Circular motionAQA A-Level Physics: Revision notes

Section 1

Motion in a circle

An object moving in a circle at constant speed has a velocity that changes direction continuously. Since velocity is a vector, its rate of change, the acceleration, is not zero. The acceleration is directed towards the centre of the circle, so a resultant force towards the centre is needed: the centripetal force. Without it the object would travel in a straight line.

The centripetal force is not a new kind of force: it is the name for whichever real force (tension, friction, gravitational attraction, a normal contact force) acts towards the centre. Because it is always perpendicular to the velocity, it does no work and the speed does not change.

Key termscentripetal forcecentripetal acceleration

Common mistake

Drawing a 'centrifugal force' outwards on the object. The only resultant force is inwards; the outward feeling is the object's tendency to continue in a straight line.

Section 2

Radians and angular speed

The radian is the angle subtended at the centre of a circle by an arc equal in length to the radius. An arc of length $s$ subtends an angle $\theta = s/r$ radians, so a full circle is $2\pi$ rad ( $360^\circ = 2\pi$ rad).

The angular speed $\omega$ is the angle turned through per unit time, measured in rad s⁻¹:

$\omega = \frac{\theta}{t} = \frac{v}{r} = 2\pi f = \frac{2\pi}{T}$

so the linear speed is $v = \omega r$ , where $f$ is the frequency (revolutions per second) and $T$ the time period. Direction of angular velocity is not required.

Key termsradianangular speedfrequency

Exam tip

Set your calculator to radians for any calculation involving ω, and convert rpm to revolutions per second by dividing by 60.

Section 3

Centripetal acceleration

For an object moving at speed $v$ in a circle of radius $r$ the centripetal acceleration is

$a = \frac{v^2}{r} = \omega^2 r$

directed towards the centre. The derivation is not examined. The two forms are linked by $v = \omega r$ . Choose $v^2/r$ when the speed is given and $\omega^2 r$ when the frequency or period is given.

Common mistake

Using v/r or v²r. Check units: v²/r gives m s⁻².

Section 4

Centripetal force

From $F = ma$ the resultant force needed to keep an object of mass $m$ in a circle is

$F = \frac{mv^2}{r} = m\omega^2 r$

Doubling the speed needs four times the force; doubling the radius at the same speed halves it. To find the real force involved, identify what acts towards the centre: tension for a stone on a string, friction for a car on a flat bend, a normal contact force for a rider against a wall. In the vertical direction the forces balance if the object does not accelerate vertically.

Exam tip

Draw the forces first, then say which one (or which component) is the resultant towards the centre.

Section 5

Worked example

A car of mass 1200 kg takes a flat bend of radius 60 m at a constant 18 m s⁻¹. Find the friction force needed.

$a = v^2/r = 18^2 / 60 = 5.4$ m s⁻².
$F = ma = 1200 \times 5.4 = 6.5 \times 10^3$ N, towards the centre of the bend.

At 20 m s⁻¹ the force would be $1200 \times 20^2/60 = 8.0 \times 10^3$ N, which shows the effect of $v^2$ .

Must Know

Circular motion at constant speed needs a resultant force towards the centre
$\omega = v/r = 2\pi f$ ; $v = \omega r$
$a = v^2/r = \omega^2 r$
$F = mv^2/r = m\omega^2 r$
Centripetal force does no work: it is perpendicular to the velocity

That's the notes covered.

Carry on to the next subtopic.

Exam questions on Circular motion

A stone of mass 0.15 kg is whirled in a horizontal circle of radius 0.80 m on a light string. It moves at a constant speed and completes 2.5 revolutions every second.
Explain why the stone is accelerating although its speed is constant.2 marks
A car of mass 1200 kg travels at a constant speed of 18 m s⁻¹ round a flat circular bend of radius 60 m. The only horizontal force on the car is the sideways friction between the tyres and the road.
The maximum friction force that the tyres can provide on a wet road is 8.0 kN. Calculate the maximum constant speed at which the car can take the bend.2 marks
A laboratory centrifuge spins sample tubes about a vertical axis at 6000 revolutions per minute. The base of each tube is 0.12 m from the axis of rotation.
Calculate the angular speed of the tubes and the speed of the base of a tube.3 marks

See the full worksheet

Written by the Exaim team, led by Shaun Daswani (Head of Upper Secondary, Improve ME Institute; MSc Financial Mathematics, Imperial College London; BSc, UCL) and Jason Daswani (operational lead, Improve ME Institute; LSE).