Lagrange's equations: What is $\dot{q}_j$? I'm looking at the solutions to a problem about a uniform thin disk. For the sake of this question, I start with
$$L=\frac{1}{2}m\left( r\omega \right)^2$$
Then we plug it into Lagrange's equations:
$$\begin{align*}
\frac{d}{dt} \frac{\partial L}{\partial \dot{q}_j} - \frac{\partial L}{\partial q_j} &= Q\\
\frac{d}{dt} \frac{\partial \frac{1}{2}m\left( r\omega \right)^2}{\partial \left( r\omega \right)} - \frac{\partial \frac{1}{2}m\left( r\omega \right)^2}{\partial q_j} &= Q
\end{align*}$$
How is it that $\dot{q}_j = r\omega$?
What is $q_j$ then, as well? I'm thinking along the lines of
$$\begin{align*}
\dot{q}_j &= r\omega\\
\dot{q}_j &= r\frac{d\theta}{dt}\\
q_j &= r\theta
\end{align*}$$
 A: You have it backwards. When trying to describe some system, you first need to find kinematical parameters that describe possible configurations the system could be in. Here, in the most simple picture, the disk is characterized by a single parameter $q \equiv \phi$, the angle it has rotated (since the beginning of the experiment, say) and the corresponding velocity parameter $\dot q = \dot \phi = \omega$. Once you have your parameters, you can try to express kinetic and potential energy in terms of them and finally you can write a Lagrangian $L = T(\dot \phi) - V(\phi)$. Here you have only kinetic energy and it corresponds (contrary to what you write) to a particle constrained to a circle of radius $r$ (the actual disk would carry a different factor in front of $mv^2 / 2 = m (r \omega)^2 / 2$ due to the fact that its center of mass is at the position $r_0 < r$).
Okay, with that out of the way, you have Lagrange's equations
$${{\rm d} \over {\rm d} t} {\partial L \over \partial {\omega}} - {\partial L \over \partial {\phi}} = 0.$$
Can you take it from here?
