Question

An equilateral triangular wire frame is made up of three thin uniform rods of mass per unit length $\lambda ,2\lambda$ and $3\lambda$ as shown in figure. Assuming the origin to be at point $\text{O}$, which of the following options is(are) correct regarding the position of $COM$ of system?

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

Answer 1

Answer: (A), (D)

$y_{cm}>0$

For equilateral triangular wire frame shown in the given figure.


Let the length of each side be $l$.
$\therefore$ Mass of rod with density $\lambda$
$=m_1=\lambda l$
Mass of rod with density $2\lambda$
$=m_2=2\lambda l$
Mass of rod with density $3\lambda$
$=m_3=3\lambda l$

Assuming origin to be at point $\text{O}$,
$COM$ of rod with density $\lambda$
$=(x_1,y_1)=(0,0)$
COM of rod with density $2\lambda$
$=(x_2,y_2)=(\frac{-l}{4},\frac{\sqrt{3}l}{4})$
COM of rod with density $3\lambda$
$=(x_3,y_3)=(\frac{l}{4},\frac{\sqrt{3}l}{4})$
Therefore, for the equilateral triangular wire frame.
$x_{cm}=\frac{m_1{x}_1+m_2{x}_2+m_3{x}_3}{m_1+m_2+m_3}$
$=\frac{\lambda l\left(0\right)+2\lambda l\left(\frac{-l}{4}\right)+3\lambda l\left(\frac{l}{4}\right)}{6\lambda l}=\frac{l}{24}>0$
and
$y_{cm}=\frac{m_1{y}_1+m_2{y}_2+m_3{y}_3}{m_1+m_2+m_3}$
$=\frac{\lambda l\left(0\right)+2\lambda l\left(\frac{\sqrt{3}l}{4}\right)+3\lambda l\left(\frac{\sqrt{3}l}{4}\right)}{6\lambda l}=\frac{5\sqrt{3}l}{24}>0$​​​​