Klaynos Posted January 15, 2008 Posted January 15, 2008 OK, I know this is simple and I'm just having an issue tonight. A simple pendulum, is damped, Theta is the angle from vertical hanging. There should be a change to either the KE or PE of the system due to the damping... But I can't remember what it is, I *think* the PE should change to: [math]U=-mglCos\theta + A\dot\theta[/math] Where A is some constant of friction... The reason I ask in this way is because I'm trying to do this using Lagrangian mechanics...
Severian Posted January 16, 2008 Posted January 16, 2008 That is not how you should be doing it. You cannot write a disapative force in terms of a potential - in fact, iirc that is the definition of a conservative force, that it can be written as a potential. The Euler-Lagrange equations as [math]\frac{d}{dt} \left( \frac{\partial {\cal L}}{\partial \dot{\theta}} \right) - \frac{\partial {\cal L}}{\partial \theta} = 0[/math] are not valid for your system. Instead you need to write [math]\frac{d}{dt} \left( \frac{\partial {\cal L}}{\partial \dot{\theta}} \right) - \frac{\partial {\cal L}}{\partial \theta} = W[/math] where [math]W[/math] is the work done by the disapative force, in your case [math]W=-A \dot{\theta}[/math], where [math]A[/math] is a positive constant.
Klaynos Posted January 16, 2008 Author Posted January 16, 2008 Thanks Severian That'd be why it was giving me so many conceptual problems trying to think about how it should work Every time I sat down and thought about it I just bit a brick wall thinking it can't work any way I thought of...
Severian Posted January 16, 2008 Posted January 16, 2008 Sorry, I am getting my terms mixed up. W isn't the work done - it is the disapative force. The integral over it is the work done.
Resha Caner Posted March 1, 2008 Posted March 1, 2008 Didn't someone later modify Lagrangian mechanics to add non-conservative forces? I seem to remember someone doing that. Or is that just my heritage with engineering kluges? Caner
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