Lagrangian of double pendulum

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  • Thread starter Andreas C
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  • #1
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Main Question or Discussion Point

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:

T1=(dθdt)^2/2

T2=((dθdt)^2+(dθdt+dαdt)^2)/2+(dθdt)*(dθdt+dαdt)*cosα

I just have no idea how to get them.
 

Answers and Replies

  • #2
Since there is no t in these
actually there is ##\alpha=\alpha(t),\quad \theta=\theta(t)##
with a weight mass m=1kg
this is not a good manner: you deprive yourself from opportunity of using the dimensions of physics quantities to check your formulas
 
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  • #3
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actually there is α=α(t),θ=θ(t)
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.

this is not a good manner: you deprive yourself from opportunity of using the dimensions of physics quantities to check your formulas
Well, I'm not the one who made the problem :)
 
  • #4
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Anyway, the question still stands, I still can't figure out what to do here.
 
  • #5
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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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