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 ?

From a disc of radius R and mass M , a circular hole of diameter R , w

1.	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 ?
Answer» 11 `MR^(2) //32` `9 MR^(2)// 32` `15 MR^(2)//32` `13 MR^(2)//32` Solution :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)`