Circular motion and angular velocityEdexcel A-Level Physics: Revision notes

An object moving in a circle sweeps out an angular displacement $\theta$, measured from a reference line. Angles can be measured in degrees or radians.

One radian is the angle subtended at the centre of a circle by an arc equal in length to the radius. For an arc of length $s$ on a circle of radius $r$:

$\theta = \dfrac{s}{r}$

A full circle is $2\pi$ radians, so $360^\circ = 2\pi$ rad, $180^\circ = \pi$ rad and $1\ \text{rad} \approx 57.3^\circ$. To convert degrees to radians multiply by $\pi/180$; to convert radians to degrees multiply by $180/\pi$.

The angular velocity $\omega$ is the rate of change of angular displacement:

$\omega = \dfrac{\Delta\theta}{\Delta t}$

It is measured in rad s⁻¹. In one period $T$ the object turns through $2\pi$ rad, so

$\omega = \dfrac{2\pi}{T} = 2\pi f \qquad \text{i.e.} \qquad T = \dfrac{2\pi}{\omega}$

The linear speed of an object moving in a circle of radius $r$ is related to ω by

$v = \omega r$

so objects further from the axis of a rotating body move faster, but all have the same angular velocity.

An object moving round a circle at constant speed has a velocity that changes direction all the time. Velocity is a vector, so a change in direction is a change in velocity, and the object is accelerating even though its speed is constant.

The velocity is always along the tangent to the circle. The change in velocity points towards the centre of the circle, so the acceleration, called the centripetal acceleration, is directed towards the centre, perpendicular to the velocity.

Consider an object moving at constant speed $v$. In time $\Delta t$ it moves from one point on the circle to another, turning through a small angle $\Delta\theta$. Its velocity changes from $v_1$ to $v_2$, both of magnitude $v$.

Draw $v_1$ and $v_2$ tail to tail. The change in velocity $\Delta v = v_2 - v_1$ is the third side of an isosceles triangle with angle $\Delta\theta$ between the sides. For a small angle, the third side is almost an arc of a circle of radius $v$, so

$\Delta v = v\,\Delta\theta$

Then $a = \dfrac{\Delta v}{\Delta t} = v\dfrac{\Delta\theta}{\Delta t} = v\omega$. Using $\omega = v/r$:

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

As $\Delta\theta$ gets smaller, $\Delta v$ becomes perpendicular to the velocity, pointing towards the centre.

The magnitude of the centripetal acceleration is

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

Choose the form that matches the data: use $\omega^2 r$ when you have ω, T or f, and $v^2/r$ when you have the speed.

Worked example. A centrifuge rotor spins at 6000 rpm, with a sample 0.080 m from the axis. $f = 100$ Hz, $\omega = 2\pi \times 100 = 628$ rad s⁻¹, $v = \omega r = 50$ m s⁻¹, and $a = \omega^2 r = 628^2 \times 0.080 = 3.2 \times 10^4$ m s⁻².

For a fixed radius, doubling ω gives four times the acceleration, as $a \propto \omega^2$.