Topic: Engineering Mechanics
A thin wire of length L and uniform linear mass density rho is bent into a circular loop with centre O as shown in figure. The moment of inertia of the loop about the axis XX' (tangential to the loop) is
- A. rho L³ / 16 pi²
- B. rho L³ / 8 pi²
- C. 5 rho L³ / 16 pi²
- D. 3 rho L³ / 8 pi²
Answer: D. 3 rho L³ / 8 pi²
Explanation: The length of the wire forms the circumference, so L = 2 * pi * R, giving the radius R = L / (2 * pi). The total mass is M = rho * L. The moment of inertia for a ring about a central diametrical axis is I_c = (M * R²) / 2. By applying the parallel axis theorem to find the moment of inertia about the tangential axis XX', we get I_XX = I_c + M * R² = (3/2) * M * R². Substituting M and R into this formula gives I_XX = (3/2) * (rho * L) * (L / (2 * pi))^2 = (3 * rho * L³) / (8 * pi²).