Tutorial 7 : Motion in a circle
1. A satellite moves at constant speed in a circular orbit around the Earth. State and explain
whether the following physical quantities of the satellite is constant or varying.
(a) angular velocity
(b) linear velocity
(c) linear momentum
(d) acceleration
(e) force
(f) kinetic energy
(a) angular velocity
(constant)
Since the satellite is moving at a constant speed, thus for a revolution around Earth, the
time taken should be the same, i.e. angular velocity should be constant (same magnitude
and direction)
(b) linear velocity
(varying)
Since the direction of velocity is always changing, the linear velocity is varying.
(c) linear momentum
(varying)
Since linear momentum is the product of mass and linear velocity, the linear momentum
is varying (with constant mass and varying linear velocity)
(d) acceleration
(varying)
Since the centripetal acceleration is always pointing towards the centre of circle, the
direction of (centripetal) acceleration will be varying at different points of the circular path.
Thus acceleration is varying.
(e) force
(varying)
Since centripetal acceleration is varying at different points of the circular path, the
centripetal force is varying too.
(f) kinetic energy
(constant)
Since kinetic energy only depends on the mass and the speed, the kinetic energy of a
constant mass satellite with constant speed will be constant.
2. The Singapore Flyer is 150 metres in diameter and makes one complete revolution in
30 min. There are 28 capsules traveling with constant angular velocity.
(a) Find the angle (in degrees and radian) subtended at the centre by two neighboring
capsules.
(b) Find the angular velocity for the wheel.
(c) Find the linear velocity for the capsule.
(d) Find the centripetal acceleration on one capsule.
(a)  = 360o / 28 = 12.9° = 0.224 rad
(b)  
2
2

 3.5  103 rad s-1
T
30  60
(c) v  r   75  3.49  103  0.262 m s-1
(d) ac 
v 2 0.2622

 9.14  104 m s-2
r
75
3. In the movie “2001: A Space Odyssey”, a rotating spacecraft pro-vides artificial gravity.
The people would be walking inside the circle; their feet toward the exterior and their
head toward the center; the floor and ceiling would curve upwards.
The radius of the outer rim of the space ‘wheel’ is 1.0 km.
(i) Calculate the angular velocity for the spacecraft to generate an artificial g equals to
9.81 m s-2.
(ii) What is its equivalent no. of revolutions per min?
(i) ac = r ω2 = g
(1000) ω2 = 9.81
ω = 0.0990 rad s-1
(ii) ω = 2πf
0.0990 = 2πf
f = 0.0158 s-1
f = 0.0158 × 60 = 0.946 r.p.m.
4. [N86/I/5]
A straight length of tape winds on to a roll rotating about a
fixed axis with constant angular velocity, the radius of the roll
increasing at a steady rate.
Which one of the graphs below correctly shows how the speed v
at which the tape moves towards the roll varies with time?
Ans : D
 is constant and r/t is constant
Since v = r 
v/t is constant. Therefore, linear graph.
5. [N06/P1/9]
o
Singapore is on the Equator. Cambridge is at a latitude 52 N, as shown in the diagram.
Cambridge
o
52
Equator
A student in Singapore has a centripetal acceleration as because of the Earth’s rotation about
its axis. The centripetal acceleration of another student at Cambridge is ac.
What are the magnitudes of the centripetal acceleration for each student?
Radius of Earth
Angular velocity of Earth about its own axis
= 6.4  106 m
= 7.3  10-5 rad s-1
(i) at Equator
2


ac  R 2  6.4  106  
 24  3600 
2
Cambridge
Rcos52o
= 0.0338 m s-2
o
52
(ii) at Cambridge
Equator
R
ac = Rcos2
2


 6.4  10 cos 52  

 24  3600 
2
6
= 0.0208 m s-2
6. The diagram represents a cyclist making a left turn on a rough road
surface at a constant speed v as viewed from behind. The total mass of the
bicycle and rider is m and their combined centre of gravity is at G.
If R is the resultant force on the normal reaction and frictional force, which
vector diagram represents the directions of the forces acting on the bicycle
and its rider?
R
R
R
R
G
G
G
A
mv 2
r
G
mv 2
r
mv 2
r
mg
mv 2
r
mg
B
mg
C
Ans : A
FBD only show the real forces. It does not include the centripetal force.
mg
D
The centripetal force comes from the horizontal component of R, the force exerted on the
bicycle by the ground.
7. [J85/II/8 (part)]
An aircraft is traveling at a constant speed of 180 m s-1 in a horizontal circle of radius 20 km.
A plumbline, attached to the roof of the cabin, settles at an angle θ to the true vertical while
the aircraft is turning.
(a) Find the centripetal acceleration of the aircraft.
(b) Name the forces which act on the bob of the plumbline and draw a labeled diagram to
show the directions of these forces and of their resultant (Indicate the centre of circle on
your sketch)
(c) Find the angle θ.
(d) Shows by means of a simple sketch of the cross-section of the aircraft and its cabin how
the plumbline is oriented with respect to the aircraft.
2
2
(a) ac = v = (180) 3 = 1.62 m s-2
r
20  10
(b)
Centre of circle
T : Tension
W : Weight

T
W
(c) Since the bob is undergoing a circular motion, a centripetal force must be acting on the
bob, causing it to move in a circle.
Resolving forces in x-direction
Fc = mac
T sin θ = m × 1.62
--- (1)
Another thing to note is the circular motion is a horizontal circular motion, i.e. the
centripetal acceleration must be acting horizontal. Thus there is no acceleration in the
vertical direction, i.e. no resultant force in the vertical direction.
Resolving forces in y-direction
∑ Fy = 0
T cos θ + (-mg) = 0
T cos θ = mg
--- (2)
(2)/(1): tan  
1.62 1.62

g
9.81
 = 9.37°
(d)
Centre of circle

The plumbline would appear vertical wrt to the passengers in the cabin.
8. [N06/P3/3(part)]
(a) A theme park ride is illustrated in the figure below. The carriages accelerate down the
slope and then loop the loop on a circular section of the track. The radius of the circular
section of track is 8.6 m.
(i) Find the minimum speed of the carriages at the top of the circular track so that the
carriages remain in contact with the track.
(ii) In practice, it is essential for designers to build in a considerable safety margin. Each
carriage and its passengers has a total mass 800 kg. At the top of the loop, the carriage
travels at 17 m s-1. Calculate the force that the track exerts on the carriage at the top of
the loop.
(b) For the carriage in (a) to have a speed of 17 m s-1 at the top of the loop, it must have
fallen from a height of at least x.
(i) Deduce a value for x.
(ii) State an assumption that you made in making this deduction.
(a) (i) At the top of the circular track,
Fc = mac
mv 2
W N 
r
For the car to just remain in contact with track, N ≈ 0,
mv 2
mg  0 
r
2
v
9.81 
8.6
v = 9.19 m s-1
W N
(ii) With a considerable safety margin, N > 0.
800  17 2
800

9
.
81

N



8.6
N = 1.90 × 104 N
(b) (i) Using the Principle of Conservation of Mechanical Energy.
KEx + PEx + W = KEtop + PEtop
0 + mg x + 0 = ½ m v2 + mg(2R)
(9.81)x = ½ (17.0)2 + (9.81 × 2  8.6)
x = 31.9 m
(ii) There is no energy lost from the system due to resistive forces such as air-resistance
nor friction.
More challenging problem.
9. A skier starts at rest at the top of a large hemispherical hill. Neglecting friction, show that
the skier will leave the hill below and become airborne at a distance d = R/3 below the
top of the hill. [Serway]
Resolving forces along the radius,
N
Fc = mac
mv 2
R
The skier just leaves the hill (ie becomes airborne) when N = 0
mv 2
mg cos   0 
R
2
--- (1)
mgR cos  mv
W cos    N  
Using the Principle of C.O.M.E.,
KEtop + PEtop + W = KEx + PEx
0 + mgR = ½ m v2 + mg (Rcos)
Sub Eqn (1),
mgR = ½ mgRcos + mg (Rcos)
1 = ½ cos + cos
= 3/2 cos
cos = 2/3
Distance dropped d
= R - Rcos
= R (1-cos)
= R (1-2/3) = R/3
(shown)
Rcos

R 
W