Find the center of mass of three particles at the vertices of an equilateral triangle. The masses of the particles are $100 g$ , $150 g$ and $200 g$ respectively. Each side of the equilateral triangle is $0.5 m$ long.

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Solution

Let
$\begin{matrix} m_{1} = 100 g, m_{2} = 150 g, m_{3} = 200 g a = 0.5 m = 50 c m \end{matrix}$
Center of mass coordinates:-
$\begin{matrix} x = \frac{[\frac{- a}{2} \times m_{1} + 0 \times m_{2} + \frac{a}{2} \times m_{3}]}{m_{1} + m_{2} + m_{3}} = \frac{(m_{3} - m_{1}) a}{2 (m_{1} + m_{2} + m_{3})} = \frac{(200 - 100) 50}{2 \times 450} = \frac{50}{9} c m y = \frac{(0 \times m_{1} + \frac{\sqrt{3}}{2} \times a \times m_{2} + 0 \times m_{3})}{m_{1} + m_{2} + m_{3}} = \frac{\frac{\sqrt{3}}{2} \times a \times m_{2}}{m_{1} + m_{2} + m_{3}} = \frac{\frac{\sqrt{3}}{2} \times 50 \times 150}{450} = \frac{25}{\sqrt{3}} c m \end{matrix}$

$y$ coordinates of the centroid of the triangle
$= \frac{1}{\sqrt{3}} \times \frac{\sqrt{3}}{2} \times a = \frac{25}{\sqrt{3}} c m$

$∴$ the center mass is $\frac{50}{9} c m$ .

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Find the center of mass of three particles at the vertices of an equilateral triangle. The masses of the particles are 100 g, 150 g and 200 g respectively. Each side of the equilateral triangle is 0.5 m long.

Find the center of mass of three particles at the vertices of an equilateral triangle. The masses of the particles are $100 g$ , $150 g$ and $200 g$ respectively. Each side of the equilateral triangle is $0.5 m$ long.