A-Level — physicsHard4 min read

Circular Motion and Centripetal Force

ω = 2πf; a = v²/r; F = mv²/r; applications to orbits, banked curves

Tutor AI Team
Tutor AI Team

# Circular Motion and Centripetal Force — A-Level Physics

When an object moves in a circle, even at constant speed, it is accelerating because its direction is constantly changing. This acceleration requires a force directed towards the centre — the centripetal force. Understanding circular motion is crucial for topics from orbital mechanics to particle accelerators.

1. Angular Quantities

Angular displacement (θ\theta): measured in radians (rad)

  • Full circle = 2π2\pi rad = 360°
  • θ\theta (rad) = θ\theta (degrees) × π/180\pi/180

Angular velocity (ω\omega): rate of change of angular displacement ω=θt=2πT=2πf\boxed{\omega = \frac{\theta}{t} = \frac{2\pi}{T} = 2\pi f}

Where:

  • ω\omega = angular velocity (rad/s)
  • TT = period (s)
  • ff = frequency (Hz)

Relationship between linear and angular speed: v=rω\boxed{v = r\omega}

2. Centripetal Acceleration

An object moving in a circle at constant speed has an acceleration directed towards the centre of the circle:

a=v2r=rω2\boxed{a = \frac{v^2}{r} = r\omega^2}

This is centripetal acceleration — it changes the direction of velocity, not the speed.

3. Centripetal Force

By Newton's Second Law:

F=mv2r=mrω2\boxed{F = \frac{mv^2}{r} = mr\omega^2}

Centripetal force is not a separate force — it is the resultant force directed towards the centre, provided by whatever is keeping the object in a circle:

Scenario What provides centripetal force
Ball on a string Tension in the string
Car on a curve Friction between tyres and road
Satellite in orbit Gravitational force
Electron in atom Electrostatic force
Clothes in washing machine Normal force from drum
Banked track Component of normal force

4. Common Scenarios

Object on a Horizontal Turntable

Friction provides centripetal force: f=mrω2f = mr\omega^2

Maximum speed before sliding: μmg=mv2/r\mu mg = mv^2/r → vmax=μgrv_{\text{max}} = \sqrt{\mu gr}

Car on a Flat Curve

μmg=mv2r  ⟹  vmax=μgr\mu mg = \frac{mv^2}{r} \implies v_{\text{max}} = \sqrt{\mu gr}

Car on a Banked Curve (no friction)

The horizontal component of normal force provides centripetal force: Nsin⁡θ=mv2rN\sin\theta = \frac{mv^2}{r} Ncos⁡θ=mgN\cos\theta = mg tan⁡θ=v2rg\tan\theta = \frac{v^2}{rg}

Vertical Circle (e.g., ball on a string)

At the top: T+mg=mv2/rT + mg = mv^2/r → T=mv2/r−mgT = mv^2/r - mg

At the bottom: T−mg=mv2/rT - mg = mv^2/r → T=mv2/r+mgT = mv^2/r + mg

Minimum speed at top (string just taut, T=0T = 0): vmin=grv_{\text{min}} = \sqrt{gr}

5. Conical Pendulum

A mass on a string swings in a horizontal circle, with the string tracing out a cone:

Tsin⁡θ=mv2r=mrω2T\sin\theta = \frac{mv^2}{r} = mr\omega^2 Tcos⁡θ=mgT\cos\theta = mg tan⁡θ=rω2g\tan\theta = \frac{r\omega^2}{g}

Worked Example: Basic Circular Motion

Problem

A 2 kg mass moves in a circle of radius 0.5 m at 3 m/s. Find the centripetal force and acceleration.

a=v2/r=9/0.5=18 m/s2a = v^2/r = 9/0.5 = 18 \text{ m/s}^2 F=ma=2×18=36 NF = ma = 2 \times 18 = 36 \text{ N}

Solution

Worked Example: Angular Velocity

Problem

A CD spins at 500 rpm. Calculate ω\omega and the speed of a point 5 cm from the centre.

f=500/60=8.33f = 500/60 = 8.33 Hz ω=2πf=52.4\omega = 2\pi f = 52.4 rad/s v=rω=0.05×52.4=2.62v = r\omega = 0.05 \times 52.4 = 2.62 m/s

Solution

Worked Example: Banked Curve

Problem

A road is banked at 15° for a curve of radius 200 m. Find the design speed (no friction needed).

v=rgtan⁡θ=200×9.81×tan⁡15°v = \sqrt{rg\tan\theta} = \sqrt{200 \times 9.81 \times \tan15°} =200×9.81×0.268=525.6=22.9 m/s= \sqrt{200 \times 9.81 \times 0.268} = \sqrt{525.6} = 22.9 \text{ m/s}

Solution

Worked Example: Vertical Circle

Problem

A 0.5 kg ball on a 1.2 m string swings in a vertical circle at 6 m/s at the bottom. Find the tension.

T−mg=mv2/rT - mg = mv^2/r T=0.5(9.81)+0.5(36)/1.2=4.91+15.0=19.9 NT = 0.5(9.81) + 0.5(36)/1.2 = 4.91 + 15.0 = 19.9 \text{ N}

Solution

7. Practice Questions

    1. A satellite orbits Earth at 7500 m/s at a height where r=6.8×106r = 6.8 \times 10^6 m. Calculate its period. (3 marks)
    1. A car takes a flat bend of radius 50 m. The coefficient of friction is 0.6. Find the maximum safe speed. (3 marks)
    1. A ball on a 0.8 m string completes a vertical circle. Find the minimum speed at the top for the string to remain taut. (2 marks)
    1. A conical pendulum has a string of length 0.6 m making 25° with the vertical. Find the period of revolution. (4 marks)

    Answers

    1. ω=v/r=7500/6.8×106=1.103×10−3\omega = v/r = 7500/6.8 \times 10^6 = 1.103 \times 10^{-3} rad/s. T=2π/ω=5700T = 2\pi/\omega = 5700 s = 95 minutes.

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Summary

  • ω=2π/T=2πf\omega = 2\pi/T = 2\pi f; v=rωv = r\omega
  • Centripetal acceleration: a=v2/r=rω2a = v^2/r = r\omega^2 (towards centre)
  • Centripetal force: F=mv2/rF = mv^2/r (provided by tension, gravity, friction, etc.)
  • Not a separate force — always identify what provides it
  • Banked curves: tan⁡θ=v2/(rg)\tan\theta = v^2/(rg)
  • Vertical circles: tension varies (max at bottom, min at top)
Tags:
#a-level#physics#mechanics#circular-motion#centripetal-force

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