Answer:
About center ,[tex]I_o=\dfrac{1}{12}ML^2[/tex]
About an end ,[tex]I=\dfrac{1}{3}ML^2[/tex]
Explanation:
Given that
Mass =m
Length = L
The moment of inertia of rod about center given as
[tex]I_o=\dfrac{1}{12}ML^2[/tex]
We know that the moment of inertia about a parallel axis which at a distance d from the center given as
I=Io+ m d²
The distance of one end from center
[tex]d=\dfrac{L}{2}[/tex]
[tex]I=I_o+md^2[/tex]
[tex]I=\dfrac{1}{12}ML^2+M\times \dfrac{L^2}{4}[/tex]
[tex]I=\dfrac{1}{12}ML^2+ \dfrac{1}{4}ML^2[/tex]
[tex]I=\dfrac{1}{3}ML^2[/tex]
