Physics - Center of Mass Question with Solution | TestHub

Let the original disc be $M_1$ with mass $M$ and center at $x_1=0$. Its area is $A_1=\pi R^2$. Let the cut-out hole be $M_2$ with mass $m^{\prime }$ and center at $x_2=R/2$. Its area is $A_2=\pi (R/2{)}^2=\pi R^2/4$.

The mass of the hole $m^{\prime }$ is proportional to its area. If $\sigma$ is the surface mass density, $M=\sigma \pi R^2$ and $m^{\prime }=\sigma \pi (R/2{)}^2=M/4$. The remaining portion has mass $M_{rem}=M-m^{\prime }=M-M/4=3M/4$. Let its center of mass be $X_{CM}$.

We can treat the original disc as a combination of the remaining portion and the cut-out hole. The center of mass of the original disc is at $0$.$$X_{CM,original}=\frac{M_{rem}{X}_{CM}+m^{\prime }{x}_2}{M_{rem}+m^{\prime }}=0$$ $$0=\frac{(3M/4)X_{CM}+(M/4)(R/2)}{M}$$ $$(3M/4)X_{CM}+(M/4)(R/2)=0$$ $$(3/4)X_{CM}=-(1/4)(R/2)$$ $$X_{CM}=-\frac{R}{6}$$

The magnitude of the position of the center of mass of the remaining portion from the center of the original disc is $R/6$.