Find the centre 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 centre 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.
With the $x$ -and $y$ -axes chosen as shown in Flg.
$7.9$, the coordinates of potnts $O. A $ and $B$ forming the equilateral triangle are respectively $(0,0)$ $(0.5,0),(0.25,0.25 \sqrt{3}) .$ Let the masses $100 g$
$150 g$ and $200 g$ be located at $O , A$ and $B$ be respectively. Then,
$X=\frac{m_{1} x_{1}+m_{2} x_{2}+m_{3} x_{3}}{m_{1}+m_{2}+m_{3}}$
$=\frac{100(0)+150(0.5)+200(0.25)}{(100+150+200) g}$
$=\frac{75+50}{450} m =\frac{125}{450} m =\frac{5}{18} m$
$Y=\frac{100(0)+150(0)+200(0.25 \sqrt{3})}{450 g}$
$=\frac{50 \sqrt{3}}{450} m =\frac{\sqrt{3}}{9} m =\frac{1}{3 \sqrt{3}} m$
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