Question

Moment of inertia of the rod about an axis passing through the centre and perpendicular to its length is '$I_1$'. The same rod is bent into a ring and its moment of inertia about the diameter is '$I_2$'. Then $I_1/I_2$ is ______.

• A. $\frac{3\pi^2}{2}$
• B. $\frac{2\pi^2}{3}$
• C. $\frac{\pi^2}{3}$
• D. $\frac{\pi^2}{9}$

Hint:

Read carefully! It asks for the MOI of the ring about its diameter ($\frac{1}{2}MR^2$), NOT its central axis ($MR^2$). This factor of $1/2$ is the most common trap in this classic problem.

Answer 1

The Correct Option is B

Step 1: Understanding the Question:
A rigid rod is reshaped into a circular ring. We must compare its original moment of inertia (as a rod) to its new moment of inertia (as a ring) by taking their ratio.

Step 2: Detailed Explanation:
Let the uniform rod have a total mass $M$ and a total length $L$.
1. Moment of Inertia of the Rod ($I_1$):
The formula for the MOI of a rod about its transverse central axis is:
$I_1 = \frac{ML^2}{12}$
2. Bending the Rod into a Ring:
When the rod is bent into a full circle, its entire length $L$ forms the circumference of the new ring.
Let the radius of the newly formed ring be $R$.
Circumference $= 2\pi R = L$
Therefore, the radius $R$ of the ring is:
$R = \frac{L}{2\pi}$
3. Moment of Inertia of the Ring ($I_2$):
The standard MOI of a ring about its central geometric axis is $MR^2$.
By the Perpendicular Axis Theorem, the MOI of a ring about its diameter is exactly half of that:
$I_2 = \frac{1}{2} M R^2$
Substitute $R = \frac{L}{2\pi}$ into this equation:
$I_2 = \frac{1}{2} M \left( \frac{L}{2\pi} \right)^2$
$I_2 = \frac{1}{2} M \left( \frac{L^2}{4\pi^2} \right)$
$I_2 = \frac{ML^2}{8\pi^2}$
4. Calculate the Ratio $I_1 / I_2$:
$\text{Ratio} = \frac{I_1}{I_2} = \frac{ \frac{ML^2}{12} }{ \frac{ML^2}{8\pi^2} }$
The $ML^2$ terms cancel out completely:
$\text{Ratio} = \frac{1}{12} \times \frac{8\pi^2}{1}$
$\text{Ratio} = \frac{8\pi^2}{12}$
Divide numerator and denominator by 4 to simplify:
$\text{Ratio} = \frac{2\pi^2}{3}$

Step 3: Final Answer:
The ratio is $\frac{2\pi^2}{3}$, matching option (b).