Question

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

Answer 1

Given that,

$m_1=100g$

$m_2=150g$

$m_3=200g$

a = 0.5 m = 50 cm

Now, according to diagram

Center of mass coordinates

$X_{cm}=\frac{0\times 100+150\times 50+200\times 25}{450}$

$X_{cm}=\frac{7500+5000}{450}$

$X_{cm}=27.8cm$

$X_{cm}=0.28m$

Now,

$Y_{cm}=\frac{0\times 100+0\times 150+\sqrt{3}\times 25\times 200}{450}$

$Y_{cm}=\frac{8660.3}{450}$

$Y_{cm}=19.2cm$

$Y_{cm}=0.2m$

Hence, the center of mass of three particles at the vertices of an equilateral triangle is $X_{cm}=0.28cm$ and $Y_{cm}=0.2cm$ respectively