Two point masses M each are placed at (L, 0) | vedclass

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(D) Let the positions of the three masses be $m_1 = M$ at $(L, 0)$, $m_2 = M$ at $(-L, 0)$, and $m_3 = M$ at $(L \cos 	heta, L \sin 	heta)$.The coor

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$x^2 + y^2 = L^2/9$

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(D) Let the positions of the three masses be $m_1 = M$ at $(L, 0)$, $m_2 = M$ at $(-L, 0)$, and $m_3 = M$ at $(L \cos 	heta, L \sin 	heta)$.The coordinates of the center of mass $(X, Y)$ are given by:$X = \frac{m_1 x_1 + m_2 x_2 + m_3 x_3}{m_1 + m_2 + m_3} = \frac{M(L) + M(-L) + M(L \cos 	heta)}{3M} = \frac{L \cos 	heta}{3}$$Y = \frac{m_1 y_1 + m_2 y_2 + m_3 y_3}{m_1 + m_2 + m_3} = \frac{M(0) + M(0) + M(L \sin 	heta)}{3M} = \frac{L \sin 	heta}{3}$Squaring and adding the equations for $X$ and $Y$:$X^2 + Y^2 = (\frac{L \cos 	heta}{3})^2 + (\frac{L \sin 	heta}{3})^2 = \frac{L^2}{9} (\cos^2 	heta + \sin^2 	heta) = \frac{L^2}{9}$Thus, the path traced by the $COM$ is $x^2 + y^2 = L^2/9$.

Two point masses $M$ each are placed at $(L, 0)$ and $(-L, 0)$. $A$ third point mass $M$ is uniformly rotating on the circle $x^2 + y^2 = L^2$. The equation of the path traced by the $COM$ of the $3$ point masses is:

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