Further Mechanics 3 - Simple Circular Motion


Motion in a Circle

The rules of circular motion help us to describe:

We have so far applied Newtons 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?


Question 1

What is the path of an object which is thrown horizontally and allowed to fall?   Sketch the path of the falling object. 



The path is a parabola.


Now let us suppose that the force applied at 90o to the object was always at 90o.  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.


Question 2

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.


Make sure that you set your calculator to radians.  It's up to you make sure that you know how to do this.


Question 3

A train is travelling at 50 m/s round a curve of radius 6000 m.  What is its angular velocity?


Question 4

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 bear-trap is to fail to convert revolutions per minute to radians per second.  Divide the rpm by 60, then multiply the answer by 2p.



Centripetal Acceleration

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.

Question 5

If a = w2r, show that a = v2/r




 Where there is acceleration, there is a force.   We call the force centripetal force (NOT centrifugal force!), which is described by the formula:

F = mv2


The force acts towards the centre of the circle.


This is the circular motion version of Newton II, since F = ma and a = v2/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:

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.

Free body diagram:

a = v2/r = (3.45 m/s)2 0.8 m = 14.9 m/s2

            F = ma = 0.300 kg 14.9 m/s2 = 4.46 N. 

The string pulls the weight towards the middle of the circle.



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.

Question 6

Does the mass of the car matter?



Centrifugal force

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.