From a circular disc of radius R and mass 9M, 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. `4M^(2)`
B. `40/9MR^(2)`
C. `10MR^(2)`
D. `37/9MR^(2)`

Answer

A

Explanation

Mass per unit area of disc`=(9M)/(piR^(2))`
`therefore` Mass of removed portion `=(9M)/(piR^(2))xxpi(R/3)^(2)=M`
Let moment of inertia of removed portion = `I_(1)`
`thereforeI_(1)=M/2(R/3)^(2)+M((2R)/3)^(2)`, by theorem of parallel or `I_(1)=(MR^(2))/2`
Let `I_(2)` = Moment of inertia of the whole disc
`I_(2)=(9MR^(2))/2`
`therefore` Let I = Moment of inertia of remaining disc
`thereforeI=I_(2)-I_(1)`
or `I=(9MR^(2))/2-(MR^(2))/2=(8MR^(2))/2=4MR^(2)`
or `I=4MR^2`