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

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Question

(A) The axis $BB'$ passes through the center of the square and is perpendicular to the plane of the square.Each sphere has a mass $M$ and radius $a$ (

Vedclass · Accepted Answer

$4M \left[ \frac{2}{5}a^2 + \frac{b^2}{2} \right]$

Vedclass · Answer

(A) The axis $BB'$ passes through the center of the square and is perpendicular to the plane of the square.Each sphere has a mass $M$ and radius $a$ (since diameter is $2a$).The moment of inertia of each sphere about its own center of mass axis is $I_{CM} = \frac{2}{5}Ma^2$.The distance of each sphere from the axis $BB'$ is $r = \frac{b}{\sqrt{2}}$.Using the parallel axis theorem for each sphere,the moment of inertia of one sphere about $BB'$ is $I = I_{CM} + Mr^2 = \frac{2}{5}Ma^2 + M(\frac{b}{\sqrt{2}})^2 = \frac{2}{5}Ma^2 + \frac{Mb^2}{2}$.Since there are four such spheres,the total moment of inertia is $I_{BB'} = 4 	imes (\frac{2}{5}Ma^2 + \frac{Mb^2}{2}) = 4M [\frac{2}{5}a^2 + \frac{b^2}{2}]$.

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 $BB'$.

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