1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Analytical mechanics problem

  1. Sep 23, 2006 #1
    As in the attached picture.
    A cylindrical shell with mass M can roll without gliding on a horizontal plane

    In the cylindrical shell a particle ,p, with mass m can glide without friction.

    At the begining there is no motion and the angle to the particles position is [tex] \phi=\frac{\pi}{2} [/tex]

    I am suposed to find the movement of the center of the circle as a function of the angle [tex] \phi [/tex]

    Im not sure how I should start.

    The potential energy of the system is, if I place the plane of reference on the level of the horizontal plane.

    [tex] V=mgR(1-cos\phi) [/tex]

    Now this problem obviously only has one degree of freedom and that is the angle [tex] \phi [/tex]

    But if I want to construct a lagrangian I must use two degres of freedom. The rotation angle of the cylinder [tex] \alpha [/tex] and the angle to the particle p [tex] \phi [/tex]. Because I dont know how [tex] \alpha [/tex] and [tex] \phi [/tex] are connected. Is this the correct thinking?

    In that manner I get this equation

    [tex] L=T-V=\frac{3}{4}MR^2\dot{\alpha}^2+\frac{1}{2}m[\dot{\alpha}^2 R^2 + R^2\dot{\phi}^2 + \dot{\alpha}^2 R^2 sin\phi ] - mgR(1-cos\phi ) [/tex]

    Should I use this and solve the two lagrande equations

    [tex] \frac {d}{dt} \frac{dL}{d\dot{\alpha}}-\frac{dL}{d\alpha}=0 [/tex]

    [tex] \frac {d}{dt} \frac{dL}{d\dot{\phi}}-\frac{dL}{d\phi}=0 [/tex]

    Im not sure if this will give me any answere though? It feels like I should express [tex] \alpha [/tex] in [tex] \phi [/tex] before I even do the lagrangian. Am I on the right track?

    Attached Files:

    Last edited: Sep 23, 2006
  2. jcsd
  3. Sep 23, 2006 #2
    I assume that I in the same manner can get the hamiltonian since its a conservative force situation and holonomic constrains.

    [tex] H=T+V=\frac{3}{4}MR^2\dot{\alpha}^2+\frac{1}{2}m[\dot{\alpha}^2 R^2 + R^2\dot{\phi}^2 + \dot{\alpha}^2 R^2 sin\phi ] + mgR(1-cos\phi ) [/tex]

    But Im not quite sure what I would do with that one either. I know that

    [tex] \dot{\alpha}=\frac{dH}{dP_{\alpha}}[/tex]
    [tex] -\dot{P_{\alpha}}=\frac{dH}{d\alpha} [/tex]

    But that doesnt seem to be much help either??
    Last edited: Sep 23, 2006
  4. Sep 28, 2006 #3
    I don't know...I wish I could help...

    It seems like you really do need two generalized coordinates. I mean, the particle is free to move; it doesn't seem like it is constrained in any way such that [tex]\phi[/tex] and [tex]\alpha[/tex] would be related.
  5. Sep 28, 2006 #4
    I think that there is an error in your Lagrangian: Since you are interested in the movement of the cylinder and watch it in the "labatory system", so to say, you can't say that the velocity of the particle in y-direction is [itex] R \dot \phi \cos(\phi)[/itex]. [itex] \phi [/itex] is fixed in the "cylinder system", so you have you add the y-velocity of the cylinder. If you square this, you should get a "mixed term" so that the both velocities are related.
    Last edited: Sep 28, 2006
  6. Sep 28, 2006 #5
    I managed to solve it. my expression for the Y position of the particle was

    [tex] Rsin\phi + R\alpha [/tex]

    So velocity was

    [tex] R\dot{\phi}cos\phi + R\dot{\alpha}[/tex]

    I used the fact that the conjugated momentum is constant for the alpha coordinate since there is no alpha, just alpha dot, dependance :) tricky problem.
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?

Similar Discussions: Analytical mechanics problem
  1. Analytic Problem (Replies: 3)

  2. Analytical Mechanics (Replies: 0)

  3. Analytical mechanics (Replies: 6)

  4. Analytical mechanics (Replies: 1)