Lagrange equations of a spinning parabola

1. Apr 16, 2013

tehdiddulator

1. The problem statement, all variables and given/known data

Consider a bead of mass m sliding without friction on a wire that is bent in the shape of a parabola and is being spun with constant angular velocity ω about its vertical axis. Use cylindrical polar coordinates and let the equation of the parabola be $z = kρ^{2}$. Write down the lagrangian in terms of ρ as the generalized coordinate. Find the equation of motion.

2. Relevant equations

$$L = T-U = \frac{1}{2}mv^2-mgy$$

3. The attempt at a solution

I just need a hint on how to set up the kinetic energy of the parabola, maybe someone can explain it in a different way that would push me in the right direction.

Also, sorry if this isn't considered advanced physics. This is a classical mechanics class (upper level at my school) and it seems like a gray area for me.
Thanks!

Last edited by a moderator: Apr 16, 2013
2. Apr 16, 2013

sunrah

you need to change your coordinates to suit the problem better

3. Apr 16, 2013

vela

Staff Emeritus
A straightforward method is to write down expressions for x(t), y(t), and z(t) and then differentiate them with respect to time. For example, $x(t) = \rho\cos \phi$, so
$$\dot{x}(t) = \dot{\rho}\cos \phi - \rho\sin\phi \,\dot{\phi}.$$ When you have all three, plug them into $v^2 = \dot{x}(t)^2 + \dot{y}(t)^2 + \dot{z}(t)^2$.

By the way, I think the potential term in the Lagrangian should be mgz, not mgy.

It's in the right place.

4. Apr 16, 2013

tehdiddulator

Ah, I suppose I should add in the major fact that they want it in terms of the generalized coordinate rho. I can try and find/make a picture of the graph if it helps.

and yes, you are correct, the potential would be mgz.

5. Apr 16, 2013

vela

Staff Emeritus
The expression you end up with will be in terms of $\rho$.

6. Apr 16, 2013

tehdiddulator

Everything worked out properly! Thanks a bunch for your help.