The moment of inertia of a uniform rod of mas | vedclass

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(B) The rod is bent at the center,so each half has mass $m = M/2$ and length $l = L/2$.The axis of rotation passes through the point of bending $O$ an

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$\frac{1}{12} ML^2$

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(B) The rod is bent at the center,so each half has mass $m = M/2$ and length $l = L/2$.The axis of rotation passes through the point of bending $O$ and is perpendicular to the plane of the bent rod.For a rod of mass $m$ and length $l$ rotating about an axis passing through one of its ends,the moment of inertia is $I = \frac{1}{3} ml^2$.Here,we have two such rods,each of mass $M/2$ and length $L/2$,rotating about the common end $O$.The total moment of inertia $I_{total}$ is the sum of the moments of inertia of the two halves:$I_{total} = I_1 + I_2 = \frac{1}{3} (M/2) (L/2)^2 + \frac{1}{3} (M/2) (L/2)^2$$I_{total} = 2 	imes \left[ \frac{1}{3} \cdot \frac{M}{2} \cdot \frac{L^2}{4} ight]$$I_{total} = 2 	imes \left[ \frac{ML^2}{24} ight] = \frac{ML^2}{12}$.

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