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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

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