Using Lagrange to solve rotating parabolic motion and equilibrium

[relativespeak]

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

 

[voko]

If the bead remains fixed, what can be said about the derivatives of its polar radius?

 

[relativespeak]

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?

 

[voko]

Have you studied any equilibrium conditions/criteria?

 

[paranormal]

)-2k2p(dp/dt)2

To find equilibrium, we can set the right side of the equation to zero, which gives us:

dp/dt=0 and p=0 or dp/dt=2pw/3k and p=1/2pw2/3k2

These values of p represent positions where the bead can remain fixed without sliding up or down the spinning wire. We can determine the stability of these equilibrium positions by looking at the second derivative of the potential energy, which is related to the curvature of the parabola. If the second derivative is positive, the equilibrium position is stable, and if it is negative, the equilibrium position is unstable.

In this case, the second derivative of the potential energy is positive for both equilibrium positions, indicating that they are both stable. This is because the curvature of the parabola increases as p increases, making it more difficult for the bead to slide up or down the wire.

Overall, using Lagrange to solve for the equation of motion and determine equilibrium positions in rotating parabolic motion is a useful tool for analyzing and understanding the dynamics of this system.