From a uniform circular disc of radius R and mass 9M, a small disc of radius `(R )/(3)` is removed as shown in the figure. The moment of inertia of the remaining disc about an axis perpendicular to the plane of the disc and passing through centre of disc is :

Options

A. `4MR^(2)`
B. `40/4 MR^(2)`
C. `10MR^(2)`
D. `37/9MR^(2)`

Answer

A

Explanation

Let `sigma` be the mass per unit area. The total mass of the disc `=sigmaxxpiR^(2)=9M`

`M=`Mass of the circular disc cut
`=sigmaxxpi(R/3)^(2)`
`=sigmaxx(piR^(2))/9=M`
`R` Let us consider the above system as a complete disc of mass `9M` and a negative mass `M` superimposed on it.
Moment of inetia `(I_(1))` of the complete disc `=9MR^(2)//2` about an axis passing through `O` and perpendicular to the plane of the disc.
MI of the cutout portion about an axis passing through `O'` and perpendicular to the plane of disc is
`1/2xxMxx(R/3)^(2)`
Therefore MI`(I_(2))` of the cutout portion about an axis passing through `O` and perpendicular to the plane of disc is
`[1/2xxMxx(R/3)^(2)+Mxx((2R)/3)^(2)]`
[Using perpendicular axis theorem]
Therefore the total `MI` of the system about an axis passing through `O` and perpendicular to the plane of the disc is
`I=I_(1)+I_(2)`
`=1/29MR^(2)-[1/2xxMxx(R/3)^(2)+Mxx((2R)/3)^(2)]^(2)`
`=1/2 9 MR^(2)-MR^(2)[1/18+4/9]`