# How to calculate jerk in uniform circular motion?

We can calculate the centripetal acceleration in circular motion by the equation v^2/r. But how do we calculate the jerk (which is acceleration over time)?

• Jerk is "rate of change of acceleration over time", not "acceleration over time" (which doesn't really make sense). Commented Aug 5 at 21:58

Consider the trajectory $$\vec{r} = r \begin{pmatrix} \sin(\omega t) \\ \cos(\omega t) \end{pmatrix}$$ Differentiate once $$\vec{v} = \omega r\begin{pmatrix} \cos(\omega t) \\ -\sin(\omega t) \end{pmatrix}$$ Differentiate twice $$\vec{a} = -\omega^2 r\begin{pmatrix} \sin(\omega t) \\ \cos(\omega t) \end{pmatrix} = - \omega^2 \vec{r} = -\frac{v^2}{r} \hat r$$ Differentiate thrice $$\vec{j} = - \omega^3 r\begin{pmatrix} \cos(\omega t) \\ -\sin(\omega t) \end{pmatrix} = - \omega^2 \vec{v} = - \frac{v^3}{r^2} \hat v$$ and you got your jerk. The hat indicates a normalized vector. Also note (thank you @JEB) that since $$\vec{a} = const. \vec{r}$$, the derivatives repeat themselves periodically modulo a multiplicative constant $$-\omega^2$$ or, if you like normalized vectors, $$\frac{v^2}{r}$$. The $$2n^{\text{th}}$$ derivative thus reads $$\frac{\partial^{2n}}{\partial t^{2n}} \vec r = \left( -\omega^2 \right)^n \vec r = ( -1)^n \frac{v^{2n}}{r^{2n-1}} \hat r$$ and the $$(2n+1)^{\text{st}}$$ $$\frac{\partial^{2n+1}}{\partial t^{2n+1}} \vec r = \left( -\omega^2 \right)^n \vec v = ( -1)^n \frac{v^{2n+1}}{r^{2n}} \hat r$$
• you missed a change to note that the n-th derivative of position goes like $v^n/r^{n-1}$.