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.
(A)327MR2
(B)329MR2
(C)3217MR2
(D)3213MR2✓ Correct
Answer: (D)
Solution
The full disc has mass M, radius R, and moment of inertia about the central axis Ifull=21MR2.
The removed piece is a disc of radius R/2 whose centre lies at distance R/2 from O, tangent to the rim of the original disc as shown. Since mass is proportional to area for a uniform disc, its mass is
m=M(RR/2)2=4M.
Its moment of inertia about its own centre is 21m(2R)2=21⋅4M⋅4R2=32MR2.
Shifting to the axis through O with the parallel axis theorem (offset d=R/2):