Moment of inertia of the rod about an axis pa | vedclass

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(B) Let the mass of the rod be $M$ and its length be $L$. The moment of inertia of the rod about an axis passing through its center and perpendicular

Vedclass · Accepted Answer

$\frac{2 \pi^2}{3}$

Vedclass · Answer

(B) Let the mass of the rod be $M$ and its length be $L$. The moment of inertia of the rod about an axis passing through its center and perpendicular to its length is $I_1 = \frac{ML^2}{12}$.When the rod is bent into a ring of radius $R$,the circumference of the ring is equal to the length of the rod,so $2\pi R = L$,which implies $R = \frac{L}{2\pi}$.The moment of inertia of a ring of mass $M$ and radius $R$ about its diameter is $I_2 = \frac{1}{2}MR^2$.Substituting $R = \frac{L}{2\pi}$ into the expression for $I_2$,we get $I_2 = \frac{1}{2}M(\frac{L}{2\pi})^2 = \frac{1}{2}M(\frac{L^2}{4\pi^2}) = \frac{ML^2}{8\pi^2}$.Now,calculating the ratio $I_1 / I_2 = \frac{ML^2/12}{ML^2/8\pi^2} = \frac{8\pi^2}{12} = \frac{2\pi^2}{3}$.

Moment of inertia of the rod about an axis passing through the centre and perpendicular to its length is $I_1$. The same rod is bent into a ring and its moment of inertia about the diameter is $I_2$. Then $I_1 / I_2$ is

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