A thin rod of length L and mass M is bent at | vedclass

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(D) The rod is bent at the midpoint into two equal segments,each of length $l = \frac{L}{2}$ and mass $m = \frac{M}{2}$.Each segment acts as a rod of

Vedclass · Accepted Answer

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

Vedclass · Answer

(D) The rod is bent at the midpoint into two equal segments,each of length $l = \frac{L}{2}$ and mass $m = \frac{M}{2}$.Each segment acts as a rod of length $l$ rotating about one of its ends.The moment of inertia of a rod of mass $m$ and length $l$ about an axis passing through one end and perpendicular to its length is $I = \frac{1}{3}ml^2$.Since there are two such segments,the total moment of inertia $I_{	ext{total}}$ about the axis passing through the bending point $O$ and perpendicular to the plane is the sum of the moments of inertia of the two segments:$I_{	ext{total}} = I_1 + I_2 = \frac{1}{3} \left(\frac{M}{2}ight) \left(\frac{L}{2}ight)^2 + \frac{1}{3} \left(\frac{M}{2}ight) \left(\frac{L}{2}ight)^2$$I_{	ext{total}} = 2 	imes \left[ \frac{1}{3} 	imes \frac{M}{2} 	imes \frac{L^2}{4} ight]$$I_{	ext{total}} = 2 	imes \left[ \frac{ML^2}{24} ight] = \frac{ML^2}{12}$

$A$ thin rod of length $L$ and mass $M$ is bent at its mid-point into two halves so that the angle between them is $90^o$. The moment of inertia of the bent rod about an axis passing through the bending point and perpendicular to the plane defined by the two halves of the rod is

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