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

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(A) $1$. The total mass $M$ of the wire is $M = \lambda L$.$2$. The circumference of the ring is $2 \pi R = L$,so the radius $R = \frac{L}{2 \pi}$.$3$

Vedclass · Accepted Answer

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

Vedclass · Answer

(A) $1$. The total mass $M$ of the wire is $M = \lambda L$.$2$. The circumference of the ring is $2 \pi R = L$,so the radius $R = \frac{L}{2 \pi}$.$3$. The moment of inertia of a ring about its central axis (perpendicular to the plane) is $I_{cm} = MR^2$.$4$. By the perpendicular axis theorem,the moment of inertia about a diameter is $I_{diameter} = \frac{1}{2} MR^2$.$5$. Using the parallel axis theorem,the moment of inertia about a tangential axis in the plane is $I = I_{diameter} + MR^2$.$6$. $I = \frac{1}{2} MR^2 + MR^2 = \frac{3}{2} MR^2$.$7$. Substituting $M = \lambda L$ and $R = \frac{L}{2 \pi}$:$I = \frac{3}{2} (\lambda L) (\frac{L}{2 \pi})^2 = \frac{3}{2} \lambda L \cdot \frac{L^2}{4 \pi^2} = \frac{3 \lambda L^3}{8 \pi^2}$.

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

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