## Engineering Mechanics: Statics & Dynamics (14th Edition)

$\bar{y}=1.78m$ $I_G=4.45Kgm^2$
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: $I_G=4.45Kgm^2$