The Lagrangian for a simple pendulum is given by L = 12mL^2dot ^2 - mgL( 1 - cos ) Hamiltonian equations are given by

dot p_ = - mgLsin ,dot = p_ mL^2

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Classical Mechanics

The Lagrangian for a simple pendulum is given by $$L = \frac{1}{2}m{L^2}{{\dot \theta }^2} - mgL\left( {1 - \cos \theta } \right)$$
Hamiltonian equations are given by

A) $${{\dot p}_\theta } = - mgL\,\sin \theta ,\,\dot \theta = \frac{{{p_\theta }}}{{m{L^2}}}$$ B) $${{\dot p}_\theta } = mgL\,\sin ,\,\dot \theta = \frac{{{p_\theta }}}{{m{L^2}}}$$ C) $${p_\theta } = - m\dot \theta ,\,\dot \theta = \frac{{{p_\theta }}}{m}$$ D) $${{\dot p}_\theta } = - \left( {\frac{g}{L}} \right)\theta ,\,\dot \theta = \frac{{{p_\theta }}}{{mL}}$$

Answer & Solution

Correct Answer: Option A

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The Lagrangian for a simple pendulum is given by L =...

The Lagrangian for a simple pendulum is given by $$L = \frac{1}{2}m{L^2}{{\dot \theta }^2} - mgL\left( {1 - \cos \theta } \right)$$ Hamiltonian equations are given by

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Related Questions on Classical Mechanics

The Lagrangian for a simple pendulum is given by $$L = \frac{1}{2}m{L^2}{{\dot \theta }^2} - mgL\left( {1 - \cos \theta } \right)$$
Hamiltonian equations are given by