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(D) The length of the rod is $L$,and its linear mass density is $\lambda$. Therefore,the total mass of the rod is $M = \lambda L$.When the rod is bent

Vedclass · Accepted Answer

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

Vedclass · Answer

(D) The length of the rod is $L$,and its linear mass density is $\lambda$. Therefore,the total mass of the rod is $M = \lambda L$.When the rod is bent into a ring of radius $R$,the circumference of the ring is equal to the length of the rod: $2 \pi R = L$,which gives $R = \frac{L}{2 \pi}$.The moment of inertia of a ring of mass $M$ and radius $R$ about its diameter is given by the formula $I = \frac{1}{2} M R^2$.Substituting the values of $M$ and $R$ into the formula:$I = \frac{1}{2} (\lambda L) \left( \frac{L}{2 \pi} \right)^2$$I = \frac{1}{2} (\lambda L) \left( \frac{L^2}{4 \pi^2} \right)$$I = \frac{\lambda L^3}{8 \pi^2}$.

$A$ rod of linear mass density $\lambda$ and length $L$ is bent to form a ring of radius $R$. The moment of inertia of the ring about any of its diameters is:

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