Lagrange equations of a spinning parabola

AI Thread Summary
The discussion focuses on deriving the Lagrangian for a bead sliding on a parabolic wire while spinning with constant angular velocity. The equation of the parabola is given as z = kρ², and the Lagrangian is expressed in terms of the generalized coordinate ρ. Participants suggest setting up the kinetic energy by expressing the coordinates x(t), y(t), and z(t) in cylindrical polar coordinates and differentiating them to find the velocity. There is clarification that the potential energy term should be mgz instead of mgy. The conversation emphasizes the importance of correctly transforming coordinates to solve the problem effectively.
tehdiddulator
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Homework Statement



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.

Homework Equations



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

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!
 
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you need to change your coordinates to suit the problem better
 
tehdiddulator said:

Homework Statement



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.

Homework Equations



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

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.
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.

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!
It's in the right place.
 
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.
 
The expression you end up with will be in terms of ##\rho##.
 
Everything worked out properly! Thanks a bunch for your help.
 
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