Calculate the moment of inertia of a rod of mass M, and length l about an axis perpendicular to it passing through one of its ends.

Explanation

For the rod of mass M and length l the moment of inertia of the rod about an axis AB passing through its centre of mass is given `I_(AB)=(Ml^(2))/(12)`

According to the parallel axes theorem
`l_(CD)=l_(AB)+M((l)/(2))^(2)=(Ml^(2))/(12)+(Ml^(2))/(4)=(Ml^(2)+3Ml^(2))/(12)=(4Ml^(2))/(12)=(Ml^(2))/(3)`
`l_(CD)=(Ml^(2))/(3)`