15 | June | 2023 | Physics Town

Structure and Interpretation of Classical Mechanics

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

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

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Daily Archives: June 15, 2023

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