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Homework Help: Hamiltonian Cylinder!

  1. Feb 11, 2012 #1
    1. The problem statement, all variables and given/known data
    Particle of mass m constrained to move on the surface of a cylinder radius R, where [itex]R^2 = x^2 + y^2[/itex]. Particle subject to force directed towards origin and related by F = -kx

    2. Relevant equations
    L = T - U
    H = T + U

    3. The attempt at a solution
    So I have the solution, but not sure why they did a step. Here goes:
    They find that [tex]L = \frac{1}{2}m(R^2\dot{\theta^2} + \dot{z^2} - \frac{1}{2}k(R^2 + z^2),[/tex] which I agree with.

    They then find that [tex] p_{\theta} = mR^2\dot{\theta},[/tex] and [tex] p_z = m\dot{z}, [/tex] and then then since H = T+U they state that [tex] H = \frac{1}{2}m(R^2(\frac{p_{\theta}}{mR^2})^2 + (\frac{p_z}{m})^2 - \frac{1}{2}kz^2. [/tex] My question is where did the [itex] \frac{1}{2}mR^2[/itex] in the potential energy go? Why did they cancel it? I know that it is constant, and everything else seems to have a varying component, but why does that mean you can simply do away with it? Is this always the case?

    Last edited by a moderator: Feb 14, 2012
  2. jcsd
  3. Feb 14, 2012 #2


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    The last term of H has the wrong sign.

    It looks like they dropped the term because it's a constant, and yes, you can always throw it out. The equations of motion depend in the derivatives of H, so a constant term won't make a difference.
  4. Feb 14, 2012 #3
    Yeah, your right, the sign is wrong. I understand that the equations of motion are based on the derivatives, and thus all constant terms are thrown out, but I assumed that if the question asked for the Hamiltonian and no further, that these constant terms would still be included. If it doesn't matter though, then thanks for the clarification!

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