Three thin metal rods,each of mass M and leng | vedclass

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(A) For a single rod of mass $M$ and length $L$,the moment of inertia about its center of mass is $I_{cm} = \frac{ML^2}{12}$.The distance $d$ from the

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$\frac{1}{2} ML^2$

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(A) For a single rod of mass $M$ and length $L$,the moment of inertia about its center of mass is $I_{cm} = \frac{ML^2}{12}$.The distance $d$ from the center of the equilateral triangle to the center of each rod is $d = \frac{L}{2\sqrt{3}}$.Using the parallel axis theorem,the moment of inertia of one rod about the axis passing through the center of the triangle and perpendicular to its plane is:$I_{rod} = I_{cm} + Md^2 = \frac{ML^2}{12} + M\left(\frac{L}{2\sqrt{3}}ight)^2$$I_{rod} = \frac{ML^2}{12} + M\left(\frac{L^2}{12}ight) = \frac{ML^2}{12} + \frac{ML^2}{12} = \frac{2ML^2}{12} = \frac{ML^2}{6}$.Since there are three such rods,the total moment of inertia $I_{total}$ is:$I_{total} = 3 	imes I_{rod} = 3 	imes \frac{ML^2}{6} = \frac{ML^2}{2}$.

Three thin metal rods,each of mass $M$ and length $L$,are welded to form an equilateral triangle. The moment of inertia of the composite structure about an axis passing through the centre of mass of the structure and perpendicular to its plane is

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