Question

using method of integration, show that the moment of inertia of a uniform thin rod of length L about an axis perpendicular to the rod at a point L/4 from one of its ends can be written as 7/48ML2

Answer 1

$I=\frac{mL^2}{12}+m(\frac L4)^2=\frac{mL^2}{12}+\frac{mL^2}{16}=\frac{4mL^2}{48}+\frac{3mL^2}{48}=\frac{7mL^2}{48}.$