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(D) Let the square be $ABCD$ with side length $b$. Four spheres,each of mass $M$ and radius $a$,are placed at the corners $A, B, C,$ and $D$.We calcul

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$\frac{2}{5} M (4a^2 + 5b^2)$

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(D) Let the square be $ABCD$ with side length $b$. Four spheres,each of mass $M$ and radius $a$,are placed at the corners $A, B, C,$ and $D$.We calculate the moment of inertia about one side,say $BC$,as the axis.The moment of inertia of a sphere of mass $M$ and radius $a$ about its diameter is $I_{cm} = \frac{2}{5}Ma^2$.For spheres at $B$ and $C$,the axis $BC$ passes through their centers. By the parallel axis theorem,their moment of inertia about $BC$ is $I_B = I_C = \frac{2}{5}Ma^2$.For spheres at $A$ and $D$,the distance from the axis $BC$ is $b$. By the parallel axis theorem,$I_A = I_D = I_{cm} + Mb^2 = \frac{2}{5}Ma^2 + Mb^2$.The total moment of inertia $I = I_A + I_B + I_C + I_D$.$I = (\frac{2}{5}Ma^2 + Mb^2) + \frac{2}{5}Ma^2 + \frac{2}{5}Ma^2 + (\frac{2}{5}Ma^2 + Mb^2)$.$I = \frac{8}{5}Ma^2 + 2Mb^2 = \frac{2}{5}M(4a^2 + 5b^2)$.

Four spheres,each of mass $M$ and radius $a$,are placed at the four corners of a square of side $b$. Calculate the moment of inertia of the system about one of the sides of the square as the axis.

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