Find the moment of inertia of thin and massless rod about an axis passing through its centre of mass of rod and pair of mass is suspended on both end of this rod.

Explanation

A system of a ligh rod of length l with a pair of small masses is shown as in figure. Total mass of system is M.

System is rotating about an axis through the centre of mass of system and perpendicular to the rod.
Suppose C is the centre of mass. So distance of small mass from its ends is `(l)/(2)`.
Momentum of inertia of masses about mention axis is `((M)/(2))((l)/(2))^(2)` :
`therefore` Total moment of inertia of a system
`I=((M)/(2))((l)/(2))^(2)+((M)/(2))((l)/(2))^(2)`
`=2((M)/(2))((l)/(2))^(2)`
`=(Ml^(2))/(4)`