[ Mass Grouping Theorem ]

From a circle $\mathrm{S}$ of radius 1, a circle $\mathrm{S}^{\prime}$ of radius $1 / 2$ was cut out, the boundary of which passes through the center of the original circle. Determine where the center of gravity of the resulting figure F is located.

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If the "hole" of figure F is filled, the center of gravity will move to the center of circle S.

## Solution

Let the centers of circles $\mathrm{S}$ and $\mathrm{S}^{\prime}$ be denoted as $\mathrm{O}$ and $\mathrm{O}^{\prime}$, respectively, and the center of gravity of figure $\mathrm{F}$ as M. From symmetry, it follows that M lies on the line OO'. If we denote the area of circle $\mathrm{S}$ as 1, then the area of circle $\mathrm{S}^{\prime}$ will be $1 / 4$, and the area of figure $\mathrm{F}$ will be 3/4. Introducing coordinates on the line OO', taking point O as the origin; the positive direction will be the direction of the ray OO'. Point O' will then have a coordinate of $1 / 2$. Let point M have a coordinate of $m$. According to the theorem of mass regrouping, to determine the center of mass of the combined figures S' and F based on the positions of the centers of mass O' and M of figures S' and F, we need to assign the mass (area) of figure S' to point O', and the mass of figure F to point M, and then find the center of mass of the resulting system of two point masses. Thus, let there be a mass of $1 / 4$ at point O' with coordinate $1 / 2$, and a mass of 3/4 at point M with coordinate $m$. The center of mass of the resulting system is located at the point with coordinate $((1 / 4) * (1 / 2) + (3 / 4) m) / ((1 / 4) + (3 / 4)) = 1 / 8 + (3 / 4) m$. However, the union of figures S' and F is the circle S with its center at the origin. Therefore, $1 / 8 + (3 / 4) m = 0$ and $m = -1 / 6$. Thus, the center of gravity of figure $\mathrm{F}$ is located on the line OO' outside the ray $\mathrm{OO}^{\prime}$ at a distance of $1 / 6$ from point O.