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What force keeps an object moving in a circle, and how is it related to its speed and radius?

Circular motion: angular displacement and angular velocity, the period and frequency of circular motion, centripetal acceleration, and the centripetal force needed to maintain circular motion.

A focused answer to the OCR H556 circular motion content, covering angular displacement in radians, angular velocity and its link to period and frequency, the relationship between linear and angular speed, centripetal acceleration, and the centripetal force required to keep an object moving in a circle.

Generated by Claude Opus 4.811 min answer

Reviewed by: AI editorial process; not yet individually human-reviewed

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  1. What this dot point is asking
  2. The answer
  3. Examples in context
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What this dot point is asking

OCR wants you to define angular displacement in radians and angular velocity, relate angular velocity to the period and frequency, link linear and angular speed, define and calculate centripetal acceleration, and find the centripetal force needed to keep an object moving in a circle.

The answer

Angular displacement and angular velocity

Linear and angular speed

Centripetal acceleration

Centripetal force

Examples in context

A satellite is held in orbit by gravity acting as the centripetal force, which is why orbital radius and period are linked. On a banked race track or aircraft turn, a component of the normal (or lift) force supplies the centripetal force, allowing higher speeds without relying on friction. A centrifuge spins samples so that the required centripetal force separates components of different density. Fairground rides such as the rotor use the wall's normal force to provide centripetal force.

Try this

Q1. Convert an angular velocity of 3.03.0 revolutions per second to rad s−1\text{rad s}^{-1}. [1 mark]

  • Cue. ω=2πf=2π×3.0=19 rad s−1\omega = 2\pi f = 2\pi \times 3.0 = 19\ \text{rad s}^{-1}.

Q2. A 0.50 kg0.50\ \text{kg} mass moves in a circle of radius 1.2 m1.2\ \text{m} at 4.0 m s−14.0\ \text{m s}^{-1}. Find the centripetal force. [2 marks]

  • Cue. F=mv2r=0.50×4.021.2=8.01.2=6.7 NF = \frac{mv^2}{r} = \frac{0.50 \times 4.0^2}{1.2} = \frac{8.0}{1.2} = 6.7\ \text{N}.

Q3. Explain why an object moving at constant speed in a circle is accelerating. [2 marks]

  • Cue. Velocity is a vector; although the speed is constant, the direction is continuously changing, so the velocity changes and the object accelerates towards the centre.

Exam-style practice questions

Practice questions written in the style of OCR exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

OCR 20194 marksA car of mass 1200 kg1200\ \text{kg} travels around a flat circular bend of radius 50 m50\ \text{m} at a steady speed of 15 m s−115\ \text{m s}^{-1}. Calculate the centripetal force required, and state what provides it.
Show worked answer →

Centripetal force: F=mv2r=1200×15250=1200×22550=270 00050=5400 NF = \frac{mv^2}{r} = \frac{1200 \times 15^2}{50} = \frac{1200 \times 225}{50} = \frac{270\,000}{50} = 5400\ \text{N}.

The friction between the tyres and the road provides this centripetal force, directed towards the centre of the bend.

Markers reward F=mv2rF = \frac{mv^2}{r}, the value 5400 N5400\ \text{N}, and identifying friction as the source of the centripetal force.

OCR 20224 marksA satellite orbits the Earth once every 9090 minutes at a radius of 6.8×106 m6.8 \times 10^{6}\ \text{m}. Calculate its angular velocity and its centripetal acceleration.
Show worked answer →

Period in seconds: T=90×60=5400 sT = 90 \times 60 = 5400\ \text{s}.

Angular velocity: ω=2πT=2π5400=1.16×10−3 rad s−1\omega = \frac{2\pi}{T} = \frac{2\pi}{5400} = 1.16 \times 10^{-3}\ \text{rad s}^{-1}.

Centripetal acceleration: a=ω2r=(1.16×10−3)2×(6.8×106)=(1.35×10−6)(6.8×106)=9.2 m s−2a = \omega^2 r = (1.16 \times 10^{-3})^2 \times (6.8 \times 10^{6}) = (1.35 \times 10^{-6})(6.8 \times 10^{6}) = 9.2\ \text{m s}^{-2}.

Markers reward converting the period to seconds, ω=2πT\omega = \frac{2\pi}{T}, and the acceleration about 9.2 m s−29.2\ \text{m s}^{-2} (close to surface gravity, as expected for a low orbit).

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