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Classical mechanics - Lagrange multipliers

  1. Dec 15, 2008 #1
    1. The problem statement, all variables and given/known data
    A disk moves on an inclined plane, with the constraint that it's velocity is always at the same direction as it's plane (similar to an ice skate, maybe). In other words: If [tex]\hat{n}[/tex] is a vector normal to the disk's plane, we have at all times: [tex]\hat{n} \cdot \vec{v} = 0[/tex]. Also, it's free to move without friction, and always perpendicular to the plane. (as seen in the figure.)

    I need to get and solve the equations of motion for certain initial conditions that I'll write promptly. We set an x-y coordinate system at the top-right corner of the plane with the y axis going downwards, and denote that angle between [tex]\hat{n}[/tex] as [tex]\varphi[/tex].

    2. Relevant equations

    The constraint is:

    [tex]c_1 = \dot{x}\cos\varphi + \dot{y}\sin\varphi[/tex]

    and accordingly the Lagrangian is:

    [tex]L = \frac{1}{2}m(\dot{x}^2 + \dot{y}^2) + \frac{1}{2}I\dot{\varphi}^2 + mgy + \lambda(\dot{x}\cos\varphi + \dot{y}\sin\varphi)[/tex]

    The initial conditions that were given are that at t=0:

    [tex]x=0, y=0, \dot{x}=0, \dot{y}=0, \varphi = 0, \dot{\varphi} = \omega_0 [/tex]

    3. The attempt at a solution
    The obvious way of solving is to use Euler-lagrange and get the equations of motion. The problem is that I can't solve them! They're too damn complicated. There is a hint that I should try to find constants of motion by setting t=0 in the equations, but I can't seem to find them.

    I would appreciate any help...

    Thanks in advance!

    Attached Files:

  2. jcsd
  3. Dec 18, 2008 #2
    ok well if [itex] y=0 [/atex] and [itex] \dot y=0 [/itex] then [itex] y=0 \forall t [/itex] which will simplfy your Lagrangian
  4. Dec 18, 2008 #3
    The same would happen for [itex] x [/itex] so I get [itex] L = \frac{1}{2} I {\omega_0}^{2} [/itex] Then solving the Euler-Lagrange I get [itex] I \ddot \psi = 0 [/itex]. This could be massively wrong but I though I'd give it a shot anyway.
  5. Dec 23, 2008 #4
    Thanks, I solved it by variating the action integral - the TA said that Lagrange multipliers give here the wrong answer, and he doesn't know why. I guess it's a question for mathematicians to answer.
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