Ex 1.29 A particle of mass m slides off a horizontal cylinder, 1.3

Posted on by

Structure and Interpretation of Classical Mechanics

.

…

\displaystyle{\begin{aligned} \dot \theta^2 &= \pm \frac{2 g (\beta \sin \theta - \cos \theta) }{R (\beta^2 + 1)} + \pm \frac{2 g }{R (\beta^2 + 1)} e^{-\beta \theta} \\ &= \pm \frac{2 g }{R (\beta^2 + 1)} \left[ \beta \sin \theta - \cos \theta + e^{-\beta \theta} \right] \\ \end{aligned}}

Let us reconsider the equation of motion along the normal direction:

\displaystyle{\begin{aligned} m g \cos \theta - F_R &= \frac{m v^2}{R} \\ \end{aligned}},

where \displaystyle{F_R} is the normal reaction force.

.

When the mass m leaves the cylinder with \theta=\theta_1, F_R = 0.

\displaystyle{\begin{aligned} m g \cos \theta_1 &= \frac{m R^2 \dot \theta^2}{R} \\ \dot \theta^2 &= \frac{g}{R} \cos \theta_1 \\ \\ \frac{g}{R} \cos \theta_1 &= \pm \frac{2 g }{R (\beta^2 + 1)} \left[ (\beta \sin \theta - \cos \theta) + e^{-\beta \theta} \right] \\ (1+ \beta^2) \cos \theta_1 &= 2 \left| ( \beta \sin \theta_1 - \cos \theta_1 + e^{-\beta \theta_1} ) \right| \\ \end{aligned}}

If there is no air resistance, \beta = 0,

\displaystyle{\begin{aligned} (1+ \beta^2) \cos \theta_1 &= 2 \left| ( \beta \sin \theta_1 - \cos \theta_1 + e^{-\beta \theta_1} ) \right| \\ \cos \theta_1 &= 2 ( 1 - \cos \theta_1) \\ \cos \theta_1 &= \frac{2}{3} \\ \end{aligned}}

— Me@2023-05-23 11:02:25 AM

.

.

2023.06.15 Thursday (c) All rights reserved by ACHK