Question

A thin wire of length \(L\) and uniform linear mass density \(\rho\) is bent into a circular loop with centre at \(O\) as shown. The moment of inertia of the loop about the axis \(XX'\) is: center

center

• A. \(\dfrac{\rho L^3}{8\pi^2}\)
• B. \(\dfrac{\rho L^3}{16\pi^2}\)
• C. \(\dfrac{5\rho L^3}{16\pi^2}\)
• D. \(\dfrac{3\rho L^3}{8\pi^2}\)

Hint:

For a ring: \[ I_{\text{diameter}}=\frac12 MR^2 \] and for tangent axis in the plane: \[ I_{\text{tangent}}=\frac32 MR^2 \] using Parallel Axis Theorem.

Answer 1

The Correct Option is D

Concept: The moment of inertia of a ring about a tangent axis in its plane is obtained using the Parallel Axis Theorem. For a circular ring: \[ I_{\text{diameter}}=\frac12 MR^2 \] Using Parallel Axis theorem: \[ I_{\text{tangent}}=I_{\text{diameter}}+MR^2 \] Hence, \[ I=\frac12 MR^2+MR^2 \] \[ =\frac32 MR^2 \]

Step 1: Express mass in terms of linear density. Mass of the wire: \[ M=\rho L \] Circumference: \[ L=2\pi R \] Thus, \[ R=\frac{L}{2\pi} \]

Step 2: Substitute into moment of inertia formula. \[ I=\frac32 MR^2 \] \[ =\frac32 (\rho L)\left(\frac{L}{2\pi}\right)^2 \] \[ =\frac32 (\rho L)\frac{L^2}{4\pi^2} \] \[ =\frac{3\rho L^3}{8\pi^2} \] Therefore, \[ \boxed{\frac{3\rho L^3}{8\pi^2}} \]