Question

What will be the moment of inertia of a thin rod of mass M and length l along the axis perpendicular to its length and passing through one point from a distance l /4 from one end.

Answer 1

Using parallel axis theorem, $I=I_{cm}+m{a}^2$

$I=I_{cm}+M{\left(\frac{l}{4}\right)}^2$

$I=\frac{1}{12}M{l}^2+\frac{M{l}^2}{16}$

$I=M{l}^2(\frac{1}{12}+\frac{1}{16})$

$I=\frac{7M{l}^2}{48}$

$\therefore$ Moment of inertia of rod at a distance of $\frac{1}{4}$ from one end is $\frac{7M{l}^2}{48}$.