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

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(B) The rod is bent at its midpoint $A$ into two segments,each of length $L/2$ and mass $M/2$.Each segment acts as a rod of length $l = L/2$ and mass

Vedclass · Accepted Answer

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

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(B) The rod is bent at its midpoint $A$ into two segments,each of length $L/2$ and mass $M/2$.Each segment acts as a rod of length $l = L/2$ and mass $m = M/2$ 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 given by $I = \frac{ml^2}{3}$.Here,for each segment,$m = M/2$ and $l = L/2$.So,the moment of inertia for one segment is $I_1 = \frac{(M/2)(L/2)^2}{3} = \frac{(M/2)(L^2/4)}{3} = \frac{ML^2}{24}$.Since there are two such segments,the total moment of inertia about the axis passing through $A$ is $I_A = I_1 + I_1 = 2 	imes \frac{ML^2}{24} = \frac{ML^2}{12}$.

$A$ thin rod of mass $M$ and length $L$ is bent at its midpoint $A$ such that it forms an angle of $60^{\circ}$. What is the moment of inertia about an axis passing through the midpoint $A$ and perpendicular to the plane of the rod?

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