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A 2 \mathrm{~kg}$ steel rod of length $0.6 \mathrm{~m}  is clamped on a table vertically at its lower end and is free to rotate in vertical plane. The upper end is pushed so that the rod falls under gravity. lgnoring the friction due to clamping at its lower end, the speed of the free end of rod when it passes through its lowest position is___________ \mathrm{ms}^{-1}.
(Take \: \: \mathrm{g}=10 \mathrm{~ms}^{-2} )
 

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best_answer


PE_{lost}= KE_{gained}
mgh= \frac{1}{2}I\omega ^{2}
Mgh= \frac{1}{2}\frac{ML^{2}}{3}\times \omega ^{2}
\frac{6gh}{L^{2}}= \omega ^{2}
V\doteq \left ( \omega L \right )= \sqrt{6gh}= \sqrt{6\times 0\cdot 6\times 10}
    = 6\frac{m}{s}

 

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vishal kumar

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