Two particles of masses m₁ and m₂ are connected to 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 length of rod and passing through the centre of mass is:

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

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

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

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

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Solution

The correct option is A
$\frac{m_{1} m_{2}}{m_{1} + m_{2}} r^{2}$

C is centre of mass of the dumb belt, $r_{1}$ and $r_{2}$ are distance of $m_{1}, m_{2}$ from C. Then M.O.I. of dumb bell about the given axis
$I =, m_{1} r_{1}^{2} + m_{2} r_{2}^{2} r = r_{1} + r_{2} m_{1} r_{1} = m_{2} r_{2} m_{1} r_{1} = m_{2} r - m_{2} r_{1} r_{1} = \frac{m_{2} r}{m_{1} + m_{2}}, r_{2} = \frac{m_{1} r}{m_{1} + m_{2}} I = m_{1} [\frac{m_{2} r}{m_{1} + m_{2}}]^{2} + m_{2} [\frac{m_{1} r}{m_{1} + m_{2}}]^{2} I = \frac{m_{1} m_{2}}{m_{1} + m_{2}} r^{2}$

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Two particles of masses m1 and m2 are connected to 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 length of rod and passing through the centre of mass is:

The correct option is A m1m2m1+m2r2 C is centre of mass of the dumb belt, r1 and r2 are distance of m1,m2 from C. Then M.O.I. of dumb bell about the given axis I=,m1r21+m2r22r=r1+r2m1r1=m2r2m1r1=m2r−m2r1r1=m2rm1+m2,r2=m1rm1+m2I=m1[m2rm1+m2]2+m2[m1rm1+m2]2I=m1m2m1+m2r2

Two particles of masses m₁ and m₂ are connected to 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 length of rod and passing through the centre of mass is: