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ProblemEdit

Rotating tube
A particle of mass m is placed in a smooth uniform tube of mass M and length l.  The tube is free to rotate about its center in a vertical plane.  The system is started from rest with the tube horizontal and the particle a distance $ r_0 $ from the center of the tube.


For what length of the tube will the particle leave the tube when $ \dot{\theta}=\omega $ is a maximum and $ \theta=\theta_m $?  Your answer should be in terms of $ \omega $ and $ \theta_m $.

SolutionEdit

Find the Lagrangian and Euler-Langrange equation of motion
$ T=\frac{1}{2}m\dot{r}^2+\frac{1}{2}mr^2\dot{\theta}^2+\frac{1}{2}I\dot{\theta}^2 $ (where $ I=\frac{1}{12}M l^2 $ is the moment of inertia of the tube about its center of mass).

$ V=-mg\frac{l}{2}\cos{\theta} $

$ L=\frac{1}{2}m\dot{r}^2+\frac{1}{2}mr^2\dot{\theta}^2+\frac{1}{2}I\dot{\theta}^2 +mg\frac{l}{2}\cos{\theta} $

so
$ \ddot{\theta}=\frac{-1}{mr^2+I}\left(2mr\dot{r}\dot{\theta}+mg\frac{l}{2}\sin{\theta}\right) $

At the moment when $ \dot{\theta} $ is at a maximum:
$ \ddot{\theta}=0 $
$ r=l/2 $
$ \dot{\theta}=\omega $
$ \theta=\theta_m $

so

$ \dot{r}_m\omega=\frac{-g}{2}\sin{\theta_m} $

or

$ \dot{r}_m^2=\frac{g^2}{4\omega^2}\sin^2{\theta_m} $

where $ \dot{r}_m $ is the radial component of velocity when $ \theta=\theta_m $. $ \dot{r}_m $ can be found by conservation of energy, namely

$ E_o=0=E_f $

$ E_f=\frac{1}{2}m\dot{r}_m^2+\frac{1}{2}mr_{m}^2\omega^2+\frac{1}{2}I\omega^2-mgr_m\cos{\theta_m}=0 $

subbing $ r_m=\frac{l}{2} $

$ \dot{r}_m^2=gl\cos{\theta_m}-\left(\frac{l}{2}\right)^2\omega^2-\frac{I\omega^2}{m} $

Therefore,

$ \frac{g^2}{4\omega^2}\sin^2{\theta_m}=gl\cos{\theta_m}-\left(\frac{l}{2}\right)^2\omega^2-\frac{I\omega^2}{m} $

or

$ \left(\frac{3m+M}{3m}\right)\omega^2l^2-(4gl\cos{\theta_m})+\frac{g^2}{\omega^2}\sin^2{\theta_m}=0 $

which is a quadratic equation in $ l $, which can be solved for in terms of g, m, M, $ \omega $, and $ \theta_m $