# Can potential be velocity dependent?

In the lagrangian solution for the equation of motion, there's a seemingly out of place $$\frac{\mathrm{d} }{\mathrm{d} t}\frac{\partial V}{\partial \dot{q_j}}$$

term. Potential energy is usually a function of the set of $x_i$ or position only. If $x_i$ can all be rewritten as functions of only $q_i$ and $t$, and $q_i$ can be varied without having to change $\dot{q_i}$. Then we're left with this term being precisely $0$

So at least for conservative forces, this term should equal zero. But where do we find cases where it isn't? Magnetic forces? Is it frictional forces (what is potential for a frictional force anyway? And does Lagrange's equation even work for inelastic systems, considering energy is not conserved?)

By Hamilton's Principle we know that a system with a coordinate $q(t)$ that follows second-order differential equations on $t$ can be equivalently described by the minimization of the functional $$S[q(t)]=\int_{t_1}^{t_2}\mathrm{d}t\,L(q,\dot{q},t),$$ since a necessary condition for writing $\delta S=0$ is itself a second order differential equation for $q(t)$, given by the Euler-Lagrange equation: $$\frac{\partial{L}}{\partial q}-\frac{\mathrm{d}}{\mathrm{d}t}\frac{\partial{L}}{\partial \dot q}=0.$$ After this is defined, we can begin describing interactions.
Suppose our system can be described through a specific Lagrangian, and the intensity of the interactions is parametized by a continuous parameter $\lambda$ (this can be charge, mass, etc.) such that $L(\lambda)$ for $\lambda=0$ describes the motion of the system without any interaction. If we expand $L$ around $\lambda=0$, we get $$L=L(0)+\sum_{n=1}^{\infty}\lambda^n\frac{\partial^nL}{\partial\lambda^n},$$ and defining $L_{\mathrm{free}}=L(0)$, $L_{\mathrm{int}}=\sum_{n=1}^{\infty}\lambda^n\frac{\partial^nL}{\partial\lambda^n}$, we can write the Hamiltonian (i.e. the energy) of our system as $$H=-\{(L_{\mathrm{free}}+L_{\mathrm{int}})-\dot q\frac{\partial}{\partial\dot q}(L_{\mathrm{free}}+L_{\mathrm{int}})\}\\H=(\dot q\frac{\partial}{\partial\dot q}-1)L_{\mathrm{free}}+(\dot q\frac{\partial}{\partial\dot q}-1)L_{\mathrm{int}}.$$ From this we define the kinetic energy $K$ as the first term, and the interaction potential $U$ as the second term. In the specific case where the potential does not depend on the velocity, we see that $U=-L_{int}$, which is what we would use for the usual Lagrangian definition. The usual definition also has $K=L_{\mathrm{free}}$, however, which is only the case in nonrelativistic mechanics, which by solving the differential equation for $L_{\mathrm{free}}$ gives us $L_{\mathrm{free}}\propto\dot q^2$. In the case of frictional forces and inelastic systems, however, what we have is a time dependant Lagrangian, resulting in a time dependant energy and therefore dissipation.