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Finding the Hamiltonian of this system

  1. Nov 4, 2015 #1
    1. The problem statement, all variables and given/known data

    I am asked to find the Hamiltonian of a system with the following Lagrangian:

    ##L=\frac{m}{2}[l^2\dot\theta^2+\dot{\tilde{y}}^2+2l\dot{\tilde{y}}\dot{\theta}\sin{\theta}]-mg[\tilde{y}-l\cos{\theta}]##

    2. Relevant equations

    ##H = \dot{q_i}\frac{\partial L}{\partial \dot{q_i}}-L=\dot{\theta}\frac{\partial L}{\partial \dot{\theta}}-L##

    ##p_i=\frac{\partial L}{\partial \dot{q_i}}\implies p_{\theta}=\frac{\partial L}{\partial \dot{\theta}}##

    3. The attempt at a solution

    ##H = \dot{\theta}\frac{\partial L}{\partial \dot{\theta}}-L\\
    ~~~=\dot{\theta}[ml^2\dot{\theta}^2+ml\dot{\tilde{y}}\sin{\theta}]-L\\
    ~~~=\frac{m}{2}[2l^2\dot{\theta}^2+2ml\dot{\tilde{y}}\dot{\theta}\sin{\theta}]-L\\
    ~~~=\frac{m}{2}[l^2\dot{\theta}-\dot{\tilde{y}}^2]+mg[\tilde{y}-l\cos{\theta}]\\\\##
    ##p_{\theta}=\frac{\partial L}{\partial \dot{\theta}}=ml^2\dot{\theta}+ml\dot{\tilde{y}}\sin{\theta}\\
    \implies \dot{\theta}=\frac{p_{\theta}-ml\dot{\tilde{y}}\sin{\theta}}{ml^2}\\~~~~~~~~~~~~~=\frac{p_{\theta}}{ml^2}-\frac{\dot{\tilde{y}}}{l}\sin{\theta}\\\\##
    After plugging this into the above form of the Hamiltonian and a little simplifying I get:

    ##H=\frac{m}{2}[\frac{p_{\theta}^2}{m^2l^2}-\frac{2p_{\theta}\dot{\tilde{y}}}{ml}-\dot{\tilde{y}}^2\cos{\theta}^2]+mg[\tilde{y}-l\cos{\theta}]##

    I would like to know if this is correct and in particular, if the method is correct.We have not been taught how to take the Legendre transformation of the Lagrangian to get the Hamiltonian so I am unsure of the first equation I have listed in "Relevant equations". Thank you for any help!
     
  2. jcsd
  3. Nov 4, 2015 #2

    TSny

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    That all looks good to me.
     
  4. Nov 4, 2015 #3

    Henryk

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    It's been a while since I did this, but it looks to me you have two generalized coordinates, ##\theta## and ##y## and you should also consider ##p_y##
     
  5. Nov 4, 2015 #4

    TSny

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    The background information for the Lagrangian is not given here. This thread is a continuation of a recent thread https://www.physicsforums.com/threa...range-equation-for-a-special-pendulum.841136/
    where ##y(t)## is given to be a specific function of time. So ##y## is not a degree of freedom.
     
  6. Nov 5, 2015 #5
    Thanks a lot for the help guys, particularly TSny!

    And the Hamiltonian will not be conserved in this case right ? We can use ##\frac{dH}{dt}=\frac{\partial H}{\partial t}## which comes out to something ugly and doesn't look like it will reduce to zero... alternatively we can just logically deduce that it won't be conserved since energy is being put into the system to force the support at a specific frequency?
     
  7. Nov 5, 2015 #6

    TSny

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    The Hamiltonian does not always represent the total energy when the transformation equations from Cartesian coordinates to the generalized coordinates depend explicitly on on time. You can check that your Hamiltonian does not equal the total energy of the pendulum.

    A separate question is whether or not the energy will be conserved. As you say, the driving of the pendulum by the moving support will cause the total energy to vary with time.
     
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