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I A weird answer -- Lagrangian mechanics

  1. Aug 9, 2016 #1
    Just refer to my profile picture to see what the issue is! :biggrin:

    Here's the problem: a ball of mass m is connected to a vertical pole with an inextensible, massless string of length r. The angle between the string and the pole is θ. The pole rotates around the z axis with a constant angular velocity $$\dotφ$$

    We can easily determine the following equations:
    $$x=r*sinθ*cosφ$$ $$y=r*sinθ*sinφ$$ $$z=r(1-cosθ)$$

    Now I determined that the pesky Lagrangian is equal to the following:
    $$L=\frac{mr^2}{2}(\dot\theta^2+\dot\phi^2sin^2\theta)+mgr(1-cos\theta)$$

    So, where's the issue? Well, solving the Euler-Lagrange equations we find this:
    $$mr^2\ddot\phi sin\theta+2mr^2\dot\phi\dot\theta cos\theta sin\theta=0$$

    Which means that
    $$\ddot\phi=-\frac{2\dot\phi\dot\theta}{tan\theta}$$

    That's a really odd result. φ' is supposed to be constant, so φ''=0. But here we can see that φ'' can only be 0 if either φ' or θ' are 0, or if θ=π/4. Weird.

    Am I missing something (I obviously am, it's a rhetorical question)?
     
  2. jcsd
  3. Aug 9, 2016 #2

    vanhees71

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    You have to set
    $$x=r \cos (\omega t) \sin \theta, \quad y=r \sin(\omega t) \sin \theta, \quad z=r (1-\cos \theta),$$
    evaluate ##T=m \dot{\vec{x}}^2/2## and then derive the equation of motion for ##\theta## via the Euler-Lagrange equations.
     
  4. Aug 9, 2016 #3
    So I don't have to solve the Euler-Lagrange equations for φ and ##\dotφ## all?
     
  5. Aug 9, 2016 #4

    vanhees71

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    I've reformulated my suggestion a bit, but it's of course finally the same. If I understand your problem right, it's forced to have a rotation with constant angular velocity ##\omega=\dot{\phi}## around the ##z## axis. That's a constraint rather than a degree of freedom.
     
  6. Aug 9, 2016 #5
    Ah ok! Thanks!
     
  7. Aug 9, 2016 #6
    Oh whoops, in my original post I meant "if θ was π/2", not π/4.
     
  8. Aug 9, 2016 #7

    vanhees71

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    As I said, you must enforce that ##\dot{\phi}=\text{const}##!
     
  9. Aug 9, 2016 #8
    Yeah, yeah, I did that, and I got a result that makes sense. I was just saying...
     
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