- #1

JyJ

- 18

- 0

## Homework Statement

A circular hoop of radius R rotates with angular frequency ω about a vertical axis coincident with its diameter. A bead of mass m slides frictionlessly under gravity on the hoop. Let θ be the bead’s angular position relative to the vertical (so that θ = 0 corresponds to the bead being at the bottom of the hoop).

Under what conditions can the bead sit at a fixed angle θ_0 not equal to 0 or π?

Show that this situation corresponds to a stable equilibrium, and determine the frequency of small oscillations about this stable equilibrium.

## Homework Equations

In cylindrical coordinates:

$$\vec{v} = \dot{r}\vec{e_r} + r\dot{\theta}\vec{e_\theta} + rsin(\theta)\dot{\phi}\vec{e_\phi}$$

## The Attempt at a Solution

By using Lagrangian methods I found that the equation of motion is:

$$mR^2\ddot{\theta} + mgRsin(\theta)-mR^2(\omega)^2sin(\theta)cos(\theta)=0$$ which agrees with the solution given.

By plugging:

$$\ddot{\theta} = \dot{\theta} = 0$$

For the first question I got the following as a condition on θ_0:

$$\omega = \sqrt{\frac{g}{cos(\theta_0)}}$$

I also then computed the frequency of small deviations from equilibrium by using Taylor expansion with:

$$\theta = \theta_0 + a$$

where a is the small deviation of order h from stable equilibrium. Them by getting rid off terms of order h

^{2}or higher I found the frequency to be:

$$R^2\ddot{a} + gRsin(a+\theta_0) - \frac{gR^2}{cos(\theta_0)}sin(a+\theta_0)cos(a+\theta_0)=0

\\ R^2\ddot{a} + gR(a+\theta_0)-\frac{gR^2}{cos(\theta_0)}(a+\theta_0) = 0$$

which leads to frequency equal to:

$$\frac{\frac{Rg}{cos(\theta_0)}-g}{R}$$

This is how I think the frequency would be found.

Now, the only bit that is left to prove is to show that θ=θ_0 indeed corresponds to the stable equilibrium. I am having difficulties with this part as I am not sure how to show it using the information I have. Typically I would be able to estimate stable equilibrium from the minimum of the potential but in this case my potential is dependent on both y and x and so I am unsure what to do:

$$ U = -mgRcos(\theta) = -mgRcos(arctan\frac{y}{x}) $$

I would appreciate any comments and advice!