From a circular disc of radius R and mass 9 M , a small disc of radius R/3 is removed from the disc. The moment of inertia of the remaining disc about an axis perpendicular to the plane of the disc and passing through O 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]`