H2 Physics Circular Motion
Circular Motion
Multiple Choice Questions
1 A pilot executes a vertical dive, then follows a circular arc until it is going straight up. The
path taken is shown in the figure below.
Just as the plane is at its lowest point, the force on him is
A less than mg, and pointing up
B less than mg, and pointing down
C more than mg, and pointing up
D more than mg, and pointing down
2 A spacecraft is moving in a circular orbit around the Earth when its engine is fired and
produces a force on the spacecraft exactly equal and opposite to that exerted by the
Earth’s gravitational field.
How would the spacecraft move?
A Along a spiral path towards the Earth’s surface.
B Along the line joining it to the centre of the Earth (i.e. radially)
C Along a tangent to the orbit.
D In a circular orbit with a longer period.
3 In a ride at an entertainment park, a person sits in a cage which moves in a vertical circle
at a constant speed.
At the instant shown, what is the direction of the force exerted by the cage on the person?
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H2 Physics Circular Motion
D
B
4 A turntable consists of a rotating horizontal disc moving at a fixed rotational speed. Two
small objects P and Q, are placed on the turntable as shown, where P is at a distance r 1
from the centre O and Q is at a distance r 2 from the centre. The mass of P is twice that of
Q.
Q r O r
2 P
1
Turntable rotating on a horizontal plane
Which of the following correctly relates P and Q’s linear speeds, angular speeds and their
centripetal forces?
Linear speed Angular speed Centripetal force
A same for P and Q same for P and Q same for P and Q
B greater for P greater for P greater for P
C greater for P same for P and Q greater for P
D greater for Q same for P and Q same for P and Q
5 In an amusement park ride, a brave person sat in a roller coaster carriage without
strapping himself down with a seatbelt. At three different portions of the ride, P, Q and R,
he maintained contact with his seat while moving with speed v as shown below. The
magnitude of the normal contact force at P,Q and R is the same but the speeds v are
different.
P v
Q
v
vv
R
Point P inner Point Q
Moving on the Moving on the inner
track of radius R. At track
the highest point
of radius R. on
At the highest point on
the inner track of the outer track of
radius r radius r
Point R
At the extreme right
on the inner track of
radius r
2
H2 Physics Circular Motion
Which of the following statements is true with regards to the speed?
A The speed is the largest at point R.
B The speed is the same at P, Q and R.
C The speed is the largest at point P and the least at point Q.
D The speed is the largest at point Q and the least at point P.
6 A point Q on the circumference of a turntable, rotating at 33.3 revolutions per minute,
makes an arc RS (see Fig. 12) in time t. Find the value of t.
R
Q S
12.4º
P
P
Fig. 12
A 0.00103 s B 0.0621 s C 2.07 s D 3.72 s
7 A space laboratory is rotating to create artificial gravity as shown in the figure below. Its
period of rotation is chosen so that the outer ring (ro = 2150 m) simulates the acceleration
due to gravity on Earth (9.81 ms-2). What should be the approximate radius r1 of the inner
ring, so that it simulates the acceleration due to the gravity on the surface of Mars
(3.72ms-2)?
r1 ro
A 700 m B 800 m C 900 m D 1000 m
8 A weight is suspended from the middle of a rope whose ends are at the same level. In
order for the rope to be perfectly horizontal, the forces applied to the ends of the rope
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H2 Physics Circular Motion
A must be equal to the weight.
B must be greater than the weight.
C must be equal to half of the weight.
D must be infinite.
9 An object is traveling in a circle of radius r with angular velocity and speed v. Which
expression gives the centripetal acceleration?
A r B v C v/r D v / r2
10 A simple pendulum consists of a bob of mass m at the end of a light and inextensible
thread of length L. The other end of the thread is fixed at C. The bob swings through
point B with velocity v and just reaches A, where the string is just taut.
A C
v B
What is the tension in the thread as the bob passes point B?
A mg B 2 mg C 3 mg D 4 mg
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H2 Physics Circular Motion
11 An engineer is working in Shanghai which has a latitude of 31.28 o N. Assuming the Earth
to be a sphere of radius 6380 km, find the engineer’s linear velocity due to the rotation of
the Earth about its axis.
Earth’s axis
equator
Shanghai
31.28o
A 241 ms-1 B 337 ms-1 C 397 ms-1 D 464 ms-1
12 A radar tower 0.50 km tall is built at the equator on a hill of height 2.5 km. As a result of
the Earth’s rotation, what is the difference in the speed between a man at the top of the
tower and someone at sea level?
A 0 m s-1
B 0.22 m s-1
C 5.2 m s-1
D 19 m s-1
13
H
A B
B
The diagram shows part of the route of a roller-coaster in an amusement park. The cart
descends from H, completes a circular loop A and moves to B. If the cart is to complete
the central circular track safely, what is the minimum speed of the cart at the bottom of
the circular track A? Assume that there is no friction between the cart and the track.
A 9.81 m s-1 B 19.6 m s-1 C 22.1 m s-1 D 24.5 m s-1
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H2 Physics Circular Motion
14 In a particular ride in an amusement park, a passenger of mass 55 kg travels upside
down in a carriage at the top of the circle, of diameter
14 m, at a speed of 14 m s-1 as shown in Fig. 7.1 below. What is the force exerted
by the carriage on the passenger?
Fig.7.1
A 1000 N B 1450 N C 2000 N D 2750 N
15 Two blocks of 10.0 g and 21.0 g are tied together and performing a uniform horizontal
circular motion on a smooth table, at an angular speed of 6.28 rad s -1, as shown in Fig.
11.
10.0 g
T
21.0 g 2
T1 5.0 cm
15.0 cm Fig. 11
Centre of
circle
What is the difference in magnitude of the Tension T 1, the tension in the string connecting
the 21.0 g mass to the centre and T 2, the tension in the string connecting the 10.0 g mass
to the 21.0 g mass?
A 0N
B 0.104 N
C 0.124 N
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H2 Physics Circular Motion
D 0.166 N
16 Singapore Flyer is the world’s largest observation wheel. Standing at a stunning
165 m from the ground, the Flyer offers breathtaking, panoramic views of the Marina Bay,
our island city and beyond. This memorable ride takes 30 minutes to complete a cycle.
Figure 10
What is the angular velocity of the Flyer?
A 3.5 x 10-3 rad s-1
B 0.21 rad s-1
C 0.58 rad s-1
D 35 rad s-1
17 A car of mass m moving at a constant speed v passes over a humpback bridge of radius
of curvature r. Given that the car remains in contact with the road, what is the magnitude
of the net force R exerted by the car on the road when it is at the top of the bridge?
A R = mg + Error! C R = mg Error!
B R = Error! D R = Error! mg
18 A car is moving in a circular path. Which diagram shows the resultant acceleration acting
on the car at the instant when it is slowing down?
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H2 Physics Circular Motion
19 A car of mass m is moving with constant speed v on a banked circular track of angle θ
and the car is at a distance d from the centre of the track. The normal reaction force
between the car and the ground is R. Assume the car does not slide sideways.
What is the relationship between m and R ?
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H2 Physics Circular Motion
20 A disc is rotating about an axis through its centre and perpendicular to its
plane. Points P and Q are on the disc at distances of x and 2x respectively
angular velocity of P
from the centre of the circle. What is the ratio angular velocity of Q
?
A 4
B 2
C 1
D ½
21 Yu Han noticed that her clock is not accurate. She observed that the minute hand of the
clock travels at an angular velocity of 1.744 × 10−3 rad s−1. Will her clock be behind or
ahead of time and by how much if the minute hand travelled a total of 48π radians? (Take
π = 3.142)
A 0.134 seconds ahead of time B 1.28 seconds behind time
C 8.04 seconds ahead of time D 77.0 seconds behind time
22 One end of an inextensible light string is attached to the handle of a pail of water which
has a total weight of 2.5 N. The other end is held by a teacher. The teacher swings the
pail of water such that its trajectory is a circle in a vertical plane with a fixed centre (i.e.
the hand that holds the string remains at the same position as it swings the pail of water
in the vertical circle) and the pail of water undergoes uniform circular motion.
If the maximum tension of the string is 10.0 N, the minimum tension of the string is
A 10.0 N
B 7.5 N
C 5.0 N
D 0.0 N
Structured Questions
1 Domestic washing machines often incorporate washing, rinsing, spinning and drying
of clothes. This question is about the spin-dry function of a washing machine.
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H2 Physics Circular Motion
(a) The inner drum of the machine into which the clothes are placed has quite
large holes in it.
Explain how, when the clothes are being spin-dried, the water gets out from
the clothes and through the holes.
[3]
(b) One of the spin speeds in one model of washing machine was listed as
1000 rpm (revolutions per minute).
Calculate the largest resultant force that could be exerted on a wet jacket of
mass 0.500 kg given that the radius of the spinning drum is 12.5 cm.
largest resultant force =................................... N [3]
(c) If clothes are unevenly distributed in the machine, it vibrates slightly as it
rotates. The outer drum within which the spinning drum rotates is attached
to the rest of the framework of the washing machine by springs.
Suggest what the purpose of these springs is.
[2]
(MI 08 P2 Q2)
2 In the 2008 Tour de France, Frenchman Sylvain Chavanel made a left turn on a rough
level road surface at a constant speed v, as viewed from behind (see Fig. 1.1). The total
mass of the bicycle and rider is m and their combined centre of gravity is at G.
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H2 Physics Circular Motion
Fig. 1.1
(a) On Fig. 1.1 above, draw and label all the forces acting on the system of the
rider and his bicycle. Ignore forces parallel to the direction of motion.
[3]
(b) If the rider is negotiating a turn with a radius of curvature of 60 m, the total mass of
the rider and bicycle is 85 kg, and the friction provided by the road surface is 70 N,
calculate the velocity with which he is turning.
velocity = ……………….. m s-1 [ 2 ]
(c) The rider now makes the same left turn on a rough surface banked at 20 o to the
horizontal as shown in Fig. 1.2 below.
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H2 Physics Circular Motion
20o
Fig. 1.2
Assuming that the frictional forces remain as 70 N, and radius of curvature is still
60 m,
(i) Explain how the banked surface assists the rider in travelling around
the corner at a higher speed.
[2]
(ii) Calculate the new maximum velocity with which the rider may
negotiate the turn.
velocity = ………………….. m s-1 [ 2 ]
(SAJC 08 P2 Q1)
3 The maximum friction that can be provided by the Formula One race track is 5.80 kN. The
maximum speed of a 605 kg Formula One car (including driver) is 360 km h -1. An
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H2 Physics Circular Motion
engineer suggested that banking of the road is necessary for safety purposes at a curved
section of the road with radius 1000 m.
(a) Explain the meaning of ‘banking’ in the above passage and state the purpose of
banking.
(b) Show by calculation why the engineer thinks it is necessary to bank the road.
(c) Show by calculation, whether a banking of 20° is sufficient for the car to turn safely.
(SRJC 08 P2 Q3)
4 (a) Explain why a body traveling with uniform speed in a circle has acceleration.
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H2 Physics Circular Motion
[2]
(b) A toy consists of a small wedge that has an acute angle and a rod as shown in
Fig. 2.1. The wedge is spun by rotating the vertical rod that is firmly attached to the
bottom of the wedge such that mass m moves in a horizontal circle. The angular
velocity of the wedge is such that frictional force can be neglected.
L m
wedge
rod
Fig. 2.1
(i) Draw a free body diagram showing the forces acting on mass m.
[2]
(ii) Write an equation, in terms of the forces indicated in your diagram, for the
sum of forces in the vertical direction.
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H2 Physics Circular Motion
[1]
(iii) Show that, when the mass is at a constant distance L up along the slope,
the speed of the mass must be
v= ,g L sin
[3]
(TPJC 08 P2 Q2)
5 A boy whirls a stone of mass 54 g at constant speed in a horizontal circle 1.8 m above the
ground by means of a string 1.2 m long. The string itself is inclined at 3 0 to the horizontal
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H2 Physics Circular Motion
(a) Draw a free body diagram showing the forces acting on the stone as it is whirled in
a horizontal circle. State the nature of each force drawn.
Write down an equation that shows the relationship between the forces drawn. [4]
(b) After some time the string breaks, and the stone flies off horizontally, striking the
ground 9.1 m away from the breakaway point.
(i) Show that the time interval from the string breaking to the stone striking
the ground is 0.61 s. [1]
(ii) Calculate the centripetal acceleration on the stone during its circular
motion. [2]
(iii) Prove that the vertical component of the velocity of the stone when it
strikes the ground is 5.9 m s-1. [1]
vy2
(c) Find the vertical force exerted by the ground on the stone if, upon landing on the
ground, the stone comes a total stop in 0.10 s.
[2] d
50.0 cm
2.5 kg
(NYJC 08 P3 Q1)ω
6 A 2.5 kg pendulum bob is attached to a 15.0 cm inextensible string and fixed
to point B of a ‘L’ shape structure. The structure is mounted on a rotating disc A. The
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A
H2 Physics Circular Motion
spinning causes the pendulum bob to swing at an angle of θ to the vertical.
C
13.6 cm s-1
Fig. 7.1 Fig. 7.2
(a) Assuming that the structure rotates with a constant angular velocity such that point
C on the circumference of the disk has speed of 13.6 cms -1, find the rate of
rotation of the disc A in terms of rev s-1 if the diameter of the disc is 6.00 cm.
rate of rotation = ______________ rev s-1 [2]
(b) (i) As the structure rotates, the pendulum bob swings and maintains a
constant angle, θ with the vertical, draw a free body diagram for the
pendulum bob.
[2]
(ii) Given that θ = 30o, determine the tension in the string and hence calculate
the length d.
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H2 Physics Circular Motion
tension = __________ N
d = ______________m [3]
(iii) Explain qualitatively whether angle θ will increase or decrease if length d
increases.
[3]
(iv) Explain how the angle θ will vary if the mass of the pendulum bob doubles.
[1]
(PJC 08 P3 Q7 part)
7 A bob is tied to one end of an inextensible string of negligible mass and the other end of
the string is fixed to a point at the top of a vertical rod. The motor rotates the vertical rod
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H2 Physics Circular Motion
at a constant angular velocity. As a result, the bob moves in a horizontal circle as shown
in Fig. 1.4.
(i) Derive an expression for the angle θ in terms of ω, l and g , where ω is the
angular velocity of the vertical rod, l is the length of the string and g is the
acceleration of free fall.
[4]
(ii) Explain what happens to the angle θ when the length of the string is shortened but
the angular velocity of the vertical rod is maintained.
[1]
(HCI 08 P3 Q1 part)
Answers (MCQs)
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H2 Physics Circular Motion
1. No ans
2. C
3. A
4. C
5. C
6. B
7. B
8. D
9. B
10. C
11. C
12. B
13. C
14. A
15. C
16. A
17. C
18. B
19. C
20. C
21. NA
22. C
Answers (Structured)
1 (a) The rotating drum exerts a normal reaction force on the clothes and water which
acts towards the centre of the rotation. B1
No such force is possible at the holes and so the water over each hole has nothing to keep it
moving in a circle and thus flies off at a tangent to the drum’s motion. B1 While the water can
pass through the holes in the drum but the clothes are too large to do this so water is removed
from the clothes during the spin-dry cycle. B1
(b) = 1000 rev per min = 1000 x 2 / 60 = 105 rad s1 C1
F = m r 2 = 0.500 x 0.125 x 104.722 = 685 N C1, A1 (allow ecf on )
(c) When the spinning drum vibrates, it will rock the outer drum. If there were no
springs and the outer drum was to be attached to the rest of the machine, the shock of the
vibrations would not be damped B1 and would damage the machine.B1
2 (a)
Normal
Reaction
Friction
[ 1 ] for each labeled force
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H2 Physics Circular Motion
Weight of rider
(b) Frictionand
= Centripetal
bicycle force required for turning
mv 2
f [1]
r
70
85 v 2
60
v = 7.03 m s-1 [1]
(c) (i) For a surface which is banked, the horizontal component of the normal
reaction is an additional source of the centripetal force.
[1]
mv 2
Since Centripetal Force = , an increase in the centripetal force would
r
allow the rider to turn the corner at a
higher speed without slipping. [1]
(ii) Considering forces acting on the rider/bicycle in the horizontal direction:
mv 2
N sin 20 f cos 20 ---------------------- (1)
r
For the equilibrium in the vertical direction: [1]
Ncos20 mg f sin 20 --------------------(2)
Solving (1) & (2),
mv 2
tan 20mg f sin 20 f cos 20
r
max velocity during turning v = 16.3 m s-1 [1]
3 (a) Banking is the tilting of the road surface, such that the road slopes upwards away
from the centre of the curvature and that it will allow cars to turn at a higher speed
without sliding or overturning.
(b)
mv 2 605(1002 )
= = 6.05 kN > 5.8 kN
r 1000
The maximum friction from the road is less than the required centripetal force, hence,
it is necessary to bank the road.
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H2 Physics Circular Motion
(c)
For vertical forces, resultant vertical force is zero, hence,
Ncosθ = mg [1]
Therefore, for horizontal forces,
mg 605(9.81)
Nsinθ + Fcosθ = cosθ
sinθ + Fcosθ = cos20°
sin20° + 5800cos20° = 7.6 kN
The horizontal forces is more than 6.05 kN and hence, it would be sufficient for the car
to turn safely.
4 (a) Since acceleration is defined as the rate of change of velocity , a change in
direction of velocity constitutes an acceleration just as does a change in
magnitude of velocity. Thus, an object revolving in a circle is continuously
accelerating, even when the speed remains constant.
(b) (i) [2]
Normal reaction, N
Weight,
mg
(ii) N cos = mg
(iii) ‘L cos ’ give 1m for radius if stated explicitly
N cos = mg ---(1)
N sin = mv2 /L cos ---(2)
(2) / (1):
tan = mv2 / mg L cos
sin = v2 /gL
v= ,g L sin (Shown)
5 (a)
T T is the pull of the string on the stone
30
W is the gravitational attraction of the earth on the stone
W 22
T sin = W
T sin 30 = W
H2 Physics Circular Motion
(b) (i) [1]
y = uy t + ½ ayt2
1.8 = 0 + ½ (9.81) t2
t = 0.61 s
(ii) [2]
Working backwards fr projectile Using force equations
x = ux t T sin 30 = 0.054 (9.81)
OR
9.1 = ux (0.61) T cos 30 = 0.054 a
ux = 15 m s-1`
r = 1.2 cos 30 = 1.2 m a = 190 m s-22`
(iii) [1]
a= = = 190 m s-2`
vy2 vy2 = uy2 + 2 ay y using v = u + at
OR
vy = 0 + 2 (9.81) (1.8) = 0 + 9.81 (0.61)
vy = 5.9 m s-1` = 5.9 m s-1
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H2 Physics Circular Motion
(e) Find the vertical force exerted by the ground on the stone if, upon landing on the
ground, the stone comes a total stop in 0.10 s. [2]
m(v y - u y ) mv dp
Fground on stone = accept F = or F = or similar
t t dt
0.054 (0 - 5.9)
=
0.10
= -3.2 N
6 (a) (i) ωA = v/r = 0.136 / 0.03 = 4.53 rads-1
4.53
rate of rotation = =0.721 revs-1
2
(b)(i)
T
θ
mg
(b) (ii) Free body diagram for the pendulum bob,
T cos θ = mg ------------------------------(1)
T sin θ = m ω2(d+0.15 sin θ) -----------(2)
mg
When θ = 30o , T = = 28.32 N
cos(30 o )
Eqn (2)/(1)
o
2 [d 0.15(sin 30 o )]
tan (30 ) =
g
(using ω = 4.53 rads-1)
d = 0.20 m
(b) (iii) From equation (2),
T sin θ = m ω2(d+0.15 sin θ)
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H2 Physics Circular Motion
As d increases (without changing ω, the pendulum bob will effectively be
further away from the centre of rotation (having a greater radius of
rotation) and hence require a greater centripetal force. (Fc =mR1ω2).
Hence it requires a greater horizontal component of tension force.
( T sin Fc )
The product of T and sin θ increases. Hence, angle θ increases, so as to
give a greater value of sin θ (the function of sin θ is an increasing function
with θ).
(b) (iv) The angle θ is independent of the mass of the pendulum bob.
this is apparent from the following equation
2 [d 0.15(sin )]
tan (θ) =
g
7 (i)
(ii) the angle θ will increase when length l of the string is shortened.
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