The Lagrangian of a particle' of mass m is L =...

Home / Engineering Physics / Classical Mechanics / Question

Examveda

The Lagrangian of a particle' of mass m is $$L = \frac{m}{2}\left[ {{{\left( {\frac{{dx}}{{dt}}} \right)}^2} + {{\left( {\frac{{dy}}{{dt}}} \right)}^2} + {{\left( {\frac{{dz}}{{dt}}} \right)}^2}} \right] - \frac{V}{2}\left( {{x^2} + {y^2}} \right) + W\sin \omega t$$
where V, W and ω are constants, then conserved quantities are

A. energy and z-component of linear momentum

B. energy and z-component of angular momentum

C. z-component of both linear and angular momentum

D. energy and z-component of both angular and linear momentum

Answer: Option A

Related Questions on 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

View Answer

A. $$\frac{c}{3}$$

B. $$\frac{{\sqrt 2 }}{3}c$$

C. $$\frac{c}{2}$$

D. $$\frac{{\sqrt 3 }}{2}c$$

View Answer

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. $$\frac{{{p_x}^2 + {p_y}^2}}{{4ab}} + kxy$$

View Answer

A. circular

B. elliptical

C. parabolic

D. hyperbolic

View Answer