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I Does small angle mean small angular velocity?

  1. Aug 4, 2016 #1
    Why is the term ##(\sin\theta\,\dot{\theta})^2## fourth order in the small ##\theta##, as claimed by the sentence below (5.108)?

    By small-angle approximation, ##(\sin\theta\,\dot{\theta})^2\approx\theta^2\,\dot{\theta}^2##.

    For this to be fourth order, it seems like we must have ##\theta=\dot{\theta}##. Why is this true? What are the conditions for this to be true?

    Screen Shot 2016-08-04 at 5.14.16 pm.png
     
  2. jcsd
  3. Aug 4, 2016 #2
    why quantities of different dimension must be equal? They must not. David Morin wrote a good book but definitely not for high school
     
  4. Aug 4, 2016 #3
    So is he right or wrong?
     
  5. Aug 4, 2016 #4
    My advice is that you should first think about your own understanding whether it is right or wrong.
     
  6. Aug 4, 2016 #5
    My understanding is that ##(\sin\theta\,\dot{\theta})^2## is second order in ##\theta##. But that would mean that it cannot be ignored, since (5.109) contains second-order terms.
     
  7. Aug 4, 2016 #6
    The degree of the term ## \theta^2\dot \theta^2## is equal to 2+2=4. It is a degree of polynomial ##P(\theta,\dot \theta)= \theta^2\dot \theta^2##. Both quantities ## \theta,\dot \theta## are assumed to be small. The approximation of the Lagrangian up to the second order terms corresponds to linearization of the Lagrange equations in the vicinity of equilibrium
     
    Last edited: Aug 4, 2016
  8. Aug 4, 2016 #7

    mfb

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    Staff: Mentor

    Here it is important that θ is small, but the timescale is not. ##\dot \theta## is "something with θ divided by time", which is small of θ is small.
     
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