A thin uniform wire of mass m and linear dens | vedclass

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(A) $1$. The mass of the ring is $m$ and its linear density is $\rho$. The circumference of the ring is $L = \frac{m}{\rho}$.$2$. Since $L = 2 \pi R$,

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

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(A) $1$. The mass of the ring is $m$ and its linear density is $\rho$. The circumference of the ring is $L = \frac{m}{\rho}$.$2$. Since $L = 2 \pi R$, the radius of the ring is $R = \frac{m}{2 \pi \rho}$.$3$. The moment of inertia of a ring about its diameter is $I_{diam} = \frac{1}{2} m R^2$.$4$. According to the parallel axis theorem, the moment of inertia about a tangent parallel to the diameter is $I = I_{diam} + m R^2 = \frac{1}{2} m R^2 + m R^2 = \frac{3}{2} m R^2$.$5$. Substituting $R = \frac{m}{2 \pi \rho}$ into the expression: $I = \frac{3}{2} m \left( \frac{m}{2 \pi \rho} \right)^2 = \frac{3}{2} m \left( \frac{m^2}{4 \pi^2 \rho^2} \right) = \frac{3 m^3}{8 \pi^2 \rho^2}$.

$A$ thin uniform wire of mass $m$ and linear density $\rho$ is bent in the form of a circular ring. The moment of inertia of the ring about a tangent parallel to its diameter is

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