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## Homework Statement

Consider a bead of mass m that is confined to move on a circular hoop of radius r. The axis of symmetry of the hoop is horizontal, and the hoop is rotating about a vertical axis at a uniform rate [itex]\omega[/itex]. Neglect friction and assume a constant gravitational acceleration of magnitude g. Let the angle [itex]\phi[/itex] specify the position of the bead on the hoop, with [itex]\phi = 0[/itex] corresponding to the bead lying at the bottom of the hoop.

Derive the following Lagrangian:

[tex] L = T-V = \frac{1]{2}mr^2(\dot{\phi}^2 + \omega^2\sin^2\phi) + mgr\cos\phi [/tex]

Since the Lagrangian doesn't depend explicitly on time, there is a conserved quantity for this system, [itex]E = \dot{\phi}\frac{\partial L}{\partial \dot{\phi}}[/itex]. Compute E and show that it is constant. Is the constant the total energy of the system T+V?

## Homework Equations

Most relevant equations are there

## The Attempt at a Solution

I thought I computed the Lagrangian just fine. My logic was that there were two sources of kinetic energy: that of the bead moving in circular motion ([itex]mr^2\dot{\phi}^2/2[/itex]) and that of the bead rotating. My thinking was that as the bead rotated, if it was at an angle [itex]\phi[/itex], then it was tracing a circle of radius [itex]\r\sin\phi[/itex], so its kinetic energy would be [itex]mr^2\omega^2\sin^2\phi/2[/itex].

I think the potential energy is purely gravitational. If h=0 corresponds to the height of the center of the hoop, then the potential energy is [itex]-mgr\cos\phi[/itex]. This gives the correct Lagrangian.

However, it does not give the correct energy. When I compute E, it comes out to be [itex]1/2 mr^2(\dot{\phi}^2 - \omega^2\sin^2\phi - mgr\sin\phi[/itex]. That middle term has the wrong sign, but I can't figure out why. Moreover, this version of E is not conserved. The only conclusion I can reach is that somehow, what I took as rotational kinetic energy is actually potential energy, but that doesn't make sense. Any thoughts? Thank you!