Sunday, 9 February 2020

classical mechanics - Friction in Lagrangian formulation


We know the Lagrange equations are: $$\frac{\partial \mathcal{L}}{\partial q_i}-\frac{d}{dt}\left(\frac{\partial \mathcal{L}}{\partial \dot{q_i}}\right)=0.$$ Then, when we add friction in there, we rewrite it in terms of the Rayleigh dissipation function as $$\frac{\partial \mathcal{L}}{\partial q_i}-\frac{d}{dt}\left(\frac{\partial \mathcal{L}}{\partial \dot{q_i}}\right)=\frac{\partial \mathcal{F}}{\partial \dot{q_i}}.$$ However, this is assuming that you can write the friction force as $F_f=-k\dot{q_i}$. How would you do it for a case where the friction force is not proportional to velocity? For example, in the case of a block sliding on a flat surface the friction would be $F_f=-m\ddot{q}$, which is proportional to the acceleration not the velocity.




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