Let
σ be the mass per unit area.
The total mass of the disc
=σ×πR2=9M The mass of the circular disc cut
=σ×π()2=σ×=M Let us consider the above system as a complete disc of mass
9M and a negative mass
M super imposed on it.
Moment of inertia
(I1) of the complete disc
= 9MR2 about an axis passing through
O and perpendicular to the plane of the disc.
M.I. of the cut out portion about an axis passing through
O′ and perpendicular to the plane of disc
=×M×()2 ∴ M.I.
(I2) of the cut out portion about an axis passing through
O and perpendicular to the plane of disc
=×M×()2 ∴ M.I.
(I2) of the cut out portion about an axis passing through
O and perpendicular to the plane of disc
=[×M×()2+M×()2][Using perpendicular axis theorem]
∴ The total
M.I. of the system about an axis passing through
O and perpendicular to the plane of the disc is
I=I1+I2 =9MR2−[×M×()2 +M×()2] =−= =4MR2