A thin rod MN, free to rotate in the vertical plane aboutthe fixed end N, is held horizontal. When the end M is released the speed of this end, when the rod makes an angle \alpha with the horizontal, will be proportional to : (see figure)

Rotational Motion PYQ — JEE Main 2018 · 15 Apr · Shift 2 · Q54 with Solution

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Correct answer: A

A thin rod of length $L$ and mass $m$ is pivoted at one end $N$ and released from the horizontal position.

We need the speed of the free end $M$ when the rod has rotated through an angle such that it makes angle $\alpha$ with the horizontal.

Initially, the rod is horizontal and at rest.

When the rod rotates downward by angle $\alpha$ , its center of mass falls vertically by

\Delta h = \frac{L}{2}\sin\alpha

So the loss in gravitational potential energy is

mg\left(\frac{L}{2}\sin\alpha\right)

This becomes rotational kinetic energy:

\frac{1}{2}I\omega^2 = mg\left(\frac{L}{2}\sin\alpha\right)

For a thin rod about one end,

I = \frac{1}{3}mL^2

Substitute:

\frac{1}{2}\cdot \frac{1}{3}mL^2\omega^2 = mg\left(\frac{L}{2}\sin\alpha\right)

\frac{1}{6}mL^2\omega^2 = \frac{1}{2}mgL\sin\alpha

Cancel $m$ and one $L$ :

\frac{1}{6}L\omega^2 = \frac{1}{2}g\sin\alpha

\omega^2 = \frac{3g}{L}\sin\alpha

Thus,

\omega \propto \sqrt{\sin\alpha}

The free end is at distance $L$ from the pivot, so

v_M = L\omega

Hence,

v_M \propto \omega \propto \sqrt{\sin\alpha}

So the correct option is A.