On the last page we said the centripetal force causes an acceleration towards the centre — but we never worked out how big that acceleration is. Here we do. The surprising part is that an object can accelerate while its speed never changes, purely because its direction keeps turning. That direction-only acceleration is the centripetal acceleration, and it always points to the centre of the circle.
📘 What you need to know
An object in uniform circular motion has an acceleration directed towards the centre of the circle
This centripetal acceleration is always perpendicular to the velocity
Its magnitude is a = v2/r = ω2r
Using the period, it can also be written a = 4π2r/T2
The acceleration exists because the direction of the velocity changes, even though its size stays constant
By Newton’s second law, this acceleration is caused by the centripetal force: F = ma
Accelerating without speeding up
This sounds like a contradiction, so let’s clear it up straight away. Acceleration is the rate of change of velocity — and velocity is a vector, carrying both a size and a direction. Change either one and the velocity has changed, which means there’s an acceleration.
In uniform circular motion the size of the velocity (the speed) is constant, but the direction swings round continuously. So the object is accelerating the whole time, even though a speedometer would read a steady value. And that acceleration points straight to the centre of the circle.
The velocity (red) is tangent to the circle; the centripetal acceleration (purple) points inward to the centre. They meet at a right angle at every point on the path.
Here’s an everyday feel for it. When a car turns a corner at steady speed, you’re thrown towards the outside of the bend — but that’s your body trying to keep going straight. The car (and your seat) is actually accelerating inward, and it drags you round with it. That inward acceleration is exactly the centripetal acceleration.
Where the equation comes from
You don’t have to memorise the derivation, but seeing it makes the equation stick. Picture the object at two points a short moment apart, with velocity vectors v1 and v2. The speed is the same at both points, so the two arrows are the same length — only their direction differs.
Left: two velocity arrows tangent to the circle a moment apart. Right: drawn from a common point, they form an isosceles triangle whose short side is the velocity change Δv — and it points inward, towards the centre.
Here’s the clever bit. The triangle formed by v1, v2 and Δv has the same shape as the triangle formed by the two radii and the small arc between the points — they’re similar triangles. Matching up their sides gives Δv/v = Δs/r, where Δs is the little bit of arc travelled. Dividing through by the time Δt and remembering that Δs/Δt = v, the whole thing tidies up into a beautifully simple result.
Centripetal accelerationa = v2 ÷ r
And because the arrow Δv points inward, so does the acceleration — confirming it’s directed to the centre.
The other forms of the equation
Using v = rω from the angular-velocity page, you can rewrite the acceleration in terms of angular speed:
Using angular speeda = ω2r
And since ω = 2π/T, substituting gives a version in terms of the period T — handy whenever you know how long one revolution takes:
Using the perioda = 4π2r ÷ T2
All three describe the same acceleration — you just pick whichever matches the quantities the question gives you: a linear speed, an angular speed, or a period.
Watch out for how the radius behaves in the two main forms. In a = v2/r the radius is on the bottom, so for a fixed speed, a tighter circle means more acceleration. But in a = ω2r the radius is on top, so for a fixed angular speed, a bigger circle means more acceleration. Both are correct — they just hold different things constant. Read the question to see which quantity is fixed.
Linking back to force
Newton’s second law ties this straight back to the centripetal force from the previous page. Since F = ma, multiplying the acceleration by the mass gives the force:
Force from accelerationF = ma = mv2/r = mrω2
So the centripetal force equation you met before is simply the centripetal acceleration with a mass attached. Both point to the centre; both describe the same inward “turning” of the motion.
🛠️ Choosing the right form
Given a linear speed v? Use a = v2/r.
Given an angular speed ω? Use a = ω2r.
Given a period T or frequency f? Use a = 4π2r/T2 (or find ω = 2πf first).
Need the force too? Just multiply by the mass: F = ma.
WE 1
A car rounds a bend of radius 45 m at a steady 15 m s−1. Find its centripetal acceleration.
Step 1 — we have a speed, so use a = v²/r
a = v² ÷ r
Step 2 — substitute the valuesa = 15² ÷ 45 = 225 ÷ 45a = 5.0 m s⁻²Directed towards the centre of the bend — about half of g.
WE 2
A fairground ride spins riders in a horizontal circle of radius 6.0 m, completing one turn every 5.0 s. Find the centripetal acceleration and compare it to g.
Step 1 — find the angular speed from the periodω = 2π ÷ T = 2π ÷ 5.0 = 1.26 rad s⁻¹Step 2 — use a = ω²ra = 1.26² × 6.0a = 9.5 m s⁻²That’s about 0.97g — nearly the strength of gravity, which is why the ride feels so intense.
WE 3
A person stands on the equator, carried round by the Earth’s rotation. The Earth’s radius is 6.37 × 106 m and one rotation takes 24 hours. Find their centripetal acceleration.
Step 1 — we have a radius and a period, so use a = 4π²r/T²
T = 24 × 3600 = 86400 s
Step 2 — substitute the valuesa = 4π² × (6.37 × 10⁶) ÷ 86400²a = 0.034 m s⁻²Tiny — only about 0.3% of g — which is why you don’t feel yourself being whirled around by the spinning Earth.
WE 4
A ball on a string moves in a circle of radius 1.5 m with an angular speed of 3.5 rad s−1. By what factor does the centripetal acceleration change if both the radius and the angular speed are doubled?
Step 1 — look at a = ω²r
a depends on r once, and on ω squared
Step 2 — apply the doublingdoubling r → ×2; doubling ω → ×2² = ×4a increases by a factor of 8Numbers check: original a = 3.5² × 1.5 = 18.4; new a = 7.0² × 3.0 = 147 m s⁻², which is 8× bigger.
💡 Top tips
Direction, not speed. The acceleration exists because the velocity’s direction changes — it’s real even at constant speed.
Match the form to the data — v2/r for a speed, ω2r for an angular speed, 4π2r/T2 for a period.
Mind where r sits. It’s on the bottom in v2/r but on top in ω2r — check which quantity is held fixed.
Force is one step away. Multiply the acceleration by the mass and you have the centripetal force.
Quick recap: Centripetal acceleration points to the centre and comes from the velocity’s changing direction, not its size. Its magnitude is a = v2/r = ω2r = 4π2r/T2. Multiply by mass and you get the centripetal force, F = ma.
⚠ Common mistakes
Thinking there’s no acceleration because the speed is constant — the direction changes, so there is
Drawing the acceleration along the velocity (tangent) rather than towards the centre
Forgetting the speed is squared in v2/r
Confusing the two forms — putting r on the wrong side because you mixed up v2/r and ω2r
Using an angle in degrees when finding ω — circular motion needs radians
So far every circle we’ve drawn has been at constant speed — uniform circular motion, with the acceleration always pointing dead centre. But real circles aren’t always so tidy: swing a ball on a string in a vertical loop and gravity speeds it up at the bottom and slows it at the top. That’s next: Non-Uniform Circular Motion, where the speed itself changes as you go round.
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