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

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(A) The length of the wire is $l = 2\pi r$,so the radius is $r = \frac{l}{2\pi}$.The mass of the loop is $m = \rho l$.The moment of inertia of a ring

Vedclass · Accepted Answer

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

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(A) The length of the wire is $l = 2\pi r$,so the radius is $r = \frac{l}{2\pi}$.The mass of the loop is $m = \rho l$.The moment of inertia of a ring of mass $m$ and radius $r$ about its diameter is $I_{diam} = \frac{1}{2}mr^2$.Using the parallel axis theorem,the moment of inertia about a tangent axis $XX'$ is $I_{XX'} = I_{diam} + mr^2 = \frac{1}{2}mr^2 + mr^2 = \frac{3}{2}mr^2$.Substituting $m = \rho l$ and $r = \frac{l}{2\pi}$ into the expression:$I_{XX'} = \frac{3}{2}(\rho l)(\frac{l}{2\pi})^2 = \frac{3}{2}(\rho l)(\frac{l^2}{4\pi^2}) = \frac{3\rho l^3}{8\pi^2}$.

$A$ thin wire of length $l$ and uniform linear mass density $\rho$ is bent into a circular loop with centre $O$ and radius $r$ as shown in the figure. The moment of inertia of the loop about the axis $XX'$ is

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