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(D) The moment of inertia of a solid sphere about its diameter is $I_{cm} = \frac{2}{5} m r^2$.For the three spheres whose centers lie on the axis of

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(D) The moment of inertia of a solid sphere about its diameter is $I_{cm} = \frac{2}{5} m r^2$.For the three spheres whose centers lie on the axis of rotation $A-A'$,the moment of inertia of each about this axis is $I_1 = \frac{2}{5} m r^2$.For the two spheres whose centers are at a distance $r$ from the axis of rotation $A-A'$,we use the parallel axis theorem: $I = I_{cm} + m d^2$,where $d = r$.Thus,$I_2 = \frac{2}{5} m r^2 + m r^2 = \frac{7}{5} m r^2$.The total moment of inertia of the system is $I_{total} = 3 	imes I_1 + 2 	imes I_2$.$I_{total} = 3 \left( \frac{2}{5} m r^2 ight) + 2 \left( \frac{7}{5} m r^2 ight) = \frac{6}{5} m r^2 + \frac{14}{5} m r^2 = \frac{20}{5} m r^2 = 4 m r^2$.

Five solid spheres,each of mass $m$ and radius $r$,are arranged as shown in the figure. The axis of rotation $A-A'$ passes through the centers of three spheres. Calculate the moment of inertia of the system about the axis of rotation $A-A'$. (in $m r^2$)

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