Answer
$I_z=0.113Kgm^2$
Work Step by Step
We can find the required moment of inertia as follows:
The mass of the thin plate is given as
$m=\sigma A$
$\implies m=10(0.4)^2$
$\implies m=1.6Kg$
and the mass of the thin disk is given as
$m=\sigma (-A)$
$\implies m=10(-\pi (0.1)^2)=-0.314Kg$
Now $I_z=\Sigma (I+md^2_x)$
We plug in the known values to obtain:
$I_z=\frac{1.6(0.4)^2}{12}+\frac{1.6[(0.4)^2+(0.4)^2]}{12}+1.6(0.2)^2+\frac{-0.31(0.1)^2}{4}+\frac{-0.314(0.1)^2}{2}+0.314(-\pi(0.2)^2)$
This simplifies to:
$I_z=0.113Kgm^2$
