Q. Masses $8,\ 2,\ 4,\ 2\ kg$ are placed at the corners $A,\ B,\ C,\ D$ respectively of a square $ABCD$ of diagonal $80\ cm$. The distance of centre of mass from $A$ will be
A) $20\ cm$
B) $30\ cm$
C) $40\ cm$
D) $60\ cm$
Explanation
(b) According to figure let $A$ is the origin and co-ordinates of centre of mass be $(x,y)$ then,
$x = \frac{m_{1}x_{1} + m_{2}x_{2} + m_{3}x_{3} + m_{4}x_{4}}{m_{1} + m_{2} + m_{3} + m_{4}}$
$= \frac{0 + 2 \times \frac{80}{\sqrt{2}} + 4 \times \frac{80}{\sqrt{2}} + 0}{16} = \frac{30}{\sqrt{2}}$
Similarly $y = \frac{30}{\sqrt{2}}$ so, $r = \sqrt{x^{2} + y^{2}} = 30cm$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.