Question

from a disc of radius R amd mass M, a circular hole of radius R/4 is removed from a distance 3R/4 from centre. The moment of inertia of remaining part of disc about an axis perpendicular to plane and Passing through centre of original disc is?? ​

Answer 1

Answer:

13MR2/32

Explanation:

Mass removed m=M/4

radius of mass removed=r=R/2

M.I of whole disc about its center=MR2/2

M.I of mass removed about its center=mr2/2=MR2/32

By parallel axis theorem,

M.I of mass removed about center of disc=mr2/2+mr/2=3MR2/32

M.I of remaining mass about center of disc=MR2/2−3MR2/32

=13MR2/32