A thin wire of length L and uniform linear ma | vedclass

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(A) The total mass of the wire is $M = \lambda L$. Since the wire is bent into a circular coil of radius $R$,the circumference is $2 \pi R = L$,so $R

Vedclass · Accepted Answer

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

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(A) The total mass of the wire is $M = \lambda L$. Since the wire is bent into a circular coil of radius $R$,the circumference is $2 \pi R = L$,so $R = \frac{L}{2 \pi}$.The moment of inertia of a circular ring about its diameter is $I_{diam} = \frac{1}{2} M R^2$.Using the parallel axis theorem,the moment of inertia about a tangential axis in the plane of the coil is $I = I_{diam} + M R^2$.$I = \frac{1}{2} M R^2 + M R^2 = \frac{3}{2} M R^2$.Substituting $M = \lambda L$ and $R = \frac{L}{2 \pi}$:$I = \frac{3}{2} (\lambda L) \left( \frac{L}{2 \pi} \right)^2 = \frac{3}{2} \lambda L \left( \frac{L^2}{4 \pi^2} \right) = \frac{3 \lambda L^3}{8 \pi^2}$.

$A$ thin wire of length $L$ and uniform linear mass density $\lambda$ is bent into a circular coil. The moment of inertia of this coil about a tangential axis in the plane of the coil is:

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