Derivation of centripetal acceleration

Problem

Starting from the vector definition of acceleration $\vv{a} = \dd\vv{v}/\dd t$ and considering two instants separated by a small interval $\Delta t$ along a circle, show that in uniform circular motion the magnitude of the acceleration is $a_c = v^2/r = \omega^2 r$ and is directed towards the centre.

Solution

Set-up. In uniform circular motion the magnitude of the velocity $v$ is constant, but the vector $\vv{v}$ continuously changes direction: it is always tangent to the circle. There is therefore acceleration, because the direction of $\vv{v}$ varies.

Two nearby instants. Consider the position at time $t$ and at time $t+\Delta t$. In the interval $\Delta t$ the radius sweeps out a small angle $\Delta\theta$. Since $\vv{v}$ is perpendicular to the radius at every point, the two velocity vectors $\vv{v}_1$ and $\vv{v}_2$ form between them the same angle $\Delta\theta$ as the two radii.

Velocity triangle. Draw $\vv{v}_1$ and $\vv{v}_2$ from the same origin. They have the same magnitude $v$ and form the angle $\Delta\theta$; the change vector $\Delta\vv{v} = \vv{v}_2 - \vv{v}_1$ is the base of an isosceles triangle. For small angles the length of the base is the arc of radius $v$: $\abs{\Delta\vv{v}} \approx v\,\Delta\theta$ This is entirely analogous to the geometric relation on the circle, where the arc length travelled is $\Delta s = r\,\Delta\theta$.

Magnitude of the acceleration. Dividing by $\Delta t$ and taking the limit $\Delta t\to 0$: $a_c = \lim_{\Delta t\to 0}\frac{\abs{\Delta\vv{v}}}{\Delta t} = v\,\lim_{\Delta t\to 0}\frac{\Delta\theta}{\Delta t} = v\,\omega$ where $\omega = \dd\theta/\dd t$ is the angular velocity. Using the fundamental relation $v = \omega r$: $a_c = v\,\omega = (\omega r)\,\omega = \omega^2 r \qquad\text{and also}\qquad a_c = v\,\omega = v\cdot\frac{v}{r} = \frac{v^2}{r}$

$\ev{a_c = \dfrac{v^2}{r} = \omega^2 r}$

Direction. In the limit $\Delta t\to 0$ the angle $\Delta\theta\to 0$: the isosceles triangle becomes infinitely thin and the base $\Delta\vv{v}$ tends to become perpendicular to $\vv{v}$, i.e. directed along the radius. Looking at the sign, $\Delta\vv{v}$ points from the point inwards: the acceleration is centripetal, directed towards the centre of the circle. It is the acceleration that continually “curves” the trajectory without changing the magnitude of the velocity.

Links

Topics: Dynamics Concepts: Uniform circular motion · Centripetal acceleration · Angular velocity Skills: Symbolic set-up · Vector resolution · Use of trigonometry