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A solid sphere A of radius R and mass M is attached at a point to a smaller solid sphere B of radius r < R and mass m < M. Assume that the line joining their centres lies along the horizontal. The moment of inertia of the system calculated about a vertical axis passing through the centre of A is $I_A$ and that calculated about a vertical axis passing through the centre of B is $I_B$. The difference $I_A-I_B$ is:
Detailed Solution
$I_A=\frac{2}{5}MR^2+\frac{2}{5}mr^2+m(R+r)^2$ and $I_B=\frac{2}{5}mr^2+\frac{2}{5}MR^2+M(R+r)^2$. $I_A-I_B=m(R+r)^2-M(R+r)^2=(m-M)(R+r)^2$.