# 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.