Angular velocity and the period of circular motion, centripetal acceleration and centripetal force, and examples such as orbits, conical pendulums and banked tracks.

What this dot point is asking

CCEA wants you to define angular velocity and the period of circular motion, derive and use the centripetal acceleration, identify what real force acts as the centripetal force in different situations, and apply the ideas to orbits, conical pendulums and banked or unbanked tracks. Numerical questions are common, so you must be fluent with the equations and confident converting between $v$, $\omega$, $T$ and $f$.

The answer

Angular velocity and period

One full revolution is $2\pi$ radians, so in time $T$ the radius turns through $2\pi$, giving $\omega = 2\pi / T$. The tangential speed follows from the circumference: $v = 2\pi r / T = \omega r$. The object's speed is constant, but its velocity changes direction continuously, so it is always accelerating.

Centripetal acceleration and force

The two forms are linked by $v = \omega r$: substituting into $a = v^2 / r$ gives $a = \omega^2 r$. Choose the form that matches the data you are given. If a question quotes a period or a rotation rate, use the $\omega$ form; if it quotes a tangential speed, use the $v$ form.

Where the centripetal force comes from

The centripetal force is supplied by a different real force in each context:

• Satellite or planet: gravity provides the centripetal force, so $\frac{GMm}{r^2} = \frac{mv^2}{r}$.
• Conical pendulum or whirling string: the horizontal component of the tension, $T\sin\theta = \frac{mv^2}{r}$.
• Car on a flat bend: friction between the tyres and the road.
• Banked track: the horizontal component of the normal reaction (assisted by friction), which is why race tracks are banked so a bend can be taken faster without relying on friction alone.

Worked example: conical pendulum

Examples in context

Example 1. A geostationary satellite over the equator. A communications satellite must orbit with a period of exactly one sidereal day so it stays above the same point. Here $\omega = 2\pi / T = 2\pi / 86400 = 7.27 \times 10^{-5}\ \text{rad s}^{-1}$. Gravity supplies the centripetal force, $\frac{GMm}{r^2} = m\omega^2 r$, which sets the orbital radius at about $4.2 \times 10^{7}\ \text{m}$. The satellite is in continuous free fall, accelerating towards Earth at $\omega^2 r \approx 0.22\ \text{m s}^{-2}$, yet never gets closer because it moves sideways fast enough.

Example 2. A car on a banked motorway slip road. Engineers bank curves so the horizontal component of the normal reaction provides the centripetal force, reducing reliance on tyre friction in wet weather. For a design speed $v$ on a bend of radius $r$ banked at angle $\theta$, the ideal frictionless condition is $\tan\theta = \frac{v^2}{rg}$. A bend of radius $80\ \text{m}$ designed for $20\ \text{m s}^{-1}$ needs $\tan\theta = \frac{400}{80 \times 9.81} = 0.510$, so $\theta = 27^\circ$.

Try this

Q1. A $0.20\ \text{kg}$ ball is whirled in a horizontal circle of radius $0.50\ \text{m}$ at $3.0\ \text{m s}^{-1}$. Calculate the centripetal force. [2 marks]

• Cue. $F = \frac{mv^2}{r} = \frac{0.20 \times 3.0^2}{0.50} = 3.6\ \text{N}$.

Q2. State what provides the centripetal force for a satellite orbiting the Earth. [1 mark]

• Cue. The gravitational attraction of the Earth.

Q3. A turntable rotates at $45\ \text{rpm}$. Calculate the angular velocity and the centripetal acceleration of a point $0.12\ \text{m}$ from the centre. [3 marks]

• Cue. $\omega = \frac{45 \times 2\pi}{60} = 4.71\ \text{rad s}^{-1}$; $a = \omega^2 r = 4.71^2 \times 0.12 = 2.7\ \text{m s}^{-2}$.