A circular hole of radius R 4 is made in a thin uniform disc having mass and radius R , as shown in figure. The moment of inertia of the remaining portion of the disc about an axis passing through the point O and perpendicular to the plane of the disc is-

Step-by-step solution

I D = m r 2 2 Moment of inertia of removed portion about the axis. I r e m o v e d = 1 2     m 16     r 2 16 + m 16     9 r 2 16       (by parallel axis theorem) = m r 2 + 18 m r 2 512 = 19 m r 2 512 I r e m a i n i n g = m r 2 2 - 19 512   m r 2 = 237 512   m r 2