$\newcommand{\dd}{\mathrm{d}}$ I am trying to work out the Lagrangian mechanics for a pendulum problem in order to animate it. I'm working on one of the examples in the Wikipedia page on Lagrangian mechanics where a pendulum is attached to a block as in the picture below

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I understand how most of this works with the exception of the friction between the block and the surface it resides on.

I've solved the problem with just the Kinetic + Potential energy of the block

$$ \frac{\partial L}{\partial x} - \frac{\dd}{\dd t}\left(\frac{\partial L}{\partial \dot{x}}\right) = 0;$$ $$ \frac{\partial L}{\partial \theta} - \frac{\dd}{\dd t}\left(\frac{\partial L}{\partial\dot\theta}\right) = 0;$$

From another example from my class notes of a disk rolling down an inclined plane in 2D the friction can be accounted for by adding a constraint that relates the two generalized coordinates. Do I need to do something similar for friction in this case? How would I do that?

  • 1
    $\begingroup$ Google friction, Lagrange, multiplier and you will get answers. $\endgroup$ – Lewis Miller Mar 5 '16 at 0:28
  • $\begingroup$ In the past, I've set the Euler La Grange equation for the x variable equal to the non-conservative forces acting in that direction instead of equal to 0. $\endgroup$ – Mephistopheles Mar 5 '16 at 0:35
  • $\begingroup$ So, @Mephistophales, it should be dL/dx - (d/dt)(dL/dx') = frictionCoefficient * Mg;...? $\endgroup$ – Nick Mar 5 '16 at 0:39
  • $\begingroup$ Friction coefficient times the NORMAL force $\endgroup$ – Mephistopheles Mar 5 '16 at 0:43
  • $\begingroup$ You will have to do this piecewise, with a friction "potential" term equal to $\pm \mu F_{n} x$, solve the EOM for both signs, and then piece together a solution so that $\dot x$ and $x$ are both continuous at the turnaround points where $\dot x = 0$ $\endgroup$ – Jerry Schirmer Mar 5 '16 at 1:17

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