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 a perpendicular axis , passing through the centre ?

Options

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

Answer

D

Explanation

Mass per unit area of disc = `(M)/(pi R^(2))`
Mass of removal portion of disc ,
`M. = (M)/(pi R^(2)) xx pi ((R)/(2))^(2) = (M)/(4)`

Moment of interia of removed portion about an axis passing through centre of disc O and perpendicular to the plane of disc ,
`I._(O) = I_(O.) + M.d^(2) = (1)/(2) xx (M)/(4) xx ((R)/(2))^(2) + (M)/(4) xx ((R)/(2))^(2)`
`= (MR^(2))/(32) + (MR^(2))/(16) = (3 MR^(2))/(32)`
The moment of inertia of complete disc about centre O is `I_(O) = (1)/(2) MR^(2)`
So, moment of inertia of the disc with removed portion is
`I = I_(O) - I._(O) = (1)/(2) MR^(2) - (3MR^(2))/(32) = (13 MR^(2))/(32)`