Calculate the moment of inertia of a rod of mass 2 kg and length 5 m about an axis perpendicular to it and 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)=(MI^(2))/(12)`

According to the parallel axes theorem,
`I_(CD)=I_(AB)+M((1)/(2))^(2)=(Ml^(2))/(12)+(Ml^(2))/(4)`
`=(Ml^(2)+2Ml^(2))/(12)=(4Ml^(2))/(12)=(Ml^(2))/(3)`
`I_(CD)=(Ml^(2))/(3)=(2(5)^(2))/(3)=(50)/(3) "kg-m"^(2)`.