Question

Three particles of masses $50\, g, 100\, g$ and $150\, g$ are placed at the vertices of an equilateral triangle of side $1\, m$ (as shown in the figure). The $(x, y)$ coordinates of the centre of mass will be :

• A. $\bigg(\frac{7}{12} m , \frac{\sqrt{3}}{8}\bigg)$
• B. $\bigg(\frac{\sqrt{3}}{4} m , \frac{\sqrt{5}}{12}\bigg)$
• C. $\bigg(\frac{7}{12} m , \frac{\sqrt{3}}{4}\bigg)$
• D. $\bigg(\frac{\sqrt{3}}{8} m , \frac{\sqrt{7}}{12}\bigg)$

Answer 1

Answer: (C)

$\bigg(\frac{7}{12} m , \frac{\sqrt{3}}{4}\bigg)$

Answer 2

The co-ordinates of the centre of mass $\vec{r}_{cm} \frac{0 + 150 \times \bigg(\frac{1}{2} \hat{i} + \frac{\sqrt{3}}{2} \hat{j}\bigg) + 100 \times \hat{i}}{300}$ $\vec{r}_{cm} \, = \, \frac{7}{12} \hat{i} + \frac{\sqrt{3}}{4} \hat{j}$ $\therefore \,$Co- ordinate $\bigg(\frac{7}{12} , \frac{\sqrt{3}}{4}\bigg)m$