Answer
$\bar{y}=1.78m$
$I_G=4.45Kgm^2$
Work Step by Step
We know that
$\bar{y}=\frac{\Sigma m\bar{y}}{\Sigma m}$
We plug in the known values to obtain:
$\bar{y}=\frac{3(1)+5(2+0.25)}{3+5}$
$\implies \bar{y}=1.78m$
Now the required moment of inertia can be determined as:
$I_G=\Sigma (I^{\prime}+Ad^2)$
We plug in the known values to obtain:
$I_G=\frac{3(2)^2}{12}+3(1.78-1)^2+\frac{5(1)^2}{12}+5(2.251)^2$
This simplifies to:
$I_G=4.45Kgm^2$
