# Ex 1.29 A particle of mass m slides off a horizontal cylinder, 1.1

Structure and Interpretation of Classical Mechanics

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A particle of mass $m$ slides off a horizontal cylinder of radius $R$ in a uniform gravitational field with acceleration $g$. If the particle starts close to the top of the cylinder with zero initial speed, with what angular velocity does it leave the cylinder?

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Along the tangential direction,

$\displaystyle{m \frac{dv}{dt} = m g \sin \theta - f_a - f}$

Assuming there is only air friction,

$\displaystyle{m \frac{dv}{dt} = m g \sin \theta - f_a}$

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If the air resistance $\displaystyle{f_a}$ equals $\displaystyle{\frac{\beta m v^2}{2 R}}$,

$\displaystyle{ m \frac{dv}{dt} = m g \sin \theta - \frac{\beta m v^2}{2 R} }$

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Along the normal direction,

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

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

So

\displaystyle{\begin{aligned} m R \frac{d \dot \theta}{dt} &= m g \sin \theta - \frac{\beta}{2} \left( m g \cos \theta - F_R \right) \\ R \ddot \theta &= g \sin \theta - \frac{\beta}{2} \left( g \cos \theta - F_R \right) \\ \end{aligned}}

This equation is not useful yet, because $\displaystyle{F_R(\theta(t))}$ is still not known. So we keep using the original equation:

\displaystyle{\begin{aligned} m \frac{dv}{dt} &= m g \sin \theta - \frac{\beta m v^2}{2 R} \\ R \frac{d^2 \theta}{dt^2} &= g \sin \theta - \frac{\beta R \dot \theta^2}{2} \\ \end{aligned}}

Let

\displaystyle{\begin{aligned} u &= \dot \theta^2 \\ \end{aligned}}

— Me@2023-05-23 11:02:25 AM

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