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: [figure]
- 3mL38
- 3mL28 2
- 3mL38 2
- 3mL28
Answer
(3) 3mL38 2
Solution
Mass of the ring M = mL; circumference 2 R = L ⇒ R = L2
About a diameter (perpendicular-axis theorem): Id = MR22
yy' is a tangent in the plane of the ring, parallel to a diameter at distance R (parallel-axis theorem):
I = MR22 + MR2 = 32MR2
I = 32(mL) L24 2
I = 3mL38 2
Answer: (3)