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:

\vec{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:

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

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

Finding the Acceleration

Differentiating the velocity gives us acceleration:

\begin{align*} \vec{a}(t) &= \frac{d\vec{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\vec{r}(t) \end{align*}

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