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Mathematical model of pendulum

  1. May 17, 2009 #1
    1. The problem statement, all variables and given/known dataW
    We know the model of an ideal pendulum at rest is given by
    [tex]L \ddot{\theta} + g sin \theta =0, t\geq0[/tex]
    [tex] \dot{\theta}(0)=0 [/tex]
    [tex] \theta(0)=\theta_0 [/tex]
    where [tex]\theta(t)[/tex] is the pendulum angle at time t, L is the length of the pendulum, and g is gravity.
    Now, consider the total energy per unit mass of the pendulum given by
    [tex]E(\theta,\dot{\theta}) = \frac{1}{2}L \dot{\theta}^2 - g cos \theta[/tex]
    Show that this equation is constant along the previously given solutions (the first 3 equations initially given, lets call them EQ [1] ), [tex]\forall t\geq0[/tex]. Use this result to show that the solution of [1] should satisfy
    [tex]\frac{1}{2}L\dot{\theta}^2-g cos \theta + g cos \theta_0 = 0, \forall t\geq0[/tex]
    (lets call this EQN [2])
    2. Relevant equations

    we know that if E(theta, thetadot) is a constant, then
    [tex]\frac{d E(\theta,\dot{\theta})}{dt} = 0 [/tex]

    3. The attempt at a solution
    so, taking the derivative of the energy, we have:

    [tex]\frac{d E(\theta,\dot{\theta})}{dt} = \frac{d}{dt}[\frac{1}{2} L \dot{\theta}^2 - g cos \theta][/tex]

    [tex] = L \ddot{\theta} +\dot{\theta}g sin\theta[/tex]
    however, this does not seem to equal [1], as [tex]\dot{\theta}gsin\theta [/tex] never appears in the solution. what am I doing wrong?
    or do we not take the time derivative of the second term in the energy equation (the sin(theta) term), thus giving
    [tex]\frac{d E(\theta,\dot{\theta})}{dt} =L \ddot{\theta} +g sin\theta[/tex] which is equal to zero by definition of [1], thus if the derivative of the energy equation is equal to zero, then by definition of constant, energy is constant!
    now, i am in confusion as how to use this result to prove EQN [2]?
     
    Last edited: May 18, 2009
  2. jcsd
  3. May 18, 2009 #2

    tiny-tim

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    Hi limofunder! :smile:

    (have a theta: θ :wink:)
    hmm :rolleyes: … d/dt (Lθ') = Lθ''

    but d/dt (Lθ'2) = … ? :smile:
     
  4. May 18, 2009 #3

    gabbagabbahey

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    [tex] \frac{d}{dt}\left(\frac{1}{2} L \dot{\theta}^2\right)=L\dot{\theta}\ddot{\theta}[/tex]

    :wink:

    EDIT: Tinytim beat me to it
     
  5. May 18, 2009 #4
    ok, so then that gives way to
    dE/dt = θ' (Lθ''+g sinθ)
    the second term being by definition of [1] >= 0 for all time.
    Thank you,
    so that still begs the question, I do not see how this forces
    [1] to satisfy
    1/2 Lθ' - g cosθ + g cosθ0
     
  6. May 18, 2009 #5

    tiny-tim

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    Because E = constant is a solution, so just call the constant gcosθ0 :smile:
     
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