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Lagrange equations of a spinning parabola

  1. Apr 16, 2013 #1
    1. The problem statement, all variables and given/known data

    Consider a bead of mass m sliding without friction on a wire that is bent in the shape of a parabola and is being spun with constant angular velocity ω about its vertical axis. Use cylindrical polar coordinates and let the equation of the parabola be ##z = kρ^{2}##. Write down the lagrangian in terms of ρ as the generalized coordinate. Find the equation of motion.


    2. Relevant equations

    $$L = T-U = \frac{1}{2}mv^2-mgy$$


    3. The attempt at a solution

    I just need a hint on how to set up the kinetic energy of the parabola, maybe someone can explain it in a different way that would push me in the right direction.

    Also, sorry if this isn't considered advanced physics. This is a classical mechanics class (upper level at my school) and it seems like a gray area for me.
    Thanks!
     
    Last edited by a moderator: Apr 16, 2013
  2. jcsd
  3. Apr 16, 2013 #2
    you need to change your coordinates to suit the problem better
     
  4. Apr 16, 2013 #3

    vela

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    A straightforward method is to write down expressions for x(t), y(t), and z(t) and then differentiate them with respect to time. For example, ##x(t) = \rho\cos \phi##, so
    $$\dot{x}(t) = \dot{\rho}\cos \phi - \rho\sin\phi \,\dot{\phi}.$$ When you have all three, plug them into ##v^2 = \dot{x}(t)^2 + \dot{y}(t)^2 + \dot{z}(t)^2##.

    By the way, I think the potential term in the Lagrangian should be mgz, not mgy.

    It's in the right place.
     
  5. Apr 16, 2013 #4
    Ah, I suppose I should add in the major fact that they want it in terms of the generalized coordinate rho. I can try and find/make a picture of the graph if it helps.

    and yes, you are correct, the potential would be mgz.
     
  6. Apr 16, 2013 #5

    vela

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    The expression you end up with will be in terms of ##\rho##.
     
  7. Apr 16, 2013 #6
    Everything worked out properly! Thanks a bunch for your help.
     
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