Four spheres each of diameter 2a and mass M a | vedclass

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(B) The moment of inertia $(I_{cm})$ of a solid sphere about its center of mass is given by $I_{cm} = \frac{2}{5}Ma^2$.Let the axis of rotation be alo

Vedclass · Accepted Answer

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

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(B) The moment of inertia $(I_{cm})$ of a solid sphere about its center of mass is given by $I_{cm} = \frac{2}{5}Ma^2$.Let the axis of rotation be along one side of the square. Two spheres have their centers on this axis,so their distance from the axis is $0$. The other two spheres have their centers at a distance $b$ from the axis.Using the parallel axis theorem,$I = I_{cm} + Md^2$,where $d$ is the distance from the axis.For the two spheres on the axis: $I_1 = 2 	imes (\frac{2}{5}Ma^2 + M(0)^2) = \frac{4}{5}Ma^2$.For the two spheres at distance $b$ from the axis: $I_2 = 2 	imes (\frac{2}{5}Ma^2 + Mb^2) = \frac{4}{5}Ma^2 + 2Mb^2$.The total moment of inertia of the system is $I = I_1 + I_2 = \frac{4}{5}Ma^2 + \frac{4}{5}Ma^2 + 2Mb^2 = \frac{8}{5}Ma^2 + 2Mb^2$.

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

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