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(D) The moment of inertia of a single rod of mass $M$ and length $L$ about an axis passing through its center of mass and perpendicular to its length

Vedclass · Accepted Answer

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

Vedclass · Answer

(D) The moment of inertia of a single rod of mass $M$ and length $L$ about an axis passing through its center of mass and perpendicular to its length is $I_{CM} = \frac{M L^2}{12}$.Using the parallel axis theorem,the moment of inertia of one rod about an axis passing through the centroid of the triangle and perpendicular to the plane is $I_{rod} = I_{CM} + M x^2$,where $x$ is the distance from the center of the rod to the centroid of the triangle.For an equilateral triangle of side $L$,the distance from the centroid to the midpoint of any side is $x = \frac{L}{2 \sqrt{3}}$.Thus,$I_{rod} = \frac{M L^2}{12} + M \left( \frac{L}{2 \sqrt{3}} ight)^2 = \frac{M L^2}{12} + \frac{M L^2}{12} = \frac{2 M L^2}{12} = \frac{M L^2}{6}$.Since there are three such rods,the total moment of inertia is $I_{total} = 3 	imes I_{rod} = 3 	imes \frac{M L^2}{6} = \frac{M L^2}{2}$.

Three rods are arranged in the form of an equilateral triangle. Calculate the moment of inertia about an axis passing through the centroid and perpendicular to the plane of the triangle. (Each rod has mass $M$ and length $L$)

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