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Jerk times velocity cross position

  1. Aug 29, 2013 #1

    d2x

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    The problem asked me to prove the following (where [itex] \vec{r}, \vec{v}, \vec{a} [/itex] are the position, velocity and acceleration vectors of a moving particle):

    [itex]\frac{d}{dt} [\vec{a} \cdot (\vec{v} \times \vec{r})] = \dot{\vec{a}} \cdot (\vec{v} \times \vec{r}) [/itex]

    I already did so, but my question is what does the right hand side of the equation say about the motion of the particle? What is the physical meaning of this expression?
     
  2. jcsd
  3. Sep 1, 2013 #2

    lightgrav

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    Homework Helper

    There is NO general meaning for this - you can tell because it includes the location vector, which in general would be from an arbitrary origin. That would have arbitrary length in an arbitrary direction, so the scalar end result (even before taking the time-derivative) would be arbitrary.
    Interesting special-case would be a 3-d simple pendulum (mass on a stick) at constant distance from origin.
     
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