Using Lagrange to solve rotating parabolic motion and equilibrium

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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?
 
To solve this, I first used the units to work out that a= m* a/m, i.e. t=z/λ. This would allow you to determine the time duration within an interval section by section and then add this to the previous ones to obtain the age of the respective layer. However, this would require a constant thickness per year for each interval. However, since this is most likely not the case, my next consideration was that the age must be the integral of a 1/λ(z) function, which I cannot model.
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