A light rod of length l has two masses $m_1$and $m_2$ attached to its two ends. The moment of inertia of the system about an axis perpendicular to the rod and passing through the centre of mass is

If O is the centre of mass of m₁ and m₂ separated by a distance ‘l’, then ‘r₁’ distance of COM from m₁ , r₁ $=\frac{m_2\mathrm{l}}{m_1+m_2}$

and from ${\mathrm{m}}_2,{\mathrm{r}}_2=\frac{{\mathrm{m}}_1}{{\mathrm{m}}_1+{\mathrm{m}}_2}$

${\mathbf{I}}_{\mathrm{COM}}={\mathrm{m}}_1{\mathrm{r}}_1^2+{\mathrm{m}}_2{\mathrm{r}}_2^2=$

${\mathrm{m}}_1{\left(\frac{{\mathrm{m}}_2\mathrm{I}}{{\mathrm{m}}_1+{\mathrm{m}}_2}\right)}^2+{\mathrm{m}}_2{\left(\frac{{\mathrm{m}}_1\mathrm{I}}{{\mathrm{m}}_1+{\mathrm{m}}_2}\right)}^2$

$I_{\text{com}}=\left(\frac{m_1{m}_2}{m_1+m_2}\right)I^2$