Using Lagrange to solve rotating parabolic motion and equilibrium

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The discussion focuses on using Lagrangian mechanics to analyze a bead sliding on a parabolic wire while spinning with constant angular velocity. The Lagrangian is expressed in terms of the generalized coordinate p, leading to the derived equation of motion. Equilibrium positions occur when the bead remains fixed, specifically when the condition w² = 2kg is satisfied, indicating potential equilibrium at p = 0. The stability of these equilibrium positions requires further analysis of the second derivative of the potential energy with respect to p. Understanding these concepts is crucial for solving the problem effectively.
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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 w about its vertical axis. Use cylindrical polar coordinates and let the equation of the parabola be z = kp2. Write down the Lagrangian in terms of p as the generalized coordinate. Find the equation of motion of the bead and determine whether there are positions of equilibrium, that is, values of p at which the bead can remain fixed, without sliding up or down the spinning wire. Discuss the stability of any equilibrium positions you find.

Homework Equations

L=E-U
v1=pw
v2=dp/dt
v3=2kp(dp/dt)
U=mgkp^2
E=mv^2/2

The Attempt at a Solution



I found the equation of motion using the lagrangian, which matched the answer in the back of the book. I just don't know how to use Lagrance to find equilibrium, and I've tried several things such as setting d2p/dt2=0 and d2U/dp2=0.The equation of motion is: (g=accel. due to gravity)

pw2-4k2p(dp/dt)2-2kgp=(d2p/dt2)(1+4k2p2
 
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If the bead remains fixed, what can be said about the derivatives of its polar radius?
 
dp/dt=0, d2p/dt2=0

so equil. occurs when w2=2kg, or when p=0. Is that right? And then how can I find which is stable/unstable?
 
Have you studied any equilibrium conditions/criteria?
 

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