Answer:
the answer is below
Explanation:
The diagram of the problem is given in the image attached.
Mass of rod AB = (0.4 m + 0.4 m) * 3 kg/m = 2.4 kg
Mass of rod OC = (1.5 m) * 3 kg/m = 4.5 kg
Mass of plate = 10 kg/m²[ (π* 0.3²) - (π* 0.1²)] = 2.513 kg
The center of mass is:
[tex]\hat{y}[2.4+2.513+4.5]=(0.75*4.5)+(2.513*0.5)\\\\\hat{y}=0.839\\\\I_{AB}=\frac{1}{12}*2.4*(0.4+0.4)^2= 0.128\ kg/m^2\\\\I_{OC}=\frac{1}{12}*4.5*(1.5)^2= 3.375\ kg/m^2\\\\I_{Gplate}=\frac{1}{2}*(\pi*0.3^2*10)*(0.3)^2-\frac{1}{2}*(\pi*0.1^2*10)*(0.1)^2= 0.126\ kg/m^2\\\\I_{plate}=I_{Gplate}+md^2=0.126+2.513(1.8^2)=8.27\ kg/m^2\\\\I_o=I_{plate}+I_{AB}+I_{OC}\\\\I_{o}=0.128+3.375+8.27=11.773\ kg/m^2 \\\\I_G=I_o-m_{tot}\hat{y}^2\\\\I_G=11.773-(9.413*0.839^2)\\\\I_G=5.147\ kg/m^2[/tex]
