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Question:

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?

15MR2/32

9MR2/32

13MR2/32

11MR2/32

Solution:

ITotal disc = MR2/2
As mass is proportional to area,
MRemoved = M/4
Now, about the same perpendicular axis:
IRemoved = (M/4)(R/2)2/2 + (M/4)(R2)2 = 3MR2/32
⇒IRemaining Disc = ITotal - IRemoved = MR2/2 - 3MR2/32 = 13MR2/32