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

Answer 1

The correct option is D $(\frac{7}{12}m,\frac{\sqrt{3}}{4}m)$

$x_{cm}=\frac{m_1{x}_1+m_2{x}_2+m_3{x}_3}{m_1+m_2+m_3}$
$=\frac{0+100\times 1+150\times 0.5}{50+100+150}=\frac{175}{300}=\frac{7}{12}m$
$y_{cm}=\frac{m_1{y}_1+m_2{y}_2+m_3{y}_3}{m_1+m_2+m_3}$
$=\frac{0+0+150\times 0.5\times \tan {60}^{\circ }}{300}=\frac{75}{300}\times \sqrt{3}=\frac{\sqrt{3}}{4}m$
$CM=(x,y)=(\frac{7}{12},\frac{\sqrt{3}}{4})$