Question

\( I_1 \) represents moment of inertia of a thin, uniform rod about an axis perpendicular to its length and passing through its centre of mass. The same rod is bent into the shape of a ring. If \( I_2 \) is moment of inertia of ring about an axis that is tangent to the ring and perpendicular to its plane, then \(\frac{I_1}{I_2}=\):

• A. \( \frac{\pi^2}{6} \)
• B. \( \frac{\pi}{6} \)
• C. \( 6\pi^2 \)
• D. \( 6\pi \)

Hint:

Always relate the parameters of the transformed shape (like radius \(R\)) back to the original dimensions (like length \(L\)) immediately.

Answer 1

The Correct Option is A

Step 1: Rod's moment of inertia.
A thin rod of mass $M$, length $L$, about its centre and perpendicular to its length, gives $I_1 = \dfrac{ML^2}{12}$.
Step 2: Reshape into a ring.
The rod's length becomes the ring's circumference, $2\pi R = L$, so $R = \dfrac{L}{2\pi}$, keeping mass $M$ unchanged.
Step 3: Ring about its central axis.
A ring about the axis through its centre, perpendicular to its plane, has $I_{centre} = MR^2$.
Step 4: Move to the tangent axis.
By the parallel axis theorem, $I_2 = MR^2 + MR^2 = 2MR^2$.
Step 5: Substitute $R$.
$I_2 = 2M\left(\dfrac{L}{2\pi}\right)^2 = \dfrac{ML^2}{2\pi^2}$.
Step 6: Divide to get the ratio.
\[ \frac{I_1}{I_2} = \frac{ML^2/12}{ML^2/(2\pi^2)} = \frac{2\pi^2}{12} = \frac{\pi^2}{6} \]
\[ \boxed{\dfrac{\pi^2}{6}} \]