The moment of inertia of a thin rod of mass 'M' and length 'l' about an axis passig through its centre is `(Ml^(2))/(12)`. Calculate the moment of inertia about a parallel axis through end of rod.

Explanation

According to parallel axis theorem,
`I = I_(0) + mx^(2)`
`= (Ml^(2))/(12) + M ((l)/(2))^(2)`
`= (Ml^(2))/(12) + (Ml^(2))/(4) = Ml^(2)[(1)/(12) + (1)/(4)] = Ml^(2) [(1)/(3)]`
`I = (Ml^(2))/(3)`