Answer
$I=29.4Kgm^2$
Work Step by Step
We can determine the required moment of inertia as follows:
The mass of the hemisphere is given as
$m=\rho V$
$\implies m=7.85\frac{2}{3}(300)^3$
$\implies m=443.91Kg$
The mass of the big cone is $m=\rho V$
$\implies m=7.85(\frac{1}{3})\pi (675)(300)^2$
$\implies m=500Kg$
and the mass of the small cone is
$m=\rho V$
$\implies m=785(-\frac{1}{3}\pi (225)(100)^2)$
$\implies m=-18.5Kg$
Now $I=\Sigma I^{\prime}$
$\implies I=\frac{2}{5}(443.91)(0.3)^2+\frac{3}{10}(500)(0.3)^2+\frac{3}{10}(-18.5)(0.1)^2$
This simplifies to:
$\implies I=29.4Kgm^2$
