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Derivative of a cross product

  1. Dec 26, 2007 #1
    [SOLVED] derivative of a cross product

    1. The problem statement, all variables and given/known data
    In some lecture notes I'm reading they jump straight from [tex] \frac{d}{d\mathbf{r}}( \frac{m}{2} |\mathbf{\omega}\times\mathbf{r}|^2)[/tex]
    to
    [tex]\mathbf{r}\omega^2-\mathbf{\omega}(\mathbf{\omega}.\mathbf{r})[/tex]


    2. Relevant equations



    3. The attempt at a solution

    It's easy to check this by writing out each component but this is messy so there's probably an easier way to do it. Any help appreciated.
    Thanks
     
    Last edited: Dec 26, 2007
  2. jcsd
  3. Dec 26, 2007 #2

    HallsofIvy

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    Surely not! There must be an m/2 in there!
    Ignoring the constant m/2, [itex]|\omega\times r|^2= (\omega\times r)\cdot(\omega\times\r)[/itex]. Take the derivative of that, apply the product rule- which is true, for vectors, for both dot product and cross product: [itex](\vec{u}\cdot\vec{v})'= \vec{u} '\cdot\vec{v}+ \vec{u}\cdot\vec{v} '[/itex] and [itex](\vec{u}\times\vec{v})'= \vec{u} '\times\vec{v}+ \vec{u}\times\vec{v} '[/itex].
     
  4. Dec 26, 2007 #3
    opps I forgot an m in the answer: sorry.

    Thanks, I've solved it now.
     
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