Rod is released from horizontal position. Find the angular velocity of rod, when it becomes vertical. 10 kg mass is fixed on the other end of the rod as shown in figure. The rod has mass 20kg and length 10 meter :

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$\sqrt{\frac{7}{5}} rad / s$

$\sqrt{\frac{12}{5}} rad / s$

$\sqrt{\frac{3}{5}} rad / s$

2 rad/s

answer is C.

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Detailed Solution

Let angular velocity in the by using COME
Loss in PE = Gain in KE
$\begin{array}{l} (10) (10) (10) + (20) (10) (5) = \frac{1}{2} [20 (\frac{10^{2}}{3}) + 10 (10)^{2}] ω^{2} \\ \Rightarrow ω = \sqrt{\frac{12}{5}} rad / s \end{array}$

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