Problem 6.8 in Morin's book on Classical Mechanics has this setup:
A massless stick pivots at its end in a horizontal plane with constant angular velocity $\omega$, while a frictionless bead of mass $m$ slides along it.
The goal is to compute the Lagrangian $L$ and the Hamiltonian $p\cdot \dot{q}-L=H$.($\ast$)
The solution makes sense to me when, using polar coordinates, the Lagrangian winds up being just the kinetic energy $L=\frac12 m\dot r^2+\frac12mr^2\dot\theta^2=\frac12 m\dot r^2+\frac12mr^2\omega^2$
because $\dot\theta=\omega$
Then to evaluate $H=\frac{\partial L}{\partial \dot\theta}\dot\theta+\frac{\partial L}{\partial\dot r}\dot r – L$, I think that
$\frac{\partial L}{\partial \dot\theta}=mr\dot\theta$ and $\frac{\partial L}{\partial\dot r}=m\dot r$, so that the terms before $-L$ in $H$ above become
$mr\dot\theta^2+m\dot r^2$
and therefore I thought $H=\frac12 m\dot r^2 + \frac12mr^2\dot\theta^2=\frac12 m\dot r^2 + \frac12mr^2\omega^2$.
But according to the solution, this ought to come out to just $m\dot r^2$ (as opposed to also having $mr\dot\theta^2$) and $H$ is supposed to be $\frac12 m\dot r^2 – \frac12mr^2\omega^2$.
What it looks like to me, is that rather than continue to use the form of $L$ with $\dot \theta$ in it to compute the conjugate momenta, $\dot\theta$ was immediately replaced with the constant $\omega$ so that the term $\frac12mr^2\omega^2$ became zero in in $\frac{\partial L}{\partial \dot\theta}$ (being a constant with respect to $\dot\theta$.)
So that is what I'm asking about: why should I believe we have license to substitute $\omega$ for $\dot\theta$ into $L$ before computing a partial derivative? I do not recall any explicit discussion of that but it may be in there: I only have the sample chapter, not the whole book. Obviously it does not yield the same answer if you substitute at the very end!
($\ast$) I guess probably I shouldn't call it the Hamiltonian because the book doesn't do that. The Hamiltonian is supposed to be a function of the coordinates and the conjugate momenta. But as I understand it, $H$ and $L$ are supposed to be related this way via the Legendre transformation.
Best Answer
"why should I believe we have license to substitute ω for $\dot{\theta}$ into L before computing a partial derivative?"
Good question. The reason is that here, unlike the more common case, $\theta$ is completely determined externally, even though it's not constant. The premise of the setup is that no matter what the bead does, its $\theta$-value is completely predetermined by the movement of the rod. For that reason it's an external field, not a coordinate.
The fundamental assumption of Lagrangian mechanics is that the system picks the path that minimizes the action (or at least a path that satisfies first-order conditions). For your system, you could surely lower the action still further if you could choose a different value for $\theta(t)$, but you're given that you cannot. All you can do is minimize over the coordinates that are given, which in this case is just $r(t)$.