Question

A thin wire of length ‘L’ and linear mass density ‘m’ is bent into a circular ring (in x-y plane) with centre ‘C’ as shown in figure. The moment of inertia of the ring about an axis yy′ will be: ____.

• A. 3mL²/8π
• B. 3mL²/8π²
• C. 3mL³/8π
• D. 3mL³/8π²

Hint:

For a ring, $I$ about an axis through center (perpendicular to plane) is $MR^2$. About a diameter, it's half of that ($\frac{1}{2}MR^2$). About a tangent in the plane, it's $\frac{3}{2}MR^2$.

Answer 1

The Correct Option is D

Step 1: Understanding the Concept:
We need to find the moment of inertia ($I$) of a ring about a tangential axis in its own plane. We will use the Theorem of Parallel Axes.

Step 2: Key Formula or Approach:
1. Total mass $M = m \times L$ 2. Circumference $L = 2\pi R \implies R = L / (2\pi)$ 3. $I_{\text{diameter}} = \frac{1}{2} MR^2$ 4. $I_{\text{tangent}} = I_{\text{diameter}} + MR^2$ (Parallel axis theorem)

Step 3: Detailed Explanation:
1. $I_{\text{tangent}} = \frac{1}{2} MR^2 + MR^2 = \frac{3}{2} MR^2$ 2. Substitute $M = mL$ and $R = \frac{L}{2\pi}$: \[ I = \frac{3}{2} (mL) \left( \frac{L}{2\pi} \right)^2 \] \[ I = \frac{3}{2} \cdot mL \cdot \frac{L^2}{4\pi^2} \] \[ I = \frac{3mL^3}{8\pi^2} \]

Step 4: Final Answer:
The moment of inertia about axis yy′ is 3mL³/8π².