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(D) The rod is bent at the midpoint $O$ at an angle of $90^{\circ}$. This divides the rod into two segments,$OA$ and $OB$,each of length $l = L/2$ and

Vedclass · Accepted Answer

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

Vedclass · Answer

(D) The rod is bent at the midpoint $O$ at an angle of $90^{\circ}$. This divides the rod into two segments,$OA$ and $OB$,each of length $l = L/2$ and mass $m = M/2$.The axis of rotation passes through point $O$ and is perpendicular to the plane of the bent rod. For a rod of length $l$ and mass $m$ rotating about an axis passing through one of its ends,the moment of inertia is $I = \frac{1}{3} ml^2$.For segment $OA$,the moment of inertia about the axis through $O$ is:$I_A = \frac{1}{3} \left( \frac{M}{2} \right) \left( \frac{L}{2} \right)^2 = \frac{1}{3} \left( \frac{M}{2} \right) \left( \frac{L^2}{4} \right) = \frac{ML^2}{24}$.Similarly,for segment $OB$,the moment of inertia about the axis through $O$ is:$I_B = \frac{1}{3} \left( \frac{M}{2} \right) \left( \frac{L}{2} \right)^2 = \frac{ML^2}{24}$.The total moment of inertia of the system is the sum of the moments of inertia of the two segments:$I = I_A + I_B = \frac{ML^2}{24} + \frac{ML^2}{24} = \frac{2ML^2}{24} = \frac{ML^2}{12}$.

$A$ thin rod of length $L$ and mass $M$ is bent at its midpoint $O$ at an angle of $90^{\circ}$. The moment of inertia of the bent rod about an axis passing through the point $O$ and perpendicular to the plane of the bent rod is:

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