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 $$\begin{align*} \dot{q}_j &= r\omega\\ \dot{q}_j &= r\frac{d\theta}{dt}\\ q_j &= r\theta \end{align*}$$


1 Answer 1


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?

  • $\begingroup$ A generalized parameter $q_j$ (or $\dot{q}_j$ for that matter) is something that remains constant with respect to the particle whatever changes are made to $\mathbf{r}$ (the particle's radius vector with respect to some arbitrary origin). The usefulness of $q_j$ is that it is symmetrical wherever the particle is (or more formally, whatever $\mathbf{r}$ is). Am I correct in my understanding? $\endgroup$
    – Kit
    Commented Jul 19, 2011 at 8:09
  • $\begingroup$ @Kit: no, that is actually a complete opposite of the correct picture. As you vary $\mathbf r$, $q_j$s have to vary as well (otherwise they would not be good parameters). Consider a particle constrained to an unit circle. Its position is ${\mathbf r} = (x,y) = (\cos \phi, \sin \phi)$. You can see that as you vary $\mathbf r$ (respecting the constraint) the $\phi$ changes too. $\endgroup$
    – Marek
    Commented Jul 19, 2011 at 8:23
  • $\begingroup$ More generally, consider an arbitrary curved surface in 3D. This has no symmetry at all. But you can still introduce a coordinate chart on it (think e.g. about lattitude and longitude for Earth's surface). Again, for every point on the surface you have an unique pair of numbers that describe it. $\endgroup$
    – Marek
    Commented Jul 19, 2011 at 8:27

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.