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(C) The moment of inertia of a single rod $AB$ about an axis passing through its center $P$ and perpendicular to its length is $I_C = \frac{1}{12}ML^2

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$\frac{4}{3}M{L^2}$

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(C) The moment of inertia of a single rod $AB$ about an axis passing through its center $P$ and perpendicular to its length is $I_C = \frac{1}{12}ML^2$.Using the parallel axis theorem,the moment of inertia of rod $AB$ about an axis passing through the center $O$ of the square is:$I = I_C + Md^2 = \frac{1}{12}ML^2 + M\left(\frac{L}{2}ight)^2 = \frac{1}{12}ML^2 + \frac{1}{4}ML^2 = \frac{1+3}{12}ML^2 = \frac{4}{12}ML^2 = \frac{1}{3}ML^2$.Since there are four such rods,the total moment of inertia $I'$ about the axis passing through the center $O$ and perpendicular to the plane is:$I' = 4 	imes I = 4 	imes \left(\frac{1}{3}ML^2ight) = \frac{4}{3}ML^2$.

Four rods are arranged in the form of a square. Calculate the moment of inertia about an axis passing through the center and perpendicular to the plane. (Mass $M$ and length $L$ for each rod)

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