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 in figure. The moment of inertia of the loop about the axis \(XX'\) is

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

Hint:

Find the radius from the wire length, then use the parallel axis theorem from the diameter to the tangent.

Answer 1

The Correct Option is D

Step 1: Understanding the Concept
The axis XX' touches the loop at one point and lies in its plane, so it is a tangent to the circle.

Step 2: Mass and Radius
Mass \(M=\rho L\). The circumference is \(L=2\pi R\), so \(R=\dfrac{L}{2\pi}\).

Step 3: Parallel Axis Theorem
About a diameter, \(I_d=\dfrac{1}{2}MR^{2}\). About the tangent, \(I=\dfrac{1}{2}MR^{2}+MR^{2}=\dfrac{3}{2}MR^{2}\).

Step 4: Substitute
\(I=\dfrac{3}{2}\rho L\cdot\dfrac{L^{2}}{4\pi^{2}}=\dfrac{3\rho L^{3}}{8\pi^{2}}\).

Final Answer:
The moment of inertia about XX' is 3 rho L cubed over 8 pi squared. \[ \boxed{\text{(D) }\dfrac{3\rho L^{3}}{8\pi^{2}}} \]