A thin rod of length `l` and mass m per unit length is rotating about an axis passing through the mid-point of its length and perpendicular to it. Prove that its kinetic energy =` (1)/(24) m omega^(2)l^(3), omega`= angular velocity of the rod.

Explanation

Kinetic energy of the rod `(1)/(2)Iomega^(2)`
According to the problem,
`I=(1)/(12)Ml^(2)`[ M= mass of the rod = ml]
`=(1)/(12)xxmlxxl^(2)= (ml^(3))/(12)`
`:.` Kinetic energy of the rod
`=(1)/(2)xx(ml^(3))/(12)xxomega^(2)=(1)/(24)momega^(2)l^(3).`