Four identical thin rods,each of mass M and l | vedclass

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(D) Consider one rod of the square frame,say rod $AB$. The moment of inertia of a rod of mass $M$ and length $l$ about an axis passing through its cen

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$\frac{4}{3}Ml^2$

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(D) Consider one rod of the square frame,say rod $AB$. The moment of inertia of a rod of mass $M$ and length $l$ about an axis passing through its center and perpendicular to its length is $I_{cm} = \frac{Ml^2}{12}$.The distance $d$ from the center of the rod $AB$ to the center of the square frame is $d = \frac{l}{2}$.Using the parallel axis theorem,the moment of inertia of one rod about the axis passing through the center of the square and perpendicular to its plane is:$I_{rod} = I_{cm} + Md^2 = \frac{Ml^2}{12} + M\left(\frac{l}{2}ight)^2 = \frac{Ml^2}{12} + \frac{Ml^2}{4} = \frac{Ml^2 + 3Ml^2}{12} = \frac{4Ml^2}{12} = \frac{Ml^2}{3}$.Since there are four identical rods forming the square frame,the total moment of inertia $I$ is:$I = 4 	imes I_{rod} = 4 	imes \frac{Ml^2}{3} = \frac{4}{3}Ml^2$.

Four identical thin rods,each of mass $M$ and length $l$,form a square frame. The moment of inertia of this frame about an axis passing through the centre of the square and perpendicular to its plane is

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