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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
$(m_1 + m_2)l^2$
B
$\sqrt{m_1m_2}\,l^2$
C
$\frac{m_1m_2}{m_1 + m_2}l^2$
D
$\frac{m_1 + m_2}{m_1m_2}l^2$
Explanation
$I = \mu l^2$ with reduced mass μ.
Detailed Solution
$r_1 = \frac{m_2l}{m_1 + m_2}$, $r_2 = \frac{m_1l}{m_1 + m_2}$
$I_{cm} = m_1r_1^2 + m_2r_2^2 = \frac{m_1m_2}{m_1 + m_2}l^2$
$I_{cm} = m_1r_1^2 + m_2r_2^2 = \frac{m_1m_2}{m_1 + m_2}l^2$