Three particles of masses 50, g,100, g,and 15 | vedclass

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(D) The coordinates of the three particles are:$m_1 = 50\, g$ at $(0, 0)$$m_2 = 100\, g$ at $(1, 0)$$m_3 = 150\, g$ at $(0.5, \frac{\sqrt{3}}{2})$The

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$\left( \frac{7}{12}\,m, \frac{\sqrt{3}}{4}\,m \right)$

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(D) The coordinates of the three particles are:$m_1 = 50\, g$ at $(0, 0)$$m_2 = 100\, g$ at $(1, 0)$$m_3 = 150\, g$ at $(0.5, \frac{\sqrt{3}}{2})$The $x$-coordinate of the center of mass is:$X_{cm} = \frac{m_1x_1 + m_2x_2 + m_3x_3}{m_1 + m_2 + m_3} = \frac{50(0) + 100(1) + 150(0.5)}{50 + 100 + 150} = \frac{100 + 75}{300} = \frac{175}{300} = \frac{7}{12}\, m$The $y$-coordinate of the center of mass is:$Y_{cm} = \frac{m_1y_1 + m_2y_2 + m_3y_3}{m_1 + m_2 + m_3} = \frac{50(0) + 100(0) + 150(\frac{\sqrt{3}}{2})}{300} = \frac{75\sqrt{3}}{300} = \frac{\sqrt{3}}{4}\, m$Thus,the coordinates of the center of mass are $\left( \frac{7}{12}\,m, \frac{\sqrt{3}}{4}\,m \right)$.

Three particles of masses $50\, g$,$100\, g$,and $150\, g$ are placed at the vertices of an equilateral triangle of side $1\, m$ (as shown in the figure). The $(x, y)$ coordinates of the centre of mass will be

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