Radial Acceleration Formula: aᵣ = v²/r = ω²r with Examples

The radial acceleration formula is $a_r = \omega^2 r$ , which you can also write as $a_r = \frac{v^2}{r}$ using the speed. It is the part of a body's acceleration that lies along the radius of its curved path and points in toward the centre of curvature.

For motion on a circle, radial acceleration and centripetal acceleration are the same thing. The word radial is used when a problem splits acceleration into two perpendicular parts: radial (along r) and tangential (along the path). This split is how JEE and AP problems on non uniform circular motion are set up.

Quick facts

Formula
$a_r = \omega^2 r = \frac{v^2}{r}$
Direction
Along the radius, toward the centre
SI unit
m/s²
Dimensional formula
[M⁰ L¹ T⁻²]
Partner
Tangential $a_t = r\alpha$
Total
$a = \sqrt{a_r^2 + a_t^2}$

Using the radial acceleration formula

Use $\omega^2 r$ when the question gives a spin rate and a radius. Use $\frac{v^2}{r}$ when it gives a speed. They agree because $v = \omega r$ . On a rigid spinning body all points share ω, so $a_r$ grows in direct proportion to r: the rim of a disc has the largest radial acceleration.

Same quantity, three inputs
Known values	Expression
ω, r	$\omega^2 r$
v, r	$\frac{v^2}{r}$
v, ω	$v\omega$

Worked examples

Example 1: a point on a spinning disc

A disc spins at 50 rad/s. Find the radial acceleration of a point 6 cm from its centre.

r = 0.06 m. $a_r = \omega^2 r = 50^2 \times 0.06 = 2500 \times 0.06 = 150\ \text{m/s}^2$

Example 2: a pendulum bob at the lowest point

A 0.4 kg bob on a 1.5 m string passes the lowest point at 3 m/s. Find its radial acceleration and the tension in the string.

$a_r = \frac{v^2}{r} = \frac{9}{1.5} = 6\ \text{m/s}^2$ , pointing up toward the pivot.

Newton's second law along the string: $T - mg = ma_r$ , so $T = 0.4(9.8 + 6) = 6.32\ \text{N}$ .

Example 3: radial plus tangential

A wheel of radius 0.5 m has ω = 4 rad/s and α = 6 rad/s² at one instant. Find the total acceleration of a rim point.

$a_r = 4^2 \times 0.5 = 8\ \text{m/s}^2$ ; $a_t = r\alpha = 0.5 \times 6 = 3\ \text{m/s}^2$ .

$a = \sqrt{8^2 + 3^2} = \sqrt{73} \approx 8.54\ \text{m/s}^2$

Radial and tangential parts

When the speed changes along a curve, the acceleration vector tilts away from the centre. Split it: the radial part $a_r$ changes direction and the tangential acceleration $a_t$ changes speed. The angle φ between the total acceleration and the radius obeys $\tan\phi = \frac{a_t}{a_r}$ . In Example 3 that is $\tan^{-1}(3/8) \approx 20.6^\circ$ .

Multiply the radial part by mass and you get the net inward force, the centripetal force. Multiply the tangential part by mass and you get the force along the path.

The vertical circle

A stone whirled in a vertical circle is the standard case where both parts matter. At the top, gravity points toward the centre and helps: $T + mg = \frac{mv^2}{r}$ . The least speed at the top that keeps the string tight is $\sqrt{gr}$ . At the bottom, gravity points away from the centre: $T - mg = \frac{mv^2}{r}$ . Between them, gravity also has a tangential part, which is why the speed is not constant.

For the stone to complete the loop on a string, it needs at least

\sqrt{5gr}

at the bottom.

Common mistakes

Calling radial acceleration outward. It points to the centre of curvature.
Using rev/s as ω. Multiply by 2π first.
Treating radial and tangential parts as one number by simple addition.
Taking r as the length of the arc instead of the radius.
Forgetting that on a non circular path, such as an ellipse, r means the local radius of curvature, which changes from point to point.

A quick way to judge an answer: a person walking briskly round a 5 m bend at 1.5 m/s has only about 0.45 m/s² of radial acceleration, while a point on a drill bit can reach thousands of m/s². If your figure for a gentle everyday motion is many times g, recheck the units of ω and r.

The angular velocity formula gives ω from rpm or a period, and the circular motion section lists the related pages.

Frequently asked questions

What is the formula for radial acceleration?

Radial acceleration is aᵣ = ω²r, which equals v²/r. It points along the radius toward the centre of the circle.

Is radial acceleration the same as centripetal acceleration?

On a circular path, yes. Radial is the name used when you also discuss the tangential part, especially for curves whose speed changes.

Can radial acceleration be zero?

Only if the path is straight or the body is momentarily at rest. Any curved motion with nonzero speed needs an inward acceleration.

How do radial and tangential accelerations combine?

They are perpendicular, so the total is √(aᵣ² + aₜ²) and its angle to the radius has tangent aₜ / aᵣ.