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(C) Let the three rods be $OA$,$OB$,and $AB$,each of mass $m$ and length $a$.$1$. The center of mass $(CM)$ of rod $OA$ (along the x-axis from $(0,0)$

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$\frac{a}{3}, \frac{a}{3}$

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(C) Let the three rods be $OA$,$OB$,and $AB$,each of mass $m$ and length $a$.$1$. The center of mass $(CM)$ of rod $OA$ (along the x-axis from $(0,0)$ to $(a,0)$) is at $(a/2, 0)$.$2$. The center of mass $(CM)$ of rod $OB$ (along the y-axis from $(0,0)$ to $(0,a)$) is at $(0, a/2)$.$3$. The center of mass $(CM)$ of rod $AB$ (connecting $(a,0)$ and $(0,a)$) is at the midpoint of the line segment,which is $(a/2, a/2)$.Now,calculate the coordinates of the center of mass of the system:$X_{cm} = \frac{m(a/2) + m(0) + m(a/2)}{m + m + m} = \frac{ma}{3m} = \frac{a}{3}$$Y_{cm} = \frac{m(0) + m(a/2) + m(a/2)}{m + m + m} = \frac{ma}{3m} = \frac{a}{3}$Thus,the coordinates of the center of mass are $(a/3, a/3)$.

Three rods of equal mass are arranged as shown in the figure. Find the coordinates of the center of mass of the system.

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