Question

Rod is hanging vertically in state of rest. Hinge reaction on the rod is

Answer 1

can't understand the question XD

Answer 2

T =5/2mg.

Explanation:

When the rod of length l is released from the horizontal position and it becomes vertical, the center of mass (C) of the rod moves down by 1/2.

Loss in potential energy = 1/2mg.

Gain in kinetic energy =1/2 Iω²

= $\frac{1}{2} (\frac{ml2}{3})$ ω²

=\frac{1}{2} (\frac{ml²}{3}) ω²

According to the principle of conservation of mechanical energy, gain in KE= loss in PE

=1/2(ml²/3)ω²

=1/2mg.

 ω² = mg X    $\frac{l X 6}{2ml²}$

ω = √3g/l

In the vertical position, the reaction of the hinge is only in the vertical direction. Let it be T.

From the free-body diagram of the rod, we find

T - mg = mω²(1/2)

T -mg = 1/2m ($\frac{3g}{l}$)

= $\frac{3mg}{2}$

T = \frac{3mg}{2} + mg

T = 5/2 mg.