Physics - Center of Mass Question with Solution | TestHub

Let $\sigma$ be the mass per unit area of the uniform disc. The mass of the original disc is $M=\sigma (\pi R^2)$, and its center of mass is $C_D=(0,0)$. The square hole has side length $a=R/2$. Its area is $A_{hole}=a^2=(R/2{)}^2=R^2/4$. The mass of the square hole is $m_{hole}=\sigma A_{hole}=\frac{M}{\pi R^2}\cdot \frac{R^2}{4}=\frac{M}{4\pi }$. Since one corner of the square is at the origin and its sides are parallel to the axes, its center of mass is $C_{hole}=(a/2,a/2)=(R/4,R/4)$. The mass of the remaining portion is $M_{rem}=M-m_{hole}=M-\frac{M}{4\pi }=M\frac{4\pi -1}{4\pi }$. Using the principle of center of mass for a removed part: $M_{rem}{\vec{R}}_{CM}=M{\vec{R}}_D-m_{hole}{\vec{R}}_{hole}$. For the $x$ -coordinate: $M\frac{4\pi -1}{4\pi }X=M(0)-\frac{M}{4\pi }(\frac{R}{4})$. This gives $X=-\frac{R}{4(4\pi -1)}$. Similarly for the $y$ -coordinate: $M\frac{4\pi -1}{4\pi }Y=M(0)-\frac{M}{4\pi }(\frac{R}{4})$. This gives $Y=-\frac{R}{4(4\pi -1)}$. Thus, the center of mass of the remaining portion is $(-\frac{R}{4(4\pi -1)},-\frac{R}{4(4\pi -1)})$.