In uniform circular motion, an object moves around a circle of radius $r$ with constant speed $v$. Its speed is constant, but its velocity is not: the velocity vector is always tangent to the circle, so its direction continuously changes.

A changing velocity means there is an acceleration even though the speed does not change.

Period, frequency, and angular speed

If one revolution takes a time $T$, the period is $T$. The frequency is the number of revolutions per unit time,

$$f=\frac1T.$$

One revolution corresponds to $2\pi$ radians, so the angular speed is

$$\omega=\frac{2\pi}{T}=2\pi f.$$

The tangential speed is

$$v=\omega r.$$

Why the acceleration points inward

Consider two nearby points on the circular path separated by a small angle $\Delta\theta$. The velocity vectors at those points have the same magnitude $v$ but differ in direction by the same small angle. For small $\Delta\theta$,

$$|\Delta\mathbf v|\approx v,\Delta\theta.$$

During the same interval,

$$\Delta\theta\approx\frac{v,\Delta t}{r}.$$

Therefore

$$a=\lim_{\Delta t\to0}\frac{|\Delta\mathbf v|}{\Delta t} =\frac{v^2}{r}.$$

The change in velocity points toward the center of the circle, so the acceleration does too. This inward acceleration is the centripetal, or radial, acceleration:

$$a_c=\frac{v^2}{r}=\omega^2r.$$

It is perpendicular to the instantaneous velocity. Consequently it changes the direction of motion without changing the speed.

Example

A point on a rotor moves in a circle of radius $0.20,\mathrm m$ at $5.0$ revolutions per second. Then

$$\omega=2\pi f=10\pi,\mathrm{rad/s},$$

$$v=\omega r\approx6.28,\mathrm{m/s},$$

and

$$a_c=\omega^2r\approx197,\mathrm{m/s^2}.$$

That is about twenty times Earth's surface gravitational acceleration. Large accelerations can therefore arise purely from rapid changes in the direction of velocity.

Centripetal acceleration is a kinematic requirement of circular motion. It does not by itself identify what physical force produces that acceleration.