Tangential Acceleration Formula: aₜ = rα with Worked Examples

The tangential acceleration formula is $a_t = r\alpha$ : the distance from the axis multiplied by the angular acceleration. Put another way, $a_t = \frac{dv}{dt}$ , the rate at which the speed along a curved path changes. It points along the tangent to the path, in the direction of motion when the body speeds up and against it when the body slows.

A body moving on a circle can have two separate accelerations. The inward one, $\frac{v^2}{r}$ , turns its direction. The tangential one changes how fast it goes. A car pulling away from traffic lights while still in a curve has both, and so does a point on a fan blade as the fan winds up. Read this page with the centripetal acceleration formula for the full picture.

Quick facts

Formula
$a_t = r\alpha$
Calculus form
$a_t = \frac{dv}{dt}$
SI unit
m/s²
Dimensional formula
[M⁰ L¹ T⁻²]
Direction
Along the tangent to the path
Total acceleration
$a = \sqrt{a_t^2 + a_c^2}$
Zero when
Speed is constant (uniform circular motion)

Where the tangential acceleration formula comes from

For a point at fixed distance r from the axis, its speed is $v = \omega r$ . Take the rate of change of both sides. r is constant, so $\frac{dv}{dt} = r\frac{d\omega}{dt}$ , which is $a_t = r\alpha$ . The radius converts an angular rate (rad/s²) into a linear one (m/s²).

Every point on a rigid wheel shares one α, but points at larger r have larger $a_t$ . The rim of a 0.3 m wheel has three times the tangential acceleration of a point 0.1 m from the axle. The angular acceleration formula gives α if the question supplies spin rates and a time.

Worked examples

Example 1: a disc winding up

A disc of radius 0.2 m has an angular acceleration of 5 rad/s². Find the tangential acceleration of a point on its rim.

$a_t = r\alpha = 0.2 \times 5 = 1.0\ \text{m/s}^2$

Example 2: a car speeding up on a curve

A car on a bend of radius 50 m speeds up steadily from 10 m/s to 20 m/s in 4 s. Find its tangential acceleration and its total acceleration at the moment its speed is 20 m/s.

Tangential: $a_t = \frac{20 - 10}{4} = 2.5\ \text{m/s}^2$ .

Centripetal at 20 m/s: $a_c = \frac{20^2}{50} = 8\ \text{m/s}^2$ .

$a = \sqrt{2.5^2 + 8^2} = \sqrt{70.25} \approx 8.38\ \text{m/s}^2$

The total points mostly inward, at $\tan^{-1}(2.5/8) \approx 17^\circ$ ahead of the radius.

Example 3: finding α from a rim reading

The edge of a 30 cm diameter grinding wheel has a tangential acceleration of 1.2 m/s². Find α.

r = 0.15 m (half the diameter). $\alpha = \frac{a_t}{r} = \frac{1.2}{0.15} = 8\ \text{rad/s}^2$ .

Tangential, centripetal and total acceleration

Two perpendicular parts of acceleration on a curve
Part	Formula	Direction	What it changes
Tangential	$a_t = r\alpha = \frac{dv}{dt}$	Along the path	Speed
Centripetal (radial)	$a_c = \frac{v^2}{r} = \omega^2 r$	Toward the centre	Direction
Total	$\sqrt{a_t^2 + a_c^2}$	In between	Both

As the two components are perpendicular, you combine them like the sides of a right triangle, never by simple addition. The radial acceleration page covers the inward part in more depth, and the circular motion hub ties all three together.

Special cases to recognise

Uniform circular motion: speed is constant, so aₜ = 0 and only the inward part remains.
Straight line motion: the radius is infinite, the inward part vanishes, and aₜ is the whole acceleration.
Vertical circle: gravity has a component along the path, mg sin θ, so a stone on a string has aₜ = g sin θ at angle θ from the lowest point.
Rolling without slipping: the centre of a wheel of radius R has linear acceleration a = Rα, the same relation.

Units and errors to avoid

m × rad/s² gives m/s², because the radian drops out. So aₜ shares both its unit and its dimensions, [M⁰ L¹ T⁻²], with every linear acceleration.

Using diameter in place of radius.
Writing $a_t = \frac{v^2}{r}$ . That is the inward part, not the tangential one.
Adding aₜ and ac as plain numbers. Use the square root of the sum of squares.
Assuming aₜ is zero on every circle. Only constant speed makes it zero.

Quick test: if the speed on the speedometer is changing, there is a tangential acceleration. If the steering wheel is turned, there is a centripetal one.

Frequently asked questions

What is the formula of tangential acceleration?

It is aₜ = rα, the radius times the angular acceleration. Equivalently aₜ = dv/dt, the rate of change of speed along the path.

What is the difference between tangential and centripetal acceleration?

Tangential acceleration lies along the path and changes the speed. Centripetal acceleration points to the centre and changes the direction. They are perpendicular.

How do you find total acceleration in circular motion?

Combine the two perpendicular parts: a = √(aₜ² + ac²), where ac = v²/r.

Is tangential acceleration zero in uniform circular motion?

Yes. The speed does not change, so aₜ = 0 and the only acceleration is the inward v²/r.

What unit does tangential acceleration have?

Metres per second squared, m/s², like any linear acceleration.