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

$I_y=0.144Kgm^2$
We can determine the required moment of inertia as follows: The mass of the thin plate is given as $m=\sigma A$ $\implies m=10(0.4)^2$ $\implies m=1.6Kg$ and the mass of thin disk is given as $m=\sigma (-A)$ $\implies m=10(-\pi (0.1)^2)$ $\implies m=-0.314Kg$ Now $I_y=\Sigma (I^{\prime}_y+md^2_x)$ We plug in the known values to obtian: $I_y=2(\frac{1.6(0.4)^2}{12})+1.6(0.2)^2+2(\frac{-0.314(0.1)^2}{4}-0.314(0.2)^2)$ This simplifies to: $I_y=0.144Kgm^2$