Question

A wire of mass $m\text{kg}$ with uniform cross section is bent in the shape as shown in figure. If origin is taken at $\text{O}$, find the co-ordinates of the center of mass of the given system (in metres).

• A. $(\frac{15}{14},\frac{12}{14})$
• B. $(\frac{15}{7},\frac{12}{7})$
• C. $(2,2)$
• D. $(\frac{12}{7},\frac{15}{7})$

Answer 1

The correct option is B $(\frac{15}{7},\frac{12}{7})$

Given, mass of wire $=m\text{kg}$
Total length of wire $\left(L\right)=14\text{meters}$
$\therefore$ Mass of each $4\text{m}$ wire $=\frac{m}{14}\times 4=\frac{2m}{7}\text{kg}$
Mass of $2\text{m}$ wire $=\frac{m}{14}\times 2=\frac{m}{7}\text{kg}$


The co-ordinates of $COM$ of the sides are
$C_1=(x_1,y_1)=(0,2)$
$C_2=(x_2,y_2)=(2,0)$
$C_3=(x_3,y_3)=(4,2)$
$C_4=(x_4,y_4)=(3,4)$

To find $COM$ of system:
$x_{com}=\frac{m_1{x}_1+m_2{x}_2+m_3{x}_3+m_4{x}_4}{m_1+m_2+m_3+m_4}$
$=\frac{\frac{2m}{7}\left(0\right)+\frac{2m}{7}\left(2\right)+\frac{2m}{7}\left(4\right)+\frac{m}{7}\left(3\right)}{m}$
$=\frac{15}{7}\text{m}$
and $y_{com}=\frac{m_1{y}_1+m_2{y}_2+m_3{y}_3+m_4{y}_4}{m_1+m_2+m_3+m_4}$
$=\frac{\frac{2m}{7}\left(2\right)+\frac{2m}{7}\left(0\right)+\frac{2m}{7}\left(2\right)+\frac{m}{7}\left(4\right)}{m}$
$=\frac{12}{7}\text{m}$