A uniform circular disc of radius ' R ' and mass ' M ' is rotating about an axis perpendicular to its plane and passing through its centre. A small circular part of radius R/2 is removed from the original disc as shown in the figure. Find the moment of inertia of the remaining part of the original disc about the axis as given above. [figure]
- 1732MR2
- 1332MR2
- 932MR2
- 732MR2
Answer
(2) 1332MR2
Solution
Remaining disc = full disc - removed disc (same axis through O).
[figure]
m= M4 (area), Ifull= 12MR2
Removed disc (parallel axis, distance R2):
I= 12m ( R2 )2+m ( R2 )2
= 38mR2= 332MR2
I= 12MR2- 332MR2= 1332MR2
Answer: (2)