The moment of inertia of a thin uniform rod about a perpendicular axis passing through one of its ends is…
The moment of inertia of a thin uniform rod about a perpendicular axis passing through one of its ends is 'I'. Now, the rod is bent in a ring and its moment of inertia about diameter is ' $\mathrm{I}_{1}^{\prime}$. Then $\frac{\mathrm{I}}{\mathrm{I}_{1}}$ is
$\frac{8 \pi^{2}}{3}$
$\frac{11 \pi^{2}}{3}$
$\frac{4 \pi^{2}}{3}$
$\frac{\pi^{2}}{3}$
Solution
If $M$ is the mass of the rod and $L$ is its length, then $I=\frac{M L^{2}}{3}$ Radius of the circular ring is given by
$\begin{aligned}
2 \pi r=L &=\frac{L}{2 \pi} \\
\therefore I_{1} &=\frac{M r^{2}}{2}=\frac{M}{2} \cdot \frac{L^{2}}{4 \pi^{2}} \\
\therefore \frac{1}{I_{1}} &=\frac{8 \pi^{2}}{3}
\end{aligned}$
.