The moment of inertia (MI) of a thin uniform rod, of mass M and length L, about a transverse axis through its centre of mass is
`I_(CM)=(ML_(2))/(12)" "`.....(i)
Let I be its MI about a transverse axis through its end. By the theorem of parallel axis, `I=I_(CM)+Mh^(2)" "`.......(2)
In this case, h=distance between the parallel axes`=(L)/(2)`.
`:. I=(ML^(2))/(12)+M((L)/(2))^(2)=(ML^(2))/(12)+(ML^(2))/(4)`
`=(ML^(2))/(12)(1+3)=(ML^(2))/(3)" "`.....(3)