Physics - Center of Mass Question with Solution | TestHub

Let the four corners of the square be $(a/2,a/2)$, $(-a/2,a/2)$, $(-a/2,-a/2)$, and $(a/2,-a/2)$. The center of the square is at $(0,0)$. If the mass at $(a/2,a/2)$ is removed, the remaining three masses are at $m_1(-a/2,a/2)$, $m_2(-a/2,-a/2)$, and $m_3(a/2,-a/2)$. The x-coordinate of the center of mass is $X_{CM}=\frac{m(-a/2)+m(-a/2)+m(a/2)}{3m}=\frac{-a/2}{3}=-a/6$. The y-coordinate of the center of mass is $Y_{CM}=\frac{m(a/2)+m(-a/2)+m(-a/2)}{3m}=\frac{-a/2}{3}=-a/6$. Thus, the center of mass is at $(-a/6,-a/6)$.