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I Lagrangian of double pendulum

  1. May 23, 2016 #1
    Ok, I'm reading up on Lagrangian mechanics, and there is a problem that I don't really understand: the double pendulum (in this case, without a gravitational field). So, I want to take it step by step to make sure I understand all of it.

    We've got a pendulum (1) with a weight mass m=1kg attached to a rod of length r=1m making an angle of θ with the vertical, and another identical pendulum (2) attached on the weight of pendulum (1), making an angle α with it. Transforming these coordinates to Cartesian ones, we get x(1)=sinθ, y(1)=cosθ and x(2)=sinθ+sin(α+θ) and y(2) = cosθ+cos(α+θ).

    Ok. So far so good. Now I am supposed to compute the time derivatives of the Cartesian velocity components in terms of the angles to compute the kinetic energy. How exactly am I supposed to do that? Since there is no t in these, I can't directly find their time derivatives, unless I use the equations that describe the motions of pendulums that are already known, but aren't these wrong in the case of a double pendulum?

    I know that the results I am supposed to get for the kinetic energies are these:



    I just have no idea how to get them.
  2. jcsd
  3. May 23, 2016 #2
    actually there is ##\alpha=\alpha(t),\quad \theta=\theta(t)##
    this is not a good manner: you deprive yourself from opportunity of using the dimensions of physics quantities to check your formulas
    Last edited: May 23, 2016
  4. May 23, 2016 #3
    What I meant was that it's not explicitly shown. Of course there is, but I don't know what the value of the function α(t) is.

    Well, I'm not the one who made the problem :)
  5. May 23, 2016 #4
    Anyway, the question still stands, I still can't figure out what to do here.
  6. May 23, 2016 #5
    Ok, I found the answer, here it is for anyone who might be interested:

    Using the chain rule, we write say dx1/dt=sinθ as dθ/dt ⋅ d(sinθ)/dθ = dθ/dt ⋅ cosθ. If we plug this (and y1) into the equation for kinetic energy in this case (which is T1=(x^2+y^2)/2), and do some algebra, we eventually get the anticipated solution of T1=(dθ/dt)^2/2. It's the same thing for T2, only a bit more complicated. The point is that you're meant to apply the chain rule for derivatives.
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