From a disc of radius R and mass m, a circular hole of diamter 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. `13 MR^(2)//32`
B. `11 MR^(2)//32`
C. `9 MR^(2)//32`
D. `15 MR^(2)//32`

Answer

A

Explanation

The moment of inertia of the disc about its centre of mass, `I_("disc")=(MR^(2))/(2)`
As the disc is a planar object its mass can be assumed to be uniformly distributed proportional to area. Area of original disc is four times to that of the hole.
Mass of removed part, `M.=(M)/(4)`
Moment of inertia of the removed part about the same axis,
`I.=(M)/(4)((R//2)^(2))/(2)+(M)/(4)((R)/(2))^(2)=(3MR^(2))/(32)`
The moment of inertia of the remaining disc =I-I.
`=(MR^(2))/(2)=(3)/(32)MR^(2)=(13)/(32)MR^(2)`