28 | June | 2023 | Physics Town

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?

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Kinetic energy:

$\displaystyle{\begin{aligned} T(r(t), \theta(t)) &= \frac{1}{2} m \left( v_x^2 + v_y^2 \right) \\ &= \frac{1}{2} m \left[ \left( \frac{d}{dt} r \sin \theta \right)^2 + \left( \frac{d}{dt} r \cos \theta \right)^2 \right] \\ &= \frac{1}{2} m \left[ \left( \dot r \sin \theta + r \dot \theta \cos \theta \right)^2 + \left( \dot r \cos \theta - r \dot \theta \sin \theta \right)^2 \right] \\ &= \frac{1}{2} m \left[ \dot r^2 + r^2 \dot \theta^2 \right] \\ \end{aligned}}$

Potential energy:

$\displaystyle{\begin{aligned} V(r(t), \theta(t)) &= mgr \cos \theta \\ \end{aligned}}$

Constraint:

$\displaystyle{\begin{aligned} \phi(t) &= r(t) - R \\ &= 0 \\ \end{aligned}}$

The Lagrangian:

$\displaystyle{\begin{aligned} L(r(t), \theta(t)) &= T - V + \lambda \phi \\ &= \frac{1}{2} m ( \dot r^2 + r^2 \dot \theta^2 ) - mg r \cos \theta + \lambda (r - R) \\ \end{aligned}}$

The Lagrange equations:

$\displaystyle{ \begin{aligned} \frac{d}{dt} \left( \frac{\partial L}{\partial \dot r} \right) - \frac{\partial L}{\partial r} &= 0 \\ \frac{d}{dt} \left( \frac{\partial L}{\partial \dot \theta} \right) - \frac{\partial L}{\partial \theta} &= 0 \\ \frac{d}{dt} \left( \frac{\partial L}{\partial \dot \lambda} \right) - \frac{\partial L}{\partial \lambda} &= 0 \\ \end{aligned}}$

$\displaystyle{ \begin{aligned} \frac{d}{dt} \left( \frac{\partial L}{\partial \dot r} \right) - \frac{\partial L}{\partial r} &= 0 \\ \frac{d}{dt} m \dot r - \left( m r \dot \theta^2 - mg \cos \theta + \lambda \right) &= 0 \\ m \ddot r &= m r \dot \theta^2 - mg \cos \theta + \lambda \\ \end{aligned}}$

$\displaystyle{ \begin{aligned} \frac{d}{dt} \left( \frac{\partial L}{\partial \dot \theta} \right) - \frac{\partial L}{\partial \theta} &= 0 \\ \frac{d}{dt} \left( m r^2 \dot \theta \right) - mgr \sin \theta &= 0 \\ m \left( 2 r \dot r \dot \theta + r^2 \ddot \theta \right) - mgr \sin \theta &= 0 \\ \end{aligned}}$

$\displaystyle{ \begin{aligned} \frac{d}{dt} \left( \frac{\partial L}{\partial \dot \lambda} \right) - \frac{\partial L}{\partial \lambda} &= 0 \\ - ( r - R ) &= 0 \\ r(t) &= R \\ \end{aligned}}$