What is the minimum moment of inertia of a system of particles of masses $m_{1}$ and $m_{2}$ being interconnected by a rigid light bar of length $l$ , above an axis perpendicular to the rod and passing through it?

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Solution

Center of mass of the system $x_{c m} = \frac{m_{1} x_{1} + m_{2} x_{2}}{m_{1} + m_{2}}$
Let the position of $m_{1}$ to be origin i.e. $x_{1} = 0$
So, $x_{2} = l$
Thus $X_{c m} = \frac{m_{2} l}{m_{1} + m_{2}}$
Distance of COM from $m_{2}$ , $l - x_{c m} = \frac{m_{1} l}{m_{1} + m_{2}}$
Moment of inertia is minimum about an axis passing through centre of mass.
So, $I_{m i n} = m_{1} \times (\frac{m_{2} l}{m_{1} + m_{2}})^{2} + m_{2} \times (\frac{m_{1} l}{m_{1} + m_{2}})^{2}$
$⟹ I_{m i n} = \frac{m_{1} m_{2}}{(m_{1} + m_{2})} l^{2}$

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What is the minimum moment of inertia of a system of particles of masses m1 and m2 being interconnected by a rigid light bar of length l, above an axis perpendicular to the rod and passing through it?

Center of mass of the system xcm=m1x1+m2x2m1+m2Let the position of m1 to be origin i.e. x1=0So, x2=lThus Xcm=m2lm1+m2Distance of COM from m2, l−xcm=m1lm1+m2Moment of inertia is minimum about an axis passing through centre of mass.So, Imin=m1×(m2lm1+m2)2+m2×(m1lm1+m2)2⟹ Imin=m1m2(m1+m2)l2

What is the minimum moment of inertia of a system of particles of masses $m_{1}$ and $m_{2}$ being interconnected by a rigid light bar of length $l$ , above an axis perpendicular to the rod and passing through it?