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)

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$\sqrt{\sin α}$

$\sin α$

$\sqrt{\cos α}$

$\cos α$

answer is A.

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Detailed Solution

When the rod forms an angle α with the formula displacement of center of mass = $\frac{1}{2} \sin α$

$mg \frac{1}{2} sinα = \frac{1}{2} {Iω}^{2}$

$mg \frac{1}{2} sinα = \frac{{ml}^{2}}{6} ω^{2}$

$ω = \sqrt{\frac{3 gsinα}{l}}$

$speed of end = ω \times l = \sqrt{3 gsinαl}$

$Thus ω is proportional to \sqrt{sinα}$

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