A crucial part of a piece of machinery starts as a flat uniform cylindrical disk of radius R0 = 0.62 m and mass M = 24.29 kg . It then has a circular hole of radius R1 = 0.05 m drilled into it. The hole’s center is a distance h = 0.09 m from the center of the disk. Find the moment of inertia of this disk (with Âoff-center hole) when rotated about its center, in kg*m^2. Solution Given :- Ro = 0.62 m M = 24.29 kg R1 = 0.05 m h = 0.09 m Moment of inertia of the disk I = 1/2*M/R 0 2 [R 0 4 - R 1 4 - 2 R 1 2 h 2 ] I = 1/2 * 24.29/0.62 2 [0.62 4 - 0.05 4 - 2*0.05 2 0.09 2 ] I = 4.67 kg.m 2 .