# Lagrangian: Bead on a rotating hoop with mass

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1. May 12, 2015

### raggle

1. The problem statement, all variables and given/known data
'Consider the system consisting of a bead of mass m sliding on a smooth circular wire hoop of mass 2m and radius R in a vertical plane, and the vertical plane containing the hoop is free to rotate about the vertical axis. Determine all relative equilibria of the bead.'

2. Relevant equations
$$T_r= \frac{1}{2}I\omega^2$$
The kinetic energy of the bead is
$$T=\frac{1}{2}m(\dot\theta^2+\dot\phi^2\sin^2\theta) + \frac{1}{2}mR^2\sin^2\theta\dot\phi^2$$
The kinetic energy of the hoop is
$$T=\frac{1}{2}mR^2\dot\phi^2$$
3. The attempt at a solution
First of all I said the bead is at position
$$r=R(\sin\theta\cos\phi, \sin\theta\sin\phi, 1-\cos\theta)$$
the equation for the z coordinate looks a bit odd but I think it's right (using this mean that the bead is at height 0 when it's at the bottom of the hoop and at 2R when it's at the top of the hoop). Going through the usual steps to get the Lagrangian gives (I can go into more detail on this if anyone wants)
$$L=\frac{1}{2}[\dot\theta^2+\dot\phi^2(1+2\sin^2\theta)]+mgR\cos\theta$$

Since I'm looking for solutions where $\theta$ is constant I rewrite this as
$$L=\frac{1}{mR^2}[\frac{1}{2}\dot\theta^2 -V]$$
where
$$V=-\frac{1}{mR^2}[\frac{1}{2}\dot\phi^2(1+2\sin^2\theta)+mgR\cos\theta]$$

This gives
$$\ddot\theta=-\frac{\partial V}{\partial\theta}$$
Setting this equal to zero gives
$$2\dot\phi^2\sin\theta\cos\theta-mgR\sin\theta=0$$

Giving 3 equilibria, one at $\theta=0$, one at $\theta=\pi$ and one at $\cos\theta=\frac{g}{2R\dot\phi^2}$.
I'm not sure about the third solution at all. The closest I could find was a similar problem in Tong's lecture notes, where the third solution is $\cos\theta=\frac{g}{R\dot\phi^2}$ (page 27). The factor of 2 is bothering me, is it from me including the moment of inertia of the hoop?

2. May 12, 2015

### BvU

For the first equation: typo? or error in the dimensions ?
For te second: wasn't the ring mass equal to 2m ?

Since you made the invitation "I can go into more detail on this if anyone wants", why not show the steps ?

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3. May 12, 2015

### raggle

You're right about the first equation, it should be
$$T=\frac{1}{2}mR^2(\dot\theta^2+\dot\phi^2\sin^2\theta) + \frac{1}{2}mR^2\sin^2\theta\dot\phi^2$$

You're also right about the second equation, I thought the moment of inertia of a hoop was $\frac{1}{2}mR^2$ but it's actually $mR^2$.

I figured out where I was going wrong, I accidentally count the rotational kinetic energy of the bead twice. That gives me the factor of 2 in the equation of motion. The actual Lagrangian is
$$L=\frac{1}{2}[\dot\theta^2+\dot\phi^2(1+\sin^2\theta)]+mgR\cos\theta$$

4. May 12, 2015

### BvU

the moment of inertia of a hoop is $\tfrac{1}{2}mR^2$ for rotation about a middle line. But the mass is 2mbead.