Answer
$I_G=118slug\cdot ft^2$
Work Step by Step
We can determine the required moment of inertia as follows:
The mass of the big cylinder is given as
$m=\gamma V$
$\implies m=90\pi(2.5)^2(1)$
$\implies m=1767lb$
The mass of the cylinder number 1 is given as
$m=\gamma(-V)$
$\implies m=90[-\pi(2)^2\times(0.375)]$
$\implies m=-424lb$
The mass of the cylinder 2 is
$m=\gamma(-V)$
$\implies m=90[-\pi(1)^2(0.25)]$
$\implies m=-70.69lb$
and the mass of the cylinder number 3 is given as
$m=\gamma(-V)$
$\implies m=90[-\pi(2)^2(0.375)]$
$\implies m=-424lb$
Now $I_G=\Sigma I^{\prime}$
We plug in the known values to obtain:
$I_G=\frac{1767(2.5)^2}{2}+\frac{-424(2)^2}{2}+\frac{-70.69(1)^2}{2}+\frac{-424(2)^2}{2}$
This simplifies to:
$I_G=3791lb ft^2$
$\implies I_G=118slug\cdot ft^2$
