Question

Four particles of masses $1kg$, $2kg,$ $3kg$ and $4kg$ are placed at the four vertices $A$, $B$, $C$ and $D$, respectively of a square of side $1m$. Find the position of centre of mass of the particles.

Answer 1

Assuming $D$ as the origin. $DC$ as x-axis and $DA$ as y-axis we have
$m_1=1kg,(x_1,y_1)=(0,1m)$
$m_2=2kg,(x_2,y_2)=(1m,1m)$
$m_3=3kg,(x_3,y_3)=(1m,0)$
$m_4=4kg,(x_4,y_4)=(0,0)$
Coordination of their $CM$ are calculated as follows.
$x_{CM}=\frac{m_1{x}_1+m_2{x}_2+m_3{x}_3+m_4{x}_4}{m_1+m_2+m_3+m_4}$
$=\frac{\left(1\right)\left(0\right)+2\left(1\right)+3\left(1\right)+4\left(0\right)}{1+2+3+4}$
$\frac{5}{10}=\frac{1}{2}m=0.5m$
Similarly, $y_{CM}=\frac{m_1{y}_1+m_2{y}_2+m_3{y}_3+m_4{y}_4}{m_1+m_2+m_3+m_4}$
$=\frac{\left(1\right)\left(1\right)+2\left(1\right)+3\left(0\right)+4\left(0\right)}{1+2+3+4}$
$=\frac{3}{10}m=0.3m$
$\therefore (x_{CM},yCM)=(0.5m,0.3m)$