Circular motion
Structured Questions (<8 marks)
(Level 1)
*Code: 12L1B001, Total marks: 6
A wheel has a radius of 0.3 m and is rotating at 600 revolutions per minute.
(a) What is the period of the motion of the wheel in seconds?
(2 marks)
(b) What is the angular speed of the wheel?
(2 marks)
(c)
What is the linear speed of a point on the edge of the wheel?
(2 marks)
Answer:
60
= 0.1 s.
600
2π 2π

(b) The angular speed is
= 62.83 ≈ 62.8 rad s−1.
T
0.1
(c) The linear speed is rω = 0.3 × 62.83 ≈ 18.8 m s−1.
(a) The period is
(1M+1A)
(1M+1A)
(1M+1A)
*Code: 12L1B002, Total marks: 4
A potter’s wheel rotates steadily at 180 revolutions per minute (rpm).
(a) What is the frequency of rotation (in Hz)?
(2 marks)
(b) What is the centripetal acceleration of a point 10 cm from the rotation axis?
(2 marks)
Answer:
(a) The frequency is
180
= 3 Hz.
60
(b) The centripetal acceleration is r 2 
(1M+1A)
10
 2π  3  1.88 m s  2 .
100
(1M+1A)
*Code: 12L1B003, Total marks: 4
An aeroplane travelling at a constant speed of 100 m s−1 flies in a horizontal circle of
a radius of 3000 m.
(a) How long does it take the aeroplane to complete one revolution?
(2 marks)
(b) If the total mass of the aeroplane is 12 000 kg, what is the centripetal force
needed?
(2 marks)
Answer:
2 π  3000
≈ 188 s.
100
mv 2
(b) Applying F 
, the centripetal force is
r
12 000  100 2
= 40 000 N.
3000
(a) The time is
(1M+1A)
(1M+1A)
*Code: 12L1B004, Total marks: 3
A powerboat of total mass 3800 kg makes a sharp turn with a uniform speed of
20 m s−1. The path describes an arc of a circle of a radius of 24 m.
(a) What is the centripetal force provided to the powerboat during its turn?
(2 marks)
(b) With the force in (a) unchanged, would the boat trace out an arc of a larger or
smaller radius if the speed becomes higher?
(1 mark)
Answer:
mv 2 3800  20 2

 6.33  10 4 N .
r
24
(b) The powerboat will trace out an arc of a larger radius.
(a) The centripetal force is F 
(1M+1A)
(1A)
*Code: 12L1B005, Total marks: 6
A puck of mass 0.5 kg, attaching to one end of a string, is moving with a uniform
speed in a horizontal circle of a radius of 1.5 m on a smooth table. The string breaks if
the tension T exceeds 100 N.
(a) What is the maximum value of v before the string breaks? What is the
corresponding angular speed ω?
(b)
(4 marks)
The string suddenly breaks. Describe the subsequent motion of the puck.
(2 marks)
Answer:
mv 2
(a)
Applying F 
, the maximum value, vmax is
r
2
0.5  v max
100 
(1M)
1.5
v max  17.32
The maximum value of v is 17.3 m s−1.
(1A)
v 17.32
The angular speed ω is 
≈ 11.5 rad s−1.
(1M+1A)
r
1.5
(b) The puck moves uniformly (1A) along the direction just before the string breaks
(tangential to the circular path at the point of breaking) (1A).
*Code: 12L1B006, Total marks: 4
A roller coaster car of total mass 320 kg travels around a horizontal circular track of a
radius of 32 m with a uniform speed of 16 m s−1.
(a) What is the magnitude and direction of the resultant force acting on the car by
the track?
(4 marks)
(b) Hence, or otherwise, determine the angle the track should be banked such that no
frictional force is needed by the car to travel around the track.
(1 mark)
Answer:
(a) The horizontal component of the force acting on the car by the track provides the
mv 2 320  16 2
centripetal force,
(1M)

 2560 N .
r
32
The vertical component of the force acting on the car by the track balances the
weight of the car, mg = 320 × 10 = 3200 N.
(1M)
Therefore the magnitude of the resultant force is
2560 2  3200 2 ≈ 4.10 × 103 N.
 3200 
The force points to tan 1 
 ≈ 51.3° above the horizontal direction.
 2560 
(b) The angle is 51.3° above the horizontal direction.
(1A)
(1A)
(1A)
*Code: 12L1B007, Total marks: 6
A car moving at a uniform speed of 10 m s−1 turns around a corner, which is an arc of
a circle of a radius of 25 m. The maximum frictional force developed between the car
tyre and the road surface is 0.7 times the weight of the car.
(a) Will the car skid as it turns around the corner?
(3 marks)
(b)
What is the maximum speed vmax of the car before it skids?
(1 mark)
Answer:
(a) Let m kg be the mass of the car.
mv 2 m  10 2

= 4m.
r
25
The maximum frictional force is 0.7mg = 0.7 × m × 10 = 7m.
Therefore the car will not skid.
The centripetal force needed is
(1M)
(1M)
(1A)
(b) The speed vmax can be determined by
mv 2
 0.7mg
r
2
v max
 0.7  10
25
vmax  13.2 m s 1
(1A)
*Code: 12L1B008, Total marks: 5
An aeroplane of mass 12 000 kg is flying at a constant speed of 90 m s−1 in a
horizontal circle of a radius of 2400 m. The lift force U is perpendicular to the wings.
(a) Find the centripetal force needed by the aeroplane.
(2 marks)
(b) At what angle to the vertical, θ must the wings be banked?
(3 marks)
Answer:
(a) Applying F 
mv 2
, the centripetal force needed is
r
12 000  90 2
= 40 500 N.
2400
(b) The horizontal component of U provides the centripetal force.
The vertical component of U balances the weight.
Consider the horizontal and vertical components of U.
40 500
tan  
12 000  10
  18.6
(1M+1A)
(1M)
(1M)
(1A)
Structured Questions (<8 marks)
(Level 2)
*Code: 12L2B001, Total marks: 5
A stone of mass m = 50 g is whirled in a horizontal circle at a steady speed as shown.
The length of the string is l = 0.8 m and makes an angle θ with the vertical.
stone
(a) Suppose θ = 60°. Find the angular speed of the stone.
(3 marks)
(b) If the experiment is carried out on the surface of the Moon where the
acceleration due to gravity is smaller, should the stone be whirled faster or
slower to keep the same angle θ? Briefly explain your answer.
(2 marks)
Answer:
(a) Let T be the tension of the string.
The vertical component of T is equal to the weight of the stone, i.e.
T cos   mg
50
 10
1000
T 1N
The horizontal component of T provides the centripetal force, i.e.
T sin   mr 2
T cos 60 
(1M)
50
(1M+1A)
 0.8  sin 60   2
1000
  5 rad s 1
(b) From (a), we have
T sin   mL sin   2
mg
sin   mL sin   2
(1A)
cos 
g

L cos 
Therefore, where g becomes smaller, the stone have to be whirled slower to keep
the same angle θ.
(1A)
1  sin 60 
*Code: 12L2B002, Total marks: 7
In an experiment, a rubber bung is connected to a weight via a glass tube by an
inextensible string. The rubber bung is then whirled around horizontally at a steady
angular speed ω. The mass of the rubber bung and the weight are M and m
respectively. The length of the string extended from the glass tube to the rubber bung
is L. The angle between the string and the vertical is θ.
glass tube
rubber bung
weight
(a)
Derive an expression to describe the relation between M, m, ω and L.
(3 marks)
(b) The bung completes 20 revolutions in 12 s.
(i) Find ω.
(2 marks)
(ii) Suppose L = 1.2 m. Find θ.
(2 marks)
Answer:
(a) The tension in the string T is equal to mg.
The horizontal component of the tension, i.e. mg sin θ provides the centripetal
force.
(1M)
2
The centripetal force provided is Mω L sin θ.
(1M)
Therefore,
mg sin   M 2 L sin 
(1A)
mg  M 2 L
2 π  20
(b) (i)  
= 10.47 ≈ 10.5 rad s−1
(1M+1A)
12
(ii) Consider the horizontal and vertical components of the tension in the string.
M 2 L sin 
tan  
Mg
1
10.472  1.2

cos 
10
  85.6
(1M+1A)
*Code: 12L2B003, Total marks: 5
A car of mass 1500 kg moving at a uniform speed of 10 m s−1 turns around a corner,
which is an arc of a circle of a radius of 50 m.
(a) Suppose the road is level such that the centripetal force is solely provided by the
frictional force f acting on the car by the road surface. Find f.
(2 marks)
(b) Suppose the road is banked such that the centripetal force is solely provided by
the normal reaction R of the car. Find the angle θ at which the road is banked.
(3 marks)
Answer:
mv2
1500  102
, f 
(1M+1A)
 3000 N .
50
r
(b) The normal reaction R is equal to component of the weight perpendicular to the
road surface, i.e.
mg  R sin 90   
(a) Applying F 
1500  10  R cos 
15 000
R
(1)
(1M)
cos
In this case, the horizontal component of the normal reaction is equal to 3000 N,
i.e.
3000  R cos90   
(2)
(1M)
R sin   3000
Substitute (1) into (2),
15 000  sin 
 3000
cos 
(1A)
tan   0.2
  11.3
Structured Questions (≥ 8 marks)
(Level 1)
*Code: 12L1C001, Total marks: 8
A toy aeroplane of mass 0.5 kg flies in a horizontal circle. The lifting force U = 10 N
is perpendicular to the wings. The aeroplane completes one revolution in 6 s.
(a) Find the angular speed of the aeroplane.
(2 marks)
(b) Find θ, the angle between the vertical and the lifting force.
(2 marks)
(c) Find the radius of the circle r.
(2 marks)
(d) The lifting force U remains unchanged. How does θ change if
(i) the mass of the aeroplane becomes smaller.
(ii) the aeroplane travels at a higher speed but the radius of the circle remains
unchanged.
(2 marks)
Answer:
2 π 2

 1.047  1.05 rad s 1 .
(1M+1A)
T
6
(b) The vertical component of the lifting force balances the weight of the aeroplane,
i.e.
U cos  mg
(a) The angular speed is
(1M+1A)
10  cos  0.5  10
  60
(c) The horizontal component of the lifting force provides the centripetal force, i.e.
U sin   mr 2
10  sin 60  0.5  r  1.047 
r  15.8 m
(d) (i) increases
(ii) increases
2
(1M+1A)
(1A)
(1A)
Structured Questions (≥ 8 marks)
(Level 2)
*Code: 12L2C001, Total marks: 9
A ball is whirled in a horizontal circle of a radius of 0.4 m on a plane 1.2 m above the
ground. It is suddenly released and it lands on a position 2 m away as shown.
(a) What is the centripetal acceleration of the ball just before it is released?
(4 marks)
Find the angle between the string and vertical just before the ball is released.
(3 marks)
(c) How do (1) the time of flight in the air and (2) the horizontal distance travelled
by the ball change in the following cases.
(i) using a heavier ball without changing the angular speed
(ii) increasing the whirling speed without changing the initial height of the ball
from the ground
(2 marks)
(b)
Answer:
(a) Let t be the time of flight of the ball and u be the horizontal speed of the ball.
Consider the horizontal motion of the ball.
(1M)
ut  2
Consider the vertical motion of the ball.
1
1.2  0   10  t 2
2
t  0.4899 s
Thus u = 4.082 m s−1.
(1M)
2
v
Applying a 
, the centripetal acceleration is
r
4.082 2
(1M+1A)
 41.67  41.7 m s 2
0.4
(b) Let m be the mass of the ball.
The vertical component of the tension in the string balances the weight of the
ball, i.e. mg = 10m.
(1M)
The horizontal component of the tension provides the centripetal force, i.e.
41.67m.
(1M)
 41.67m 
The angle is tan 1 
(1A)
  76.5 .
 10m 
(c) (i) (1) unchanged
(2) unchanged
(1A)
(ii) (1) unchanged
(2) increases
(1A)
*Code: 12L2C002, Total marks: 9
There is an amusement ride called the Rotor. It consists of a large barrel rotating at
about 33 revolutions per minute (rpm). When the barrel attains its full speed, the
riders have already been pinned against the wall. Consider the barrel below. A boy of
mass 45 kg has already been pinned against the wall.
(a)
Sketch a free-body diagram of the boy.
−2
(b)
The centripetal acceleration is about 1.5g, i.e. 15 m s . Find r.
(c)
What is the linear speed of the boy?
(2 marks)
(2 marks)
(2 marks)
(d) While the boy is still pinned against the wall, the barrel rotates uniformly at a
lower speed. The resultant force acting on the boy now becomes 450 N. Find the
linear speed of the boy again.
(3 marks)
Answer:
(a)
(1A for friction, normal reaction and weight, 1A for correct directions)
f = friction
N = normal reaction from the wall
W = weight of the boy
2 π  33
= 1.1π.
60
Applying a = rω2, we have
15  r  1.1
(b) The angular speed is
r  4.340 589
 4.34 m
(c) The linear speed is v = rω = 4.340 589 × 1.1π = 15 m s−1.
(d) Applying F = ma, we have
450  50  a
a  10 m s 2
Applying a 
(1M+1A)
(1M+1A)
(1M)
v2
, we have
r
v2
10 
4.340 589
v  6.59 m s 1
(1M+1A)