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 express variables of the new shape in terms of the original shape's dimensions.

Answer 1

The Correct Option is A

Concept: We need to determine the moment of inertia for both the rod and the resulting ring, then find their ratio.

Step 1: Moment of inertia of a rod (\( I_1 \)).
For a rod of mass \( M \) and length \( L \): $$ I_1 = \frac{ML^2}{12} $$

Step 2: Moment of inertia of a ring (\( I_2 \)).
The rod of length \( L \) becomes a ring of circumference \( L = 2\pi R \), so \( R = \frac{L}{2\pi} \). The moment of inertia of a ring about an axis passing through the center and perpendicular to the plane is \( MR^2 \). Using the parallel axis theorem for an axis tangent to the ring and perpendicular to the plane: $$ I_2 = MR^2 + MR^2 = 2MR^2 = 2M\left(\frac{L}{2\pi}\right)^2 = 2M\frac{L^2}{4\pi^2} = \frac{ML^2}{2\pi^2} $$

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