Q. Two particles of masses $m_{1}$ and $m_{2}$ are connected by a rigid massless rod of length $r$ to constitute a dumb-bell which is free to move in the plane. The moment of inertia of the dumb-bell about an axis perpendicular to the plane passing through the centre of mass is
A) $\frac{m_{1}m_{2}r^{2}}{m_{1} + m_{2}}$
B) $\left( m_{1} + m_{2} \right)r^{2}$
C) $\frac{m_{1}m_{2}r^{2}}{m_{1} - m_{2}}$
D) $\left( m_{1} - m_{2} \right)r^{2}$
Explanation
(a) $x_{1} + x_{2} = r$ …(i)
$m_{1}x_{1} = m_{2}x_{2}$
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From Eqs. (i) and (ii)
$x_{1} = \frac{m_{2}r}{m_{1} + m_{2}}$
and $x_{2} = \frac{m_{1}r}{m_{1} + m_{2}}$
$\therefore\ \ I_{AB} = m_{1}x_{1}^{2} + m_{2}x_{2}^{2}$
$= \frac{m_{1}m_{2}r^{2}}{m_{1} + m_{2}}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.