# Generalized momentum conjugate and potential $U(q, \dot q)$

On Goldstein's "Classical Mechanics" (first ed.), I have read that

if $$q_j$$ is a cyclic coordinate, its generalized momentum conjugate $$p_j$$ is costant.

He obtained that starting from Lagrange's equation: $$\frac {d}{dt} \frac {\partial L}{\partial \dot q_j}- \frac {\partial L}{\partial q_j}=0.$$

But this Lagrange's equation refers to a conservative system. What would happen if I considered a system in which the potential is $$U(q, \dot q)$$?

OP wrote(v1):

What would happen if I considered a system in which the potential is [velocity-dependent] $U(q, \dot q)$?

Well, if OP already knows that the generalized force$^1$

$$\tag{1} Q_j~=~\frac {d}{dt} \frac {\partial U}{\partial \dot q^j}- \frac {\partial U}{\partial q^j}$$

is given in terms of a velocity-dependent potential $U=U(q, \dot q, t)$, this means that Lagrange's equations

$$\tag{2}\frac {d}{dt} \frac {\partial T}{\partial \dot q^j}- \frac {\partial T}{\partial q^j}~=~Q_j,$$

can be written as

$$\tag{3}\frac {d}{dt} \frac {\partial L}{\partial \dot q^j}- \frac {\partial L}{\partial q^j}~=~0, \qquad L~=~T-U.$$

As long as one has Lagrange's equations (3), then it is still true that if $q^j$ is a cyclic coordinate $\frac {\partial L}{\partial q^j}=0$, then its generalized momentum conjugate $p_j:=\frac {\partial L}{\partial \dot q^j}$ is a constant of motion.

$^1$ Here we consider for simplicity a system with only one type of generalized force. In practice, there may be several types of forces (e.g. gravity force, Lorentz force, etc.). The generalization is straightforward.