Physics/Option Modules/Engineering Physics/Centripetal Acceleration
Physics · Engineering Physics · A2 Level

Centripetal Acceleration

A component on a spinning centrifuge can move at a perfectly constant speed and still be accelerating hard — a fact that trips up almost everyone the first time they meet it. Understanding why is the key to everything involving circular motion in engineering, from bearing loads to the stresses inside a flywheel's rim.

7 min read

Why moving in a circle always means accelerating

Acceleration is the rate of change of velocity — and velocity is a vector, with both a magnitude (speed) and a direction. An object moving in a circle at absolutely constant speed still has its velocity constantly changing, because the direction of travel is continuously rotating to keep following the curved path. Since velocity is changing, the object is, by definition, accelerating — even though its speed never changes at all.

This acceleration is directed toward the centre of the circular path at every instant, which is why it's called centripetal acceleration (from Latin, "centre-seeking"). It's the acceleration responsible for continuously bending what would otherwise be straight-line motion into a circular path.

The misconception this lesson exists to correct

Constant speed does not mean zero acceleration. It only means zero tangential acceleration (no speeding up or slowing down along the direction of travel). An object moving at constant speed around any curved path still has a nonzero acceleration directed toward the centre of curvature, because its direction of motion is changing. Circular motion at constant speed is accelerated motion.

Deriving the centripetal acceleration formula

The magnitude of centripetal acceleration can be found from the geometry of how the velocity vector itself rotates as an object moves round a circular path.

DerivationCentripetal acceleration: $a = v^2/r = \omega^2 r$
1
Consider an object moving at constant speed $v$ around a circle of radius $r$. Over a short time interval $\Delta t$, it sweeps through a small angle $\Delta\theta$, moving from point A to point B.
Setting up the small-angle geometry needed for the derivation.
2
The velocity vector at A and at B both have magnitude $v$, but point in different directions — the direction has rotated through the same small angle $\Delta\theta$ as the position vector.
Velocity is always tangent to the circular path, so it rotates in step with the position.
3
Drawing the two velocity vectors tip-to-tail forms an isosceles triangle with two sides of length $v$ and an included angle $\Delta\theta$; for small $\Delta\theta$, the change in velocity has magnitude $\Delta v \approx v\Delta\theta$.
Small-angle approximation: the chord of a triangle with two sides $v$ and included angle $\Delta\theta$ has length $\approx v\Delta\theta$ when $\Delta\theta$ is small.
4
Divide by $\Delta t$: $\dfrac{\Delta v}{\Delta t} \approx v\dfrac{\Delta\theta}{\Delta t} = v\omega$
This ratio is the magnitude of the acceleration, and $\Delta\theta/\Delta t$ is the angular velocity $\omega$.
5
So $a = v\omega$. Using $v = \omega r$ (from the linear–angular relation) gives two equivalent forms.
Substitute $v = \omega r$ into $a = v\omega$, or substitute $\omega = v/r$ instead, to get both versions.
6
$a = \dfrac{v^2}{r} = \omega^2 r$
Both forms are equivalent and useful depending on which quantities — $v$ or $\omega$ — are given in a problem.
Derived

As $\Delta t \to 0$, the direction of this acceleration points exactly toward the centre of the circle, confirming it as centripetal. Both forms of the result are used constantly in engineering contexts: $a = \omega^2 r$ is convenient when a component's rotational speed is known directly (as for a flywheel or rotor), while $a = v^2/r$ is convenient when a linear speed is known instead (as for a vehicle on a curved track).

Worked exampleCentripetal acceleration in a centrifuge
1
A laboratory centrifuge spins a sample at radius $r = 0.10$ m with angular velocity $\omega = 3140$ rad s$^{-1}$ (about 30,000 rpm). Find the centripetal acceleration of the sample.
$\omega$ is given directly, so use $a = \omega^2 r$.
2
$a = \omega^2 r = (3140)^2 \times 0.10 \approx 9.86 \times 10^5$ m s$^{-2}$
Substitute directly; note how large this is compared with $g \approx 9.81$ m s$^{-2}$.
3
Express this as a multiple of $g$.
Centrifuge performance is often quoted in "g-force" for comparison with everyday experience.
4
$\dfrac{9.86\times10^5}{9.81} \approx 1.0 \times 10^5\ g$
A realistic figure for a high-speed laboratory centrifuge, illustrating why centripetal acceleration matters for engineering design.
Result

Key takeaways

  • Any object moving in a circle is accelerating, even at constant speed, because its direction of travel is continuously changing.
  • This acceleration, called centripetal acceleration, is always directed toward the centre of the circular path.
  • The small-angle geometry of the changing velocity vector gives $a = v\omega$, which combines with $v=\omega r$ to give $a = v^2/r = \omega^2 r$.
  • Use $a = \omega^2 r$ when angular velocity is known; use $a = v^2/r$ when linear (tangential) speed is known.
  • Constant speed does not mean zero acceleration — this is the single most important idea in this lesson and a very common source of exam errors.
  • Centripetal acceleration is essential in engineering design wherever components rotate at speed, from centrifuges to flywheels.
  • The next lesson uses this result to find the centripetal force needed to produce this acceleration in real rotating components.
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