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In summary, the conversation discusses a problem involving a bead sliding on a circular hoop that is rotating at a fixed angular velocity. The kinetic and potential energy of the bead are calculated, and the equation of motion is derived using the Lagrangian. The stability of the equilibrium position is then analyzed, with the condition for stability being \omega^{2}a < g. If this condition is not met, a new equilibrium position is found and the angular frequency of small oscillations around this position is calculated.

## Homework Statement

Consider a bead of mass m that can slide freely on a circular hoop of radius a which lies in a vertical plane. The hoop rotates about a vertical axis through its centre at fixed angular velocity $$\omega$$. The angle $$\theta$$ is the polar angle of the bead measured from the downward vertical.

1. Find the kinetic and potential energy of the bead.

2. Find the bead's equation of motion.

3. Show that for $$\omega^{2}a < g$$ the bead can remain in a stable equilibrium at the bottom of the hoop, otherwise this position is unstable.

4. In the case $$\omega^{2}a > g$$ find the new position $$\theta^{*}$$ of stable equilibrium and calculate the angular frequancy $$\Omega$$ of small oscillations about this position.

## Homework Equations

I worked out the answers for the first 2 questions myself (see below)

## The Attempt at a Solution

1. The kinetic energy is $$\stackrel{1}{2}ma^{2}(\omega^{2}sin^{2}(\theta)+ \stackrel{.}{\theta}^{2})$$

The potential energy is mga$$(1-cos(\theta))$$

The above are just from the geometry of the situation

2. The equation of motion of the bead is $$\stackrel{..}{\theta} = (\omega^{2}cos(\theta) - g/a)sin(\theta)$$

This just comes from using the Lagrangian L=T+V and treating it as a function of $$\theta$$ and $$\stackrel{.}{\theta}$$

3 and 4. This is where I am getting stuck! When I set to zero $$dv/d\theta$$ I get $$mgasin(\theta)=0$$ which does not relate to $$\omega$$! Would it be necessary to use minimise an effective potential instead? I've tried this but the algebra is not going anywhere and a little insight would be helpful.

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I don't think your kinetic energy is right. You should have three terms since the kinetic energy ought to be 1/2*m*v.v, where v velocity of the bead.

I agree with his Kinetic energy expression - the velocities are confined to different planes- this should mean the dot product between is zero. Or have I screwed my geometry up there something rotten?

The KE is correct. It's the sum of two orthogonal velocities, the part along the hoop and the part related to the rotation of the hoop.

As for the rest of the problem, you can think of the right side of the theta eom as a 'force'. The equiibrium points are where it vanishes. Take the theta=0 point, if you expand to first order in theta, you get
$$\stackrel{..}{\theta} = (\omega^{2} - g/a) \theta$$

Now you can see that if the factor in front of the theta is positive, then if theta moves away from zero it will keep accelerating away. If it's negative, it will accelerate back. It must be obvious where the other equilibrium point is, right? Play the same game there and write an expansion in terms of dtheta, where dtheta is the difference from equilibrium. To get the oscillation frequency, notice these expansions look like harmonic oscillators.

Yes, sorry. I read the question wrong!

## What is the purpose of a bead on a hoop in mechanics?

The purpose of a bead on a hoop in mechanics is to study the motion and forces acting on the bead as it moves along the hoop.

## How does the position of the bead affect its motion on the hoop?

The position of the bead along the hoop will determine the direction and magnitude of the forces acting on it, which in turn affects its motion.

## What is the role of friction in the mechanics of a bead on a hoop?

Friction plays a crucial role in the mechanics of a bead on a hoop as it affects the speed and direction of the bead's motion. Friction between the bead and the hoop can either increase or decrease the speed of the bead, and can also cause it to change direction.

## How does the weight of the bead affect its motion on the hoop?

The weight of the bead affects its motion on the hoop by creating a downward force that is counteracted by the normal force exerted by the hoop. This results in the bead moving in a circular path along the hoop.

## What factors can affect the stability of a bead on a hoop?

The stability of a bead on a hoop can be affected by factors such as the shape and size of the hoop, the weight and size of the bead, the speed and direction of its motion, and the amount of friction present between the bead and the hoop.