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Centripetal acceleration

Forces and motion · level 7, lesson 1 of 6

A question to start

You stand on the equator of Mars. The planet's spin carries you round a circle 21,000 km long every day. Your speed is steady, but your direction keeps changing. Are you accelerating? By how much?

The idea

Velocity has a direction, so moving in a circle at a steady speed means a changing velocity: an acceleration. It points to the centre of the circle, and it is called centripetal ("centre-seeking") acceleration.

Its size: in a short time \(\Delta t\), an object at speed \(v\) moves through a small angle \(\Delta\theta = v\Delta t/r\). Its velocity arrow, always at right angles to the radius, turns through the same angle. The change in velocity is a short arrow of length \(v\,\Delta\theta\), pointing inwards. So

\[a = \frac{\Delta v}{\Delta t} = \frac{v\,\Delta\theta}{\Delta t} = v \cdot \frac{v}{r} = \frac{v^2}{r}.\]
rvvvΔv, inwards

It is often easier to use the angular speed \(\omega\), the angle turned per second, in radians per second. One turn is \(2\pi\) radians, so \(\omega = 2\pi/T\) for a period \(T\). With \(v = \omega r\):

\[a = \frac{v^2}{r} = \omega^2 r.\]
Standing on the equator Mars Earth
Radius 3,396 km 6,378 km
Time for one turn (sidereal day) 24.62 h 23.93 h
\(\omega\) \(7.09 \times 10^{-5}\) rad/s \(7.29 \times 10^{-5}\) rad/s
Centripetal acceleration \(\omega^2 r\) 0.017 m/s² 0.034 m/s²
As a share of \(g\) 0.46% 0.35%

So the spin makes you a tiny bit lighter on the equator: part of gravity is used to keep you going round.

Worked examples

1. Mars's equator. \(\omega = 2\pi / 88{,}640 \text{ s} = 7.09 \times 10^{-5}\) rad/s. \(a = (7.09 \times 10^{-5})^2 \times 3.396 \times 10^6 = 0.0171\) m/s².

2. A car on a roundabout. \(v = 10\) m/s, \(r = 25\) m. \(a = 100/25 = 4\) m/s².

3. A centrifuge. Spinning at 3,000 turns a minute, radius 0.1 m: \(\omega = 2\pi \times 50 = 314\) rad/s, \(a = 314^2 \times 0.1 = 9{,}870\) m/s², about 1,000 g.

Try it

  1. Which way does centripetal acceleration point?
  2. A runner goes round a bend of radius 40 m at 8 m/s. Find her acceleration.
  3. Phobos orbits Mars at 9,376 km from the centre in 7 h 39 min. Find its centripetal acceleration.
  4. If you double the speed on the same circle, what happens to \(a\)?
Answers
  1. To the centre of the circle.
  2. \(64 / 40 = 1.6\) m/s².
  3. \(T = 27{,}540\) s, \(\omega = 2.28 \times 10^{-4}\) rad/s, \(a = \omega^2 r = 0.49\) m/s². (This is the pull of Mars's gravity at that distance.)
  4. It goes up four times.

Common mistake

Using degrees for \(\omega\). \(v = \omega r\) and \(a = \omega^2 r\) work only with \(\omega\) in radians per second.

Where it leads