A solid sphere $A$ of radius $R$ and mass $M$ is attached at a point to a smaller solid sphere $B$ o

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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:

Accepted Answer

Sol. [Key Concept: Moment of inertia of a composite body] Using $I_{C} = \frac{2}{5}mR^{2}$ and Parallel axis theorem, moment of inertia of system$$\begin{array}{r} I_{A} = \frac{2}{5}MR^{2} + \frac{2}{5}mr^{2} + m(R + r)^{2}\#(1) \end{array}$$Similarly,$$\begin{array}{r} I_{B} = \frac{2}{5}MR^{2} + M(R + r)^{2} + \frac{2}{5}mr^{2}\#(2) \end{array}$$Now, subtracting (1) and (2)$$I_{A} - I_{B} = (m - M)(R + r)^{2}$$

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