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I Partial differentiation

  1. Aug 4, 2016 #1
    I believe there is a mistake in the second equation of (5.139).

    Screen Shot 2016-08-05 at 2.39.45 am.png

    The equation is obtained from (5.138) using the Euler-Lagrange equation
    ##\frac{d}{dt}\frac{\partial L}{\partial\dot{\theta}}=\frac{\partial L}{\partial\theta}.##

    LHS##\,\,=\frac{d}{dt}\frac{\partial L}{\partial\dot{\theta}}=\frac{d}{dt}(mr^2\dot{\theta}-mr\dot{x}\cos\theta)##
    ##=mr^2\ddot{\theta}-mr\dot{x}(-\sin\theta)\dot{\theta}-mr\ddot{x}\cos\theta\,\,\,\,\,\,\,\,\,\,## (Note that ##\dot{r}## terms are ignored.)

    RHS##\,\,=mgr\sin\theta##

    Equating LHS and RHS, and dividing by ##m## and ##r##, we have
    ##r\ddot{\theta}+\dot{x}\sin\theta\dot{\theta}-\ddot{x}\cos\theta=g\sin\theta##.

    Am I right?
     
  2. jcsd
  3. Aug 4, 2016 #2
    There's a ##\theta##-dependent term that you have forgotten to differentiate inside the parentheses.
     
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