Question

A square frame is made using four thin uniform rods of length $l$ each having linear mass densities $\lambda ,2\lambda ,3\lambda$ and $4\lambda$ as shown in the figure. Which of the following is correct?

• A. $x_{cm}=0,y_{cm}=0$
• B. $x_{cm}>0,y_{cm}>0$
• C. $x_{cm}<0,y_{cm}<0$
• D. $x_{cm}=0,y_{cm}<0$

Answer 1

The correct option is D $x_{cm}=0,y_{cm}<0$

Mass of rods will be $m_1=\lambda l$, $m_2=2\lambda l$, $m_3=3\lambda l$, $m_4=4\lambda l$ whose centre of mass are shown in the figure given below :-


Position of x coordinate of centre of mass :
$x_{cm}=\frac{m_1{x}_1+m_2{x}_2+m_3{x}_3+m_4{x}_4}{m_1+m_2+m_3+m_4}=\frac{\lambda l\left(\frac{l}{2\sqrt{2}}\right)+2\lambda l(-\frac{l}{2\sqrt{2}})+3\lambda l(-\frac{l}{2\sqrt{2}})+4\lambda l\left(\frac{l}{2\sqrt{2}}\right)}{\lambda l+2\lambda l+3\lambda l+4\lambda l}$
$=\frac{\frac{\lambda }{2\sqrt{2}}(l-2l-3l+4l)}{10\lambda }=0y_{cm}=\frac{m_1{y}_1+m_2{y}_2+m_3{y}_3+m_4{y}_4}{m_1+m_2+m_3+m_4}=\frac{\lambda l\left(\frac{l}{2\sqrt{2}}\right)+2\lambda l\left(\frac{l}{2\sqrt{2}}\right)+3\lambda l\left(\frac{-l}{2\sqrt{2}}\right)+4\lambda l\left(\frac{-l}{2\sqrt{2}}\right)}{10\lambda l}$
$=\frac{\frac{\lambda }{2\sqrt{2}}(l+2l-3l-4l)}{10\lambda }=\frac{-l}{5\sqrt{2}}$
So, $x_{cm}=0$ and $y_{cm}<0$