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

  1. Jan 16, 2016 #1
    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. jcsd
  3. Jan 16, 2016 #2

    haruspex

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    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}##.
     
  4. Jan 16, 2016 #3
    Thank you for pointing out that I forgot to do a basic differentiation. It is much appreciated.
     
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