Answer
$I_O=282slug\cdot ft^2$
Work Step by Step
We can determine the required moment of inertia as follows:
The mass of the big cylinder is given as
$m=\gamma V$
$\implies m=(90)\pi(2.5)^2(1)=1767lb$
The mass of the cylinder number 1 is given as
$m=\gamma (-V)$
$\implies m=90(-\pi)(2)^2(0.375)=-424lb$
The mass of the cylinder number 2 is
$m=\gamma(-V)$
$\implies m=90(-\pi) (1)^2(0.25)=-70.69lb$
and the mass of the cylinder number 3 is given as
$m=\gamma(-V)$
$\implies m=90(-\pi)(2)^2(0.375)=-424lb$
Now $I_O=\Sigma (I^{\prime}+md^2)$
We plug in the known values to obtain:
$I_O=\frac{1767(2.5)^2}{2}+1767(2.5)^2+\frac{-424(2)^2}{2}-424(2.5)^2+\frac{-70.69(1)^2}{2}-70.69(2.5)^2+\frac{(-424.1)(2)^2}{2}-424(2.5)^2$
This simplifies to:
$I_O=9092lb ft^2$
$\implies I_O=282slug\cdot ft^2$
