Rotation 1 Question 16
19. A thin wire of length $L$ and uniform linear mass density $\rho$ is bent into a circular loop with centre at $O$ as shown. The moment of inertia of the loop about the axis $X X^{\prime}$ is (2000)
(a) $\frac{\rho L^{3}}{8 \pi^{2}}$
(b) $\frac{\rho L^{3}}{16 \pi^{2}}$
(c) $\frac{5 \rho L^{3}}{16 \pi^{2}}$
(d) $\frac{3 \rho L^{3}}{8 \pi^{2}}$
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Answer:
Correct Answer: 19. (d)
Solution:
- Mass of the $\operatorname{ring} M=\rho L$. Let $R$ be the radius of the ring, then
or
$$ \begin{aligned} & L=2 \pi R \\ & R=\frac{L}{2 \pi} \end{aligned} $$
Moment of inertia about an axis passing through $O$ and parallel to $X X^{\prime}$ will be
$$ I _0=\frac{1}{2} M R^{2} $$
Therefore, moment of inertia about $X X^{\prime}$ (from parallel axis theorem) will be given by
$$ I _{X X^{\prime}}=\frac{1}{2} M R^{2}+M R^{2}=\frac{3}{2} M R^{2} $$
Substituting values of $M$ and $R$
$$ I _{X X^{\prime}}=\frac{3}{2}(\rho L) \frac{L^{2}}{4 \pi^{2}}=\frac{3 \rho L^{3}}{8 \pi^{2}} $$
