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)

$\left[\frac{7b}{12}, \frac{3\sqrt{3b}}{12}, 0\right]$

Answer 2

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, $X_{CM}=\frac{1\times0+2\times b+3\left(b / 2\right)}{1+2+3}=\frac{7b}{12}$ $Y_{CM}=\frac{1\times0+3\sqrt{3}\left(b / 2\right)}{1+2+3}=\frac{3\sqrt{3}b}{12}$ So, the coordinates of centre of mass are $\left[\frac{7b}{12}, \frac{3\sqrt{3b}}{12}, 0\right]$