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Mathematical model of pendulum

  • Thread starter limofunder
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  • #1
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Homework Statement

W
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])

Homework Equations



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

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:

Answers and Replies

  • #2
tiny-tim
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Hi limofunder! :smile:

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

but d/dt (Lθ'2) = … ? :smile:
 
  • #3
gabbagabbahey
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[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]
[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
 
  • #4
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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
 
  • #5
tiny-tim
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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
Because E = constant is a solution, so just call the constant gcosθ0 :smile:
 

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