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

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(D) The linear mass density of the wire is $\rho$. The total mass of the wire of length $L$ is $M = \rho L$. Since the wire of length $L$ is bent into

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

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(D) The linear mass density of the wire is $\rho$. The total mass of the wire of length $L$ is $M = \rho L$. Since the wire of length $L$ is bent into a circular loop,its circumference is $2\pi R = L$,which gives the radius $R = \frac{L}{2\pi}$. The moment of inertia of a circular ring about its diameter is $I_{diam} = \frac{1}{2}MR^2$. Using the parallel axis theorem,the moment of inertia about the axis $XX'$ (which is tangent to the loop and parallel to the diameter) is $I = I_{cm} + MR^2$,where $I_{cm} = I_{diam} = \frac{1}{2}MR^2$. Thus,$I = \frac{1}{2}MR^2 + MR^2 = \frac{3}{2}MR^2$. Substituting $M = \rho L$ and $R = \frac{L}{2\pi}$: $I = \frac{3}{2}(\rho L)\left(\frac{L}{2\pi}\right)^2 = \frac{3}{2}\rho L \left(\frac{L^2}{4\pi^2}\right) = \frac{3\rho L^3}{8\pi^2}$.

$A$ thin wire of length $L$ and uniform linear mass density $\rho$ is bent into a circular loop (as shown in the figure) with center $O$. The moment of inertia of the loop about the axis $XX'$ is:

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