Lagrange's Undetermined Multipliers

[tshafer]

Homework Statement

We have a hoop of radius r and mass m resting on a cylinder of radius R which rolls without slipping on the cylinder under the influence of gravity. If the hoop begins rolling from rest at the top of the cylinder, at what point does the hoop fall off of the cylinder?

Homework Equations

Hoop fixed on cylinder: f_1 = (R+r)-\xi = 0, \xi is the distance from the center of the cylinder to center of the hoop.
Rolling w/out slipping: f_2 = Rd\theta - rd\phi = 0

L = \frac 1 2 m \left( \dot{\xi}^2 + \xi^2\dot{\theta}^2 + r^2\dot{\phi}^2 \right) - mg\xi\cos\theta

The Attempt at a Solution

\frac{d}{dt}\frac{\partial L}{\partial\dot{\xi}} - \frac{\partial L}{\partial\xi} \implies -m\xi\dot\theta^2+mg\cos\theta = -\lambda_1

\frac{d}{dt}\frac{\partial L}{\partial\dot{\theta}} - \frac{\partial L}{\partial\theta} \implies m\xi^2\ddot{\theta}-mg\xi\sin\theta=R\lambda_2

\frac{d}{dt}\frac{\partial L}{\partial\dot{\phi}} - \frac{\partial L}{\partial\phi} \implies mr^2\ddot\phi = -r\lambda_2

I'm pretty sure the Lagrangian/E-L equations are wrong... I end up with a nonsensical expression for \lambda_2. Am I thinking about the problem incorrectly?

 

[gabbagabbahey]

L = \frac 1 2 m \left( \dot{\xi}^2 + \xi^2\dot{\theta}^2 + r^2\dot{\phi}^2 \right) - mg\xi\cos\theta

Why are you only including the energy of the hoop? Is the cylinder fixed or is it free to rotate? Also, I assume the cylinder is resting on the ground?