A circular hole of radius `R//2` is cut from a circular disc of radius `R`. The disc lies in the `xy`-plane and its centre coincides with the origin. If the remaining mass of the disc is `M`, then
a. determine the initial mass of the disc and
b. determine its moment of inertia about the `z`-axis.

Explanation

Let `sigma` be the surface mass density of the plate given by
`sigma=M/(piR^(2)-pi(R/2))^(2)=4/3 M/(piR^(2))`
Now, initial mass of the plate `m_(1)=sigma(piR^(2))=4/3M`
mass of the plate removed `m_(2)=sigma((piR^(2))/4)=M/3`
The given arrangement may be considered as a combination of a circular plate of mass `(4//3)M` and circular plate of negative mass `(M//3).`

The moment of inertia of the big disc of radius `R` and mass `m_(1)` about `O` and parallel to the `z`-axis is
`I_(O_1)=(m_(1)R^(2))/2`
The moment of inertia of the small disc of mass `m_(2)` and radius `r/2` about `O `and parallel to `z`-axis.
`I_(O_(1))=(m_(2)(R/2)^(2))/2=(m_(2)R^(2))/8`
The moment of inertia of the small disc about `O` (using parallel axis theorem).
`I_(O_(2))=(m_(2)R^(2))/8+(m_(2)R^(2))/4=(3m_(2)R^(2))/8`
As the part of the disc on mass `m_(2)` is removed from disc `m_(1)`, hence the net moment of inertia of structure is
`I_(O)=I_(O_(1))=I(O_(2))`
`I_(O)=(m_(1)R^(2))/2-(3m_(2)R^(2))/8`
substituting the values of `m_(1)` and `m_(2)` we get
`I_(0)=13/24MR^(2)`