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Lagrangian of inverted pendulum

  1. Sep 24, 2008 #1
    An inverted pendulum consists of a particle of mass [tex]m[/tex] supported by a rigid massless rod of length [tex]l[/tex] . The pivot [tex]O[/tex] has a vertical motion given by [tex]z=Asin\omega t[/tex]. Obtain the Lagrangian and find the differential equation of motion.


    I'm not sure how to obtain the kinetic and potential energies. For the potential energy, would it just be
    [tex]V=mglcos\theta+Asin\omega t[/tex]?

    And is the kinetic energy
    [tex]T=\frac{1}{2}m(l^{2}\dot{\theta}^{2}+A^{2}\omega^{ 2}cos^{2}\omega t)[/tex]?

    Since the Lagrangian wouldn't be time-independent, would this in any way affect the Euler-Lagrange equation, or would it remain the same?

    Thanks, all.
     
  2. jcsd
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