If I_1 is the moment of inertia of a thin rod | vedclass

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Question

(B) The moment of inertia of a thin rod of mass $M$ and length $l$ about an axis passing through its centre and perpendicular to its length is given b

Vedclass · Accepted Answer

$I_1:I_2=\pi^2:3$

Vedclass · Answer

(B) The moment of inertia of a thin rod of mass $M$ and length $l$ about an axis passing through its centre and perpendicular to its length is given by $I_1 = \frac{Ml^2}{12}$.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 gives $R = \frac{l}{2\pi}$.The moment of inertia of the ring about an axis passing through its centre and perpendicular to its plane is $I_2 = MR^2$.Substituting the value of $R$,we get $I_2 = M \left(\frac{l}{2\pi}\right)^2 = \frac{Ml^2}{4\pi^2}$.Now,calculating the ratio $I_1 : I_2$:$\frac{I_1}{I_2} = \frac{Ml^2/12}{Ml^2/4\pi^2} = \frac{4\pi^2}{12} = \frac{\pi^2}{3}$.Thus,$I_1 : I_2 = \pi^2 : 3$.

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