JEE Main · PhysicsHard

$A thin uniform rod of mass $M$ and length $L$ is hinged at one end and released from a horizontal position. When the rod makes an angle $\theta$ with the vertical, the magnitude of the hinge reaction force is:$

Show correct answer & step-by-step solution

Correct answer: D — $\frac{Mg}{4}\sqrt{\sin^2\theta + 81\cos^2\theta}$

Solution

The hinge reaction components are

F_{x} = M ω^{2} (L /2)

and

F_{y} = M g - M α (L /2)

. Using energy conservation for

ω^{2}

and torque for

α

, the resultant force is

F_{x}^{2} + F_{y}^{2}