Question

A uniform rod of length $l$ is held vertically on a horizontal floor. Fixing its lower end, the rod is allowed to fall on to the ground. The angular velocity with which it falls just before coming in contact with the ground is :(Given acceleration due to gravity $=g$)

• A. $\sqrt{\frac{5g}{l}}$
• B. $\sqrt{\frac{3g}{l}}$
• C. $\sqrt{\frac{7g}{l}}$
• D. $None$

Answer 1

Answer: (B)

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

Work done by gravity $=\frac{mgl}{2}$
$I=\frac{m{l}^2}{3}$
$KE=\frac{1}{2}I{\omega }^2$
by Work Energy theorem
$mg\frac{l}{2}=\frac{1}{2}\times \frac{m{l}^2}{3}\times {\omega }^2$
$\sqrt{\frac{3g}{l}}=\omega$