Why the Force on a Body Undergoing Circular Motion is Centripetal

When observing objects in circular motion, from planets orbiting the sun to electrons in particle accelerators, we find they all require a force directed toward the center of their circular path. This centripetal force is not a new type of force, but rather the net force required to maintain circular motion. Let's derive why this must be the case using vector calculus.

Setting Up the Problem

Consider a particle moving in a circular path of radius $r$ with constant angular velocity $\omega$. We can describe its position using the vector:

$$\mathbf{r}(t)=r\cos (\omega t)\hat{i}+r\sin (\omega t)\hat{j}$$

where $\hat{i}$ and $\hat{j}$ are unit vectors in the $x$ and $y$ directions respectively.

Finding the Velocity

To find the velocity, we differentiate the position vector with respect to time:

$$\mathbf{v}(t)=\frac{d\mathbf{r}}{dt}=-r\omega \sin (\omega t)\hat{i}+r\omega \cos (\omega t)\hat{j}$$

Note that the magnitude of velocity is $|\mathbf{v}|=r\omega$, which is constant, confirming uniform circular motion.

Finding the Acceleration

Differentiating the velocity gives us acceleration:

$$\begin{array}{rl}\mathbf{a}(t)& =\frac{d\mathbf{v}}{dt}\\ & =-r{\omega }^2\cos (\omega t)\hat{i}-r{\omega }^2\sin (\omega t)\hat{j}\\ & =-{\omega }^2[r\cos (\omega t)\hat{i}+r\sin (\omega t)\hat{j}]\\ & =-{\omega }^2\mathbf{r}(t)\end{array}$$

, i.e. the acceleration vector is proportional to the negative of the position vector.