Two particles having masses $m_{1}$ and $m_{2}$ are situated in a plane perpendicular to line AB at a distance of $r_{1}$ and $r_{2}$ respectively as shown. Find the moment of inertia of the entire setup about an axis passing through system's centre of mass and perpendicular to line joining $m_{1}$ and $m_{2}$ .

$\frac{m_{1} m_{2}}{m_{1} + m_{2}} (r_{1} + r_{2})^{2}$

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$\frac{m_{1} m_{2}}{m_{1} + m_{2}} (r_{1} - r_{2})^{2}$

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$\frac{m_{1} m_{2}}{m_{1} + m_{2}} (r_{1}^{2} + r_{2}^{2})$

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$\frac{m_{1} m_{2}}{m_{1} + m_{2}} (r_{1}^{2} - r_{2}^{2})$

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Solution

The correct option is A
$\frac{m_{1} m_{2}}{m_{1} + m_{2}} (r_{1} + r_{2})^{2}$

Centre of mass of system $r_{c m}$ = $m_{2} \frac{r_{1} + r_{2}}{m_{1} + m_{2}}$ = distance of centre of mass from $m_{1}$

Distance of centre of mass from mass $m_{2}$ = $\frac{m_{1} (r_{1} + r_{2})}{(m_{1} + m_{2})}$

So moment of inertia about its centre of mass = $I_{C M}$ = $m_{1} (\frac{m_{2} (r_{1} + r_{2})}{m_{1} + m_{2}})^{2} + m_{2} (\frac{m_{1} (r_{1} + r_{2})}{m_{1} + m_{2}})^{2}$

$I_{c m}$ = $\frac{m_{1} m_{2} (r_{1} + r_{2})^{2}}{m_{1} + m_{2}}$

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