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

Structure and Interpretation of Classical Mechanics


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?


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}


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}    }


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.


\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}}


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

— Me@2023-05-23 11:02:25 AM



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