How to Pass SQA Advanced Higher Physics 9781398312227, 1398312223

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How to Pass ADVANCED HIGHER

Physics

How to Pass

ADVANCED HIGHER

Physics

98312227_HTP_Advanced_Higher_Physics_TXT_DESIGN.indd 1

Paul Chambers Mark Ramsay

29/07/2021 12:39

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The Publishers would like to thank the following for permission to reproduce copyright material. Acknowledgements The publishers are grateful for the use of material on pp.v–vi, 1, 6, 14, 20, 27, 31, 40, 46, 50, 56, 61, 69, 73, 83, 93, 104–108 taken from the SQA Advanced Higher Physics Course Specification and on pp.110–115 taken from the 2019 SQA Advanced Higher Physics Question Paper, both Copyright © Scottish Qualifications Authority. Questions on p.94 and corresponding answers on p.121 are taken from the 2015 SQA Specimen Question Paper for Advanced Higher for CfE Physics, and used by permission, Copyright © Scottish Qualifications Authority. All other questions, answers and worked examples do not emanate from SQA material. Every effort has been made to trace all copyright holders, but if any have been inadvertently overlooked, the Publishers will be pleased to make the necessary arrangements at the first opportunity. Although every effort has been made to ensure that website addresses are correct at time of going to press, Hodder Gibson cannot be held responsible for the content of any website mentioned in this book. It is sometimes possible to find a relocated web page by typing in the address of the home page for a website in the URL window of your browser. Hachette UK’s policy is to use papers that are natural, renewable and recyclable products and made from wood grown in well-managed forests and other controlled sources. The logging and manufacturing processes are expected to conform to the environmental regulations of the country of origin. Orders: please contact Hachette UK Distribution, Hely Hutchinson Centre, Milton Road, Didcot, Oxfordshire, OX11 7HH. Telephone: +44 (0)1235 827827. Email [email protected]. Lines are open from 9 a.m. to 5 p.m., Monday to Friday. You can also order through our website: www.hoddereducation.co.uk. If you have queries or questions that aren’t about an order, you can contact us at [email protected] © Mark Ramsay and Paul Chambers 2021 First published in 2021 by Hodder Gibson, an imprint of Hodder Education An Hachette UK Company 211 St Vincent Street Glasgow, G2 5QY Impression number Year



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All rights reserved. Apart from any use permitted under UK copyright law, no part of this publication may be reproduced or transmitted in any form or by any means, electronic or mechanical, including photocopying and recording, or held within any information storage and retrieval system, without permission in writing from the publisher or under licence from the Copyright Licensing Agency Limited. Further details of such licences (for reprographic reproduction) may be obtained from the Copyright Licensing Agency Limited, www.cla.co.uk Cover photo © Valdis Torms – stock.adobe.com Illustrations by Aptara, Inc. Typeset in Cronos Pro Light 13/15 pts by Aptara, Inc. Printed in Spain A catalogue record for this title is available from the British Library. ISBN: 978 1 3983 1222 7

We are an approved supplier on the Scotland Excel framework. Schools can find us on their procurement system as: Hodder & Stoughton Limited t/a Hodder Gibson.

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Contents Introduction iv Area 1 Rotational motion and astrophysics Topic 1 Kinematic relationships

1

Topic 2 Angular motion

6

Topic 3 Rotational dynamics

14

Topic 4 Gravitation

20

Topic 5 General relativity 

27

Topic 6 Stellar physics

31

Area 2 Quanta and waves Topic 7 Introduction to quantum theory

40

Topic 8 Particles from space

46

Topic 9 Simple harmonic motion

50

Topic 10 Waves

56

Topic 11 Interference

61

Topic 12 Polarisation of light

69

Area 3 Electromagnetism Topic 13 Fields

73

Topic 14 Circuits

83

Topic 15 Electromagnetic radiation

93

Area 4 Units, prefixes and uncertainties

95

Appendices 1 Your project

104

2 Your exam

107

3 Data sheet and relationships sheet

110

Answers 116

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Introduction Welcome to How to Pass SQA Advanced Higher Physics. This book has been written to help you achieve the highest possible grade in your Advanced Higher Physics examination. It is written around the information given in the SQA Advanced Higher course specification: www.sqa.org.uk/files_ccc/AHPhysicsCourseSpec.pdf

How to use this book The content of this book is designed to help you prepare for the SQA Advanced Higher Physics exam. It contains a summary of the Advanced Higher course, with the following features to support your understanding and recall of important points.

Summary notes These are concise, illustrated notes that explain and expand on the key points for each topic. You should read and review these several times.

Key points !

There is a key points list at the start of each topic, detailing the knowledge and skills that will be assessed in the exam. Once you have worked through each topic and completed the Study questions, review the key points to ensure you have gained the knowledge you need of the topic.

Examples

f

Worked examples show how to answer exam-style questions. Before reading through the solution provided, attempt to answer the question yourself, using the information in the topic. Then check your answer against the worked example.

Study questions

?

At the end of each topic you will find Study questions. These questions have been modelled on those in previous SQA Advanced Higher exams and should be used to test your knowledge and understanding of the topic. You should try to answer these yourself before checking the answers provided.

Answers Answers to Study questions are provided at the end of the book. Mark breakdowns show where marks are awarded in the exam.

iv

Hints & tips



These tips help you achieve the best marks by highlighting important points to remember and pitfalls to avoid.

Key terms Key terms are technical terms that you should be familiar with for the topic. They appear in the list of key points and are further defined in key term boxes. You should study these key terms and their definitions carefully and review them at the end of a topic.

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Introduction

The Advanced Higher Physics course The Advanced Higher Physics course comprises four areas, which are covered in the four sections of this How to Pass guide. Rotational motion and astrophysics

Quanta and waves

Electromagnetism

Units, prefixes and uncertainties

Kinematic relationships Angular motion Rotational dynamics Gravitation General relativity Stellar physics

Introduction to quantum theory Particles from space Simple harmonic motion Waves Interference Polarisation of light

Fields Circuits Electromagnetic radiation

Units, prefixes and scientific notation Uncertainties Data analysis Evaluation and significance of experimental uncertainties

Advanced Higher Physics assessment There are two parts to the Advanced Higher Physics assessment: ●● the examination question paper ●● the project.

Examination question paper The question paper has 155 marks. This is scaled by SQA to 120 marks, to represent 75% of the overall marks for the course assessment. The question paper contains restricted-response and extended-response questions. A data sheet and a relationships sheet will be provided. This written exam lasts for 3 hours.

Project The project has 30 marks. This is scaled by SQA to 40 marks, to represent 25% of the overall marks for the course assessment. The project allows you to plan and carry out an in-depth investigation of a physics topic of your choice. You are required to produce a project report of between 2500 and 4500 words. For more detailed information and advice on the project and exam, see Appendices 1 and 2 on pages 104–109. More information about the course and the exam is available on the SQA website – www.sqa.org.uk

v

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Introduction

Relationships required for Advanced Higher Physics A data sheet and a relationships sheet are provided in the exam – see Appendix 3 on pages 110–115.

Derivations You are required to recreate the following derivations: ●● Derivation of the equations of motion v = u + at and s   =  ut   +   1 at 2 , using calculus methods 2

2GM r

●●

Derivation of the relationship for escape velocity v =

●●

Derivation of the relationships for velocity and kinetic energy for simple harmonic motion v =   ± ω ( A2 − y 2 )  and Ek = 21 mω 2 ( A2 − y 2 )

●●

Derivation of the thin film interference relationship d = λ 4n

●●

Derivation of the relationship for Brewster’s angle n = tan iP

vi

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Area 1 Rotational motion and astrophysics Topic 1

Kinematic relationships

Key points !

✽✽ Knowledge that differential calculus notation is used to represent rate of change. ✽✽ Knowledge that velocity is the rate of change of displacement with time, acceleration is the rate of change of velocity with time, and acceleration is the second differential of displacement with time. ✽✽ Use of calculus methods to calculate instantaneous displacement, velocity and acceleration for straight line motion with a constant or varying acceleration. ✽✽ Use of appropriate relationships to carry out calculations involving displacement, velocity, acceleration and time for straight line motion with constant or varying acceleration. ✽✽ Knowledge that the gradient of a curve (or a straight line) on a motion–time graph represents instantaneous rate of change, and can be found by differentiation. ✽✽ Knowledge that the gradient of a curve (or a straight line) on a displacement–time graph is the instantaneous velocity and that the gradient of a curve (or a straight line) on a velocity–time graph is the instantaneous acceleration. ✽✽ Knowledge that the area under a line on a graph can be found by integration. ✽✽ Knowledge that the area under an acceleration–time graph between limits is the change in velocity and that the area under a velocity–time graph between limits is the displacement. ✽✽ Determination of displacement, velocity or acceleration by the calculation of the gradient of the line on a graph or the calculation of the area under the line between limits on a graph.

Kinematic relationships and calculus methods In order to explain more clearly the relationship between acceleration, velocity and displacement, we need to review our present understanding of certain concepts. In Advanced Higher we explore the ‘rate of change’ concept in a more mathematical way. 1

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Area 1  Rotational motion and astrophysics

This involves the use of calculus in order to explain and describe the relationships. The level of calculus is comparable to that used in Higher Mathematics. We use these relationships to derive the equations of motion we are familiar with. dv a = and v = ds dt dt dv a= dt a∫0t dt = ∫uv dv a[ t ]0 = [v ]u t

v

at = v − u v = u + at v = ds = u + at dt

∫ d s = ∫ (u + at)dt s 0

t 0

t  1 2   [ s]0 =  ut + 2 at    s

0

s = ut + 1 at 2 2

v = u + at   ∴ t =  v −a u    Substituting for t in s = ut + 12 at 2 u(v − u) 1  v − u  s= + a  a 2  a 

2

Multiplying all terms by 2a and opening out the brackets gives 2as = 2uv − 2u2 + v2 + u2 − 2uv v2 = u2 + 2as The purpose of the derivations is to show that given a relationship between displacement or velocity with time, we can use calculus to calculate s, v or a at some point in that object’s motion. This mathematical analysis and representation is important in more advanced physics. dy The gradient of a slope is often given as and this relates to the rate of dx change of one variable with another. This represents the instantaneous rate of change, which can be the instantaneous velocity at a point or instantaneous acceleration at a point.

2

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Topic 1  Kinematic relationships

In Higher Physics you learned that the area under a velocity–time graph is equivalent to the displacement and the area under an acceleration–time graph is equivalent to the change in velocity. These concepts can be applied to other areas of physics such as ●● area under a current/time graph ●● area under a charge/voltage graph. The interpretation of gradient and area under a graph is treated slightly differently in advanced physics. In Higher Physics you calculated the area (within certain limits) of a velocity–time graph in order to determine the displacement of a moving body. You also may have calculated the area (within limits) of an acceleration–time graph in order to determine the change in velocity of a body. These ‘areas’ would be composed of triangles, squares and rectangles. In Advanced Higher you may be given a formula for the displacement or velocity of a body (compared to a graph) with respect to time and you are expected to use calculus to answer the question.

Example

f

1 An object, initially at rest, accelerates in a straight line and its

velocity at time, t, is represented by the equation v = 0 .245t2 + 2 .15t a) Determine its acceleration at 4 . 50 s. b) Determine its displacement from its original position after 6 .75 s.

Solution a) a =

dv dt

d a = dt (0 .245t2 + 2.15t) a = 0 . 49t + 2.15 at t = 4 . 50 s a = 0 . 49 × 4 . 50 + 2 .15 a = 4. 36 m s-2 b) v = ds = 0.245t 2 + 2.15t dt

∫

s 0

ds =

∫

t 0

(0.245t 2 + 2.15t)dt

 .  t 3 + 1 2.15t 2  [ s]0 = 0 245 2 3 0 3 2 s = 0 . 0817t + 1. 08t at t = 6 . 75 s s = 0 . 0817 × 6 . 753 + 1 . 08 × 6 .752 s = 74 .1 m t

s

Hints & tips



Remember that there are only two questions to be asked about a graph: 3 What is the gradient? The units of the gradient are those of the y-axis divided by those of the x-axis and may give a clue to what the gradient tells us. For example, the gradient of a velocity– time graph has units of ms−2 and so is the acceleration. 3 What is the area of the graph? The units of the gradient are those of the y-axis multiplied by those of the x-axis and may give a clue to what the area tells us. For example, the area of a velocity– time graph has units of m and so is the displacement. 3

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Area 1  Rotational motion and astrophysics

f 2 The acceleration of a particle is given by the expression a = 2.8t.

Example

At time t = 0 s the displacement of the particle is 0.0 m and its velocity is 3.6 m s−1. a) Show that the velocity of the particle at time, t, is given by v = 1.4t2 + 3.6. b) Calculate the displacement of the particle when its velocity is 5.0 m s−1.

Solution a) a =

dv dt t

v = ∫ 0 a dt v = ∫ t 2 . 8t dt 0 .8 2 2 v= t +c 2 v = 1 . 4t2 + 3 . 6    as v = 3 . 6 when t = 0 b) v = 1 . 4t2 + 3 . 6  5.0 = 1 . 4t2 + 3 . 6 .  t = 5 0 −. 3.6 14   t = 1 . 0 s  v = ds dt    ds = ∫ t 1. 4t 2 + 3 . 6dt 0 dt . . s = 134 t 3 + 326 t 2 + c s = 0 . 47t3 + 1 . 8t2    as s = 0 when t = 0 at t = 1 . 0 s, s = 0 . 47 × 1 . 03 + 1 . 8 × 1 . 02 s = 0 . 47 + 1.8 s = 2 . 3 m

Example

f

3 A train on a long straight track accelerates from rest. The train’s run

begins at time t = 0. Its velocity, v, at time t, is given by the equation v = 0.18t2 + 0.75t. Using calculus methods: a) determine the acceleration of the train at t = 10.0 s b) determine the displacement of the train from its original position at this time.

4

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Topic 1  Kinematic relationships

Solution a) a = dv

dt a = d (0 .18t 2 + 0 .75t) dt a = 0 .36t + 0 .75 at t = 10 s, a = 0 .36 × 10 + 0 .75 a = 4 .35 a = 4.4 m s-2 b) ds = t (0 .18t 2 + 0 .75t)dt dt ∫ 0 . 0 .75 2 s = 0 18 t 3 + t +c 2 3 s = 0 . 06t3 + 0 .375t2    as s = 0 when t = 0 at t = 10 . 0 s, s = 0.06 × 10 . 03 + 0 . 375 × 10 . 02 s = 60 + 37 . 5 s = 97 . 5 m s = 98 m

Study questions

?

1 a) The acceleration of a particle moving in a straight line is given by a = dv dt where the symbols have their usual meaning. (i) Show, by integration, that when a is constant v = u + at. 3 2 2 2 (ii) Show that when a is constant v = u + 2as. b) The path taken by a short track speed skater is shown in the figure below. The path consists of two straights each of length 29.8 m and two semicircles each of radius 8.20 m. 29.8 m Z

8.20 m

X

Y

Figure 1.1 Speed skating track



Starting at point X, half way along the straight, the skater accelerates uniformly from rest. She reaches a speed of 9.64 m s−1 at point Y, the end of the straight. (i) Calculate the acceleration of the skater. (ii) The skater exits the curve at point Z with a speed of 10.9 m s−1. Calculate the change in velocity of the skater between Y and Z.

4 2 5

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Area 1  Rotational motion and astrophysics

Topic 2

Angular motion

Key points !

✽✽ Use of the radian as a measure of angular displacement. Conversion between degrees and radians. ✽✽ Use of appropriate relationships to carry out calculations involving angular displacement, angular velocity, angular acceleration and time. ✽✽ Use of appropriate relationships to carry out calculations involving angular and tangential motion. ✽✽ Use of appropriate relationships to carry out calculations involving constant angular velocity, period and frequency. ✽✽ Knowledge that a centripetal (radial or central) force acting on an object is necessary to maintain circular motion, and results in centripetal (radial or central) acceleration of the object. ✽✽ Use of appropriate relationships to carry out calculations involving centripetal acceleration and centripetal force.

Angular motion and radians Angular motion considers objects that travel in a curve or a circle, for example, a planet orbiting a star, a vehicle following a curving road or a person on a park roundabout. In these cases, we are concerned with how an object travels along the perimeter of a path or how many rotations it makes. Angles are measured in degrees typically, but this is an arbitrary unit. In physics, a radian is a better measure of an angle as radians are dimensionless. In a circle, radians are related to the radius of that circle. Radians link the arc of a circle to the radius of the circle and their use also allows the linear displacement to be calculated more easily. In Figure 1.2, the angle θ (theta) is given by the ratio of the length of the arc to the radius.  θ = rs   The circumference of a circle is 2πr. Therefore, the number of radians in a complete circle is 2π r = 2π . r 2π radians is equivalent to 360 degrees. You may be asked to convert between radians and degrees. Radian measure also makes it simpler to convert from angular displacement to linear displacement.

6

s

q

r

Figure 1.2 Angle θ shown as the ratio of length of arc, s, to the radius, r

Key term Radian: An alternative unit of measurement for angles. 2π radians is equivalent to 360°. The radian is used as a measure of angular displacement.

Hints & tips



We must use radians in our calculations as they are dimensionless. Make sure your calculator is set to radians!

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Topic 2  Angular motion

f 1 An object follows a circular path of radius 25.7 m. It travels 94.2 m.

Example

Calculate its angular displacement in degrees.

Solution

Circumference of circle of radius 25.7 m = 2 × π × 25.7 = 161.5 m. . 94 2. =  0 . 583 161 5 0.583 × 360 = 209.88 = 210°. To calculate the angular displacement in radians is .  θ = rs    = 94 .2 = 3. 66 radians . 25 7 Angular motion can be treated in the same way as linear motion. Linear displacement, velocity and acceleration have angular equivalents. Linear displacement, s, in m Linear velocity, v, in m s−1 Linear acceleration, a, in m s−2

Angular displacement, θ, in radians Angular velocity, ω, in rad s−1 Angular acceleration, α, in rad s−2

As with linear motion this leads to dθ ω  = dt 2 dω d θ α= = dt dt 2 Applying calculus methods, as we did earlier with linear motion, we arrive at ω = ω0 + αt ω = ω 0 + 2αθ 2

2

θ = ω 0t + 1 α t 2 2

Example

for constant angular acceleration only

f

2 A rotor is accelerated from rest to 12 500 rpm in 25 s. Its acceleration

is constant. a) Calculate its angular acceleration. b) How many revolutions has it turned to reach this velocity?

Solution a) 12500rpm = 12500 × 2 × π = 25000 π/min = 25000π = 1309rads −1

60

Initial angular velocity is zero. Using ω = ω 0 + α t 1309 = 0 + α × 25 α = 1309 = 52.4 rads −2 25 2 1 2 1 b) θ = ω 0t + 2 α t = 0 + 2 × 52. 4 × 25 = 16375 rad 2π rad = 1 revolution, so

16375 2610 revolutions 2π = 7

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Area 1  Rotational motion and astrophysics

Angular and tangential motion In order to describe more fully the motion of a rotating body, we have to consider the relationship between its angular motion and the linear motion it experiences as it travels along the perimeter of the circle. This type of linear motion is referred to as tangential, as the object is always travelling at right angles to the centre of the circle or axis of rotation. In the example of a rotating object shown in Figure 1.3, the object is rotating around the centre of the circle at a constant angular velocity, ω. Its tangential velocity, v, is continually changing direction. The relationship between tangential motion and angular motion is represented by s = rθ.  From this we get v = rω and at = rα where s is the tangential displacement, v is the tangential velocity and at is the tangential acceleration. V A key point here is that when a rotating object increases its angular velocity, this will increase the tangential velocity leading to a tangential acceleration. When a rotating object is rotating at a steady angular velocity, the tangential velocity is also changing but due only to its change in direction. This change in direction is due to the centripetal force acting on the body in the direction of the centre. This is referred to as the centripetal (sometimes referred to as radial or central) acceleration, not tangential.

Example

Key terms Tangential: A line touching at one point. Acting at 90° to a particular point. Centripetal: Moving towards or in a central direction. V

r q r

Figure 1.3 Rotating object with constant angular velocity, ω, and tangential velocity, v, with continually changing direction

f

3 A child is on a roundabout spinning with a steady rotation of 1 revolution every 6.5 seconds.

The child is 1.4 m from the centre of the roundabout. a) Calculate the angular velocity of the roundabout. b) Calculate the angular velocity of the child. c) Calculate the tangential velocity of the child. The child moves out to the edge of the roundabout at a distance of 2.3 m from the centre. d) Determine the new angular velocity and tangential velocity.

Solution

a) One rotation every 6.5 s equates to 2π every 6.5 s.



ω=

dθ dt

2π = 0.97rad s −1 6 .5 b) As the child is on the roundabout, the angular velocity of the child will be 0.97 rad s−1. c) v = rω = 1.4 × 0.97 = 1.36 = 1.4 m s−1 d) ω = 0.97 rad s−1. As the roundabout still rotates at the same angular velocity

v = rω = 2.3 × 0.97 = 2.2 m s−1

8

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Topic 2  Angular motion

A key point here is that in these cases, the child has to be in good contact with the roundabout by holding on to a bar. It is the bar that provides the central force that keeps the child spinning. If the child loses contact, they will move off tangentially and appear to be thrown from the roundabout. This may cause the misconception that a ‘force’ is causing the child to slide off. It is not. This is just the direction the child would move if no forces acted upon them. The central force pulls them towards the centre in order to maintain the rotation.

Example

f

4 The Moon orbits the Earth once in 28·3 days. The Moon has a mass

of 7.35 × 1022 kg and the mean radius of its orbit about the Earth is 3.84 × 108 m. Calculate the angular velocity of the Moon.

Solution ω = dθ

dt t = 28.3 × 24 × 60 × 60 t = 2.45 × 106 s θ = 2π radians, that is, one revolution

ω = . 2π 6 2 45 × 10 ω = 2.57 × 10−6 rad s−1

f 5 A disc rotating at 180 rad s

Example

–1

decelerates at a rate of 3.5 rad s–2

for 7.0 s. a) Determine the angular velocity at the end of the seven seconds. b) What is the angular displacement in this time?

Solution a) ω = ω0 + α t

ω = 180 − 3.5 × 7.0 ω = 155.5 ω = 156 rad s−1 b) θ = ω0t + 1 α t 2 2

. 1 . . 2 θ = 180 × 7 0 − 2 × 3 5 × 7 0 θ = 1174 rad θ = 1170 rad

9

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Area 1  Rotational motion and astrophysics

f 6 The angular velocity of an engine is increased from 800 rpm to

Example

3000 rpm in 8.0 s. a) Determine the angular acceleration. You may assume this is uniform. b) Find the total angular displacement. c) Calculate how many revolutions the engine makes during this 8.0 s.

Solution a) ω = ω0 + αt

3000 = 800 + α8.0 α = 3000 .− 800 80 α = 275 α = 280 rad s−2 b) θ = ω0t + 1 α t 2 2

θ = 800 × 8.0 + 1 × 280 × (8 . 0)2 2

   θ = 15 360    θ = 15 000 rad c) 1 revolution = 2π radians revs = 15000 2π revs = 2389 revs = 2400 revolutions

Example

f

7 A wheel accelerates uniformly from rest at 3.0 rad s−2 for 5.0 s. a) Find (i) the final angular velocity after 5 . 0 s (ii) the angular displacement after 5 . 0 s. b) The wheel has a radius of 1.50 m. Determine the linear velocity at

a point on its rim at the end of the 5.0 s.

Solution a) (i)  ω = ω0 + αt

b) v = ω r

ω = 0 + 3.0 × 5.0   ω = 15 rad s−1  

v = 15 × 1.50 v = 22.5 v = 23 m s−1

(ii) θ =  ω0t + 1 α t 2 2

θ = 0 × 5.0 +   θ = 37.5   θ = 38 rad

10

1 2

× 3 . 0 × (5 . 0)2

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Topic 2  Angular motion

Example 2 above leads us into situations where we are dealing with objects that have constant angular velocities. Examples where objects may be spinning at high rates, such as hard drives and electric motors, often have their velocities described in rps (revolutions per second) or similar. When dealing with such examples, the terms period and frequency are often used to describe the motion of a spinning object, and you should be able to relate these to the angular velocity and convert from period, frequency, revolutions per minute, and so on. From Higher Physics you should be aware that f = 1 T A frequency of 5 Hz means 5 revolutions per second. This equates to 5 × 2π radians per second.

Key terms Period (T): The time for one revolution measured in seconds. Frequency (f): The number of revolutions per second measured in hertz.

ω = 10π rad s−1 = 2πf rad s−1

ω = 2π f =

2π rad s−1 T

Example

f

8 A proton in a synchrotron travels around a circular path of radius

65 m. Its tangential velocity is 2.4 × 108 m s−1. a) Calculate its period and frequency. b) Calculate its angular velocity.

Solution a) Period is time for one rotation.

T = s = 2 × π × 65 = 1.7 × 10 −6 s v 2. 4 × 10 8 ms −1

f = 1 = . 1 −6 = 5.9 × 105 Hz T 1 7 × 10

b) ω = 2π f = 2 × π × 5.9 × 105 = 3.7 × 106 rad s−1

f 9 The following table gives data about three planets orbiting the Sun.

Example

Planet

Radius, R, of orbit around the Sun/109 m

Venus Mars Jupiter

108 227 780

Orbit period, T, around the Sun/years

0.62 1.88 12.0

Use all of the data to show that T2 is directly proportional to R3 for these three planets.

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Solution This is a typical SQA number-processing question. Either calculate the 3 ratio of R 2 for each planet or plot a graph of R3 vs T2 and show that the T best fit line is both straight and passes through the origin. Planet

R3

T2

Venus Mars

1.3 × 106 1.2 × 107

3.8 × 10-1 3.5

3.3 × 106 3.3 × 106

Jupiter

4.7 × 108

1.4 × 102

3.3 × 106

R3 T2

The minimum number of significant figures in the question is two. 3 The ratio of R 2 is constant and so T2 is directly proportional to R3. T

Centripetal force and acceleration When an object is travelling in a circular path it is continually changing direction. In order for it to change direction, it must experience a force. This force acts in the direction of the centre of the circle. The direction of this centripetal force continually changes as the object moves along its path. As stated earlier, the change in direction of the tangential velocity means this velocity is continually changing. This leads to the fact that there is a constant centripetal acceleration.

Hints & tips



Remember that the force causing the rotation is called centripetal and acts along the radius vector towards the centre of rotation. Do not use the word ‘centrifugal’.

The centripetal or radial acceleration is given by 2 ar = vr = rω 2 This leads to the centripetal force being given by

mv 2 F = r = mrω 2 The derivation is not required, but it is not too complicated mathematically.

12

Figure 1.4 Changing direction of tangential velocity

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Topic 2  Angular motion

Example

f

10 A ball of mass 0.15 kg, attached to string, is spinning horizontally in a

circle of radius 0.45 m at a tangential velocity of 3.5 m s−1. a) Calculate its period and frequency. b) Calculate its angular velocity. c) Calculate the centripetal acceleration of the ball. d) Calculate the magnitude of the central force.

Solution .

. a) T = 2πr = 2π × 0 45 v 3 . 5 = 0 81s f = 1 = .1 = 1.2 Hz T 0 81 b) ω = 2πf = 7.5 rad s–1 c) a = rω 2 = 0.45 × 7.52 = 25.3 m s–2 d) F = mrω 2 = 0.15 × 25.3 = 3.795 = 3.8 N

Study questions

?

1 a) A student places a radio-controlled car that has a mass of 2.5 kg on a horizontal circular track as shown in Figure 1.5.

4.0 m

Figure 1.5 Horizontal circular track for model car



She uses the radio control to make the car travel with a constant speed of 6.0 m s−1 around a circular path of radius 4.0 m. (i) Calculate the car’s radial acceleration. (ii) The radial friction between the car’s tyres and the track has a maximum value of 23 N. Show that this force is sufficient to prevent the car skidding off the track. b) The student now places the car on a banked track as shown in Figure 1.6.

3 3

4.0 m

Figure 1.6 Banked circular track for model car



She again uses the radio control to make the car follow a circular path of radius 4.0 m. Explain why the car can now travel much faster than 6.0 m s−1 without skidding off the track.

2

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Area 1  Rotational motion and astrophysics

Topic 3

Rotational dynamics

Key points !

✽✽ Knowledge that an unbalanced torque causes a change in the angular (rotational) motion of an object. ✽✽ Definition of moment of inertia of an object as a measure of its resistance to angular acceleration about a given axis. ✽✽ Knowledge that moment of inertia depends on mass and the distribution of mass about a given axis of rotation. ✽✽ Use of an appropriate relationship to calculate the moment of inertia for a point mass. ✽✽ Use of an appropriate relationship to calculate the moment of inertia for discrete masses. ✽✽ Use of appropriate relationships to calculate the moment of inertia for rods, discs and spheres about given axes. ✽✽ Use of appropriate relationships to carry out calculations involving torque, perpendicular force, distance from the axis, angular acceleration and moment of inertia. ✽✽ Use of appropriate relationships to carry out calculations involving angular momentum, angular velocity, moment of inertia, tangential velocity, mass and its distance from the axis. ✽✽ Statement of the principle of conservation of angular momentum. ✽✽ Use of the principle of conservation of angular momentum to solve problems. ✽✽ Use of appropriate relationships to carry out calculations involving potential energy, rotational kinetic energy, translational kinetic energy, angular velocity, linear velocity, moment of inertia and mass.

Rotational motion and torque In referring to rotational motion we need to examine what will cause an object to rotate. A force applied to a free-standing body will cause that body to accelerate in the direction of the force. The magnitude of the F . The term, m, is the mass of the body and acceleration is given by a = m the mass is a measure of the body’s resistance to motion. An object with greater mass is more difficult to start and stop. In order to cause something to rotate, we have to apply a force perpendicular to a point or centre. In Figure 1.7 there is a force applied to a circular object causing it to rotate. This could be a cycle pedal or someone turning a wheel. This force acting in a perpendicular direction is called a torque and its magnitude is calculated using τ = Fr. F is the magnitude of the applied force and r is the perpendicular distance from the force to the centre. 14

Figure 1.7 Torque applied to an object

Key term Torque: A force acting in a perpendicular direction that causes an object to rotate about an axis.

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Topic 3  Rotational dynamics

The unit of torque is N m. (Force × distance) A greater torque can be applied by administering a larger force or applying it at a greater distance from the centre. You may be familiar with using a longer spanner or thicker screwdriver when trying to tighten or remove a screw or nut. You were increasing the torque you could apply by using these larger tools, which increased the distance, r. In rotational dynamics a torque causes an object to start or stop its rotation. This applied torque causes the object to rotate more quickly. This is referred to as angular acceleration. A linear force causes a change in linear velocity. A torque causes a change in angular velocity. Whereas mass is a measure of the object’s resistance to change in linear terms (inertia), there is a similar concept when objects are made to rotate. This ‘equivalent’ to mass is referred to as the moment of inertia, I, and is not only the amount of mass but the distribution of mass around its axis of rotation. It is a complex concept and the determination of an object’s moment of inertia can be difficult. For a point mass at a distance from its point of rotation I = mr². For simple shapes the moment of inertia can be determined using the following relationships:

Key terms Angular acceleration: The time rate of change of angular velocity of a rotating body, denoted by the Greek letter alpha (α). Inertia: A property of matter which resists change in motion or movement. Moment of inertia: A measure of an object's resistance to angular acceleration about a given axis.

Rod about centre

I = 121 ml 2

Hints & tips

Rod about end

I = 1 ml 2

Disc about centre

I = 12 mr 2

You do not have to memorise the equations for moment of inertia – look them up on the SQA relationships sheet in Appendix 3.

3

Sphere about centre I = 2 mr 2 5

When a torque is applied to an object, and we attempt to determine its angular velocity or acceleration, we need to find the moment of inertia rather than mass alone. Objects can have the same mass but dramatically different moments of inertia.



Consider a metre stick with two 100 g masses slotted at the centre of the stick. Hold the stick at the centre and rotate the metre stick from side to side. If you now move the masses, putting one at either end of the stick, and try to rotate the stick you will find that it is much more difficult, even though the mass of the system has not changed.

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Area 1  Rotational motion and astrophysics

f 1 A flywheel consisting of a solid, uniform disc is free to rotate about

Example

a fixed axis. The disc has a mass of 16 kg and a radius of 0.30 m. Calculate the moment of inertia of the flywheel.

Solution For a disc about its centre:

I = 12 mr 2 I = 12 × 16 × 0 .30 2  I  = 0.72 kg m2

f 2 A student attempts to determine the moment of inertia of a solid

Example

sphere about an axis through its centre and measures the mass of the sphere to be 3.8 kg and the radius to be 0.053 m. Calculate the moment of inertia of the sphere.

Solution For a sphere about its centre:

I = 25 mr 2 I = 25 × 3 . 8 × 0 . 0532  I  = 0.00426968  I  = 4.3 × 10-3 kg m2

Angular momentum The comparison of moment of inertia and mass needs to be revisited slightly when dealing with the momentum of a rotating object. Momentum = m × v Angular momentum is calculated in a similar way but using moment of inertia and angular velocity L = Iω The angular momentum is the product of these two quantities. For a point mass, m, a distance, r, from its centre of rotation, this relationship can be expanded:   L = Iω = mr²ω = mvr  remember ω = vr    16

Key term Angular momentum: The quantity of rotation of a body. It is given as the product of the moment of inertia and angular velocity. If no external torque acts on a system, the angular momentum of the system will remain constant. This is known as the principle of conservation of angular momentum.

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Topic 3  Rotational dynamics

f 3 A horizontal, flat disc of mass 6.0 kg and radius 0.50 m is allowed to

Example

rotate freely about its central axis. Calculate the moment of inertia of the disc.

Solution For a disc about its centre:

I = 12 mr 2 I = 12 × 6 . 0 × 0 . 502 I = 0.75 kg m2

f 4 The same disc is rotating with an angular velocity of 12 rad s . A cube

Example

−1

of mass 2.0 kg is dropped onto the disc. It lands and remains at a distance 0.40 m from the axis of rotation. a) Determine the total moment of inertia of the disc and cube. b) Calculate the angular velocity of the disc after the cube lands.

Solution a) For the cube at a radius:

I = mr2 I = 2. 0 × 0 .402 I = 0 .32 kg m2 For the system of disc and mass together: Itotal = Idisc + Imass Itotal = 0 .75 + 0 .32 Itotal = 1.1 kg m2 b) Lbefore = Lafter Ioωo = I1ω1 0 .75 × 12 = 1.1 × ω1 ω1 = 8 .2 rad s−2



In linear motion the kinetic energy of a moving object is given by Ek = 12 mv 2. An object which is rotating has a rotational kinetic energy which is given by Ek( rot ) = 1 I ω 2. 2

Very large amounts of energy can be stored (and retrieved) using rotating objects and they are replacing batteries in some systems, for example, a Kinetic Energy Recovery System. Along with braking, which converts kinetic energy to heat, a mechanical flywheel is caused to rotate by the car while slowing. The energy is stored as rotational kinetic energy, which is then used while accelerating to give additional power. 17

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Study questions

?

1 a) A mass m rotates at a distance r from a fixed axis. State the expression for its moment of inertia. b) A toy gyroscope consists of an axle, a narrow ring and spokes as shown in Figure 1.8.

1

axle narrow ring

spoke

Figure 1.8 Toy gyroscope



The mass of the axle and spokes is negligible compared to the mass of the narrow ring. The narrow ring has a mass of 1.5 kg and an average radius of 0.20 m. Show that the moment of inertia of the gyroscope is 0.060 kg m2.  c) The axle of the gyroscope has a radius of 4.0 mm. The gyroscope is made to spin using a thin cord. A 0.50 m length of thin cord is wound round the axle, as shown in Figure 1.9.

1

thin cord

Figure 1.9 Gyroscope with thin cord attached



18

The cord is pulled with a steady horizontal force of 25 N. A constant frictional torque of 0.070 N m opposes the motion of the gyroscope. (i) Calculate the resultant torque acting on the gyroscope.  (ii) Calculate the angular acceleration of the gyroscope.  (iii) Show that the angular displacement of the gyroscope is 125 radians just as the cord fully unwinds.  (iv) Calculate the maximum angular velocity of the gyroscope.  (v) After the cord has fully unwound, the frictional torque remains constant. How long does it take for the angular velocity of the gyroscope to decrease to 4.2 rad s−1? 

3 3 3 3 4

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Topic 3  Rotational dynamics

2 The front wheel of a racing bike can be considered to consist of five spokes and a rim, as shown in Figure 1.10. rim spokes

0.60 m

Figure 1.10 Spokes and rim of a racing bike wheel

The mass of each spoke is 0.040 kg and the mass of the rim is 0.24 kg. The wheel has a diameter of 0.60 m. a) (i) Each spoke can be considered as a uniform rod. Calculate the moment of inertia of a spoke as the wheel rotates. 3 4 (ii) Show that the total moment of inertia of the wheel is 2.8 × 10-2 kg m2. b) The wheel is placed in a test rig and rotated as shown in Figure 1.11.

brake

Figure 1.11 Racing bike wheel in brake test rig

(i) The tangential velocity of the rim is 19.2 m s−1. Calculate the angular velocity of the wheel. (ii) The brake is now applied to the rim of the wheel, bringing it uniformly to rest in 6.7 s. (A) Calculate the angular acceleration of the wheel. (B) Calculate the torque acting on the wheel.

3 3 3

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Area 1  Rotational motion and astrophysics

Topic 4

Gravitation

Key points !

✽✽ Conversion between astronomical units (AU) and metres and between light-years (ly) and metres. ✽✽ Definition of gravitational field strength as the gravitational force acting on a unit mass. ✽✽ Sketch of gravitational field lines and field line patterns around astronomical objects and astronomical systems involving two objects. ✽✽ Use of an appropriate relationship to carry out calculations involving gravitational force, masses and their separation. ✽✽ Use of appropriate relationships to carry out calculations involving period of satellites in circular orbit, masses, orbit radius and satellite speed. ✽✽ Definition of the gravitational potential of a point in space as the work done in moving unit mass from infinity to that point. ✽✽ Knowledge that the energy required to move mass between two points in a gravitational field is independent of the path taken. ✽✽ Use of appropriate relationships to carry out calculations involving gravitational potential, gravitational potential energy, masses and their separation. ✽✽ Definition of escape velocity as the minimum velocity required to allow a mass to escape a gravitational field to infinity, where the mass achieves zero kinetic energy and maximum (zero) potential energy. ✽✽ Use of an appropriate relationship to carry out calculations involving escape velocity, mass and distance.

Gravitational attraction This section deals with how objects with mass behave when they interact with each other gravitationally. In gravitational terms, every object interacts with every other object in the Universe. All objects exert an attractive force on each and every other object. This force is proportional to the mass of each object and inversely proportional to the distance (separation) between them. where G is the universal gravitational This can be expressed as F   = GMm 2 r constant. It is gravitational attraction that determines the structure of our Solar System, the formation of stars and planets and the shape and movement of our galaxies. It acts over infinite distances, although it is extremely small when distances become large. The distances between planetary objects can be extremely large and we incorporate other units to help appreciate the scale of the distances. 20

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Topic 4  Gravitation

The SI unit of distance is the metre and is familiar to most. When dealing with planetary objects in our Solar System another unit, the astronomical unit (AU), is commonly used. It is nominally the distance between the Sun and the Earth, although this distance varies since the orbit of the Earth around the Sun is not a perfect circle. Its current definition is 149 597 870 700 m. This is taken as 1.5 × 1011 m. Use of the astronomical unit allows comparison of orbits of other planets or objects to be made more easily. Earth’s distance from the Sun Venus’s distance from the Sun Saturn’s distance from the Sun Neptune’s distance from the Sun Kuiper Belt distance from the Sun Proxima Centauri (next nearest star) distance from the Sun

Key term Astronomical unit (AU): Roughly the mean distance between the Earth and the Sun (1.5 × 1011 m).

1 AU 0.72 AU 10 AU 30 AU 30–50 AU 268 700 AU

The Voyager space probes were launched in 1977 and are the most distant man-made objects from us. Voyager 1 is now 145 AU from us. When describing distances between galaxies we require a larger unit. After the Sun, our next nearest star is approximately 270 000 AU away. Other stars are many orders of magnitude beyond that. A light-year is used in these cases to describe the distance. It is the distance travelled by light through space in one year (365.25 days). This equates to 3.0 × 108 × 60 × 60 × 24 × 365.25 = 9.5 × 1015 m.

Hints & tips



You need to be able to convert light-years to metres. The equation is not provided on the SQA relationships sheet.

Gravitational fields The concept of a field is used to describe and explain what occurs when forces act upon various bodies. We describe phenomena using electric fields, magnetic fields, and so on. Gravitational fields are used to describe the force a mass would experience when placed near another mass. Field lines are used to give an indication of the magnitude and direction of the force that the mass would experience when placed in the field. 104 km

Key term Gravitational field: A region where an object experiences a force due to its mass.

g-field, equipotential surfaces

Earth

g-field, lines of force

Figure 1.12 Field lines and equipotentials in a radial field

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Area 1  Rotational motion and astrophysics

At the surface of the Earth, however, these lines will appear vertical and parallel due to the large radius of the Earth. This is often described as a uniform field. g-field, lines of force

Earth’s surface

Figure 1.13 A uniform gravitational field at the Earth’s surface

This means that at the surface of the Earth objects will fall vertically downwards but, more accurately, objects are falling towards the centre of the mass of the Earth. The gravitational field strength, g, at a point on the surface (say) is F given by g   =   mg where Fg is the gravitational force experienced by a body of mass, m, at that point. The unit of g is N kg−1. On Earth g is 9.8 N kg−1, on Mars g is 3.7 N kg−1 and on Jupiter g is 23 N kg−1. From this we can easily note that an object with the same mass will have a different weight on the surface of different planets.

Determination of g at the Earth’s surface Fg g= = m

GMe m r 2 = GMe m r2

Mass of Earth Radius of Earth G

g=

6.0 × 1024 kg 6.4 × 106 m 6.67 × 10−11 m3 kg−1 s−2

6.67 × 10−11 × 6.0  × 1024 = 9.77 = 9.8 N kg –1 6 2 . (6 4 × 10 )

Interaction of gravitational fields The gravitational field lines around a mass show the direction of the gravitational force and also the magnitude of that force. The greater the number of field lines per unit area, the greater the strength of the force in that region.

22

Key term Gravitational field strength: The gravitational force acting on a unit mass.

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Topic 4  Gravitation

An interesting result of this is what happens when we consider the interaction of two astronomical bodies and their associated gravitational fields. Consider a planet and a smaller moon.

moon

planet

Figure 1.14 A planet and a smaller moon

The density of field lines closer to the planet seen in Figure 1.15 indicates that the field is stronger closer to the planet. An object, such as an asteroid, passing through space will experience attractive forces from various astronomical objects as it travels. If it passes close to the planet and is travelling relatively slowly, it may be diverted towards the planet. The interaction of the gravitational fields leads to the lines curving slightly when they reach the area where the force attracting an object to the planet, in this case Earth, is similar to the force attracting the object to the Moon.

P

Figure 1.15 Interaction of gravitational forces between the Earth and Moon with the point of balance, P

The interaction of the field lines shows that, at point P, the gravitational forces of attraction for the object due to the Earth and Moon are balanced and in opposite directions. At this point the object would experience no resultant force due to the balancing of the gravitational forces. This point is obviously not midway between the Earth and Moon. The mass of the Earth is around 100 times that of the Moon and, accordingly, the balance point is roughly 90% of the distance from the Earth to the Moon. 23

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Area 1  Rotational motion and astrophysics

Satellites Satellites are objects which orbit larger masses. A satellite needs to be at the correct height and velocity such that it orbits the larger mass and stays in orbit almost continually. The Moon is a natural satellite, but we have thousands of artificial satellites which orbit the planet for such purposes as communication, remote sensing, navigation and weather forecasting.

Key term Satellite: An object which orbits a larger mass.

In order for the satellite to remain in orbit, it must experience a force of attraction or central force which will maintain the satellite in a circular orbit. This force of attraction is given by F=

GMm mv 2 and this is equivalent to the central force = r . r2 2

2   GMm = ω 2 and as ω = 2π , F = mr  2π  . mr This leads to F = 2 = mv r r T  T

Satellites orbit the Earth at a specific height and tangential velocity. We can place satellites at various heights depending on the data we want them to gather but they all must travel around the Earth at high velocities. The International Space Station (ISS) orbits the Earth at a height of 409 km. In order to maintain this height, it has to travel at a velocity of 7600 m s−1. This maintains the ISS in orbit and also freefall. If its velocity was to reduce, it would gradually reduce its height above the Earth and, if this was not corrected, it would ultimately be destroyed by the air friction. Huge amounts of heat would be generated as it entered the atmosphere. The opposite would also hold. If the ISS were to increase its velocity, it would gradually increase its height and possibly leave Earth’s orbit.

Gravitational potential In dealing with masses within a gravitational field, the concept of gravitational potential (V) is used. It is defined as the gravitational potential energy at a certain point within a gravitational field per unit mass. Gravitational potential is also defined as the work done in moving a unit mass from infinity to that point. This may seem unusual at first, but it is the movement within the gravitational field that results in the transfer or transformation of energy.

24

Key term Gravitational potential: The energy an object of mass has at a point in relation to another massive object due to gravity.

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Topic 4  Gravitation

104 km g-field, equipotential surfaces

X

Y Earth

g-field, lines of force

Figure 1.16 Energy required to move objects within a gravitational field

In a gravitational field, the energy required to move a mass between two points is dependent only on the position relative to the field; it is independent of the path taken. Looking at Figure 1.16, consider a mass at point X on an equipotential line. When that mass moves along the line it remains at that potential. No work is being done in moving it as it is at the same potential. When it moves from point Y to X, it is moving to a point of different potential within the field. This is when energy is required. If the mass was to move clockwise or anticlockwise, the amount of energy required would be the same as the only concern is the movement of the mass between different points in the field. In a uniform gravitational field, such as at the surface of a planet, the change in the gravitational potential is determined by ∆Vg = g ∆ℎ

(At the surface we take g to be constant)

At a larger scale, in a radial field the gravitational potential is given by V = − GM r The potential energy of a mass at a point in that field is given by Ep = Vm = − GMm r

Escape velocity How can an object escape the gravitational field of a planet? There are a number of ways to think about this. You are familiar with throwing an object upwards. Once thrown it will slow down, stop and return to Earth. If you throw it with a great velocity, it will go higher but ultimately return. Is there a velocity it can be thrown with that will cause it to be able to overcome the planet’s attraction and leave?

Key term Escape velocity: Minimum velocity of a body which will allow it to escape a gravitational field to infinity.

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Area 1  Rotational motion and astrophysics

In energetic terms we could supply the object with enough kinetic energy so that its energy is greater than that due to its potential. In simplistic terms the final energy must equal zero. This is equivalent to the object being at a point at infinity. Etotal = Ek + Ep = 0 1 mv 2 2

− GMm r =0

1 mv 2 2

= GMm r

v=

Hints & tips



If you recreate this derivation you must start with Etotal = 0.

2GM r

Study questions

?

1 a) The gravitational field strength, g, on the surface of Mars is 3.7 N kg−1. The mass of Mars is 6.4 × 1023 kg. Show that the radius of Mars is 3.4 × 106 m. b) (i) A satellite of mass, m, has an orbit of radius R. Show that the angular velocity, ω, of the satellite is given by the expression

ω=

3

GM 3 R

where the symbols have their usual meanings. 3 (ii) A satellite remains above the same point on the equator of Mars as the planet spins on its axis. Figure 1.17 shows this satellite orbiting at a height of 1.7 × 107 m above the Martian surface.

1.7 × 107 m

Figure 1.17 A satellite orbiting above the Martian surface (not to scale)

Calculate the angular velocity of the satellite. (iii) Calculate the length of one Martian day.

26

3 3

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Area 1  Rotational motion and astrophysics

Topic 5

General relativity

Key points !

✽✽ Knowledge that special relativity deals with motion in inertial (nonaccelerating) frames of reference and that general relativity deals with motion in non-inertial (accelerating) frames of reference. ✽✽ Statement of the equivalence principle (that it is not possible to distinguish between the effects on an observer of a uniform gravitational field and of a constant acceleration) and knowledge of its consequences. ✽✽ Consideration of spacetime as a unified representation of three dimensions of space and one dimension of time. ✽✽ Knowledge that general relativity leads to the interpretation that mass curves spacetime, and that gravity arises from the curvature of spacetime. ✽✽ Knowledge that light or a freely moving object follows a geodesic (the path with the shortest distance between two points) in spacetime. ✽✽ Representation of world lines for objects which are stationary, moving with constant velocity and accelerating. ✽✽ Knowledge that the escape velocity from the event horizon of a black hole is equal to the speed of light. ✽✽ Knowledge that, from the perspective of a distant observer, time appears to be frozen at the event horizon of a black hole. ✽✽ Knowledge that the Schwarzschild radius of a black hole is the distance from its centre (singularity) to its event horizon. ✽✽ Use of an appropriate relationship to solve problems relating to the Schwarzschild radius of a black hole.

This part of the course is assessed using mostly descriptive, extendedprose responses rather than the more traditional numerical answers to algebraic-type questions. The majority of the key points relate to knowledge and so you must be able to replicate and apply the descriptions of physics in the following paragraphs.

Inertial and non-inertial reference frames A frame of reference is an origin and set of co-ordinates (x, y, z) that can be used to determine positions and velocities of objects in that frame. There is no universal frame of reference against which all positions and velocities are measured. Different observers may select or define different frames of reference and these move relative to each another. If the observer (and their frame of reference) is stationary, or moving with constant speed, with respect to the background then they are in an inertial frame of reference. In this case, Einstein’s laws of Special Relativity are applicable.

Key terms Frame of reference: This refers to a co-ordinate and reference system by which we can determine position and velocities. Inertial frame of reference: One where the observer is stationary or moving at a constant speed. 27

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Area 1  Rotational motion and astrophysics

If the observer (and their frame of reference) is accelerating with respect to the background then they are in a non-inertial frame of reference. In this case, Einstein’s laws of General Relativity are applicable. General relativity states that the effects of both a uniform gravitational field and constant acceleration on an observer are indistinguishable. This is known as the equivalence principle. For example, this means that an astronaut in a spacecraft, moving with constant acceleration, would not be able to tell if they begin to fall to the floor because the spacecraft started accelerating or because they are now in a gravitational field.

Spacetime

Key terms Non-inertial frame of reference: One where the observer is accelerating. Equivalence principle: It is indistinguishable for an observer to tell if they are in an accelerating object or in a gravitational field. Spacetime: A model where the three dimensions of space are combined with the dimension of time.

Spacetime combines the reference frame, that is, three dimensions of space, with the fourth dimension of time. Spacetime is useful in describing the effects of general relativity, such as why different observers perceive differently where and when events occur and the effect of massive objects on the observed motion of objects. A massive object (such as a star) will cause spacetime to distort such that it becomes curved – much like a sheet of rubber is deformed when a mass is placed in its centre. This curvature means that the shortest distance between two points in spacetime may not be a straight line. An object or ray of light moving between these two points would follow a geodesic. This path would appear curved to a distant observer in the same way as the orbit of a planet around a star is curved. The observed motion of the planet may be explained Unusually, the vertical by either the existence of a gravitational force or by the curvature of spacetime by the mass of the star. axis of the spacetime diagram has units of time. A spacetime diagram may be used to represent where and when The horizontal (plane) has events occur in spacetime relative to an observer. The vertical axis units of distance. has units of time (often measured in units of ct) and the horizontal axis has units of dimension in space. A world line connects a series of observed events for an object. The slope of any world constant line is the inverse of the velocity of the object experiencing light ct stationary velocity accelerated those events. For Figure 1.18, such a line has a minimum slope of c, the speed of light, as no object may travel faster than that. As can be seen from Figure 1.18: ●● A stationary object has a vertical world line. ●● An object moving with constant velocity has a sloped x world line (slope > 45°). ●● An accelerating object has a curved world line. Figure 1.18 A spacetime diagram

Hints & tips

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Topic 5  General relativity 

Occasionally, such diagrams are used to attempt to represent more than one dimension in spacetime. This diagram creates a lightcone within which all possible past events (below the dimension plane) and future events (above the dimension plane) for the observer located at the origin are contained. Such a diagram is shown in Figure 1.19.

Escape velocity and black holes For a very massive object, spacetime will become so curved that the escape velocity may become larger than the speed of light. Objects this massive create a black hole in spacetime. The event horizon of a black hole is where the escape velocity is equivalent to the speed of light. The distance from the centre of the black hole to the event horizon is known as the Schwarzschild radius.

ct

future

event

The Schwarzschild radius is given by

now y

2GM r

v=

past

For light to escape, this means that c=

2GM r

rsch =

2GM c2

x

Figure 1.19 A spacetime diagram with more than one dimension

Key terms Black hole: An incredibly massive object which curves spacetime so that the escape velocity is greater than the speed of light. Event horizon: The boundary of a black hole at which the escape velocity is equal to the speed of light. An event horizon can be defined as the point from which nothing can return. From the perspective of a distant observer, time appears to be frozen at the event horizon of a black hole. Schwarzchild radius: The Schwarzchild radius of a black hole is the distance from its centre (singularity) to its event horizon.

f 1 A star of mass 4.5 × 10  kg collapses to form a black hole. Calculate

Example

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the Schwarzschild radius of this black hole.

Solution rsch = 2GM c2 2 × 6.67 × 10 –11 × 4.5 × 10 31 rsch = (3 × 10 8 )2 r   = 6.67 × 104 m sch

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Area 1  Rotational motion and astrophysics

Study questions

?

1 a) The world lines for three objects A, B and C are shown in Figure 1.20.

C

time

A

0

B

distance

Figure 1.20 World lines for three objects A, B and C



To which of these objects does the General Theory of Relativity apply? Explain your choice. b) A rocket ship is accelerating through space. Clocks P and Q are at opposite ends of the ship as shown in Figure 1.21. An astronaut inside the rocket ship is beside clock P and can also observe clock Q.

direction of acceleration

2

clock P

clock Q

Figure 1.21 Rocket with two clocks, P and Q



What does the astronaut observe about the passage of time for these clocks? Justify your answer. c) Part of an astronaut’s training is to experience the effect of ‘weightlessness’. This can be achieved inside an aircraft that follows a path as shown in Figure 1.22.

2

Figure 1.22 Weightless flight path



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Use the equivalence principle to explain how this ‘weightlessness’ is achieved.

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Area 1  Rotational motion and astrophysics

Topic 6

Stellar physics

Key points !

✽✽ Use of appropriate relationships to solve problems relating to luminosity, apparent brightness, b, distance between the observer and the star, power per unit area, stellar radius and stellar surface temperature. (Using the assumption that stars behave as black bodies.) ✽✽ Knowledge that stars are formed in interstellar clouds when gravitational forces overcome thermal pressure and cause a molecular cloud to contract until the core becomes hot enough to sustain nuclear fusion, which then provides a thermal pressure that balances the gravitational force. ✽✽ Knowledge of the stages in the proton–proton chain (p–p chain) in stellar fusion reactions which convert hydrogen to helium. ✽✽ Knowledge that Hertzsprung–Russell (H–R) diagrams are a representation of the classification of stars. ✽✽ Classification of stars and position in Hertzsprung–Russell (H–R) diagrams, including main sequence, giant, supergiant and white dwarf. ✽✽ Use of Hertzsprung–Russell (H–R) diagrams to determine stellar properties, including prediction of colour of stars from their position in an H–R diagram. ✽✽ Knowledge that the fusion of hydrogen occurs in the core of stars in the main sequence of a Hertzsprung–Russell (H–R) diagram. ✽✽ Knowledge that hydrogen fusion in the core of a star supplies the energy that maintains the star’s outward thermal pressure to balance inward gravitational forces. When the hydrogen in the core becomes depleted, nuclear fusion in the core ceases. The gas surrounding the core, however, will still contain hydrogen. Gravitational forces cause both the core and the surrounding shell of hydrogen to shrink. In a star like the Sun, the hydrogen shell becomes hot enough for hydrogen fusion in the shell of the star. This leads to an increase in pressure which pushes the surface of the star outwards, causing it to cool. At this stage, the star will be in the giant or supergiant regions of a Hertzsprung–Russell (H–R) diagram. ✽✽ Knowledge that, in a star like the Sun, the core shrinks and will become hot enough for the helium in the core to begin fusion. ✽✽ Knowledge that the mass of a star determines its lifetime. ✽✽ Knowledge that every star ultimately becomes a white dwarf, a neutron star or a black hole. The mass of the star determines its eventual fate.

Stars This topic deals with the formation, growth and ultimate demise of stars. It describes how we classify stars and how their appearance is related to the processes of energy transfer.

Hints & tips



You need to be able to (briefly) describe the lifecycle of a star. 31

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Area 1  Rotational motion and astrophysics

Stars are formed from the accretion of space dust. They form within relatively dense clouds of dust and gases, often referred to as nurseries. At extreme cold temperatures, around 10 K, the gases may exist as molecules. In certain regions of these dust clouds, the matter pulls together and the core of this matter forms massive fragments and these, in turn, form protostars. These have a large mass, many times the mass of our Sun, and as such have a strong gravitational field around them. The collapsing cloud rotates as it is drawn to the centre of the protostar and, at the centre, becomes more massive. The area around it may form into a disc shape with some planetary formation within that. Larger clouds may separate and create regions where a number of stars form relatively close together. As the protostar becomes more dense, atoms at the core begin to fuse together and energy is released. When the surrounding gas and dust has been absorbed or expelled, the protostar is considered to be a pre-main sequence star. The protostar will continue to collapse under its own gravitational attraction and create more fusion until the thermal pressure (outwards) from the fusion reaction balances the gravitational pressure (inwards). This prevents more contraction and at this point it becomes a stable main sequence star. This star will continue to radiate energy for a very long period of time, dependent upon its size. Our Sun will be stable for approximately 10 000 000 000 years. Interestingly, the photons generated in the core of the Sun travel slowly through the various zones and take thousands of years to reach the surface. They then take only 500 s to reach the Earth.

Classification of stars We classify stars based on a number of observable phenomena.

Size As stars are spherical objects we can classify them in terms of their radius, R. The radius of our Sun is 6.96 × 108 m. Betelgeuse has a radius of around 1800 times that of our Sun, while some dwarf stars have a radius less than 0.01 times the radius of our Sun.

Temperature Black-body radiation refers to energy emitted from a black or nonreflective body. It has a characteristic spectrum of wavelengths which is dependent upon the surface temperature of the object. Stars are considered to be spherical, perfect black-body radiators. That is, they neither absorb nor reflect radiation, only emitting radiation as a result of their thermal energy. 32

Key term Protostar: The embryonic early phase of a star.

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power redirected per m2 of a black body at a particular wavelength (arbitrary units)

Topic 6  Stellar physics

10 9 8 7 visible

6

6000 K

5 4

5000 K

3

4000 K 3000 K

2 1 100

500

1000 1500 2000 wavelength/nm

2500

Figure 1.23 Curves showing the spectra measured from objects at different surface temperatures

At higher temperatures more energy is emitted (not surprisingly). The total amount of energy emitted per second (power) is equivalent to the area under the graph of the spectral diagram. Obviously, a high temperature object radiates more energy than a lower temperature object. The peak of the spectrum is related to the temperature. A peak at a shorter wavelength indicates a higher temperature of the object. A peak in the red area of the spectrum indicates a temperature of around 4000 K. A peak in the blue-green area indicates a temperature of around 6000 K.

Stefan’s law This states that the power radiated, P, by a black body with surface area, A, is given by P = AσT4 sometimes written as P = σ T 4 where A σ = Stefan–Boltzmann constant, 5.67 × 10−8 W m−2 K−4 T = surface temperature in Kelvin. This relationship allows us to determine the amount of energy emitted by a star. Often this is related to brightness but that would not take into account energy emitted in the forms of X-rays or infrared, for example. We use the term luminosity, and this refers to the total power emitted by a star across all wavelengths in all directions. Luminosity L = 4πr 2P/A and since P = AσT4, L = 4πr2σT4

Key term Luminosity: The total amount of electromagnetic energy (of all forms) radiated by a star.

The luminosity of the Sun can be determined using L = 4πr2σT4 = 4 × π × (6.96 × 108)2 × 5.67 × 10-8 × (5800)4. = 3.9 × 1026 W This power output is relatively middling for a star; some are a thousand times smaller and others over a thousand times larger.

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Area 1  Rotational motion and astrophysics

The classification or description of a star based on its brightness is more subjective than its luminosity but still has some value as brighter objects are more visible to us and easier to detect. Brightness is however dependent upon the luminosity of the star and also how distant the star is from us, so we use the term apparent brightness. Stars which are dimmer but closer to the Earth can have the same apparent brightness as stars which are further away but brighter. Apparent brightness is given by b = L 2 where d is the distance from 4πd the star to us. This is obviously an ‘inverse square’ relationship with which you should be familiar. This gives the apparent brightness of the Sun as L 3.9 × 10 26 −2 b= = 2 = 1379 Wm . 4πd 2 4 × π × (1.5 × 1011)

Example

f

1 The table below shows some of the physical properties of a star

called Rigel. Property

Surface temperature Radius Mass Distance to Earth

Value

1.20 × 104 K 5.49 × 1010 m 18 solar masses 773 light-years

a) Calculate the luminosity of Rigel. b) State the assumption made in your calculation. c) Calculate the absolute brightness of Rigel.

Solution a) L = 4πr 2σT4

L = 4π(5.49 × 1010)2σ(1.20 × 104)4 L = 4.45 × 1031 W b) Rigel is a perfect black-body radiator (like all other stars) L c) b = 4πr 2 Convert light-years to metres r = 773 × 365 × 24 × 60 × 60 × (3 × 108) = 7.31 × 1018 m 4.45 × 10 31 b= 2 4π(7.31 × 1018) b = 6.63 × 10-8 W m-2

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Key term Apparent brightness: A slightly more subjective description of the luminosity of a star as it refers to the amount of energy reaching the observer which depends also on the distance of the star from us.

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Topic 6  Stellar physics

Spectral classes Upon observing a great number of stars, it became apparent to scientists that they could be classified into different types or classes dependent upon their temperature, colour and absorption lines. Spectral class Intrinsic colour Temperature/K

Prominent absorption lines

O B A F G K M

He+, He, H He, H H (strongest) ionised metals Ionised metals Ionised and neutral metals Neutral metals Neutral metals, TiO

Blue Blue Blue-white White Yellow-white Orange Red

25 000–50 000 11 000–25 000 7500–11 000 6000–7500 5000–6000 3500–5000