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

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(D) Let the four spheres be $A, B, C,$ and $D$ located at the corners of a square of side $b$. Let the axis of rotation be the side $AB$.$1$. The mome

Vedclass · Accepted Answer

$\frac{8}{5} M a^2+2 M b^2$

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(D) Let the four spheres be $A, B, C,$ and $D$ located at the corners of a square of side $b$. Let the axis of rotation be the side $AB$.$1$. The moment of inertia of a solid sphere of mass $M$ and radius $a$ about an axis passing through its center is $I_{cm} = \frac{2}{5} M a^2$.$2$. For spheres $A$ and $B$,the axis of rotation passes through their centers. Thus,the moment of inertia for each is $I_A = I_B = \frac{2}{5} M a^2$.$3$. For spheres $C$ and $D$,the axis of rotation is at a perpendicular distance $b$ from their centers. Using the parallel axis theorem,$I = I_{cm} + M d^2$,where $d = b$. Thus,$I_C = I_D = \frac{2}{5} M a^2 + M b^2$.$4$. The total moment of inertia of the system is $I_{total} = I_A + I_B + I_C + I_D = \frac{2}{5} M a^2 + \frac{2}{5} M a^2 + (\frac{2}{5} M a^2 + M b^2) + (\frac{2}{5} M a^2 + M b^2) = \frac{8}{5} M a^2 + 2 M b^2$.

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

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