From a circular disc of radius R and mass 9M, a small disc of mass M and radius R3 is removed concentrically. The moment of inertia of the remaining disc about an axis perpendicular to the plane of the disc and passing through its centre is
- 409MR2
- MR2
- 4MR2
- 49MR2
Answer
(1) 409MR2
Solution
Full disc: I1 = 12(9M)R2 = 92MR2
Removed disc: I2 = 12M ( R3 )2 = MR218
I = I1 - I2
= 81MR2 - MR218
= 8018MR2 = 409MR2
Answer: (1)