Question

Find out the centre of mass of a system of three particles of $1kg,2kg,3kg$ respectively kept at the vertices of an equilateral triangle of side $1m$ as shown.

• A. $\frac{2}{3}m$,$\frac{\sqrt{3}}{6}m$
• B. $\frac{4}{3}m$,$\frac{\sqrt{4}}{6}m$
• C. $\frac{5}{3}m$,$\frac{\sqrt{5}}{6}m$
• D. $\frac{2}{3}m$,$\frac{\sqrt{8}}{6}m$

Answer 1

The correct option is A $\frac{2}{3}m$,$\frac{\sqrt{3}}{6}m$

R.E.F image

$X_{com}=\frac{0\times 1+\frac{1}{2}\times 2+3\times 1}{3+2+1}=\frac{4}{6}=\frac{2}{3}m$

$Y_{com}=\frac{0\times 1+\frac{\sqrt{3}}{2}\times 2+0\times 3}{3+2+1}=\frac{\sqrt{3}}{6}m$

So, center of mass of the system is at $(\frac{2}{3}m,\frac{\sqrt{3}}{6}m)$

Ans is (A)