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Derivative of cross product

  1. May 16, 2010 #1
    1. The problem statement, all variables and given/known data

    http://damtp.cam.ac.uk/user/dt281/dynamics/two.pdf" [Broken]

    Looking at page 5, equations (2.19) and (2.20)

    3. The attempt at a solution

    I cannot understand how they derived the (2.20), at first from comparing the solutions I had assumed r(dot)' had disappeared as we were differentiating with respect to dr'.

    I then went about the derivative of the cross product:

    Omega X r(dot)' + .... But in the solution we find Omega x (Omega X r').

    Could anyone please help clear this up for me.
    Last edited by a moderator: May 4, 2017
  2. jcsd
  3. May 16, 2010 #2

    D H

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    What that paper means by [itex]\partial L/\partial \mathbf r'[/itex] is

    [tex]\frac{\partial L}{\partial{\mathbf r}'} \equiv
    \frac{\partial L}{\partial x'}\hat x' +
    \frac{\partial L}{\partial y'}\hat y' +
    \frac{\partial L}{\partial z'}\hat z'[/tex]

    It might be easier for you to see how (2.20) follows from (2.19) by using (2.19) in its first form,

    L = \frac 1 2 m \Bigl((\dot x - \omega y')^2 + (\dot y + \omega x')^2 + \dot z^2\Bigl)

    The result in (2.20) is a much more general result. It applies to any rotation vector [itex]\boldsymbol \omega[/itex], not just the pure z rotation used in that example.
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