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APPENDIX 1
UNIFORM CIRCULAR MOTION
When an object is moving in a circle or circular arc (part of a circle) with constant
speed it is said to be in uniform circular motion. There are many everyday
examples of this type of motion: a car travelling around a corner or curve, a person
on a merry-go-round or a satellite orbiting the Earth. In each case the motion is the
same. Although the speed of the object does not change the velocity quite clearly
does since the direction is constantly changing (remember velocity is a vector). This
means that the object is accelerating. The figure below illustrates this for the points p
and q that lie on a circular path. Note that the direction of the acceleration is towards
the centre of the circle. This is always true for all uniform circular motion.
vp
q
p
vq
vq

v
vp

Fig. 1. The change in velocity in uniform circular motion.
We can see that the magnitude of the acceleration must be constant since the motion
is circular and so has complete symmetry. The magnitude of the acceleration is given
by
a =
v2
r
where v is the speed and r the radius of the motion. This is called the centripetal
acceleration. To produce this acceleration there must be a force which is termed the
centripetal force and is given by Newton’s second law
F = ma =
mv 2
r
The direction of the centripetal force is also towards the centre of the circular path.
Now the centripetal force is not another type of force it is simply the name we give to
the force that is responsible for uniform circular motion. To see this let us consider a
few examples.
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A car taking a curve. The force that makes the car turn in a circle is the force of
friction between the tyres and the road. What makes you turn in a circle if you are in
the car? Well initially you don’t go in a circle. Newton’s first law says you will
travel in a straight line unless there is a net force on you. But while you are trying to
move in a straight line the car is turning. Eventually you meet with the car (the back
rest of the seat or a door) and this contact force makes you turn too.
A satellite orbiting the Earth. There is a force due to gravity on the satellite. This
force is directed towards the centre of the Earth, which is the centre of the circle of
motion.
Q.
A certain string can withstand a tension of 40 N before breaking. A
child ties a stone of mass 0.5 kg to the end of the string and whirls the
stone in a vertical circle of radius 0.7 m, gradually increasing the
speed. (a) Where is the stone on its path when the string breaks? (b)
What is the speed of the stone as the string
breaks?
T
W
T
W
A.
There are 2 external forces acting on the stone. The tension in the
string and the weight of the stone. As the stone swings around the
direction of the weight force always points down but the direction of
the tension changes. We have drawn free-body diagrams for 2
instances, when the stone is at the top of the swing and when it is at
the bottom.
At the top of the swing the 2 forces are in the same direction and their
magnitudes add to give the net force. While at the bottom of the
swing the magnitudes subtract to give the net force.
When solving problems in uniform circular motion it is useful to take
co-ordinate axes radially (towards the centre of the circle of motion)
and tangentially to the motion. This means the direction is actually
always changing but this is alright since we consider what happens at
any instant.
(a) Applying Newton’s second law in the radial direction (inwards is
positive) we have
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F
F
r
= T + W = ma
for the instant at the top of swing
r
= T W = ma
for the instant at the bottom of swing
Our equation for uniform circular motion says that the acceleration
on the stone must be
v2
a =
r
towards the centre of the circle
From this it can be seen that at the bottom of the swing the speed will
be lower when the string breaks. At the top of the swing T and W add
to give a larger value for v.
(b) Solving for v we have
v2
T  W = m
r
r
v = (T  W )
m
=
(40  (0.5 kg)(9.8 m.s-2 )
= 7.0 m.s-1
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0.7 m
0.5 kg
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