Centripetal forceEdexcel A-Level Physics: Revision notes

Velocity is a vector, so an object moving in a circle at constant speed is accelerating, because the direction of its velocity is always changing. By Newton's second law there must be a resultant force in the direction of the acceleration.

This force acts towards the centre of the circle and is always at right angles to the velocity. It is called the centripetal force. Because it is perpendicular to the motion it does no work, so it changes the direction of the velocity but not the speed.

For an object of mass $m$ moving in a circle of radius $r$ at speed $v$:

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

The two forms are linked by $v = \omega r$, where $\omega$ is the angular speed in rad s⁻¹, and by $\omega = \dfrac{2\pi}{T} = 2\pi f$. The centripetal acceleration is $a = v^2/r = r\omega^2$.

Because $F \propto v^2$, doubling the speed at a fixed radius needs four times the resultant force. Because $F \propto \omega^2 r$, doubling the radius at a fixed angular speed needs twice the force.

The force towards the centre is supplied by whatever real force or combination of forces points that way:

• A puck or conker on a string: the tension in the string
• A car on a flat bend: sideways friction between the tyres and the road
• A rider pressed against the wall of a rotor ride: the normal reaction of the wall
• A car at the top of a hump-back bridge: the difference between the weight and the contact force

If the force available is smaller than $mv^2/r$ the object cannot stay on the circle, for example a car skids outwards when the maximum friction is less than the force needed.

In a vertical circle the weight is one of the forces, so the other force must make up the difference. Take the direction towards the centre as positive.

• At the top of a hump-back bridge the centre is below, so $mg - R = \dfrac{mv^2}{r}$ and $R = mg - \dfrac{mv^2}{r}$. The contact force is less than the weight.
• At the bottom of a dip the centre is above, so $R - mg = \dfrac{mv^2}{r}$ and $R = mg + \dfrac{mv^2}{r}$. The contact force is greater than the weight.

Contact is lost at the top when $R = 0$, which gives $v = \sqrt{gr}$.

A car of mass 1200 kg takes a flat bend of radius 25 m at 14 m s⁻¹.

Resultant force needed: $F = \dfrac{mv^2}{r} = \dfrac{1200 \times 14^2}{25} = 9400$ N, supplied by sideways friction.

If the maximum friction available is 8.0 kN, which is less than 9.4 kN, the car cannot stay on the bend and skids outwards. At what speed could it just stay on? $v = \sqrt{Fr/m} = \sqrt{8000 \times 25 / 1200} = 12.9$ m s⁻¹.