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Find the Lagrangian of an unwinding pendulum

  • Thread starter mishima
  • Start date
517
31
Problem Statement
See image
Relevant Equations
L=T-V, ke=1/2mv^2
242045


I think my confusion on this is where the best origin for polar coordinates is. I've tried the center of the circle, and note the triangle made from the r coordinate reaching out to ##m, a,## and ##l+a\theta##. Then

$$r=\sqrt{a^2+(l+a\theta)^2}$$
$$\dot r = \frac {a(l+a\theta)} {\sqrt{a^2+(l+a\theta)^2}}$$

but my book has a much simpler expression for T where ##\dot r = 0##...I am missing some crucial insight.
 

Orodruin

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I think my confusion on this is where the best origin for polar coordinates is.
You should not be using polar coordinates. All you need to know is the kinetic and potential energies of the system expressed in terms of ##\theta## and its time derivative. You can do this by writing down the position of the bead as a function of ##\theta## in Cartesian coordinates and differentiating it to find the velocity.
 
517
31
242165

From that I was able to get

$$x=(l+a\theta) sin\theta + a cos\theta$$
$$y=-(l+a\theta) cos\theta + a sin\theta$$

and so the ##(\dot x^2 + \dot y^2)## term in the kinetic energy was just ##(l+a\theta)^2\dot \theta^2##!

Thanks so much.
 

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