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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?

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

    vela

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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!

    Ari
     
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