SCHOLAR Study Guide

SQA Advanced Higher Physics Unit 1: Mechanics

Andrew Tookey Heriot-Watt University

Campbell White Tynecastle High School

Heriot-Watt University Edinburgh EH14 4AS, United Kingdom.

First published 2001 by Heriot-Watt University. This edition published in 2013 by Heriot-Watt University SCHOLAR. Copyright © 2013 Heriot-Watt University. Members of the SCHOLAR Forum may reproduce this publication in whole or in part for educational purposes within their establishment providing that no profit accrues at any stage, Any other use of the materials is governed by the general copyright statement that follows. All rights reserved. No part of this publication may be reproduced, stored in a retrieval system or transmitted in any form or by any means, without written permission from the publisher. Heriot-Watt University accepts no responsibility or liability whatsoever with regard to the information contained in this study guide.

Distributed by Heriot-Watt University. SCHOLAR Study Guide Unit 1: SQA Advanced Higher Physics 1. SQA Advanced Higher Physics ISBN 978-1-906686-06-2 Printed and bound in Great Britain by Graphic and Printing Services, Heriot-Watt University, Edinburgh.

Acknowledgements Thanks are due to the members of Heriot-Watt University's SCHOLAR team who planned and created these materials, and to the many colleagues who reviewed the content. We would like to acknowledge the assistance of the education authorities, colleges, teachers and students who contributed to the SCHOLAR programme and who evaluated these materials. Grateful acknowledgement is made for permission to use the following material in the SCHOLAR programme: The Scottish Qualifications Authority for permission to use Past Papers assessments. The Scottish Government for financial support. All brand names, product names, logos and related devices are used for identification purposes only and are trademarks, registered trademarks or service marks of their respective holders.

i

Contents 1 Kinematic relationships 1.1 Introduction . . . . . . . . . . . . . . 1.2 Derivation of kinematic relationships 1.3 Motion in one dimension . . . . . . . 1.4 Motion in two dimensions . . . . . . 1.5 Summary . . . . . . . . . . . . . . . 1.6 End of topic test . . . . . . . . . . .

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1 2 2 4 9 14 14

2 Relativistic motion 2.1 Introduction . . . . . . . 2.2 Relativistic dynamics . . 2.3 Relativistic energy . . . 2.4 Other relativistic effects 2.5 Summary . . . . . . . . 2.6 End of topic test . . . .

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3 Angular velocity and acceleration 3.1 Introduction . . . . . . . . . . . . . . . . . 3.2 Angular displacement and radians . . . . 3.3 Angular velocity and acceleration . . . . . 3.4 Kinematic relationships for angular motion 3.5 Tangential speed and angular velocity . . 3.6 Summary . . . . . . . . . . . . . . . . . . 3.7 End of topic test . . . . . . . . . . . . . .

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29 30 30 33 35 39 43 44

4 Centripetal force 4.1 Introduction . . . . . . . 4.2 Centripetal acceleration 4.3 Centripetal force . . . . 4.4 Applications . . . . . . . 4.5 Summary . . . . . . . . 4.6 End of topic test . . . .

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5 Rotational dynamics 5.1 Introduction . . . . . . . . . . . . . . . . . . . 5.2 Torque and moment . . . . . . . . . . . . . . 5.3 Newton's laws applied to rotational dynamics 5.4 Angular momentum and kinetic energy . . . . 5.5 Summary . . . . . . . . . . . . . . . . . . . . 5.6 End of topic test . . . . . . . . . . . . . . . .

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ii

CONTENTS

6 Gravitational force and field 6.1 Introduction . . . . . . . . . 6.2 Newton's law of gravitation 6.3 Weight . . . . . . . . . . . . 6.4 Gravitational fields . . . . . 6.5 Summary . . . . . . . . . . 6.6 End of topic test . . . . . .

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85 86 86 89 92 96 97

7 Gravitational potential and satellite motion 7.1 Introduction . . . . . . . . . . . . . . . . . . 7.2 Gravitational potential and potential energy 7.3 Satellite motion . . . . . . . . . . . . . . . . 7.4 Escape velocity . . . . . . . . . . . . . . . . 7.5 Summary . . . . . . . . . . . . . . . . . . . 7.6 End of topic test . . . . . . . . . . . . . . .

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99 100 100 105 110 114 115

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117 118 118 121 126 129 134 136 137

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139 140 140 146 151 152

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155 156 156 162 163 164

8 Simple harmonic motion 8.1 Introduction . . . . . . . . . 8.2 Defining SHM . . . . . . . . 8.3 Equations of motion in SHM 8.4 Energy in SHM . . . . . . . 8.5 Applications and examples 8.6 Damping . . . . . . . . . . 8.7 Summary . . . . . . . . . . 8.8 End of topic test . . . . . .

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9 Wave-particle duality 9.1 Introduction . . . . . . . . . . . . 9.2 Wave-particle duality of waves . 9.3 Wave-particle duality of particles 9.4 Summary . . . . . . . . . . . . . 9.5 End of topic test . . . . . . . . . 10 Introduction to quantum mechanics 10.1 Introduction . . . . . . . . . . . . 10.2 Atomic models . . . . . . . . . . 10.3 Quantum mechanics . . . . . . . 10.4 Summary . . . . . . . . . . . . . 10.5 End of topic test . . . . . . . . .

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11 Mechanics end-of-unit assessment

165

Glossary

168

Hints for activities

172

Answers to questions and activities 1 Kinematic relationships . . . . . . 2 Relativistic motion . . . . . . . . . 3 Angular velocity and acceleration 4 Centripetal force . . . . . . . . . .

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182 182 185 186 189

© H ERIOT-WATT U NIVERSITY

CONTENTS

5 6 7 8 9 10 11

Rotational dynamics . . . . . . . . . . . . . Gravitational force and field . . . . . . . . . Gravitational potential and satellite motion Simple harmonic motion . . . . . . . . . . Wave-particle duality . . . . . . . . . . . . Introduction to quantum mechanics . . . . Mechanics end-of-unit assessment . . . .

© H ERIOT-WATT U NIVERSITY

iii

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191 193 194 195 198 199 200

1

Topic 1

Kinematic relationships

Contents 1.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.2 Derivation of kinematic relationships . . . . . . . . . . . . . . . . . . . . . . . .

2 2

1.3 Motion in one dimension . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.3.1 Horizontal motion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

4 4

1.3.2 Vertical motion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

6

1.4 Motion in two dimensions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.5 Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

9 14

1.6 End of topic test . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

14

Prerequisite knowledge • Knowledge of the difference between vector and scalar quantities. • Calculus - definite integrals. • Some familiarity with the kinematic relationships would be useful. Learning Objectives By the end of this topic, you should be able to: • derive the three kinematic relationships from the calculus definitions of acceleration and velocity; • apply these equations to describe the motion of a particle with uniform acceleration moving in one dimension; • find the horizontal and vertical components of the velocity of a body moving in two dimensions; • apply the kinematic relationships to describe the motion of a particle with uniform acceleration moving in two dimensions.

2

TOPIC 1. KINEMATIC RELATIONSHIPS

1.1

Introduction

This topic deals with the motion of particles moving with a uniform acceleration. You should already be familiar with the kinematic relationships describing this motion, and with the vector nature of the quantities displacement, velocity and acceleration. For an object moving in a straight line with uniform acceleration, the average velocity over a period of time is given by (u + v) 2 where u and v are the initial and final velocities, respectively. The displacement s after a time t is then given by (u + v) ×t s= 2 This relationship is, of course, equivalent to the scalar equation d = v × t, where d = distance and v = average speed. There are three more kinematic relationships used to describe motion with uniform acceleration. We will begin by using calculus to derive these relationships, starting from the definitions of instantaneous acceleration and velocity. These relationships will then be applied to motion in one and two dimensions.

1.2

Derivation of kinematic relationships





Learning Objective To derive the three kinematic relationships from the calculus definitions of acceleration and velocity.



The equations describing motion with a constant acceleration can be derived from the definitions of acceleration and velocity. We will start with the definition of instantaneous acceleration: acceleration is the rate of change of velocity a=

dv dt

We are looking at motion where a is a constant. To find the velocity after time t, we will integrate this expression over the time interval from t = 0 to t = t. 

v u

 dv =

t t=0

 adt = a

t

dt t=0

Carrying out this integration: © H ERIOT-WATT U NIVERSITY

TOPIC 1. KINEMATIC RELATIONSHIPS

3

[v]vu =a [t]t0 ∴ v − u =at

(1.1)

∴ v =u + at .......................................... Equation 1.1 gives us the velocity v after time t, in terms of the acceleration a and the initial velocity u. Velocity is defined as the rate of change of displacement. We will now use this definition to derive the second of the kinematic relationships: v=

ds dt

We can substitute for v in Equation 1.1 using this expression. ds = u + at dt Again, we will integrate over the time interval from 0 to t.

 t ds = (u + at)dt s=0 t=0   1 2 t s ∴ [s]0 = ut + at 2 0 1 2 ∴ s = ut + at 2 

s

(1.2) .......................................... Equation 1.2 gives us the displacement s after time t, in terms of the acceleration and the initial velocity. To obtain the third kinematic relationship, we first rearrange Equation 1.1. t=

v−u a

Substituting this expression for t into Equation 1.2 gives us     1 v−u 2 v−u + s =u a 2 a ∴ 2as =2vu − 2u2 + v 2 − 2vu + u2 2

2

2

∴ 2as = − 2u + v + u

∴ v 2 =u2 + 2as .......................................... © H ERIOT-WATT U NIVERSITY

(1.3)

4

TOPIC 1. KINEMATIC RELATIONSHIPS

We have obtained three equations relating displacement s, time elapsed t, acceleration a and the initial and final velocities u and v. Note that with the exception of t, these are all vector quantities. 1 s = ut + at2 2

v = u + at

v 2 = u2 + 2as

Although we are dealing with vector quantities, we are only considering the special case of motion in a straight line. It is vital to ensure we assign the correct +ve or -ve sign to each of the quantities s, u, v and a. We can show graphically how the acceleration, velocity and displacement all vary with time. With these three equations, along with the equation in the Introduction relating displacement to average velocity and time, we can solve all the problems we will encounter during this topic.

Acceleration, velocity and displacement At this stage there is an online activity. 20 min

This simulation shows how the quantities acceleration a, velocity v and displacement s vary as functions of time. The program allows you to set values for a and the initial velocity u, and then calculates how a, v, and s evolve in time. You should be able to explain how the velocity-time graph relates to the equation v = u + at and how the displacement-time graph relates to the equation s = ut + 12 at2 . ..........................................

1.3 

Motion in one dimension



Learning Objective To apply the kinematic relationships to describe the motion of a particle moving in one dimension.



Both horizontal and vertical motion are covered in this section.

1.3.1

Horizontal motion

The kinematic relationships can be used to solve problems of motion in one dimension with constant acceleration. In all cases we will be ignoring the effects of air resistance. Example A car accelerates from rest at a rate of 4.0 m s-2 . 1. What is its velocity after 10 s? © H ERIOT-WATT U NIVERSITY

TOPIC 1. KINEMATIC RELATIONSHIPS

5

2. How long does it take to travel 72 m? 3. How far has it travelled after 8.0 s? We can list the data given to us in the question: u = 0 m s-1 (the car starts from rest) a = 4.0 m s-2 1. We are told that the time elapsed t = 10 s and we wish to find v. So with u, a and t known and v unknown, we will use the equation v = u + at.

L

L=?

K=0

J =10

J v = u + at

∴ v = 0 + (4.0 × 10) ∴ v = 40 m s−1 2. In part 2 we know u, a and s, and t is the unknown, so we use s = ut + 12 at2 . u = 0 m s-1 a = 4.0 m s-2 s = 72 m t=?

I

I = 72

J =?

© H ERIOT-WATT U NIVERSITY

J

6

TOPIC 1. KINEMATIC RELATIONSHIPS

s = ut + 12 at2   ∴ 72 = 0 + 12 × 4.0 × t2

∴ 72 = 2t2 ∴ t2 = 36

∴ t = 6.0 s 3. Finally in part 3 we are asked to find s, given u, a and t, so we will again use s = ut + 12 at2 , this time using different data. u = 0 m s-1 a = 4.0 m s-2 t= 8.0 s s=? s = ut + 12 at2   ∴ s = 0 + 12 × 4 × 82

∴ s = 128 m

.......................................... The same strategy should be used in solving all of the problems in this topic. Firstly, list the data given to you in the question. This will ensure that you use the correct values when you perform any calculations, and should also make it clear to you which of the kinematic relationships to use. It is often useful to make a sketch diagram with arrows, to ensure that any vector quantities are being measured in the correct direction.

Horizontal Motion 20 min

Suppose a car is being driven at a velocity of 12.0 m s -1 towards a set of traffic lights, which are changing to red. The car driver applies her brakes when the car is 30.0 m from the stop line. What is the minimum uniform deceleration needed to ensure the car stops at the line? At this stage there is an online activity where you try it out. Always use the same procedure to solve a kinematics problem in one dimension - sketch a diagram, list the data and select the appropriate kinematic relationship. ..........................................

1.3.2

Vertical motion

When dealing with freely-falling bodies, the acceleration of the body is the acceleration due to gravity, g = 9.8 m s-2 . If any force other than gravity is acting in the vertical plane, the body is no longer in free-fall, and the acceleration will take a different value. Problems should be solved using exactly the same method we used to solve horizontal motion problems.

© H ERIOT-WATT U NIVERSITY

TOPIC 1. KINEMATIC RELATIONSHIPS

7

Example A student drops a stone from a second floor window, 15 m above the ground. 1. How long does it take for the stone to reach the ground? 2. With what velocity does it hit the ground? When dealing with motion under gravity, we must take care with the direction we choose as the positive direction. Here, if we take a as a positive acceleration, then v and s will also be positive in the downward direction.

L

I

L=?

I = 15

K=0

J =?

J

We are told that u = 0 m s-1 a = g = 9.8 m s-2 s = 15 m 1. To find t, we use s = ut + 12 at2 . Putting in the appropriate values s = ut + 12 at2   ∴ 15 = 0 + 12 × 9.8 × t2

∴ 15 = 4.9t2 15 ∴ t2 = 4.9 ∴ t2 = 3.06 ∴ t = 1.7 s

2. To find v, we use v 2 = u2 + 2as. Again, we put the appropriate values into this kinematic relationship v 2 = u2 + 2as ∴ v 2 = 0 + (2 × 9.8 × 15) ∴ v 2 = 294 ∴ v = 17 m s−1 © H ERIOT-WATT U NIVERSITY

J

8

TOPIC 1. KINEMATIC RELATIONSHIPS

.......................................... Difficulties sometimes occur when the initial velocity is directed upwards, for example if an object is being thrown upwards from the ground. To solve such a problem, it is usual to take the vertical displacement as being positive in the upwards direction. The velocity vector is then positive when the object is travelling upwards, and negative when it is returning to the ground. In this situation the acceleration is always negative, as it is always directed towards the ground.

Quiz 1 Motion in one dimension 20 min

First try the questions. If you get a question wrong or do not understand a question, there are 'Hints'. The hints are in the same order as the questions. If you read the hint and still do not understand then ask your tutor. All references in the hints are to online materials. Useful data: acceleration due to gravity g

9.8 m s-2

Q1: An object is moving with a uniform acceleration of 5 m s -2 . A displacement-time graph showing the motion of this object has a gradient which a) b) c) d) e)

increases with time. decreases with time. equals 5 m s-1 . equals 5 m s-2 . equals 0. ..........................................

Q2: A car accelerates from rest with a uniform acceleration of 2.50 m s -2 . How far does it travel in 6.00 s? a) b) c) d) e)

7.50 m 15.0 m 18.75 m 45.0 m 112.5 m ..........................................

Q3: Neglecting air resistance, a stone dropped from the top of a building 125 m high hits the ground after a) b) c) d) e)

1.25 s. 4.00 s. 5.05 s. 12.7 s. 25.5 s.

© H ERIOT-WATT U NIVERSITY

TOPIC 1. KINEMATIC RELATIONSHIPS

9

.......................................... Q4: A diver jumps upwards with an initial vertical velocity of 3.00 m s -1 from a diving board which is 8.00 m above the swimming pool. With what vertical velocity does he enter the pool? a) b) c) d) e)

3.0 m s-1 3.6 m s-1 9.0 m s-1 13 m s-1 24 m s-1 ..........................................

Q5: A stone is dropped from a window 28.0 m above the ground. What is the velocity of the stone when it is 10.0 m above the ground? a) b) c) d) e)

10.0 m s-1 14.0 m s-1 18.8 m s-1 23.4 m s-1 98.1 m s-1 ..........................................

1.4 

Motion in two dimensions



Learning Objective To apply the kinematic relationships to an object moving in two dimensions.



Because we are dealing with vector quantities, we can also solve problems of motion in two dimensions. Usually this involves separating the motion into x- and y-components, which can be treated independently. Most problems we come across involve projectile motion, where horizontal and vertical components are studied separately. For example, consider a parcel dropped from a train window 3 m above the ground, when the train is travelling at 40 m s -1 along a straight track. Neglecting air resistance, there is no horizontal force acting on the parcel, and hence no acceleration. The parcel will continue to move at the same horizontal velocity in the same direction as the train. In the vertical plane, the parcel is falling under gravity, and will accelerate downwards at a rate of g m s -2 . We can calculate the time taken for the parcel to hit the ground by considering the vertical motion only. Once we know this time, we can find out the distance between the point when the parcel was released and where it hits the ground by considering the horizontal motion only. Vertically uy = 0 m s-1 , ay = g = 9.8 m s-2 , sy = 3 m, t = ? © H ERIOT-WATT U NIVERSITY

10

TOPIC 1. KINEMATIC RELATIONSHIPS

sy = uy t + 12 ay t2   ∴ 3 = 0 + 12 × 9.8 × t2 3 ∴ t2 = 4.9 ∴ t2 = 0.612 ∴ t = 0.782 s Since there is no acceleration horizontally, ux = vx = 40 m s-1 , and we can use the simple equation s x = vx × t. sx = vx × t ∴ sx = 40 × 0.782 sx = 31 m Example Consider another example - a golf ball struck with an initial velocity of 24 m s -1 at an angle of 30 ◦ to the ground. How far does it travel horizontally before striking the ground again? We must first separate the initial velocity into its orthogonal components u x and uy .

K θ

KO KN

Using the trigonometric relationships, we can see that ux = u cos θ and uy = u sin θ The initial velocity of the particle is given by u2 = u2x + u2y in the direction given by the angle θ. Now we can study the x- and y-components of the motion separately, and as in the previous example, we need to work out the time-of-flight from the vertical motion in order to calculate the horizontal range. © H ERIOT-WATT U NIVERSITY

TOPIC 1. KINEMATIC RELATIONSHIPS

11

Vertically, we want to know the time taken until s y returns to zero. Taking upwards as the positive direction, we have uy = 24 × sin30 = 12 m s-1 ay = -g = -9.8 m s-2 sy = 0 m t=? s = ut + 12 at2 ∴ 0 = (24 × sin 30 × t) +

1

∴ 0 = 12t − 4.9t2

2

× −9.8 × t2



One solution to this equation, of course, is t = 0 s, since the displacement is 0 when t = 0 s. Ignoring this solution, we can divide both sides of the above equation by t to give us 0 = 12 − 4.9t 12 ∴t= 4.9 ∴ t = 2.45 s Knowing this time-of-flight, we can use the simple s x = vx × t relation for the horizontal motion to calculate s x sx = vx × t ∴ sx = (24 × cos 30) × 2.45 ∴ sx = 20.78 × 2.45 ∴ sx = 51 m .......................................... There are a couple of important points brought up in this example. First, make sure you understand the sign convention that has been used for the vertical motion. We have taken upwards as the positive direction, so the initial velocity uy is a positive vector quantity, and the acceleration a y is a negative vector quantity. The second point to note is that when the golf ball has returned to Earth, its vertical displacement s y is zero. The displacement tells us how far, and in what direction, an object is from its starting point, not the total distance it has travelled. In this case, when the ball hits the ground its vertical displacement is exactly the same as it was at t = 0, so sy = 0. By considering the x- and y-components, we can work out the angle of trajectory required to give the maximum horizontal range for a given initial projectile speed. To solve this problem, we first consider the horizontal motion. sx = u cos θ × t So for a given u, the range depends on θ and t. Let us look now at the vertical motion. The value of t that we are considering is the time taken for the projectile to return to Earth. This means we require s y , the vertical displacement, to return to zero. © H ERIOT-WATT U NIVERSITY

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TOPIC 1. KINEMATIC RELATIONSHIPS

1 sy =uy t + ay t2 2 1 ∴ 0 =u sin θt − gt2 2 Ignoring the t = 0 solution, we can divide by t and then rearrange this equation in terms of t. 2u sin θ t= g We can substitute this expression for t into our equation for horizontal motion. sx =u cos θ × t 2u sin θ ∴ sx =u cos θ × g u2 × 2 sin θ cos θ ∴ sx = g Now we can use the trig relation sin 2θ = 2 sin θ cos θ in the above equation to give us sx =

u2 sin 2θ g

So for any value of u, the maximum value of s x is obtained when sin2θ is at a maximum. This occurs when sin2θ = 1, and hence 2θ = 90 ◦ . Therefore the maximum range is obtained when θ = 45 ◦ . Putting this value of θ back into the above equation, we can see that u2 g .......................................... sx,max =

(1.4)

Motion in two dimensions 25 min

A motorcycle stunt rider is trying to jump a 41.0 m gap. The take-off ramp launches him at an angle of 20.0 ◦ , and the landing ramp is at the same height as the take-off ramp. Use the kinematic relationships to calculate the minimum speed at take-off that will get the rider safely to the other side. You can test your result out in the online activity. In all problems which involve motion in two dimensions, the motion must be split into xand y-components. ..........................................

© H ERIOT-WATT U NIVERSITY

TOPIC 1. KINEMATIC RELATIONSHIPS

13

Quiz 2 Motion in two dimensions First try the questions. If you get a question wrong or do not understand a question, there are 'Hints'. The hints are in the same order as the questions. If you read the hint and still do not understand then ask your tutor. All references in the hints are to online materials. Useful data: acceleration due to gravity g

9.8 m s-2

Q6: A truck is travelling at 20 m s-1 on a straight horizontal road. A man in the truck fires a gun straight up in the air. Ignoring air resistance, where will the bullet land? a) b) c) d) e)

In front of the truck. Behind the truck. On the truck. Depends on the vertical speed of the bullet. Depends on the value of g. ..........................................

Q7: What is the maximum horizontal range of a football kicked with an initial velocity of 12.0 m s-1 , if air resistance is negligible? a) b) c) d) e)

1.73 m 3.83 m 8.49 m 12.0 m 14.7 m ..........................................

Q8: An arrow is fired from a bow at 24.0 m s -1 towards a target 30.0 m away. If the bow and the bull's-eye of the target are at the same height, what is the minimum angle of elevation needed to ensure the arrow hits the bull's-eye? a) b) c) d) e)

15.3◦ 30.7◦ 45.0◦ 53.1◦ 74.6◦ ..........................................

Q9: A boy kicks a ball at an angle of elevation of 75 ◦ and velocity 8.00 m s-1 . With what velocity must he run over level ground to catch the ball before it bounces? a) b) c) d) e)

0.11 m s-1 2.07 m s-1 4.00 m s-1 7.73 m s-1 8.00 m s-1 ..........................................

© H ERIOT-WATT U NIVERSITY

20 min

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TOPIC 1. KINEMATIC RELATIONSHIPS

Q10: An aeroplane flying horizontally at an altitude of 900 m releases a package when it is a horizontal distance of 1600 m from its target. How fast is the plane travelling, if the package lands on its target? a) b) c) d) e)

1.78 m s-1 118 m s-1 144 m s-1 900 m s-1 1600 m s-1 ..........................................

1.5

Summary

The material covered in this topic should have added to your understanding of motion of bodies moving with uniform acceleration. By the end of this topic you should be able to: • derive the three kinematic relationships 1 v = u + at s = ut + at2 2

v 2 = u2 + 2as

from the calculus definitions of acceleration and velocity. • apply these equations to describe the motion of a particle with uniform acceleration moving in one dimension. • find the horizontal and vertical components of the velocity of a body moving in two dimensions. • apply the kinematic relationships to describe the motion of a particle with uniform acceleration moving in two dimensions.

1.6

End of topic test





Learning Objective topic



End of topic test 30 min

At this stage there is an end of topic test available online.. If however you do not have access to the internet you may try the questions which follow. The following data should be used when required: acceleration due to gravity g = 9.8 m s -2 Q11: A car accelerates uniformly from rest, travelling 55 m in the first 8.5s. Calculate its acceleration in m s -2 . © H ERIOT-WATT U NIVERSITY

TOPIC 1. KINEMATIC RELATIONSHIPS

15

.......................................... Q12: The brakes on a car can provide a deceleration of up to 4.9 m s

-2

.

Calculate the minimum stopping distance (in m) if the brakes are applied when the car is initially travelling at 20 m s -1 . .......................................... Q13: A toy rocket is fired vertically upwards with an initial velocity of 16 m s -1 . Calculate how long (in s) the rocket stays in the air before it returns to the ground. .......................................... Q14: A piano is being raised to the third floor of a building using a rope and pulley. The piano is 15 m above the ground, moving upwards at 0.25 m s -1 , when the rope snaps. Calculate how much time (in s) elapses before the piano hits the ground. .......................................... Q15: A cricket ball is thrown with an initial velocity 17 m s -1 at an angle of elevation 48 ◦ above the horizontal. Calculate how much time (in s) passes before the ball hits the ground again. .......................................... Q16: In a film stunt, a car is driven off a cliff with a horizontal velocity of 14 m s cliff face is vertical, and the cliff is 46 m high.

-1

. The

Calculate the distance in m from the base of the cliff to the point where the car strikes the ground. .......................................... Q17: A golfer strikes a ball with an initial velocity 30 m s-1. The ball lands on the green, which is on the same horizontal plane 45 m away. Calculate the minimum angle of elevation θ at which the golf ball should be struck in order for it to land on the green. ..........................................

© H ERIOT-WATT U NIVERSITY

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TOPIC 1. KINEMATIC RELATIONSHIPS

© H ERIOT-WATT U NIVERSITY

17

Topic 2

Relativistic motion

Contents 2.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.2 Relativistic dynamics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

18 18

2.3 Relativistic energy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.4 Other relativistic effects . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

21 25

2.5 Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

25

2.6 End of topic test . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

26

Prerequisite knowledge • Knowledge of Newtonian mechanics, such as the equation for the kinetic energy of an object. Learning Objectives By the end of this topic, you should be able to: • Understand that the properties of objects moving at very large speeds are subject to relativistic phenomena; • State that the speed of light in a vacuum is constant, and that no object can travel faster than the speed of light; • Calculate the relativistic mass and energy of an object.

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TOPIC 2. RELATIVISTIC MOTION

2.1

Introduction

Probably the most well-known of all physics equations is E = mc 2 . Almost everyone has heard or seen this equation, even if most people don't understand its meaning or implications. In this Topic we will be studying relativity. By the end of it you should understand what this equation means and when it should be applied. The concept of relativity was first proposed by Albert Einstein in 1905 in his Special Theory of Relativity. It is something that we are not often aware of in everyday life, and in some ways requires a new way of looking at things. This Topic only scratches the surface of the subject, as many of its harder concepts require a lot of advanced mathematics. One consequence of the principle of relativity is that two people measuring the same thing may not necessarily get the same result if they are moving with respect to each other. The greater their relative speed, the greater the discrepancy in their measurements. The laws of mechanics that you are used to applying are called Newtonian mechanics, since they are based on Newton's laws of motion. In this Topic we will see that when an object is moving with a speed close to the speed of light, Newtonian mechanics can no longer be used to describe its motion.

2.2

Relativistic dynamics





Learning Objective To state that the relativistic mass of an object increases as its speed approaches the speed of light.



Consider an object of mass m travelling at speed v. We could use a balance to find the value of m, a ruler to find how far it has travelled and a stopwatch to find how long it takes to travel this distance. We could also apply Newtonian mechanics to calculate the momentum and kinetic energy of the object, and the time it takes to travel between any two points. When an object is moving at a very high speed, the laws of Newtonian mechanics are no longer valid. To an observer who is stationary compared to the moving object, the mass, distance travelled, time taken and other related properties are subject to relativistic mechanics and are calculated using a different set of equations. Since relativity is such a large topic we will concentrate on calculating relativistic mass and kinetic energy. For completeness, relativistic effects on time and length are also discussed at the end of the topic. The relativistic mass m of an object moving with speed v is given by the equation m=

m0

2

1 − v c2 ..........................................

(2.1)

© H ERIOT-WATT U NIVERSITY

TOPIC 2. RELATIVISTIC MOTION

19

In Equation 2.1 m0 is the rest mass of the object (its mass when it is stationary) and c is the speed of light in a vacuum (3.00 × 10 8 m s-1 ). So to an observer who is stationary, the mass of the object is m rather than m 0 . Example An electron (rest mass 9.11 × 10-31 kg) is accelerated to a velocity of 5.00 × 107 m s-1 . What is the relativistic mass of the electron? Putting the values of rest mass and speed into Equation 2.1, m=

m0

2

1 − v c2 9.11 × 10−31 ∴m=  7 2 1 − (5.00×108 )2 (3.00×10 )

9.11 × 10−31 ∴m= 15 ) 1 − (2.50×10 (9.00×1016 ) ∴m=

9.11 × 10−31 1 − 2.5 90

9.11 × 10−31 √ 0.972 9.11 × 10−31 ∴m= 0.986 ∴ m = 9.24 × 10−31 kg ∴m=

The mass of the electron has increased to 9.24 × 10 -31 kg, for a stationary observer making measurements or calculations on a moving electron. .......................................... This example shows that even at a speed of 5.00 × 10 7 m s-1 , the relativistic increase in mass is still only around 1% of the rest mass. As the speed of the object gets closer and closer to c, so the relativistic mass gets larger and larger. There are a couple of important points to be aware of when calculating the relativistic mass of an object. Firstly, relativistic effects only become noticeable when an object is moving with a speed 2

of around 0.1c or greater. If the speed is less than this, the factor v c2 is extremely small, and the value of the square root term in Equation 2.1 is very close to 1, hence m ≈ m0 . A second point to note is that an object's mass would become infinite if it were travelling at the speed of light. If v = c in Equation 2.1, the denominator becomes zero and hence m becomes infinite. To accelerate an object to the speed of light would therefore require an infinite-sized force to be applied to the object, which is impossible. This means that the speed of light is "nature's speed limit" as it is impossible for an object with non-zero rest mass to reach this speed. (It can also be shown that an object with zero rest mass, such as a photon, travels at speed c.) © H ERIOT-WATT U NIVERSITY

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TOPIC 2. RELATIVISTIC MOTION

Quiz 1 Relativistic dynamics 20 min

First try the questions. If you get a question wrong or do not understand a question, there are Hints. The hints are in the same order as the questions. If you read the hint ans still do not understand then ask your tutor. All references in the hints are to online materials. Useful data: rest mass of an electron me rest mass of a proton mp speed of light in a vacuum c

9.11 × 10-31 kg 1.67 × 10-27 kg 3.00 × 108 m s-1

Q1: A proton is accelerated to a speed of 1.00 × 10 8 m s-1 . What is the relativistic mass of the proton? a) b) c) d) e)

1.77 × 10-27 1.88 × 10-27 5.01 × 10-27 1.50 × 10-26 5.01 × 10-26

kg kg kg kg kg ..........................................

Q2: a) b) c) d) e)

Einstein's theory of relativity tells us that electrons can be accelerated to the speed of light. the relativistic mass of an object is less than its rest mass. objects moving at very high speeds do not obey the laws of Newtonian mechanics. any particle can travel with a speed greater than c. only objects with a non-zero rest mass can travel at the speed of light. ..........................................

Q3: At approximately what speed is an object travelling when relativistic effects become important? a) b) c) d) e)

340 m s-1 (speed of sound) 0.1 c c 10c c2 ..........................................

Q4: The relativistic mass of a particle is equal to twice its rest mass. At what speed is the particle travelling? a) b) c) d) e)

7.50 × 107 1.50 × 108 2.12 × 108 2.25 × 108 2.60 × 108

m s-1 m s-1 m s-1 m s-1 m s-1 .......................................... © H ERIOT-WATT U NIVERSITY

TOPIC 2. RELATIVISTIC MOTION

21

Q5: When the value of v << c, the value of a) b) c) d) e)

2

1 − v c2 is approximately

0 1 v c ∞ ..........................................

2.3

Relativistic energy



Learning Objective To state the relativistic energy of an object, and calculate the rest energy, kinetic energy and total energy of an object.



One important consequence of Einstein's work on special relativity is the principle of mass-energy equivalence. This principle states that the mass of an object is a measure of its total energy. This is summed up in the famous equation E = mc2

(2.2)

.......................................... Here E is the total energy of an object, and m is its relativistic mass. A stationary object therefore has energy E0 = m0 c2

(2.3)

.......................................... E0 is called the rest energy of the object. A moving object has kinetic energy E k in addition to its rest energy, so E = Ek + E0 ∴ Ek = E − E0 ..........................................

(2.4)

We can substitute for the values of E and E 0 in Equation 2.4 to find the kinetic energy of an object moving at a relativistic speed Ek = E − E0 ∴ Ek = mc2 − m0 c2 m0 ∴ Ek = c2 − m0 c2

2 1 − v c2 ⎞ ⎛ 1 ∴ Ek = m0 c2 ⎝ − 1⎠ 2

v 1 − c2 © H ERIOT-WATT U NIVERSITY

(2.5)

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TOPIC 2. RELATIVISTIC MOTION

.......................................... This equation looks very different from the equation for kinetic energy that you will have seen before, Ek = 12 mv 2 . However it can be shown that when v is much smaller than c, and relativistic effects are negligible, Equation 2.5 does give us the values we would expect.

Relativistic and Newtonian kinetic energy (optional) 20 min

This is an OPTIONAL activity for students wishing to show that the relativistic expression for kinetic energy is the same as the Newtonian expression for an object moving at low speed. Equation 2.5 gives the kinetic energy of an object moving at relativistic speeds. Does this mean that the kinetic energy equation derived from Newtonian mechanics is incorrect? Well luckily it doesn't, unless we are moving at very high speeds. In this exercise we will show that the two expressions are equivalent at low speeds. To do this we will make a binomial expansion of the relativistic expression for kinetic energy. The binomial theory allows us to expand any expression which is of the general form (1+x)n , n (n − 1) x2 n (n − 1) (n − 2) x3 + + .......... 2! 3! ..........................................

(1 + x)n = 1 + nx +

(2.6)

We need to rearrange Equation 2.5 to allow us to expand the term in brackets as a binomial, ⎞

⎛ 1

− 1⎠ 2

v 1 − c2 (2.7)   1  −2 2

−1 1 − v c2 ∴ Ek = m0 c2 .......................................... − 12  2

term in Equation 2.7 to Equation 2.6, we have a binomial Comparing the 1 − v c2

2 with x = −v c2 and n = − 12 . Performing this expansion we obtain, Ek = m0 c2 ⎝

(1 + x)n = 1 +  ∴  ∴

v2 1− 2 c v2 1− 2 c

− 12 − 12

nx +  1   2 − 2 × −v =1+ + c2 =1+

v2 2c2

+

n (n − 1) (n − 2) x3 n (n − 1) x2 + + ... 2! 3!  1  3 − 2 × − 2 × v4 +... 2c4 3v 4 8c4

+...

© H ERIOT-WATT U NIVERSITY

TOPIC 2. RELATIVISTIC MOTION

23

When v is much less than c, all terms in the expansion become extremely small, except the first two. So under these conditions, 

v2 1− 2 c

− 12

≈1+

v2 2c2

If we substitute this expression back into Equation 2.7, we obtain 

v2 Ek = m0 c 1+ 2 2c 2 v ∴ Ek = m0 c2 × 2 2c 1 ∴ Ek = m0 v 2 2 2



 −1

This is the Newtonian expression for kinetic energy. So as long as v is much less than c, we can use the standard expression for kinetic energy. Using a binomial expansion it can be shown that the equations for Newtonian and relativistic kinetic energy are equivalent for an object moving at low speed. ..........................................

Relativistic Mass and Energy At this stage there is an online activity. This simulation plots the relativistic mass and energy of an object as its speed increases from 1 × 107 m s-1 .

15 min

The mass of an object increases the closer to the speed of light c that the object is travelling at. Kinetic energy also increases with increasing speed, but the relativistic kinetic energy becomes much larger than the Newtonian kinetic energy as the speed becomes closer to c. ..........................................

Quiz 2 Relativistic energy First try the questions. If you get a question wrong or do not understand a question, there are Hints. The hints are in the same order as the questions. If you read the hint ans still do not understand then ask your tutor. All references in the hints are to online materials. Useful data: rest mass of an electron me rest mass of a proton mp speed of light in a vacuum c

© H ERIOT-WATT U NIVERSITY

9.11 × 10-31 kg 1.67 × 10-27 kg 3.00 × 108 m s-1

20 min

24

TOPIC 2. RELATIVISTIC MOTION

Q6: A stationary nucleus has mass 6.64 × 10 -27 kg. What is the rest energy of this nucleus? a) b) c) d) e)

5.98 × 10-10 1.99 × 10-18 3.58 × 10-19 3.97 × 10-36 1.32 × 10-44

J J J J J ..........................................

Q7: a) b) c) d) e)

The total relativistic energy of a moving object is made up of kinetic energy only. rest energy and light energy. kinetic energy and light energy. rest energy and kinetic energy. rest energy only. ..........................................

Q8: A particle with a rest mass of 2.0 × 10-26 kg emerges from a particle accelerator, travelling at such a speed that its relativistic mass is 2.7 × 10-26 kg. What is the total energy of the particle? a) b) c) d) e)

6.3 × 10-10 J 1.8 × 10-9 J 2.4 × 10-9 J 4.2 × 10-9 J 4.9 × 10-9 J ..........................................

Q9: How does the kinetic energy of an object moving at a speed close to the speed of light, calculated using a relativistic calculation, compare with the kinetic energy calculated using a Newtonian calculation? a) b) c) d) e)

The relativistic calculation gives a lower value of kinetic energy. The relativistic calculation gives a higher value of kinetic energy. Both calculations give the same value of kinetic energy. Impossible to say without knowing the exact speed of the object. Impossible to say without knowing the rest mass of the object. ..........................................

Q10: A proton is accelerated to a speed of 2.0 × 10 8 m s-1 . Using a relativistic calculation, find the kinetic energy of the proton. a) b) c) d) e)

3.8 × 10-11 5.1 × 10-11 1.2 × 10-10 1.9 × 10-10 2.0 × 10-10

J J J J J .......................................... © H ERIOT-WATT U NIVERSITY

TOPIC 2. RELATIVISTIC MOTION

2.4 

Other relativistic effects

25



Learning Objective To describe some of the other concepts of relativity



The theory of relativity does not just deal with the effects on mass and energy of travelling at very high speeds. Other effects become apparent too. For example, length contraction occurs, which means that to a stationary observer, an object travelling at high speeds appears to be shorter than it does when it is at rest. Another relativistic effect, time dilation, means that time passes more slowly for an object moving at high speeds than it does for a stationary object. An example of this is the creation of muons in the upper atmosphere by cosmic radiation. Muons are subatomic particles with extremely short lifetimes (∼10-6 s) and are typically created at an altitude of over 10 km above the Earth's surface. Even travelling at close to the speed of light, these particles would disintegrate before reaching the Earth's surface, yet detectors on Earth can detect muons. The reason that they survive long enough to be detected is because the time dilation increases their lifetime.

Time dilation At this stage there is an online activity. A spaceship is flying a distance of 5 light hours, for example from Earth to the planet Pluto. The speed can be regulated with the upper buttons. The animation demonstrates that the clock in the spaceship goes more slowly than the two clocks of the system in which Earth and Pluto are motionless. Time dilation occurs for an object travelling at close to the speed of light, meaning that a stationary clock on Earth would run faster than an identical clock placed in a rocket ship travelling at high speed. ..........................................

2.5

Summary

A stationary observer will find that measurements made on the motion of an object moving at a very high speed will not be consistent with the laws of Newtonian mechanics. An object moving at a speed close to the speed of light will undergo relativistic effects. In this topic we have concentrated on the relativistic mass and energy of an object. The relativistic mass of an object is greater than its rest mass, and increases with increasing speed. The total energy of such an object consists of rest energy and kinetic energy, where the kinetic energy is larger than that calculated using Newtonian mechanics.

© H ERIOT-WATT U NIVERSITY

15 min

26

TOPIC 2. RELATIVISTIC MOTION

By the end of this topic you should be able to: • use the equation m=

m0

2

1 − v c2 to calculate the relativistic mass of an object of rest mass m0 ;

• state that the speed of light in a vacuum is the maximum speed in nature, and that an object with non-zero rest mass cannot attain this speed; • state that the rest energy of an object is given by the equation E = m 0 c2 and the relativistic energy of a particle is given by the equation E = mc2 ; • use the equation

⎞

⎛ 1

− 1⎠ 2

v 1 − c2 to find the kinetic energy of a particle travelling at relativistic speed. Ek = m0 c2 ⎝

2.6

End of topic test

End of topic test 30 min

At this stage there is an end of topic test available online. If however you do not have access to the internet you may try the questions which follow. the following data should be used when required. speed of light in a vacuum c

3.00 × 10 8 m s -1

rest mass of an electron m e

9.11 × 10 -31 kg

rest mass of a proton m p

1.67 × 10 -27 kg

Q11: Calculate the relativistic mass (in kg) of a proton moving at speed 1.65 × 10 8 m s -1 . .......................................... Q12: A electron, rest mass m relativistic mass is 2.1 m e . At what speed, in m s

-1

e

, has been accelerated to such a speed that its

, is the electron travelling? .

.......................................... Q13: The nucleus of an atom has rest mass 7.07 × 10 -26 kg. Calculate the rest energy (in J) of this nucleus. .......................................... Q14: An electron is travelling at speed 2.53 × 10 8 m s -1 . Calculate the relativistic energy, measured in J, of the electron. © H ERIOT-WATT U NIVERSITY

TOPIC 2. RELATIVISTIC MOTION

.......................................... Q15: A proton travelling at high speed has relativistic mass 2.15 × 10 -27 kg. Calculate the kinetic energy of the proton, measured in J. .......................................... Q16: An electron is travelling at a speed of c . Calculate the total relativistic energy (in J) of the electron. ..........................................

© H ERIOT-WATT U NIVERSITY

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TOPIC 2. RELATIVISTIC MOTION

© H ERIOT-WATT U NIVERSITY

29

Topic 3

Angular velocity and acceleration

Contents 3.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.2 Angular displacement and radians . . . . . . . . . . . . . . . . . . . . . . . . .

30 30

3.3 Angular velocity and acceleration . . . . . . . . . . . . . . . . . . . . . . . . . . 3.4 Kinematic relationships for angular motion . . . . . . . . . . . . . . . . . . . . .

33 35

3.5 Tangential speed and angular velocity . . . . . . . . . . . . . . . . . . . . . . .

39

3.6 Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.7 End of topic test . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

43 44

Prerequisite knowledge • Kinematic relationships (Mechanics Topic 1). • Calculus - definite integrals. Learning Objectives By the end of this topic, you should be able to: • express angles and angular displacement in radians; • calculate the angular velocity, periodic time and angular acceleration of an object moving in a circle; • derive the kinematic relationships for angular motion, and use the relationships to solve problems; • relate angular velocity and acceleration to the tangential speed and acceleration.

30

TOPIC 3. ANGULAR VELOCITY AND ACCELERATION

3.1

Introduction

In the Kinematics topic we studied the motion of objects travelling in a straight line. Now we will move on to study circular motion, which has a wide range of applications. These include the motion of planets around the Sun, the disc spinning in a CD player and a car taking a corner. We begin here by introducing angular displacement, angular velocity and angular acceleration, and introducing the units in which they are measured. We will use these quantities to describe the motion of a point object moving in a circular path, or a solid object rotating about an axis. We have already used kinematic relationships involving linear displacement, velocity and acceleration. We will derive the angular equivalent of these relationships, and look at the relationship between angular and linear quantities.

3.2

Angular displacement and radians





Learning Objective To use radians to measure angles and angular displacement.



In the first Topic of the course we investigated motion in one and two dimensions. We are going to apply many of the ideas we met in that Topic to describe the motion of an object moving in a circle. We will see in the next Section that instead of using velocity and acceleration vectors, we will be using angular velocity and angular acceleration. Firstly we will look at angular displacement, which replaces the linear displacement we are used to dealing with. Imagine a disc spinning about a central axis, as shown in Figure 3.1. We can draw a reference line along the radius of the disc. The angular displacement after time t is the angle through which this line has swept in time t. Figure 3.1: Angular displacement

At J = 0

After time interval J

θ

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

© H ERIOT-WATT U NIVERSITY

TOPIC 3. ANGULAR VELOCITY AND ACCELERATION

31

The angular displacement is given the symbol θ, and is measured in radians (rad). Throughout this Topic, radians will be used to measure angles and angular displacement. A brief explanation of radian measurement, and how radians and degrees are related, will be given next. Figure 3.2: Radian measurement

s θ

r

.......................................... With reference to Figure 3.2, the angle θ, measured in radians, is equal to s/r. Since the radian is defined as being one distance (s) divided by another (r) then strictly speaking it is a dimensionless quantity, however the radian is regarded as a supplementary SI unit. To compare radians with degrees, consider an angular displacement of one complete circle, equivalent to a rotation of 360 ◦ . In this case, the distance s in Figure 3.2 is equal to the circumference of the circle. Hence s r 2πr ∴θ= r ∴ θ = 2π rad .......................................... θ=

(3.1)

So 360◦ is equivalent to 2π rad, and this relationship can be used to convert from radians to degrees, and vice versa. It is useful to remember that π rad is equivalent to 180◦ and π/2 rad is equivalent to 90 ◦ . For the sake of neatness and clarity, it is common to leave an angle as a multiple of π rather than as a decimal, so the equivalent of 30◦ is usually expressed as π/6 rad rather than 0.524 rad.

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Quiz 1 Radian measurement 15 min

First try the questions. If you get a question wrong or do not understand a question, there are Hints. The hints are in the same order as the questions. If you read the hint and still do not understand then ask your tutor. All references in the hints are to online materials. Q1: a) b) c) d) e)

Convert the angle 145 ◦ into radians. 0.395 rad 0.405 rad 2.53 rad 23.1 rad 911 rad ..........................................

Q2: a) b) c) d) e)

What is the equivalent in degrees ( ◦ ) to 1.20 radians? 3.76◦ 7.54◦ 68.75◦ 138◦ 432◦ ..........................................

Q3: a) b) c) d) e)

Express 120◦ in radians. π/ 120 rad π/ rad 4 π/ rad 3 2π/ rad 3 3π/ rad 2 ..........................................

Q4: An object moves through 3 complete rotations about an axis. What is its total angular displacement? a) b) c) d) e)

π/ rad 6 π/ rad 3 3π rad 6π rad 2π 3 rad ..........................................

Q5: An object moves 5.00 cm around the circumference of a circle of radius 24.0 cm. What is the angular displacement of the object? a) 0.208 rad b) 0.208π rad c) 4.80 rad © H ERIOT-WATT U NIVERSITY

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d) 4.80π rad e) 9.60π rad ..........................................

3.3

Angular velocity and acceleration



Learning Objective To define the angular velocity, angular acceleration and periodic time of an object moving in a circle, and to calculate these quantities.



Now that we have introduced the concept of angular displacement, it follows that the angular velocity ω is equal to the rate of change of angular displacement, just as the linear velocity v is the rate of change of linear displacement s. ω=

dθ dt

v=

ds dt

ω is measured in radians per second (rad/s or rad s -1 ). The average angular velocity over a period of time t is the total angular displacement in time t divided by t. Example It takes the Moon 27.3 days to complete one orbit of the Earth. Assuming the Moon travels in a circular orbit at constant angular velocity, what is the angular velocity of the Moon? The Moon moves through 2π rad in 27.3 days. (27.3 × 24) = 655.2 hours (655.2 × 60) = 39312 minutes (39312 × 60) = 2358720 s

Now, 27.3 days is equal to which is equal to which is equal to So the angular velocity ω is given by

angular displacement time elapsed 2π ∴ω= 2358720 ∴ ω = 2.66 × 10−6 rad s−1 ω=

.......................................... You should also be aware of two other useful ways of describing the rate at which a body is moving in circular motion. One is to use the periodic time (or period) T , which © H ERIOT-WATT U NIVERSITY

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TOPIC 3. ANGULAR VELOCITY AND ACCELERATION

is the time taken for one complete rotation. The other is to express the rate in terms of revolutions per second, which is the inverse of the periodic time. In the example above, the periodic time T for the Moon orbiting the Earth is 27.3 days, or 2.36 × 10 6 s 1 −7 revolutions per in SI units. The rate of rotation is equal to 1/T = 2.36×10 6 = 4.24 × 10 second. The relationship between periodic time and angular velocity is 2π T .......................................... ω=

(3.2)

Orbits of the planets 20 min

Following on from the above example, calculate the periodic times and angular velocities of the motion of Mercury, Venus and the Earth around the Sun. Fill in the gaps in the table below. Planet

Orbit radius (m) Period (days)

Mercury Venus Earth

5.79 × 1010 1.08 × 1011 1.49 × 1011

Period (s)

Angular velocity (rad s-1 )

88.0 225 365

The angular velocity and periodic time of an object moving in a circle are related by the equation ω = 2π/T . .......................................... Having defined angular displacement θ and angular velocity ω, it should be clear that if ω is changing then we have an angular acceleration. The instantaneous angular acceleration α is the rate of change of angular velocity, measured in rad s -2 . dω dt .......................................... α=

(3.3)

The average angular acceleration over time t is the total change in angular velocity divided by t.

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TOPIC 3. ANGULAR VELOCITY AND ACCELERATION

3.4

35

Kinematic relationships for angular motion





Learning Objective To derive and use the angular kinematic relationships.



In the Kinematics topic we derived the kinematic relationships for linear motion with constant acceleration from the definitions of instantaneous acceleration and velocity. We can use the same technique to derive the equations of circular motion with constant angular acceleration by starting from the definitions of angular velocity and angular acceleration. We begin with the definition of instantaneous angular acceleration α=

dω dt

Remember that we are only considering motion where α is a constant. Integrating the above expression over the time interval from 0 to t gives us 

ω

 dω =

ω0



t

αdt = α

t=0

t

dt t=0

Carrying out this integration: [ω]ωω0 = α [t]t0 ∴ ω − ω0 = αt

(3.4)

∴ ω = ω0 + αt .......................................... Equation 3.4 gives us the angular velocity after time t in terms of the angular velocity at t = 0 and the angular acceleration. Now we can substitute for ω = dθ/dt in this equation dθ = ω0 + αt dt Integrating this equation over the same time interval 

θ



t

(ω0 + αt) dt   1 2 t ∴ θ = ω0 t + αt 2 0 1 2 ∴ θ = ω0 t + αt 2 .......................................... θ=0

dθ =

t=0

(3.5)

Equation 3.5 gives us the angular displacement after time t, again in terms of the angular velocity at t = 0 and the angular acceleration. Finally we use Equation 3.4, rearranged as follows © H ERIOT-WATT U NIVERSITY

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TOPIC 3. ANGULAR VELOCITY AND ACCELERATION

t=

ω − ω0 α

Substituting this expression for t into Equation 3.5 gives us   1 ω − ω0 2 θ = ω0 + α 2 α 1 ∴ αθ = ω0 (ω − ω0 ) + (ω − ω0 )2 2 ∴ 2αθ = 2ω0 (ω − ω0 ) + (ω − ω0 )2 

ω − ω0 α



(3.6)

∴ 2αθ = 2ω0 ω − 2ω02 + ω 2 − 2ω0 ω + ω02 ∴ 2αθ = −ω02 + ω 2 ∴ ω 2 = ω02 + 2αθ .......................................... We now have a set of three kinematic relationships that describe circular motion with constant angular acceleration. If you have trouble remembering them, you should be able to work them out by comparison with their linear equivalents: Linear motion v = u + at s = ut + 12 at2

v 2 = u2 + 2as

Circular motion ω = ω0 + αt θ = ω0 t + 12 αt2 ω 2 = ω02 + 2αθ

We can now use these equations to solve problems involving motion with constant angular acceleration. Example An electric fan has blades that rotate with angular velocity 80 rad s -1 . When the fan is switched off, the blades come to rest after 12 s. What is the angular deceleration of the fan blades? We follow the same procedure as we used to solve problems in linear motion - list the data and select the appropriate kinematic relationship. Here we are told ω 0 = 80 rad s-1 ω = 0 rad s-1 t = 12 s α=? With ω 0 , ω and t known and α unknown, we use ω = ω 0 + αt to find α.

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ω = ω0 + αt ∴ ω − ω0 = αt ω − ω0 ∴α= t 0 − 80 ∴α= 12 ∴ α = −6.7 rad s−2 So the angular acceleration is -6.7 rad s -2 , equivalent to a deceleration of 6.7 rad s -2 . ..........................................

Will the car stop before the lights? In the Kinematics topic we used a simulation of a car stopping before a set of traffic lights, and calculated the deceleration needed to stop the car. We have the same problem here, only this time you are asked to calculate the angular deceleration required to stop the wheels turning before the car passes through the lights.

25 min

We previously worked out the linear deceleration required for a car travelling at 12.0 m s-1 to stop before a set of traffic lights 30.0 m away. At 12.0 m s-1 , the car wheels are rotating at 27.0 rad s -1 . What is the angular deceleration required for the car wheels to stop turning before the traffic lights? You should be able to solve this problem in the same way that linear problems are solved, using the angular version of the kinematic relationships. ..........................................

Quiz 2 Angular velocity and angular kinematic relationships First try the questions. If you get a question wrong or do not understand a question, there are Hints. The hints are in the same order as the questions. If you read the hint and still do not understand then ask your tutor. All references in the hints are to online materials. Q6: A compact disc is spinning at a rate of 8.00 revolutions per second. What is the angular velocity of the disc? a) b) c) d) e)

0.020 rad s-1 0.785 rad s-1 1.27 rad s-1 25.1 rad s-1 50.3 rad s-1

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.......................................... Q7: A turntable rotates through an angle of 1.8π rad in 4.0 s. What is the average angular velocity of the turntable? a) b) c) d) e)

0.45π rad s-1 0.90π rad s-1 2.2π rad s-1 4.0π rad s-1 7.2π rad s-1 ..........................................

Q8: A disc is spinning about an axis through its centre with constant angular velocity 7.50 rad s-1 . What is the angular displacement of the disc in 8.00 s? a) b) c) d) e)

0.938 rad 1.07 rad 9.55 rad 60.0 rad 377 rad ..........................................

Q9: At the top of a hill, the wheels of a bicycle are rotating at angular velocity 5.0 rad s-1 . When the cyclist reaches the bottom of the hill, the angular velocity has increased to 12 rad s-1 . If it took the cyclist 10 s to cycle down the hill, what was the angular acceleration of the wheels? a) b) c) d) e)

0.35 rad s-2 0.70 rad s-2 1.4 rad s-2 1.7 rad s-2 5.95 rad s-2 ..........................................

Q10: A car parked on a slope begins to slowly roll forward. The angular acceleration of the wheels from rest is 0.20 rad s-2 . How long does it take for the wheels to rotate through one complete revolution? a) b) c) d) e)

3.5 s 5.0 s 7.9 s 12.6 s 31 s ..........................................

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TOPIC 3. ANGULAR VELOCITY AND ACCELERATION

3.5

39

Tangential speed and angular velocity





Learning Objective To relate the angular velocity of an object to its tangential speed in m s -1 .



Suppose an object is moving in a circle of radius r with angular velocity ω. What is the relationship between ω and v, the velocity of the object measured in m s -1 ? We can answer this question by considering the way the radian is defined. Looking back to Figure 3.2, the angle θ, in radians, is given by θ=

s r

Differentiating this equation with respect to t gives us d s dθ = dt dt r Since r is constant, this equation can be rearranged 1 ds dθ = dt r dt 1 ∴ω = ×v r ∴ v = rω ..........................................

(3.7)

Equation 3.7 gives us the relationship between the speed of the object (in m s -1 ) and its angular velocity (in rad s-1 ). We shall see in the next Topic that this speed is not the same as the linear velocity of the object, since the object is not moving in a straight line, and the velocity describes both the rate and direction at which an object is travelling. (Remember that velocity is a vector quantity.) Figure 3.3: Tangential speed at several points around a circle L

ω L H

L

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.......................................... Figure 3.3 shows that the speed v at any point is the tangential speed, and is always perpendicular to the radius of the circle at that point. If we imagine that Figure 3.3 shows a mass on a string being whirled in a circle, what would happen if the string broke? The mass would continue to travel in a straight line in the direction of the linear speed arrow, that is, it would travel at a tangent to the circle. We will explore this question more fully in the next topic. Equation 3.7 also shows that there is an important difference between v and ω - that two objects with the same angular velocity can be moving with different tangential speeds. This point is illustrated in the next worked example. Example Consider a turntable of radius 0.30 m rotating at constant angular velocity 1.5 rad s -1 . Compare the tangential speeds of a point on the circumference of the turntable and a point midway between the centre and the circumference. The point on the circumference has r = 0.30 m, so v1 = r1 ω ∴ v1 = 0.30 × 1.5 ∴ v1 = 0.45 m s−1 The point midway between the centre and the circumference is moving with the same angular velocity, but the radius of the motion is only 0.15 m. v2 = r2 ω ∴ v2 = 0.15 × 1.5 ∴ v2 = 0.225 m s−1 The point on the circumference is moving at twice the speed (in m s -1 ) of the point with the smaller radius. ..........................................

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41

This difference in tangential speeds is emphasised in Figure 3.4. Figure 3.4: Points moving at the same angular velocity, but with different tangential velocities

ω

=

>

?

I=

θ

I> I?

=

>

?

.......................................... The points a, b and c all lie on the same radius of the circle, which is rotating at angular velocity ω about the centre of the circle. The radius moves through angle θ in time δt. In this same time δt, the points a, b and c move through distances s a , sb and sc respectively, where sa < sb < sc . The tangential speeds at each point on the radius are therefore sa/δt, sb/δt and sc/δt where sa/δt < sb/δt < sc/δt.

Compact disc player At this stage there is an online activity. If however you do not have access to the internet you may try the question which follows. This activity shows how the information on a CD is read by the laser in a CD player. Q11: As the disc begins playing, the laser is focussed at a point 2.50 cm from the centre of the disc. Calculate the angular velocity, if the disc is moving over the beam with speed 1.25 m s-1 . .......................................... Q12: As the disc nears the end, the beam is focused at a point 5.70 cm from the centre of the disc. Calculate the new angular velocity, bearing in mind that the tangential speed is the same. .......................................... Q13: If the disc takes a total of 55 mins (3300 s) to play, what is the average angular deceleration of the disc while it plays? ..........................................

© H ERIOT-WATT U NIVERSITY

20 min

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The tangential speed of a point on a disc depends on both the angular velocity and the distance from the centre of the disc.

We can use Equation 3.7 to find an expression relating the angular acceleration to the tangential acceleration, the rate at which the speed of an object is changing. Starting with Equation 3.7 v = rω We can differentiate this equation with respect to time d dv = (rω) dt dt dω dv =r ∴ dt dt ∴ a = rα ..........................................

(3.8)

As with the tangential speed, the tangential acceleration depends on the radius of the motion as well as the rate of change of angular velocity. This acceleration is always perpendicular to the radius of the circle.

Quiz 3 Angular velocity and tangential speed 20 min

First try the questions. If you get a question wrong or do not understand a question, there are Hints. The hints are in the same order as the questions. If you read the hint and still do not understand then ask your tutor. All references in the hints are to online materials. Q14: A mass on the end of a 1.20 m length of rope is being swung round in a circle at an angular velocity of 4.00 rad s -1 . What is the tangential speed of the mass? a) b) c) d) e)

0.300 m s-1 3.33 m s-1 4.80 m s-1 3.33π m s-1 4.80π m s-1 ..........................................

Q15: An object is moving in a circular path with angular velocity 12.5 rad s -1 . If the speed of the object is 20.0 m s -1 , find the radius of the path. a) b) c) d) e)

0.255 m 0.625 m 1.60 m 10.1 m 250 m © H ERIOT-WATT U NIVERSITY

TOPIC 3. ANGULAR VELOCITY AND ACCELERATION

.......................................... Q16: What is the speed of an object moving in a circle of radius 1.75 m with periodic time 2.40 s? a) b) c) d) e)

0.729 m s-1 1.50 m s-1 2.29 m s-1 4.20 m s-1 4.58 m s-1 ..........................................

Q17: A spinning disc slows down from ω = 4.50 rad s -1 to 1.80 rad s-1 in 5.00 s. If the radius of the disc is 0.200 m, find the tangential deceleration of a point on the circumference of the disc. a) b) c) d) e)

0.108 m s-2 0.370 m s-2 0.540 m s-2 2.70 m s-2 3.39 m s-2 ..........................................

Q18: An object is moving in a circle of radius 0.25 m with a constant angular velocity of 6.4 rad s-1 . Through what distance does the object travel in 2.0 s? a) b) c) d) e)

1.6 m 3.2 m 10.2 m 12.8 m 51.2 m ..........................................

3.6

Summary

This topic is an introduction to circular motion. In this topic we have described circular motion in terms of angular displacement, velocity and acceleration, and related these quantities to the linear distance travelled and the tangential speed and acceleration. By the end of this topic, you should be able to: • measure angles and angular displacement in radians, and convert an angle measured in degrees into radians, and vice versa; • use angular displacement, angular velocity and angular acceleration to describe motion in a circle;

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TOPIC 3. ANGULAR VELOCITY AND ACCELERATION

• apply the equation

2π ω relating the periodic time to the angular velocity; T =

• derive the angular kinematic relationships, and apply them to solving problems of circular motion with uniform angular acceleration; • relate the angular velocity to the tangential (linear) speed of a body moving in a circle, and derive the equation v = rω; • apply the expression a = rα relating tangential and angular accelerations.

3.7

End of topic test

End of topic test 30 min

At this stage there is an end of topic test available online. If however you do not have access to the internet you may try the questions which follow. The following data should be used when required: acceleration due to gravity g = 9.8 m s -2 Q19: How many radians are equivalent to 68 ◦ ? .......................................... Q20: A flywheel is rotating with constant angular velocity 14 rad s

-1 .

Calculate the time taken (in s) for the flywheel to complete exactly 7 revolutions. .......................................... Q21: A moon orbiting a distant planet has a period of 46 days. Calculate the angular velocity of the planet, in rad s

-1

.

.......................................... Q22: An electric fan is switched from a "low" setting (angular velocity rad 3.5 s "high" setting (angular velocity 12.0 rad s -1 ). If the angular acceleration is 4.5 rad s reach the higher angular velocity.

-2

-1

) to a

, calculate the time taken (in s) for the fan to

.......................................... Q23: As the brakes are applied to a bicycle wheel, the angular velocity of the wheel changes from 23.5 rad s -1 to 5.00 rad s -1 . If the wheel turns through exactly 10 revolutions whilst the brakes are being applied, calculate the magnitude of the angular deceleration of the wheel, in rad s -2 . .......................................... Q24: Passengers in a Ferris Wheel sit in cars 13 m from the centre of the wheel. The wheel takes 200 s to complete one revolution, travelling with uniform angular velocity. © H ERIOT-WATT U NIVERSITY

TOPIC 3. ANGULAR VELOCITY AND ACCELERATION

Calculate the speed in m s

-1

of a passenger car.

.......................................... Q25: A cyclist travelling around a circular track of radius 32 m increases his speed from 12 m s -1 to 20 m s -1 in 4.0 s. Calculate the average angular acceleration in rad s

-2

of the cyclist.

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

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Topic 4

Centripetal force

Contents 4.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.2 Centripetal acceleration . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

48 48

4.3 Centripetal force . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.3.1 Object moving in a horizontal circle . . . . . . . . . . . . . . . . . . . . .

52 54

4.3.2 Vertical motion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

55

4.4 Applications . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.4.1 Conical pendulum . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

58 58

4.4.2 Cars cornering . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.5 Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

60 63

4.6 End of topic test . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

63

Prerequisite knowledge • Understanding of circular motion in terms of angular displacement and angular velocity (Mechanics Topic 3). • Addition and subtraction of vectors. • Application of Newton's laws of motion to static and dynamic situations, including the use of free-body diagrams to solve problems. Learning Objectives By the end of this topic, you should be able to: • derive expressions for the centripetal acceleration in terms of tangential or angular velocity; • derive expressions for the centripetal force in terms of tangential or angular velocity; • use these expressions to calculate centripetal force and acceleration in practical situations.

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TOPIC 4. CENTRIPETAL FORCE

4.1

Introduction

In the previous topic we discussed circular motion, and concentrated on the kinematic relationships for circular motion. In this topic we will shift our attention to how Newton's laws of motion apply to circular motion. We will see that the velocity of an object moving in a circle is constantly changing, hence there is a constant acceleration, requiring a force that acts towards the centre of the circle. This acceleration is quite separate from the tangential acceleration (a) we have met previously. After discussing Newton's laws we will move on to look at some practical examples of circular motion. We will be paying particular attention to the different forces that are involved, using free-body diagrams to determine the magnitude and direction of these forces.

4.2

Centripetal acceleration





Learning Objective To derive expressions for the centripetal acceleration in terms of tangential or angular velocity.



Consider an object moving in a circle of radius r with constant angular velocity ω. We know that at any point on the circle, the object will have tangential velocity v = rω. The object moves through an angle Δθ in time Δt. Figure 4.1 shows the velocity vectors at two points A and B on the circumference of the circle. The tangential velocities va and vb at these points have the same magnitude, but are clearly pointing in different directions. If the angular displacement between A and B is Δθ, then the angle between the vectors v a and vb is also Δθ. Check you can prove this before you proceed. Figure 4.1: Velocity vectors at two points on a circle

ω

Δθ L>

H

)

Δθ

*

L=

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TOPIC 4. CENTRIPETAL FORCE

49

.......................................... The change in velocity Δv is equal to v b - va . We can use a ‘nose-to-tail' vector diagram to determine Δv, as shown in Figure 4.2. Both vb and va have magnitude v, so the vector XY represents vb and vector YZ represents -va . Figure 4.2: Determination of the change in velocity

Z

-va

Δθ

Δv

Y

vb

X .......................................... In the limit where Δt is small, Δθ tends to zero. In this case the angle ZXY tends to 90◦ , and the vector Δv is perpendicular to the velocity, so Δv points towards the centre of the circle. The velocity change, and hence the acceleration, is directed towards the centre of the circle. To distinguish this acceleration from any tangential acceleration that may occur, we will denote it by the symbol a ⊥ , since it is perpendicular to the velocity vector. To calculate the magnitude of a ⊥ , we use the equation a⊥ =

Δv Δt

Since Δθ is small, Δv = v Δθ, so long as Δθ is measured in radians. The acceleration is therefore a⊥ =

vΔθ Δt

Taking the limit where Δt approaches zero, this equation becomes vdθ dt ∴ a⊥ = vω a⊥ =

Finally we can substitute for v = rω to get v2 = rω 2 r .......................................... a⊥ =

© H ERIOT-WATT U NIVERSITY

(4.1)

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TOPIC 4. CENTRIPETAL FORCE

This acceleration a⊥ is called the centripetal acceleration. It is always directed towards the centre of the circle, and it must not be confused with the tangential acceleration a we met in the previous Topic, which occurs when an orbiting object changes its tangential speed. The centripetal acceleration occurs whenever an object is moving in a circular path, even if its tangential speed is constant. (In some textbooks it is referred to as the radial acceleration, as it is always directed along a radius of the circle.) Example Find the centripetal acceleration of an object moving in a circular path of radius 1.20 m with constant tangential speed of 4.00 m s -1 . The centripetal acceleration is calculated using the formula v2 r 4.002 ∴ a⊥ = 1.20 ∴ a⊥ = 13.3 m s−2 a⊥ =

.......................................... Work through the next example to make sure you understand the difference between tangential and centripetal acceleration. Example A model aeroplane on a rope 10 m long is circling with angular velocity 1.2 rad s -1 . If this speed is increased to 2.0 rad s-1 over a 5.0 s period, calculate 1. the angular acceleration; 2. the tangential acceleration; 3. the centripetal acceleration at these two velocities. 1. The data in the question tells us ω 0 = 1.2 rad s-1 , ω = 2.0 rad s-1 , t = 5.0 s and α is unknown. Using the relationship ω = ω0 + αt we can calculate the angular acceleration α. ω = ω0 + αt ∴ αt = ω − ω0 ω − ω0 ∴α= t 2.0 − 1.2 ∴α= 5.0 ∴ α = 0.16 rad s−2 © H ERIOT-WATT U NIVERSITY

TOPIC 4. CENTRIPETAL FORCE

51

2. To calculate the tangential acceleration a, use the relationship between tangential and angular acceleration derived in the previous Topic. a = rα ∴ a = 10 × 0.16 ∴ a = 1.6 m s−2 3. We use the formula a ⊥ =rω 2 to calculate the centripetal acceleration. When ω = 1.2 rad s-1 a⊥ = rω 2 ∴ a⊥ = 10 × 1.22 ∴ a⊥ = 14.4 m s−2 When ω = 2.0 rad s-1 a⊥ = rω 2 ∴ a⊥ = 10 × 2.02 ∴ a⊥ = 40 m s−2 ..........................................

Quiz 1: Centripetal acceleration First try the questions. If you get a question wrong or do not understand a question, there are Hints. The hints are in the same order as the questions. If you read the hint and still do not understand then ask your tutor. All references in the hints are to online materials. Q1: A disc of radius 0.50 m is rotating with angular velocity 4.1 rad s -1 . What is the centripetal acceleration of a point on its circumference? a) b) c) d) e)

2.1 m s-2 4.2 m s-2 8.4 m s-2 17 m s-2 34 m s-2 ..........................................

Q2: Which one of the following statements is true? a) b) c) d) e)

The centripetal acceleration is always at a tangent to the circle. The centripetal acceleration is always directed towards the centre of the circle. There is only a centripetal acceleration when the tangential speed is changing. The centripetal acceleration does not depend on the angular velocity. The centripetal acceleration does not depend on the radius of the circular motion.

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20 min

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TOPIC 4. CENTRIPETAL FORCE

.......................................... Q3: A car takes a corner at 15 m s-1 . If the radius of the corner is 40 m, what is the centripetal acceleration of the car? a) b) c) d) e)

0.18 m s-2 5.6 m s-2 107 m s-2 600 m s-2 9000 m s-2 ..........................................

Q4: A particle moves in a circular path of radius 0.10 m at a constant rate of 6.0 revolutions per second. What is the centripetal acceleration of the particle? a) b) c) d) e)

3.6 m s-2 3.8 m s-2 14 m s-2 36 m s-2 140 m s-2 ..........................................

Q5: An object is moving in a circular path with speed v m s -1 . If the radius of the path doubles whilst the speed v remains constant, what happens to the centripetal acceleration? a) b) c) d) e)

The centripetal acceleration halves in value. The centripetal acceleration doubles in value. The new centripetal acceleration equals the square of the original value. The new centripetal acceleration equals the square root of the original value. The centripetal acceleration remains constant. ..........................................

4.3

Centripetal force



Learning Objective To derive expressions for the centripetal force in terms of tangential or angular velocity.



Newton's second law of motion tells us that if an object is undergoing acceleration, then a net force must be acting on the object in the direction of the acceleration. Since we have a centripetal acceleration acting towards the centre of the circle, there must be a centripetal force acting in that direction. Newton's law can be summed up by the equation F = ma © H ERIOT-WATT U NIVERSITY

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2

If the centripetal acceleration is given by a⊥ = v r = rω 2 , then the centripetal force acting on a body of mass m moving in a circle of radius r is

mv 2 = mrω 2 r .......................................... F =

(4.2)

Figure 4.3: Centripetal force

ω

H

.

L

.......................................... This is the force which must act on a body to make it move in a circular path. If this force is suddenly removed, the body will move in a straight line at a tangent to the circle with speed v, since there will be no force acting to change the velocity of the body. Example Compare the centripetal forces required for a 2.0 kg mass moving in a circle of radius 40 cm if the velocity is: 1. 3.0 m s-1 ; 2. 6.0 m s-1 . 1. Using Equation 4.2, the force required is mv 2 r 2.0 × 3.02 ∴F = 0.40 ∴ F = 45 N F =

2. In this case mv 2 r 2.0 × 6.02 ∴F = 0.40 ∴ F = 180 N F =

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So doubling the velocity means that the centripetal force required increases by a factor of four, since F ∝ v 2 . ..........................................

4.3.1

Object moving in a horizontal circle

Consider the case shown in Figure 4.4(a) of an object moving at constant angular velocity ω in a horizontal circle, such as a mass being whirled overhead on a string. Figure 4.4: (a) An object moving in a horizontal circle; (b) horizontal force acting on the object =

H

6

ω =

>

.......................................... In the horizontal direction, Figure 4.4(b) shows there is an acceleration a⊥ acting towards the centre of the circle. The only force acting in the horizontal direction is the tension in the string, so this tension must provide the centripetal force mrω 2 . So in this case T = mrω 2

(4.3)

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

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4.3.2

55

Vertical motion

Let us now consider the same object being whirled in a vertical circle. Figure 4.5 shows the object at three points on the circle, with the forces acting at each point. Figure 4.5: Object undergoing circular motion in a vertical plane

mg T a a mg T

ω

a T mg .......................................... If we consider the forces acting when the object is at the top of the circle, we find that both the weight and the tension in the string are acting downwards. Thus the resultant force acting towards the centre of the circle is Ttop + mg = mrω 2 This resultant force must provide the centripetal force to keep the object moving in its circular path. Hence T top = mrω 2 − mg

(4.4)

.......................................... On the other hand, when the object reaches the lowest point on the circle, the tension and the weight are acting in opposite directions, so the resultant force acting towards the centre of the circle is T bottom − mg = mrω 2 Again, this force supplies the centripetal force so in this case Tbottom = mrω 2 + mg .......................................... © H ERIOT-WATT U NIVERSITY

(4.5)

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Equation 4.4 and Equation 4.5 give us the minimum and maximum values for the tension in the string. When the string is horizontal, there is no component of the weight acting towards the centre of the circle, so the tension in the string provides all the centripetal force, Thoriz. = mrω 2

(4.6)

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

Motion in a vertical circle This is a paper-based calculation involving motion in a vertical circle. 15 min

A man has tied a 1.2 kg mass to a piece of rope 0.80 m long, which he is twirling round in a vertical circle, so that the rope remains taut. He then starts to slow down the speed of the mass. 1. At what point of the circle is the rope likely to go slack? 2. What is the speed (in m s-1 ) at which the rope goes slack?

You should be able to explain why the rope will go slack when the speed of the mass drops below a critical value, and at which point in the circle it will go slack. ..........................................

Quiz 2 Horizontal and vertical motion 20 min

First try the questions. If you get a question wrong or do not understand a question, there are Hints. The hints are in the same order as the questions. If you read the hint and still do not understand then ask your tutor. All references in the hints are to online materials. Q6: What is the centripetal force acting on a 2.5 kg mass moving in a circle of radius 4.0 m when its speed is 8.0 m s -1 ? a) b) c) d) e)

5.0 N 6.3 N 40 N 102 N 640 N

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.......................................... Q7: A centripetal force of 25 N causes a 1.4 kg mass to move in a circular path of radius 0.50 m. What is the angular velocity of the mass? a) b) c) d) e)

1.8 rad s-1 3.0 rad s-1 6.0 rad s-1 8.9 rad s-1 36 rad s-1 ..........................................

Q8: Consider an object of fixed mass m moving in a circle with a constant tangential speed v. Which one of the following statements is true? a) b) c) d) e)

Since the tangential speed is constant, there is no centripetal force. The centripetal force increases if the radius increases. The centripetal acceleration increases if the centripetal force decreases. If the radius of the circle is increased, the centripetal force decreases. There is no centripetal acceleration. ..........................................

Q9: A 1.20 kg mass on a 2.00 m length of string is being whirled in a horizontal circle. If the maximum tension in the string is 125 N, what is the maximum possible speed of the mass? a) b) c) d) e)

8.66 m s-1 14.4 m s-1 17.3 m s-1 20.4 m s-1 208 m s-1 ..........................................

Q10: A mass on a string is being whirled around in a vertical circle. At which point in its motion is the string most likely to snap? a) b) c) d) e)

When the mass is at the bottom of the circle. When the mass is at the top of the circle. When the string is horizontal. When the string is at 45◦ to the horizontal. There is an equal chance of the string snapping everywhere around the circle. ..........................................

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4.4

Applications



Learning Objective To calculate the centripetal force and acceleration in some practical situations.



The centripetal force and acceleration is calculated in this section for two situations: • conical pendulum • cars cornering

4.4.1

Conical pendulum

The next situation we will study is the conical pendulum - a pendulum of length l whose bob moves in a circle of radius r at a constant height. Figure 4.6 shows such a pendulum, with a free-body diagram of all the forces acting on the bob. Figure 4.6: (a) Conical pendulum moving in a horizontal circle (b) free-body diagram showing the forces acting on the bob

φ

l

φ

T

T

mg

a

r

mg

a

ω

=

>

.......................................... The string of the pendulum makes an angle φ with the vertical, such that sin φ = r/l. To calculate this angle, we use the free-body diagram in Figure 4.6(b) and resolve vertically and horizontally. Vertically, the system is in equilibrium, so T cos φ = mg

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Horizontally, we have an acceleration towards the centre of the circle, so there must be a centripetal force which is provided by the horizontal component of the tension, T sinφ, so T sin φ =mrω 2 ∴ T sin φ =ml sin φω 2 ∴ T =mlω 2 We can substitute for T = mlω 2 in the first equation, which gives us mlω 2 cos φ =mg ∴ lω 2 cos φ =g

(4.7)

g ∴ cos φ = 2 lω .......................................... We have an inverse square dependence of cosφ on the angular velocity. If ω increases, cosφ decreases and hence φ increases. The faster the bob is moving, the closer to horizontal the string becomes.

Conical Pendulum At this stage there is an online activity. This simulation shows a practical example of a conical pendulum. As the speed increases, a larger centripetal force is required, meaning that the horizontal component of the tension in the string must increase. .......................................... Example Consider a conical pendulum of length 1.0 m. Compare the angle the string makes with the vertical when the pendulum completes exactly 1 revolution and 2 revolutions per second (ω = 2π rad s-1 and ω = 4π rad s-1 ). In the first instance, when ω = 2π rad s-1 cos φ =

g lω 2

9.8 1.0 × (2π)2 9.8 ∴ cos φ = 2 4π ∴ cos φ = 0.248 ∴ cos φ =

∴ φ = 76◦ © H ERIOT-WATT U NIVERSITY

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When ω = 4π rad s-1 cos φ =

g lω 2

9.8 1.0 × (4π)2 9.8 ∴ cos φ = 16π 2 ∴ cos φ = 0.062 ∴ cos φ =

∴ φ = 86◦ Clearly the angle φ approaches 90 ◦ as the speed of the pendulum bob gets higher and higher. ..........................................

4.4.2

Cars cornering

When a car takes a corner, the frictional force between the car wheels and the road provides the centripetal force. If there is insufficient friction the car will skid. If we know the size of the frictional force, we can calculate the maximum speed at which a car can safely negotiate a corner. Figure 4.7: (a) A car taking a corner; (b) the forces acting on the car

4

=

H

.

B

9

=

>

.......................................... Figure 4.7(a) shows a car of mass m taking a corner. The radius of the bend is r. A free-body diagram of the forces acting on the car is shown in Figure 4.7(b). The centripetal force is provided by the frictional force Ff . Resolving vertically, using Newton's third law of motion, the normal reaction force R is equal to the weight W = mg. The maximum value of the static frictional force Ff,max is given by the relationship Ff,max = μs × R, where μs is the coefficient of static friction. Since this frictional force provides the centripetal force acting on the car, the maximum speed at which the car can take the corner is given by

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2 mvmax r 2 mvmax ∴ μs R = r 2 mvmax ∴ μs mg = r √ ∴ vmax = μs gr ..........................................

Fmax =

(4.8)

Notice that the maximum speed does not depend on the mass of the car. The coefficient of friction is the important quantity, and this coefficient becomes extremely small when the road is icy.

Banked corners At this stage there is an online activity. This simulation shows how a vehicle can take a corner when there is no frictional force present. The example shown here is a bobsleigh going around a banked corner. The horizontal component of the normal reaction force provides the centripetal force. The simulation begins by showing a free-body diagram of a mass on a smooth slope to help you visualise this horizontal force.

20 min

On a banked corner, a component of the normal reaction force provides the centripetal force. A steeper bank allows the corner to be taken at a greater speed. ..........................................

Quiz 3 Conical pendulum and cornering First try the questions. If you get a question wrong or do not understand a question, there are Hints. The hints are in the same order as the questions. If you read the hint and still do not understand then ask your tutor. All references in the hints are to online materials. Q11: The string of a conical pendulum makes an angle φ with the vertical, and the tension in its string is T. What is the centripetal force acting on the bob, in terms of T and φ? a) b) c) d) e)

T ×φ T × cos φ T/ cos φ T × sin φ T/ sin φ

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.......................................... Q12: Consider a conical pendulum of length 50 cm. What angle does the string make with the vertical when the angular speed of the pendulum is 6.0 rad s -1 ? a) b) c) d) e)

0.55◦ 33◦ 57◦ 87◦ 89◦ ..........................................

Q13: The conical pendulum in question 11 is rotating at constant angular velocity. What is the period of the pendulum? g a) 2π l cos φ b) 2π g/l cos φ l cos φ/ c) g φ d) 2π l cos g e) 2π l cos φ/g

.......................................... Q14: For a car cornering on a flat (unbanked) road, the maximum speed at which the car can take the corner without slipping depends on a) b) c) d) e)

the coefficient of static friction between the car and the road. the mass of the car. the weight of the car. the radius of the car wheels. the centre of mass of the car. ..........................................

Q15: A car can take a corner on a banked road at a higher speed than on a horizontal road because a) b) c) d) e)

the weight of the car increases. less centripetal force is required. the radius of the corner decreases. a component of the normal reaction force contributes to the centripetal force. the mass of the car decreases. ..........................................

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4.5

63

Summary

The centripetal acceleration and centripetal force of an object moving in a circular path depend on the speed or angular velocity of the object and the radius of the circle. Without this force, the object would move off in a straight line at a tangent to the circle. We have looked at several situations involving circular motion. In each case a free-body diagram has been used to calculate the tensions and other forces that supply the centripetal force. Before carrying out the self-assessment test for this topic you should make sure you understand the forces involved in each of the examples detailed in Section 4.4. Can you think of other situations involving circular motion, and the forces that provide the centripetal force? By the end of this topic you should be able to: • derive the expressions for centripetal motion in terms of v and ω; • calculate the centripetal acceleration and centripetal force of an object undergoing circular motion; • use free-body diagrams to calculate centripetal forces; • describe a number of real-life situations in which the centripetal force plays an important role.

4.6

End of topic test

End of topic test At this stage there is an end of topic test available online. If however you do not have access to the internet you may try the questions which follow. The following data should be used when required: acceleration due to gravity g = 9.8 m s -2 Q16: A cyclist is riding in a circle of radius 18 m at an average speed of 10.5 m s

-1

.

Calculate the cyclist's centripetal acceleration, in m s -2 . .......................................... Q17: A disc is rotating at constant angular velocity about its centre. At the edge of the disc, the centripetal acceleration is 6.4 m s -2 , whilst at a point 5.0 cm from the centre of the disc, the centripetal acceleration is 5.6 m s -2 . Calculate the radius of the disc, in cm. .......................................... Q18: A 1.25 kg mass is moving in a circular path of radius 1.72 m, with periodic time 1.65 s. Calculate the centripetal force in N acting on the mass. .......................................... © H ERIOT-WATT U NIVERSITY

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Q19: A model aeroplane (mass 0.80 kg) can fly at up to 26 m s on a guideline which has a breaking tension of 140 N.

-1

in a horizontal circle

Calculate the minimum radius (in m) in which the plane could be flown if it is travelling at its maximum speed. .......................................... Q20: A conical pendulum consists of a string of length 1.4 m and a bob of mass 0.25 kg. The string makes an angle of 25 ◦ with the vertical. 1. Calculate the tension in the string, in N. 2. Calculate the angular velocity of the pendulum, in rad s

-1

.

.......................................... Q21: A motorcyclist is approaching a hump-backed bridge, as shown in the diagram. The bridge forms part of a circle of radius r = 18 m. The combined mass of the motorcycle and rider is 155 kg.

Calculate the maximum speed in m s the road at the top of the bridge.

-1

at which the rider could travel without leaving

.......................................... Q22: The static frictional force between a car of mass 2000 kg and a dry road has a maximum value of 1.71 × 10 4 N. Calculate the maximum speed (in m s -1 ) at which the car could take an unbanked bend of radius 40.0 m without skidding. ..........................................

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Topic 5

Rotational dynamics

Contents 5.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.2 Torque and moment . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

66 66

5.3 Newton's laws applied to rotational dynamics . . . . . . . . . . . . . . . . . . . 5.3.1 Torques in equilibrium . . . . . . . . . . . . . . . . . . . . . . . . . . . .

70 70

5.3.2 Angular acceleration and moment of inertia . . . . . . . . . . . . . . . .

72

5.4 Angular momentum and kinetic energy . . . . . . . . . . . . . . . . . . . . . . . 5.5 Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

78 82

5.6 End of topic test . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

83

Prerequisite knowledge • Applying Newton's laws of motion. • Angular velocity and acceleration (Mechanics topics 3 and 4). • Understanding of the principle of conservation of linear momentum. Learning Objectives By the end of this topic, you should be able to: • describe what is meant by the torque or moment of a force, and calculate simple torques; • understand what is meant by moment of inertia, and calculate the moment of inertia of a body or system of bodies; • apply Newton's laws of motion to rotational dynamics; • describe what is meant by angular momentum, calculate the angular momentum of a rotating body, and apply the principle of conservation of angular momentum; • state the difference between translational and rotational kinetic energy, and calculate both for a rolling body.

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5.1

Introduction

In the previous topic we derived kinematic relationships for describing circular motion which compared directly to the linear kinematic relationships. We will be performing a similar exercise in this topic for other aspects of rotational dynamics. We will begin by looking at the turning effect (torque) of a force being used to make an object rotate about an axis. This will lead us to the concept of a moment of inertia. Using the moment of inertia we can find equations for angular momentum, Newton's second law, and kinetic energy, which again compare directly to the linear equations with which you should be familiar.

5.2

Torque and moment



Learning Objective To understand what is meant by the torque or moment of a force, and to be able to calculate simple torques.



We will begin by studying the turning effect of a force, and the example we shall look at is a force being used to push a door open. The axis of rotation is the door hinge. What happens if you push the door in the same place with different forces? Figure 5.1: (a) Small force, and (b) large force used to push open a door

.......................................... Since F = ma, the door opens more quickly the larger the force you use, something you can easily check for yourself! The turning effect increases when the force is increased. Now think about what happens if you apply the same force near the hinge or near the edge of the door.

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Figure 5.2: Same force being applied (a) far from the axis, and (b) close to the axis

.......................................... The force has a greater turning effect when it is applied near the edge of the door - it is harder to open the door when you push it near the hinge. So the turning effect also depends on how far from the axis of rotation the force is being applied. The turning effect of a force is called the torque T or moment, and is defined by T = Fr

(5.1)

.......................................... In this equation F is the size of the force and r is the distance between the axis and the point where the force is being applied. Since the torque is producing a circular motion, r is the radius of the motion. T has the units N m. Looking at Figure 5.3 below, we can see that Equation 5.1 does not quite tell the full story. This equation is only valid when the force is applied at right angles to the radius. Figure 5.3: Same force applied (a) perpendicular to the door, and (b) at an angle to the door

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The dashed line in Figure 5.3 represents the line of action of the force. When calculating the torque, the important distance is not where the force is applied, but the perpendicular distance from the line of action to the axis of rotation. If F makes an angle φ to the line r joining the axis to the point where the force is applied, then the torque is given by T = F r sin φ

(5.2)

.......................................... Example A gardener picks up her wheelbarrow to wheel it along by applying a perpendicular force of 50 N to the handles. If the handle grips are 1.2 m from the axis of the wheel, what is the torque that she applies? The situation is shown schematically in Figure 5.4 Figure 5.4: Torque applied to the wheelbarrow

.......................................... Since a perpendicular force is applied, we can use Equation 5.1 T = Fr ∴ T = 50 × 1.2 ∴ T = 60 N m ..........................................

Quiz 1 Torques 20 min

First try the questions. If you get a question wrong or do not understand a question, there are Hints. The hints are in the same order as the questions. If you read the hint © H ERIOT-WATT U NIVERSITY

TOPIC 5. ROTATIONAL DYNAMICS

and still do not understand then ask your tutor. All references in the hints are to online materials. Q1: Which of the following sets of dimensions could be used to measure a torque? a) b) c) d) e)

kg m kg m2 N kg m2 N m-1 Nm ..........................................

Q2: What is the torque when a 35 N force is applied at a distance 1.4 m from an axis, perpendicular to the radius? a) b) c) d) e)

0.04 N m 25 N m 35 N m 49 N m 69 N m ..........................................

Q3: A 16.0 N force produces a 60.0 N m torque when applied at right angles to the radius of rotation. How far from the axis is the force applied? a) b) c) d) e)

3.75 m 26.7 m 48.0 m 60.0 m 960 m ..........................................

Q4: A force of 12.0 N is applied 0.750 m from the axis to produce a torque. The force is applied at an angle of 70 ◦ to the radius of motion. What is the size of the torque? a) b) c) d) e)

0.156 N m 3.08 N m 8.46 N m 9.00 N m 630 N m ..........................................

Q5: A torque of 48.0 N m is produced when a force is applied 0.500 m from an axis. If the force makes an angle of 60 ◦ with the radius, what is the magnitude of the force? a) b) c) d) e)

20.8 N 83.1 N 96.0 N 111 N 192 N ..........................................

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5.3

Newton's laws applied to rotational dynamics



Learning Objective To apply Newton's laws of motion to rotational dynamics. To calculate the moment of inertia of a body or system of bodies.



We are used to the idea of forces in equilibrium producing no acceleration, and a resultant force producing an acceleration in the direction of the force. A similar situation exists when we are applying torques instead of forces. A system of torques in equilibrium will produce no net turning effect, whilst a resultant torque produces an angular acceleration about the axis.

5.3.1

Torques in equilibrium

To consider the equilibrium situation, let us return to the problem of the wheelbarrow. We will assume the barrow stays horizontal, and we will remove its support so that the gardener has to apply a perpendicular force to keep the barrow horizontal. Suppose the barrow is loaded with 20 kg of rubble, and the centre of mass of this rubble is 50 cm from the wheel axis. If we can neglect the mass of the barrow, what force is necessary to keep it horizontal? Figure 5.5: Torques in equilibrium

.......................................... For the above system to be in equilibrium, the clockwise torque (due to the mass of the rubble) must equal the anti-clockwise torque (due to the gardener applying a force). Therefore

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TOPIC 5. ROTATIONAL DYNAMICS

Ta−c = Tc ∴ F × 1.2 = mg × 0.50 ∴ F × 1.2 = 20 × 9.8 × 0.50 ∴ F × 1.2 = 98 ∴ F = 82 N The problem is similar to a Newton's first law problem, and the method of solving it is the same: draw a free-body diagram to show all the forces acting. The difference is that we are no longer considering point objects, so we must also include where the forces are acting on the diagram. In the above problem, we have been told to ignore the weight of the barrow. In a situation where this cannot be ignored, the weight acts through the centre-of-mass of the object, and the distance from the axis to the centre-of-mass must be known to calculate the torque. The next example illustrates how to solve more complicated problems. Example A serving table of mass 2.5 kg and length 80 cm is hinged to a wall, and is supported by a chain which makes an angle of 50 ◦ with the horizontal table top. The chain is attached to the edge of the table furthest from the wall (r c = 80 cm). The table is uniform, so its centre-of-mass is rt = 40 cm from the wall. A full serving dish of mass 1.0 kg is placed on the table r d = 60 cm from the wall. Calculate the tension in the chain. Figure 5.6: Free-body diagram showing the forces acting on the table

.......................................... © H ERIOT-WATT U NIVERSITY

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In equilibrium, the anti-clockwise torques must balance the clockwise torques. This means the torques due to the two masses are balanced by the torque due to the tension in the chain. F × sin φ × rc = (md × g × rd ) + (mt × g × rt ) ∴ F × 0.766 × 0.80 = (1.0 × 9.8 × 0.60) + (2.5 × 9.8 × 0.40) ∴ 0.6128F = 5.88 + 9.8 ∴ 0.6128F = 15.68 ∴ F = 25.6 N .......................................... If you try this calculation again with the dish at a different position, you will find that the tension in the chain depends on whereabouts on the table the dish is positioned. If the dish is placed on the edge of the table, (r d = 80 cm), then the tension F = 28.8 N, whereas placing the dish at r d = 20 cm from the wall reduces F to 19.2 N. Clearly, when we are considering torques, we are not just interested in the mass of an object or objects, but how that mass is distributed relative to the axis.

Torque and static equilibrium At this stage there is an online activity which allows you to calculate and balance torques. 20 min

You should be able to calculate torques about an axis, and balance clockwise and anticlockwise torques. The distance of a force or a mass from the axis of rotation is just as important as the magnitude of the force. ..........................................

5.3.2

Angular acceleration and moment of inertia

Let us now consider a Newton's second law-type problem. We are used to a force producing an acceleration, and the relationship F = ma. We will now see if there is an equivalent expression involving torques and angular acceleration. The problem we will look at is that of an object moving in a horizontal circle of fixed radius r at an angular velocity ω. Figure 5.7: (a) Object moving around a circle; (b) a tangential force applied to the object .

ω

H

=

>

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TOPIC 5. ROTATIONAL DYNAMICS

73

.......................................... If we apply an anticlockwise force F to the object at a tangent to the circle, as shown in Figure 5.7(b), the object will accelerate. We can use Newton's second law F = ma We can express this acceleration as an angular acceleration using the expression a = rα. Hence F = mrα Since the force is being applied at a distance r from the axis of rotation, perpendicular to the radius, we can substitute for F using T = F r T = mrα r ∴ T = mr 2 α

(5.3)

∴ T = Iα .......................................... We have used the symbol I to represent mr 2 to give us an expression in Equation 5.3 which looks similar to F = ma . The quantity I is called the moment of inertia, and for a point mass m rotating with radius r about an axis. I = mr 2

(5.4)

.......................................... I has the units kg m2 . Throughout this topic we will be using I in our calculations, as it takes into account not only the mass of an object, but also how far the mass is from the axis of rotation. There are some important properties about the moment of inertia that you must make sure you understand. First, we cannot state a fixed value of I for any object. The value of I depends not only on the mass of the object, but the position of the axis. The axis must be specified when giving the moment of inertia of an object. Although there are similarities between mass and moment of inertia, there is a major difference in that the mass of a body is constant but the moment of inertia changes for different axes of rotation. Secondly, the equation

I = mr 2

is only valid for a point mass with a known position. For a collection of masses m 1 , m2 , m3 ...mi , the total moment of inertia is Itotal = m1 r12 + m2 r22 + m3 r32 + ..... =

 i

mi ri2

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(5.5)

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Calculating the moment of inertia 20 min

This is a paper-based calculation. Different masses are placed on a rotating platform to study the moment of inertia of a system. Investigate how the same moment of inertia can be obtained with different combinations of mass and position. We will be considering a platform of negligible mass, with an axis through its centre. The radius of the platform is 50 cm. Q6: What is the moment of inertia when a 1.0 kg mass is placed 40 cm from the axis as shown?

.......................................... Q7: If we remove the 1.0 kg mass, where should a 4.0 kg mass be placed on the platform to produce the same moment of inertia? .......................................... Q8: Suppose both masses are placed on the platform, with the 4.0 kg mass 30 cm from the centre and the 1.0 kg mass 20 cm from the centre. What is the moment of inertia now? .......................................... Q9: Calculate the moment of inertia of the following system shown: a 1.0 kg mass 10 cm from the centre, a 2.5 kg mass 20 cm from the centre and a 2.0 kg mass 40 cm from the centre.

.......................................... Q10: Finally, consider the masses in question 4. If they were combined to form a single object of mass 5.5 kg, where should it be placed to give the same moment of inertia as obtained in question 4? © H ERIOT-WATT U NIVERSITY

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.......................................... You should understand what is meant by the moment of inertia I of a single or manybody system, and be able to calculate I for different systems.

Calculations of the moment of inertia of solid bodies are complicated. Using a calculus approach, the body is divided into infinitely small sections. I is calculated for each section and then an integration is carried out to find the total moment of inertia. Such a calculation is beyond the scope of this course. For a number of standard shapes, the moments of inertia are well-known, and will be given to you when you need them. Example A Catherine wheel consists of a light rod 0.64 m in length, pivoted about its centre. At either end of the rod a 0.50 kg firework is attached, each of which provides a force of 25 N perpendicular to the rod when lit. Calculate the angular acceleration of the Catherine wheel when the fireworks are lit. The angular acceleration is calculated using the equation T = Iα , so we need to calculate the moment of inertia of the Catherine wheel and the total torque provided by the two fireworks. Figure 5.8: Catherine wheel diagram

.

 

 

.

.......................................... The total moment of inertia is I=



mr 2

    ∴ I = 0.50 × 0.322 + 0.50 × 0.322 ∴ I = 0.1024 kg m2 The total torque is the sum of the torques provided by each firework T =



Fr

∴ T = (25 × 0.32) + (25 × 0.32) ∴ T = 16 N m © H ERIOT-WATT U NIVERSITY

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The angular acceleration can now be calculated α=

T I

16 0.1024 ∴ α = 160 rad s−2 .......................................... ∴α=

Extra Help: Solving problems with more than 1 torque There is an online activity to provide practice. ..........................................

Quiz 2 Equilibrium and moment of inertia 20 min

First try the questions. If you get a question wrong or do not understand a question, there are Hints. The hints are in the same order as the questions. If you read the hint and still do not understand then ask your tutor. All references in the hints are to online materials. Q11: What size of force applied 1.60 m from an axis would be needed to maintain equilibrium against an oppositely-directed 6.50 N force applied 2.40 m from the axis? a) b) c) d) e)

0.103 N 4.33 N 6.50 N 9.75 N 25.0 N ..........................................

Q12: A girl of mass 35 kg sits on one side of a see-saw, 1.8 m from the fulcrum. Where on the other side of the fulcrum should a 45 kg boy sit to balance the see-saw? a) b) c) d) e)

0.43 m from the fulcrum 0.71 m from the fulcrum 1.4 m from the fulcrum 1.8 m from the fulcrum 2.3 m from the fulcrum

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.......................................... Q13: Consider the hinged table in the diagram below:

The weight of the table is 15 N, and acts through the centre-of-mass 50 cm from the axis. A load of 40 N sits 20 cm from the axis. The chain, 1.0 m from the axis, makes an angle of 60◦ with the table. Calculate the tension F in the chain. a) b) c) d) e)

15.5 N 18 N 25 N 31 N 55 N ..........................................

Q14: An object of mass 2.20 kg is moving in a circle of radius 1.30 m. What is the moment of inertia of the mass about the centre of the circle? a) b) c) d) e)

1.30 kg m2 1.69 kg m2 2.20 kg m2 2.86 kg m2 3.72 kg m2 ..........................................

Q15: Masses A (1.20 kg) and B (2.00 kg) are placed on a platform of negligible mass, which can rotate about an axis XY. A is positioned 10.0 cm from XY and B sits 25 cm from XY. Calculate the total moment of inertia about the axis XY. a) b) c) d)

0.137 kg m2 0.264 kg m2 0.392 kg m2 0.620 kg m2

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e) 3.20 kg m2 ..........................................

5.4

Angular momentum and kinetic energy



Learning Objective To calculate the angular momentum and rotational kinetic energy of a rotating object.



The angular momentum L of a body rotating with moment of inertia I and angular velocity ω is defined as L = Iω

(5.6)

.......................................... L is measured in units of kg m2 s-1 . Equation 5.6 is analogous to the expression p = mv for linear momentum, and the same conservation of momentum principle applies. In this case, we can state that the angular momentum of a rotating rigid body is conserved unless an external torque acts on the body. If we want to describe the angular momentum at a point, then we can substitute for I = mr 2 and ω = v/r in Equation 5.6 L = Iω ∴ L = mr 2 ω

L = mr 2 × v/r

∴ L = mvr

The conservation of angular momentum has some interesting consequences. Since angular momentum depends on the moment of inertia, then it depends on the distribution of mass about the axis. If the moment of inertia increases, the angular velocity must decrease to keep the angular momentum constant, and vice versa. This effect is used by ice skaters to speed up or slow down as they spin, and the same effect can be seen if you spin on a swivel chair. Flinging out your arms and legs moves more of your body mass away from the axis of rotation. Your moment of inertia increases, so your angular velocity must decrease. (This works even better if you have a heavy book in each hand!) If you then hunch up on the chair, your moment of inertia decreases, and hence your angular velocity increases. Example An ice skater is spinning with his arms extended at an angular velocity of 7.0 rad s -1 . What is his angular velocity if he pulls his arms in to his sides? The moment of inertia of the skater is 4.7 kg m2 when his arms are extended, and 1.8 kg m 2 when they are by his sides. The angular momentum of the skater must be conserved so the initial angular momentum must equal the final angular momentum © H ERIOT-WATT U NIVERSITY

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Li = Lf ∴ Ii ω i = If ω f ∴ 4.7 × 7.0 = 1.8 × ωf 4.7 × 7.0 ∴ ωf = 1.8 ∴ ωf = 18 rad s−1 Due to the conservation of angular momentum the skater's angular velocity increases when he pulls his arms in, since he has reduced his moment of inertia about the axis of rotation. .......................................... There are also expressions for calculating work done and kinetic energy in rotational motion, that are again analogous to expressions used to describe linear motion. You should be familiar with the expression work done = force × displacement. In the case of rotational work, the equivalent expression is work done = T × θ

(5.7)

.......................................... In Equation 5.7, T is the torque and θ is the angular displacement. This equation tells us the work done when a torque T applied to an object produces an angular displacement θ of the object. A spinning body has kinetic energy. The rotational kinetic energy of a body is given by KErot = 12 Iω 2

(5.8)

.......................................... Note that this energy is due to the rotation about an axis. If we consider a wheel spinning about its axis, then this is the total kinetic energy. What if the wheel is rolling along the ground? In this case the wheel has both rotational kinetic energy, due to its turning motion, and translational kinetic energy due to its moving along the ground. The total kinetic energy in this case is KEtot = KErot + KEtrans ∴ KEtot = 12 Iω 2 + 12 mv 2 Note that the translational speed is the same as the tangential speed at the circumference of the wheel, so long as the wheel does not slip. If you cannot see why, imagine the wheel rotating through one complete circle. A point on the circumference moves through a distance 2πr in one revolution. The wheel must have rolled the same distance, and it has done so in the same period of time. The translational speed must therefore be equal to the tangential speed at the circumference.

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Example A solid sphere of radius r and mass m has moment of inertia I = 25 mr 2 about any axis though its centre. Calculate the translational and rotational kinetic energies of a sphere of mass 0.40 kg and radius 20 cm rolling (without slipping) along a horizontal surface with translational speed 5.0 m s -1 . The translational kinetic energy is KEtrans = 12 mv 2 ∴ KEtrans =

1 2

× 0.40 × 5.02

∴ KEtrans = 5.0 J To find the rotational kinetic energy, we need to find the moment of inertia I and angular velocity ω of the sphere. If the sphere has a translational speed of v m s -1 , then a point on the circumference of the sphere must also have speed v. The angular velocity is therefore equal to v/r , which in this case is

ω=

5.0 v = = 25 rad s−1 r 0.20 .

The rotational kinetic energy is KErot = 12 Iω 2 ∴ KErot = ∴ KErot =

1 2 1 5

× 25 mr 2 × ω 2

× 0.40 × 0.202 × 252

∴ KErot = 2.0 J ..........................................

Total energy of a rolling body At this stage there is an online activity. 30 min

This activity allows the user to plot out the energy changes that occur for different objects rolling down a slope. A rolling body has both translational and rotational kinetic energy. When a body rolls down a slope, the gravitational potential energy of the body at the top of the slope is converted into rotational and translational kinetic energy as it rolls down. If you know the moment of inertia of the body, you should be able to calculate the two kinetic energies of the body. .......................................... © H ERIOT-WATT U NIVERSITY

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Quiz 3 Angular momentum and rotational kinetic energy First try the questions. If you get a question wrong or do not understand a question, there are Hints. The hints are in the same order as the questions. If you read the hint and still do not understand then ask your tutor. All references in the hints are to online materials. Q16: A solid object with moment of inertia 2.40 kg m 2 is rotating at 3.00 rad s-1 . What is its angular momentum? a) b) c) d) e)

0.800 kg m2 s-1 1.25 kg m2 s-1 1.92 kg m2 s-1 7.20 kg m2 s-1 17.3 kg m2 s-1 ..........................................

Q17: A disc of mass m and radius r spinning about an axis through its centre has moment of inertia I = 12 mr 2 . What is the angular momentum of a disc (m = 0.400 kg, r = 0.25 m) spinning at 20 rad s -1 ? a) b) c) d) e)

0.05 kg m2 s-1 0.25 kg m2 s-1 5.0 kg m2 s-1 8.0 kg m2 s-1 250 kg m2 s-1 ..........................................

Q18: A horizontal turntable is rotating on a frictionless mount at constant angular velocity. What happens if a lump of clay is dropped onto the turntable, sticking to it? a) b) c) d) e)

The turntable would slow down. The turntable would speed up. The turntable would continue rotating at the same speed. The lump of clay would rotate in the opposite direction to the turntable. The turntable would stop rotating but the lump of clay would continue to rotate. ..........................................

Q19: A square sheet of mass m and side length l has a moment of inertia I = 13 ml2 when rotating about one edge. What is the rotational kinetic energy of a square sheet of mass 1.0 kg and side 0.50 m rotating at 0.60 rad s -1 about one edge? a) b) c) d) e)

9.0 × 10-4 J 4.2 × 10-3 J 0.015 J 0.030 J 0.18 J ..........................................

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Q20: A rotating platform is spinning at a rate of 1.50 revolutions per second. If the moment of inertia of the platform is 4.20 kg m 2 , what is the rotational kinetic energy of the platform? a) b) c) d) e)

4.73 J 19.8 J 39.6 J 47.3 J 187 J ..........................................

5.5

Summary

Using the moment of inertia allows us to describe the angular motion of rigid bodies in the same way that we describe linear motion of point objects. The same conservation rules apply, and the following tables should clarify the different quantities and relationships: Table 5.1: Kinematic relationships linear motion v = u + at v 2 = u2 + 2as s = ut + 12 at2

angular motion ω = ω0 + αt ω 2 = ω02 + 2αθ θ = ω0 t + 12 αt2

.......................................... Table 5.2: Linear and angular motion linear motion Displacement s Force F Velocity v Acceleration a

angular motion Angular displacement θ Torque T Angular velocity ω Angular acceleration α

.......................................... Table 5.3: Newton's second law and kinetic energy linear motion Newton's second law: F = ma Momentum p = mv Work done: W = F s Translational kinetic energy KEtrans = 12 mv 2

angular motion Newton's second law: T = Iα Angular momentum L = Iω Work done: W = T θ Rotational kinetic energy KErot = 12 Iω 2

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

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The moment of inertia of a rigid body depends on the mass of the body, and how that mass is distributed about the axis of rotation. The axis must be specified when a moment of inertia is being calculated. For a collection of point masses, the total moment of inertia is the sum of the individual moments. By the end of this topic, you should be able to: • state what is meant by the moment of a force, and to state and apply the equation T = F r; • state that an unbalanced torque acting on an object produces an angular acceleration, and that this angular acceleration depends on the moment of inertia of the object; • explain that the moment of inertia of an object about an axis depends on the mass of the object, and the distribution of the mass about the axis; • state that the moment of inertia I of an object of mass m at a distance r from a fixed axis is given by the equation I = mr 2 • state and apply the equation T = Iα; • state that the angular momentum L of a rigid body is given by the equation L = Iω and state that in the absence of external torques, L is conserved; • state the expression KErot = 12 Iω 2 for the rotational kinetic energy of a rigid body, and carry out calculations using this relationship.

5.6

End of topic test

End of topic test At this stage there is an end of topic test available online. If however you do not have access to the internet you may try the questions which follow. The following data should be used when required: acceleration due to gravity g

9.8 m s -2

moment of inertia of a solid cylinder or disc about an axis through 1 / Mr2 2 its centre moment of inertia of a hollow cylinder or disc about an axis through Mr 2 its centre moment of inertia of a solid sphere about an axis through its centre 2 /5 Mr 2 © H ERIOT-WATT U NIVERSITY

30 min

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Q21: A spanner, length 40 cm, is being used to try to undo a nut stuck on a bolt. A force of 60 N is applied at the end of the spanner. Calculate the moment (in Nm) which is applied. .......................................... Q22: A torque of 64 Nm is applied to a solid flywheel of moment of inertia 30 kg m 2 . Calculate the resulting angular acceleration, in rad s

-2 .

.......................................... Q23: Three objects are arranged on a circular turntable of negligible mass, which is able to rotate about an axis through its centre. Mass A ( 0.46 kg) is 20 cm from the axis, mass B ( 0.63 kg) is 40 cm from the axis, and mass C ( 0.22 kg) is 70 cm from the axis. Calculate the total moment of inertia of the loaded turntable about the central axis, in kg m 2. .......................................... Q24: A solid sphere, radius 20.0 cm and mass 1.06 kg, is rotating about an axis through its centre with angular velocity 5.15 rad s -1 . Calculate the rotational kinetic energy of the sphere, in J. .......................................... Q25: A flat horizontal disc of moment of inertia 1.4 kg m 2 is rotating at 4.5 rad s -1 about a vertical axis through its centre. A 0.13 kg mass is dropped onto the disc, landing without slipping 1.7 m from the centre. Calculate the new angular velocity of the disc, in rad s

-1 .

.......................................... Q26: A solid cylinder of radius 0.81 m and mass 4.7 kg is at rest. A 4.0 N m torque is applied to the cylinder about an axis through its centre. Calculate the angular velocity of the cylinder, in rad s applied for 2.0 s.

-1

, after the torque has been

.......................................... Q27: A solid cylinder of mass 2.7 kg and radius 1.5 m is rolling without slipping along a horizontal road. The translational speed of the cylinder is 5.4 m s -1 . 1. Calculate the angular velocity of the cylinder, in rad s

-1 .

2. Calculate the total kinetic energy of the cylinder, in J. ..........................................

© H ERIOT-WATT U NIVERSITY

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Topic 6

Gravitational force and field

Contents 6.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.2 Newton's law of gravitation . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

86 86

6.2.1 The Cavendish-Boys experiments to determine G . . . . . . . . . . . . 6.3 Weight . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

88 89

6.4 Gravitational fields . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

92

6.5 Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.6 End of topic test . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

96 97

Prerequisite knowledge • The prerequisite for this topic is an understanding of Newton's laws of motion. Learning Objectives By the end of this topic, you should be able to: • describe the gravitational force that exists between any two bodies; • understand the concept of a gravitational field.

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6.1

Introduction

In this topic we will be investigating gravity and the gravitational force that exists between any two objects. As well as calculating the force, the concept of a gravitational field will be introduced. Throughout this topic, some approximations will be made with regard to planets and their orbits. In all the examples and exercises in this topic, it will be assumed that the Sun, the Moon, the Earth and other planets are all spherical objects. Furthermore, we will be treating their orbits as circular.

6.2

Newton's law of gravitation



Learning Objective To state and apply Newton's Universal Law of Gravitation.



Working in the 17th century, Sir Isaac Newton discovered the Universal Law of Gravitation. He used his own observations, along with those of Johannes Kepler, who had formulated a set of laws that described the motion of the planets around the Sun. Kepler's laws will be studied in the next topic. Newton's Law of Gravitation states that there is a force of attraction between any two objects in the universe. The size of the force is proportional to the product of the masses of the two objects, and inversely proportional to the square of the distance between them. This law can be summed up in the equation Gm1 m2 r2 .......................................... F =

(6.1)

In Equation 6.1, m1 and m2 are the masses of the two objects, and r is the distance between them. The constant of proportionality is G, the gravitational constant. The value of G is 6.67 × 10-11 N m2 kg-2 . A simple example will show us the order of magnitude of this force. Example Consider two point masses, each 2.00 kg, placed 1.20 m apart on a table top. Calculate the magnitude of the gravitational force between the two masses. Using Newton's Law of Gravitation Gm1 m2 r2 6.67 × 10−11 × 2.00 × 2.00 ∴F = 1.202 −10 2.668 × 10 ∴F = 1.44 ∴ F = 1.85 × 10−10 N F =

© H ERIOT-WATT U NIVERSITY

TOPIC 6. GRAVITATIONAL FORCE AND FIELD

The gravitational force between these two masses is only 1.85 × 10 -10 N. This is an extremely small force, one which is not going to be noticeable in everyday life. .......................................... We rarely notice the gravitational force that exists between everyday objects as it is such a small force. You do not have to fight against gravity every time you walk past a large building, for example, as the gravitational force that the building exerts on you is too small to notice. Because the constant G in Newton's Law of Gravitation is so small, the gravitational force between everyday objects is usually negligible. The force only really becomes important when we are dealing with extremely large masses such as planets. The gravitational force is always attractive, and always acts in the direction of the straight line joining the two objects. According to Newton's third law of motion, the gravitational force is exerted on both objects. As the Earth exerts a gravitational force on you, so you exert an equal force on the Earth. Most of the work we will be doing on gravitation concerns the forces acting between planets and stars. So far we have only considered point objects, so do we need to adapt Newton's Law of Gravitation when we are dealing with larger bodies? The answer is no - for spherical objects (or more accurately, objects with a spherically symmetric mass distribution), the gravitational interaction is exactly the same as it would be if all the mass was concentrated at the centre of the sphere. Remember, we will assume in all our calculations that the planets and stars we are dealing with are spherical. Example The Earth has a radius of 6.38 × 10 6 m and a mass 5.97 × 10 24 kg. What is the gravitational force due to the Earth acting on a woman of mass 60.0 kg standing on the surface of the Earth? The solution is obtained by calculating the force between two point objects placed 6.38 × 106 m apart, if we treat the Earth as a uniform sphere. Hence Gm1 m2 r2 6.67 × 10−11 × 5.97 × 1024 × 60.0 ∴F = (6.38 × 106 )2 2.389 × 1016 F = 4.070 × 1013 F = 587 N F =

The gravitational force acting on the woman is 587 N, and this force is directed towards the centre of the Earth. If you calculate the force due to gravity acting on the woman by another method, using F =m × g, you should find you get a very similar answer, 588 N. The slight difference comes from using a value of g = 9.8 m s -2 . As we shall see later in this Topic, the value of g obtained by assuming the Earth to be a sphere of uniform density is slightly less than this value. © H ERIOT-WATT U NIVERSITY

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

6.2.1

The Cavendish-Boys experiments to determine G

An accurate measurement of G can be carried out using the Cavendish-Boys method. Cavendish first performed this experiment in the late 18th century and the accurate determination was performed nearly 100 years later by Boys. The experiment uses a torsion balance to measure the gravitational force between lead spheres. Figure 6.1: Cavendish-Boys experiment, viewed from (a) the side, and (b) the top

.......................................... Two identical masses (m1 ) are held close to two smaller masses (m2 ). These smaller masses are attached to the ends of a light rod of length l suspended from a torsion wire in a draft-free chamber. By reflecting a narrow beam of light from a mirror attached to the torsion wire, the angular deflection of the wire can be measured as the masses m 1 are brought close to the masses m2 . In equilibrium, the rotation caused by the gravitational force between m1 and m2 is balanced by the restoring torque (turning force) in the wire. The equipment must be calibrated first to determine the torsional constant c of the wire. The restoring torque is then equal to c × θ when the angular displacement of the rod is θ rad (see Figure 6.1(b)). Hence in equilibrium, m1 m2 l = cθ r2 .......................................... G

(6.2)

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TOPIC 6. GRAVITATIONAL FORCE AND FIELD

6.3

89

Weight





Learning Objective To calculate the weight of an object and the acceleration due to gravity.



The weight W of an object of mass m can be defined as the gravitational force exerted on it by the Earth. GmE m 2 rE .......................................... W = Fgrav =

(6.3)

In Equation 6.3, mE and rE are the mass and radius of the Earth. The acceleration due to gravity, g, is found from Newton's second law of motion F = ma ∴ W = mg We can substitute for F in this equation GmE m 2 rE GmE ∴g= 2 rE .......................................... mg =

(6.4)

So the acceleration of an object due to gravity close to the Earth's surface does not depend on the mass of the object. In the absence of friction, all objects fall with the same acceleration. Equation 6.4 is a specific equation for calculating g on the Earth's surface. In general, at a distance r from the centre of a body (a star or a planet, say) of mass m, the value of g is given by Gm r2 .......................................... g=

(6.5)

The mass of an object is constant; it is an intrinsic property of that object. The weight of an object tells us the magnitude of the gravitational force acting upon it, so it is not a constant.

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Example Compare the values of g on the surfaces of the Earth (m E = 5.97 × 1024 kg, rE = 6.38 × 106 m) and the Moon (m M = 7.35 × 1022 kg, rM = 1.74 × 106 m). To solve this problem, use Equation 6.5 with the appropriate values g= ∴g=

GmE 2 rE

g=

6.67 × 10−11 × 5.97 × 1024

(6.38 × ∴ g = 9.78 m s−2

106 )2

∴g=

GmM 2 rM

6.67 × 10−11 × 7.35 × 1022

(1.74 × 106 )2 ∴ g = 1.62 m s−2

The value for g on the surface of the Earth is 9.78 m s -2 , whilst on the surface of the Moon the value of g is 1.62 m s -2 . ..........................................

Quiz 1 Gravitational force 20 min

First try the questions. If you get a question wrong or do not understand a question, there are Hints. The hints are in the same order as the questions. If you read the hint and still do not understand then ask your tutor. All references in the hints are to online materials. Useful data: Gravitational constant G Mass of the Moon mM Radius of the Moon r M Mass of Venus mV Radius of Venus rV

6.67 × 10-11 N m2 kg-2 7.35 × 1022 kg 1.74 × 106 m 4.87 × 1024 kg 6.05 × 106 m

Q1: Two snooker balls, each of mass 0.25 kg, are at rest on a snooker table with their centres 0.20 m apart. What is the magnitude of the gravitational force that exists between them? a) b) c) d) e)

2.1 × 10-11 4.3 × 10-11 8.3 × 10-11 1.0 × 10-10 4.2 × 10-10

N N N N N

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TOPIC 6. GRAVITATIONAL FORCE AND FIELD

.......................................... Q2: The Sun exerts a gravitational force F S on the Earth. The Earth exerts a gravitational forceFE on the Sun. Which one of these statements about the magnitudes of FS and FE is true? FS = FE FS < FE FS > FE FS/ = massS/ FE mass E 2 F (mass ) S S e) /F = (massE )2 E

a) b) c) d)

.......................................... Q3: What is the weight of a 5.00 kg mass placed on the surface of the Moon? a) b) c) d) e)

0.123 N 1.42 N 1.62 N 8.10 N 40.5 N ..........................................

Q4: An object is taken from sea level to the top of Mount Everest. Which one of the following statements is true? a) b) c) d) e)

Its mass remains constant but its weight increases. Its mass remains constant but its weight decreases. Its weight remains constant but its mass increases. Its weight remains constant but its mass decreases. Neither its mass nor its weight alter. ..........................................

Q5: What is the value of the acceleration due to gravity on the surface of the planet Venus? a) b) c) d) e)

0.887 m s-2 5.37 m s-2 6.67 m s-2 8.29 m s-2 8.87 m s-2 ..........................................

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6.4 

Gravitational fields



Learning Objective To describe the concept of a gravitational field.



The region of space around an object A, in which A exerts a gravitational force on another object B placed in that region, is called the gravitational field of A. The concept of a field is used in many situations in physics, such as the electric field surrounding a charged particle, or the magnetic field around a bar magnet. The gravitational field strength g at a point in a gravitational field is defined as the gravitational force acting on a unit mass placed at that point in the field. At a distance r from a point object of mass m, the gravitational field strength g is given by Gm r2 .......................................... g=

(6.6)

The units of g are N kg-1 . These units are equivalent to m s -2 , so the gravitational field strength at a point in a gravitational field is equal to the acceleration due to gravity at that point, and Equation 6.6 is identical to Equation 6.5. This is why the same symbol g is used for both the gravitational field strength and the acceleration due to gravity. Example The Earth orbits the Sun with a mean radius of 1.50×10 11 m. What is the gravitational field strength on Earth due to the Sun, given that the mass of the Sun is 1.99×10 30 kg. In the Sun's gravitational field, the field strength is given by g=

GmS r2

So, at the location of the Earth, the field strength is g= ∴g=

GmS r2 6.67 × 10−11 × 1.99 × 1030 (1.50 × 1011 )2

∴ g = 5.90 × 10−3 N kg−1 .......................................... The gravitational field around a body is often shown diagramatically by drawing field lines. Figure 6.2 shows the gravitational field lines around a point object. The pattern of the field lines is symmetrical in three dimensions. This pattern would be exactly the same outside an object with a spherically symmetric mass distribution. The lines are symmetrical about the centre of the object and show the direction of the force exerted on any mass placed in the field. The closer together the field lines are, the greater the field strength. © H ERIOT-WATT U NIVERSITY

TOPIC 6. GRAVITATIONAL FORCE AND FIELD

Figure 6.2: Gravitational field lines around a point mass

.......................................... The gravitational field is an example of a conservative field. This is a field in which the work done in moving from one point to another in the field is independent of the path taken. Conservative fields are studied in more detail in the Gravitational Potential topic. The field lines representing the gravitational field caused by two or more objects can also be sketched. The gravitational field due to two point masses is shown in Figure 6.3. Figure 6.3: Gravitational field lines around two identical point masses

.......................................... To calculate the gravitational field strength at any point in the field due to two point masses or spheres, we take the vector sum of the two individual fields. © H ERIOT-WATT U NIVERSITY

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Example The mean distance between the centre of the Earth and the centre of the Moon is 3.84 × 108 m. Given that the mass of the Earth is 5.97 × 10 24 kg and the mass of the Moon is 7.35 × 10 22 kg, at what distance from the centre of the Earth is the point where the total gravitational field strength due to the Earth and the Moon is zero? A sketch is useful in solving this sort of problem, as it can clearly show the direction of the field vectors. Figure 6.4: Earth - Moon system

gE

gM

x

E

M

rE-M d

rE-M -d

.......................................... At position X, the point where the total gravitational field strength is zero, the gravitational field strength gE of the Earth is balanced by gM , the gravitational field strength of the Moon. Position X is a distance d from the centre of the Earth. To find the distance d, we must solve the equation gE = gM gE = gM GmE GmM ∴ = 2 d (rE−M − d)2 mE mM ∴ 2 = d (rE−M − d)2 √ √ mE mM = ∴ d (rE−M − d) √ √ ∴ mE × (rE−M − d) = mM × d √ √ √ ∴ mE × rE−M = d × ( mM + mE )   ∴ 2.44 × 1012 × 3.84 × 108 = d × 2.71 × 1011 + 2.44 × 1012 9.37 × 1020 2.71 × 1012 ∴ d = 3.46 × 108 m

∴d=

The combined gravitational field strength is zero at a point 3.46×10 8 m from the centre of the Earth. This point is very much closer to the Moon than the Earth because the Earth is much more massive than the Moon. ..........................................

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Gravitational fields At this stage there is an online activity. The gravitational field lines around an isolated object are plotted. A second object can be added to the system, and the field lines are re-plotted if the separation between the objects or their masses are changed.

20 min

With this simulation you can see the pattern of the gravitational field lines around a single object or a collection of objects. You can also view the gravitational potential, which is covered in Topic 7. Note that masses are always added at the same position, so if you do not move the masses they will all be located at the same point. This simulation shows you how the gravitational field between two objects depends on the two masses and their separation. Make sure you understand how this ties in with the equation for gravitational field strength. ..........................................

Quiz 2 Gravitational fields First try the questions. If you get a question wrong or do not understand a question, there are Hints. The hints are in the same order as the questions. If you read the hint and still do not understand then ask your tutor. All references in the hints are to online materials. Useful data: Gravitational constant G

6.67 × 10-11 N m2 kg-2

Q6: The planet Jupiter has a mass of 1.90 × 10 27 kg and a radius of 6.91 × 10 7 m. What is the gravitational field strength on the surface of Jupiter? a) b) c) d) e)

0.040 N kg-1 1.83 N kg-1 9.81 N kg-1 26.5 N kg-1 168 N kg-1 ..........................................

Q7: Which one of the following units is equivalent to the units used to express gravitational field strength? a) b) c) d) e)

m s-2 N m s-2 kg m-2 kg N-2 N s-2

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.......................................... Q8: The gravitational field strength at a distance 5.0 × 10 5 m from the centre of a planet is 8.0 N kg-1 . At what distance from the centre of the planet is the field strength equal to 4.0 N kg -1 ? a) b) c) d) e)

7.1 × 105 m 9.0 × 105 m 1.0 × 106 m 2.0 × 106 m 2.5 × 1011 m ..........................................

Q9: P is a point mass (mass 9m) and Q is a point mass (mass m). P and Q are separated by 4.0 m. At what point on the line joining P to Q is the net gravitational field strength zero? a) b) c) d) e)

0.44 m from P 1.33 m from P 3.0 m from P 3.6 m from P 3.95 m from P ..........................................

Q10: Astronomical observations tell us that the radius of the planet Neptune is 2.48 × 107 m, and the gravitational field strength at its surface is 11.2 N kg -1 . What is the mass of Neptune? a) b) c) d) e)

3.66 × 103 kg 4.59 × 104 kg 4.16 × 1018 kg 1.03 × 1026 kg 1.16 × 1027 kg ..........................................

6.5

Summary

A gravitational force exists between any two objects of finite mass. This force is always attractive, and the magnitude of the force can be calculated using Newton's Law of Gravitation. Newton's Law of Gravitation allows us to calculate the weight of an object and the acceleration due to gravity at any point in a gravitational field. The acceleration due to gravity is the same for any object, whatever its mass, placed at the same point in a gravitational field. The gravitational field strength can be calculated at any point in a combined field by taking the vector sum of the individual field strengths due to all the objects contributing to the field. © H ERIOT-WATT U NIVERSITY

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By the end of this topic, you should be able to: • state and apply the equation Gm1 m2 r2 to calculate the gravitational force between two objects; F =

• calculate the weight of an object using Newton's Law of Gravitation; • state what is meant by a gravitational field, and calculate the gravitational field strength at a point in the field; • calculate the value of the acceleration due to gravity at a point in a gravitational field, given the local conditions; • describe what is meant by a conservative field; • sketch the field lines around one or two point masses.

6.6

End of topic test

End of topic test At this stage there is an end of topic test available online. If however you do not have access to the internet you may try the questions which follow. The following data should be used when required: Universal constant of gravitational G

6.67 × 10-11 N m2 kg-2

Mass of the Earth mE

5.97 × 1024 kg

Mass of the Moon mM

7.35 × 1022 kg

Mass of the Moon mS Radius of the EarthrE Radius of the Moonr M

1.99 × 1030 kg 6.38 × 106 m 1.74 × 106 m

Q11: A distant planet has mass 5.45 × 10 25 kg. A moon, mass 3.04 × 10 22 kg, orbits this planet with an orbit radius of 7.06 × 10 8 m. Calculate the size of the gravitational force that exists between the moon and the planet. .......................................... Q12: Two identical solid spheres each have mass 0.653 kg and diameter 0.365 m. Find the gravitational force between them when they are touching. .......................................... Q13: Given that the mass of the planet Neptune is 1.03 × 10 26 kg and its radius is 2.48 × 107 m, calculate the weight of a 5.14 kg mass on the surface of Neptune. .......................................... © H ERIOT-WATT U NIVERSITY

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Q14: The value of the acceleration due to gravity is not constant, decreasing with height above the surface of the Earth. What is the value of the acceleration due to gravity in the ionosphere at a height 1.43 × 105 m above the Earth's surface? .......................................... Q15: On the surface of the Earth, a particular object has a weight of 24 N. Calculate its weight on the surface of the Moon. .......................................... Q16: A planet in a distant galaxy has mass 6.07 × 10 25 kg and radius 4.02 × 10 7 m. Calculate the value of the gravitational field strength on the surface of this planet. .......................................... Q17: The gravitational field strength at a distance 2.18 × 10 6 m from the centre of a planet is 5.15 N kg-1 . Calculate the field strength at a distance 8.72 × 10 6 m from the centre of the planet. .......................................... Q18: In a distant solar system, a planet (mass 2.22 × 10 28 kg) is orbiting the sun (mass 5.01 × 1030 kg) with an orbit radius of 4.44 × 10 11 m. Calculate the magnitude of the net gravitational field strength midway between the planet and the sun. ..........................................

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Topic 7

Gravitational potential and satellite motion

Contents 7.1 Introduction . . . . . . . . . . . . . . . . . . 7.2 Gravitational potential and potential energy 7.2.1 Gravitational potential . . . . . . . . 7.2.2 Gravitational potential energy . . . . 7.3 Satellite motion . . . . . . . . . . . . . . . . 7.3.1 Satellite motion . . . . . . . . . . . . 7.3.2 Geostationary satellites . . . . . . . 7.3.3 Kepler's laws . . . . . . . . . . . . . 7.4 Escape velocity . . . . . . . . . . . . . . . . 7.4.1 Escape velocity . . . . . . . . . . . . 7.4.2 Black holes . . . . . . . . . . . . . . 7.5 Summary . . . . . . . . . . . . . . . . . . . 7.6 End of topic test . . . . . . . . . . . . . . .

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100 100 100 102 105 105 107 108 110 110 112 114 115

Prerequisite knowledge The prerequisites for this topic are: • Gravitational fields and forces (Mechanics Topic 6). • Circular motion - centripetal force, periodic time (Mechanics Topics 3 and 4). Learning Objectives By the end of this topic, you should be able to: • understand what is meant by the gravitational potential at a point in a gravitational field; • describe and perform calculations on satellite motion; • calculate the escape velocity for an object in a gravitational field.

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7.1

Introduction

The material in this topic builds on the work on gravitational forces and fields in the previous topic. To begin with, we will look at the potential energy of a body in a gravitational field, so that the work done in moving a body around in a gravitational field can be calculated. Satellite motion will then be studied in some depth. We will again be treating the motion of satellites as circular, and we will see how the centripetal force due to the gravitational interaction determines the speed and period of a satellite. The topic ends with a section dealing with escape velocity - how fast a rocket needs to be travelling when it takes off from Earth in order to escape from the Earth's gravitational field. The subject of black holes is also dealt with in this section.

7.2

Gravitational potential and potential energy





Learning Objective To describe what is meant by the gravitational potential at a point in a gravitational field.



The two sections covered in this part are: • Gravitational potential • Gravitational potential energy.

7.2.1

Gravitational potential

The concept of gravitational potential energy should already be familiar to you from Newtonian mechanics. If a mass m is raised through a height h, it gains potential energy mgh. If the mass is then allowed to fall back down, this potential energy is converted to kinetic energy as it is accelerated downwards. When an object moves through large distances in a gravitational field, we can no longer use this simple expression for the change in potential energy, as the value of the gravitational field strength g is not constant. The gravitational potential is used to describe how the potential energy of an object changes with its position in a gravitational field, and how much work is done in moving an object within the field. The gravitational potentialU at a point in a gravitational field is defined as the work done by external forces in moving a unit mass from infinity to that point. Suppose we are considering the field around a mass m, and moving a unit mass from infinity to a point a distance r from m. We cannot use the simple expression work done = force×distance to calculate the work done against the gravitational force, as the gravitational force acting on the unit mass increases as it moves closer to m. Instead, we have to use a calculus approach, in which we consider the small amount of work dU done in moving a unit mass a distance dr in the field. Integrating over the range from ∞ to r

© H ERIOT-WATT U NIVERSITY

TOPIC 7. GRAVITATIONAL POTENTIAL AND SATELLITE MOTION

 U=

101

r

F dr ∞

Using Newton's law of gravitation for the force acting on the body  U=

r

Gm1 m2 dr r2

∞

In this expression, the values of m1 and m2 are m and 1 kg  U=

r

∞

Gm dr r2

Performing this integration 

1 dr 2 ∞ r  r −1 ∴ U = Gm r ∞    1 1 ∴ U = Gm − − − r ∞   1 1 − ∴ U = Gm ∞ r Gm ∴U =− r .......................................... U = Gm

r

(7.1)

The gravitational potential U is measured in J kg -1 . Remember that there is a minus sign in Equation 7.1. The zero of gravitational potential is at an infinite distance from m. The potential becomes lower and lower the closer we get to m, so it must be a negative number. The gravitational potential at a distance r from the Earth is plotted in Figure 7.1. Figure 7.1: Gravitational potential around the Earth

Earth

rE U∝1/r E U=- Gm rE © H ERIOT-WATT U NIVERSITY

r

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TOPIC 7. GRAVITATIONAL POTENTIAL AND SATELLITE MOTION

.......................................... Example The planet Pluto orbits at a mean distance of 5.92 × 10 12 m from the Sun. What is the gravitational potential due to the Sun's gravitational field at this distance? We will use Equation 7.1, remembering that m here represents the mass of the Sun, since we are calculating the potential due to the Sun's gravitational field. The mass of the Sun is 1.99 × 10 30 kg, so Gm r 6.67 × 10−11 × 1.99 × 1030 ∴U =− 5.92 × 1012 ∴ U = −2.24 × 107 J kg−1 U =−

The gravitational potential due to the Sun's gravitational field is -2.24 × 10 7 J kg-1 . ..........................................

7.2.2

Gravitational potential energy

Equation 7.1 tells us the gravitational potential at a point in a gravitational field per unit mass. To find the gravitational potential energy of an object of mass m 2 placed at that point in the field around a mass m 1 , we simply multiply by m2 . Gm1 m2 r .......................................... P E = U × m2 = −

(7.2)

We can now calculate the work done in moving an object in a gravitational field using this equation. The work done is equal to the change in potential energy of the object. When carrying out these calculations, it is important to take care to get the sign correct. Example A rocket ship, mass 4.00 × 105 kg, is travelling away from the Moon. The ship's rockets are fired when the ship is at a distance of 3.00 × 10 6 m from the centre of the Moon. If the mass of the Moon is 7.35 ×10 22 kg, how much work is done by the rockets in moving the ship to a distance 3.20 × 10 6 m from the Moon's centre? The rocket ship has an initial gravitational potential energy of GmM ms r 6.67 × 10−11 × 7.35 × 1022 × 4.00 × 105 ∴ P E1 = − 3.00 × 106 11 ∴ P E1 = −6.54 × 10 J P E1 = −

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The final value of the potential energy is GmM ms r 6.67 × 10−11 × 7.35 × 1022 × 4.00 × 105 ∴ P E2 = − 3.20 × 106 ∴ P E2 = −6.13 × 1011 J P E2 = −

The change in potential energy ΔPE is ΔP E = P E2 − P E1

  ∴ ΔP E = −6.13 × 1011 − −6.54 × 1011 ∴ ΔP E = 4.1 × 1010 J The potential energy of the rocket ship has increased by 4.1 × 10 10 J, so the amount of work done by the rockets must also be equal to 4.1 ×10 10 J. .......................................... The gravitational field is a conservative field, which means that the work done in moving a mass between two points in the field is independent of the path taken. In the above example, all that we are concerned with are the initial and final locations in the gravitational field. We do not need any details about the path taken between these two locations. (An example of doing non-conservative work would be the work done against friction in sliding a heavy mass between two points. Obviously we do less work if we slide the mass directly from one point to the other rather than taking a longer route.)

Quiz 1 Gravitational potential First try the questions. If you get a question wrong or do not understand a question, there are Hints. The hints are in the same order as the questions. If you read the hint and still do not understand then ask your tutor. All references in the hints are to online materials. Useful data: Gravitational constant G

6.67 ×10-11 N m2 kg-2

Mass of the Earth Radius of the Earth Mass of the Moon Radius of the Moon

5.97 × 1024 kg 6.38 × 106 m 7.35 × 1022 kg 1.74 × 106 m

Q1: Consider the gravitational potential at a point A in the Earth's gravitational field. The value of the gravitational potential depends on a) b) c) d)

the mass of an object placed at A. the speed of an object passing through A. the distance of A from the centre of the Earth. the density of an object placed at A.

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e) the mass of the Earth only. .......................................... Q2: What is the gravitational potential on the surface of the Moon, due to the Moon's gravitational field? a) b) c) d) e)

-1.62 J kg-1 -2.82 × 106 J kg-1 -2.29 × 108 J kg-1 -1.19 × 1023 J kg-1 -2.07 × 1029 J kg-1 ..........................................

Q3: A satellite orbits the Earth at an altitude of 3.00 × 10 5 m above the Earth's surface. What is the gravitational potential at this altitude? a) b) c) d) e)

-8.92 J kg-1 -4.42 × 103 J kg-1 -5.96 × 107 J kg-1 -1.33 × 109 J kg-1 -5.33 × 1024 J kg-1 ..........................................

Q4: What is the gravitational potential energy of a satellite of mass 800 kg, moving in the Earth's gravitational field with orbit radius 6.60 ×10 6 m from the centre of the Earth? a) b) c) d) e)

-7.30 ×103 J -6.03 ×107 J -6.24 ×107 J -2.45 ×1010 J -4.83 ×1010 J ..........................................

Q5: The gravitational potential energy of a satellite orbiting the Earth changes from -5.0 × 109 J to -7.0 ×109 J. Which one of the following statements could be true? a) b) c) d) e)

The satellite has moved closer to the Earth. The satellite has moved further away from the Earth. The mass of the satellite has decreased only. The orbit and mass have stayed the same, but the satellite is moving faster. The orbit and mass have stayed the same, but the satellite is moving slower. ..........................................

© H ERIOT-WATT U NIVERSITY

TOPIC 7. GRAVITATIONAL POTENTIAL AND SATELLITE MOTION

7.3 

105

Satellite motion



Learning Objective To describe and perform calculations on satellite motion.



Man-made satellites orbiting the Earth are used extensively by the telecommunications industry. Other artificial satellites are used for applications such as meteorological observations. The Earth does have a natural satellite too, of course, namely the Moon. In this section we will be studying the motion of satellites, investigating the relationship between orbit radius and orbit period.

7.3.1

Satellite motion

The following interactive activity provides a good starting point for an investigation of satellite motion. A projectile fired from a high mountain will fall to Earth. If the mountain is high enough or the projectile has a great enough speed, the curvature of the Earth must be taken into account when trying to predict where the projectile will land.

Newton's cannon At this stage there is an online activity. 20 min

If it is projected from high altitude with a large enough horizontal velocity, a ‘falling' projectile will not land on the Earth at all. .......................................... If we ignore air resistance, then the only force acting on a projectile is the gravitational force exerted on it by the Earth. If the speed of the projectile is increased, the projectile travels further and further horizontally before hitting the ground, until eventually it does not hit the ground at all, but arrives back at its starting point having travelled in a circular path around the Earth. This is the motion of a satellite in a circular orbit around the Earth. We can think of the satellite as ‘falling' under the influence of gravity. If the speed is increased still further, with the projectile being released from the same point, it will move in an elliptical orbit around the Earth. Even greater initial speed, and the projectile will reach escape velocity, which is discussed later in this Topic. If a satellite is orbiting the Earth with a uniform speed, then there must be a constant centripetal force acting on it. We have seen that the only force acting on the satellite (mass m) is the gravitational force exerted on it by the Earth (mass mE ). This force acts towards the centre of the Earth and is the centripetal force. Thus mv 2 GmE m = r2 r Gm E ∴ v2 = r GmE ∴v= r ..........................................

© H ERIOT-WATT U NIVERSITY

(7.3)

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Equation 7.3 shows that the speed of the satellite is determined by the radius of the satellite's orbit. The inverse relationship means that the speed increases when the orbit radius decreases. The mass of the satellite does not affect the speed. It is this fact that leads to the effect of ‘weightlessness' experienced by astronauts. An astronaut in a satellite can be considered to be a satellite himself, and will be orbiting at the same speed as the satellite. Since there will be no force pushing the astronaut against the floor or the walls, the astronaut experiences an apparent weightlessness relative to the satellite. A common misconception is that the weightlessness an astronaut feels is due to a lack of gravity. In fact, it is because the astronaut and the satellite are both ‘falling' with the same acceleration. The period T of an object performing circular motion with radius r is related to the speed v of the object by the equation v=

2πr T

(You should remember that 2πr is the circumference of a circle.) Substituting this expression into Equation 7.3, we can derive an expression for the period T of a satellite orbiting the Earth. 

GmE r  GmE 2πr = ∴ T r  r T = ∴ 2πr GmE  r3 ∴ T = 2π GmE .......................................... v=

(7.4)

Again, the periodic time does not depend on the mass of the satellite. Example A meteorological satellite orbits the Earth at an altitude of 2.50 ×10 5 m above the Earth's surface. What are the speed and periodic time of the satellite? Before we can calculate the speed, we have to find the radius of the satellite's orbit, which is equal to the radius of the Earth plus the altitude of the satellite above the Earth's surface. r = rE + altitude     ∴ r = 6.38 × 106 + 2.50 × 105 ∴ r = 6.63 × 106 m

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The speed can now be calculated using Equation 7.3.  GmE v= r  6.67 × 10−11 × 5.97 × 1024 ∴v= 6.63 × 106 ∴ v = 7.75 × 103 m s−1 To calculate the period T , we can put this value of v into the circular motion equation. 2πr T 2πr ∴T = v 2π × 6.63 × 106 ∴T = 7.75 × 103 ∴ T = 5.38 × 103 s v=

The satellite orbits with a speed of 7.75 × 10 3 m s-1 and a period of 5380 s (about 90 minutes). .......................................... The total energy of a satellite in orbit around a planet is made up of kinetic energy due to its motion and potential energy due to its position in the gravitational field of the planet. The following activity will show how these two components of the total energy change if the orbit of the satellite changes.

Changing the orbit of a satellite At this stage there is an online activity. This simulation shows how the energy of a satellite changes when its orbit changes. A satellite has kinetic and potential energy, and both of these will change if the radius of the satellite's orbit changes. ..........................................

7.3.2

Geostationary satellites

A satellite orbiting the Earth above the equator with a periodic time exactly equal to one day will appear to be fixed above that point on the Earth. That is to say, the satellite will always appear to be in the same position in the sky to an observer on Earth. Such a satellite is called a geostationary satellite. Communications satellites are all in geostationary orbits, so that a fixed receiver dish back on Earth can receive a signal. A geostationary satellite must be orbiting the Earth with the same angular velocity as the Earth's rotation about its axis. This orbit is sometimes referred to as a parking orbit. The angular velocity ω is equal to 2π/T . A satellite of mass m in a geostationary orbit of radius r around the Earth requires a centripetal force given by the equation © H ERIOT-WATT U NIVERSITY

20 min

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F = mrω 2



∴ F = mr × ∴F =

2π T

2

4π 2 mr T2

This centripetal force is provided by the gravitational force exerted by the Earth on the satellite, hence the centripetal force is equal to the gravitational force calculated using Newton's Law of Gravitation. GmE m 4π 2 mr = 2 T r2 GmE T 2 ∴ r3 = 2 4π 2 3 GmE T ∴r= 2 4π ..........................................

(7.5)

(Note that Equation 7.5 is equivalent to Equation 7.4 rearranged to make r the subject of the equation). We can evaluate the orbital radius of a geostationary satellite as follows. The Earth rotates about its axis every 24 hours, so the periodic time T of this motion is (24 × 60 × 60) = 86400 s. Since G and m E are constants, the radius of the geostationary orbit is fixed. Putting the values of T, G and m E into Equation 7.5, the radius of orbit equals 4.22 × 10 7 m. The radius of the Earth is 6.38 × 10 6 m, so a geostationary satellite orbits (4.22 × 107 ) - (6.38 × 106 ) = 3.58 × 107 m above the Earth's surface.

7.3.3

Kepler's laws

The three laws that govern the motion of planets were discovered by Johannes Kepler in the early 17th century. The laws were based on the astronomical observations of Tycho Brahe, and are known as Kepler's laws. These laws state that: 1. The planets move in elliptical orbits, with the Sun at one focus of the ellipse. 2. The line joining the Sun and a planet sweeps out equal areas in equal times. 3. For each planet, the square of the periodic time of its orbit is proportional to the cube of its mean distance from the Sun. In the simple analysis we have carried out in this Topic, we have assumed circular orbits, which is a special case of the first two laws. The mathematics of ellipses and elliptical motion are beyond the scope of this course. For uniform circular motion, the second law should be obvious to you. We have already encountered the third law in Equation 7.4. Newton also made use of the third law when deriving his universal law of gravitation; he suspected that the gravitational force between two bodies had an inverse-square dependence on the distance between the bodies. Making this assumption, he then showed that this was consistent with Kepler's third law. © H ERIOT-WATT U NIVERSITY

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Quiz 2 Satellite motion First try the questions. If you get a question wrong or do not understand a question, there are Hints. The hints are in the same order as the questions. If you read the hint and still do not understand then ask your tutor. All references in the hints are to online materials. Useful data: Gravitational constant G

6.67 × 10-11 N m2 kg-2

Mass of the Earth Radius of the Earth

5.97 × 1024 kg 6.38 × 106 m

Q6: A satellite of mass 1200 kg is orbiting the Earth with an orbit radius of 6.80 × 106 m. What is the speed of the satellite? a) b) c) d) e)

102 m s-1 7650 m s-1 1.03 × 104 m s-1 2.65 × 105 m s-1 5.86 × 107 m s-1 ..........................................

Q7: Which one of the following statements is false? a) The speed of a satellite orbiting a planet depends on the mass of the planet. b) The period of a satellite orbiting a planet depends on the mass of the planet. c) Two satellites of different mass, orbiting the same planet with the same orbit radius, will have the same period. d) A satellite's speed has an inverse-square-root dependence on its radius of orbit. e) The period of a satellite orbiting the Earth depends on the mass of the satellite. .......................................... Q8: A geostationary satellite orbits at a height of 3.58 × 10 7 m above the Earth's surface. Calculate the speed at which the satellite is travelling. a) b) c) d) e)

417 m s-1 488 m s-1 3070 m s-1 3330 m s-1 9.42 × 106 m s-1 ..........................................

Q9: If the orbit radius of a satellite decreases, a) b) c) d) e)

its potential and kinetic energies both decrease. its potential and kinetic energies both increase. its potential energy increases but its kinetic energy decreases. its potential energy decreases but its kinetic energy increases. its potential energy decreases but its kinetic energy remains constant.

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.......................................... Q10: Astronomical observations of the period T and orbit radius r of a moon around a planet allow us to calculate the mass m p of the planet. Which of these equations correctly expresses mp in terms of T and r? 3

a) mp = 2πr GT 2 3

b) mp = 4πr GT 2 2 2

c) mp = 4π r GT 3 2 3

d) mp = 4π r GT 2 e) mp = 4πr/GT

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

7.4

Escape velocity



Learning Objective To calculate the escape velocity for an object in a gravitational field.



The section covers: • Escape velocity • Black holes.

7.4.1

Escape velocity

At the beginning of the section on satellite motion, you should have carried out an interactive activity in which a projectile was launched into orbit with different speeds. If the speed was sufficiently great, the projectile did not complete an orbit, but escaped from the Earth's gravitational field. We will now calculate what minimum speed is required for a projectile to escape from the Earth's gravitational field, and then generalise it to all other gravitational fields. To escape from the gravitational field, an object must have sufficient kinetic energy. We know that at an infinite distance from Earth, the potential energy of a body will be zero. We can also work out the (negative) potential energy of the body when placed at the Earth's surface. If the body has sufficient kinetic energy to raise its total energy above zero J, then it can escape from the gravitational field. The gravitational potential energy of an object such as a rocket of mass m at a point on the Earth's surface can be calculated using Equation 7.2. PE = −

GmE m rE

To escape the Earth's gravitational field, the work done by the rocket must equal the potential difference between infinity and the point on the Earth's surface. © H ERIOT-WATT U NIVERSITY

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−GmE m −GmE m − ∞ rE GmE m ∴ ΔP E = 0 + rE GmE m ∴ ΔP E = rE ΔP E =

The rocket must have an initial kinetic energy at least equal to this, so that its velocity does not drop to zero before it has escaped from the field. We can use the equation for kinetic energy to calculate the initial velocity of the rocket. 1 GmE m mv 2 = 2 rE 2Gm E ∴ v2 = rE  2GmE ∴v= rE ..........................................

(7.6)

This velocity is the minimum velocity required. Now we will use the expression for the acceleration due to gravity g, equivalent to the gravitational field strength at the Earth's surface, which is given by the equation GmE 2 rE GmE ∴ grE = rE g=

Substituting this expression into the equation for the escape velocity gives us 

2GmE r  E ∴ v = 2grE .......................................... v=

(7.7)

The escape velocity for a rocket fired from Earth is given by Equation 7.7. Putting in the values of g = 9.8 m s-2 and rE = 6.38 × 106 m, the escape velocity has a value v= ∴v=

 

2grE 2 × 9.8 × 6.38 × 106

∴ v = 1.1 × 104 m s−1 or 11 km s−1 You should note that the escape velocity does not depend on the mass of the rocket - the escape velocity is the same for any object launched from the Earth's surface. In general, the escape velocity from the gravitational field around a body of mass m, starting from a point r from the centre of the field, is given by the following equation  vesc = © H ERIOT-WATT U NIVERSITY

2Gm r

(7.8)

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.......................................... Example What is the escape velocity for a lunar probe taking off from the surface of the Moon? (mM = 7.35 × 1022 kg, rM = 1.74 × 106 m) Using Equation 7.8 with the values given in the question, the escape velocity from the Moon is  2GmM vesc = r  M 2 × 6.67 × 10−11 × 7.35 × 1022 vesc = 1.74 × 106  vesc = 5.635 × 106 vesc = 2.37 × 103 m s−1 The escape velocity from the Moon's gravitational field is 2.37 × 10 3 m s-1 , or 2.37 km s-1 . ..........................................

7.4.2

Black holes

The escape velocity can be expressed in terms of the density of the body. Taking a star as a sphere of uniform density, we will find the escape velocity from the surface of a star of mass m and radius r in terms of its density ρ ρ=

m V

The volume V of a sphere is equal to 43 πr 3 , so the density of the sphere can be written in terms of r as ρ=

m 4 3 3 πr

3m 4πr 3 m 3 ∴ρ= r 4πr 2 4πr 2 ρ m = ∴ r 3 ∴ρ=

Substituting into Equation 7.8, the escape velocity in terms of the density of the star is  vesc =

4πr 2 ρ 3  2 8Gπr ρ = 3

∴ vesc = ∴ vesc

2Gm r 

2G ×

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So the escape velocity depends on the density and radius of the star. Now, we know that there is a maximum speed, the speed of light c, which cannot be exceeded. If the density of a star is great enough, the escape velocity would have to be greater than c, and not even photons will be able to escape. Such a star is called a black hole. Black holes tend to be formed when a star ‘collapses' upon itself. When this happens, the radius of the star decreases and the density increases, dramatically increasing the escape velocity until not even photons can escape.

Quiz 3 Escape velocity First try the questions. If you get a question wrong or do not understand a question, there are Hints. The hints are in the same order as the questions. If you read the hint and still do not understand then ask your tutor. All references in the hints are to online materials. Useful data: Gravitational constant G

6.67 × 10-11 N m2 kg-2

Mass of the Earth Radius of the Earth

5.97 × 1024 kg 6.38 × 106 m

Q11: The escape velocity of an object taking off from Earth is the minimum velocity required to a) b) c) d) e)

place the object in a geostationary orbit. place the object in a non-geostationary orbit. reach the point where the combined field of the Earth and the Moon is zero. escape from the Earth's atmosphere. escape from the Earth's gravitational field. ..........................................

Q12: The escape velocity of an object from the Earth depends on a) b) c) d) e)

the masses of the Earth and the object. the mass and radius of the Earth. the mass and density of the object. the mass of the object and the radius of the Earth. the mass and radius of the Earth, and the mass of the object. ..........................................

Q13: What would be the escape velocity of a Martian spacecraft of mass 900 kg takingoff from the surface of Mars, if the mass and radius of Mars are 6.42 × 10 23 kg and 3.40 × 106 m? a) b) c) d) e)

2510 m s-1 3550 m s-1 5020 m s-1 1.06 × 105 m s-1 1.51 × 105 m s-1 ..........................................

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Q14: A black hole is formed when a) b) c) d) e)

the escape velocity becomes zero for all objects. the escape velocity is greater than the speed of light. a star rapidly expands its volume. the density of a star becomes extremely small. the photon density becomes extremely large. ..........................................

7.5

Summary

The change in potential energy caused by moving a mass in a gravitational field can be calculated using the gravitational potential of the field. The change in potential energy depends only on the initial and final positions in the field, as the gravitational field is a conservative field. Satellite motion has been described in terms of circular orbits, in which the gravitational force exerted by the central body on the orbiting satellite provides the centripetal force. The speed and period of the satellite are independent of its mass. By calculating the gravitational potential, the escape velocity for an object in a gravitational field can be calculated. By the end of this topic you should be able to: • state the expression Gm r and use it to calculate the gravitational potential U at a point in a gravitational field; U =−

• calculate the potential energy of a mass in a gravitational field and calculate the change in the potential energy when a mass is moved between points in the field; • calculate the period and speed of a satellite; • explain what is meant by a geostationary satellite; • derive the expression

 vesc =

2Gm r

and use it to calculate the escape velocity; • state that a black hole is formed when the density of a body is so great that the escape velocity is greater than the speed of light.

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115

End of topic test

End of topic test At this stage there is an end of topic test available online. If however you do not have access to the internet you may try the questions which follow. The following data should be used when required: universal constant of gravitation G

6.67 × 10-11 N m2 kg-2

Mass of the Earth M E Radius of the Earth R E Mass of the Sun M S

5.97 × 1024 kg 6.38 × 106 m 1.99 × 1030 kg

Q15: Calculate the gravitational potential (in J kg -1 ) at a distance 2.25 × 10 7 m from the centre of the Earth, due to the Earth's gravitational field. .......................................... Q16: A spaceship, mass 4.7 × 10 4 kg is travelling through the solar system. At one point in its journey, the spaceship passes near the planet Jupiter taking photographs at a distance 9.2 × 107 m from the centre of Jupiter. If the mass of Jupiter is 1.9 × 1027 kg, calculate the potential energy (in J) of the spaceship at this point. .......................................... Q17: The gravitational potential at a distance 2.36 × 10 7 m from the centre of a planet of radius 8.55 × 106 m is - 6.04 × 107 J kg -1 . Calculate the gravitational potential at a distance 4.72 × 10 7 m (twice the original distance) from the centre of the planet. .......................................... Q18: A meteorological satellite is orbiting the Earth. The mass of the satellite is 4.25 × 103 kg and it orbits at a height 1.35 × 10 5 m above the Earth's surface. Calculate the gravitational potential energy (in J) of the satellite. .......................................... Q19: Scientists wish to launch a satellite (mass 2.4 × 10 4 kg) which will orbit the Earth once every 5200 seconds. At what height in m above the Earth's surface should the satellite be placed? .......................................... Q20: The planet Zaarg has mass 7.74 × 10 25 kg and radius 2.58 × 10 7 m. Calculate the escape velocity in m s

-1

of a rocket ship launched from the planet Zaarg.

.......................................... Q21: Astronomers observing a distant solar system have noticed a planet orbiting a star with a period 6.57 × 10 7 s. The distance from the planet to the star is 2.55 × 10 11 m. Calculate the mass of the star, in kg. .......................................... © H ERIOT-WATT U NIVERSITY

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Topic 8

Simple harmonic motion

Contents 8.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8.2 Defining SHM . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

118 118

8.3 Equations of motion in SHM . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8.4 Energy in SHM . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

121 126

8.5 Applications and examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

129

8.5.1 Mass on a spring - vertical oscillations . . . . . . . . . . . . . . . . . . . 8.5.2 Simple pendulum . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

129 131

8.5.3 Loudspeaker cones and eardrums . . . . . . . . . . . . . . . . . . . . . 8.6 Damping . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

133 134

8.7 Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8.8 End of topic test . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

136 137

Prerequisite knowledge • Newton's laws of motion. • Radian measurement of angles and angular velocity (Mechanics topic 3). • Differentiation of trig functions. Learning Objectives By the end of this topic, you should be able to: • understand what is meant by simple harmonic motion (SHM); • apply the equations of motion in different SHM systems; • understand how energy is conserved in an oscillating system; • explain what is meant by damping of an oscillating system.

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8.1

Introduction

In most of the previous Mechanics topics, we have been studying motion where the acceleration is constant, whether it be linear motion (Topic 1) or circular motion (Topics 3 - 5). In this topic we are looking at a specific situation in which the acceleration is not constant. We will be studying simple harmonic motion (SHM), in which the acceleration of an object depends on its displacement from a fixed point. The topic begins by defining what is meant by simple harmonic motion, followed by derivations of the equations of motion for SHM. We then consider the kinetic energy and potential energy of an SHM system, as well as examples of different SHM systems. We will return to the subject of energy in the final section of the Topic, when we consider what happens to a ‘damped' system.

8.2

Defining SHM



Learning Objective To state what is meant by simple harmonic motion.



We will begin this topic by looking at the horizontal motion of a mass attached to a spring. Figure 8.1: Mass attached to a horizontal spring

.......................................... The mass rests on a frictionless surface, and the spring is assumed to have negligible mass. The spring obeys Hooke's law, so the force F required to produce an extension y of the spring is F = ky

(8.1)

.......................................... k is the spring constant, measured in N m -1 . This expression generally holds true for a spring so long as the extension is not large enough to cause a permanent deformation of the spring. If we pull the mass back a distance y and hold it (Figure 8.2), Newton's © H ERIOT-WATT U NIVERSITY

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first law tells us the tension T in the spring (the restoring force) must also have magnitude F . In fact, T = -F , since force is a vector quantity, and T and F are in opposite directions. Hence, T = −ky

(8.2)

.......................................... Figure 8.2: Mass pulled back and held

6

.

O .......................................... If we now release the mass (Figure 8.3), it is no longer in equilibrium, and we can apply Newton's second law F = ma. In this case the force acting on the block is T , so T = ma d2 y dt2 2 d y k ∴ 2 =− y dt m .......................................... ∴ −ky = m

Figure 8.3: Mass accelerated by the tension in the extended spring

acceleration

6

O .......................................... Equation 8.3 tells us that the acceleration of the mass is proportional to its displacement, since m and k are constants. The minus sign means that the © H ERIOT-WATT U NIVERSITY

(8.3)

TOPIC 8. SIMPLE HARMONIC MOTION

acceleration and displacement vectors are in the opposite direction. This equation holds true whatever the extension or compression of the spring, since the spring also obeys Hooke's law when it is compressed (Figure 8.4). Figure 8.4: Mass accelerated by the force in the compressed spring

acceleration

6

O .......................................... The mass will oscillate around the equilibrium (T = 0) position, with an acceleration always proportional to its displacement, and always in the opposite direction to its displacement vector. Such motion is called simple harmonic motion (SHM). A plot of displacement against time for the mass on a spring shows symmetric oscillations about y = 0. The maximum value of the displacement of an object performing SHM is called the amplitude of the oscillations. Figure 8.5: Displacement against time for an object undergoing SHM

   O

120

   J

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

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Example A 1.0 kg mass attached to a horizontal spring (k = 5.0 N m -1 ) is performing SHM on a horizontal, frictionless surface. What is the acceleration of the mass when its displacement is 4.0 cm? Using Equation 8.3, the acceleration is k y m 5.0 × 0.040 ∴a=− 1.0 ∴ a = −0.20 m s−2 a=−

The magnitude of the acceleration is 0.20 m s -2 . Normally we do not need to include the minus sign unless we are concerned with the direction of the acceleration. .......................................... Equation 8.3 describes a specific example of SHM, that of a mass m attached to spring of spring constant k. In general, for an object performing SHM d2 y ∝ −y dt2 d2 y or 2 = −ω 2 y dt ..........................................

(8.4)

The constant of proportionality is ω 2 , where ω is called the angular frequency of the motion. We are used to using ω to represent angular velocity in circular motion, and it is no coincidence that the same symbol is being used again. We will see now how simple harmonic motion and circular motion are related.

8.3

Equations of motion in SHM





Learning Objective To derive and apply the SHM equations of motion.



To try to help us understand SHM, we can compare the motion of an oscillating object with that of an object moving in a circle. Let us consider an object moving anti-clockwise in a circle of radius a at a constant angular velocity ω with the origin of an x-y axis at the centre of the circle. We will be concentrating on the y-component of the motion of this object. Assuming the object is at position y = 0 at time t = 0, then after time t the angular displacement of the object will be θ = ωt.

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Figure 8.6: Object moving in a circle

J=J =

O

θ

ω

J=0

.......................................... From Figure 8.6, we can see that the y displacement at time t is equal to a sin θ = a sin ωt Hence y = a sin ωt Differentiating this equation, the velocity in the y-direction is v=

dy = ωa cos ωt dt

We can differentiate again to get the acceleration in the y-direction. dv d2 y = −ω 2 a sin ωt = 2 dt dt d2 y ∴ 2 = −ω 2 y dt ..........................................

(8.5)

Equation 8.5 gives us the relationship between acceleration and displacement, and is exactly the same as Equation 8.4 for an object undergoing SHM. (Remember it is the projection of the object on the y-axis that is undergoing SHM, rather than the object itself.) The angular velocity of the circular motion is therefore equivalent to the angular frequency of the SHM. As with circular motion, we can talk about the period of the SHM as the time taken to complete one oscillation. The relationship between the periodic time T and ω is T =

2π ω

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.......................................... The frequency f of the SHM is the number of complete oscillations performed per second, so ω 1 = T 2π ∴ ω = 2πf .......................................... f=

(8.7)

Frequency is measured in hertz Hz, equivalent to s -1 . Another solution to Equation 8.4 is y = a cos ωt The proof that y = a cos ωt represents SHM is given in the following example. Example Prove that the equation y = a cos ωt describes simple harmonic motion. 2

For an object to be performing SHM, it must obey the equation d y dt2 = −ω 2 y . We can differentiate our original equation twice

y dy ∴ dt d2 y ∴ 2 dt d2 y ∴ 2 dt

= a cos ωt = −ωa sin ωt = −ω 2 a cos ωt = −ω 2 y

So y = a cos ωt does represent SHM. In this case, the displacement is at a maximum when t = 0, whereas the displacement is 0 at t = 0 when y = a sin ωt. ..........................................

Displacement, velocity and acceleration in SHM At this stage there is an online activity. This simulation allows you to make a thorough investigation of the mass-on-a-spring system. The acceleration, velocity and displacement of an object undergoing SHM all vary with time. You should know at which point in the cycle each of these quantites have their maximum and minimum values, and when they are equal to zero. .......................................... © H ERIOT-WATT U NIVERSITY

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There is one more equation we need to derive. We have expressed acceleration as a function of displacement (Equation 8.4 and Equation 8.5). We can also find an expression showing how the velocity varies with displacement. In deriving Equation 8.5 we used the expressions y = a sin ωt and v = ωa cos ωt. Rearranging both of these equations gives us sin ωt =

v y and cos ωt = a ωa

Now, using the trig identity sin2 θ + cos2 θ = 1 , we have sin2 θ + cos2 θ = 1 ∴ sin2 (ωt) + cos2 (ωt) = 1 y2 v2 + =1 a2 ω 2 a2 ∴ y 2 ω 2 + v 2 = ω 2 a2

∴

∴ v 2 = ω 2 a2 − y 2 ω 2   ∴ v 2 = ω 2 a2 − y 2  ∴ v = ω (a2 − y 2 ) This equation tells us that for any displacement other than y = a, there are two possible values for v, since the square root can give positive or negative values, meaning that the velocity could be in two opposite directions. In fact, to emphasise this point the equation is usually written as,

v = ±ω

 a2 − y 2 (8.8)

.......................................... Example An object is performing SHM with amplitude 5.0 cm and frequency 2.0 Hz. What is the maximum value of the velocity of the object? From Equation 8.8 and the on-screen animation, you should have realised that the velocity has its maximum value when the displacement is zero. So, remembering that ω = 2πf

∴ vmax



a2 − y 2  = 2πf × a2 − 0

v = ±ω

∴ vmax = 2πf × a ∴ vmax = 2π × 2.0 × 0.050 ∴ vmax = 0.63 m s−1 .......................................... © H ERIOT-WATT U NIVERSITY

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Quiz 1 Defining SHM and equations of motion First try the questions. If you get a question wrong or do not understand a question, there are Hints. The hints are in the same order as the questions. If you read the hint and still do not understand then ask your tutor. All references in the hints are to online materials. Q1: For an object to be performing simple harmonic motion, the net force acting on it must be proportional to its a) b) c) d) e)

frequency. displacement. velocity. mass. amplitude. ..........................................

Q2: An object is performing SHM with ω = 8.00 rad s -1 and amplitude a = 0.150 m. What is the frequency of the oscillations? a) b) c) d) e)

1.27 Hz 2.55 Hz 25.1 Hz 50.3 Hz 120 Hz ..........................................

Q3: An object is performing SHM oscillations. continuously changing as the object oscillates? a) b) c) d) e)

Which one of these properties is

Period Mass Amplitude Acceleration Frequency ..........................................

Q4: A mass on a spring oscillates with frequency 5.00 Hz and amplitude 0.100 m. What is the maximum acceleration of the mass? a) b) c) d) e)

0.063 m s-2 2.50 m s-2 10.0 m s-2 24.7 m s-2 98.7 m s-2

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.......................................... Q5: An object is performing SHM with ω = 4.2 rad s -1 and amplitude 0.25 m. What is the speed of the object when its displacement is 0.15 m from its equilibrium position? a) b) c) d) e)

0.17 m s-1 0.42 m s-1 0.84 m s-1 1.1 m s-1 3.5 m s-1 ..........................................

8.4

Energy in SHM



Learning Objective To describe how energy is conserved in an oscillating system.



Let us return to the first example of SHM we looked at, a mass attached to a spring, oscillating horizontally on a smooth surface. We will now consider the energy in this system. The principle of conservation of energy means that the total energy of the system must remain constant. As the displacement y of the mass increases, the velocity decreases, and so the kinetic energy of the mass decreases. To keep the total energy of the system constant, this ‘lost' energy is stored as potential energy in the spring. The more the spring is stretched or compressed, the greater the potential energy stored in it. We can derive equations to show how the kinetic and potential energies of the system vary with the displacement of the mass.  We already know that the velocity of an object performing SHM is v = ±ω a2 − y 2 . We also know that the kinetic energy of a object of mass m is KE = 12 mv 2 . So   v 2 = ω 2 a2 − y 2   1 ∴ KE = mω 2 a2 − y 2 2 ..........................................

(8.9)

To obtain the potential energy, consider the work done against the spring's restoring force in moving the mass to a displacement y. We cannot use a simple work done = force×distance calculation since the force changes with displacement. Instead we must use a calculus approach, calculating the work done in increasing the displacement by a small amount dy and then performing an integration over the limits of y. The work done in moving a mass to a displacement y is equal to 

y

F dy y=0

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This work done is stored as potential energy in the system, so 

y

PE =

F dy y=0

2

Now, the general equation for SHM is d y dt2 = −ω 2 y, and Newton's second law of motion tells us that force = mass × acceleration, so

F =m

d2 y = −mω 2 y dt2

Substituting for F in the potential energy equation  PE = ∴ PE =

y

F dy y=0  y

mω 2 y dy

y=0

Note that the minus sign has been dropped. This is because work is done on the spring in expanding it from y = 0 to y = y. The potential energy stored in the spring is positive, since the PE is at its minimum value (zero) when y = 0 

y

mω 2 y dy  y y dy ∴ P E = mω 2 PE =

y=0

(8.10)

y=0

1 ∴ P E = mω 2 y 2 2 ..........................................

Energy in simple harmonic motion This is a paper-based graph-sketching exercise, showing how energy varies with displacement and time. Consider an SHM system of a 0.25 kg mass oscillating with ω = π rad s -1 and amplitude a = 1.2 m. Sketch graphs showing 1. PE, KE and total energy against displacement; 2. PE, KE and total energy against time. In both cases sketch the energy over three cycles, assuming the displacement y = 0 at t = 0. The kinetic and potential energies vary between zero and their maximum value, while the total energy remains constant.

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.......................................... In summary then, the kinetic energy is at a maximum when the speed of an oscillating object is at a maximum, which is when the displacement y equals zero. Note that at y = 0, the force acting on the object is also zero, and hence the potential energy is zero. At the other extreme of the motion, when y = a, the velocity is momentarily zero, and so the kinetic energy is zero. The restoring force acting on the object is at a maximum at this point, and so the potential energy of the system is at a maximum. At all other points in the motion of the object, the total energy of the system is partly kinetic and partly potential, with the sum of the two being constant.

Energy of a mass on a spring At this stage there is an online activity. 10 min

The on-screen animation shows the same SHM system of a mass oscillating vertically on the end of a spring that was used in the earlier simulation. This time it is used to see how the energy varies as the mass oscillates. The kinetic and potential energies of a real oscillating system behave in the way predicted in the earlier graph sketching activity. ..........................................

Quiz 2 Energy in SHM 20 min

First try the questions. If you get a question wrong or do not understand a question, there are Hints. The hints are in the same order as the questions. If you read the hint and still do not understand then ask your tutor. All references in the hints are to online materials. Q6: A 2.0 kg block is performing SHM with angular frequency 1.6 rad s -1 . What is the potential energy of the system when the displacement of the block is 0.12 m? a) b) c) d) e)

0.023 J 0.037 J 0.074 J 0.19 J 0.38 J ..........................................

Q7: When the kinetic energy of an SHM system is 25 J, the potential energy of the system is 60 J. What is the potential energy of the system when the kinetic energy is 40 J? a) b) c) d) e)

15 J 25 J 37.5 J 45 J 60 J .......................................... © H ERIOT-WATT U NIVERSITY

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Q8: A mass attached to a spring is oscillating horizontally on a smooth surface. At what point in its motion does the kinetic energy have a maximum value? a) b) c) d) e)

When its displacement from the rest position is at its maximum. When the mass is midway between its rest position and its maximum displacement. When its displacement from the rest position is zero. The kinetic energy is constant at all points in its motion. Cannot tell without knowing the amplitude and frequency. ..........................................

Q9: A mass-spring system is oscillating with amplitude a. What is the displacement at the point where the kinetic energy of the system is equal to its potential energy? a) b) c) d) e)

±a/4 ±a/2

±a √2 √ ±a 2 ±2a ..........................................

Q10: The total energy of a system oscillating with SHM is 100 J. What is the kinetic energy of the system at the point where the displacement is half the amplitude? a) b) c) d) e)

8.66 J 10 J 50 J 75 J 100 J ..........................................

8.5

Applications and examples





Learning Objective To describe some practical examples of SHM.



The applications covered in this section are: • Mass on a spring - vertical oscillations • Simple pendulum • Loudspeaker cones and eardrums.

8.5.1

Mass on a spring - vertical oscillations

We have been using a mass on a spring to illustrate different aspects of SHM. In Section 8.2 we analysed the horizontal mass-spring system. We will look now at the © H ERIOT-WATT U NIVERSITY

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vertical mass-spring system. This is the system we used in the on-screen activities earlier in this Topic. Here we will analyse the forces acting on the mass to show that its motion is indeed SHM. Figure 8.7: Mass hanging from a spring

l

e

(a)

m

y

(b)

m

acceleration

(c)

.......................................... The spring has a natural length l, spring constant k and negligible mass Figure 8.7(a). When a mass m is attached to the spring Figure 8.7(b), it causes an extension e, and the system is at rest in equilibrium. Hence the tension T in the spring (k×e) acting upwards on the mass must be equal in magnitude and opposite in direction to the weight mg. If the mass is now pulled down a distance y and released Figure 8.7(c), the tension in the spring is k(y+e). The resultant force acting upwards is equal to T W . Hence the resultant force is F = k(y + e) − mg ∴ F = ky + ke − mg But ke = mg, so the resultant force F = ky. Using Newton's second law, this resultant force must be equal to (mass × acceleration), so ky = −m

d2 y dt2

d2 y k ∴ 2 =− y dt m ..........................................

(8.11)

The minus sign appears because the displacement and acceleration vectors are in 2

opposite directions. Equation 8.11 has the form d y dt2 = −ω 2 y showing that the motion is SHM, with  ω=

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131

Simple pendulum

Another SHM system is the simple pendulum. A simple pendulum consists of a bob, mass m, attached to a light string of length l. The bob is pulled to one side through a small angle θ and released. The subsequent motion is SHM. Figure 8.8: Simple pendulum

θ

l

l

y mg

mgsinθ

.......................................... The restoring force on the bob acts perpendicular to the string, so no component of the tension in the string contributes to the restoring force. The restoring force is mg sin θ, the component of the weight acting perpendicular to the string. Now for small θ, when θ is measured in radians, sinθ ≈ θ. Also, since we are measuring θ in radians, θ = y/l. Hence the restoring force acting on the bob is mg sin θ = mg × y/l. Now, using Newton's second law d2 y y = −m 2 l dt 2 d y g ∴ 2 =− y dt l .......................................... mg

(8.12)

Again, the minus sign appears because the displacement and acceleration vectors are in the opposite directions. So we have SHM with ω = g/l. Note that the angular frequency, and hence the period, are both independent of the mass of the bob. Note also that we have assumed small θ and hence small displacement y, in deriving Equation 8.12. The simple pendulum only approximates to SHM because of the approximation sin θ ≈ θ.

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Simple pendulum At this stage there is an online activity. 10 min

This activity explores the oscillations of a simple pendulum and demonstrates why this is SHM. You should understand why the motion of a simple pendulum is SHM. Note that the displacement, velocity and acceleration vectors behave in the same way as they do with the mass-spring system. You should see that the energy relationships are also as before. ..........................................

Extra Help: Understanding, significance and treatment of uncertainties At this stage there is an online activity. This activity is set in the context of an experimental determination of g using a simple pendulum. However, the activity is relevant to all of the units of Advanced Higher Physics and to the Investigation which is part of this course. The activity is designed to help you understand how to analyse and interpret uncertainties at the level of Advanced Higher. .......................................... Example Two simple pendula are set in motion. Pendulum A has string length p and a bob of mass q. Pendulum B has string length 4p and the mass of its bob is 5q. What is the ratio TA/T of the periodic times of the two pendula? B

A simple pendulum has ω =



g/ , so the periodic time l 2π = 2π T = ω

 l g

The ratio of periodic times is therefore

2π lgA TA = TB 2π lgA √ TA lA ∴ =√ TB l √B p TA ∴ =√ TB 4p 1 TA = ∴ TB 2 So the ratio of periodic times depends on the square root of the ratio of string lengths. © H ERIOT-WATT U NIVERSITY

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Remember, the mass of the bob does not affect the periodic time of a pendulum. ..........................................

8.5.3

Loudspeaker cones and eardrums

A practical situation in which SHM occurs is in the generation and detection of sound waves. Sound waves are longitudinal waves, in which air molecules are made to oscillate back and forth due to high and low pressure regions being created. In a loudspeaker, a cardboard cone is rigidly attached to an electric coil which sits in a magnetic field. A fluctuating electric current in the coil causes the coil and the cone to vibrate back and forth. A pure note causes SHM vibrations of the cone (other notes cause more complicated vibrations). The oscillations of the cone cause the air molecules to oscillate, creating the high and low pressure regions that cause the wave to travel. The opposite process occurs in the ear, where the oscillations of air molecules cause the tympanic membrane (commonly called the eardrum) to oscillate. These oscillations are converted to an electrical signal in the inner ear, and are transmitted to the brain via the auditory nerve.

Quiz 3 SHM Systems First try the questions. If you get a question wrong or do not understand a question, there are Hints. The hints are in the same order as the questions. If you read the hint and still do not understand then ask your tutor. All references in the hints are to online materials. Q11: An SHM oscillator consists of a 0.500 kg block suspended by a spring (k = 2.00 N m-1 ), oscillating with amplitude 8.00 × 10 -2 m. What is the period T of this system? a) b) c) d) e)

0.318 s 1.25 s 2.00 s 3.14 s 12.6 s ..........................................

Q12: What is the frequency f of a simple pendulum consisting of a 0.250 kg mass attached to a 0.300 m string? a) b) c) d) e)

0.0278 Hz 0.910 Hz 0.997 Hz 5.20 Hz 35.9 Hz

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.......................................... Q13: A 0.19 kg mass is oscillating vertically, attached to a spring. The period of the oscillations is 0.45 s. What is the spring constant of the spring? a) b) c) d) e)

2.7 N m-1 5.9 N m-1 17 N m-1 28 N m-1 37 N m-1 ..........................................

Q14: A simple pendulum is oscillating with period 0.75 s. If the pendulum bob, mass m, is replaced with a bob of mass 2m, what is the new period of the pendulum? a) b) c) d) e)

0.375 s 0.75 s 1.06 s 1.5 s 3.0 s ..........................................

Q15: A simple pendulum has frequency 5 Hz. How would you increase its frequency to 10 Hz? a) b) c) d) e)

Decrease its length by a factor of 4. Decrease its length by a factor of 2. Increase its length by a factor of 2. Increase its length by a factor of 4. Increase its length by a factor of 10. ..........................................

8.6

Damping





Learning Objective To explain what is meant by the damping of an oscillating system.



In the examples we have looked at so far, we have ignored any external forces acting on the system, so that the oscillator continues to oscillate with the same amplitude and no energy is lost from the system. In reality, of course, this does not happen. For our horizontal mass-spring system, for example, friction between the mass and the horizontal surface would mean that some energy would be ‘lost' from the system with every oscillation.

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At the maximum displacement, all of the system's energy is potential energy, given by Equation 8.10 P E = 12 mω 2 y 2 If the energy of the system is decreasing, the amplitude must also be decreasing to satisfy the above equation. The damping of the system describes the rate at which energy is being lost, or the rate at which the amplitude is decreasing. A system in which the damping effects are small is described as having light damping. A plot of displacement against time for such an oscillator is shown in Figure 8.9

displacement

Figure 8.9: Lightly-damped system

J

.......................................... In a heavily damped system, the damping is so great that no complete oscillations are seen, and the ‘oscillating' object does not travel past the equilibrium point.

  

Figure 8.10: Heavily-damped and critically-damped systems

       J

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A critically damped system is one with an amount of damping that ensures the oscillator comes to rest in the minimum possible time. Another way of stating this is that critical damping is the minimum amount of damping that completely eliminates the oscillations. An important application of damping is in the suspension system of a car. The shock absorber attached to the suspension spring consists of a piston in a reservoir of oil which moves when the car goes over a bump in the road. Underdamping (light damping) would mean that the car would continue to wobble up and down as it continued along the street. Overdamping (heavy damping) would make the suspension system ineffective, and produce a bumpy ride. Critical damping provides the smoothest journey, absorbing the bumps without making the car oscillate.

8.7

Summary

A system is oscillating with simple harmonic motion if its acceleration is proportional to its displacement, and is always directed towards the centre of the motion. We have analysed several SHM systems, and looked at some applications of SHM. Conservation of energy has been discussed, and the effects of damping have been demonstrated. By the end of this topic, you should be able to: • define what is meant by simple harmonic motion, and be able to describe some examples of SHM; • state the equation d2 y = −ω 2 y dt2 and explain what each term in this equation means; • show that y = a cos ωt and y = a sin ωt are solutions of the above equation, and show that  v = ±ω a2 − y 2 in both of these cases; • derive expressions for the potential and kinetic energies of an SHM system; • explain what is meant by damping of an oscillating system.

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8.8

137

End of topic test

End of topic test At this stage there is an end of topic test available online. If however you do not have access to the internet you may try the questions which follow. The following data should be used when required: acceleration due to gravity g = 9.8 m s -2 Q16: A 0.65 kg mass is attached to a light spring ( k = 40 N m -1 ) and rests on a smooth horizontal surface. The mass is pulled back a distance of 3.7 cm and released. Calculate the size of the acceleration of the mass, in m s

-2

, at the instant it is released.

.......................................... Q17: A 0.36 kg mass attached to a spring ( k = 26 N m smooth horizontal surface.

-1

) is performing SHM on a

Calculate the periodic time of these oscillations, in s. .......................................... Q18: An SHM system is oscillating with angular frequency ω = 3.9 rad s -1 and amplitude a = 0.18 m. 1. Calculate the maximum value of the acceleration of the oscillator, in m s 2. Calculate the maximum value of the velocity of the oscillator, in m s

-1

-2 .

.

.......................................... Q19: An SHM system consists of a 1.64 kg mass oscillating with amplitude 0.217 m. If the angular frequency of the oscillations is 8.00 rad s the system, in J.

-1

, calculate the total energy of

.......................................... Q20: A body of mass 0.25 kg is performing SHM with amplitude 0.25 m and angular frequency 5.5 rad s -1 . Calculate the displacement of the body, in m, when the potential energy of the system is equal to its kinetic energy. .......................................... Q21: A simple pendulum has a bob of mass 0.35 kg and a string of length 0.42 m. Calculate the frequency of the SHM oscillations of this pendulum, in Hz. .......................................... Q22: A simple pendulum, consisting of a 0.15 kg bob on a 0.72 m string, is oscillating with amplitude 48 mm. Calculate the maximum kinetic energy of the pendulum, in J. ..........................................

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Q23: An SHM system consists of an oscillating 1.4 kg mass. The system is set into motion, with initial amplitude 0.82 m. Damping of the system means that the mass continues to oscillate with an angular frequency of 4.0 rad s -1 , but 10% of the system's energy is lost in work against friction with every oscillation. Calculate the amplitude, in m, of the second oscillation after the system is set in motion. ..........................................

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Topic 9

Wave-particle duality

Contents 9.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9.2 Wave-particle duality of waves . . . . . . . . . . . . . . . . . . . . . . . . . . .

140 140

9.2.1 The photoelectric effect . . . . . . . . . . . . . . . . . . . . . . . . . . . 9.2.2 Compton scattering . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

140 143

9.3 Wave-particle duality of particles . . . . . . . . . . . . . . . . . . . . . . . . . .

146

9.3.1 Diffraction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9.4 Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

147 151

9.5 End of topic test . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

152

Prerequisite knowledge • Momentum and collisions. • Wavelength, frequency and speed of a light wave. • Diffraction and diffraction gratings. Learning Objectives By the end of this topic, you should be able to: • state the conditions under which waves and particles exhibit wave-particle duality; • carry out calculations involving the wave-particle duality of electromagnetic waves and electrons.

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9.1

Introduction

Early in the 20th century, physicists realised that there were certain phenomena which could not be explained using the physical laws known at the time. Principally, phenomena involving light (thought to be composed of electromagnetic waves) could not be explained using this wave model. This topic begins with a look at some of the experiments that required a new theory - quantum theory - to explain their results. This theory explains that in some circumstances a beam of light can be considered to be made up of a stream of particles called photons. Light is said to exhibit ‘wave-particle duality', as both the wave model and the particle model can be used to describe the behaviour of light. In the second part of the topic, this idea of wave-particle duality is explored further when it is shown that particles can sometimes exhibit behaviour associated with waves. Concentrating mainly on electrons, it will be shown that a stream of electrons can act in the same way as a beam of light passing through a narrow aperture.

9.2

Wave-particle duality of waves



Learning Objective To describe some situations in which waves exhibit particle-like properties.



This section looks at the photoelectric effect and Compton scattering.

9.2.1

The photoelectric effect

The photoelectric effect was first observed in experiments carried out around 100 years ago. An early experiment carried out by Hallwachs showed that a negativelycharged insulated metal plate lost its charge when exposed to ultraviolet radiation. Other experiments, using equipment such as that shown in Figure 9.1, showed that electrons can be emitted from the surface of a metal plate when the plate is illuminated with light. This phenomenon is called the photoelectric effect.

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Figure 9.1: Apparatus for photoelectric effect experiments

u-v radiation anode

-

cathode

A

power supply .......................................... The equipment shown in Figure 9.1 can be used to observe the photoelectric effect. The cathode and anode are enclosed in a vacuum, with a quartz window to allow the cathode to be illuminated using an ultraviolet lamp (quartz is used because ordinary glass does not transmit ultraviolet light). A sensitive ammeter records the current in the circuit (the photocurrent). The potentiometer can be used to provide a ‘stopping potential' to reduce the photocurrent to zero. Under these conditions, the stopping potential provides a measure of the kinetic energy of the electrons emitted from the cathode (the photoelectrons‘'), since when the photocurrent drops to zero, even the most energetic of the photoelectrons does not have sufficient energy to reach the anode. In classical wave theory, the irradiance (power per unit area) of a beam of light is proportional to the square of the amplitude of the light waves. This means that the brighter a beam of light, the more energy is falling per unit area in any period of time, as you would expect. Yet experiments showed that the speed or kinetic energy of the emitted electrons did not depend on the irradiance of the beam. In fact the energy of the photoelectrons depended on the frequency (or wavelength) of the light. Below a certain frequency, no electrons were emitted at all, whatever the irradiance of the beam. These observations cannot be explained using the wave theory of light. In 1905 Einstein proposed a quantum theory of light, in which a beam of light is considered to be a stream of particles ('quanta') of light, called photons, each with energy E, where E = hf © H ERIOT-WATT U NIVERSITY

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.......................................... In Equation 9.1, f is the frequency of the beam of light, and h is a constant, called Planck's constant after the German physicist Max Planck. The value of h is 6.63 × 10-34 J s. The equivalent expression in terms of the speed c and wavelength λ of the light waves is E=

hc λ

Example Calculate the energy of a photon in the beam of light emitted by a helium-neon laser, with wavelength 633 nm. The frequency of the photon is given by the equation c f= λ 3.00 × 108 ∴f = 6.33 × 10−7 ∴ f = 4.74 × 1014 Hz The photon energy E is therefore E = hf = 6.63 × 10−34 × 4.74 × 1014 ∴ E = 3.14 × 10−19 J .......................................... The quantum theory explains why the kinetic energy of the photoelectrons depends on the frequency, not the irradiance of the incident radiation. It also explains why there is a ‘cut-off frequency', below which no electrons are emitted. When a photon is absorbed by the cathode, its energy is used in exciting an electron. A photoelectron is emitted when that energy is sufficient for an electron to escape from an atom. The cut-off frequency corresponds to the lowest amount of energy required for an electron to overcome the attractive electrical force and escape from the atom. The conservation of energy relationship for the photoelectric effect is given by Einstein's photoelectric equation hf = hf0 + 12 me ve2

(9.2)

.......................................... In Equation 9.2, hf is the energy of the incident photon and 12 me ve2 is the kinetic energy of the photoelectron. hf 0 is called the work function of the material, and is the minimum photon energy required to produce a photoelectron. Example Monochromatic ultraviolet radiation of frequency 1.06 × 10 15 Hz is focused onto a magnesium cathode in a photoelectric effect experiment. The kinetic energy of the photoelectrons produced is 1.13 × 10 -19 J. What is the work function of magnesium? © H ERIOT-WATT U NIVERSITY

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The photon energy is E = hf = 6.63 × 10−34 × 1.06 × 1015 ∴ E = 7.03 × 10−19 J Rearranging Einstein's equation, the work function is hf0 = hf − 12 me ve2     ∴ hf0 = 7.03 × 10−19 − 1.13 × 10−19 ∴ hf0 = 5.90 × 10−19 J ..........................................

9.2.2

Compton scattering

Experiments carried out in 1923 by the American physicist Arthur Compton provided further evidence to support the photon theory. He studied the scattering of a beam of X-rays fired at a thin sheet of graphite and found that some of the emergent beam was scattered into a wide arc. Furthermore the scattered X-rays had a slightly longer wavelength than the incident beam. Compton explained his observations in terms of a collision between a photon and an electron, as shown in Figure 9.2. Figure 9.2: Compton scattering from the collision of a photon with an electron

scattered electron

incident

electron

photon

θ scattered photon

.......................................... The incident X-rays have frequency f , so an individual photon has energy E = hf . It was found that the scattered photons had frequency f  , where f  < f . We can compare the Compton scattering experiment to a moving snooker ball colliding with a stationary snooker ball. The X-ray is acting like a particle colliding with another © H ERIOT-WATT U NIVERSITY

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particle, rather than a wave. Assuming the electron is initially at rest, the equation of conservation of energy is hf = hf  + KErecoil The conservation of energy relationship shows that f  < f , since the recoil energy of the electron KErecoil is greater than zero. The wavelength of the scattered photon is λ = c/f  , so the scattered wavelength is larger than the incident wavelength. Compton also used the conservation of linear momentum in his analysis. Momentum is a property associated with particles, rather than waves, so Compton scattering is further proof of the dual nature of light. Using Einstein's energy-mass relationship from special relativity, the energy of a particle is related to its mass by the equation E = mc2 We can also express the energy of a photon in terms of its frequency f or wavelength λ E = hf =

hc λ

Combining these two equations hc λ h ∴ mc = λ h ∴p= λ .......................................... mc2 =

(9.3)

The momentum p of a photon is given by Equation 9.3. Using the conservation of energy and (relativistic) momentum, Compton showed theoretically that the shift in wavelength Δλ is given by the equation Δλ =

h (1 − cos θ) mc

Δλ depends on the scattering angle θ. Compton performed experiments that confirmed this wavelength shift, which cannot be explained using a wave model of light. Compton scattering is therefore another example of the particle nature of light. Equation 9.3 is a particularly important equation, as it demonstrates the wave-particle duality of light. The equation relates one property associated with waves, the wavelength λ, to a property associated with particles, namely the momentum p.

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Quiz 1 Wave-particle duality of waves First try the questions. If you get a question wrong or do not understand a question, there are Hints. The hints are in the same order as the questions. If you read the hint and still do not understand then ask your tutor. All references in the hints are to online materials. Useful data: 6.63 × 10-34 J s

Planck's constant h Speed of light c

3.00 × 108 m s-1

Q1: An argon laser produces monochromatic light with wavelength 512 nm (5.12 × 10-7 m). What is the energy of a single photon emitted by an argon laser? a) b) c) d) e)

3.88 × 10-19 1.29 × 10-27 1.02 × 10-31 3.39 × 10-40 1.13 × 10-48

J J J J J ..........................................

Q2: The work function of aluminium is 6.53 × 10 -19 J. What is the minimum frequency of light which can be shone on a piece of aluminium to produce photoelectrons? a) b) c) d) e)

1.01 × 10-15 Hz 9.85 × 1014 Hz 2.95 × 1023 Hz Cannot say without knowing the wavelength of the light. Cannot say without knowing the kinetic energy of the electrons. ..........................................

Q3: In a Compton scattering experiment, a) b) c) d) e)

the momentum of the photons is unchanged after scattering. the kinetic energy of the photons is unchanged after scattering. the wavelength of the scattered photons decreases. the wavelength of the scattered photons increases. the wavelength of the scattered photons is unchanged. ..........................................

Q4: A source of light contains blue (λ = 440 nm) and red (λ = 650 nm) light. Which one of the following statements is true? a) The blue photons have greater photon energy and greater photon momentum. b) The blue photons have greater photon energy, but the red photons have greater photon momentum. c) The red photons have greater photon energy, but the blue photons have greater photon momentum. d) The red photons have greater photon energy and greater photon momentum. e) All photons have the same photon energy. © H ERIOT-WATT U NIVERSITY

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.......................................... Q5: Calculate the momentum of a single microwave photon emitted with wavelength 25.0 mm. 7.96 × 10-24 7.96 × 10-27 2.65 × 10-32 2.65 × 10-35 5.53 × 10-44

a) b) c) d) e)

kg m s-1 kg m s-1 kg m s-1 kg m s-1 kg m s-1 ..........................................

9.3

Wave-particle duality of particles





Learning Objective To describe some situations in which particles exhibit wave-like properties



We have seen that light can be described in terms of a beam of particles (photons) rather than as transverse waves, under certain circumstances. The equation λ = h/p (Equation 9.3) relates a property of waves (the wavelength λ) to a property of particles (the momentum p). In 1924, Louis de Broglie (pronounced ‘de Broy') proposed that this equation could also be applied to particles. That is to say, he suggested that under certain circumstances particles would behave as if they were waves, with a wavelength given by the above equation. Within three years, experiments with electron beams had proved that this was indeed the case. The wavelength given by λ = h/p for a particle is now known as the de Broglie wavelength. As the following example shows, in most everyday cases the de Broglie wavelength is extremely small. Example Find the de Broglie wavelength of: 1. an electron (me = 9.11 × 10-31 kg) travelling at 4.00 × 105 m s-1 ; 2. a golf ball (mb = 0.120 kg) travelling at 20 m s-1 . 1. The momentum p e of the electron is pe = me ve = 9.11 × 10−31 × 4.00 × 105 = 3.64 × 10−25 kg m s−1 The de Broglie wavelength of the electron is therefore λe =

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2. The momentum p b of the golf ball is pb = mb vb = 0.120 × 20 = 2.40 kg m s−1 The de Broglie wavelength of the golf ball is therefore λb =

h 6.63 × 10−34 = 2.76 × 10−34 m = pb 2.40

The de Broglie wavelength we have calculated for the golf ball is extremely small, so small that we do not observe any wave-like behaviour for such an object. .......................................... Whilst Equation 9.3 can be used to calculate the de Broglie wavelength, it should be noted that if a particle is travelling at greater than about 0.1 c, the momentum must be determined using a relativistic calculation. Such a calculation is beyond the scope of this course.

9.3.1

Diffraction

Since we can now calculate the wavelength associated with a moving particle, under what circumstances can we expect a particle to exhibit wave-like behaviour? There are several aspects of the behaviour of light which can only be explained using a wave model. Interference, for example, occurs when two waves overlap in time and space, and the irradiance of the resultant wave depends on whether they interfere constructively or destructively. Figure 9.3: (a) Constructive and (b) destructive interference of two sine waves

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.......................................... Constructive interference (Figure 9.3(a)) of two identical light waves would result in very bright illumination at the place where they overlap. Destructive interference (Figure 9.3(b)) results in darkness - the two waves ‘cancel each other out.' Such an effect cannot usually be seen with particles - if two people kick footballs at you, you will feel the effects of both of them hitting you - they won't cancel each other out! Another property of waves, which particles do not usually exhibit is diffraction. If a wave passes through an aperture, the size of which is about the same as the wavelength of the light, the light ‘spreads out' as it emerges from the aperture. This effect is shown in Figure 9.4, where the wavefronts of two sets of waves approaching two different apertures are shown. Figure 9.4

λ

@



@  λ

λ



@∼λ

.......................................... In Figure 9.4(a), the diameter d of the aperture is much greater than the wavelength λ, and the waves incident on the aperture pass through almost unaffected. In Figure 9.4(b), the waves have a wavelength that is of the same order of magnitude as the aperture diameter, so the waves are diffracted and spread out as they emerge from the aperture. An analogy for particles would be a train travelling through a tunnel. The width of the tunnel is only slightly larger than the width of the train, so one might expect to observe diffraction. Of course this doesn't happen - the train passes through the tunnel and © H ERIOT-WATT U NIVERSITY

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continues travelling along the track. We don't observe diffraction because the de Broglie wavelength associated with the train is many orders of magnitude smaller than the tunnel width, as we have already seen with the example of the golf ball. Our earlier example showed that for electrons, the de Broglie wavelength is much larger, although still a very small number. But this wavelength (∼10 -9 to 10-10 m) is about the same as the distance between atoms in a crystalline solid. The first proof of de Broglie's hypothesis came in 1927 when the first electron diffraction experiments were performed independently by Davisson and Germer in the United States and G. P. Thomson in Aberdeen. Figure 9.5: Electron diffraction (a) schematic diagram of the experiment; (b) diffraction pattern recorded on the photographic plate

electron beam

crystal

(a)

photographic plate (b)

.......................................... The spacing between atoms in a solid is typically around 10 -10 m, so an electron travelling with a suitable momentum between two atoms in a crystal could be diffracted by the atoms. In a crystalline solid the atoms are arranged in a regular pattern or array, and act like a diffraction grating. Strong diffraction occurs in specific directions, which are determined by the atomic spacing within the crystal and the wavelength of the incident beam. Figure 9.5 shows the experimental arrangement and a typical diffraction pattern obtained. If the wavelength of the beam is known, the diffraction pattern can be used to determine the atomic spacing and crystal structure.

9.3.1.1 The electron microscope In an optical microscope a sample is illuminated with white light and viewed through a series of lenses. The resolving power, or resolution, of the microscope is limited by the wavelength of the light. A microscope allows us to view a magnified image of very small objects, or view the fine structure of a sample. The resolving power tells us the smallest particles that can be distinguished when viewing through a microscope. The resolution is diffraction-limited, because light will be spread out when passing between two objects placed about one wavelength apart, making the two objects indistinguishable. An electron microscope has a much smaller wavelength than visible light. Typically, the de Broglie wavelength of electrons in an electron microscope is around 10 -10 m whilst the wavelength of visible light is around 10 -7 m. Using an electron beam instead of a light source to ‘illuminate' the sample means smaller features of the sample can be © H ERIOT-WATT U NIVERSITY

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distinguished.

Quiz 2 Wave-particle duality of particles 20 min

First try the questions. If you get a question wrong or do not understand a question, there are Hints. The hints are in the same order as the questions. If you read the hint and still do not understand then ask your tutor. All references in the hints are to online materials. Useful data: Planck's constant h Speed of light c

Q6: a) b) c) d) e)

6.63 × 10-34 J s 3.00 × 108 m s-1

de Broglie's theory suggests that particles exhibit wave-like properties. particles have momentum and kinetic energy. only waves can be diffracted. photons always have zero momentum. photons and electrons have the same mass. ..........................................

Q7: What is the de Broglie wavelength of an electron (mass m e = 9.11 × 10-31 kg) travelling with velocity 6.40 × 106 m s-1 ? a) b) c) d) e)

1.75 × 10-15 m 2.43 × 10-12 m 2.47 × 10-12 m 1.14 × 10-10 m 1.56 × 10-7 m ..........................................

Q8: A particle has a speed of 3.0 × 10 6 m s-1 . Its de Broglie wavelength is 1.3 × 10-13 m. What is the mass of the particle? a) b) c) d) e)

1.5 × 10-38 kg 7.8 × 10-32 kg 1.7 × 10-27 kg 5.1 × 10-23 kg 5.1 × 10-21 kg ..........................................

Q9: The demonstrations of electron diffraction by Davisson & Germer and G P Thomson showed that a) b) c) d)

electrons have momentum. particles can exhibit wave-like properties. light cannot be diffracted. electrons have zero kinetic energy. © H ERIOT-WATT U NIVERSITY

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e) electrons have zero rest mass. .......................................... Q10: An electron microscope can provide higher resolution than an optical microscope because a) b) c) d) e)

electrons cannot undergo diffraction. the electrons have a shorter wavelength than visible light. electrons are particles whereas light is waves. photons cannot be focused properly. photons have a smaller momentum than electrons. ..........................................

9.4

Summary

Wave-particle duality tells us that waves can exhibit particle-like behaviour and particles can exhibit wave-like behaviour. A beam of light can be thought of as a stream of photons, each with a distinct photon energy and momentum. The photon energy E is given by the relationship E = hf and its momentum p is given by p = h/λ. These relationships also hold true for particles, the latter giving the de Broglie wavelength associated with the particle. A beam of particles can exhibit wave-like phenomena such as interference and diffraction. The electron microscope is an example of an instrument which depends on the wave-like properties of a beam of electrons in its operation. By the end of this topic you should be able to: • describe evidence which shows that electromagnetic waves can exhibit waveparticle duality; • describe evidence which shows that moving particles, such as a beam of electrons, can exhibit wave-particle duality; • state the expression p = h/λ, and use it to calculate the momentum associated with an electromagnetic wave and the wavelength associated with a moving particle; • state that the above equation relates the wave and particle models; • state that the de Broglie wavelength of a particle is usually extremely small compared to any physical system, other than on an atomic or sub-atomic scale.

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9.5

End of topic test

End of topic test 30 min

At this stage there is an end of topic test available online. If however you do not have access to the internet you may try the questions which follow. The following data should be used when required: Planck's constant h Speed of light in a vacuumc

6.63 × 10-34 J s

rest mass of an electron me

9.11 × 10-31

rest mass of an electron mp

1.67 × 10-27

3.00 × 108 m s-1

Q11: Photons emitted by a monochromatic light source have energy 3.73 × 10 -19 J. Calculate the frequency in Hz of the light. .......................................... Q12: Photons in a beam of monochromatic infrared radiation each have individual photon energy 6.35 × 10 -20 J. What is the wavelength (in metres) of the infrared beam? .......................................... Q13: Light with a maximum frequency 8.35 × 10 14 Hz is shone on a piece of metal. The work function of the metal is 3.25 × 10 -19 J. Calculate the maximum kinetic energy of the electrons produced due to the photoelectric effect, measured in J. .......................................... Q14: A laser emits monochromatic light with wavelength 455 nm. Calculate the momentum (in kg m s

-1

) of a single photon emitted by the laser.

.......................................... Q15: A filament lamp emits light with a wavelength range of 375 to 680 nm. Calculate the largest value of photon momentum (in kg m s by the lamp.

-1

) of the photons emitted

.......................................... Q16: A proton and an electron are both travelling at the same velocity. 1. Which particle has the greater momentum? A) the proton B) the electron C) neither: both have the same momentum 2. Which particle has the greater de Broglie wavelength? A) the proton B) the electron © H ERIOT-WATT U NIVERSITY

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C) neither: both have the same de Broglie wavelength .......................................... Q17: In a television tube, electrons are accelerated by a high voltage, and strike the screen with an average velocity of 5.95 × 10 6 m s -1 . 1. On average, what is the momentum (measured in kg m s before it strikes the screen?

-1

) of an electron just

2. Calculate the average de Broglie wavelength (in m) of the electrons. ..........................................

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Topic 10

Introduction to quantum mechanics

Contents 10.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10.2 Atomic models . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

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10.2.1 The Bohr model of the hydrogen atom . . . . . . . . . . . . . . . . . . . 10.2.2 Atomic spectra . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

157 160

10.3 Quantum mechanics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

162

10.3.1 Limitations of the Bohr hydrogen model . . . . . . . . . . . . . . . . . . 10.3.2 Quantum mechanics . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

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10.4 Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10.5 End of topic test . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

163 164

Prerequisite knowledge • Rutherford scattering. • Circular motion (Mechanics topics 3 and 4). • Angular momentum (Mechanics topic 5). Learning Objectives By the end of this topic, you should be able to: • describe the Bohr model of the atom; • understand some of the basic principles of quantum mechanics.

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10.1

Introduction

Towards the end of the 19th century, physical phenomena were described in terms of ‘classical' theory, as either particles or waves. However, some new discoveries (such as the photoelectric effect) could not be explained using classical theory. As we have seen, such phenomena required a theory that included a particle-like description of light. We will see in this Topic that a quantum approach was also used to answer questions about the structure of the atom. The hydrogen atom has the simplest structure, and we will look at a model of the hydrogen atom based on quantum theory and wave-particle duality. We will see that the emission spectrum of hydrogen is due to the electron moving between its allowed orbits, which can be determined by treating the electron as a wave. Whilst the model of the atom dealt with here gives good agreement with experiments performed on the hydrogen atom, it is found to be unsuitable for larger atoms and molecules. The theory of quantum mechanics, which is used to describe such atoms, is introduced at the end of the topic.

10.2

Atomic models

The widely accepted view at the beginning of the 20th century was that an atom consisted of a large positively-charged mass with negatively-charged electrons embedded in it at random positions. This ‘plum pudding' model (suggested by J J Thomson) was consistent with experimental data available at the time. But the Rutherford scattering experiments first performed early in the 20th century were not consistent with that model, and required a major rethink. Rutherford interpreted the results of his experiments as evidence that the atom consisted of a relatively massive positively charged nucleus with electrons of far lower mass orbiting around it. The study of atomic spectra also gave interesting results. The emission spectra of elements such as hydrogen, shown in Figure 10.1, consist of a series of discrete lines, rather than a continuous spectrum. In terms of photons, there was no explanation as to why a particular element would only emit photons with certain energies. Figure 10.1: Emission spectrum of hydrogen



     

}

}

}

         

  

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.......................................... The next stage in the development of atomic theory was to try to build up a picture in which these experimental results could be explained.

10.2.1

The Bohr model of the hydrogen atom

In 1913 the Danish physicist Neils Bohr proposed an alternative model for the hydrogen atom. In the Bohr model, the electron orbits the nucleus (consisting of one proton) in a circular path, as shown in Figure 10.2. (This figure is not drawn to scale). Figure 10.2: Simple model of the hydrogen atom

L

ep+

.......................................... Bohr assumed that the electron could only have certain allowed values of energy, the total energy being made up of kinetic energy (due to its circular motion) and potential energy (due to the electrical field in which it was located). Each value of energy corresponded to a unique electron orbit, so the orbit radius could only take certain values. Using classical and quantum theory, Bohr calculated the allowed values of energy and radius, matching the gaps between electron energies to the photon energies observed in the line spectrum of hydrogen. Since the electron is moving with speed v in a circle of radius r, its centripetal 2

acceleration is equal to v r . This presents a problem, as classical physics theory states that an accelerating charge emits electromagnetic radiation. A ‘classical' electron would therefore be losing energy, and would spiral into the nucleus, rather than continue to move in a circular path. Bohr suggested that certain orbits in which the electron had an allowed value of angular momentum were stable. The angular momentum L of any particle of mass m moving with speed v in a circle of radius r is L = mvr. According to Bohr, so long as the angular momentum of the electron is a multiple of h/2π , the orbit is stable (h is Planck's constant). Thus the Bohr model proposed the concept of quantisation of angular momentum of the electron in a hydrogen atom. That is to say, the electron must obey the condition

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nh 2π .......................................... angular momentum =

(10.1)

where n is an integer. This quantisation of angular momentum fitted in with the predicted energy levels, but left a crucial question unanswered. Why were these particular orbits allowed? In other words, what made this value of angular momentum so special, and why did having angular momentum of nh/2π make the orbit stable? The answer to this came when de Broglie's ideas on wave-particle duality were published a decade later. Treating the electron as a stationary wave, let us suppose the smallest allowed circumference of the electron's orbit corresponds to one wavelength λ. The next allowable orbit corresponds to 2λ, and so on. The orbit will be stable, then, if the circumference is equal to nλ. A stationary wave does not transmit energy, so if the electron is acting as a stationary wave then all its energy is confined within the atom, and the problem of a classical electron radiating energy does not occur. Figure 10.3: Standing wave orbit of the electron in a hydrogen atom

λ

p+

.......................................... Figure 10.3 shows an electron in the n = 6 orbit. There are six de Broglie wavelengths in the electron orbit. The circumference of a circle of radius r is 2πr, so the condition for a stable orbit is nλ = 2πr

(10.2)

.......................................... Rearranging the de Broglie equation p = h/λ λ=

h h = p mv © H ERIOT-WATT U NIVERSITY

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Now, we can substitute this expression for λ into Equation 10.2. nλ = 2πr nh = 2πr ∴ mv nh ∴ = mvr 2π nh ∴ mvr = 2π ..........................................

(10.3)

Thus the angular momentum mvr is a multiple of h/2π , as predicted by Bohr (Equation 10.1). Treating the electron as a stationary wave gives us the quantisation of angular momentum predicted by Bohr. The unit of allowed angular momentum is sometimes given the symbol  ('h bar'), where  = h/2π . It should be pointed out that this result gives a good model for the hydrogen atom, but does not give us the complete picture. For instance, the electron is actually moving in three dimensions, whereas the Bohr atom only considers two dimensions. We will see later in this Topic that a full 3-dimensional wave function is used nowadays to describe an electron orbiting in an atom. Example In the n = 2 orbit of the hydrogen atom, the electron can be considered as a particle travelling with speed 1.09 × 10 6 m s-1 . Calculate: 1. the angular momentum of the electron; 2. the de Broglie wavelength of the electron; 3. the radius of the electron's orbit. 1. Since we have n = 2 nh 2π 2 × 6.63 × 10−34 ∴L= 2π ∴ L = 2.11 × 10−34 kg m2 s−1 L=

2. The de Broglie wavelength is λ=

h h = p mv

6.63 × 10−34 9.11 × 10−31 × 1.09 × 106 ∴ λ = 6.68 × 10−10 m

∴λ=

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3. The angular momentum is quantised, so we can use Equation 10.3 to find r nh 2π 2 × 6.63 × 10−34 ∴r= 2πmv 1.326 × 10−34 ∴r= 2π × 9.11 × 10−31 × 1.09 × 106 ∴ r = 2.13 × 10−10 m

mvr =

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

The Bohr model of the hydrogen atom At this stage there is an online activity. 15 min

This simulation allows you to view the electron as a particle or wave in the Bohr atom. This animation allows you to view the electron as a particle or as a wave. Make sure you understand why the wave pattern looks different for each value of the quantum number n. ..........................................

10.2.2

Atomic spectra

The Bohr model tells us the allowed values of angular momentum for the electron in a hydrogen atom in terms of the quantum number n. For each value of n, the electron has a specific value of angular momentum L and total electron energy E. The Bohr model of the hydrogen atom allows E to be calculated for any value of n. When the electron moves between two orbits, it either absorbs or emits energy (in the form of a photon) in order to conserve energy, as shown in Figure 10.4. Figure 10.4: (a) Emission and (b) absorption of a photon

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The line spectrum produced by atomic hydrogen allows us to calculate the difference between energy levels. When an electron moves from a large n orbit to a lower n orbit it loses energy, this energy being emitted in the form of a photon (Figure 10.4(a)). Similarly, absorbtion of a photon raises the electron to a higher n orbit (Figure 10.4(b)). The important point here is that only photons with the correct energy can be emitted or absorbed. The photon energy must be exactly equal to the energy difference between two allowed orbits. This is why the emission spectrum of hydrogen consists of a series of lines rather than a continuous spectrum.

Hydrogen line spectrum At this stage there is an online activity which explores how some of the lines in the visible part of the hydrogen spectrum are produced.

20 min

This simulation shows how some of the lines in the visible part of the hydrogen spectrum are produced. Make sure you understand why we see a series of lines. ..........................................

Quiz 1 Atomic models First try the questions. If you get a question wrong or do not understand a question, there are Hints. The hints are in the same order as the questions. If you read the hint and still do not understand then ask your tutor. All references in the hints are to online materials. Useful data: Planck's constant h mass of an electron me

6.63 × 10-34 J s 9.11 × 10-31 kg

Q1: In Bohr's model of the hydrogen atom, a) b) c) d) e)

photons orbit the atom. the electron's angular momentum is quantised. the electron constantly emits electromagnetic radiation. the electron constantly absorbs electromagnetic radiation. the angular momentum is constantly changing. ..........................................

Q2: What is the angular momentum of an electron orbiting a hydrogen atom, if the quantum number n of the electron is 4? a) b) c) d) e)

1.67 × 10-32 8.33 × 10-33 8.44 × 10-34 4.22 × 10-34 2.64 × 10-35

kg m2 kg m2 kg m2 kg m2 kg m2

s-1 s-1 s-1 s-1 s-1

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.......................................... Q3: If an electron is orbiting a hydrogen atom with a de Broglie wavelength of λ and quantum number n, the radius r of the orbit is given by the equation a) r = 2πnλ b) r = nλ/h c) r = 2π/nλ d) r = h/nλ e) r = nλ/2π .......................................... Q4: a) b) c) d) e)

The emission spectrum of hydrogen consists of a series of lines. This is because photons can only be emitted with specific energies. photons are re-absorbed by other hydrogen atoms. the photons have zero momentum. hydrogen atoms only have one electron. once a photon is emitted it is immediately re-absorbed. ..........................................

10.3

Quantum mechanics

The two sections covered in this part are: • Limitations of the Bohr hydrogen model • Quantum mechanics

10.3.1

Limitations of the Bohr hydrogen model

The Bohr model gives an accurate prediction of the spectrum of atomic hydrogen. It also works well for hydrogen-like ions which have a single electron orbiting a nucleus, such as the helium (He+ ) and lithium (Li2+ ) ions. In these cases the larger central charge means that the electrons have different energies associated with the n = 1,2,3... levels, but the underlying principle is the same - the electron has discrete energy levels, and the spectra of these ions are series of lines. The model soon becomes inadequate when more electrons are added to the system. The motion of one electron is affected not only by the static electric field due to the nucleus, but also by the moving electric field due to the other electrons. The picture becomes even more complicated when we start dealing with molecules such as carbon dioxide. A more sophisticated theory is required to describe atoms and molecules with more than one electron.

10.3.2

Quantum mechanics

Nowadays quantum mechanics is used to describe atoms and electrons. In this theory the motion of an electron is described by a wave function Ψ. The idea of using a © H ERIOT-WATT U NIVERSITY

TOPIC 10. INTRODUCTION TO QUANTUM MECHANICS

wave function was proposed by the German physicist Erwin Schrodinger, and Ψ is often referred to as the Schrodinger wave function. The motion is described in terms of probabilities, and the wave function is used to determine the probability of finding an electron at a particular location in three dimensional space. The electron cannot be thought of as a point object at a specific position; instead we can calculate the probability of finding the electron within a certain region, within a certain time period. Another German physicist, Werner Heisenberg, was also amongst the first to propose this concept. How does this relate to the Bohr hydrogen atom? The radii corresponding to different values of n predicted by Bohr are the positions of maximum probability given by quantum mechanics, so for the hydrogen atom the two theories are compatible. The allowed energy and angular momentum values calculated using the Bohr model are still correct. As we have mentioned already though, more complicated atoms and molecules require quantum mechanics.

10.4

Summary

A simple model of the hydrogen atom, the Bohr model, proposes that the electron is only allowed to orbit the nucleus with certain values of angular momentum. This theory states that the electron can only have certain allowed orbit radii and energies. Applying de Broglie's theory to the orbiting electron, treating it as a wave, confirms that angular momentum is quantised in units of h/2π . These theoretical concepts match experimental measurements of the hydrogen line spectrum. The lines in the spectrum match the energy differences between allowed electron energy levels. A photon with energy equal to a particular energy difference is absorbed or emitted when an electron moves between allowed orbits (and hence allowed energy levels). Whilst the Bohr model gives accurate results for hydrogen, more complicated atoms and molecules require a more sophisticated theory. That theory is quantum mechanics, in which a wave function is used to determine probabilities of an electron's motion and position. By the end of this topic you should be able to: • state that the angular momentum of an electron about a nucleus is quantised in units of h/2π ; • state the equation nh 2π and perform calculations using this equation; mvr =

• qualitatively describe the Bohr model of the atom; • state that quantum mechanics is used to provide a wider-ranging model of the atom than the Bohr model, and state that quantum mechanics can be used to determine probabilities. © H ERIOT-WATT U NIVERSITY

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10.5

End of topic test

End of topic test 15 min

At this stage there is an end of topic test available online. If however you do not have access to the internet you may try the questions which follow. The following data should be used when required: 6.63 × 10-34 J s 3.00 × 108 m s-1 9.11 × 10-31 kg

Planck's constant h speed of light in a vacuum c rest mass of an electron me rest mass of an proton mp

Q5:

1.67 × 10-27 kg

Consider an electron orbiting in a Bohr hydrogen atom.

Calculate the angular momentum in kg m of the electron is 6.

2

s -1 of the electron, if the quantum number n

.......................................... Q6: An electron in the n = 1 orbit of the hydrogen atom moves in a circle of radius 5.29 × 10-11 m around the nucleus. Calculate the de Broglie wavelength of the electron, in m. .......................................... Q7: An electron in a hydrogen-like atom in the n = 3 orbital has a de Broglie wavelength of 1.85 × 10-10 m. Calculate the orbit radius of the electron, in m. .......................................... Q8: An electron orbiting in the n = 4 level of a hydrogen-like atom drops to the n = 2 level, emitting a photon. The energy difference between the two levels is 2.29 × 10 -19 J. 1. What is the energy, measured in J, of the emitted photon? 2. Calculate the wavelength in m of the emitted photon. ..........................................

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Topic 11

Mechanics end-of-unit assessment

Contents

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TOPIC 11. MECHANICS END-OF-UNIT ASSESSMENT

End-of-unit assessment 30 min

At this stage the end of unit test is available online. If however you do not have access to the internet you may try the questions which follow. The following data should be used when required: acceleration due to gravity g gravitational constant G

9.8 m s-2 6.67 × 10-11 N m2 kg-2

mass of the earth

5.97 × 1024 kg

radius of the earth

6.38 × 106 m

Planck's constant h speed of light in a vacuum c

6.63 × 10-34 J s

rest mass of an electron m e

9.11 × 10-31 kg

rest mass of a proton m p

1.67 × 10-27 kg

3.00 × 108 m s-1

Q1: 1. A stone dropped from the roof of a building hits the ground travelling at speed 22.5 m s-1 . How tall is the building? 2. An electron is travelling at speed v. The relativistic mass of an electron travelling at this speed is 9.43 × 10-31 kg. Calculate the value of v. .......................................... Q2: A mass of 0.48 kg is being rotated by a string in a vertical circle of radius 1.1 m at a constant speed ω m s-1 . The tension in the string when the mass is at the bottom of the circle is 12 N. 1. Calculate ω. 2. Calculate the tension when the mass is at the top of the circle. 3. Calculate the tension when the string is horizontal. .......................................... Q3: A disc of moment of inertia 32 kg m 2 is made to rotate about an axis through its centre by a torque of T. The disc starts from rest, and after 14 s has kinetic energy 580 J. 1. Calculate the angular velocity after 14 s. 2. Calculate the magnitude of the torque T. 3. The torque is removed when the disc is rotating at 8.0 rad s -1 . An opposing torque of 12.5 N m is applied to slow the disc down. Calculate the number of revolutions the disc makes after the second torque is applied, before it comes to rest.

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TOPIC 11. MECHANICS END-OF-UNIT ASSESSMENT

.......................................... Q4: A satellite of mass 1060 kg is orbiting the earth at a height of 5.75 × 10 5 m above the earth's surface. 1. Calculate the gravitational force that the earth exerts on the satellite. 2. Calculate the gravitational potential at this height. .......................................... Q5: An object of mass 0.21 kg is performing simple harmonic oscillations, with amplitude 16 mm and periodic time 1.8 s. 1. Calculate the maximum value of the object's acceleration. 2. Calculate the maximum value of the object's velocity. .......................................... Q6: A proton in a particle accelerator is travelling with velocity 1.45 × 10 5 m s-1 . Calculate the de Broglie wavelength of the proton, measured in m. .......................................... An on-line assessment is provided to help you review this unit. ..........................................

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GLOSSARY

Glossary Acceleration The rate of change of velocity. Acceleration is a vector quantity, measured in m s-2 . Amplitude The maximum displacement of an oscillating object from the zero displacement (equilibrium) position. Angular acceleration The rate of change of angular velocity, measured in rad s -2 . Angular displacement The angle, measured in radians, through which a point or line has been rotated about an axis, in a specified direction. angular momentum The product of the angular velocity of a rotating object and its moment of inertia about the axis of rotation, measured in kg m 2 s-1 . Angular velocity The rate of change of angular displacement, measured in rad s -1 . black hole An object in space, usually formed when a star collapses under its own gravitational forces, which has an escape velocity that exceeds the speed of light c. Bohr model A model of the hydrogen atom, in which the electron orbits numbered n = 1,2,3....have angular momentum nh/2π . Centripetal acceleration The acceleration of an object moving in a circular path, which is always directed towards the centre of the circle. Centripetal force A force acting on an object causing it to move in a circular path. Coefficient of static friction The coefficient that determines the size of the static frictional force between two surfaces. If the frictional force is small (for a smooth mass on a polished surface, say), the coefficient is nearly equal to zero. If there is a large frictional force (for example, a cycle brake block pressing against a wheel), then the coefficient is nearly equal to one. Compton scattering The reduction in energy of a photon due to its collision with an electron. The scattering manifests itself in an increase in the photon's wavelength. © H ERIOT-WATT U NIVERSITY

GLOSSARY

Conical pendulum A pendulum consisting of a bob on a string moving in a horizontal circle. Conservative field A field in which the work done in moving an object between two points in the field is independent of the path taken. Damping A decrease in the amplitude of oscillations due to the loss of energy from the oscillating system, for example the loss of energy due to work against friction. de Broglie wavelength A particle travelling with momentum p has a wavelength λ associated with it, the two quantities being linked by the relationship λ = h/p. λ is called the de Broglie wavelength. Displacement The location of an object, in terms of its distance from an origin or reference point, and its direction. Displacement is a vector quantity, measured in m. escape velocity The minimum speed required for an object to escape from the gravitational field of another object, for example the minimum speed a rocket taking off from Earth would need to escape the Earth's gravitational field. Frequency The rate of repetition of a single event, in this case the rate of oscillation. Frequency is measured in hertz (Hz), equivalent to s -1 . Geostationary satellite A satellite which orbits the Earth above the equator, with the same periodic time as the Earth's rotation. The satellite remains directly above the same point on the Earth's surface. Gravitational field The region of space around an object in which any other object with a mass will have a gravitational force exerted on it by the first object. Gravitational field strength The gravitational field strength at a point in a gravitational field is equal to the force acting per unit mass placed at that point in the field. Gravitational potential At a particular point in a gravitational field, the gravitational potential is the work done by external forces in bringing a unit mass from infinity to that point. Horizontal range The horizontal distance travelled by a projectile when it is launched from the ground with a given speed and launch angle, before it returns to the ground. The maximum value of the range is obtained when the projectile is launched at an angle of 45 ◦ to the ground. © H ERIOT-WATT U NIVERSITY

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GLOSSARY

Moment of inertia The moment of inertia of an object about an axis is the sum of mass × distance from the axis, for all elements of the object. Newtonian mechanics The system of mechanics in which objects obey Newton's laws of motion. Periodic time The time taken for an object rotating or moving in a circle to complete exactly one revolution. Photoelectric effect The emission of electrons from a substance exposed to electromagnetic radiation. Photoelectrons An electron emitted by a substance due to the photoelectric effect. Photons A quantum of electromagnetic radiation, with energy E = hf , where f is the frequency of the radiation and h is Planck's constant. Quantum mechanics A system of mechanics, developed from quantum theory, used to explain the properties of particles, atoms and molecules. radian A unit of measurement of angle, where one radian is equivalent to 180/π degrees. One radian is defined in terms of the angle at the centre of a circle made by the sector of the circle in which the length along the circumference is equal to the radius of the circle. Relativistic mechanics The system of mechanics in which an object is travelling at close to the speed of light, and hence no longer obeys the laws of Newtonian mechanics. Rest mass The mass of an object when it is stationary with respect to an observer. Rotational kinetic energy The kinetic energy of an object due to its rotation about an axis, measured in J. A rotating object still has kinetic energy even if it is not moving from one place to another. The total kinetic energy of a rotating object is the sum of the rotational and translational kinetic energies. Simple harmonic motion (SHM) Motion in which an object oscillates around a fixed (equilibrium) position. The acceleration of the object is proportional to its displacement, and is always directed towards the equilibrium point.

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GLOSSARY

Tangential acceleration The rate of change of tangential speed, measured in m s -2 . Tangential speed The speed in m s -1 of an object undergoing circular motion, which is always at a tangent to the circle. Torque Also called the moment or the couple, the torque due to a force is the turning effect of the force, equal to the size of the force times the perpendicular distance from the point where the force is applied to the point of rotation. Torque is measured in units of N m. Torsion balance Very small forces can be measured using a torsion balance. The forces are applied to the end of a light rod suspended at its centre by a vertical thread. As the forces turn the rod, the restoring torque (turning force) in the thread increases until the turning forces balance. Universal Law of Gravitation Also known as Newton's law of gravitation, this law states that there is a force of attraction between any two massive objects in the universe. For two point objects with masses m1 and m2 , placed a distance r apart, the size of the force F is given by the equation Gm1 m2 F = r2 Velocity The rate of change of displacement. Velocity is a vector quantity, measured in m s-1 . Wave function A mathematical function used to determine the probability of finding a quantum mechanical particle within a certain region of space. Wave-particle duality The concept that, under certain conditions, waves can exhibit particle-like behaviour and particles can exhibit wave-like behaviour. Weight The weight of an object is equal to the force exerted on it by the Earth. Work function In terms of the photoelectric effect, the work function of a substance is the minimum photon energy which can cause an electron to be emitted by that substance.

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HINTS

Hints for activities Topic 1: Kinematic relationships Horizontal Motion Hint 1: List the data you are given in the question: u = ...m s-1 , v = ...m s-1 , s = ...m, a = ? Once you have the data listed, decide on the appropriate kinematic relationship.

Quiz 1 Motion in one dimension Hint 1: First sketch the graph - or see the examples in the section Motion in one dimension Hint 2: This is a straight application of the second equation of motion. Hint 3: This is a straight application of the second equation of motion. Hint 4: You must choose a positive direction 'up' or 'down' - the initial vertical velocity is up and the final displacement is down - be careful with the sign of the acceleration. Hint 5: First, work out the displacement of the stone from its starting position. Initial velocity is zero, a = g - use the third equation of motion to find v.

Motion in two dimensions Hint 1: To calculate the minimum take-off speed, we need to separate the motion into horizontal and vertical components. There are two unknowns in this problem - the takeoff speed u and the time-of-flight t. By looking at the orthogonal components, we can eliminate t and calculate u.

Quiz 2 Motion in two dimensions Hint 1: The initial horizontal velocity of the bullet is the same as the horizontal velocity of the truck. Hint 2: If air resistance can be ignored, the maximum range occurs when the angle of elevation is 45◦ - see the end of the section Motion in 2 dimensions for the proof of this. Hint 3: Use vertical motion to find an expression for the time of flight in terms of the angle of elevation - this time is equal to horizontal distance divided by the horizontal component of the arrow's velocity Hint 4: The horizontal velocity of the boy should equal the horizontal component of the velocity of the ball. © H ERIOT-WATT U NIVERSITY

HINTS

173

Hint 5: First use vertical motion to find the time for the package to fall 900 m; then use this time for the horizontal motion of the plane.

Topic 2: Relativistic motion Quiz 1 Relativistic dynamics Hint 1: This is a straight application of m=

mo 1−

v2 c2

Hint 2: See the Introduction to Relativistic motion. Hint 3: See the section titled Relativistic dynamics Hint 4: Substitute m = 2mo in m=

mo 1−

v2 c2

Hint 5: When v << c ,

v c

is a very small number

Quiz 2 Relativistic energy Hint 1: See the section titled Relativistic energy. Hint 2: See the section titled Relativistic energy. Hint 3: Use E = mc2 Hint 4: See the activity titled Relativistic energy Hint 5: The kinetic energy is equal to the total energy of the proton at this velocity minus its rest energy - see the derivation of Equation m=

© H ERIOT-WATT U NIVERSITY

m0

2

1 − v c2

174

HINTS

Topic 3: Angular velocity and acceleration Quiz 1 Radian measurement Hint 1: 360◦ = 2π radians. Hint 2: 360◦ = 2π radians. Hint 3: 360◦ = 2π radians. Hint 4: 1 complete rotation = 2π radians Hint 5: First find out the fraction of the circumference that the object has moved.

Will the car stop before the lights? Hint 1: Firstly, you need to calculate the time taken using the linear kinematic relationships. Hint 2: u = 12.0 m s-1 , v = 0 m s-1 , s = 30.0 m, a = ? Use s=

u+v ×t 2

to obtain the time taken: t = 5.00 s. Now use the appropriate angular kinematic relationship.

Quiz 2 Angular velocity and angular kinematic relationships Hint 1: ω = 2πf Hint 2: This is the angular equivalent of a speed, displacement, time calculation. Hint 3: This is the angular equivalent of a speed, displacement, time calculation. Hint 4: First work out the change in angular velocity Hint 5: The angular displacement after one complete revolution = 2π.

Quiz 3 Angular velocity and tangential speed Hint 1: This is a straight application of v = rω. Hint 2: This is a straight application of v = rω. Hint 3: First work out the circumference of the circle. Hint 4: First work out the angular deceleration then use a = rα. Hint 5: First work out the angular displacement. © H ERIOT-WATT U NIVERSITY

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Topic 4: Centripetal force Quiz 1: Centripetal acceleration Hint 1: This is a straight application of a⊥ = r 2 . Hint 2: See the section titled Centripetal acceleration. Hint 3: This is a straight application of a⊥ =

v2 r

Hint 4: First, work out either ω or v. Hint 5: Substitute r = 2r in a⊥ =

v2 r

Quiz 2 Horizontal and vertical motion Hint 1: This is a straight application of F =

mv 2 r

F =

mv 2 r

Hint 2: This is a straight application of

Hint 3: Consider each statement in turn while referring to F =

mv 2 r

Hint 4: Calculate the speed for a central force equal to the maximum tension. Hint 5: At what point is the tension in the string greatest? See the section on vertical motion.

Quiz 3 Conical pendulum and cornering Hint 1: See the section titled Conical pendulum. © H ERIOT-WATT U NIVERSITY

176

HINTS

Hint 2: See the section titled Conical pendulum for the derivation of the relationship cos ϕ =

g lω 2

cos ϕ =

g lω 2

. Remember to use SI units. Hint 3: Start from

and remember ω = 2πf Hint 4: See the section titled Cars cornering Hint 5: See the activity banked corners.

Topic 5: Rotational dynamics Quiz 1 Torques Hint 1: Consider the units on both sides of the relationship T = Fr. Hint 2: This is a straight application of T = Fr. Hint 3: This is a straight application of T = Fr. Hint 4: First, calculate the component of the force perpendicular to the radius. Hint 5: First calculate the torque and use this to find the component of the force perpendicular to the radius.

Quiz 2 Equilibrium and moment of inertia Hint 1: The magnitude of the two torques must be equal. Hint 2: The magnitude of the two torques must be equal. Hint 3: The two anticlockwise torques are balanced by the clockwise torque due to the vertical component of the tension in the chain. Hint 4: This is a straight application of I = mr2 . Hint 5: The total moment of inertia is the arithmetic sum of the moments of inertia of the individual masses.

Quiz 3 Angular momentum and rotational kinetic energy Hint 1: This is a straight application of L = Iω. Hint 2: Use the relationship given to find the moment of inertia of the disc. Then apply L = Iω.

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HINTS

177

Hint 3: Angular momentum is conserved. What effect does adding the clay have on the total moment of inertia? Hint 4: Use the relationship given to find the moment of inertia of the sheet. Then apply rotational 1 Ek = Iω 2 2 . Hint 5: First, work out the value of ω.

Topic 6: Gravitational force and field Quiz 1 Gravitational force Hint 1: This is a straight application of F =

Gm1 m2 r2

Hint 2: This is an example of Newton's Third Law which is sometimes stated as "To every action there is an equal an opposite reaction." Hint 3: Apply weight =

Gm1 m2 r2

Hint 4: What happens to the distance between the object and the centre of the planet? Hint 5: Consider the weight of a mass of 1 kg on the surface of Venus

Quiz 2 Gravitational fields Hint 1: This is a straight application of g=

Gm r2

Hint 2: Think of Newton's Second Law! Hint 3: The gravitational strength is halved - this means the distance from the centre of the planet is increased by a factor √ 2 - see the relationship g=

Gm r2

Hint 4: The two gravitational forces must be equal and opposite. Hint 5: This is a straight application of g= © H ERIOT-WATT U NIVERSITY

Gm r2

178

HINTS

Topic 7: Gravitational potential and satellite motion Quiz 1 Gravitational potential Hint 1: Consider the relationship U =−

Gm r

. Hint 2: This is a straight application of U =−

Gm r

. Hint 3: This is an application of Gm r . Remember r is the distance from the centre of the Earth. U =−

Hint 4: This is a straight application of PE = −

Gm1 m2 r

. Hint 5: Consider the options in the context of the relationship PE = −

Gm1 m2 r

.

Quiz 2 Satellite motion Hint 1: This is a straight application of  v=

GmE r

. Hint 2: Consider the options in terms of the relationships in the section titled Satellite motion. Hint 3: See the section titled Geostationary satellites. Hint 4: First, use the relationship Gm1 m2 r to figure out what happens to the potential energy - notice the negative sign!!. Re kinetic energy, consider the relationship  GmE v= r PE = −

; what happens to v when r decreases? © H ERIOT-WATT U NIVERSITY

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179

Hint 5: Rearrange the relationship at the end of the section titled Geostationary satellites.

Quiz 3 Escape velocity Hint 1: See the section titled Escape velocity. Hint 2: Consider the relationship  v=

2GmE rE

. Hint 3: This is an application of the relationship  v=

2Gmplanet rplanet

. Hint 4: See the section titled Black holes

Topic 8: Simple harmonic motion Quiz 1 Defining SHM and equations of motion Hint 1: See the section titled Defining SHM. Hint 2: ω = 2πf . Hint 3: See the section titled Defining SHM. Hint 4: See the relationships in the section titled Equations of motion in SHM. Calculate ω. Maximum acceleration occurs when displacement equals amplitude. Hint 5: See the relationships in the section titled Equations of motion in SHM.

Quiz 2 Energy in SHM Hint 1: See the relationships in the section titled Energy in SHM. Hint 2: The total energy of the system is constant Hint 3: Consider the relationship 1 Ek = mω 2 (a2 − y 2 ) 2 . © H ERIOT-WATT U NIVERSITY

180

HINTS

Hint 4: PE = KE when half of the total energy is kinetic and half is potential - find the displacement when the kinetic energy is half its maximum value. Hint 5: Consider the relationship Ek =

1 mω 2 (a2 − y 2 ) 2

with y=

1 a 2

.

Quiz 3 SHM Systems Hint 1: See the section titled Mass on a spring - vertical oscillations. First, find ω and hence find the period. Hint 2: See the section titled Simple pendulum. frequency.

First, find ω and hence find the

Hint 3: See the section titled Mass on a spring - vertical oscillations. First, find ω and hence find the spring constant. Hint 4: See the section titled Simple pendulum. Hint 5: See the section titled Simple pendulum - how does ω vary with l ?

Topic 9: Wave-particle duality Quiz 1 Wave-particle duality of waves Hint 1: This is a straight application of E = hf. Hint 2: The minimum photon energy equals the work function. Hint 3: See the section titled Compton scattering. Hint 4: E = hf and p=

h λ

p=

h λ

. Hint 5: This is a straight application of

.

Quiz 2 Wave-particle duality of particles Hint 1: See the section titled Wave-particle duality of particles. © H ERIOT-WATT U NIVERSITY

HINTS

181

Hint 2: This is a straight application of λ=

h p

λ=

h p

. Hint 3: This is a straight application of

. Hint 4: See the section titled Diffraction. Hint 5: See the section titled The electron microscope.

Topic 10: Introduction to quantum mechanics Quiz 1 Atomic models Hint 1: See the section titled The Bohr model of the hydrogen atom. Hint 2: See the section titled The Bohr model of the hydrogen atom - in particular the relationship for angular momentum. Hint 3: See the section titled The Bohr model of the hydrogen atom. Hint 4: See the section titled Atomic spectra.

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ANSWERS: TOPIC 1

Answers to questions and activities 1 Kinematic relationships Horizontal Motion (page 6) List the data you are given in the question: u = 12.0 m s-1 , v = 0 m s-1 , s = 30.0 m, a = ? The appropriate kinematic relationship is v 2 = u2 + 2as Putting the values into this equation, v 2 = u2 + 2as ∴ 02 = 12.02 + (2 × a × 30.0) ∴ 0 = 144 + 60a ∴ −60a = 144 144 ∴a=− 60 ∴ a = −2.40 m s−2 So to stop the car in exactly 10.0 m, the car must have an acceleration of -2.40 m s -2 , equivalent to a deceleration of 2.40 m s -2 .

Quiz 1 Motion in one dimension (page 8) Q1:

a) increases with time.

Q2:

d) 45.0 m

Q3:

c) 5.05 s.

Q4:

d) 13 m s-1

Q5:

c) 18.8 m s-1

Motion in two dimensions (page 12) Horizontal. The horizontal component of u = u x = u cos 20 m s-1 and is constant. Hence we can use the equation s x = ux ×t with the value of sx = - 41.0 m.

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ANSWERS: TOPIC 1

183

sx = ux × t sx ∴ ux = t 41.0 ∴ u cos 20 = t 41.0 ∴u= t cos 20 We now have an expression for u in terms of t. Vertical. In the situation where the rider just makes it to the landing ramp, we can calculate how long he has spent in the air. At the point where he lands, s y = 0 m, uy = u sin 20 m s-1 , a = g = -9.8 m s-2 and t is unknown. We use the equation

sy = uy t + 12 at2

to find an expression for t in terms of u. sy = uy t + 12 at2 ∴ 0 = (u sin 20 × t) −

9.8 2 t 2

Ignoring the t = 0 solution, we can divide by t and rearrange this equation 0 = (u sin 20 × t) − ∴ 4.9t = u sin 20 u sin 20 ∴t= 4.9 Solution. In the equation u= we can replace t by

9.8 2 t 2

41 t cos 20

u sin 20 4.9

. So

41 × 4.9 u sin 20 × cos 20 41 × 4.9 ∴ u2 = sin 20 × cos 20 2 ∴ u = 625 u=

∴ u = 25 m s−1 © H ERIOT-WATT U NIVERSITY

184

ANSWERS: TOPIC 1

Quiz 2 Motion in two dimensions (page 13) Q6:

c) On the truck.

Q7:

e) 14.7 m

Q8:

a) 15.3◦

Q9:

b) 2.07 m s-1

Q10: b) 118 m s-1

End of topic test (page 14) Q11: 1.5 m s -2 Q12: 41 m Q13: 3.3 s Q14: 1.8 s Q15: 2.6 s Q16: 43 m Q17: 15 ◦

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ANSWERS: TOPIC 2

2 Relativistic motion Quiz 1 Relativistic dynamics (page 20) Q1: a) 1.77 × 10-27 kg Q2: c) objects moving at very high speeds do not obey the laws of Newtonian mechanics. Q3: b) 0.1 c Q4: e) 2.60 × 108 m s-1 Q5: b) 1

Quiz 2 Relativistic energy (page 23) Q6: a) 5.98 × 10-10 J Q7: d) rest energy and kinetic energy. Q8: c) 2.4 × 10-9 J Q9: b) The relativistic calculation gives a higher value of kinetic energy. Q10: b) 5.1 × 10-11 J

End of topic test (page 26) Q11: 2 × 10-27 Q12: 2.6 × 108 Q13: 6.36 × 10 -9 Q14: 1.53 × 10 -13 Q15: 4.32 × 10 -11 Q16: 2.1 × 10-13

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186

ANSWERS: TOPIC 3

3 Angular velocity and acceleration Quiz 1 Radian measurement (page 32) Q1:

c) 2.53 rad

Q2:

c) 68.75◦

Q3:

d) 2π/3 rad

Q4:

d) 6π rad

Q5:

a) 0.208 rad

Orbits of the planets (page 34)

Planet

Orbit radius (m) Period (days)

Period (s)

Mercury Venus Earth

5.79 × 1010 1.08 × 1011 1.49 × 1011

7.60 × 106 1.94 × 107 3.15 × 107

88.0 225 365

Angular velocity (rad s-1 ) 8.27 × 10-7 3.24 × 10-7 1.99 × 10-7

Will the car stop before the lights? (page 37) ω 0 = 27.0 rad s-1 , ω = 0 rad s-1 , t = 5.00 s, a = ? The appropriate kinematic relationship is ω = ω 0 + αt. Putting the values into this equation, ω = ω0 + αt ∴ 0 = 27.0 + 5α ∴ −5α = 27.0 27 ∴α=− 5 ∴ α = −5.40 rad s−2 So to stop the wheels in exactly 5.00 s, they must have an angular acceleration of -5.40 rad s-2 , equivalent to an angular deceleration of 5.40 rad s -2 .

Quiz 2 Angular velocity and angular kinematic relationships (page 37) Q6:

e) 50.3 rad s-1

Q7:

a) 0.45π rad s-1

Q8:

d) 60.0 rad

Q9:

b) 0.70 rad s-2 © H ERIOT-WATT U NIVERSITY

ANSWERS: TOPIC 3

187

Q10: c) 7.9 s

Compact disc player (page 41) Q11: Use the equation v = rω, and remember to change the radius in the question into metres. v r 1.25 ∴ω= 0.025 ∴ ω = 50.0 rad s−1 ω=

Q12:

v r 1.25 ∴ω= 0.057 ∴ ω = 21.9 rad s−1 ω=

Q13: ω 0 = 50.0 rad s-1 , ω = 21.9 rad s-1 , t = 3300 s, α = ? ω = ω0 + αt ω − ω0 ∴α= t 21.9 − 50.0 ∴α= 3300 ∴ α = −8.52 × 10−3 rad s−2

Quiz 3 Angular velocity and tangential speed (page 42) Q14: c) 4.80 m s-1 Q15: c) 1.60 m Q16: e) 4.58 m s-1 Q17: a) 0.108 m s-2 Q18: b) 3.2 m

End of topic test (page 44) Q19: 1.2 Q20: 3.1 © H ERIOT-WATT U NIVERSITY

188

ANSWERS: TOPIC 3

Q21: 1.6 × 10-6 Q22: 1.9 Q23: 4.2 Q24: 0.41 Q25: 0.063

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ANSWERS: TOPIC 4

189

4 Centripetal force Quiz 1: Centripetal acceleration (page 51) Q1: c) 8.4 m s-2 Q2: b) The centripetal acceleration is always directed towards the centre of the circle. Q3: b) 5.6 m s-2 Q4: e) 140 m s-2 Q5: a) The centripetal acceleration halves in value.

Motion in a vertical circle (page 56) 1. The rope is most likely to go slack when the mass is at the top of the circle. Compare Equation 4.4 and Equation 4.5 to understand why. 2. At the top of the circle, Equation 4.4 tells us Ttop = mrω 2 − mg Expressing this in terms of the tangential speed v, Ttop =

mv 2 − mg r

When the rope goes slack, the tension in it must have dropped to zero, so 0= ∴

mv 2 − mg r

mv 2 = mg r ∴ v 2 = gr √ ∴ v = gr √ ∴ v = 9.8 × 0.80 ∴ v = 2.8 m s−1

Quiz 2 Horizontal and vertical motion (page 56) Q6: c) 40 N Q7: c) 6.0 rad s-1 Q8: d) If the radius of the circle is increased, the centripetal force decreases. Q9: b) 14.4 m s-1 Q10: a) When the mass is at the bottom of the circle. © H ERIOT-WATT U NIVERSITY

190

ANSWERS: TOPIC 4

Quiz 3 Conical pendulum and cornering (page 61) Q11: d) T × sin φ Q12: c) 57◦ Q13: e) 2π l cos φ/g Q14: a) the coefficient of static friction between the car and the road. Q15: d) a component of the normal reaction force contributes to the centripetal force.

End of topic test (page 63) Q16: 6.1 Q17: 5.7 Q18: 31.2 Q19: 3.9 Q20: 1. 2.7 2. 2.8 Q21: 13 Q22: 18.5

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ANSWERS: TOPIC 5

5 Rotational dynamics Quiz 1 Torques (page 68) Q1: e) N m Q2: d) 49 N m Q3: a) 3.75 m Q4: c) 8.46 N m Q5: d) 111 N

Calculating the moment of inertia (page 74) Q6: 0.16 kg m2 Q7: 20 cm from the centre Q8: 0.40 kg m2 Q9: 0.43 kg m2 Q10: 28 cm

Quiz 2 Equilibrium and moment of inertia (page 76) Q11: d) 9.75 N Q12: c) 1.4 m from the fulcrum Q13: b) 18 N Q14: e) 3.72 kg m2 Q15: a) 0.137 kg m 2

Quiz 3 Angular momentum and rotational kinetic energy (page 81) Q16: d) 7.20 kg m2 s-1 Q17: b) 0.25 kg m2 s-1 Q18: a) The turntable would slow down. Q19: c) 0.015 J Q20: e) 187 J

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192

ANSWERS: TOPIC 5

End of topic test (page 83) Q21: 24 Q22: 2.1 Q23: 0.23 Q24: 0.225 Q25: 3.5 Q26: 5.2 Q27: 1. 3.6 2. 59

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ANSWERS: TOPIC 6

6 Gravitational force and field Quiz 1 Gravitational force (page 90) Q1: d) 1.0 × 10-10 N Q2: a) FS = FE Q3: d) 8.10 N Q4: b) Its mass remains constant but its weight decreases. Q5: e) 8.87 m s-2

Quiz 2 Gravitational fields (page 95) Q6: d) 26.5 N kg-1 Q7: a) m s-2 Q8: a) 7.1 × 105 m Q9: c) 3.0 m from P Q10: d) 1.03 × 10 26 kg

End of topic test (page 97) Q11: 2.22 × 10 20 N Q12: 2.13 × 10 -10 N Q13: 57.4 N Q14: 9.36 m s-2 Q15: 4.0 N Q16: 2.51 N kg-1 Q17: 0.322 N kg -1 (Note - it is possible to solve this problem without having to calculate the mass of the planet.) Q18: 6.75 × 10 -3 N kg-1

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194

ANSWERS: TOPIC 7

7 Gravitational potential and satellite motion Quiz 1 Gravitational potential (page 103) Q1:

c) the distance of A from the centre of the Earth.

Q2:

b) -2.82 × 106 J kg-1

Q3:

c) -5.96 × 107 J kg-1

Q4:

e) -4.83 ×1010 J

Q5:

a) The satellite has moved closer to the Earth.

Quiz 2 Satellite motion (page 109) Q6:

b) 7650 m s-1

Q7:

e) The period of a satellite orbiting the Earth depends on the mass of the satellite.

Q8:

c) 3070 m s-1

Q9:

d) its potential energy decreases but its kinetic energy increases.

2 3

Q10: d) mp = 4π r GT 2

Quiz 3 Escape velocity (page 113) Q11: e) escape from the Earth's gravitational field. Q12: b) the mass and radius of the Earth. Q13: c) 5020 m s-1 Q14: b) the escape velocity is greater than the speed of light.

End of topic test (page 115) Q15: -1.77 × 107 Q16: -6.5 × 1013 Q17: -3.02 × 107 Q18: -2.6 × 1011 Q19: 1.1 × 105 Q20: 2 × 104 Q21: 2.27 × 10 30

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ANSWERS: TOPIC 8

195

8 Simple harmonic motion Quiz 1 Defining SHM and equations of motion (page 125) Q1: b) displacement. Q2: a) 1.27 Hz Q3: d) Acceleration Q4: e) 98.7 m s-2 Q5: c) 0.84 m s-1

Energy in simple harmonic motion (page 127) 1. Note that the sum of kinetic and potential energies is constant, so that energy is conserved in the SHM system P E + KE

  = 12 mω 2 y 2 + 12 mω 2 a2 − y 2

= 12 mω 2 a2

The total energy is independent of the displacement y.

E

1.8

KE+PE PE

KE -1.2

1.2

y

2. Again the sum of kinetic and potential energies is constant.

© H ERIOT-WATT U NIVERSITY

196

ANSWERS: TOPIC 8

E

KE+PE

1.8

KE

PE 0.0

0

1

2

3

4

5

6

t/s

Quiz 2 Energy in SHM (page 128) Q6:

b) 0.037 J

Q7:

d) 45 J

Q8:

c) When its displacement from the rest position is zero.

c) ±a √2

Q9:

Q10: d) 75 J

Quiz 3 SHM Systems (page 133) Q11: d) 3.14 s Q12: b) 0.910 Hz Q13: e) 37 N m-1 Q14: b) 0.75 s Q15: a) Decrease its length by a factor of 4.

End of topic test (page 137) Q16: 2.3 Q17: 0.74 Q18: 1. 2.7 2. 0.7 Q19: 2.47 © H ERIOT-WATT U NIVERSITY

ANSWERS: TOPIC 8

Q20: 0.18 Q21: 0.77 Q22: 2.4 × 10-3 Q23: 0.78

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198

ANSWERS: TOPIC 9

9 Wave-particle duality Quiz 1 Wave-particle duality of waves (page 145) Q1:

a) 3.88 × 10-19 J

Q2:

b) 9.85 × 1014 Hz

Q3:

d) the wavelength of the scattered photons increases.

Q4:

a) The blue photons have greater photon energy and greater photon momentum.

Q5:

c) 2.65 × 10-32 kg m s-1

Quiz 2 Wave-particle duality of particles (page 150) Q6:

a) particles exhibit wave-like properties.

Q7:

d) 1.14 × 10-10 m

Q8:

c) 1.7 × 10-27 kg

Q9:

b) particles can exhibit wave-like properties.

Q10: b) the electrons have a shorter wavelength than visible light.

End of topic test (page 152) Q11: 5.63 × 10 14 Q12: 3.13 × 10 6 Q13: 2.29 × 10 -19 Q14: 1.46 × 10 -27 Q15: 1.77 × 10 -27 Q16: 1. A 2. B Q17: 1. 5.42 × 10-24 2. 1.22 × 10-10

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ANSWERS: TOPIC 10

10 Introduction to quantum mechanics Quiz 1 Atomic models (page 161) Q1: b) the electron's angular momentum is quantised. Q2: d) 4.22 × 10-34 kg m2 s-1 Q3: e) r = nλ/2π Q4: a) photons can only be emitted with specific energies.

End of topic test (page 164) Q5: 6.33 × 10-34 Q6: 3.32 × 10-10 Q7: 8.83 × 10-11 Q8: 1. 2.29 × 10-19 2. 8.69 × 10-7

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ANSWERS: TOPIC 11

11 Mechanics end-of-unit assessment End-of-unit assessment (page 166) Q1: 1. 26 m 2. 7.75 × 107 m s-1 Q2: 1. 3.72 m s-1 2. 2.6 N 3. 7.3 N Q3: 1. 6.02 rad s-1 2. 14 N m 3. 13 Q4: 1. 8.73 × 103 N 2. -5.73 × 107 J kg-1 Q5: 1. 0.19 m s-2 2. 0.056 m s1Q6:

2.74 × 10-12

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