Q. 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
A) $\frac{m_1m_2}{m_1 + m_2}l^2$
B) $\frac{m_1 + m_2}{m_1m_2}l^2$
C) $(m_1 + m_2)l^2$
D) $\sqrt{m_1m_2}\,l^2$
Explanation
[Solution] $I = \mu l^2$ with reduced mass $\mu = \frac{m_1m_2}{m_1 + m_2}$. Answer (1).
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.