Find the moment of inertia of the rod A B about an axis yy as shown in figure. Mass of the rod is m and length is l.

Mass per unit length of the rod $=\frac{\mathrm{m}}{\mathrm{l}}$

Mass of an element P Q of the rod is, $\mathrm{dm}=\left(\frac{\mathrm{m}}{\mathrm{l}}\right)\mathrm{dx}$

Perpendicular distance of this elemental mass about y y is, $\mathrm{r}=\mathrm{xsin\alpha }$

$\therefore$ Moment of inertia of this small element of the rod (can be assumed as a point mass) about y y is,

$\mathrm{dI}=\left(\mathrm{dm}\right){\mathrm{r}}^2=\left(\frac{\mathrm{m}}{\mathrm{l}}\mathrm{dx}\right)(\mathrm{xsin\alpha }{)}^2=\left(\frac{\mathrm{m}}{\mathrm{l}}{\sin }^2\mathrm{\alpha }\right){\mathrm{x}}^2\mathrm{dx}$

$\therefore$ Moment of inertia of the complete rod.

$\mathrm{I}={\int }_{x=0}^{\mathrm{x}=\mathrm{l}}\mathrm{dl}=\frac{\mathrm{m}}{1}{\sin }^2\mathrm{\alpha }{\int }_0^l{\mathrm{x}}^2\mathrm{dx}=\frac{{\mathrm{ml}}^2}{3}{\sin }^2\mathrm{\alpha }$