Let the center of mass be at a distance $r_1$ from mass $m_1$ and $r_2$ from mass $m_2$. The position of the center of mass is given by $m_1 r_1 = m_2 r_2$. We also know that the total length of the rod is $l = r_1 + r_2$. From these two equations, we can find the distances: $r_1 = \frac{m_2 l}{m_1 + m_2}$ and $r_2 = \frac{m_1 l}{m_1 + m_2}$ The moment of inertia of the system about the center of mass is the sum of the moments of inertia of the two masses: $I = m_1 r_1^2 + m_2 r_2^2$ Substituting the values of $r_1$ and $r_2$: $I = m_1 \left( \frac{m_2 l}{m_1 + m_2} \right)^2 + m_2 \left( \frac{m_1 l}{m_1 + m_2} \right)^2$ $I = \frac{m_1 m_2^2 l^2}{(m_1 + m_2)^2} + \frac{m_2 m_1^2 l^2}{(m_1 + m_2)^2}$ $I = \frac{m_1 m_2 l^2 (m_1 + m_2)}{(m_1 + m_2)^2}$ $I = \frac{m_1 m_2}{m_1 + m_2} l^2$ This is also known as the reduced mass $\mu = \frac{m_1 m_2}{m_1 + m_2}$, so $I = \mu l^2$.