Learn7367Paper 3 · Mechanics optionCircular motionCentripetal acceleration and force

Circular motion · Subtopic 2 of 2

Centripetal Acceleration and Force

An object moving in a circle accelerates towards the centre: $a=\dfrac{v^2}r=r\omega^2$ . This needs a resultant force towards the centre, $F=\dfrac{mv^2}r=mr\omega^2$ , called the centripetal force. It isn't a new kind of force: tension, gravity, friction or a normal reaction provides it. Being perpendicular to the velocity, it changes direction but does no work.

Cambridge 9231 Paper 3Edexcel YMA01 WME03Edexcel 9FM0 OptionAQA 7367 Paper 3Cambridge 9702 Paper 4Edexcel YPH11 WPH14≈ 40 minRevision 5 · updated 2026-10-27

Section 1 of 14

Learning objectives

Use $a=\dfrac{v^2}r=r\omega^2$ and $F=\dfrac{mv^2}r$ .
Explain why a force perpendicular to velocity gives constant speed in a circle.
Identify which real force provides the centripetal force in a situation.
Solve vertical circle and banking problems.

Pause and think

When a car turns sharply, you feel "thrown outwards" against the door. Is there an outward force on you?

Section 2 of 14

Core idea

Start hereConstant speed, yet accelerating

A car takes a roundabout at a steady 30 km/h. Its speedometer doesn't change, yet it's accelerating, because velocity has direction and the direction keeps turning. Something must keep pulling it inwards, or it would go straight on.

The big idea

$a=\frac{v^2}r=\omega^2r$ towards the centre; needs $F=\frac{mv^2}r$ , provided by a real force (tension, friction, gravity, normal reaction).

1. Why there is an acceleration

Velocity along the tangent, force towards the centre

position = 0.0

If the string broke, the ball would fly off along the tangent (Newton's first law), not outwards.

a=\frac{v^2}r

$\delta\theta=\omega\,\delta t$
Velocity turns through this small angle.
$|\delta\mathbf v|\approx v\,\delta\theta$
Arc of a circle of radius $v$ in velocity space; points to the centre.
$a=\frac{v\,\delta\theta}{\delta t}=v\omega$
Divide by $\delta t$ .
$=\frac{v^2}r=\omega^2r$
$\omega=v/r$ .

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

$F$: resultant force towards the centre.

2. Examples

Force needed for a 1000 kg car on a 50 m radius bend

Doubling speed needs four times the force. Above the crossing the road cannot supply enough friction.

Worked exampleCar on a bend

1000 kg, 15 m s $^{-1}$ , radius 50 m; $\mu=0.8$ .

Goal: Friction needed; maximum speed.
Strategy: Friction supplies $mv^2/r$ .

$F=\frac{1000\times15^2}{50}=4500\text{ N}$
Required.
$\mu mg=\frac{mv^2}r$
At the limit.
$v=\sqrt{\mu gr}=\sqrt{0.8\times9.81\times50}=19.8\text{ m s}^{-1}$
$m$ cancels.

Answer

4500

19.8

m s

^{-1}

✓ CheckAvailable friction

7850

N >

4500

N at 15 m s

^{-1}

✓.

Worked exampleVertical circle

Ball on a 0.8 m string.

Goal: Minimum speed at the top.
Strategy: At the top both $T$ and $mg$ point down.

$T+mg=\frac{mv^2}r$
Top.
$T=0\Rightarrow v=\sqrt{gr}$
String just slack.
$\sqrt{9.81\times0.8}=2.8\text{ m s}^{-1}$
Evaluate.

Answer

2.8

m s

^{-1}

✓ CheckAt the bottom

T-mg=\frac{mv^2}r

: tension largest there ✓.

Common misconceptionCentrifugal force No outward force acts: the passenger tends to go straight, and the door pushes them inwards.

Summary

$a=v^2/r$ to the centre.
A real force supplies $mv^2/r$ .

Centripetal: Directed towards the centre.

HistoryHuygens and Newton. Christiaan Huygens derived

\dfrac{v^2}r

1659

while studying pendulum clocks. Newton independently found it and used it with Kepler's third law to show that the force holding the planets must fall off as

\dfrac1{r^2}

, the step that led to universal gravitation.

Fun factCentrifuges in medicine. A laboratory centrifuge spinning blood samples at

5000

r.p.m. with radius

10

cm gives a centripetal acceleration of about

27\,000

m s⁻², nearly

3000g

. That's enough to separate red cells from plasma in minutes.

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