From a circular ring of mass 'M' and radius 'R' an arc corresponding to a 90 sector is removed. The moment of inertia of the remaining part of the ring about an axis passing through the center of the ring and perpendicular to the plane of the ring is 'K' times 'MR2'. Then the value of 'K' is :
- 18
- 34
- 78
- 14
Answer
(2) 34
Solution
A 90 arc is 90 360 = 14 of the ring.
Mass of the remaining part, M' = M - M4 = 3M4
Every particle of the remaining arc is at distance R from the axis,
I = M'R2 = 34MR2
K = 34
Answer: (2)