The moment of inertia of a circular ring of mass M and diameter r about a tangential axis lying in the plane of the ring is :
- 38Mr2
- 2Mr2
- 12Mr2
- 32Mr2
Answer
(1) 38Mr2
Solution
Radius of the ring R = r2
About a diameter (perpendicular axis theorem): Id = 12MR2
Tangent in the plane is parallel to a diameter at distance R (parallel axis theorem):
I = 12MR2 + MR2 = 32MR2
= 32M ( r2 )2 = 38Mr2
Answer: (1)