Centripetal acceleration and force

Why there is an acceleration

Velocity is a vector. In circular motion at constant speed, the direction of the velocity changes continuously, so the velocity changes and there is an acceleration. The acceleration points towards the centre of the circle: it is called centripetal acceleration.

Deriving a = v² / r

Draw velocity vectors v_1 and v_2 a short time Δt apart; they differ in direction by a small angle Δθ. The vector Δv = v_2 - v_1 points towards the centre and has size Δv ≈ vΔθ. So a = Δv / Δt = v(Δθ / Δt) = vω. Using v = ωr:

a = v² / r = rω²

Centripetal force

By Newton's second law, a resultant force must act towards the centre:

F = mv² / r = mrω²

This is not a new kind of force. It is provided by an existing force: tension in a string, friction on a car's tyres, gravity on a satellite, or the normal force on a rider in a loop. There is no outward 'centrifugal force' acting on the object.

Vertical circles

For a ball on a string at the top of a vertical circle, both tension and weight point down to the centre: T + mg = mv² / r. The string stays taut only if v² ≥ gr at the top.

Worked example

A 0.20 kg mass on a 0.50 m string moves in a horizontal circle at 4.0 m/s on a smooth table. Find the tension.

1. The tension provides the centripetal force.
2. T = mv² / r = 0.20 x 4.0² / 0.50

Answer: T = 6.4 N

Check yourself

1. A car goes round a bend of radius 40 m at 12 m/s. Find its centripetal acceleration in m/s².
   Show answer

   3.6 m/s². a = v² / r = 144 / 40 = 3.6 m/s².

2. A 900 kg car goes round a bend of radius 50 m at 15 m/s. Find the centripetal force in N.
   Show answer

   4050 N. F = mv² / r = 900 x 225 / 50 = 4050 N.

3. The maximum friction force on a 1200 kg car is 6000 N. Find the maximum speed at which it can go round a bend of radius 30 m, in m/s, to 3 s.f.
   Show answer

   12.2 m/s. 6000 = 1200v² / 30, so v² = 150 and v = 12.2 m/s.

4. A 0.50 kg mass moves in a circle of radius 0.80 m at 5.0 rad/s. Find the centripetal force in N.
   Show answer

   10 N. F = mrω² = 0.50 x 0.80 x 25 = 10 N.

5. A 0.30 kg ball on a 0.60 m string moves in a vertical circle. At the top it moves at 4.0 m/s. Find the tension there in N, to 3 s.f.
   Show answer

   5.06 N. T + mg = mv² / r: T = 0.30 x 16 / 0.60 - 0.30 x 9.81 = 8.00 - 2.94 = 5.06 N.

6. Find the minimum speed at the top of that 0.60 m vertical circle for the string to stay taut, in m/s, to 3 s.f.
   Show answer

   2.43 m/s. At the minimum speed T = 0, so mg = mv² / r and v = sqrt(9.81 x 0.60) = 2.43 m/s.

7. In which direction does the resultant force on an object in uniform circular motion act?
   • Along the tangent, in the direction of motion
   • Outwards from the centre
   • Towards the centre of the circle
   • Vertically downwards
   Show answer

   Towards the centre of the circle. Centripetal means 'centre-seeking'.

8. Why is an object moving in a circle at constant speed accelerating?
   • Its speed is increasing
   • Its mass is changing
   • No resultant force acts on it
   • The direction of its velocity is changing
   Show answer

   The direction of its velocity is changing. Acceleration is the rate of change of velocity, and velocity includes direction.

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Sources

• Pearson Edexcel International Advanced Subsidiary/Advanced Level in Physics specification (XPH11/YPH11), Issue 3
• Pearson qualifications: International Advanced Levels Physics (2018)
• Pearson Edexcel IAL Physics Unit 4 (WPH14) question paper, June 2024: list of data, formulae and relationships

Last updated 7 November 2026. Spotted a mistake? Tell us.