Solving the Rotating Wire Problem

[Pushoam]

Homework Statement

A circular wire hoop rotates with constant angular velocity ! about a vertical diameter. A small bead moves, without friction, along the hoop. Find the equilibrium position of the particle and calculate the frequency of small oscillations about this position.

Homework Equations

The Attempt at a Solution

Let's take our reference inertial frame as a spherical co-ordinate system whose axis is along the axis of rotation of the who and whose origin is at the centre of the hoop.
Then position of M can be given by r, $\theta, \phi$.
Constraint: r= R,
$\dot \phi = \omega$ , constant.
There are two generalised coordinates $\phi$ and $\theta$.

L = T - U

$T = \frac 1 2 mR^2 ( {\dot \theta}^2 + \sin ^2 \theta {\dot \phi}^2)$
Taking U = 0 at the origin, U = mgR $\cos \theta$
So, L = $\frac 1 2 mR^2 ( {\dot \theta}^2 + \sin ^2 \theta {\dot \phi}^2 ) - mgR$ $\cos \theta$

Lagrange's equation of motion gives,
$\ddot \theta = \frac g R \sin {\theta} + \sin {\theta}~ \cos { \theta}~{ \dot \phi}^2$

At eqbm. $\ddot \theta = 0$
$\theta = 0, \Pi , \cos {\theta} = \frac {-g} { \omega^2 R}$

What to do next?

 

[kuruman]

What to do next?

Rewrite your equation of motion in terms of a small angular departure from equilibrium and see if you can bring the equation into the form of a harmonic oscillator.
Let $\theta \rightarrow \theta_0+\alpha$, where $\theta_0 = \arccos(-g/(\omega^2 R))$ and expand the EOM for small values of $\alpha$. Don't forget that $\sin \theta \cos \theta = \sin(2\theta)/2$.

 

[vela]

Where are you measuring $\theta$ from? If $\theta=0$ is supposed to be when the bead is at the bottom of the loop, then the sign of $U$ is wrong.

 

[kuruman]

Where are you measuring $\theta$ from? If $\theta=0$ is supposed to be when the bead is at the bottom of the loop, then the sign of $U$ is wrong.

It all makes sense if $\theta$ is measured from the top. Then the equilibrium angle is (as it should) below the horizontal diameter and its cosine is negative. U is zero when $\theta = \pi/2$.

 

[vela]

Yeah, I know. It's just that most people use the convention that $\theta=0$ is the (stable) equilibrium position when the hoop isn't spinning. Thought it was worth making sure.

 

[Pushoam]

I solved it. Thank you all for the guidance.