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A thin rod of length L is vertically kept on horizontal floor. This rod falls freely to one side without slipping of its bottom. The linear velocity of center of rod when its top end touches floor is

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$\sqrt{2 g l}$

$\sqrt{3 g l}$

$\sqrt{\frac{3 g l}{2}}$

$\sqrt{\frac{3 g l}{4}}$

answer is D.

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Detailed Solution

Using conservation of energy,

$\frac{mgL}{2} = \frac{{mv}^{2}}{2} + \frac{1}{2} \times \frac{{mL}^{2}}{3} \times ω^{2}$

$Angular velocity ω = \frac{2 v}{L}$

$\Rightarrow \frac{mgL}{2} = \frac{{mv}^{2}}{2} + \frac{1}{6} {mL}^{2} \times \frac{4 v^{2}}{L^{2}}$

$ω = \sqrt{\frac{3 g l}{4}}$

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