Question

Three identical spheres of mass $M$ each are placed at the corners of an equilateral triangle of side $2 \, m$ . Taking one of the corners as the origin as shown in the figure, the position vector of the centre of mass is

• A. $\sqrt{3}$ ( $\hat{\text{i}}$ - $\hat{\text{j}}$ )
• B. $\frac{\hat{i}}{\sqrt{3}} + \hat{j}$
• C. $\frac{\hat{\text{i}} \text{+} \hat{\text{j}}}{3}$
• D. $\hat{i} + \frac{\hat{j}}{\sqrt{3}}$

Answer 1

Answer: (D)

The $x$ coordinate of centre of mass is


$x_{c o m} = \frac{\displaystyle \sum M_{i} x_{i}}{\displaystyle \sum M_{i}}$
$=$ $\frac{M \times 0 + M \times 1 + M \times 2}{M + M + M}$ $= 1$
$y_{c o m} = \frac{\displaystyle \sum m_{i} y_{i}}{\displaystyle \sum m_{i}}$
$=$ $\frac{M \times 0 + M \times \left(\right. 2 \, sin \, 60 \left.\right) + M \times 0}{M + M + M}$
$y_{c o m} = \frac{\sqrt{3} M}{3 M} = \frac{1}{\sqrt{3}}$
Position vector of centre of mass is $\left(\hat{i} + \frac{\hat{j}}{\sqrt{3}}\right)$ .