left-hand side leads to:

[

~

+ g)

k kv

]

v = dx

Integrating this gives:

v

k

g

+-

In kv

J

+ g)

= x

.

+A

11)

where A is a constant of integration, evaluated by setting v = x = O This gives: A = - -

a

k

g

+ -

k2

a

when

+ g)

In kva

Hence: x=

a

- V k

+g a + g

g

kv

J

k

-In

)

12)

As in the example with zero resistance, the maximum height is reached when V is instantaneously instantaneously ze zero. ro. Thus, the maximum height, x m , is given by:

m

a

=

kVa)

g

k - k

In 1

13)

+g

Typically, the constant k lies between 10 10-- 3 and 10 10-- 6 see Chapter 2), so some o f the terms in equation 13) are eithe eitherr very large for example, g / F o r very small for example, kvo/g). Much care is thus required in carrying out the calculations, otherwise there will be considerable round-off error. You may recall that: for xl is

In 1

x).

f

0, whereas if initially the speed of the particle is less than the terminal speed (the case considered above), R - mg < O In each case, the sign of a, the accelera tion, ensures that the speed approaches the terminal speed for large t Figure 6.3 summarises these findings. Note that the arguments leading to the results in Figure 6.3 are independent of the form of R . This completes what we want to say about one-dimensional motion under gravity. Here is an extended example which may be considered a case study.

Example 6.5

A helicopter is stationary at a height of 1000 m above the ground when a parachutist parachu tist jumps ou out. t. When the para parachute chute is not open, air resistance can be assumed to be negligible. After the parachute is open, resistance is assumed to be proportional to the square o f speed through the formula 2 is m v R = /100, where m the mass of the parachutist and her parachute.

(a)

f

all is well, the parachutist will count 5 s, then pull the rip-cord. 135

 

GUIDE TO MECHANICS MECHANICS

v

= speed

V

speed > terminal speed

Vt

~ ; - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - ~ .

speed

«

terminal speed

Fig. 6.3 Assuming that the parachute opens instantaneously, calculate the parachutist parachu tist s velocity as a function o f time and draw a graph of v against t Calculate the terminal velocity velocity and mark this o n the graph. (Take g = 9.81 ms-2.) (b) All is not wel well, l, and the parachute only opens after a struggle at 13.5 s after the initial jump. How far from the ground is she before the parachute opens? What s her speed on reaching the ground? Will she survive? survi ve? (Assume (Ass ume that at no time must her spe ed exceed excee d 150 msms-1 , and that her speed on impact must be less than 60 ms-l.) (c) Compare the total travel times for (a) and (b). I t is instructive t o solve this problem with the axis pointing towards Earth All velocities are then positive. As origin, the helicopter s

Solution

the obvious choice. Figure 6.4 displays the set-up. The motion is in two parts: without resistance and with resistance. No resistance

Newton s second law gives: d 2x

mg = m -

2

dv

= m -

dt

=

dv v

dx

T o calculate v as a function of time, obviously we choose acceleration in

the form

dvldt:

136

 

MOTION

UN

ER

GRAVITY IN

ONE

DIMENSION

x

Fig 6 4 g =

and, since v =

°

v

dt

gt = v

hen t = 0,

we

have: v = gt

n

21)

part pa rt a), her h er fall fall w with ith zero resistance lasts for only 5 s Therefore, when

the parachute opens:

v = 9.81 x 5 = 49.05 ms-1 137

 

GUIDE T O MECHANICS

With resistance

Newton s second law gives: mg - R

=

m dv

or

dt

again using acceleration = dv/dt. The details o f integration of thi thiss equati on are not given here but the result is: t

B

=

2Vg

In

1.596

In

For part (a), when t equation (22) to find

=

=

11 10 arctanh vtv (100g) I V) = 1 - v/V( v/V(10 100g 0g)) g V(10 V(100g 0g))

1

1 0.0319v 1 - 0.0319v

I

=

3.193 arctanh 0.0319v)

5, v = 49.05. These values can be substituted in thus:

B

5

=

1.596 In (1.513)

=

2.415 2.585

=

Making v the subject of equation (22) with 0.198 v

(Note that

11

(22)

e

627t

0.00632 - 0.031ge-

- 0.0319vl

= 0.0319v

2.585 gives:

=

(23)

0627t

- 1 for this problem, since v tends to

110.0319 from above, so 1 - 0.0319v is always negative.) To complete the graph of v against t, we must know how long the parachutist remains aloft. Thus, we must solve for x against t. However,

this is impractical for this problem, and integrating equation (23) leads to an intrinsic equation for t which can only be solved numerically. A more rewarding approach is to solve the whole problem again for v against x, that is, using v dv/dx for acceleration.

This time we integrate:

No resistance

g so that

x

C

= v2/2

where C

dv

= v-

dx

=

0 since v

=

0 when x

=

O

When the

parachute opens, we have already calculated that v

=

49.05, so:

38

 

MOTION UNDER GRAVITY IN ONE DIMENSION

49.05

x 5) With resistance

2 x 9.81

With the addition of

= 122.625

R

on the left-hand side we get:

which whi ch integrates inte grates to: 50

In

I= g -  x 100

D

x

With x = 122.625 when v = 49.05, this gives D = -255.459. Making v the subject of the expression gives: v2 =

98

655

e

 0

  2r

24)

As a check, we see that, as x _ 0 0 v _ V981 = 31.321, the terminal velocity, and with x = 122.625, v = 49.05, regaining the initial condition for this part of the motion). t is clear that we can eliminate v between equations 23) and 24) to get x in terms of t, so all the information is contained in these equations. The problem is that, when x = 100 1000 0 m, v is so close to its terminal velocity that round-off error makes calculation impre cise. cise. However, Howev er, judicious u use se oflo gari thms gives gives t = 26.7 s as as the total time take n for for the drop. Using Using equation 23) for t ~ 5 and v = gt for 0 ~ t ~ 5 leads to the graph shown in Figure 6.5. v

5 parachute opens

Fig. 6.5 139

 

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the parachute does not open until this time; that is, v is given by: f

v 13.5)

= 9.81

X

t =

13.5 s, then

13.5

= 132.435

v = gt

applies until

This is the maximum velocity, since the parachute slows woman and parachute down. With resistance, the equation preceding (22) is valid, but the boundary bounda ry condition is no longer v = 49.05 when t = 5, but v = 132.435 when t = 13.5. This gives: - 12.73 = 1.596 In 1

t

Again, v

 

=

gx

is

also valid until v

x 13.5)

=

The equation preceding (24) v = 132.435, which gives:

is

v = 981

=

X

0.0319v21

1-

0 0319v

9.81

= 893.94

valid, vali d, but with conditions x 9.63

(25)

132.435, so:

132.435 2

1

X 1011

e-

 

=

893.94 when

2x

When x = 1000 this leads to v = 54.46. Since the speed never exceeds 150 ms-' , and the speed on impact is less than 60 ms l , we deduce that the parachutist survives. For (a), the total time is 26.7 s. For (b), the total time is 14.8 s v = 54.46 in equation (25)). Under the conditions prevailing in (b), the parachutist gets there faster than in (a), but they are far more risky

EXERCISES 6 5

1 Galileo II stands at the top of a 100 m tower. He not notices ices that an obje object ct dropped from the top has reached a constant speed by the time it has travelled half-way to the ground. Galileo II times the last half of the drop at 3 s. Calculate the resistance law if: (a) resistance

is

proportional to speed;

(b) resistance

is

proportional to the square of speed.

140

 

MOTION

2 Find th thee terminal velocitie velocitiess tance laws: a) rnkv 3 ; b) rna In vivo); c) rnk a constant).

UN

of

ER

GRAVITY IN

ONE

DIMENSION

falling bodies under the following resis-

4

 

7 PROJECTILES

7.1 PROJECTILES MOTION IN THE REAL WORLD Projectile motion is the free motion under gravity of a body projected in some direction which is not vertical vertical.. Examples of projectile moti motion on are the t he fl flig ight ht of a golf ball a high high jump jumper er or the motion o f a jet of water from a hose pipe. The analysis of projectile motion is one o f the most common uses of kinematics in the real world. The number of occasions in anyone day that you w wiill encounter examp examples les of projectile motion are many. f you are involved in field events in athletics o r ball games then the motion involved is lik likel ely y to be tha thatt o f a projectile. Sports scientists are particularly interested in the use of projectile motion when attempting to analyse and improve the performances of competing sports men and women. n fact a major proportion of the problems in this chapter are based on sporting applications. I f at some later stage you perform an in-depth ana analys lysis is of a particular event such as tennis shot-put football o r long jump ju mp you wi will ll find that the material of this chapter will be of considerable use. The theory of projectile motion is not only only impor important tant in sport but also anywhere that the free motion of a body is involved. n film filmss stu stunt nt men and women are required to analyse very accurately their expected motion when they perform daring leaps. Origin Originally ally the theory became impo important rtant when the use of cannons in battles made it necessary to understand the fl flig ight ht of cannon balls and many ancient technical manuals for gunners existed exist ed on the subjec subject. t. Perhaps it was then tthat hat a gre greater ater scie scienti ntific fic under standing of this area was considered to be necessary.

7.2 INDEPENDENCE OF HORIZONTAL AND VERTICAL MOTIONS The use of of v veloci elocity ty and accele acceleration ration in vector form

is

an important import ant factor in

the analysis of projectile motion.

he motion of any body which travels

42

 

PROJECTILES

- - - - - - - - - - - - - ~ - - - - - - - - - - - -  

~

Fig. 7.1

freely, so that it is subject only to the vertical acceleration due to gravity, can be analysed simply in terms of its horizontal and vertical motion. The law of vector addition, which allows us to resolve components, is a key method in the analysis of projectile motion. The independence of the two perpendicular motions is perhaps not obvious. We shall now describe an experiment which illustrates practically the independence of the horizontal and vertical motions of a projectile. Collect six identical small, heavy objects: coins are ideal. Fix a one metre

rule a t one end to a horizontal table so that it can pivot freely about that

end. Rest the six coins at 5 cm intervals, touching the rule s edge and at equal distances from the edge of the table, as shown in Figure 7.1. Draw the rule back and project it horizontally along the table, so that it rotates about the pivot, coming into contact with all the coins simultaneously. Their positions when they strike the horizontal floor are shown in Figure 7.1. Note that: (a) all six coins strike the floor at the same time; (b) their the ir positions are in a straight line whic which h passes through thr ough the th e axis axis at the pivot. What do these results tells us? The velocities rule supplies horizontal eachcoin of the coins. These are different proportional to the velocities distance oftoeach from the

pivot. They Th ey all start st art with zero vertical velocity velocity and,

as

all six coins strike the 43

 

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E TO

MECHANICS

floor at the same time, we can conclude that their vertical motions must be identical. The time of each coin s horizontal motion is the same so we can assume that the horizontal motion of each is uniform, as the horizontal distance travelled is directly proportional to its velocity at the rule. Clearly, in this case then, we can conclude that the different horizontal motions are independent of their equal vertical motions.

7 3 VELOCITY

S A VECTOR

The properties of the motion described in Section Section 7. 7.2 2 conf confirm irm that a vector approach can be used in the analysis analysis of projectil projectilee motion. We have alr already eady considered examples of velocity as a vector vector - see ffor or exam example ple the rel relati ative ve velocity problems of Chapter 1 Here we will need to use vector concepts of displacement, velocity and acceleration. As with any vector quantity in . independently two dimensions, theperpendicular motions so that they canThis be analysed in we any can tworesolve mutually directions. indepen dence was illustrated in the coin experiment and will be used in the examples that follow.

7 4 ASSUMPTIO ASSUMPTIONS NS FOR MODELLING MODELLING PROJECTILE MOTION you have have atte attempted mpted the experiment in Sec Section tion 7 7.2, .2, th then en it is unlike unlikely ly that your results will will have been entirely conclusi conclusive. ve. Such experiments experime nts highlight the shortcomings of idealised mathematical models. The term ideal condi tions wi will ll often ofte n be used. These conditions embody tthe he foll followi owing ng assump tions: f

(a) The particle model: Projectiles are particles and, as a result, are subject to no resistance forces which depend upon their size. All resulting motion will be translational. For real problems, modelling a body as a particle is a major assump tion. A long-jumper s ju jump mp may be considered as the path pat h of a projectile, but are we justified in using the particle model in such a case? Any body of finite size will rotate, and that rotational motion may appear to have an effect on the projec projectile tile path. In addition, air resistance will always act as the jumper has size, .and those forces of resistance usually depend upon the velocity and size of the body. effects of resistance beity large, but will be minimised if the timeThe of fli flight ght is kept short, thecan veloc velocity is not large and the body s

dimensions are small. Analysing these assumptions carefully can 144

 

PROJE

TILES

supply important feedback to the projectile problem. (b) The acceleration due to gravity is a constant. This is a reasonable assumption. Clearly, launching satellites from the t he Ea Eart rth h s surface cannot be modelled as a projectile. (c (c)) The mo motio tion n wi will ll be confined to two dimensions. This is not always the case in real examples o f projectile motion, as any golfer or footballer will know. There can be considerable sideways movement (swerve) created by what is known as the Magnus effect (the description of which is outside the scope of this book, but see for example texts on fluid mechanics, or S. Townend, The Mathematics o Sport Ellis Horwood, 1987). (d) The space iin n wh which ich the projec projectile tile travels is a vacuum. The Th e inclusi inclusion on of this assumption removes many of the problems of resistance already mentioned. The effects of wind and air currents would not be experi enced in a vacuum, which is a prospect that t hat would be relished by many sports men and women. In real problems, these assumptions may not be explicitly stated. t is always advisable to list any assumptions that you feel you are making in modelling a real problem and consider the significance of each in your final solution. In this way, you will become aware of which assumptions are justifiable and which are not.

Example 7.4

A woman is standing on a horizontally moving airport walkway which moves at a uniform speed of 2 ms- . She notices that the walkway passes unde r a ffix under ixed ed bar barrier rier some distance ahead of her and decides she will jump the barrier when at some strategic position. She remembers that she can jump, from a standing position, a vertical height of only 1.25 m, and realises realis es th that at this is exa exactly ctly the height of the barrier. f she jumps vertically in order to just clear the barrier, barrier , by co consid nsiderin ering g her motion to be ideal as described, determine: (a) the vertical velocity with whic which h she leaves the groun ground; d; (b) the time she will be airborne; (c) the distance from her take-off position, on the walkway, that she touches down; (d) the distance her touch-down position has moved horizontally during the motion.

(We will assume that the acceleration due to gravity is 10 ms-2. 145

 

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Solution

MECHANICS

r

2

Fig 7 2

Figure 7.2 defines the relevant quantities.

a) Consider the vertical motion: t her greatest height, the woman will have zero vertical velocity; thus:

o=

v 

-

2

x

10

x

1.25

where v is her initial vertical velocity. t follows that v = 5 ms-1 • b) Consider vertical motion: The woman will be airborne for t seconds until she is again at zero vertical height; thus:

o=

5t

- 0.5 x 10 x

t 

The two solutions here give t = 0 at take-off and t = 1 at landing. The time airborne is thus 1 s c) Consider h horizontal orizontal mo motion: tion: Bo Both th the woman and the walkway have the same horizontal velocity; she will, as a result, touch down at the same place on the belt that she takes off. d) Consider horizontal motion motion:: Hor Horizontal izontal distance = horizontal velocity x time airborne. T The he distance mo moved ved horizontally horizontally is thus 2 x 1 = 2 m.

EXERCISES 7 4

In all cases, take g

= 10

ms-1   unless otherwise stated.)

1 A car drives off a horizontal pie pierr with a spe speed ed of 45 kmh- 1 and crash crashes es into the sea 2 s after leav leaving ing the pier. Find a) the height o f the pier and b) the horizontal distance the car travels before entering the sea. 2 A high jum jumpe perr makes hi hiss jump by firs firstt running directly towards the th e

bar

with wit h speed

and then, without stopping, jumps verti vertically cally with speed

U

146

 

PROJECTILES

Find a) the greatest height he can jump and b) the di distanc stancee he needs to be from the bar at take-off. 3 A coin

is

dropped from a height of 3 m in a railway carriage travelling

with a uniform speed of

50 ms

1

•

a) Why does the coin appear appea r to fall fall vertically vertically in the carriage? b) What is the time taken for the coin to strike the floor?

7 5 MAGNITU MAGNITUDE DE AND DI DIREC RECTl[ Tl[ON ON OF THE VELOCITY OF A PROJECTILE AT A GIVEN INSTANCE The facility to resolve vectors in any two mutually perpendicular directions is common to all vector quantities. t an any y instant, in projectile motion, the velocity of the projectile is made up of two parts: its constant horizontal part u and its variable vertical part v, which usually needs to be deter mined. Completing the vector triangle for these two components, as illustrated in Figure 7.3, gives the magnitude of the velocity as u2 v2   in a direction tan- 1 v/u) with the horizontal.

L

u

~

Fig. 7.3

L

also possible to apply the process in reverse, so that, at any instant when the projectil projectilee has a velocit velocity y of magnitude V travelling in a direction 0 above the horizontal, the velocity can be resolved into horizontal and vertical components, as shown in Figure 7.4. Here, the horizontal compo nent is V co coss 0 and the vertical component is V sin 0, which we will write as t is

the vector:

V cos

V sin

Oi

O

147

 

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MECHANICS

si n

cos

e

e

Fig. 7.4

Hence, if we ret return urn to example 7.4, then the n the woman s velo velocity city at take-off is 2i Sj, or a velocity of magnitude \ 29 ms ms--1 in a direction tan- 1 (5/2) above the horizontal.

EXERCISES 7 5

1 Find the magnitude and direction of the velocity velocity with which which the car in exercise 7.4.1 strikes the sea. 2 Find the magn magnitude itude and direction of the velocity velocity of the high jump jumper er in exercise 7.4.2 at take-off. 3 Find the magnitude and direction of the actual v velocit elocity y of the coin when it strikes the carriage floor in exercise 8.4.3.

7 6 DISCUSSI DISCUSSING NG MOT MOTION ION RELATIVE TO HORIZONTAL AND VERTICAL DIRECTIONS

The motion of a projectile is often analysed in terms of (a) the greatest height it reaches above the horizontal plane and (b) the range it achieves on the horizontal plane through the point of projection. Let us determine these by considering the motion of a particle projected with velocity V at an angle e above the horizontal, as illustrated in Figure 7.5 Consider the vertical motion, with a view to determining the greatest height reached. At time = 0, the vertical velocity is V sin e The upwards acceleration is - g, whe where re g is the acceleration due to gravity. A t the greatest height h the vertical velocity is O Using constant acceleration

formulae, this gives: 148

 

PROJECTILES

height

range

Fig. 7.5

o= h

V sin 9 2 -   gh

V   sin2 9 g

Interpreting this function, we see that the vertical height attained by the particle can be increased by increa increasing sing eithe eitherr (a) tthe he spee speed d of projecti projection on o r (b) the angle of projection in the range 0° to 90°. or a given speed of project proj ection, ion, the height increases with until it achieves its maximum value when 9 = 90°. Consider vertical motion once more, this time with a view to determin ing in g the particle s range. Whe When n the particle returns to th thee horizontal plane, then the vertical distance above the plane is O This gives:

o= V

sin

- . gt   2

9t

and the time taken to achieve the range is: V t

sin g

Consider horizontal motion: The horizontal velocity is constant and its magnitude is V cos 9. The range is thus given by:

V cos 9 x

V

sin 9

g

This equation equat ion for the range is usual usually ly written in one o f the th e following forms: V 

sin 9 cos 9

or

sin 29

g

g 149

 

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TO MECHANICS

first glance, it seems a more difficult problem to determine the maximum range than the maximum height. Clearly, as the speed, V is increased, the range will increase accordingly. To study the range for a given, fixed, speed o f projection, let us consider the possible trajectories. t

Figure 7.6 shows the paths, for this speed projection from 0° to 90°.

of

projection, for angles of

Fig. 7.6 The trajectories show that there is a maximum range that appears to be at an angle o f projection of 45°. t is also apparent that all ranges less than the maximum range can be reached using two possible angles of projec tion. Examination of the fun functio ction: n: range

=

V 2 sin 28 g

reveals that tha t its maximum value occurs when sin 28 = 1 i.e. when 8 = 45° and is V2/g For any range less than this maximum value, we will be required to solve an equation in terms of sin 28, to find the necessary angle o f projection. So Solv lvin ing g the equati equation on sin Z8 equal to a positi positive ve value lless ess than unity gives two s o l u t i o ~ for 28 in the range 0° to 180°, which in turn gives two values of 8, the angle of projection, in the range of 0° t o 90°. The following example illustrates these ideas.

Example 7 6

A fielder wishes to return the ball to a catcher. The greatest distance from the catcher to any point on the boundary o f the play area is 80 m. f the fielder can throw the ball with a maximum speed of 25 ms-1   can he reach the catcher from any point in the field?

a) Throwing the ball with its maximum speed, what are the possible 150

 

PROJECTILES

angles of projection for the ball to reach the angles th e catcher without bouncing, if the distance from the fielder is 40 m? (b) In order to return the ball as quickly as possible, which angle of projection should be chosen? (Take the acceleration due to gravity to be 10 m s- 2 . olution

The maximum range will be: 25  

-

10

= 62

5m

This means that the fielder can return the ball anywhere within a circle of radiuss 62.5 radiu 62.5 m of the catcher. Hence, the catcher cannot canno t be reached from all points in the field. (a) For a range of 40 m, w e have: 25 2 sin 2e ----=40 10

to give sin 2e = 0.64. Solving this equation gives two values for the angle of projection in the range 0° to 90°. They are 19.9° and 70.1°. (b) The horizontal velocity for an angle of projection of 19.9° is 25 cos 19.9° = 23.51 ms-t, and that for an angle of projection of 70.1° is similarly 8.509 m s- 1 • This give givess the th e times of o f the ball b all s travel as 40/23.51 = 1.701 s a n d 40/8.509 = 4.701 s. Clearly, the lesser angle of projec tion, 19.9°, gives the faster return time.

EXER ISES 7 6

1 What is the smallest smallest speed of o f projection required requir ed to obtain a horizontal range of 40 m and what will be the time of flight in that case? 2 When projec projected ted at an angle angle of elevation elevation of tan-1 (3/4), a projectile falls 40 m short of a target in a horizontal plane through the point of projection. When the elevation is 45° the projectile overshoots the target by 50 m . Sh Show ow that the target t arget is at a horizontal distance of 2200 m from the point p oint of projection and a nd fi find nd the correct elevations of projection

so that the projectile hits the target. 151

 

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3 I f a particle is is proje cted cte d from the horizontal floor o f a large room whose ceiling is at a height of 5 m what wi will ll its its greatest greate st possible ran ge be if its speed of projection s a) 20 ms- 1 and b) 12 ms-1 ?

TRAJECTOf{ CTOf{Y Y 7.7 THE PATH OF A PROJECTILE THE TRAJE projectile the trajectory), it is convenient to refer t o To study the path of a projectile horizontal and vertical axes Ox and Oy through the point o f projection 0 as shown in Figure 7.7

y

~ ~ - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - ~ . x

Fig. 7.7 I f the velocity o f projection is V a t an angle a to Ox, as illustrated in Figure 7.7, then the velocity of projection can be written as the vector:

u = V cos a i

+

V sin a j

mo tion is unaffected unaffected by the accelerati acceleration on and the component The horizontal motion of velocity in that direction is constant. The upwards vertical motion is subject to an acceleration of - g . The acceleration vector is written in the

form:

a =

gj

given point r relative to 0 a t time t, on the trajector y, we c an write: write: For a given r =

u

+.

2

at 2

The vector diagram in Figure 7.8 shows the dependence o f the position

vector r on the vectors u and a. This gives, for a given point o n the path with coordinates x, y :

xi + y j = « V cos a ti + V sin a tj +

i

_gj t 2

152

 

PROJECTILES

o Fig 7 S

The set of equations: x =

V cos a t

and

=

y

V sin a t - ~ gt  

represent parametric ofthese the trajectory. t the Making the subject ofequations the first of formulae, we obtain: t=

x

V cos a

This enables us to derive: y =

as

gx  

x tan a - ~

sec2 a 2V

 

1)

the Cartesian equation for the trajectory. From this equation, we can

see that, for a given V and a the trajectory is a parabola. Having once determined this, we can use any of the properties of a parabo parabola la to disc discus usss projectile motion. The most important of these is the symmetry of the curve about a vertical line through its maximum value, its greatest height. Using this symmetry property, we can make the following observations about projectile motion: (a) The greatest height is the maximum value of y (b) The range is the value of x for which y = o (c (c)) The range is twice the x value to the greatest height. (The time to the range is twice the time to the greatest height, which follows directly.) (d) All heights are symmetrical about abo ut tthe he g greates reatestt hei height ght s horizont horizontal al position. (e) The direction of tthe he particle s motion is anti-s anti-symmet ymmetric ric about the

greatest height s horizontal positio position. n. 53

 

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MECHANICS

Example 7.7

A particle P is projec projected ted fro from m a point 0 at the to top p of a ccli liff ff,, 52 m vertically above the sea-level, and the particle moves under grav gravity ity until it str strikes ikes the sea at a point p oint S. Th Thee velo velocity city o f projection of P has horizontal and upwards vertical components of magnitude 24 ms-1 and 7 ms-1 respectively. Calcu late the magnitude and direction of the velocity o f P at the point of projection. Define horizontal and upwards vertical axes Ox and Oy through 0 and sho show w that the equa equation tion of the pa th can be written as as:: 7x

Y

=

24

5x 2 - 576

Find the horizontal distance of S from O.

Solution

At 0

the magnitude of the velocity

is:

The direction of the vel veloci ocity ty is tan- 1 7/24) above the horizontal. This gives the trajectory as: Y

=

7 x . 24 -

25 2.25)2 . 24 24 l x2

which reduces to:

required. Figure 7.9 sh show owss the traj trajectory ectory o off the particle particle,, aand nd tha thatt at S, y give:

as

5x 2

-

168x

29952

=

0

312)

=

0

=

- 5 2 to

Factorising gives: x - 96) 5x

This gives x

=

-62.4 or 96

As x

is

positive at S, then the horizontal

distance from 0 to S

is 96

m.

154

 

PROJECTILES

y

1

x

5

- 5 2 ~ - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - ~ - - - - - -

Fig. 7.9

7 8 DIRECTION OF TRAVEL AN AND D MAGNITUDE OF VELOCITY

The tangent of the ang angle le o f travel tan

e=

e at the point dY dt

I

x, y at time t is given by:

x ) = dy

dx

dt

This corresponds to a direction of travel being along the tangent at that point. From the Cartesian equation, we have: tan

dy

e=- = dx

sec2 a tan a - - '---:--V2 x

or, from the parametric equations: dy = dy

dx

dt

Ix dt

=

V sin a - g t V cos a

Example 7.8

Find the magnitude and direction of the velocity of a projectile, which is projected with speed 3 ms ms--   at an angle of 45° with the horizontal, when it has travel travelled led a horizont horizontal al d distance istance of a) 3 m and b) 5 m. Solution

The equation equat ion of the trajecto trajectory ry is is,, fr from om equatio equation n 1) 1):: x2

Y

x

9 155

 

GUIDE TO MECHANICS

Differentiating gives: y

dx

The horizontal velocity component 30

a)

f

sin

x 45

= 1

45°

is:

= 15\1 2 ms-

1

x = 30, then: y

dx

30

=1

=

45

1 3

=tane

The direction of travel is thus at an angle tanhorizontal. The magnitude of velocity is: 15\1 2

10\1 5

=

1

1/3) above the

ms 1

cos e since cos e = 3/\1 10). b)

f

x = 50, then:

1

y

=

=tane

dx

9

The direction of travel is thus tanThe magnitude o f velocity is:

1

1/9) below the horizontal.

5\1 164

3

7 9 TWO TRAJECTORIES For a given point (a, b on the trajectory of a projectile, whose speed of projection is V w e can use the Cartesian equation for the trajectory, equation equa tion 1), to gi give ve::

Using the trigonometric identity sec a = 1 2

to give:

tan a we can rearrange rearrang e this 2

156

 

PROJE

fILES

or:

This quadratic equation for tan a can be solved to give two solutions. As a result, there are two possible angles of projection for any reachable point a, b ; that is, we are able to determine two trajectories through such a point. This s consistent with the two angles o f projec projection tion for a giv given en range lesss than the maximum) on the horizont les horizontal. al. Figure 7.1 7.10 0 illus illustrates trates the two a)

b)

e)

Fig. 7.10 157

 

GUIDE TO MECHANICS

paths, in three different cases, which can be used to reach the points shown. They are: a) both angles angles abov abovee the horizontal for a point above the horizontal; b) both angles angles abo above ve the horizontal for a point below the horizontal; c) one angle above above the horizontal and one below, for a point be below low the horizontal.

Example 7 9

A body is projected from a point 0 with speed V 4ghI3) and passes through a point h, h18) where the axes are O x and Oy horizontal and vertical respectively. Find the possible angles o f projection. Solution

The equation o f the trajectory, equation equat ion 1), give gives: s: Y

using that sec2 a

=

1

=

x tan a -

3x 2 8h

1

tan ta n a 2

tan 2 a . A t the point h, hI8), this gives: 3 tan a - 8 t a n a 2

4- 0

Factorising gives: 3 t an a - 2 ) ta tan n a - 2)

and the solutions are tan a

=

23

=

0

and 2. Possible angles of projection are

then tan- 1 2/3) and tan-1 2.

EXERCISES 7 9

1 A particle is projec projected ted wi with th speed V 4ghI3) from a point 0 on a table of height h, standing on a horizontal flo floor or.. The particle reaches the floor at a point a horizontal distance 2h from O. Find the two possible angles o f projection. 2 Find the range of va values lues of the angle o f projection o f a ball which is projected projec ted wit with h speed V 2gh) in a plane perpendicular to a vertical wal walll

o f height h and a distance 2h away, so as to pass over it.

158

 

PROJECfILES

3 A particle is projec ted with with speed V ms-1 a t an angle o f elevation e from a point 0 on horizontal ground. The particle moves freely under un der gravity gravity and strikes the plane again at a point A When the particle is a t horizontal distances 30 m and 60 m from 0 its vertical heights are 8 m and 12 m respectively. Calculate: a) b) c) d) e)

the the the the the

value of tan e; value of V; time taken by the particle to reach A from ; distance OA; speed of the particle when it is 8 m above the ground.

Take the acceleration due to gravity to 4

be

10 ms-2.)

A particle is projected at an angle e above the horizontal. A t a sub sequent sequ ent time, its horizontal and vertical displacements displacements are equa equal, l, an d its o f motion is then inclined at 45° to the downwards vertical. direction Prove that tan e = 3.

from a po int 5 A particle is projected at an angle a above the horizontal from o n the edge of a table o f height h standing o n a horizontal floor. The particle reaches the floor at a point whose horizontal distance from the point of projection is 2h Show that, when it strikes the floor, the inclination e below the horizontal o f its direction of motion is given by: tan

e=

tan a

Find the speed o f projection and the time terms of g, a and h

1 of

flight

of

the particle in

7 10 ENVELOPE O OF F TRAJECTO TRAJECTORIES RIES a given given projection project ion spee d then, as the angles angles of projection are varied, a projectile can only pass through points in a finite region. Figure 7.11 shows 7 trajectories for the same speed o f projection in a plane. The angles of projection are 10°, 20°, , 160°, 170°. f we consider the envelope, that is, the equation o f the boundary of the region, of all trajectories, then we can see some of the properties it possesses. I t has a vertical axis of symmetry. Its greatest value is the greatest g reatest possible possible height that a projectile For

can reach tha t is, is, when the angle angle

of

projection is 90°). Its range

on

the

horizontal is the maximum range in each direction. The envelope is in fact a three-dimensional surface and the envelope 159

 

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E

TO MECHANICS

Fig. 7.11

shown in Figure 7.11 is a plane section through that surface. For a given point x, y) in a plane, the equation that gives the possible angles o f projection is:

Points x, y) cont contained ained within the envelope wil willl gi give ve two solutions for ta tan na from fro m th thee above equation. The T he condition for tthis his quadratic equation equa tion to have two solutions is that:

This gives the region:

The bounding equation equation of the envelope)

is:

This envelope corresponds to points with two equal solutions; that is, two equal angles of projection. Points on the envelope can be reached by just the one angle of projection.

EXERCISE 7.10 1 A footballer, when attempting a conversion kick, has to kick the

stationary ball ball from horizontal ground over a bar. The greatest horizontal distance a player can kick the ball before it first bounces is 55 m. 16

 

PROJECTILES

Determine whether or not the player could kick the ball over the bar, which is 3 m high, when the horizontal distance to the bar is 50 m.

Fig. 7.12

7 11 THE MOTION OF A PROJECTILE RELATIVE TO N INCLINED PLANE

Consider a particle projected project ed from a point on an inclin inclined ed plane up o r down it, where the motion is in the vert vertical ical plane containing the line of greatest slope. f we compare the trajectories illustrated in Figure 7.12, which are for the same projection pr ojection speed, speed , with those for trajectories over a horizontal plane, as illustrated in Figure 7.11, then we can note some significant differences; for example: (a) Symmetry is no longer apparent appar ent for any o f the angl angles es of projection of a particle over the inclined plane. (b) The greatest height no longer sseems eems important. The greatest perpendicular distance from the plane is perhaps more relevant. (c) (c) The motions up aand nd down the plane show major differ difference encess in their trajectories. Ranges up and down the plane are not equal. f course, we would would exp expect ect the range to be greatest when the project projection ion is down the plane plane.. With the velocity velocity component alon along g the plane no longer constant, it thought that the analysis of such a problem is different as well as cantly more difficult. However, inclined plane problems involving tile motion can still be tackled with respect to horizont l and

may be signifiprojecvertic l

directions. Including an inclined plan planee iin n such prob problems lems does nothing mo more re

than inter interrupt rupt the projectile s motion. The fol follow lowing ing example illus illustrates trates this. 6

 

GUIDE TO

Example

MECHANICS

7 11

A particle

is

projected with a velocity

ms-- at an angle of ms

of 4

1

3 °

to an

inclined plane in a plane containing the line o f greatest slope. The plane makes an angle of 3 ° with the horizontal. Calculate (a) the range of the particle parti cle along the plane and (b) the ang angle le tha thatt tthe he projectile s Mot Motion ion makes with the plane just before impact. Solution Removing the plane from our considerations would g ~ y e a projectile with a trajectory whose equation relative to the horizontal and vertical axes Ox and Oy is:

Y =

=

x tan

V

-

6 °

x

x

2

(2.40)2

sec2 60°

2

8

We can analyse the effect effect of the inclined plane in this proj projecti ectile le s motio motion n by considering only those points for which: y > x tan

x

3 ° =

V3

For the range, consider the point of intersection between the trajectory and the plane to be at a, b . Then: a

= aV

V3

a

2

8

The solution of this equation gives a = 160/V3. The range can then be found by simple trigonometry using the right-angled triangle in Figure 7.13: range

=

a cos 3

212

=3

m

For the direction of travel at the plane, consider: y

=

v 3 x-

dx

40

162

 

PROJECTILES

y

a

x

Fig. 7.13

x = a which gives y x = -1/v 3 and the direction of the velocity at impact as 3 ° below the horizon horizontal. tal. Again from the geome geometry try o f Figure

at

7.13 seen en plane.this is se

to

give motion at the plane in a direction

of 6 °

with the

7.12 MOTION ON A N INCLINED PLANE REFERRED TO AXES ALONG AND PERPENDICULAR TO THE PLANE In the firs firstt instance we wi will ll look at the motion when projection is up the plane in the plane containing containing the lin linee of greatest slope. The axes O X and OY up and perpendicular to the plane o f inclination n , are defined as shown in Figure 7.14.

y

Fig. 7.14 163

 

GUIDE TO MECHANICS

we define i and j up and along the plane as as shown, then the th e motion of a projectile relative to the axes O X and O Y is defined by: f

i

+

Yj = - g sin a i

Integrating with respect to

+

i

t

gives the velocity as:

+

V cos 8 - a i

}j =

+ g cos a j

sin 8 - a j -   g sin a i

+ g cos

aj t

where the projectile is at 0 at time t = 0 and has velocity Vat an angle 8 to the horizonta horizontal. l. A second integration with respect to t gives: Xi f X

+

Yj =

V cos 8 - a i

+

sin 8 - a j t - ~

is equal to the range up the inclined plane, t=

g sin a i Y =

+ g cos

aj t  

0, which gives:

2 V sin 8 - a

g cos a

2)

The range is given by: V 

sin 8 - a) cos g cos2 a

. 8 - a sm . a) - a cos a - sm

8)

which reduces to: 2 V   sin 8 - a

cos 8

2

g cos a This leads to a possible investigation problem using a microcomputer. Consider the ranges obtained for a given given speed of projec projection tion on an inclined plane of give given n inclination inclination.. From the tabula tabulated ted results, it should be possible possible to suggest a relationship between the angles of projection that have the same range. I f this is repe repeated ated for differen differentt values of angl anglee of plane, th then en a generalised result can be derived. I t is then necessary to attempt to verify the result algebraically using the relationship for the range. The results of such an investigation should also indicate a maximum range for some angle of projection both up and down the plane. f you have not performed this investigation, you should note Figure 7.12, which shows the envelope of trajectories.

The problem of ma maxim ximisi ising ng the range is agai again n impor important, tant, and as 8 va varies ries with V constant, the th e range var varies. ies. Increasing only the sp speed eed of projection, projection , 164

 

PROJECTILES

a)

08.8

'Ob

0.0 32.8

66.8

I

\ 102.8

138.8

I 174.8

210.8

- 188.9

-217.7

- 326.5

Fig. 7.15 V will always increase the range. The graph in Figure 7.15(a) shows the range as the angle o f projection 8 varies for a plane inclined at an angle of 30° to the horizontal. This also arises from the trajectories shown in Figure 7.15(b). To consider the changes in range for projection up the plane as 8 increases from a to 90° 90°, wit with h tthe he speed o f projection remainin remaining g constant,

we write the function for the range as: V

 

g cos a

(sin (28 - a - sin a

The maximum value occurs when sin (28 - a = 1 an and d the then n2 28 8 - a = 90°. This gives the maximum range up the plane as: V 

g cos a when 8 = 90°

+ a)/2.

1 - sin a

or

V 

g l

+

sin a

As with with the range o on n the horizontal plane, tthe he angle

of projection should bisect the angle between the horizontal and the

perpendicular to the plane. This is illustrated in Figure 7.16 where the given value of 8 is shown. 165

 

GUIDE TO MECHANICS

v

Fig. 7.16 A t the greatest perpendicular distance distance from the inclined plane:

t

V sin 8 - a

g cos a

which is precisely half of the value of t giv given en by equatio equ atio n 2), the time taken for the whole trajectory. The greatest perpendicular distance from the plane is given as:

V   sin 2 8 - a g cos a The greatest perpendicular distance distance and range down an inclined plane can be derived as in the case for projection up the plane, an exercise you might like t o attempt yourself. Here, we shall carry out the derivation using the results obtained for expressions up the plane.

should be clear from Figure 7.17 that, if we exchange exchan ge a for expressions already derived, then the results should follow. t

v

a

in the

Fig. 7.17 66

 

PROJECTILES

The range down the inclined plane sin 8 cos 8 g cos2 -a)

is:

sin 8 cos 8 - a ) g cos2 a

-a))

The maximum range again occurs when: 8

and

is

=

90°

-a)

2

given by:

g1

sin -a))

g 1 - sin a )

Again, the angle of projection for the maximum range bisects the angle between the horizontal and the normal to the plane, in that direction. The perpendicular distance from the plane, for a given angle of projec tion 8, follows as before and remains: V   sin2 8 2g cos a EXERCISES 7 12

1 A particle is projected down a plane inclined at 30° to the horizontal. The particle is pro projec jected ted from on the plane with veloc velocity ity 1 ms ms-- 1 at an angle 8 to the horizontal in the plane of greatest slope. f the range is 15 m, find the possible angles of projection. 2 A point 0 is at the foot of a plane which is inclined at an angle a to the horizontal. A particle is projected with speed V from 0 at an angle of elevation 8 to the horizontal, and moves in the vertical plane containing the line of greatest slope. t strikes the plane when travelling horizon tally. Express tan 8 in terms of tan a . Prove that the range in the inclined plane can be written as: sec a tan a g1

4 tan 2 a )

3 Identical particles are projecte proj ected d up and down a plane of inclination tan-1 1/3) to the horizontal, the speed of projection being the same in each

case. f the range up the plane is one-third that tha t down the plane, fi find nd the angle of projection, which is the same for each case. 167

 

GUIDE TO MECHANICS

7.13 THE DIRECTION O F TRAVEL AT THE POINT O F IMPACT OF A PROJECTILE WITH AN INCLINED PLANE In the case of motion relative relative to a horizontal plane, it proved an interesting exercise to calculate the direction direct ion o f travel for varying t or x This helped in viewing the symmetry properties of the projectile s flight. Here we have already identified the lack of symmetry, so a similar analysis would not have the same value. However, the direction o f velocity at the point o f impact with the plane is important. t should be evident that any sub sequent motion depends upon the direction of travel of the projectile immediately immedia tely before impact with the plane. Projection o f particles down planes causes no problem, as the particle after impact will always continue to travel down the plane. f we project the particle up the plane then: (a) if the direction is below the normal to the plane before impact, the particle will continue to travel up the plane after impact; (b) if the direction is above the normal before impact, the particle will travel down the plane after impact (Figure 7.18). a)

Fig. 7.18

Example 7 13

Consider a particle projected directly up a plane of inclination 30° to the horizontal. f 8 is the angle of projectio projection n measured from the pl ane, find find for which values of 8 the particle will continue to travel up the plane after impact. Solution

The time to reach the plane is given by:

o=

V sin 8)t -

( ~ g cos

3 °

t2

168

 

PROJECTILES

Fig 7 19

which gives: =

V

sin 8

gV3

The velocity at impact with the plane will then be: V

V3 V3 cos 8 - 2 sin 8)i - V sin 8j

should be noted that the magnitude of the velo velocit city y perpendicular to the plane, at impact, is equal to the magnitude of the velocity of projection perpendicular to the plane. This may seem a somewhat unusual result in the light of the lack of symmetry symmetry.. In fa fact, ct, this can be shown to to be true for all inclined planes.) f we write the velocity along the plane as: t

then for the motion down the plane after impact we have:

EXERCISES 7 13

1 A particle

is

projec pro jected ted wit with h speed

so tha thatt it strikes at right angl angles es a

plane through the point of projection inclined at Show that the range on the plane is 4 7g.

3 °

to the horizontal. 169

 

GUIDE TO MECHANICS

2 A shot is fired from a gun in a horizontal direction with a velocity o f 300 ms-l. The gun is on the side of a hill on inclination tan-1 4/5) to the horizontal. Find how far along the hill the shot will strike, and deter mine the magnitude and direction of its velo velocity city o n impact.

7 14 REAL PROBL PROBLEMS EMS WITH PROJECTILES AND INCLINED PLANES

Examples 7 14 o n a cliff 1. A shell is fired with speed V from a point cliff of heig height ht h above the sea. Find the greatest horizontal distance the shell can cover before landing in the sea.

Solution Method 1: Using range on an inclined plane. Let the point where the shell lands in the sea, which corresponds to the maximum distance, be S Then the line O S can be considered to be an inclined plane, as indicated in Figure 7.20, the distance O S being the range in that inclined plane.

5

Fig. 7.20 Using an earlier result, the distance O S has a maximum value:

g l - sin a

where sin a

= h/OS.

t

follows that:

sin a

h

=

170

 

PROJECfILES

and the corresponding value of:

=

tan a

Y V4

2V2hg

hg

The resulting triangle in Figure 7.20 gives gives h tan a as the corresponding maximum horizontal distance covered by the shell. This simplifies to give:

:' Y V2

2hg

g

Method 2: The same result can be obtained using calculus methods and considering motion relative to tthe he horizontal horizont al and vertical ax axes. es. Firs First, t, we define axes O x and Oy horizontally and vertically through O. The equation o f any trajectory when the angle of projection is e is: Y

= x tan e -

gx 2 2V2

sec2 e

Defining R as the distance distance horizontally horizontally out to sea, we have that th thee point R, - h is a point on the trajectory, so that: -h

=

R

R tan e -   g s e c2 e 2V2

Differentiating with respect to e gives: _ dR

o-

de tan e

gR

2

R sec e -

sec2

e dR de -

V2

gR

2

V 2 sec e tan e

Maximum o r minimum values o f R will occur when dR de = O This gives two solutions: R = 0 corresponds to the minimum value and: =1

gR

tan e

giving the maximum value for R by substituting in the original equation as:

v

- Y V2

g

2gh

which

is

the same result as before. 7

 

GUIDE TO MECHANICS

2

A rotary law lawn n sprinkler deli delivers vers water wit with h consta constant nt speed V and at all angles. I t is placed at a point 0 in a plane lawn, which is inclined at a to the horizontal. Defining axes Ox and Oy in the plane so that O x is up the plane, find the equation of the edge of the area that is watered. Solution For an inclin inclined ed plane, incli inclined ned at aan n angle the maximum range R up the plane is given by: R

~

to the horizonta horizontal, l,

V2

g l

+

sin

~ )

Figure 7.21 a)). f we consider a direction at angle 8 to the line o f greatest slope up the plane, then from Figure 7.21 b), we have that: s i n ~ = s i n a c o s 8

For a given point x, y on the edge o f the watered area, we have:

x = R cos 8

and

y = R sin 8

and

y - g l

or: x

V 2 cos 8

g l

+

sin a cos 8)

 

V 2 sin 8 + sin a cos 8)

a)

x

I

I I

- y

-

r:lI

Fig. 7.21 172

 

PROJECTILES

These equations can be used as parametric equations for the watered regions, o r the former equation for R allows us to study the region in polar pol ar form. The areas involved can be plotted simply using a micro. The results for a = 30°, 60° and 90° are shown in in Figure Figure 7.22 7.22 - they should should be compared with the expected result for the area o n a horizontal plane. Note that any increase in V resu results lts in an enl argement o f the figures. a)

x

;

I

b)

xl

c)

I

I y

x

y

Fig. 7 22 The regions are elliptical and the position of the sprinkler is at the focus of the ellipse. ellipse. We can see that the polar equ equati ation on is in fact fact that for

an ellipse with eccentricity sin a

MISCELL

NEOUS EXERCISE S

1 A plane is is inclined at an angle of tan-1 3/4) to the horizontal. Unit vectors i, j and k are taken horizontal to the plane, up the line o f greatest slope and perpendicular to the plane outwards respectively. Express g in terms o f i, j and k. A projectile is given a velocity of 20i + 30j + 60k ms- 1 from a point po int in the plane. Find its time of flight and the vector position of the point where it hits the plane again. 2 A small coin is proje cted along a smooth smoo th surface of a wedge with a velocity of 4 ms- 1 so that its direction is at 60° to the line of greatest slope. I f the wedge surface makes an angle o f 30° with the horizontal, find:

a) the greatest distance the coin travels up the wedge; 73

 

GUIDE TO MECH

NICS

b) the distance from the point of proj ection that the th e coin wil willl next b e at the same vertical height. 3 A tennis player volle volleys ys a ball ball from a point poi nt 0 a distance distan ce of 0.4 m below the level of the top t op of the net and a horizontal distance of 4 m from it. f the ball just clears the net, of height 1 m, when projected with speed 5 ms- I   find the possible angles of projection. f we define axes Ox and Oy horizontally and vertically upwards, find: a) the equation of the path for the shortest time to the net; b) the distance from the net that the ball hits the ground; c) the magnitude magnitude and direction of the velocity velocity with which which the ball hits the ground. 4 Mud is thrown off off the tyres of the wheels of a car ca r travelling travelling a t constant speed V. Show that mud which leaves leaves the ascending part of the tyre, at a f the wheel provided point above the wheel will be thrown clearis ogreater its height above the hubhub, when it leaves the tyre than g a / V 2, where a is the radius of the tyre. Also find the range o f values o f V for which mud will not travel above the level of the top of the wheel.

5 A ball P is projec projected ted from a point with velocity velocity U at an angle a to the horizontal; simultaneously, a ball Q is projected from a point B with velocity V at an angle ~ to the horizontal, where and B are a distance a apart on the same horizontal level. Find the condition for them to meet and the magnitude of their relative velocity when they do. 6 A diver at an indoor pool can leave leave the springboard of height 1 m above the water at any velocity V and angle a above the horizontal. Owing Owin g to the height of the ceiling, ceiling, she may not rise rise to more than tha n 12.5 12.5 m above the water, o r she would hit the roof. The diver turns a somersault in the air every 0.8 s, and must complete an odd number of half somersaults so that she hits the water head first. Find the maximum number of half-somersaults she can complete in the dive. I f in such a dive, she may not travel a horizontal distance of more than 5 m before she enters th e water, find find the maximum value value of her initial velocity and the corresponding correspondi ng value of her angle o f projection. Take g = 10 ms-2.) 7 Show that the maximum range of a projectile on the horizontal plane through the point o f projection is V 2 /g where V is the speed o f projection, and state the angle of projection. A long jumper at the instant of leaving the ground has a horizontal

speed V due to his run-up, together with a speed

at

an angle o f 8 to

174

 

PROJECTILES

the horizontal due to the jump. Show that the longest jump when e rtf3 and find the distance achieved in the jump.

s

achieved

8 A hose pipe s used to water a horizontal flow flower er bed. The wat water er leave leavess the pipe at a height of 1 m above the flower bed and at an angle of 5° with the horizontal. f the speed that water leaves the hose can be adjusted fi find nd the range of values values of the speed required in order tha thatt the whole bed s watered if the nearest and furthest points of the fl flow ower er bed are at horizontal distances of 1 m and 5 m respectively.

75

 

8 CIRCULAR MOTION

8 1 INTRODUCTION

The effects of circular motion are a common daily experience. The motion of record players, the act of cornering in a car or on a bicycle, and the motion of washing in a spin dryer are common examples, and there are many more. Circular motion is an example of two-dimensional motion. Previously, it has helped to analyse motion in two perpendicular directions, and a Cartesian coordinate system has proved ideal for this purpose. In Carte sians, the position of a body is represented parametrically, in terms of the time t, in the form x = x t), y = yet) - see for for example the case of projectile motion in Chapter 7 For the analysis of motion in a circle, it is more convenient to use polar coordinates r, 8). The advantages of using such a coordinate system will become obvious in the next section. 8 2 POLAR COORDINAT COORDINATES ES AND ANGULAR DISPLACEMENT

The particle P shown in Figure 8.1 is describin describing g a ci circle rcle with centre 0 in a plane. The radius of the circle is a metres. t a particular instant, the position of P is give given n in polar coordi coordinates nates by r = a, 8 = 0 radians), and at n / 3 radians). a, 8 some subsequent time, its position has moved to r The distance between the two positions is the most direct distance, which, using trigonometry, is a sin n/6). However, a more significant measure of the motion is the distance that P has travelled that is is,, along the arc of the circle between the two points). This is given by the expression a X n / 3 metres. More generally, if the angular coordinates of two points o n a circle, o f radius r metres, differ by an angle of 8 radians, then the distance along the

circular arc between them is r 8 metres. In Figure 8.2, a particle has travelled from a point A to point B,

on

a

176

 

CIRCU L A R MOTION

y

L

11O

a. 0

x

Fig. 8.1 circle of radius r The angle subtended at the centre of the circle by the arc B is e radians. The angle e radians is defined as the angular displacement experienced experien ced by the particle part icle in movin moving g from A t o B The distance s travelled along the arc B is the displacement of the particle and is given by the expression r o Note that equal angular displacements do not always result in equal displacements.

L ...L

---I A

Fig. 8.2

8.3 ANGULAR VELOCITY

AND

ANGULAR

ACCELERATION From Fro m Section 8.2 the equation equat ion for for a displacement

of

s metres

of

a particle 77

 

GUI

E TO MECHANICS

moving on a circle

is

s

re

where r metres is the constant radius of the circle. Differentiating both sides of this equation, with respect to the time t seconds, gives: ds dt

de

r

dt

The expression ds / dt is the linear veloci velocity, ty, v in metres per second, along the tangent tto o th thee circula circularr arc of radi radius us r, and d e d t (or e) is its equivalent angular velocity whose units are radians per second. The equation:

v

r8

connects linear and angular velocities. A second differentiation gives:

Here, dv / dt ms-z is the linear acceleration along the tangent to the arc o f the circle and has an equivalent angular acceleration of e radians per secondz (rad s-Z . 8 4 SOME OBSE OBSERVATI RVATIONS ONS OF CIRCULAR MOTION

A boy decides that he is goin going g to pr proje oject ct a small stone as far as possible. H e knows that, if he throws it from his hand, he can only achieve a short distance and he looks for an alternative method. H e has read in a book about the use of sling shots and he decides to improvise. H e attaches a string to the stone and begins twirling it in a horizontal circle, holding one end of the string while the stone is attached to the other. Here are a few questions we might ask ourselves: ( a) What Wha t forces act on the stone? (b) Why is it impo impossible ssible for the circle, described by the st stone one,, to hav havee as its centre the boy s hand? (c (c)) What happen happenss iiff the string breaks breaks?? is

of

The problem the boy has in projecting the stone similar to that hammer throwers in field athletics. f we perform an experiment o f this type, we should notice that the motion in a circle is as illustrated in Figure 178

 

CIRCUL

b)

 a)

_

K J

R

MOTION

o

 

w

c) I I

I

I I II I

I L

_

Fig. 8.3 8.3, where is a particular point on the circular path. Also notice, in the side elevation elevation shown in Figure Figure 8.3(b), that the boy s hand has not been be en shown at the centre o f the circle described by the stone. I n Figure 8.3, all the forces have been added to the diagram. The stone has no vertical motion, so upwards and downwards forces acting on the stone must balance. The weight of the stone has a fixed value W which acts vertically downwards. The weight must be balanced by an equal vertical force for ce upwards, which which can only be supplied by a component compone nt of the tension. t follows that the string cannot be horizontal if it is to have a non-zero vertical component. This answers (b). What happens when the string breaks? Perhaps you would expect the stone to t o travel outwards along along the direction of the string. Newton s first law tells us that, after the string is released, the motion of the stone will be unchanged in the horizontal direction, as the tension ceases to act and there will be no other horizontal force acting; hence, it will continue to travel along the tangent to the circle at release. Of course, the weight will continue to act vertically, so the stone travels as a projectile. The path o f the stone is in the tangent plane at the point of release, as illustrated in Figure 8.3(c). Returning to (a), it can be seen that a single force acts in the horizontal of

direction. This force is the component tension there. Newton s second law tells us that the force will produce an acceleration in the same direction. This suggests suggests that, that , in the case of non-uniform motion in a circle, circle, 79

 

GUIDE TO

MECHANICS

there will be an acceleration towards the centre of the circle. In fact the acceleration towards the centre of the circle is alw always ays pre presen sentt even when a body performs circular motion with constant angular velocity. 8 5 ACCELERATION TOWARDS THE CENTRE OF

A

CIRCLE OF MOTION

In Section Section 8.3 we sa saw w that tha t even when a body perf performs orms circular motion moti on with wit h constant ang angular ular speed it ex experience periencess an acceleration towards the centre of the cir circle cle.. Here Her e us usin ing g the definiti definition on of acceleration ffrom rom Cha Chapte pterr 1 we calculate the magnitude of the acceleration towards the centre of a circle.

r

p

rw

Fig. 8.4

In Figure 8.4 a body is movi moving ng iin n a ccirc ircle le of cen centre tre 0 and radius r metres is the with constant angular speed w radians per second. The point position of the body at some time t seconds and Q is its position at some subsequent time t, where t is small. The corresponding angular displacement in this time is 8 radians. The linear tangential speed v is given by rw ms- 1 and it follo follows ws th that at th thee change in speed along the radius towards 0 during the motion between positions and Q is: rw sin

8 t

t

- 0

In the th e limit as t tends towards zero the point Q move movess towards and as a result 8 also becomes very small. The limit will represent the acceler ation at P This gives: 180

 

CIRCUL

0) rw sin oe

I   m uo

Ot

=

rw

X

(I

Soe

m u

o

ut

0) sin

be

be

R

MOTION

x be Ot

e rwx

It

x-

dt

is usual to adopt one of the following two forms for this expression:

or

rw2

v2 r

the second being derived from the first using the relationship v

=

rw.

8 6 THE ANALYSIS OF PROBLEMS INVOLVING HORIZONTAL MOTION IN A CIRCLE has already been suggested that motion in a circle is common in the real world. Here, we analyse the circular motion in a number of real problems. It

Examples

86

1. A body of mass m lies on a rough horizontal turntable so that its distance from the turntable s centre is r metres. f the turntable com

pletesfrevolutions every minute, show that the coefficient offriction, f,t, between the body and the table must be ~ mrn?rI900 if the particle is to remain at rest relative to the turntable. I

c I

r

I I

4:

Fig 8 5

Figure 8.5, the forces of friction and normal reaction have been added to the diag diagram ram.. T There here is no motion in the vertical direction Solution

n

so that resolving in this direction gives: R

=

mg 181

 

GUIDE TO MECH

NICS

Applying Applyin g Newton's second la law w to the body in the direction of the centre of the circle gives: F

For no slipping,

we

have

F ~

=

60 2

mr 2nf R

and this gives the require required d resu result: lt:

Substituting values of 33 and 45 will give the coefficients of friction required for a typical record turntable. 2

A car of mass m kg travels on a horizontal carriage way so that, during cornering, its motion can be considered to be part of a circle circle of radius r metres. (a)

the coefficient of friction between the car's wheels and the carriage way is , find the greatest safe, steady cornering speed, v ms-I, of the car. (b) f the distance between betwee n the car's wheels is d metres and the height of its centre of gravity is h metres above the carriage way, find the maximum speed of cornering in order that the car does not overturn. f

Solution

(a)

Figure 8.6, the car is shown cornering with speed v where F and R represen representt the forc forces es of fri fricti ction on an and d normal reaction between tthe he carriage way and the car respectively. Again, resolving vertically we have: n

R

=

m

Applying Newton's second law horizontally towards the circle's centre gives: m v2 -

r

For no sliding,

F ~

R

and this leads to:

The fast fastest est cornering speed

is

then Y(Wg).

182

 

C I R C UL

R M O T IO N

a)

F r

mg

b)

e)

..... d

R

mg

o mg

Fig. 8.6

b)

the car overturns, it does sso o about it itss oute outerr wheels you m may ay need to think about this). As a result, when it is at the point of overturn ing, the normal reaction and friction forces will act on the outside wheelss only. Fro wheel From m a), we have that F = m v r and R = mg. The stability of the car is now controlled by a system of two couples. These couples have been illustrated separately separate ly in Figures 8.6 b) and c). The couple in b) h has as magnitude mg x 1I2)d and would cause an anti-clockw anti-c lockwise ise rotat rotation. ion. Th Thee couple in c) h has as magnitude mv r x h and would cause a clockwise rotation. For no overturning, an anti-clockwise rotation must be encouraged, which gives: f

 

 

Provided there is no slipping, this gives the maximum cornering speed as V rdg/2h). From this solution, it can be seen tha thatt th thee maximum speed wi will ll be increased if d increases and/or h decreases. These are both impor tantt considerations iin tan n high-speed high-speed ca carr d desi esign: gn: a) the whee wheell base

needs to be wide wide and b) the centr centree of grav gravity ity of the car must be low. Both are noticeable characteristics of formula one racing cars. 83

 

GUIDE TO MECH

NICS

EXERCISES 8 6

1 A car on a fairground rou roundab ndabout out has mass o f 1000 kg and is connected to the roundabout s central spindle by an arm of length 4 m. f the car describes a horizontal circle every 2 s, find the tension in the arm. 2 A string of length 2a is connected to a point on a horizontal table. Two particles of o f mass m are fastened to the string, one at it itss middle point and the other at its end. f the string remains in the th e same sstraight traight line, while rotating with angular speed w find the tension in each portion of the string. 3 A car of mass 1000 kg travel travelss on a rough circular track of radius 200 m. The track is banked at a constant angle of 30° to the horizontal, in order to reduce the possibility of the car skidding outwards. Find the value of the coe coeffic fficien ientt of fri frictio ction n between the car and the trac track k if slippi slipping ng occurs when the car is travelling at a speed of 80

k m h ~ l .

4 A mot motor or cy cycl clis istt corners sso o that his motion can be consider considered ed to be on a circle of radius r f the centre of gravity of the rider and his bike is at a height h when the bike is erect, find the angle that his bike makes with the vertical when cornering with speed v if he does not slip. f the motor cyclist rides on a wall of death, also of radius r with speed v, show that the angle made by his bike with the vertical is t a n ~ l rg/v2).

8 7 THE CONI CONICAL CAL PENDULUM

our earlier discussion of circular motion, we considered hammer throwing and the sling shot. t is now possible to consider a more detailed analys ana lysis is o off these problems. Thee system consi Th consists sts of a small body of mass m attached to a light inextensible string of length a metres. The other end of the string is attached to a fixed point A and the body describes a circle, about a vertical axis through A with constant angular speed w radians per second. This arrangement is that of a conical pendulum, which is illustrated in Figure 8.7, where the tension and weight acting on the body have also been indicated. The string generates the curved surface of a cone during its motion. Again, the fir first st stage iin n modelli modelling ng the p proble roblem m is to resolve vertically to n

give: mg =

cos

8

1)

184

 

C I R C UL

R MOTION

m

Fig. 8.7

where 8 is the angle between betwe en the string and the vertical. f Newton s second law is applied to the body, in the direction of the centre of the circle, this gives: ma

sin 8

OJ

2

T sin 8

which reduces to: (2)

Eliminating

T

between equations (1) and (2) gives: cos 8

.lL..

aOJ 2

Here, it can be seen that cos 8 can never be zero, although it becomes progressively smaller as OJ increases. As the angular speed, OJ, is increased, the cone generated becomes flatter, but never completely flat. This is a conical coni cal pendu lum. The development of this model has considerable potential. Here is a simple variation of the same problem.

Example

8 7

The ends of a ligh lightt inextensible string of length a are attached to two fixed points and B in the same vertical line, so that B is a distance a below A A small smooth smo oth ring of mas masss m is tied to the string at its centre C. f the ring moves in a fixed horizontal circle, with angular speed OJ, find the tension in the portion of the string Be and show that the motion is only possible if 4g/a. OJ ~

Solution n Figure 8.8, the arrangement has been illustrated and the weight mg and tensions TJ and T2   wh which ich ac actt on the body, are indicated. As 185

 

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A

c mg

B

Fig 8 8

CA then the angles made by each of the strings C and with the vertical are 60° Resolving vertically for the body gives: B = BC =

T J cos

=

60°

Tz cos

60°

BC

mg

Simplification of this expression leads to: TJ

-

 z=

2mg

(3)

Applying Newton s second law to the body, in the direction of the circle s centre, gives: T J sin 60°

T z sin 60°

=

2maw z sin 60°

Again, simplification gives: Tl

Tz

=

2maw z

(4)

Eliminating T J from equations (3) and (4) enables the tension in the string B C to be written as: Tz

=

1

2m awZ - 4g)

f this expression results in a negative value for the tension, then the string

would be slack and th would thee motion mot ion would not be possible. For the motion to be maintained, we must have that T z ~ 0, which gives W Z ~ 4g/a. 186

 

CIRCU L

EXER

R MOTION

ISES 8 7

1 A particle o f mass m is fastened by a string of length 2 to a point at a height I above a smooth table. f the particle describes a horizontal circle on the table, wit with h angular speed w find the tension in the string and the reaction between the table and the particle. Determine the greatest angular speed possible for the particle to remain in contact with the table. 2 The ends of a lig light ht string of length Sa are attached to fixed points L a n d M, in the same vertical line, so that L is 4a below M. A small bead is threaded on the string st ring and describes a circl circlee of radius 3a Determine the tension in the string and calculate the speed of the bead. 3 Two particles of masses m and 2m are connected by a light inextensible string, which is threaded through a fixed smooth ring. f the lighter particle describes a horizontal circle of radius a whi while le the othe otherr particl particlee remains stationary, find the lighter particle s speed. 4 A particle moving with a consta constant nt speed u inside a smooth spherical bowl of radius a describes a horizontal circle at a distance l/2)a below its centre. Find the value of u 5 A circular cone of semi-vertical angle a is fixed with its axis vertical and its vertex downwards. A particle of mass m is fastened to one eend nd of an inextensible string of length I the other end of which is fixed to the vertex of the cone, so that the particle can move on the smooth inner surface surf ace of the cone, with constant angular speed w Find the least value of w 2 in order that the string will remain in tension. 6 The mechanism shown in in Figure S. S.9 9 is designed to regulate the flow of steam from a boiler. t consi consists sts of four light rigi rigid d rrods ods eeach ach of length 2a togetherr wit togethe with h smal smalll spheres at and B of mass m. A D and B D are smoothly hinged at D. D . A sm smooth ooth ring of mass m is jointed to C and B C at C and can slide on the vertical spindle below D. A B and the spindle are all in the same vertical plane. When the valve is open, the ring rests on a horizontal ledge, fixed to the spindle at a distance 2a below D. The system rotates about the axis CD with angular speed w

(a) Find: (i) the tension o r thrust in the rods; (ii) (ii) the fforce orce exerte exerted d by the ledge on the rin ring. g. (b) Show tha thatt if if aw 2 2g then the ring rises, opening the valve. 87

 

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Fig 8 9

8 8 MODELLING PROB PROBLEl\l LEl\lS S OF MOTION IN A VERTICAL CIRCLE Motion in a vertical circle is very different from motion in a horizontal circle. As we have seen, in the free motion in a horizontal circle, circle, the speed s peed is constant although tthis his doe doess not al alwa ways ys have to be the case). In contrast, motion in a vertical circle has variable speed. To cause a body hanging in equilibrium by a string t o move in a vertical circle, we might start the motion by giving the body a horizontal velocity. The result will be that the body rises in a circular arc. t may complete a circle, or just swing to and fro, or it may fall from the path at some point. One noticeable factor is that the speed of the freely moving body changes as it moves along the circular path. The analysis of motion in a vertical circle is of two types: a) Moti Motion on insi inside de a circle, which occurs when a body is attached to the en d of a string or travelling on the inner surface of a cylinder. b) Motion outside a ci circl rcle, e, whi which ch oc occurs curs when a body sl slides ides on the ou outer ter surface of a cylinder. First, we consider motion inside a vertical circle. Figure 8.10 illustrates two practical examples of motion inside a vertical circle of radius a Figure 8.1O a) shows the motion of a body on the end of an inextensible string;

Figure 8.10 b) shows the motion of a small body sliding on the smooth inner surface of a cylindrical drum. The analysis of these two problems is identicall and iin identica n the diagrams we have labelled the tension in a) an and d the 88

 

CIRCUL

a)

R MOTION

b)

v

Fig 8 10

reaction in (b) as T This wil willl result in the same sa me governing equatio equ ations ns for the motion. The problem of the changing speed of the body must be considered first. f the body is projected horizontally with a speed U from its lowest position, then we need to derive an expression for the speed, V at a position where the angle made by the radius to the particle and the downwards vertical is 8. Using conservation of energy (see Chapter 4 gives:

This gives V 2 in terms

of

V2

8 as:

=

U2

+

2ga cos 8 - 1

(5)

provided Tis always positive (that is, the body remains on a circular arc). The body has an acceleration towards the centre of the circle and again Newton s second law must be considered to give: T - m g cos 8

Eliminating

VZ

=

mV2

a

between equations (5) and (6) gives the tension as:

(6)

2

T =

a

+

mg 3

(7)

cos 8 - 2)

189

 

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These expression expressionss for V 2 and T represent a model of the poss possible ible motion of the body while it is movin moving g in a vertical circ circle. le. Using these expressions, it is possible possib le to identify what ha happens ppens for various values o f U at various values of 8. Let us ask the question: question: Do Does es the body complete the circle? f the body can achieve the highest point of the motion, then, by conservation of energy, it should complete the circle. To do this, either the string must be taut, tau t, as in a), o orr the body must continue to press against the cylinder wall, as in b). This w wil illl happe happen n iiff the conditi condition on that T ~ 0 when 8 = Jt is satisfied.. Applying this condi satisfied condition tion to eq equat uatio ion n 7) giv gives: es: T

2

=

a

-

  mg ~

0

or

U2 ~

5ga

A second question is: What is the condition for the particle to describe only the lower half of the circle of the motion? Whenever a body has its motion confined to the lower half of a circle, then the need for the reaction or tension to be non-zero will always be satisfied. This time it is necessary to look at the equation for V 2 for a condition. For, provided V 2 ~ 0 when 8 = 1I2)Jt, then the particle can move on the lower half of the circle. Applying this condition to equation 5) gives: V2

=

U2

-

2ga ~

0

or

U2 ~

2ga

2ga the particle As a result, it would be expected that t hat,, in cases when U 2 will oscillate on the lower circular arc. U2 5ga? In these cases, Finally Fin ally:: What happen happenss if 2ga cases, the contact o r tension requirement will not be satisfied for some range of values of 8 in Jt

T the range 8 From the point when the particle will 1I2)Jt = a0,projectile begin to leave the vertical circl circle e and wi will ll travel freel freely y as projec tile,, until it meets the circle once more. Obviously, many paths are possible, one of which is illustrated in Figure 8.11.

reaction =

0

Fig. 8.11 190

 

CIRCUL

R

MOTION

Next, let us look at motion outside a vertical circle. circle. Exa Examples mples o f this type of motion can be analysed by cconsidering onsidering the simple proble problem m of a particle resting on the highest point of a fixe fixed d smooth sph sphere ere o f radius a In Figure 8.12, the position of the particle, whose mass is m is shown when the radius to its position makes an angle 8 with the upwards vertical. Its speed is then V. Forces of reaction, R, and the weight, mg, have also been indicated in the diagram. This motion can be started either by displacing the particle or by giving it a horizontal velocity. Certainly, we do not expect the particle to complete a circle. R

Fig 8 12

Adoptin g a simi Adopting similar lar approach as before, if U is the initial initial speed, and when the angle the particle makes with the vertical is 8 its speed is V the conservation of energy gives: . .mU 2 2

=

vz

=

. .mV 2 2

-

mga(1

cos 8)

-

and: U2

2ga(1

-

cos 8)

(8)

Newton s second law applied towards the centre of the circle gives: mg cos 8 - R

=

m

2

a

Elimination of V 2 between equations (8) and (9) gives the reaction:

(9)

R

=

m

mg(3 cos 8 - 2 - -

2

a 191

 

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clear that the particle will leave the circle at some point; in fact, it does so when R = O Solving this equation for cos e gives: t is

cos

e

=

2

U   ) 3 ga 2

-

-

As would be expected, as U increases, e decreases, which means that the particle leaves the circle at a higher position. When U 2 ~ ga it leaves immediately and travels as a projectile.

EXER ISES 8.8

1 A bead bea d li lies es insi inside de a smooth narr narrow ow circular tub tubee of radius a whose plane is vertical. f the bead, which is initially stationary, is given a velocity U show that it will complete circles if U 2 ~ 4ga. 2 A small mass m is attach attached ed to a point 0 by an inextensible string of length a The mass is held with the string taut, at the same level as 0 and released. Determine the angular velocity of the mass when the string makes an angle e wit with h the downwards vertica vertical. l. Fro From m thi thiss relationship, determine the period of one complete swing. How does this time compare with the value value of the p period eriod o off a smal smalll pendulu pendulum m of length a? 3 A sm small all body of mass m is attac attached hed by an in extensible string o off length 2a to a point o. The particle is released when the string is horizontal, and in its downwards motion it meets a peg A a distance a vertically below O. Find how high the body will rise above A in its subsequent motion. 4 A particle rests at the top of a fi fixe xed d smooth sphe sphere re of radius a which is fastened to a horizontal plane. f the particle is disturbed, find the horizontal distance that it will have travelled when it strikes the plane.

8 9 MOTION IN A CIRCLE AND CONNECTED PARTICLES

To this point, we have examined the free motion of bodies describing horizontal and vertical circles. In this section, we look at motion of connected particles in which one or other of the particles is performing

circular motion.

92

 

CIRCU L

R

MOTION

Example 8.9 Two beads P and Q of masses m and m respectively are attached to the ends of an inextensible string, whose length is greater than ]ta The string, which is taut, passes over a smooth cylinder of radius a, so tha t the plane containing the string and the beads is perpendicular to the cylinder s axis. The plane through the axis of the cylinder and P is initially horizontal. f the system is released from from rest, show that when the radius through P has turned through an angle 8 and P is still in contact with the cylinder, then:

(d8 2

3a dt

= 2g(28 - sin 8)

where 8 is the angle made by the radius to P to the horizontal. Find the tension in in the th e string string and the reaction between the bead b ead P and the cylinder. Hence, show show that th at the bead lose losess contact with the cylinder cylinder before reaching reaching the highest point.

2mg

Fig 8 13

Solution The particle s motion is illustrated in Figure 8.13. T is the tension in the string and R is the normal reaction between the particle P and the cylinder. The velocity of each particle is s and the transverse acceleration of P and linear acceleration of Q are both as Conservation of energy gives:

1

.

2m(a8)2

1

.

=

22m(a8Y

mga sin 8 - 2mga8

0

93

 

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TO MECHANICS MECHANICS

Simplification of this energy equation leads to the required equation as: 3a82 = 2g 28 - sin 8)

(10)

Differentiation of equation (10) gives:

=

6a88

2ga 28 - cos 88)

3a8 = g 2 - cos 8)

(11)

To find find the te tension, nsion, we apply Newton;s second secon d llaw aw to Q (or alternatively to P along the tangent to give: 2ma8 = 2mg - T

Using equation (10) now allows an expression for T =

2mg 1

T

to be found as:

cos 8)

3 R is found b by y applyi applying ng Newton New ton s second law in a radial directi direction on for give:

to

ma 8 2 = m g sin 8 - R

Using equation (10) gives

R

as:

R =

m g 5 sin 8 - 48)

3

the particle to leave the ci circle, rcle, its its value for some 8 must be zero. Sketches of the graphs of 48 and 5 sin 8, which are illustrated in Figure 8.14, 8.1 4, show one zero in this range, and values of 8 greate greaterr tha than n th that at rroot oot wil willl result in a negative value for R. This confirms that the particle does leave the cylinder before arriving at the highest point.

For

EXERCISES 8 9

1 Figure 8.1 8.15 5 sho shows ws a smoot smooth h narrow tu tube be in the form o off a circle, with centre 0 and radius a, which is fixed in a vertical plane. The tube contains two particles P, of mass m and Q, of mass 3 m which are connected by a light inextensible string o f length (l/2)Jta. The system is is at the level of 0 and Q is at the highest released from rest when

point of the tube. Show that, if after a time t, the line through an angle 8, then:

P

has turned

194

 

CIRCUL

R

=

R

MOTION

46

6

  1T

Fig 8 14

\

\

I I

\

1 \ a1 {

\

a 5 ' / ~ 1 ~ ,

~ /

1

I I

II

Fig 8 15 a

d8t 2

=

g 3 - 3 cos 8

sin 8)

provided that the string remains taut. Find the reaction between P and the tube. Obtain the angular acceleration in terms of g and 8. Deduce that the string becomes slack when 8 = l/4)rr. 2 Figure 8.1 8.16 6 sshow howss a smoo smooth th narrow tube in the form of a semi-c semi-circle ircle of

radius a which s fixed in a vertical plane with its diameter vertical. A particle o f mas masss m s initia initially lly at rest inside the tu tube, be, at its llowest owest point A . The particle s attached to one end of a light inextensible string which 195

 

GUIDE T O MECHANICS

B

T

Fig. 8.16 passes through the tube passes t ube and out a t the highest point B. The string is taut and its other end is pulled with a constant const ant force T. Show that, that , if a t a time t after the particle has left A the radius of the semi-circle through the particle makes an angle 8 with B A then: ma ( d8)2 dt

=

21 8 - 2mg 1

-

cos 8)

Find the reaction of the tube o n the particle and the angular acceler ation. Show that, if T = 1I2)mg, then d2 8/dt 2 = 0 when 8 = n16.

8.10 VECTOR METHODS

AND

CIRCULAR MOTION

In this section, the position vector o f a body bod y describing a circle is used and,

from it, the definition of the velocity velocity vector is derived for circular motion. The position P of a particle at time t is given by r = ret and rl = r a constant. T o use the definition o f the velocity vector v in Chapter 1, ret ~ t ) is taken to be the position vector o f the point Q as illustrated illust rated in Figure 8.17(a). For the particle moving from P a t time t to Q a t time t ~ t , the velocity vector v is written as: _

I

[ret

6t - ret ]

v- ~ t ~

0

which may be reduced to: 96

 

CIRCULAR MOTION

a)

./

f

b)

. 0

'

0< :

,

/r

r t)

p

'

t + btl ........

o Fig. 8.17 V

=

dr

dt

should be clear that th at the t he limit results in the velocity vector being along the tangent to the path at P, as was identified to be the case in Chapter l t

Figure 8,17(b) shows shows the unit vectors i andj so that the t he position vec vector tor r of P, with respect to the centre of the circle 0 can be written as:

+ r sin j

r = r cos i The velocity vector is then: v = ddtr =

r

smS S

· I

+ r cos

SS ·

l

is usual, in the case of motion where the use of polar coordinates is preferred, to write r and v in terms of the radial unit vector: t

r=

+ sin j

cos i

and the transverse unit vector:

9=

s in

i

+ cos j

This allows the position vector to be written as:

r = rr and the velocity vector as:

v= r o

The more sign signific ificant ant implications implications here are that, that , if any vector vary varying ing with t is differentiated with respect to t the result is that its direction is rotated through 1/2)31: in an anti-clockwise direction and its magnitude multiplied 197

 

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Fig. 8.18 by

O

This gives in the case dr

dt

of r a n d

. = 00 A

0: d

and

dt

=

.

-Or

8.11 VECTOR FORMULATION FOR CONSTANT ANGULAR VELOCITY Figure 8.18 the particle P is describing a circle o f centre and radius a, with constant angular speed w. The position vector of P a t any time t can be written as: In

r

=

r

Differentiating this expression leads to: v

= d r = awl dt

where a is a constant. const ant. This gives as in our ear earlie lierr analysis analysis th thee velocity as along the tangent. A second differentiation gives the acceleration vector as: f Here

as before there

= dv = dt

is

we

-wr

=

aw7

an acceleration acceleration along the radius towards the

centre

the circle w 2 • The simplicity of this this derivation should encourage the use o f vectors in these problems. of

198

 

CIRCULAR MOTION

Fig. 8.19

8.12 VECTOR FORMULATION FOR NON·CONSTAN NON·CONSTANT T ANGULAR VELOCITY Here, as in Section 8.10, the particle P is describing a circle o f radius a and centre 0 , as illustrated in Figure 8.19. This time the angu angular lar velocity is not

constant. A s before, the position vector

is

r

given by:

=

r

Differentiating to find the velocity vector gives:

v

dr =

.,

dt = aSO

which is the result already derived in Section 8.10 and t he same as would be expected for uniform speed. A second differentiation to find the acceler ation vector gives: dv

,

f = dt = aSO

as - S r

,

= aSO

. - aS2r

This acceleration acce leration vector ve ctor consists consists o f two parts: a transverse component s and a radial component a0   • These acceleration components have al ready been found in Section 8.11, but the advantage o f having them in this form is the ease with which they can be used.

8.13

CIRCULAR ORBITS

The value in using radial and transverse component vectors becomes

increasingly apparent when the for forces ces involved involved are central directed along along 99

 

GUIDE T O MECHANICS

c Fig. 8.20

the radius) o r tangential. A n important example of this is gravitational attraction. Modelling planetary motion as a circle is often very useful. The gravi tational force of attraction between two bodies o f masses M and m kilograms at a distance r metres apart is expressed as: Mm

In the case of a satellite of mass m assuming a circular orbit around the Earth of G = 6.64 X 10- 11 m3 kg- 1 and the mass of the Earth as = k to be set at M =6 X 1 M kg, this allows the value of 3.984 x 1 14 m3 In Figure 8.20, a satellite is in a circular orbit of radius r metres. Newton s second law applied to the satellite s motion towards the orbit s centre gives: S

2

21

S

I.

-mrw 

which can be written

=

km r

as:

Using this expression and the fact that the orbit is circular, so that r is constant, it follows that w is al also so constant. Thus, the satellite has a constant angular speed, and its period is:

8.14 ANGULAR VELOCITY

AS

A VECTOR

Having defined such vector quantities as displacement, linear velo velocity city and linear acceleration, we now find vector representations for angular dis2

 

C I R C UL

R M O T IO N

0

e

o Fig 8 21

placement angular ve veloci locity ty and angular acceleration. Althou Although gh the mag magni ni tude of these angular quantities is easil easily y identified the allocation of direction is not as obvious. In Figure 8.21 a point is rotating about the fixed axis 0 0 with angular speed w. We shall use vector methods to determine the linear velocity of P at any point on iits ts path path.. Clearly the velocity vector changes continuously as the rotation rota tion takes place and the velo velocity city wil illl be directed along the tangent to the circle at any time. Its magnitude will be aw where a is the radius of the circle described by P f

P is

t

is

w1 vector of at ssome time and thetrated angul angular ar Figure speed 8.2 then the the position linear velo velocity city vector vr ha has th thee direction illus illustrated in 8.21 and its magnitude is: l

r

sin 8

The velocity vector v can thus be identified as that of the vector product: v

w

X

r

where the vector 0 ) is defined as the angular velocity vector. The angular velocity vector has a direction defined by its right-handed rotation along the axis of rotation. Similar representations can be found for angular

displacement and acceleration. These ideas are developed more fully in Chapter 11 f a point is rotating about 0 0 , whose direction is defined by the unit 201

 

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vector (i 2j - 2k 2k)/ )/3, 3, w wit ith h an angu gular lar spe speed ed of 6 rad S - l , its angular velocity vector is written as: 0

=

2i

4j

4k

EXERCISES 8 14

1 Determi Det ermine ne the velo velocity city vector v for the fol follow lowing ing points with the th e position vector r and angular velocity vector 0 about an axis through the origin:

(a) r (b) r

= =

2i - j - 2k, 3i, 0 = 3j.

0

=

i - k;

2 A particle describes a cir circle cle about tthe he line wh whose ose vector equa equatio tion n iiss r = 3 + I-l)i + 2j + 1 - I-l)k wit with h angular speed radians per second. I f at time t

=

0 the particle is at the point 2k, find its velocity vector.

3 The velocity velocity vector v of a particle mov moving ing in a plane can be wri written tten as as:: v=(OXr+i

where r is the position position vector of the particle with respect to a fixed point within the plane. Show that the acceleration vector can be written as: 1

r

d r2w) ) dr

(

 

2)

r-rwr

MISCELLANEOUS EXERCISES 8

1 A large large sspace pace statio station n is built in the form of a circular tube of radius 50 m. n order to combat the problem of weightlessness, the designers decide that the tube will rotate about its axis of symmetry with a constant angular speed w radians per second. (a) How w wil illl thi thiss work fo forr th thee people who are on the space station? (b) What value will w have to take in order for the occupants of the space station to assume a weight whi which ch is the same as that due to gravity on Earth?

2 A tumble dryer is made from a cylindrical drum whose axis is horizon tal. f the radius of the drum is 30 em, determine the best strategy for drying clothes and suggest suitable rotation speeds for the drum. 202

 

9 VIBRATIONS

9 1 INTRODUCTION t is

probably safe to assume that w e all know what is meant by a vibration,

whether we think of sitting on a bus, experiencing the rattles and bumps as it struggles uphill, or the twang of a guitar string. The essential ingredient common to all vibrations is the to-and-fro motion, without there t here bein being g any overall movement in any direction. This chapter will enable us to under stand more about all types of vibrations. Engineers need to understand many things things th that at oscillate; a fe few w examples are engines, buildi buildings ngs subject to high winds, offshore structures and output from electrical circuits (for example hi-fi systems). t first sight, all of these seem very different, and very difficult to model. However, if attention is focussed on the end product, they all exhibit a to-and-fro motion, and it is this that we will use Newton s laws to describe. We will restrict our attention to motion in one dimension, or, more strictly, to a single degree of freedom.

~ 6 1 . i f Fig 9 1

Consider a mass attached to a spring on a smooth horizontal table as shown in Figure 9.1. f the spring is neither stretched nor compressed, the

mass will not be subject to any force, and therefore it will be in equilib rium. f the mass is pulled (or pushed) in the line of the spring and then released, release d, it wi will ll vibrate back and forth. The spring w wil illl alwa always ys try to resto restore re 2 3

 

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MECHANICS

the mass to its equilibrium position, but it will overshoot and, in the absence of any damping or friction, never come to rest other than instan taneously. Perfect springs springs have no damping, and a nd all th thee springs we consider consi der will be perfect, so the simple spring system shown in Figure 9.1, once set in motion, will vibrate forever. This is the system that will be examined first. Before starting on the mathematics, here are some terms with wh which ich we we need to become acquainted. We have already met the phrase equilibrium position (where the th e mass is at rest). The distance between this position position and the furthest point reached by the mass in either direction of its motion is called the amplitude of the vibration or oscillation. The words vibration and oscillation are used synonymously. The time the mass takes between leaving the left-most (say) extremity and returning there is called the period and 23t divided by this number is called the frequency of the oscillation. n the last chapt ch apter, er, motion moti on in a circle circle was discussed. discussed. f a particle is describing a horizontal circle with uniform speed, the time taken to describe the th e circle once is the period (one cycle). cycle). f the perpendicular from the particle to a diameter is drawn, the foot of this perpendicular will describe simple harmonic motion mo tion (SHM) , that tha t is, it will will oscillate. The Th e radius of the circle is the amplitude of the vibration. This provides an alternative view of vibration. Finally, a few words about springs. n reality, springs can be tricky to understand, but once attention is restricted to perfect springs, most difficulties disappear. A perfect spring, if left lying on a smooth table, has a fixed length. f it is stretched o r compressed, then freed once more to lie on the table, its length will be the same. This is called its natural length and it is the length a spring spring has has in the total tot al absence of forces forces.. There The re is, however, a second property of springs. Springs that have the same natural length can still be very different; for example, the spring found in the suspension system sys tem of a car may macould y havenot the change same natural natroles. ural length as the spring sprinisg stiffer in an electric toaster, but they The car spring (more difficult to stretch or compress) than the toaster spring, and this second property of springs is called their stiffness.

9 2

SIMPLE SIMPLE HARMONIC HARMONIC MOTION

The basic undamped vibration of Figure 9.1 is called simple harmonic motion (abbreviated to SHM). To derive an equation to describe it, we must use the second of Newton s laws of motion. Figure 9.2 is the same as Figure 9.1 but with the addition of an origin, axis, force and some labels. t is most convenient to choose as origin the

eqUilibrium eqUilibr ium position of the th e mass m Figure 9.2 shows shows this, togeth to gether er with the decisio dec ision n to t o choose the positive positive x-ax x-axis is to point po int to t o the t he right. The T he only force force acting on m the mass, is due to the spring and we assume it is stret stretched ched so 2 4

 

VIBRATIONS

l/

k I

equilibrium

osition

im

I

1

.

~ x

Fig. 9.2

that T the tension in the spring, acts as shown. Note that these choices (origin, axis, whether the spring is stretched or compressed) are, by and large, arbitrary (origins have to be fixed, x = 0 at 0 etc.). However, once the choice is made, it does not change. Applying Newton Newt on s second law (i.e. (i. e. force force = mass x acceleration) gives: 1)

To solve this equation, w e need to make some assumptions about T Clearly, the larger x is the bigger T is f yo you u stretch a spring, the further fur ther it is stretched, the greater is the force it exerts. Also, the stiffer the spring, the greater will be the value of T. (Think again of car springs, large stiffness, as opposed to toaster springs, smaller stiffness.) The simplest model, therefore, is to assume that T is directly proportional to the extension of the spring x (this is Hook Hookee s law law)) and that t hat the constant of of proportionality is the stiffness, k: T

(2)

kx

Eliminating T between equations 1) and (2) gives: (3)

Dividing by m and writing

02

= kim yields:

d x

-

dt

 

=

x

0 

(4)

where

is a number with dimensions of t i m et 1 called the natural frequency of the mass and spring system. This name arises from the fact that it is the frequency at which the mass oscillates when pulled to one side 0

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MECHANICS

and released. released. We can see this because the solution to equat equation ion 4), a second-order differential equation with constant coefficients, is: x

=

A sin

B cos

oot

5)

wt

where and B are constants obtained by using given conditions on x or specifi cificc times usually at t = 0). n example should make the dx/dt at spe determination determinati on of and B clear.

Example 9.2.1

Calculate the displacement x in terms of wan d t if, at time t the velocity is zero.

= 0, x = 3 and

The velocity of a particle that is displaced a distance x is its rate of change with respect to time, dx/dt see Chapter 1), so if x is given by equation 5), then diffe differentia rentiating ting with respect to t gives: Solution

dx

-

dt

t

time t

=

= w

cos wt -

sin wt

6)

0, we obtain:

I dx It dt

f

wB

7)

wA =

0

this is to be zero, we must have = O Setting = 0 results in the simp simplificati lification on of equ equat ation ion 5) to:

x and if x

=

3

at

t

=

= B

cos

8)

wt

:

9)

B=3

We have thus determi determined ned the particular values of and dx/dt 0 at time t O The solution is:

and B that

satisfy x = 3

=

=

x

=

3 cos

10)

wt

2 6

 

VIBRATIONS

Conditions like those in example 9.2.1 that enable us to find arbitrary constants cons tants are call called ed boundary cond conditio itions ns - we ha have ve already met them in Chapter 6 Most of such conditions met in this chapter are given at time t = 0, and they are then termed initi l conditions for obvious reasons. We see that the mass does indeed oscillate at the natural frequency J) with amplitude 3 in the case of the last example). For some problems, we know that it is more convenient to work in terms o f velocit velocity y . We also know ttha hatt accelera acceleration tion can be des described cribed by on onee o f the following alternative forms: dv

dv

or

dx

t

Using the second of thes these, e, equat equation ion 3) becomes:

mv

Or

dv dx

=

kx

rearranging: 11)

Integrating Integra ting equ equati ation on 11 11)) gi give ves: s: 1 2

+c

12)

where C is an arbitrary constant. We know that, at the extremity of an oscillation, velocity, v, is instantan instantaneously eously zero. H Henc ence, e, if we set v = 0 when x = o say), o would then be the amplitude of the oscillation. Inserting this condition into equat equation ion 12 12)) gi give ves: s: 13)

There is a lot of inform information ation in equ equati ation on 13) 13).. See examples 4.3. 4.3.1 1 for an alternative approach using energy. The right-hand side must be positive o r zero, because v is, so:  

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GUIDE TO MECHANICS

whence: 14) that is, x lies in the range between

and Xo. W e can also see from equation 13) that v is zero at the extremes x = ± xo) and has its maximum o

x o) at x = O All this is consistent with the oscillatory motion predicted by equation 5). Returning to equation 5):

value

x

=

A cos wt

B sin wt

we can rewrite this as: 15) or: x

=

V A2

B2 cos wt -

16)

< »

where: sin The quantity V A2

I>

=

B

V A2

B2

tan I>

B2 is the amplitude met earlier) and

I>

=

B

)

is the phase

new concept) of the oscillation. Equation 16) is termed the amplitudephase form o f the solution and can be more easily related to what actually happens to the mass than can the solution in the form o f equation 5). Twice the amplitude is the full extent o f the oscillation, and the phase is linked to where we take the time origin. f we start a stop-watch when the particle in Figure 9.2 is furthest from the support right-most extremity), then the phase is zero. To complete this section, we can re-derive equation 13) from equation 16). Differentiating with respect to t gives:

v=

dx dt

17)

= wX o) and equation 16) by Dividing equation 17) by wV A2 B2 V A2 B2 = x o) , squaring and adding yields: 2 8

 

VIBR

_V_ 2

2w t _

( ~ ) 2

wxo

= sin