Uniform circular motion and centripetal acceleration

Circular motion

If the speed never changes, why does a body moving in a circle accelerate?

Drag a body round a circle and change the radius and angular speed. The velocity arrow stays tangent and the same length; the acceleration arrow always points to the centre with size ω²r.

Drag the ringed points on the board.

High school
Angular speed
v = ωr
Centripetal acceleration
Centripetal force

\varphi

0.60

r

2.0

\omega

1.0

m

1.0

a = \omega^2 r = \dfrac{v^2}{r}

$v = \omega r$: 2.00
$a = \omega^2 r$: 2.00
$T = 2\pi/\omega$: 6.28
$F = m\omega^2 r$: 2.00

Guided tour

1One radian per second
With $r=2\,\text{m}$ and $\omega=1\,\text{rad/s}$ the body covers $2\,\text{m}$ of arc each second, so $v=\omega r=2\,\text{m/s}$ . One lap takes $T=2\pi\approx6.28\,\text{s}$ , and the acceleration toward the centre is $\omega^2r=2\,\text{m/s}^2$ .
2Spin twice as fast
Double $\omega$ to $2$ . The speed doubles to $4\,\text{m/s}$ , but the acceleration quadruples to $8\,\text{m/s}^2$ : it grows with $\omega^2$ .
3Same speed, tighter turn
Now $r=1$ and $\omega=2$ : the speed is back to $2\,\text{m/s}$ , as in the first step, but the acceleration is $4\,\text{m/s}^2$ , twice as much. At fixed speed $a=\dfrac{v^2}{r}$ : halve the radius, double the acceleration.
4The force that does it
Give the body a mass of $2\,\text{kg}$ at $r=2$ , $\omega=2$ . Something must pull it inward with $F=m\omega^2r=2\times4\times2=16\,\text{N}$ : a string, friction, or gravity. Cut the string and the body flies off along the blue arrow.

Explanation

1
Describing motion on a circle
The angle swept per second is the angular speed $\omega$ (rad/s). One lap is $2\pi$ radians, so the period is $T=\dfrac{2\pi}{\omega}$ and the frequency $f=\dfrac1T$ . An angle of $\varphi$ radians on a circle of radius $r$ is an arc of length $r\varphi$ , which gives the speed along the path:
$v=\omega r$
2
Turning is accelerating
Velocity is a vector: it has a direction. On the circle the direction turns all the time even though the length stays put, so the velocity changes, and a change of velocity is an acceleration. The change always points inward, toward the centre:
$a=\omega^2r=\frac{v^2}{r}$
It is at $90^\circ$ to the velocity, which is exactly why it changes the direction and not the speed.
3
Centripetal force is a job, not a new force
Newton's second law says an inward acceleration needs an inward net force, $F=m\omega^2r$ . It is always provided by a real force: string tension for a ball on a string, friction for a car on a bend, gravity for a satellite. When a question asks for the centripetal force, find which real forces add up toward the centre.

Check yourself

Q01

A wheel turns at $\omega=4\,\text{rad/s}$ . A point $0.5\,\text{m}$ from the axle moves at what speed?

Q02

A car takes a bend at constant speed. If it goes twice as fast round the same bend, the friction needed is

Q03

In uniform circular motion, which quantity is constant?

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