The moment of inertia of a thin uniform rod a | vedclass

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

Vedclass · Accepted Answer

$\frac{8 \pi^{2}}{3}$

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(A) Let $M$ be the mass of the rod and $L$ be its length. The moment of inertia of the rod about an axis passing through one end and perpendicular to its length is $I = \frac{ML^2}{3}$.When the rod is bent into a ring of radius $r$,the circumference of the ring is equal to the length of the rod: $2\pi r = L$,which implies $r = \frac{L}{2\pi}$.The moment of inertia of a ring about its diameter is given by $I_{1} = \frac{Mr^2}{2}$.Substituting the value of $r$: $I_{1} = \frac{M}{2} \left( \frac{L}{2\pi} \right)^2 = \frac{M}{2} \cdot \frac{L^2}{4\pi^2} = \frac{ML^2}{8\pi^2}$.Now,calculating the ratio $\frac{I}{I_{1}}$:$\frac{I}{I_{1}} = \frac{ML^2/3}{ML^2/8\pi^2} = \frac{ML^2}{3} \cdot \frac{8\pi^2}{ML^2} = \frac{8\pi^2}{3}$.

The moment of inertia of a thin uniform rod about a perpendicular axis passing through one of its ends is $I$. Now,the rod is bent into a ring and its moment of inertia about its diameter is $I_{1}$. Then $\frac{I}{I_{1}}$ is:

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