# What is the (higher order) time derivative of centripetal acceleration?

1. Jan 16, 2016

Just using basic dimensional analysis, it appears the time derivative of centripetal acceleration is $\vec{r} \omega^3$, but this intuitive guess would also extend to higher order time derivatives, no? Implying:

$\frac {d^n \vec{r}}{dt^n} = \vec{r} \omega^n$

It seems to follow from the general result shown in Thorton/Marion pg 390 (attached) when considering rotating bodies in a fixed frame. I assume $\vec{Q}$ is any vector, even ones that are the result of a higher order time derivative of an initial vector. The concept of finite infinite-time derivative just seems like an odd concept to me when considering real objects, but I guess the geometry of the situation allows it. But to confirm, is anything posted here incorrect?

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2. Jan 16, 2016

### haruspex

Dimensional analysis allows that as a possibility, but it does not make it true. If we start with the scalar form $r\omega^2$ we see the time derivative is $r\dot\omega^2+2r\omega\dot{\omega}$.

3. Jan 16, 2016