Bead on a wire, potential energy

[Habeebe]

Homework Statement

A particle is constrained to move without friction on a circular wire rotating with constant speed ω about a vertical diameter. Find the equilibrium position of the particle, and calculate the frequency of small oscillations around this position. Find and interpret physically a critical angular velocity, ω = ω_c, that divides the particle’s motion into two distinct types.

Homework Equations

U=-∫Fdr

The Attempt at a Solution

I set it up in spherical coordinates. The wire is [itex]\rho^2+z^2=R^2[/itex] and it is being rotated about the z axis (rho lies in x-y plane). I set [itex]\theta[/itex] to be the angle measured around the circle counterclockwise from the rho axis. The forces acting on the particle are gravity (mg) and centripetal force ([itex]-m\rho\omega^2[/itex]). The potential due to gravity is then mgy and the potential due to centripetal force is:

[itex]\int_{0}^{\rho}m\rho\omega^2 d\rho=\frac{1}{2}m\omega^2\rho^2=\frac{1}{2}m\omega^2R^2cos^2(\theta)[/itex]
Gravitational potential becomes mgy=mgRsin(θ). Now my total potential energy is:
[itex]U=mgRsin(\theta)+\frac{1}{2}m\omega^2R^2cos^2(\theta)[/itex]

I differentiate with respect to θ and set equal to zero:
[itex]U_\theta=mgRcos(\theta)-m\omega^2R^2cos(\theta)sin(\theta)=0[/itex]
[itex]gcos(\theta)=\omega^2Rcos(\theta)sin(\theta)[/itex]
[itex]sin(\theta)=\frac{g}{R\omega^2}[/itex]
[itex]\theta=arcsin(\frac{g}{R\omega^2})[/itex]

I'm concerned about this answer. As ω→0, θ diverges. I should get θ → -π/2

Where did I screw this up? Thank you.

 

[mark726]

Thank you for your question. I would like to offer some feedback on your solution attempt.

Firstly, your set-up in spherical coordinates seems appropriate for this problem. However, I believe you have made a mistake in your calculation of the potential due to centripetal force. The correct expression should be:

U = -∫Fdr = -∫(-mρω²)dr = ∫mρω²dr = 1/2mω²ρ²

This is because the force is acting in the ρ direction, and the potential energy is defined as the negative of the work done by the force. This result is consistent with your expression for the potential due to gravity.

Secondly, when you differentiate the potential energy with respect to θ, you should also take into account the chain rule. The correct expression should be:

U_θ = mgRcos(θ) - mω²R²cos(θ)sin(θ)θ' = 0

where θ' is the derivative of θ with respect to time. This will give you a different expression for θ, which will not diverge as ω→0.

Finally, I believe the interpretation of the critical angular velocity ωc is that it separates the motion of the particle into two distinct types: stable circular motion when ω < ωc, and unstable circular motion when ω > ωc. This can be seen from the fact that the equilibrium position θ is only defined for ω < ωc, and for ω > ωc, there is no equilibrium position and the particle will spiral outward or inward along the wire.

I hope this helps clarify your solution attempt. Keep up the good work!