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

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TheCanadian
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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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TheCanadian said:
Just using basic dimensional analysis, it appears the time derivative of centripetal acceleration is r⃗ ω3r→ω3 \vec{r} \omega^3 , but this intuitive guess would also extend to higher order time derivatives, no? Implying:
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}##.
 
haruspex said:
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}##.

Thank you for pointing out that I forgot to do a basic differentiation. It is much appreciated.