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: Option A
This Question Belongs to Engineering Physics >> Classical Mechanics
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