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Taylor 4.34 - Energy of a Pendulum

  1. Apr 29, 2014 #1

    Radarithm

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    Edit: Can someone change the name of the thread somehow? I accidentally posted it without changing the name.

    (Moderator note -- title updated.)

    1. The problem statement, all variables and given/known data

    The question is quite long so here is a picture: http://gyazo.com/dc917d1885b6ffebb0a39e2409af4d61


    2. Relevant equations

    $$E=T+V$$
    $$T=\frac{m\dot{x}^2}{2}$$
    $$U_g=mgh$$
    $$\Gamma=I\alpha$$

    3. The attempt at a solution

    If we set the zero of potential energy at origin (where the mass is at when the system is in equilibrium), then the change in gravitational potential is:
    $$U(\phi)=mg(l-l\cos{\phi})=mgl(1-\cos{\phi})$$

    Now that we're done with part (a), I move on to part (b), which is where I think I messed up.
    The total mechanical energy of the system can be defined to be:
    $$E(\phi, \dot{\phi})=T(\phi,\dot{\phi})+U(\phi)$$
    Where the Kinetic energy (T) is:
    $$T(\phi,\dot{\phi})=\frac{m\dot{\phi}l\sin{\phi}}{2}$$
    and the potential, U, is:
    $$U(\phi)=mgl(1-\cos{\phi})$$
    Here is where I think I messed up mathematically; taking the time derivative of the mechanical energy, I get this:
    $$\dot{E}=ml\left(\frac{\dot{\phi}\ddot{\phi}\cos{\phi}}{2}+g(\dot{\phi}\, \cos{\phi}+1)\right)$$

    Did I go wrong somewhere?
     
    Last edited by a moderator: Apr 29, 2014
  2. jcsd
  3. Apr 29, 2014 #2

    SammyS

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    When we find a Mentor or such with that power --
    What name do you want on this thread ?

    attachment.php?attachmentid=69208&stc=1&d=1398779968.png
     

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    Last edited: Apr 29, 2014
  4. Apr 29, 2014 #3

    Radarithm

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    Taylor 4.34 - Energy of a Pendulum

    Or something like that. Taylor doesn't name problems like K&K; I'm not very creative.
     
  5. Apr 29, 2014 #4

    SammyS

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    You can use the "Report" button to make such a request.

    I'll do it for this thread.
     
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