Four solid spheres,each of mass M and radius | vedclass

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(D) Let the four spheres be placed at corners $A, B, C,$ and $D$ of a square of side $a$. We want to find the moment of inertia about the side $AD$.Fo

Vedclass · Accepted Answer

${I_{AD}} = 2\left[ {\frac{4}{5}M{R^2} + M{a^2}} \right]$

Vedclass · Answer

(D) Let the four spheres be placed at corners $A, B, C,$ and $D$ of a square of side $a$. We want to find the moment of inertia about the side $AD$.For spheres at $A$ and $D$,the axis $AD$ passes through their centers. The moment of inertia of a solid sphere about its diameter is $I_{cm} = \frac{2}{5}MR^2$.Thus,$I_A = I_D = \frac{2}{5}MR^2$.For spheres at $B$ and $C$,the distance of their centers from the axis $AD$ is $a$. Using the parallel axis theorem,$I = I_{cm} + Md^2$,we get:$I_B = I_C = \frac{2}{5}MR^2 + Ma^2$.The total moment of inertia of the system about axis $AD$ is:$I_{AD} = I_A + I_D + I_B + I_C$$I_{AD} = \frac{2}{5}MR^2 + \frac{2}{5}MR^2 + (\frac{2}{5}MR^2 + Ma^2) + (\frac{2}{5}MR^2 + Ma^2)$$I_{AD} = 4(\frac{2}{5}MR^2) + 2Ma^2$$I_{AD} = \frac{8}{5}MR^2 + 2Ma^2 = 2[\frac{4}{5}MR^2 + Ma^2]$.

Four solid spheres,each of mass $M$ and radius $R$,are placed at the four corners of a square of side length $a$. Find the moment of inertia of the system about an axis coinciding with one of the sides of the square.

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