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 then, the K.E with which it falls just before coming into contact with the ground is :

• A. $\frac{1}{2}\left(\frac{m{l}^2}{3}\right){\omega }^2$
• B. $\frac{1}{4}\left(\frac{m{l}^2}{3}\right){\omega }^2$
• C. $\frac{1}{4}\left(\frac{m{l}^2}{6}\right){\omega }^2$
• D. $\frac{1}{4}\left(m{l}^2\right){\omega }^2$

Answer 1

The correct option is A $\frac{1}{2}\left(\frac{m{l}^2}{3}\right){\omega }^2$

The rod falls while its bottom is fixed. So , the axis of rotation is about the bottom

Moment of Inertia $I=I_{cm}+\frac{M{L}^2}{4}$

$I=\frac{M{L}^2}{12}+\frac{M{L}^2}{4}$

$I=\frac{M{L}^2}{3}$

Kinetic Energy of the rod $I=\frac{1}{2}I{w}^2$

$KE=\frac{1}{2}\left(\frac{M{L}^2}{3}\right)w^2$