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 disc of radius R and mass density ±  is placed. Then

Total Moment of Inertia ` I = i_1 + I_2`
`I_1 = (M_1 (2R)^2)/2`
`I_1 = (sigma pi 4 R^2 .4R^2)/2 = 8 pi sigmaR^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 ( - sigmaR^2 )R^2 " "I_2`
` = -3/2 sigmaa pi R^2`
` = - 3/2 sigma pi R^4`
Now `I = I_1 + I_2`
` I = 8 pi sigmaR^2 - 3/2 sigma pi R^4`
` I = 13/2 sigma pi R^4`