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(B) The rod is bent at the middle point $O$. Thus,the rod is divided into two equal segments,each of length $l = \frac{L}{2}$ and mass $m = \frac{M}{2

Vedclass · Accepted Answer

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

Vedclass · Answer

(B) The rod is bent at the middle point $O$. Thus,the rod is divided into two equal segments,each of length $l = \frac{L}{2}$ and mass $m = \frac{M}{2}$.The moment of inertia of a uniform rod of mass $m$ and length $l$ about an axis passing through one of its ends and perpendicular to its length is given by $I = \frac{1}{3}ml^2$.For each segment of the bent rod,the moment of inertia about the axis passing through $O$ and perpendicular to the plane of the rod is:$I_{segment} = \frac{1}{3} \left( \frac{M}{2} \right) \left( \frac{L}{2} \right)^2 = \frac{1}{3} \left( \frac{M}{2} \right) \left( \frac{L^2}{4} \right) = \frac{ML^2}{24}$.Since the axis passes through the common point $O$ for both segments,the total moment of inertia of the bent rod is the sum of the moments of inertia of the two segments:$I_{total} = I_{segment} + I_{segment} = \frac{ML^2}{24} + \frac{ML^2}{24} = \frac{2ML^2}{24} = \frac{ML^2}{12}$.

$A$ thin rod of length $L$ and mass $M$ is bent at the middle point $O$ at an angle of $60^{\circ}$ as shown in the figure. The moment of inertia of the rod about an axis passing through $O$ and perpendicular to the plane of the rod will be:

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