Question

Three particles of masses 1 kg, 2 kg and 3 kg are situated at the corners of an equilateral triangle of side $b$. The coordinates of the centre of mass are

• A. $\left[0,\frac{7b}{12},\frac{3\sqrt{3b}}{12}\right]$
• B. $\left[\frac{3\sqrt{3b}}{12},\frac{7b}{12},0\right]$
• C. $\left[\frac{7b}{12},\frac{3\sqrt{3b}}{12},0\right]$
• D. $\left[\frac{7b}{12},0,\frac{3\sqrt{3b}}{12}\right]$

Answer 1

Answer: (C)

The coordinates of points A, B and C are (0, 0, 0), (b, 0, 0) and $\left(\frac{b}{2},\frac{b\sqrt{3}}{2},0\right)$ respectively.
Now as the triangle in XY plane, i.e., Z coordinate of all the masses is zero,
so $Z_{CM}=0.$
Now, $X_{CM}=\frac{1\times 0+2\times b+3\left(b/2\right)}{1+2+3}=\frac{7b}{12}$
$Y_{CM}=\frac{1\times 0+3\sqrt{3}\left(b/2\right)}{1+2+3}=\frac{3\sqrt{3}b}{12}$
So, the coordinates of centre of mass are
$\left[\frac{7b}{12},\frac{3\sqrt{3b}}{12},0\right]$