A uniform disc having radius 2R and mass density `sigma` as shown in figure. If a small disc of radius R is cut from the disc as shown. Then find out the moment of inertia of remaining disc around the axis that passes through O and is perpendicular to the plane of the page.

Explanation

We assume that in remaning part a dise of radius R and mass density `pm sigma` is placed . Then
Total Moment of Inertia ` I = I_(1) + I_(2)`
`I_(1) = (M_(1) (2R)^(2))/(2)`
`I_(1) = (sigma pi 4R^(2) 4R^(2))/(2) = 8 pi sigma R^(4)`
To calculate `I_(2)` we use parallel axis theorem .
`I_(2) = I_(CM) + M_(2) R^(2)`
`I_(2) = (M_(2) R^(2))/(2) + M_(2) R^(2)`
`I_(2) = (3)/(2) M_(2) R^(2) = (3)/(2) (-sigma pi R^(2) ) R^(2) " " I_(2)`
` = - (3)/(2) sigma pi R^(4)`
Now ` I = I_(1) + I_(2) `
`I = 8 pi sigma R^(4) = (3)/(2) sigma pi R^(4)`
`I = (13)/(2) sigma pi R^(4)`