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 intertia about axis passing through 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 $r_{c} m = \frac{m_{2} (r_{1} + r_{2})}{(m_{1} + m_{2})}$ = 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 $I = m_{1} {[m_{2} \frac{(r_{1} + r_{2})}{(m_{1} + m_{2})}]}_{2}^{2} + m_{1} {[m_{2} \frac{(r_{1} + r_{2})}{(m_{1} + m_{2})}]}_{2}^{2} = m_{1} m_{2} \frac{(r_{1} + r_{2})^{2}}{(m_{1} + m_{2})}$

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Two particles having masses m1 and m2are situated in a plane perpendicular to line AB at a distance ofr1and r2 respectively as shown. Find the moment of intertia about axis passing through centre of mass and perpendicular to line joining m1 and m2

The correct option is A m1m2m1+m2 (r1+r2)2 Centre of mass of system rcm=m2(r1+r2)(m1+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 I=m1[m2(r1+r2)(m1+m2)]2+m1[m2(r1+r2)(m1+m2)]2=m1m2(r1+r2)2(m1+m2)

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 intertia about axis passing through centre of mass and perpendicular to line joining $m_{1}$ and $m_{2}$