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Centripetal force

Forces and motion · level 7, lesson 2 of 6

A question to start

Swing a bucket of water in a vertical circle and the water stays in. Tie a 0.5 kg ball to a 1 m string and whirl it at 4 m/s. How hard does the string pull? Is there a special "centripetal force" acting?

The idea

By the second law, an object of mass \(m\) moving in a circle needs a net force towards the centre:

\[F = m\frac{v^2}{r} = m\omega^2 r.\]

This is called the centripetal force, but it is not a new kind of force. It is whatever real forces add up to an inward pull: a tension, a friction, a normal force, gravity, or a mix.

Circular motion What supplies the inward force
A ball on a string The string's tension
A car turning on a flat road Friction between tyres and road
A train on a curved track The rails pushing sideways on the wheels
A satellite in orbit Gravity
A person in a spinning space station The floor pushing on their feet
tension Tvseen from above

Forces are unchanged by Mars gravity unless gravity or friction is the one doing the job. A string pulls just as hard on Mars as on Earth for the same ball, speed and radius. But for anything that relies on friction or weight, Mars gives less.

Worked examples

1. The ball. \(F = 0.5 \times 4^2 / 1 = 8\) N of tension, on Mars or Earth (ignoring the small downward pull).

2. A wheel's rim. A 2 kg segment of a flywheel rim at 0.5 m from the axle, spinning at 100 rad/s: \(F = 2 \times 100^2 \times 0.5 = 10{,}000\) N. The rim must hold together against forces this big.

3. The bucket at the top. At the top of a vertical circle of radius 1 m, the water needs an inward (downward) acceleration of at least \(g\), or it falls out. The slowest safe speed is \(v = \sqrt{gr}\): 1.9 m/s on Mars, 3.1 m/s on Earth.

Try it

  1. Is centripetal force a separate kind of force?
  2. What supplies the inward force for the Moon going round the Earth?
  3. A 1,000 kg car takes a bend of radius 50 m at 15 m/s. How big is the inward force?
  4. A 70 kg astronaut stands in a spinning station, 20 m from the axis, turning once every 9 s. How hard does the floor push?
Answers
  1. No: it is whatever real forces add up to an inward pull.
  2. Gravity.
  3. \(1000 \times 225 / 50 = 4{,}500\) N.
  4. \(\omega = 2\pi/9 = 0.698\) rad/s. \(F = 70 \times 0.698^2 \times 20 = 682\) N, close to their weight on Earth.

Common mistake

Drawing "centripetal force" as an extra arrow. In a free-body diagram, draw only real forces. Their sum points to the centre and equals \(mv^2/r\).

Where it leads