A thin uniform rod mass $m$ and length $l$ is hinged at the lower end to a level floor and stands vertically. It is now allowed to fall, then its upper end will strike the floor with a velocity given by

Question

Dr. Abubacker · Accepted Answer

(b) Initially rod stand vertically its potential energy $= mg\frac{l}{2}$
When it strikes the floor, its potential energy will convert into rotational kinetic energy
$mg\left( \frac{l}{2} ight) = \frac{1}{2}I\omega^{2}$
[Where, $I = \frac{ml^{2}}{3} =$ M.I. of rod about point $A$]
$	herefore mg\left( \frac{l}{2} ight) = \frac{1}{2}\left( \frac{ml^{2}}{3} ight)\left( \frac{v_{B}}{l} ight)^{2} \Rightarrow v_{B} = \sqrt{3gl}$

Q. A thin uniform rod mass $m$ and length $l$ is hinged at the lower end to a level floor and stands vertically. It is now allowed to fall, then its upper end will strike the floor with a velocity given by

Explanation

Reviewed by Dr. Abubacker

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