GATE 2021 Mechanical Engineering Q30

Question

Consider a two degree of freedom system as shown in the figure, where PQ is a rigid uniform rod of length, b and mass, m.
Assume that the spring deflects only horizontally and force F is applied horizontally at Q. For this system, the Lagrangian, L is

Accepted Answer

$\frac{1}{2}(M+m)\dot{x}^2 + \frac{1}{2}mb\dot{	heta}\dot{x}\cos	heta + \frac{1}{6}mb^2\dot{	heta}^2 - \frac{1}{2}kx^2 + mg\frac{b}{2}\cos	heta$

Answer

$\frac{1}{2}(M+m)\dot{x}^2 + \frac{1}{6}mb^2\dot{	heta}^2 - \frac{1}{2}kx^2 + mg\frac{b}{2}\cos	heta$

Answer

$\frac{1}{2}M\dot{x}^2 + \frac{1}{2}mb\dot{	heta}\dot{x}\cos	heta + \frac{1}{6}mb^2\dot{	heta}^2 - \frac{1}{2}kx^2$

Answer

$\frac{1}{2}M\dot{x}^2 + \frac{1}{2}mb\dot{	heta}\dot{x}\cos	heta + \frac{1}{6}mb^2\dot{	heta}^2 - \frac{1}{2}kx^2 + mg\frac{b}{2}\cos	heta + Fb\sin	heta$