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Effective potential and stability

  1. Mar 20, 2007 #1
    I am asked to check the stability at theta=0 of the pendulum system shown in the attachment.
    I set up the energy for the system and found it to be
    [itex]E=\frac{1}{2}m(L\sin(\theta))^{2}+\frac{1}{2}m(L\frac{d\theta}{dt})^{2}-mgL\cos(\theta)[/itex]
    which is in agreement with the books answer "A guide to physics problems, part 1"
    To check stability for theta=0, I have to take the second deriv of the effective potential energy. This leads me to my first question.
    1) What conditions require me to include a kinetic energy term as a ficticious potential energy term.
    I know I should include the first term in my effective potential, but not the second. I kind of expected that this was correct but I lack the ability to give a physical or mathematical justification for discerning this. Does it have to do with the fact that theta shows up in both the potential energy term and the first term, or does it have to do with the fact that the first term is cyclic in theta dot or something else altogether?
    At any rate I have that:
    [itex]V_{eff}=\frac{1}{2}m(L\sin(\theta))^{2}-mgL\cos(\theta)[/itex]
    The book did not do this, instead they used the lagrangian:
    [itex]L=\frac{1}{2}m(L\sin(\theta))^{2}+\frac{1}{2}m(L\frac{d\theta}{dt})^{2}+mgL\cos(\theta)[/itex]
    rewrote it as:
    [itex]L=\frac{1}{2}m(L\frac{d\theta}{dt})^{2}-\left(-\frac{1}{2}m(L\sin(\theta))^{2}-mgL\cos(\theta)\right)[/itex]
    and since [itex]L=T-U[/itex] they identified the effective potential as:
    [itex]V_{eff}=-\frac{1}{2}m(L\sin(\theta))^{2}-mgL\cos(\theta)[/itex]
    My effective potential is different than theirs by a sign on the first term. However, this makes all the difference. My plus sign results in stability for all values W, while their result yeilds W^2<g/l as the condition for stability.
    My next two questions are:
    2) Why do they use the lagrangian to determine their effective potential?
    3) Moreover, why is it apparently wrong in this case to use the eff. pot. from the equation for energy.
    These last two questions are very disturbing because in every problem I have ever worked or seen worked in a text, they have always used the energy and not the lagrangian to find the effective potential.
     

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    Last edited: Mar 20, 2007
  2. jcsd
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