Centripetal Acceleration

Suppose that we have an object moving in uniform circular motion, with the radius of the circle denoted $R$ and the constant angular velocity denoted $\omega$.

Let $x(t)$, $v(t)$, and $a(t)$ denote the position, velocity, and acceleration at time $t$. Note that all three of these are vector-valued functions.

We choose a convenient origin and $t=0$ point so that the following holds: $$x(t)=\left<R\cos(\omega t),R\sin(\omega t)\right>$$ Differentiating, $$v(t)=\left<-R\omega\sin(\omega t),R\omega\cos(\omega t)\right>$$ $$a(t)=\left<-R\omega^2\cos(\omega t),-R\omega^2\sin(\omega t)\right>$$ Notice that the acceleration is always in the opposite direction of the position.

This acceleration is called the centripetal acceleration, or the center-seeking force, since it is always pointed toward the center of rotation. It is this acceleration that keeps the object rotating.

Calculating the magnitude of the acceleration and denoting this $a$, $$\left|a(t)\right|=\left|\left<-R\omega^2\cos(\omega t),-R\omega^2\sin(\omega t)\right>\right|$$ $$\left|a(t)\right|=\sqrt{(-R\omega^2\cos(\omega t))^2+(-R\omega^2\sin(\omega t))^2}$$ $$a=\omega^2 R$$ Thus, the centripetal acceleration is equal to the angular velocity squared times the radius.

One may alternatively express $a$ as follows: $$a=\omega^2 R=\frac{v^2}{R}$$ where $v=\omega R$ is the magnitude of the velocity.