Circular motion
1. In the following examples name the force that is providing the centripetal force and draw it
on the diagram.
(a) A runner running round a circular track.
(b) A car on a rollercoaster.
2. A 2kg mass travels in a circle of radius 50cm. If the time for one revolution is 2s calculate:
(a) The angular velocity of the mass
(b) The centripetal acceleration of the mass
(c) The centripetal force of the mass
3. A ball rolls around the inside of a vertical cylinder as shown. Indentify the force that stops it
from falling down.
Formulae
ω=2π/T
F=mv2/r=mω2r
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1
1
Gravitational field
1. State Newton’s universal law of gravity.
2. Two masses are positioned as shown in the diagram.
Calculate the Force on the red one.
3. Define gravitational field strength.
4. Given that the mass of the moon is about 1/80 of the earth and its radius is ¼ estimate the
acceleration due to gravity on the surface of the moon.
Formulae
F=GMm/r2
G=6.7x10-11 m3kg-1s-2
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2
Answers to exam-style questions
Topic 6
Where appropriate, 1 ✓ = 1 mark
1 A
2 C
3 B
4 C
5 C
6 B
7 D
8 D
9 C
10 A
11 a Velocity arrow. ✓
Acceleration arrow. ✓
velocity
acceleration
2π
= 4.488 ≈ 4.5 rad s −1. ✓
1.40
The linear speed is v = ω r = 4.488 × 0.22 = 0.987 ≈ 0.99 m s −1. ✓
c At maximum distance the frictional force will be the largest possible, i.e. f max = µs N = µsmg( = 0.434 N) . ✓
µg
v2
ω 2r 2
, hence r = s 2 ✓
µsmg = m = m
r
r
ω
0.82 × 9.8
r=
= 0.399 ≈ 0.40 m ✓
4.488 2
µg
µs g
d i Using r = s 2 we find ω =
✓
r
ω
b The angular speed is ω =
0.82 × 9.8
= 6.0 rad s −1 ✓
0.22
ii The static frictional force can no longer supply the larger centripetal force required. ✓
The body will then slide and the static frictional force is now replaced by the even smaller sliding frictional
force; hence the disc will slide off the rotating platform. ✓
ω=
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ANSWERS TO EXAM-STYLE QUESTIONS – TOPIC 6
1
1
12 a From energy conservation: mv 2 = mgL so v = 2 gL , ✓
2
v = 2 × 9.8 × 2.0 = 6.26 ≈ 6.3 m s −1. ✓
v 2 6.26 2
=
= 19.6 ≈ 20 m s −2. ✓
L
2.0
c Weight vertically downwards. ✓
Larger arrow for tension upwards. ✓
d i A particle is in equilibrium if it moves with constant velocity. ✓
This particle moves on a circle and so cannot be in equilibrium. ✓
mv 2
✓
ii T − mg =
L
mv 2
5.0 × 6.26 2
T =
+ mg =
+ 5.0 × 9.8 = 147 ≈ 150 N ✓
L
2.0
mv 2
m × 2 gL
(or better: T =
+ mg =
+ mg = 3mg = 3 × 5.0 × 9.8 = 147 ≈ 150 N)
L
L
b a=
13 a Correct arrows for tension. ✓
Correct arrow for weight. ✓
tension
mg
b A particle is in equilibrium if it moves with constant velocity. ✓
This particle moves on a circle and so cannot be in equilibrium. ✓
c i The vertical component of the tension equals the weight and so T cos θ = mg , i.e. T =
The horizontal component of the tension is T sin θ and T sin θ = m
Combining gives the answer v =
mg
.✓
cos θ
v2
v2
=m
✓
r
L sin θ
gL sin 2 θ
.
cos θ
ii The angular and linear speeds are related by v = ω r = ω L sin θ . ✓
So ω =
gL sin 2 θ
cos θ . ✓
L sin θ
Which is the answer ω =
d i v=
g
.
L cos θ
9.8 × 0.45 × sin 2 60°
= 2.57 ≈ 2.6 m s −1 ✓
cos 60°
9.8
= 6.5997 ≈ 6.6 rad s −1 ✓
0.45 × cos 60°
e i The air resistance force will reduce the speed of the ball. ✓
sin 2 θ
ii A graph of
shows that because the speed decreases, the angle will also decrease. ✓
cos θ
ii θ =
2
ANSWERS TO EXAM-STYLE QUESTIONS – TOPIC 6
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iii The cosine of the angle will increase and hence the angular speed will decrease. ✓
(Note: These questions are best answered by considering the total energy of the ball:
E=
 sin 2 θ + 2 cos θ − 2 cos 2 θ 
1 2
1 gL sin 2 θ
1
mv + mgh = m
+ mgL (1 − cos θ ) = mgL 

2
2
cos θ
2
cos θ

The air resistance will reduce the total energy; graphing the total energy as a function of angle θ shows that
for the energy to decrease the angle must decrease.)
14 a Measuring distances from the top of the sphere and using energy conservation shows that:
1
0 = mv 2 − mgh where h is the vertical distance the marble falls. ✓
2
From trigonometry: h = R(1 − cos θ ). ✓ (see diagram that follows in b)
1
And so 0 = mv 2 − mgR(1 − cos θ ). ✓
2
Manipulating gives v = 2 gR(1 − cos θ ).
b The forces on the marble are the weight mg and the normal reaction force N:
N
R
Rcos θ
θ
mg
Taking components of the weight gives mg cos θ − N =
Hence N = mg cos θ −
mv 2
.✓
R
mv 2
.✓
R
Substituting the expression for the speed from above gives N = mg cos θ − 2mgR(1 − cos θ ) . ✓
And the result N = mg(3cos θ − 2) follows.
2
c The marble will lose contact when N → 0 , i.e. when cos θ = or θ ≈ 48°. ✓
3
15 a Calling this distance x we have that:
G16M
GM
✓
=
2
x
(d − x )2
16(d − x )2 = x 2 or 4(d − x ) = ± x ✓
Only the plus sign gives a positive distance and so x =
b Correct sign. ✓
Correct intersection. ✓
(The negative of this graph is also acceptable)
4d
.✓
5
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ANSWERS TO EXAM-STYLE QUESTIONS – TOPIC 6
3
g
x/d
0.2
0.4
0.6
0.8
1
c i The force is zero. ✓
ii The force from the larger mass will be larger because the particle will be closer to it. ✓
Hence the net force will be directed towards the large mass. ✓
d It will move to the left. ✓
With increasing speed and increasing acceleration. ✓
16 a i Velocity arrow. ✓
Acceleration arrow. ✓
velocity
acceleration
ii Acceleration is the rate of change of the velocity vector. ✓
Here the velocity vector is changing because its direction is so we have acceleration. ✓
b The force on the satellite is
Using v = ω r , ✓
GMm
v2
GM
i.e.
=m
= v 2. ✓
2
r
r
r
GM
= ω 2r 2. ✓
r
From which the result ω 2r 3 = GM follows.
c i Since r decreases, from ω 2r 3 = GM the angular speed will increase. ✓
GM
= v 2 , as r decrease v increases. ✓
ii From
r
ω 2r 3
d i Using ω 2r 3 = GM we find M =
✓
G
(5.31 × 10 −5 )2 × (2.38 × 108 )3
And so M =
= 5.70 × 10 26 kg . ✓
−11
6.67 × 10
gives
ii Again using ω 2r 3 = GM we find ω T2rT3 = ω E2rE3 . ✓
3
Hence ω T = ω E
Hence T =
4
 2.38 × 108 
rE3
−5
−6
−1
=
×
×
5.31
10
 1.22 × 109  = 4.58 × 10 rad s ✓
rT3
2π
2π
1.37 × 106
6
=
=
×
=
1.37
10
s
d = 15.856 ≈ 15.9 d ✓
ω T 4.58 × 10 −6
24 × 3600
ANSWERS TO EXAM-STYLE QUESTIONS – TOPIC 6
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Answers to test yourself questions
Topic 6
6.1 Circular motion
1 a The angular speed is just ω =
b The frequency is f =
2π
2π
=
= 5.07 rad s −1. The linear speed is v = ω R = 5.07 × 3.50 = 17.7 m s −1.
1.24
T
1
1
=
= 0.806 s −1.
T 1.24
2 a = 4π 2 rf 2 = 4π 2 × 2.45 × ( 3.5)2 = 1.2 × 10 3 m s −2.
∆v
. The velocity vectors at A and B and the change in the velocity ∆v
3 a The average acceleration is defined as a =
t
∆
are shown below.
2π × 2.0
The magnitude of the velocity vector is 4.0 m s −1 and it takes a time of
= 3.14 s to complete a full
4.0
3.14
revolution. Hence a time of
= 0.785 s to complete a quarter of revolution from A to B. The magnitude of
4
5.66
= 7.2 m s −2. This is
∆v is 4.0 2 + 4.0 2 = 5.66 m s −1 and so the magnitude of the average acceleration is
0.785
directed towards north-west and if this vector is made to start at the midpoint of the arc AB it is then directed
towards the center of the circle.
v 2 16.0
b The centripetal acceleration has magnitude
=
= 8.0 m s −2 directed towards the center of the circle.
r
2.0
 2π r 


4π 2 r
v
= T
= 2 = 4π 2 rf 2 . Hence
4 The centripetal acceleration is a =
r
r
T
2
2
a
=
4π 2 r
50
= 0.356 s −1 ≈ 21 min −1.
4π × 10
v2
4.00
5 a The centripetal acceleration is
=
= 10.0 m s −2. The tension is the force that provides the centripetal
r
0.400
acceleration and so T = ma = 1.00 × 10.0 = 10.0 N.
f =
2
b From T = ma = 20.0 N we have a =
v2
= 20.0 m s −2 and so v = 20 × 0.40 = 2.83 m s −1.
r
4.00 2
16.0
⇒r =
= 0.800 m
c 20.0 = 1.00 ×
r
20.0
6 With a = 9.8 m s −2 we have that a =
4π 2 r
⇒T =
T2
4π 2 × 6.4 × 106
= 5.08 × 10 3 s ≈ 85 min.
9.8
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ANSWERS TO TEST YOURSELF QUESTIONS 6
1
 2π r 


4π 2 r 4π 2 × 50.0 × 10 3
v
= 3.2 × 109 m s −2
= T
= 2 =
7 a a=
( 25.0 × 10 −3 )2
r
r
T
2
2
b The forces on the probe are (i) its weight, mg, and (ii) the normal reaction force N from the surface. Assuming
the probe to stay on the surface the net force would be
mg − N =
mv 2
mv 2
v2 

⇒ N = mg −
= m  g −  = m(8.0 × 1010 − 3.2 × 109 ) > 0.

r
r
r 
This is positive so the probe can stay on the surface.
2π R
2π × 1.5 × 1011
m s −1
= 2.99 × 104 ≈ 30 km
=
T
365 × 24 × 60 × 60
2
( 2.99 × 104 )2
v
=
=
a
= 5.95 × 10 −3 ≈ 6.0 × 10 −3 m s −2
b
11
r
1.5 × 10
8 a v=
c F = ma =
mv 2
= 6.0 × 10 24 × 5.95 × 10 −3 ≈ 3.6 × 10 22 N
r
9 The components of L are:
L x = L sin θ , L y = L cos θ
We have that
v2
L sin θ = m
R
L cos θ = mg
Dividing side by side:
v2
m
L sin θ
= R
L cos θ
mg
tan θ =
v2
gR
This gives ⇒ R =
10 a
180 2
v2
=
= 4.7 km
g tan θ 9.8 × tan 35°
friction
reaction
weight
b Let the normal reaction force from the wall be N. Then
v2
N =m
r
mg = f s
For the minimum rotation speed the frictional force must be a maximum i.e. f s = µs N . I.e.
v2
N =m
r
mg = µs N
2
ANSWERS TO TEST YOURSELF QUESTIONS 6
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16
Combining gives mg = µsm
f =
v2
i.e. v =
r
gr
=
µs
9.8 × 5.0
= 9.04 m s −1. From v = 2π rf we find
0.60
v
9.04
=
= 0.288 rev s −1 ≈ 17 rev min −1.
2π r 2π × 5.0
11 a Let v be the speed on the flat part of the road before the loop is entered. At the top the net force on the
cart is its weight and the normal reaction force from the road, both directed vertically downwards. Then,
mu 2
mu 2
⇒N =
− mg where u is the speed at the top. For the cart not to fall off the road, we must
N + mg =
R
R
1
1
2
have N > 0 i.e. u > gR . From conservation of energy, mv 2 = mg( 2R ) + mu 2 and so u 2 = v 2 − 4 gR .
2
2
Hence v 2 − 4 gR > gR , i.e. v > 5 gR = 29.7 ≈ 30 m s −1.
b For just about equal to 5 gR we get u =
gR = 13.3 ≈ 13 m s −1.
12 The tension in the string must equal the weight of the hanging mass i.e. T = Mg . The tension serves as the
Mgr
v2
v2
.
centripetal force on the smaller mass and so T = m . Hence m = Mg ⇒ v =
m
r
r
13 Let the tension in the upper string be TU and TL in the lower string. Both strings make an angle θ with the
horizontal. We have that:
TU sin θ = mg + TL sin θ
TU cos θ + TL cos θ = m
v2
r
We may rewrite these as:
TU sin θ − TL sin θ = mg
TU cos θ + TL cos θ = m
v2
r
0.50
= 0.50 ⇒ θ = 30°. Further, r = 1.0 2 − 0.50 2 = 0.866 m. Therefore the
From trigonometry, sin θ =
1.0
equations simplify to
TU − TL = 4.90
0.50 × (TU − TL ) = 2.45
or
.
TU + TL = 21.33
0.866 × (TU + TL ) = 18.48
Finally, TU = 13.1 N, TL = 8.22 N.
1 2
mv and so v = 2 gh = 2 × 9.81 × 120 = 48.9 ≈ 49 m s −1 (with this speed,
2
this amusement park should not have a licence to operate!).
14 a By conservation of energy, mgh =
b The forces on a passenger are the weight and the reaction force R both in the vertically down direction. Thus
v2
v2
R + mg = m ⇒ R = m − mg . The speed at the top is found from energy conservation as
r
r
1 2
mgH = mv + mg( 2r ) ⇒ v 2 = 9.81 × 240 − 2 × 9.81 × 60 = 1177 . Hence
2
1177
R = 60 ×
− 60 × 9.81 = 1765 ≈ 1800 N.
30
50 2
= 30 m s −2 (some passengers will be fainting
c Using v 2 = u 2 − 2as we get 0 = 49 2 − 2a × 40 and so a =
2 × 40
now, assuming they are still alive!).
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ANSWERS TO TEST YOURSELF QUESTIONS 6
3
6.2 The law of gravitation
15 a F = G
Mm
5.98 × 10 24 × 7.35 × 10 22
−11
=
.
×
×
= 1.99 × 10 20 N
6
67
10
( 3.84 × 108 )2
R2
b F =G
Mm
1.99 × 10 30 × 1.90 × 10 27
−11
=
.
×
×
= 4.17 × 10 23 N
6
67
10
(7.78 × 1011 )2
R2
c F =G
Mm
1.67 × 10 −27 × 9.11 × 10 −31
−11
=
.
×
×
= 1.0 × 10 −47 N
6
67
10
(1.00 × 10 −10 )2
R2
16 a Zero since it is being pulled equally from all directions.
b Zero, by Newton’s third law.
m(m + M )
Mm
m2
m2
=
+G
=G
F
G
c F =G
,
(d)
2
2
2
4R
4R
4R 2
4R
 GM 
 (9R )2 
g
1
17 A =
=
gB
81
 GM 
 R2 
 G 2M 
 ( 2R )2  1
g
18 A =
=
gB
2
 GM 
 R2 
19 Since star A is 27 times as massive and the density is the same the volume of A must be 27 times as large. Its radius
 G 27M 
g A  ( 3R )2 
must therefore be 3 times as large. Hence
=
= 3.
gB
 GM 
 R2 
 GM / 2 
 ( R / 2)2 
g
20 new =
=2
g old
 GM 
 R2 
21 Let this point be a distance x from the center of the Earth and let d be the center to center distance between the
earth and the moon. Then
G 81M
GM
=
2
x
(d − x )2
81(d − x )2 = x 2
9(d − x ) = x
x
9
=
= 0.9
d 10
22 a At point P the gravitational field strength is obviously zero.
b The gravitational field strength at Q from each of the masses is
g=
GM
3.0 × 10 22
−11
= 1.0 × 106 N kg −1. The net field, taking components, is directed from Q
=
.
×
×
6
67
10
2
9 2
( 2 × 10 )
R
to P and has magnitude 2 g cos 45° = 2 × 1 × 106 cos 45° = 1.4 × 106 N kg −1.
4
ANSWERS TO TEST YOURSELF QUESTIONS 6
PHYSICS FOR THE IB DIPLOMA © CAMBRIDGE UNIVERSITY PRESS 2015
18
GM
2π r
v2
GMm
4π 2 r 3
m
=
⇒ v2 =
23 We know that 2
and so we deduce that T 2 =
. Therefore
. But v =
r
r
T
r
GM
r=
3
GMT 2
=
4π 2
24 a From v 2 =
3
6.67 × 10 −11 × 6.0 × 10 24 × ( 24 × 60 × 60)2
= 4.2 × 107 m.
4π 2
GM
we calculate v =
r
6.67 × 10 −11 × 6.0 × 10 24
= 7.5828754 × 10 3 ≈ 7.6 × 10 3 m s −1.
(6.4 + 0.560) × 106
6.67 × 10 −11 × 6.0 × 10 24
b The shuttle speed is v =
= 7.5831478 × 10 3 m s −1. The relative speed of the shuttle
6
6.9595 × 10
104
−1
= 36711 s ≈ 10 hrs.
and Hubble is 0.2724 m s and so the distance of 10 km will be covered in
0.2724
25 a
Gm
2π r
v2
Gm1m2
m
=
⇒ v 2 = n − 11 . But v =
and so
2
n
r
r
T
r
4π 2 r 2 Gm1
= n −1
T2
r
2 n +1
4π r
T2 =
Gm1
2
 2π r  = Gm1 giving


T
r n −1
b For this to be consistent with Kepler’s third law we need n + 1 = 3 ⇒ n = 2
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ANSWERS TO TEST YOURSELF QUESTIONS 6
5