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 $m_{2}$ .

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Solution

The correct option is A

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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Two particles having masses m1 and m2 are situated in a plane perpendicular to line AB at a distance of r1 and r2 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 m1 and m2.

The correct option is A Centre of mass of system rcm = m2r1+r2m1+m2 = distance of centre of mass from m1 Distance of centre of mass from mass m2 = m1(r1+r2)(m1+m2) So moment of inertia about its centre of mass = ICM = m1(m2(r1+r2)m1+m2)2+m2(m1(r1+r2)m1+m2)2 Icm = m1m2(r1+r2)2m1+m2