Physics - Center of Mass Question with Solution | TestHub

The original square plate of mass $4M$ has its center of mass (COM) at $(L/2,L/2)$.

The square hole of side $L/2$ has a mass $m_{hole}=(L/2{)}^2/L^2\times 4M=M$.

Its COM is at $(L-L/4,L-L/4)=(3L/4,3L/4)$.

The COM of the plate with the hole (mass $3M$) is $x_{plate}=\frac{4M(L/2)-M(3L/4)}{4M-M}=\frac{2ML-3ML/4}{3M}=\frac{5L}{12}$ and $y_{plate}=\frac{5L}{12}$.

The rod has mass $M$ and its COM is at $(L,L/2)$. The COM of the final system (total mass $3M+M=4M$) is $x_{CM}=\frac{3M(5L/12)+M(L)}{4M}=\frac{5ML/4+ML}{4M}=\frac{9L}{16}$ and $y_{CM}=\frac{3M(5L/12)+M(L/2)}{4M}=\frac{5ML/4+ML/2}{4M}=\frac{7L}{16}$.