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I Euler-Lagrange query

  1. Dec 20, 2016 #1
  2. jcsd
  3. Dec 20, 2016 #2

    BvU

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    Hello Volican, :welcome:

    It comes from the time derivative of ##\partial {\mathcal L}\over \partial \dot\theta## $${\partial {\mathcal L}\over \partial \dot\theta }= mr^2\dot \theta $$
     
  4. Dec 20, 2016 #3
    Thanks for writting back, much appreciated. I can see where that comes from and also the -mgrsinθ but cant see how they got the 2mr˙r˙θ. Any ideas? I am thinking that I can just completley ignore the dr/dt as it is not in the euler lagrange equation? Is this wrong?
     
  5. Dec 20, 2016 #4

    BvU

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    $$
    {d\over dt} \left ( {\partial {\mathcal L}\over \partial \dot\theta } \right ) = {d\over dt} mr^2\dot \theta = m {d\over dt} \left ( r^2\right ) \dot \theta + mr^2 {d\over dt} \dot \theta = 2mr\dot r\dot\theta + mr^2\ddot\theta$$
     
  6. Dec 20, 2016 #5
    Aewsome! to differentiate r^2 did you use the chain rule?
     
  7. Dec 20, 2016 #6

    BvU

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    Yes, the chain rule is awesome :smile: $$
    {d\over dt} r^2 = {d\over dt} \left ( r \right ) r + r {d\over dt} \left ( r \right ) = 2 r\dot r $$

    Make sure you speak this jargon fluently when pursuing further studies !
     
    Last edited: Jan 11, 2017
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