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## Homework Statement

A bead of mass m slides under gravity on a smooth rod of length l which is inclined at a constant angle ##\alpha## to the downward vertical and made to rotate at angular velocity ##\omega## about a vertical axis. The displacement of the bead along the rod is r(t).

A)Find the Lagrangian and the equation of motion of the bead, and obtain the general solution.

B) Show that if the bead is released from rest at r=0, it falls off the rod after a time interval $$t = \frac{1}{\omega \sin \alpha} \cosh^{-1} \left( 1 + \frac{l \omega^2 \sin^2 \alpha}{g \cos \alpha}\right)$$

## Homework Equations

Lagrangian L = T - V

## The Attempt at a Solution

Write the kinetic and potential energy of the bead. I get: $$T = \frac{m}{2}(\dot{r}^2 + r^2 \omega^2)\,\,\,\,\,\,\,\text{and}\,\,\,\,\,\,V = -mgr\cos \alpha$$

Using Lagrange's Eqns, this gives ##\ddot{r} - \omega^2 r = g \cos \alpha## so ##r = C_1 \exp(\omega t) + C_2 \exp(- \omega t) - g \cos \alpha/\omega^2##

If the bead is released at the origin, set t=0 as well for convenience then ##r(0) = \dot{r}(0) = 0##. Solve these to obtain ##C_1, C_2## to get $$r = \frac{g \cos \alpha}{\omega^2} \left( \frac{\exp(\omega t) + \exp(-\omega t)}{2}\right) - \frac{g \cos \alpha}{\omega^2}$$

It will fall off when r=l, so solving gives $$t = \frac{1}{\omega} \cosh^{-1} \left(1 + \frac{\omega^2l}{g \cos \alpha}\right)$$

I'm not really sure where the sin term comes in. The question states that the rod is inclined at a constant angle ##\alpha## to the downward vertical. But the rod rotates so I don't see how this is possible.

Many thanks.