# Centripetal force

> CIE A-Level Physics · Unit 16: Motion in a circle
> Source: https://www.owlsprep.com/study/cie-9702-u16-centripetal-force/

This sub-topic explains what centripetal force is, how to calculate its magnitude, how to identify its physical origin in common exam scenarios, and solve problems involving uniform circular motion.

**Prerequisites:** [Uniform circular motion basics](https://www.owlsprep.com/study/cie-9702-u16-uniform-circular-motion/); [Newton's second law of motion](https://www.owlsprep.com/study/cie-9702-u04-newtons-laws-of-motion/)

## Learning objectives

- Define centripetal force and state its correct direction
- Derive and use the two common forms of the centripetal force equation
- Identify the source of centripetal force in different physical scenarios
- Solve standard exam problems involving uniform circular motion

## Definition and key properties of centripetal force

**Centripetal force** — The net resultant force that acts on an object moving in a circular path, always directed perpendicular to the object's instantaneous velocity towards the centre of the circle.

*Notation:* $F_c$

*Example:* Tension in a string spinning a attached mass provides centripetal force.

Because centripetal force is always perpendicular to the direction of motion, it does no work on the object. For uniform circular motion, this means the object's speed stays constant, only the direction of velocity changes, which maintains the circular path.

> **Common misconception**
>
> There is no 'centrifugal force' pushing outwards on the object in an inertial (stationary) frame of reference. The outward 'push' you feel turning in a car is just your inertia resisting the change in direction.

**Worked example:** A 2.0 kg mass is spun at constant speed in a horizontal circle, attached to a fixed central point by a string. State the direction of the centripetal force acting on the mass and identify what provides it.

1. By definition, centripetal force always points towards the centre of the circular path.
2. The only horizontal force acting on the mass is tension from the string, which pulls the mass towards the central fixed point.
3. Answer: Centripetal force is directed towards the central fixed point, and is provided entirely by the tension in the string.

## Equations for centripetal force

**Derivation:** Derive the magnitude of centripetal force from first principles

*Starting from:* Newton's second law $F_{net} = ma$ and the centripetal acceleration formula $a_c = \frac{v^2}{r} = \omega^2 r$

1. Centripetal force is the net force towards the centre, so substitute $a_c$ into Newton's second law:
2. $$F_c = m a_c = \frac{m v^2}{r}$$
3. To get the form in terms of angular velocity, substitute $v = \omega r$:
4. $$F_c = \frac{m (\omega r)^2}{r} = m \omega^2 r$$

*Conclusion:* We have two equivalent forms of the centripetal force equation, used depending on whether we know linear speed $v$ or angular speed $\omega$.

**Worked example:** A 1500 kg car drives around a flat circular bend of radius 40 m at a constant speed of 12 m/s. Calculate the magnitude of the centripetal force required to keep the car on the bend.

1. List known values: $m = 1500 \ \text{kg}$, $r = 40 \ \text{m}$, $v = 12 \ \text{m s}^{-1}$
2. Use the centripetal force equation for linear speed:
3. $$F_c = \frac{mv^2}{r} = \frac{1500 \times (12)^2}{40}$$
4. Calculate the result: $F_c = \frac{1500 \times 144}{40} = 5400 \ \text{N}$
5. This force is provided by friction between the car's tyres and the road surface.

## Sources of centripetal force

A critical point for exams: centripetal force is **not a new, separate force**. It is the net resultant of existing physical forces that acts towards the centre of the circle. It can be provided by a single force, or the vector sum of multiple forces.

- Tension: masses spinning on strings, ropes, or rods
- Gravity: planets orbiting stars, satellites orbiting planets
- Friction: cars and cyclists moving around flat bends
- Normal reaction: cars on banked tracks, roller coaster loop-the-loops
- Electrostatic force: electrons orbiting an atomic nucleus (Bohr model)

**Worked example:** A roller coaster car travels around a vertical circular loop of radius 10 m. At the top of the loop the car is upside down on the inside of the track. What forces contribute to the centripetal force at this point?

1. Draw a free-body diagram for the car at the top of the loop. Two vertical forces act:
2. 1. Weight of the car ($mg$), acting downwards, 2. Normal reaction from the track ($R$), also acting downwards (track is above the upside down car)
3. The centre of the loop is directly below the car at the top of the loop, so both forces point towards the centre.
4. Answer: Centripetal force is the sum of weight and normal reaction: $F_c = mg + R$, so both forces contribute.

## Common pitfalls

- **Wrong:** Treating centripetal force as an additional, separate force on a free-body diagram.
  - Why it fails: Centripetal force is just the net resultant of existing forces, not a new interaction between objects.
  - Correct: Draw all existing forces first, then sum their components directed towards the centre to get the net centripetal force.
- **Wrong:** Drawing centripetal force pointing outwards from the centre of the circle.
  - Why it fails: By definition, centripetal means 'centre-seeking', so direction is always towards the centre.
  - Correct: Always draw force components contributing to centripetal force pointing towards the centre of the circular path.
- **Wrong:** Substituting angular speed in degrees per second into $F_c = m\omega^2 r$.
  - Why it fails: The centripetal force equation is derived for angular speed measured in radians per second.
  - Correct: Always convert angular speed from degrees per second to radians per second before substitution.
- **Wrong:** Claiming centripetal force does work to keep the object moving in a circle.
  - Why it fails: Centripetal force is always perpendicular to instantaneous velocity, so work done = force × displacement in direction of force = zero.
  - Correct: State that centripetal force only changes the direction of velocity, not its magnitude, and does no work on the object.

## Cheatsheet

| Property | Formula | Key Fact |
| --- | --- | --- |
| Direction of $F_c$ | - | Always towards the centre of the circle |
| Magnitude (v known) | $F_c = \frac{mv^2}{r}$ | $v$ = instantaneous linear speed |
| Magnitude ($\omega$ known) | $F_c = m\omega^2 r$ | $\omega$ must be in radians per second |
| Nature of force | - | Not a separate force, it is the net resultant force |
| Common sources | - | Tension, gravity, friction, normal reaction |

## What's next

Centripetal force is the foundational concept for all circular motion problem-solving in CIE A-Level Physics. You will apply this core idea to more complex scenarios including vertical circular motion, conical pendulums, banked tracks, and orbital motion of satellites. Mastery of centripetal force is also required for later topics like rotational dynamics, simple harmonic motion, and gravitational fields. Practise identifying the source of centripetal force in different scenarios, as this is a very common exam question.

- [Centripetal acceleration](https://www.owlsprep.com/study/cie-9702-u16-centripetal-acceleration/)
- [Gravitational fields](https://www.owlsprep.com/study/cie-9702-u17-overview/)
- [Gravitational field concepts](https://www.owlsprep.com/study/cie-9702-u17-gravitational-field-concepts/)

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