Further Mechanics 1  Simple Circular Motion
Contents 
The rules of circular motion help us to describe:
movement of a car going round a corner,
a tethered model aeroplane;
the planets in their orbits.
We have so far applied Newton’s Laws to motion in a straight line; the forces are pushing or pulling the object along in the direction of its travel. What happens if we put a force perpendicular to the direction of motion?
What is the path of an object which is thrown horizontally and allowed to fall? 
The path is a parabola.
Now let us suppose that the force applied at 90^{o} to the object was always at 90^{o}. This time the path would be circular:
In this case, the linear speed remains constant, but the direction is always changing. This means that there is a change in velocity, hence acceleration. Along with this come concepts that are important in circular motion, such as angular velocity. Note how the force is always towards the centre of the circle.
Why does an object going around in a circular path at a constant linear speed have acceleration? 
Consider an object going round in a circle of radius r.
We
know that anything going round in a circle has a constant speed but a changing
velocity. This is because the
direction is constantly changing. If
we alter the radius, the linear speed also changes, even though we have not
changed the rate of turning. So we
have to think up another quantity that we can use to describe the rate of
turning. We use angular velocity, physics code
w
(omega, a Greek letter long ‘ō’), how big an angle is turned in one
second. We could use degrees per second, but instead we use another
kind of angular measurement, the radian.
One
radian is the angle that subtends an arc whose length is the same as the radius.
q
rad =
s/r.
We
can easily work out that 1 rad
» 57
^{o}.
1
revolution is
2p radians.
For
small angles in radians,
q
»
sin
q
»
tan
q.
This is another reason why radians are so useful.
It does not work for large angles in radians, nor does it work for degrees.
In
dimensional analysis the radian is a dimensionless unit.
In some texts, you may see it missed out altogether, although here we
will always include it.
w
rad s^{1} is the
angular
velocity.
w
=
2pf
Þ linear speed
v =
wr
=
2pfr
m s^{1}.
The
direction of the velocity is
tangential.
Make sure that you set your calculator to radians. It's up to you make sure that you know how to do this. 
The frequency and period are linked to the angular velocity by these equations:
A train is travelling at 50 m s^{1} round a curve of radius 6000 m. What is its angular velocity? 

A
washing machine spins its drum at 1200 rpm.
If the diameter of the drum is 35 cm, find:
(a)
the angular velocity of the tub;
(b) the linear speed of the rim of the tub. 
A
common beartrap is to fail to convert revolutions per minute to radians
per second.
Divide
the rpm by 60, then multiply the answer by
2p. 
We
need to distinguish between an object spinning on its axis, and an object moving
in a circular path.
We will consider the latter only.
The former situation is part of
rotational
dynamics. You
can read about the derivation of the relationship in any textbook, so we will
not cover it here.
Acceleration is always towards the centre of the circle and is given by:
a = w^{2}r
We can also express this in terms of frequency.
a = (2pf)^{2}r = 4p^{2}f^{2}r.
A very useful dodge here is that
p^{2} is approximately 10.
We can write this as:
If a = w^{2}r, show that a = v^{2}/r 
Where
there is acceleration, there is a force. We call the force
centripetal
force (NOT centrifugal force!), which is described by the formula:
The force acts towards the centre of the circle.
This is the circular motion version of Newton II, since F = ma and a = v^{2}/r.
Suppose we were to whirl a stone around on a string, the forces would be governed by the relationship above. However, if the string were to snap, the stone would fly off in a straight line at a tangent to the circle (NOT straight out from the centre). The following are other examples that obey this relationship:
Satellite orbiting the Earth (gravity provides the centripetal force)
Vehicle going a bend (friction)
Electron orbiting the nucleus (electrostatic attraction).
Problem
solving strategy
1.
State clearly what
object is being considered.
2.
Draw a free body sketch
of the object.
3.
Mark the weight of the
object
4.
Mark in any points
where the object touches anything else and the forces involved.
5.
Decide which direction
you will call positive.
6. Apply Newton II (F = ma).
Worked example A mass of 0.300 kg is moving in a circular path of radius 0.80 m on a friction free table. It is attached to a peg in the middle of the table in the centre of the circle. Draw the free body diagram of the mass and find the force exerted by the string on the mass when the mass is moving at a constant speed of 3.45 m s^{1}. 
Free body diagram:

a
= v^{2}/r
=
(3.45
m s^{1})^{2}
¸ 0.8 m = 14.9 m
s^{2}
Þ
F = ma = 0.300 kg
´
14.9 m s^{2} = 4.46 N.
The
string pulls the weight towards the middle of the circle. 
Work done in Circular Motion
When an object is moving in circular motion, no work is done. This is because the centripetal force is always at 90 degrees to the direction of movement.
W = Fs cos
cos 90 = 0
This is a favourite question in a multiple choice paper.
Going Round the Bend
When a car goes around a corner, the centripetal force is caused by the friction of the tyres on the road. When you are in a car going round a right hand bend, and you feel that you are being thrown out towards the left, in reality you are trying to go in a straight line, but the car is pushing on you, to make you go in a circle.
The centripetal force is given by:
F is the friction of the tyres on the road. It can be worked out with a simple formula,
(m is the coefficient of friction, F is the frictional force, and N is the normal force)
We can write
F = mN
Since N is the weight, we can rewrite this as
F = mmg
So we can rewrite the first equation as:
The mass terms cancel out, so we can write:
Therefore, for a particular radius of curve there is a maximum speed that depends on the coefficient of friction of the rubber on the road. Fortunately, the coefficient of friction for rubber is quite high, about 4.
Does the mass of the car matter? 
Centrifugal force does NOT exist. It is a term that is used a lot by people to explain why, for example, you feel an outwards force when you are going round a corner.
The answer, centrifugal force, is the most intuitive, but is not correct.
When a car goes round a corner, the friction of the tyres on the road causes the centripetal force. If the centripetal force were reduced to zero (e.g. the car it a patch of ice), the car would carry on in a straight line, tangential to the curved path it should have taken.
If there were centrifugal force (as a reaction force to the centripetal force), the car would slide at 90 ^{o} to its path. This does not happen.
You will find the term centrifugal force used widely among engineers. Do not try to be the clever physicist who tells them that they are wrong.
Centrifugal force does not exist. Do NOT make any mention of it in the exam. It is a physics error. 