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

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(B) The moment of inertia of a solid sphere of mass $M$ and radius $a$ about its diameter is $I_{CM} = \frac{2}{5}Ma^2$.The axis $AA'$ passes through

Vedclass · Accepted Answer

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

Vedclass · Answer

(B) The moment of inertia of a solid sphere of mass $M$ and radius $a$ about its diameter is $I_{CM} = \frac{2}{5}Ma^2$.The axis $AA'$ passes through the center of the square and lies in the plane of the square. It passes through the centers of spheres $1$ and $3$.For spheres $1$ and $3$,the axis $AA'$ is their diameter,so their contribution to the moment of inertia is $2 	imes I_{CM} = 2 	imes \frac{2}{5}Ma^2 = \frac{4}{5}Ma^2$.For spheres $2$ and $4$ (the other two corners),the perpendicular distance from the axis $AA'$ is $r = \frac{b}{\sqrt{2}}$.Using the parallel axis theorem,the moment of inertia for each of these spheres is $I = I_{CM} + Mr^2 = \frac{2}{5}Ma^2 + M\left(\frac{b}{\sqrt{2}}ight)^2 = \frac{2}{5}Ma^2 + \frac{Mb^2}{2}$.Total moment of inertia $I_{AA'} = (2 	imes I_{CM}) + 2 	imes (I_{CM} + Mr^2) = 4I_{CM} + 2Mr^2$.$I_{AA'} = 4 \left( \frac{2}{5}Ma^2 ight) + 2M \left( \frac{b^2}{2} ight) = \frac{8}{5}Ma^2 + Mb^2$.

Four spheres,each of mass $M$ and diameter $2a$,are placed at the corners of a square of side $b$ as shown in the figure. Calculate the moment of inertia about the axis $AA'$.

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