Four point masses are placed at the corners of a square of side $2 m$ as shown in the figure. Find the center of mass of the system w.r.t center of square.

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Solution

Let us consider $O$ as origin.
So, $x_{c m} = \frac{m_{1} X_{1} + m_{2} X_{2} + m_{3} X_{3} + M_{4} X_{4}}{m_{1} + m_{2} + m_{3} + m_{4}} = \frac{3 \times 0 + 2 \times 2 + 1 \times 2 + 2 \times 0}{10} = \frac{6}{10} = \frac{3}{5}$
Similarly, $y_{c m} = \frac{3 \times 0 + 2 \times 0 + 1 \times 2 + 4 \times 2}{10} = \frac{10}{10} = 1$
$∴$ COM at $(\frac{3}{5}, 1)$
or $\frac{2}{5} m$ left of center of square.

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