Four spheres of diameter 2a and mass M are pl | vedclass

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(B) Let the axis of rotation be along the side connecting the centers of spheres $A$ and $B$.$1$. For spheres $A$ and $B$,the axis of rotation passes

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$\frac{8}{5}Ma^2 + 2Mb^2$

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(B) Let the axis of rotation be along the side connecting the centers of spheres $A$ and $B$.$1$. For spheres $A$ and $B$,the axis of rotation passes through their centers. The moment of inertia of a solid sphere about its diameter is $I_{cm} = \frac{2}{5}Ma^2$. Thus,$I_A = I_B = \frac{2}{5}Ma^2$.$2$. For spheres $C$ and $D$,the axis of rotation is at a perpendicular distance $b$ from their centers. Using the parallel axis theorem,$I = I_{cm} + Md^2$,where $d = b$. Thus,$I_C = I_D = \frac{2}{5}Ma^2 + Mb^2$.$3$. The total moment of inertia of the system is $I_{total} = I_A + I_B + I_C + I_D$.$I_{total} = \frac{2}{5}Ma^2 + \frac{2}{5}Ma^2 + (\frac{2}{5}Ma^2 + Mb^2) + (\frac{2}{5}Ma^2 + Mb^2) = \frac{8}{5}Ma^2 + 2Mb^2$.

Four spheres of diameter $2a$ and mass $M$ are placed with their centers on the four corners of a square of side $b$. The moment of inertia of the system about an axis along one of the sides of the square is

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