From a disc of radius R and mass M, a circular hole of diameter R, whose rim passes through the centre is cut. What is the moment of inertia of the remaining part of the disc about a perpendicular axis, passing through the centre?
- 11MR232
- 9MR232
- 15MR232
- 13MR232
Answer
(4) 13MR232
Solution
Mass per unit area: = M R2
Removed disc: radius R2 ⇒ mass m = ( R2 )2 = M4; its centre is at distance R2 from O.
Moment of inertia of the removed part about O (parallel axis theorem):
Ihole = 12 ( M4 ) ( R2 )2 + M4 ( R2 )2
= MR232 + MR216 = 3MR232
∴ Iremaining = MR22 - 3MR232 = 13MR232
Answer: (4)