From a disc of radius R and mass M, a circular hole of diameter R, whose rim passes through the centre is cut. What is the moment of inertia of the remaining part of the disc about at perpendicular axis passing through the centre?

Options

A. `(13MR^(2))/(32)`
B. `(11MR^(2))/(32)`
C. `(9MR^(2))/(32)`
D. `(15MR^(2))/(32)`

Answer

A

Explanation

Mass of disc of radius R is M
`therefore` Mass per unit area = `(M)/(piR^(2))`
Mass of cut disc `m=(M)/(piR^(2))xx(piR^(2))/(4)=(M)/(4)`
Moment of inertia about an axis passing through point O.
`I._(0)=I_(0)+md^(2)`
`therefore I._(0)=(m((R)/(2))^(2))/(2)+m((R)/(2))^(2)`
`=(1)/(2)xx(M)/(4)xx(R^(2))/(4)+(M)/(4)xx(R^(2))/(4)`
`therefore I._(0)=(MR^(2))/(32)+(MR^(2))/(16)`
`=(3MR^(2))/(32)`
Moment of inertia of cut disc
`I=I_(0)-I._(0)`
`=(MR^(2))/(2)-(3MR^(2))/(32)`
`therefore I=(16MR^(2)-3MR^(2))/(32)`
`therefore I=(13MR^(2))/(32)`