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In the previous lesson we discovered that an object moving in a circle at constant speed has a changing velocity, because the direction of motion is continuously turning. A changing velocity means an acceleration, and by Newton's second law an acceleration requires a net force. This lesson derives the famous formula for centripetal acceleration and applies it to real situations: cars on bends, buckets swung overhead, conical pendulums and planets in orbit.
This is OCR A-Level Physics A Module 5.2 — the heart of circular motion.
The word centripetal comes from the Latin centrum (centre) + petere (to seek). A centripetal force is any force that seeks the centre — it points from the moving object towards the centre of the circle it is travelling on.
Centripetal force is not a new kind of force. It is whatever force happens to be providing the inward pull in a given situation:
| Situation | Centripetal force provided by |
|---|---|
| Ball on a string swung in a horizontal circle | Tension in the string |
| Car rounding a flat bend | Friction between tyre and road |
| Earth orbiting the Sun | Gravitational attraction |
| Electron in a Bohr orbit | Electrostatic attraction |
| Rider on a fairground wall of death | Normal contact force from wall |
Exam Tip: When an OCR question asks "What provides the centripetal force?", the correct answer is the physical force — tension, friction, gravity, normal reaction, lift. Writing "centripetal force" is never the answer and will score zero.
Consider an object moving at constant speed v in a circle of radius r. In a short time Δt it sweeps through a small angle Δθ. Its velocity vector changes from v₁ to v₂, both of magnitude v, but pointing in slightly different directions.
The change in velocity, Δv, is the vector v₂ − v₁. For small Δθ this vector points approximately towards the centre of the circle, and its magnitude is
|Δv| ≈ v × Δθ
(by the small-angle geometry of an isosceles triangle with two sides of length v and an apex angle Δθ).
The acceleration is
a = |Δv| / Δt = v × (Δθ/Δt) = v × ω
and using v = rω (from the previous lesson) we can eliminate ω to get two equivalent forms:
a = v²/r = rω²
Both formulae give the same number; you choose whichever fits the data you are given.
Because Δv points towards the centre of the circle, so does the acceleration. The centripetal acceleration is always radially inward, perpendicular to the instantaneous velocity. It has no tangential component — that is why the speed remains constant even though the velocity is changing.
graph TD
C((Centre)) --- O[Object on circle]
O -. velocity v, tangential .-> T[Tangent direction]
O -. acceleration a, radial .-> C
Applying Newton's second law F = ma to the centripetal acceleration gives
F = mv²/r = mrω²
This is the resultant force required to keep a body of mass m moving at speed v (or angular velocity ω) on a circle of radius r. It acts towards the centre of the circle.
Notice that centripetal force depends on the square of the speed. Doubling the speed quadruples the required force. This is why sharp bends become dangerous so quickly: a driver entering a corner at 60 mph instead of 30 mph needs four times as much friction.
A 0.25 kg ball is swung in a horizontal circle of radius 0.80 m at a constant speed of 4.0 m s⁻¹. Calculate (a) the centripetal acceleration and (b) the tension in the string.
(a)
a = v²/r = (4.0)² / 0.80 = 16 / 0.80 = 20 m s⁻²
(b)
F = ma = 0.25 × 20 = 5.0 N
Since the only horizontal force on the ball is the string tension, T = 5.0 N.
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