Question

Masses $8\text{kg}$, $2\text{kg}$ , $4\text{kg}$ & $2\text{kg}$ are placed at the corners $\text{A, B, C, D}$ respectively of a square $\text{ABCD}$ of diagonal $80\text{cm}$ (as shown in the figure below). The distance of the centre of mass from $A$ will be

• A. $20\text{cm}$
• B. $30\text{cm}$
• C. $40\text{cm}$
• D. $60\text{cm}$

Answer 1

The correct option is B $30\text{cm}$


Assuming length of side of square $=a\text{cm}$
$\sqrt{2}a=80\text{cm}$ (given)
$\Rightarrow$$a=\frac{80}{\sqrt{2}}\text{cm}$

From the diagram: $m_1=8\text{kg},x_1=0,y_1=0$
$m_2=2\text{kg},x_2=\frac{80}{\sqrt{2}}$, $y_2=0$
$m_3=4\text{kg},x_3=\frac{80}{\sqrt{2}},y_3=\frac{80}{\sqrt{2}}$
$m_4=2\text{kg},$$x_4=0,y_4=\frac{80}{\sqrt{2}}$

$x_{com}=\frac{m_1{x}_1+m_2{x}_2+m_3{x}_3+m_4{x}_4}{m_1+m_2+m_3+m_4}=\frac{2\times \frac{80}{\sqrt{2}}+4\times \frac{80}{\sqrt{2}}}{8+2+4+2}=\frac{30}{\sqrt{2}}\text{cm}$
Similarly, $y_{com}=\frac{m_1{y}_1+m_2{y}_2+m_3{y}_3+m_4{y}_4}{m_1+m_2+m_3+m_4}=\frac{30}{\sqrt{2}}\text{cm}$

Thus, distance of centre of mass from origin
$r=\sqrt{x^2+y^2}=30\text{cm}$