Ans:

The length of the thin rod is l and mass density i.e., mass per unit length is m. So the total mass of the rod is \( M=ml \).
The rod is rotating with angular velocity \( \omega \) about an axis passing through the middle point and perpendicular to the length of the rod.

We know that the moment of inertia of the rod about that axis is

\( I=\frac{1}{12}Ml^2 \)
[ To know the derivation of the moment of inertia of a thin rod about an axis passing through the middle point of the rod and perpendicular to the length of it ( CLICK HERE ) ]

\( or,\ I=\frac{1}{12}(ml)l^2 \)

\( or,\ I=\frac{1}{12}ml^3 \)

So the kinetic energy of the rod is

K.E.= \( \displaystyle{\frac{1}{2}I{\omega}^2} \)
[ To know the derivation of the kinetic energy of a rotating body ( CLICK HERE ) ]

= \( \frac{1}{2}\left(\frac{ml^3}{12}\right){\omega}^2 \)

\( or,\ K.E.=\displaystyle{\frac{1}{24}ml^3{\omega}^2} \)