Question

There are four rods $AB,BC,CD$ & $DA$ of same length $L$ , but different linear mass densities $\rho ,2\rho ,3\rho$ and $4\rho$ respectively. These are joined to form a square frame as shown in figure. Find the co-ordinates of the centre of mass of the structure.

• A. $\frac{3L}{5},\frac{3L}{5}$
• B. $\frac{2L}{5},\frac{2L}{5}$
• C. $\frac{L}{2},\frac{L}{2}$
• D. $\frac{2L}{5},\frac{3L}{5}$

Answer 1

The correct option is D $\frac{2L}{5},\frac{3L}{5}$

As we know,
Linear mass density $=\frac{\text{Mass}}{\text{Length}}$
Hence, $m_1=m_{AB}=\rho L,m_3=m_{CD}=\left(3\rho \right)L$
$m_2=m_{BC}=\left(2\rho \right)L,m_4=m_{DA}=\left(4\rho \right)L$

Assume, $m_1=m_{AB}=\rho L=m$;
So, $m_1=m,m_2=2m,m_3=3m,m_4=4m$
Now,
$x_{com}=\frac{m_1{x}_1+m_2{x}_2+m_3{x}_3+m_4{x}_4}{m_1+m_2+m_3+m_4}$
$x_{com}=\frac{m\left(\frac{L}{2}\right)+2m\left(L\right)+3m\left(\frac{L}{2}\right)+4m\left(0\right)}{m+2m+3m+4m}$
$\Rightarrow x_{com}=\frac{2L}{5}$

$y_{com}=\frac{m_1{y}_1+m_2{y}_2+m_3{y}_3+m_4{y}_4}{m_1+m_2+m_3+m_4}$
$\Rightarrow y_{com}=\frac{m\left(0\right)+2m\left(\frac{L}{2}\right)+3m\left(L\right)+4m\left(\frac{L}{2}\right)}{m+2m+3m+4m}$
$\Rightarrow y_{com}=\frac{3L}{5}$
Hence co-ordinates of COM will be $(\frac{2L}{5},\frac{3L}{5})$