Two particles of masses $m_{1}$ and $m_{2}$ are connected by a rigid massless rod of length $r$ to constitute a dumbbell . The moment of inertia of the dumbbell about an axis perpendicular to the rod and passing through the center of mass is:

\frac{m_{1} m_{2} r^{2}}{m_{1} + m_{2}}

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(m_{1} + m_{2}) r^{2}

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\frac{m_{1} m_{2} r^{2}}{m_{1} - m_{2}}

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(m_{1} - m_{2}) r^{2}

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Solution

The correct option is A $\frac{m_{1} m_{2} r^{2}}{m_{1} + m_{2}}$
The position of COM from $m_{1}$ mass is $\frac{m_{2} r}{(m_{1} + m_{2})}$

So MI of $m_{1}$ mass about COM is $m_{1} \times (\frac{m_{2} r}{m_{1} + m_{2}})^{2} = \frac{m_{1} m_{2}^{2} r^{2}}{(m_{1} + m_{2})^{2}}$
So MI of $m_{2}$ mass about COM is $m_{2} \times (\frac{m_{1} r}{m_{1} + m_{2}})^{2} = \frac{m_{2} m_{1}^{2} r^{2}}{(m_{1} + m_{2})^{2}}$
So total MI is
$\frac{m_{1} m_{2}^{2} r^{2}}{(m_{1} + m_{2})^{2}}$ + $\frac{m_{2} m_{1}^{2} r^{2}}{(m_{1} + m_{2})^{2}}$ = $\frac{m_{1} m_{2} r^{2}}{m_{1} + m_{2}}$

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