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

Concept: We calculate the moment of inertia for the two distinct geometries. The moment of inertia of a body is a measure of its resistance to rotational acceleration.

Step 1: Moment of inertia of the thin uniform rod (\( I_1 \)).
For a rod of mass \( M \) and length \( L \), rotated about its center of mass: $$ I_1 = \frac{ML^2}{12} $$

Step 2: Moment of inertia of the ring (\( I_2 \)).
When the rod of length \( L \) is bent into a ring, its circumference \( 2\pi R = L \), so the radius \( R = \frac{L}{2\pi} \). The moment of inertia of a thin ring about its center of mass is \( MR^2 \). Using the Parallel Axis Theorem for a tangent axis perpendicular to the ring's plane: $$ I_2 = I_{cm} + MR^2 = MR^2 + MR^2 = 2MR^2 $$ Substituting \( R = \frac{L}{2\pi} \): $$ I_2 = 2M\left(\frac{L}{2\pi}\right)^2 = 2M\frac{L^2}{4\pi^2} = \frac{ML^2}{2\pi^2} $$

Step 3: Calculate the ratio.
$$ \frac{I_1}{I_2} = \frac{\frac{ML^2}{12}}{\frac{ML^2}{2\pi^2}} $$ $$ \frac{I_1}{I_2} = \frac{1}{12} \times 2\pi^2 = \frac{\pi^2}{6} $$ $$\boxed{\frac{\pi^2}{6}}$$