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
A. increases till mass falls into hole
B. decreases till mass falls into hole
C. remains constant
D. becomes zero at radius r1, where 0 < r1 < r0
A. $$\frac{c}{3}$$
B. $$\frac{{\sqrt 2 }}{3}c$$
C. $$\frac{c}{2}$$
D. $$\frac{{\sqrt 3 }}{2}c$$
The Hamiltonian corresponding to the Lagrangian $$L = a{{\dot x}^2} + b{{\dot y}^2} - kxy$$ is
A. $$\frac{{{p_x}^2}}{{2a}} + \frac{{{p_y}^2}}{{2b}} + kxy$$
B. $$\frac{{{p_x}^2}}{{4a}} + \frac{{{p_y}^2}}{{4b}} - kxy$$
C. $$\frac{{{p_x}^2}}{{4a}} + \frac{{{p_y}^2}}{{4b}} + kxy$$
D. $$\frac{{{p_x}^2 + {p_y}^2}}{{4ab}} + kxy$$
A. circular
B. elliptical
C. parabolic
D. hyperbolic
Join The Discussion