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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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17
18
18
21
25
25
26
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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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47
48
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58
63
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65
66
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70
78
82
83
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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?
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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 =?
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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.
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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
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..........................................
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.
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..........................................
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 = ?
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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.
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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.
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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.
..........................................
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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
..........................................
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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 .
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..........................................
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.
..........................................
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TOPIC 1. KINEMATIC RELATIONSHIPS
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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)
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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
..........................................
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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)
22
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
+...
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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
27
28
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.
© H ERIOT-WATT U NIVERSITY
32
TOPIC 3. ANGULAR VELOCITY AND ACCELERATION
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
TOPIC 3. ANGULAR VELOCITY AND ACCELERATION
33
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
34
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.
© H ERIOT-WATT U NIVERSITY
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
36
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.
© H ERIOT-WATT U NIVERSITY
TOPIC 3. ANGULAR VELOCITY AND ACCELERATION
37
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
© H ERIOT-WATT U NIVERSITY
20 min
38
TOPIC 3. ANGULAR VELOCITY AND ACCELERATION
..........................................
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
..........................................
© H ERIOT-WATT U NIVERSITY
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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TOPIC 3. ANGULAR VELOCITY AND ACCELERATION
..........................................
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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TOPIC 3. ANGULAR VELOCITY AND ACCELERATION
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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TOPIC 3. ANGULAR VELOCITY AND ACCELERATION
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;
© H ERIOT-WATT U NIVERSITY
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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 3. ANGULAR VELOCITY AND ACCELERATION
© H ERIOT-WATT U NIVERSITY
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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.
48
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)
50
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
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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.
© H ERIOT-WATT U NIVERSITY
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
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TOPIC 4. CENTRIPETAL FORCE
53
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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TOPIC 4. CENTRIPETAL FORCE
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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TOPIC 4. CENTRIPETAL FORCE
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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TOPIC 4. CENTRIPETAL FORCE
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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TOPIC 4. CENTRIPETAL FORCE
..........................................
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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TOPIC 4. CENTRIPETAL FORCE
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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TOPIC 4. CENTRIPETAL FORCE
59
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◦
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TOPIC 4. CENTRIPETAL FORCE
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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TOPIC 4. CENTRIPETAL FORCE
61
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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TOPIC 4. CENTRIPETAL FORCE
..........................................
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.
..........................................
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30 min
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TOPIC 4. CENTRIPETAL FORCE
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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TOPIC 5. ROTATIONAL DYNAMICS
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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TOPIC 5. ROTATIONAL DYNAMICS
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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TOPIC 5. ROTATIONAL DYNAMICS
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
© H ERIOT-WATT U NIVERSITY
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
..........................................
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TOPIC 5. ROTATIONAL DYNAMICS
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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..........................................
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
..........................................
© H ERIOT-WATT U NIVERSITY
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TOPIC 5. ROTATIONAL DYNAMICS
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?
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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
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TOPIC 5. ROTATIONAL DYNAMICS
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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TOPIC 5. ROTATIONAL DYNAMICS
..........................................
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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TOPIC 5. ROTATIONAL DYNAMICS
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
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TOPIC 5. ROTATIONAL DYNAMICS
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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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TOPIC 5. ROTATIONAL DYNAMICS
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.
..........................................
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TOPIC 5. ROTATIONAL DYNAMICS
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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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TOPIC 5. ROTATIONAL DYNAMICS
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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83
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
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30 min
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TOPIC 5. ROTATIONAL DYNAMICS
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.
..........................................
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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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TOPIC 6. GRAVITATIONAL FORCE AND FIELD
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.
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TOPIC 6. GRAVITATIONAL FORCE AND FIELD
..........................................
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.
© H ERIOT-WATT U NIVERSITY
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TOPIC 6. GRAVITATIONAL FORCE AND FIELD
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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TOPIC 6. GRAVITATIONAL FORCE AND FIELD
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
93
94
TOPIC 6. GRAVITATIONAL FORCE AND FIELD
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.
..........................................
© H ERIOT-WATT U NIVERSITY
TOPIC 6. GRAVITATIONAL FORCE AND FIELD
95
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
© H ERIOT-WATT U NIVERSITY
20 min
96
TOPIC 6. GRAVITATIONAL FORCE AND FIELD
..........................................
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
TOPIC 6. GRAVITATIONAL FORCE AND FIELD
97
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
30 min
98
TOPIC 6. GRAVITATIONAL FORCE AND FIELD
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.
..........................................
© H ERIOT-WATT U NIVERSITY
99
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.
100
TOPIC 7. GRAVITATIONAL POTENTIAL AND SATELLITE MOTION
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
102
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 = −
© H ERIOT-WATT U NIVERSITY
TOPIC 7. GRAVITATIONAL POTENTIAL AND SATELLITE MOTION
103
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.
© H ERIOT-WATT U NIVERSITY
20 min
104
TOPIC 7. GRAVITATIONAL POTENTIAL AND SATELLITE MOTION
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)
106
TOPIC 7. GRAVITATIONAL POTENTIAL AND SATELLITE MOTION
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
© H ERIOT-WATT U NIVERSITY
TOPIC 7. GRAVITATIONAL POTENTIAL AND SATELLITE MOTION
107
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
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TOPIC 7. GRAVITATIONAL POTENTIAL AND SATELLITE MOTION
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.
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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.
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111
−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 =
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2Gm
r
(7.8)
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TOPIC 7. GRAVITATIONAL POTENTIAL AND SATELLITE MOTION
..........................................
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.
..........................................
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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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TOPIC 8. SIMPLE HARMONIC MOTION
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
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119
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
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(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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TOPIC 8. SIMPLE HARMONIC MOTION
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
(8.6)
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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.
..........................................
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TOPIC 8. SIMPLE HARMONIC MOTION
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
..........................................
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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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TOPIC 8. SIMPLE HARMONIC MOTION
..........................................
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
..........................................
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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
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TOPIC 8. SIMPLE HARMONIC MOTION
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=
k
m
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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.
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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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TOPIC 8. SIMPLE HARMONIC MOTION
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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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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TOPIC 8. SIMPLE HARMONIC MOTION
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.
140
TOPIC 9. WAVE-PARTICLE DUALITY
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
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TOPIC 9. WAVE-PARTICLE DUALITY
..........................................
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?
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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
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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.
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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 =
h
6.63 × 10−34
=
= 1.82 × 10−9 m
pe
3.64 × 10−25
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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
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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
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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.
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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
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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 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
156
156
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 . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
162
162
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
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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.
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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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TOPIC 10. INTRODUCTION TO QUANTUM MECHANICS
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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.
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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
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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.
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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.
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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.
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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
166
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.
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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
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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.
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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.
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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.
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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.
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170
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.
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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.
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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.
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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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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
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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?
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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
.
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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.
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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
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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
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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
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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.
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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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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.
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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
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ANSWERS: TOPIC 8
Q20: 0.18
Q21: 0.77
Q22: 2.4 × 10-3
Q23: 0.78
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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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200
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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