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Double pendulum

  1. Apr 11, 2005 #1
    If you have have double pendulum of two rods where the rods moves with the same frequency w at the equilibrium position and at the eq. position you hit the lower rod at a distance d from the point that connects the rods. The hit results in a motion where the upper rod moves with frequency w and the lower rod with frequency -w. If the rods lengths are l and their masses m what are the the distance d.

    I have no problem to find the equations of motion for the ordinary double pendulum. With the hit i think it gets something like

    Lagrange equations = F

    Where F is different for theta1 and theta2.
    If you find these equations you can solve for which d the resulting motion is w for the upper rod and -w for the lower rod, right?
    But how do you write F?

    Thank you
  2. jcsd
  3. Apr 11, 2005 #2
    You need to right down the lagrangian:

    L = T - U

    Where T is the kinetic energy and U is the potential energy.

    In this case, the lagrangian will be a function of four variables:

    [tex]L(\theta_1 , \theta_2 , \theta_1 ', \theta_2' )[/tex]

    Where primes denote differentiation with respect to time. So, after you have expressed the lagrangian in terms of those four variables, your equations of motion are the Euler-Lagrange equations (one for each position variable, so two total).
  4. Apr 12, 2005 #3
    Thank you.
    But as i tried to say the problem for me is to find the generalized impulse F.
    Because Lagrange equations leads to
    F = Generalized impulse = change in generalized momentum
    If I find F the problem is solved.
    and how do i get the distance d into F?
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