# A thin rod MN, free to rotate in the vertical plane about the fixed end N, is held horizontal. When the end M is released the speed of this end, when the rod makes an angle α with the horizontal, will be proportional to : (see figure) Option 1) $\sqrt{sin }\alpha$ Option 2) $sin\alpha$ Option 3) $\sqrt{cos}\: \alpha$ Option 4) $cos\alpha$

A Anuj Pandey
D Divya Saini

As we have learnt

Kinetic energy of rotation -

$K=\frac{1}{2}Iw^{2}$

- wherein

$I$ = moment of inertia about axis of rotation

$w$ = angular velocity

From energy conservation,

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

$\omega ^{2}=\frac{3g}{l}sin\alpha$ or   $\omega=\sqrt{\frac{3g}{l}}\sqrt{sin }\alpha$

speed of end m$=\omega l$

$=\sqrt{3gl}\sqrt{sin }\alpha$

$\therefore v\: \alpha \sqrt{sin }\: \alpha$

Option 1)

$\sqrt{sin }\alpha$

Option 2)

$sin\alpha$

Option 3)

$\sqrt{cos}\: \alpha$

Option 4)

$cos\alpha$

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