Q. Three masses of $2\ kg,\ 4\ kg$ and $4\ kg$ are placed at the three points $(1,\ 0,\ 0),\ (1,\ 1,\ 0)$ and $(0,\ 1,\ 0)$ respectively. The position vector of its centre of mass is
A) $\frac{3}{4}\widehat{i} + \frac{4}{5}\widehat{j}$
B) $(3\widehat{i} + \widehat{j})$
C) $\frac{2}{5}\widehat{i} + \frac{4}{5}\widehat{j}$
D) $\frac{1}{5}\widehat{i} + \frac{4}{5}\widehat{j}$
Explanation
(a) For centre of mass,
$x_{cm} = \frac{2 \times 1 + 4 \times 1 + 4 \times 0}{2 + 4 + 4} = \frac{6}{10} = \frac{3}{5}$
$y_{cm} = \frac{2 \times 0 + 4 \times 1 + 4 \times 1}{2 + 4 + 4} = \frac{8}{10} = \frac{4}{5}$
$\therefore$ Coordinate for $cm = \left( \frac{3}{5}\widehat{i},\frac{4}{5}\widehat{j} \right)$
Where $\widehat{i}$ and $\widehat{j}$ are unit vector along $x$ and $y$ axis
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.