Consider a disc of radius R and mass M . A hole of radius R 3 is created in the disk such that the center of the hole is R 3 away from centre of the disk. The moment of inertia of the system along the axis perpendicular to the disc passing through the centre of the disc is

Step-by-step solution

Mass per unit area of the disc = M π R 2 . Mass of removed portion of disc, M ' = M π R 2 × π R 3 2 = M 9 Moment of inertia of removed portion about an axis passing through centre of disc O and perpendicular to the plane of disc using parallel axis theorem is I O ' = I C O M + M ' d 2 = 1 2 × M 9 R 3 2 + M 9 R 3 2 = M R 2 162 + M R 2 81 = 3 M R 2 162 When portion of disc would not have been removed, the moment of inertia of complete disc about centre O is I O = M R 2 2 . So, mom