Centripetal Acceleration Formula: a = v²/r = ω²r Explained

The centripetal acceleration formula is $a_c = \frac{v^2}{r}$ , speed squared over radius, or equivalently $a_c = \omega^2 r$ . It applies to every object that follows a circular path, and it always points from the object toward the centre of that circle.

It can feel odd that something moving at a steady speed is accelerating. The point is that acceleration measures any change in velocity, and velocity includes direction. A runner on a curved track keeps changing direction, so she has an acceleration even if her stopwatch speed never changes. Multiply this acceleration by mass and you get the centripetal force.

Quick facts

Formula
$a_c = \frac{v^2}{r}$
Other forms
$a_c = \omega^2 r = v\omega = \frac{4\pi^2 r}{T^2}$
SI unit
m/s²
Dimensional formula
[M⁰ L¹ T⁻²]
Direction
Toward the centre, perpendicular to velocity
Also called
Radial or normal acceleration
Class
Class 11 motion in a plane, AP Physics 1

The centripetal acceleration formula and its forms

Start from $a_c = \frac{v^2}{r}$ . Because $v = \omega r$ , you can swap in ω to get $a_c = \omega^2 r$ , or mix them as $a_c = v\omega$ . With the period T, $v = \frac{2\pi r}{T}$ and the formula becomes $a_c = \frac{4\pi^2 r}{T^2}$ . All four give the same number; choose the one that uses what the question gives.

Choosing a form
Given	Form to use
speed and radius	$\frac{v^2}{r}$
ω and radius	$\omega^2 r$
speed and ω	$v\omega$
period and radius	$\frac{4\pi^2 r}{T^2}$

Watch how r behaves. At fixed speed, a bigger circle means less acceleration. At fixed ω, a bigger circle means more. Both are true; they simply hold different things constant.

Why it points to the centre: a short derivation

Let a particle move at speed v round a circle. In a short time $\Delta t$ its position vector turns through angle $\Delta\theta$ , and so does its velocity vector, because velocity is always perpendicular to the radius. The two triangles (one of positions, one of velocities) are similar, so $\frac{|\Delta \vec{v}|}{v} = \frac{|\Delta \vec{r}|}{r}$ .

Divide both sides by $\Delta t$ and let it shrink. $\frac{|\Delta\vec{r}|}{\Delta t}$ tends to v, so $a = \frac{|\Delta\vec{v}|}{\Delta t} = \frac{v \cdot v}{r} = \frac{v^2}{r}$ . The change $\Delta\vec{v}$ points across the small angle toward the centre, which gives the direction.

Worked examples

Example 1: a cyclist on a circular track

A cyclist rides at 6 m/s round a track of radius 12 m. Find the centripetal acceleration.

$a_c = \frac{6^2}{12} = \frac{36}{12} = 3\ \text{m/s}^2$

Example 2: a lab centrifuge in rpm

A centrifuge spins at 3000 rpm. A sample tube sits 10 cm from the axis. Find its acceleration and compare it with g.

$\omega = \frac{2\pi \times 3000}{60} = 100\pi \approx 314\ \text{rad/s}$ , r = 0.10 m.

$a_c = \omega^2 r = (314.2)^2 \times 0.10 \approx 9.87 \times 10^3\ \text{m/s}^2$

That is about 1000 times g, which is why centrifuges separate liquids so fast.

Example 3: the Moon around Earth

The Moon orbits at about 3.84 × 10⁸ m from Earth's centre with a period of 27.3 days. Find its centripetal acceleration.

T = 27.3 × 86400 ≈ 2.36 × 10⁶ s. $a_c = \frac{4\pi^2 r}{T^2} = \frac{39.48 \times 3.84 \times 10^8}{(2.36 \times 10^6)^2} \approx 2.72 \times 10^{-3}\ \text{m/s}^2$ .

This matches $\frac{g}{60^2}$ quite closely, because the Moon is about 60 Earth radii away and gravity falls as $\frac{1}{r^2}$ .

Units and a quick dimension check

m²/s² divided by m leaves m/s², the same unit as any other acceleration. In dimensions, $\frac{(LT^{-1})^2}{L}$ simplifies to length over time squared, with no mass at all. A check like this catches the common slip of writing $\frac{v}{r}$ , which has the unit of ω, 1/s, rather than an acceleration.

Centripetal, tangential and total acceleration

If the speed is also changing, the object has a second part of acceleration along the path, the tangential acceleration $a_t = r\alpha$ . The two parts are at right angles, so the total is $a = \sqrt{a_c^2 + a_t^2}$ . Many books also call $a_c$ the radial acceleration to stress that it lies along the radius. The circular motion hub lists the related results together.

In uniform circular motion the speed is constant but the acceleration is not: its direction keeps turning. Call it constant in magnitude only.

How big is it in everyday motion

Rough sizes, worked from a = v²/r
Motion	v and r	Acceleration, approx.
Car on a town roundabout	10 m/s, 20 m	5 m/s²
Child on a playground swing, lowest point	4 m/s, 2.5 m	6.4 m/s²
Point on Earth's equator from daily spin	465 m/s, 6.37 × 10⁶ m	0.034 m/s²
Low orbit satellite	7.67 km/s, 6.77 × 10⁶ m	8.7 m/s²

Comparing an answer with this table is a fast sense check. A bicycle turn that comes out at 300 m/s² means a unit was not converted, while a satellite value close to g is exactly what you expect, because gravity is what bends its path. The satellite case is developed on the orbital velocity page.

Errors students make

Saying the acceleration is zero because the speed is constant.
Drawing the arrow outward. Centripetal means centre seeking; it points in.
Using rpm or rev/s as ω. Convert to rad/s first.
Plugging in the diameter, or a radius in centimetres.

Frequently asked questions

What is the formula for centripetal acceleration?

It is a = v² / r. Using angular velocity, the same value is a = ω²r, and with the period it is a = 4π²r / T².

In what unit is centripetal acceleration measured?

Like every acceleration, it is measured in metres per second squared, m/s².

Can an object have centripetal acceleration at constant speed?

Yes. Its direction of motion is changing, so its velocity changes, and that change per second is the centripetal acceleration.

Is centripetal acceleration the same as radial acceleration?

For motion on a circle they are the same thing: the part of the acceleration along the radius, pointing to the centre. Radial acceleration is the more general name used when the path is any curve.

What happens to centripetal acceleration if the speed doubles?

It becomes four times larger on the same circle, because the formula depends on v squared.