Three rods each of mass m and length L are jo | vedclass

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(B) The distance of the center of mass of the equilateral triangle from each rod is $d = \frac{L}{2 \sqrt{3}}$.Using the parallel axis theorem for one

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$\frac{m L^2}{2}$

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

Three rods each of mass $m$ and length $L$ are joined to form an equilateral triangle as shown in the figure. What is the moment of inertia about an axis passing through the centre of mass of the system and perpendicular to the plane?

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