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 figure). The (x, y) coordinate of the centre of mass will be :

Engineering

Physics

Center of Mass

Question

$(\frac{7}{12} m, \frac{\sqrt{3}}{8} m)$

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

$(\frac{\sqrt{3}}{8} m, \frac{7}{12} m)$

$(\frac{\sqrt{3}}{4} m, \frac{5}{12} m)$

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m₁ = 50 g m₂ = 100 g m₃ = 150 g

x₁ = 0 x₂ = 1 m x₃ = 0.5 m

y₁ = 0 y₂ = 0 $y_{3} = \frac{\sqrt{3}}{2}$

$x_{COM} = \frac{m_{1} x_{1} + m_{2} x_{2} + m_{3} x_{3}}{m_{1} + m_{2} + m_{3}} = \frac{0 + (100) 10 + (150) (0 .5)}{300}$

$x_{COM} = \frac{100 + 75}{300} = \frac{175}{300} = \frac{7}{12} m$

$y_{COM} = \frac{m_{1} y_{1} + m_{2} y_{2} + m_{3} y_{3}}{m_{1} + m_{2} + m_{3}} = \frac{0 + (100) (0) + (150) \frac{\sqrt{3}}{2}}{300}$

$y_{COM} = \frac{75 \sqrt{3}}{300} = \frac{3 \sqrt{3}}{12} = \frac{\sqrt{3}}{4} m$

$(x_{COM}, y_{COM}) = (\frac{7}{12}, \frac{\sqrt{3}}{4})$