Question

The position of the centre of mass of a uniform bar made up of three rods each having length $2\text{m}$ and negligible area of cross-section as shown in figure is
[Assume masses of the rods are equal]

• A. $(2\text{m},2\text{m})$
• B. $(\frac{2}{3}\text{m},2\text{m})$
• C. $(1\text{m},1\text{m})$
• D. $(\frac{2}{3}\text{m},1\text{m})$

Answer 1

The correct option is D $(\frac{2}{3}\text{m},1\text{m})$

Let the mass of each rod be $m\text{kg}$


Taking the origin at the position shown in figure,
$(x_1,y_1)=$ co-ordinate of $COM$ of rod $I$ = $(1,0)$
$(x_2,y_2)=$ co-ordinate of $COM$ of rod $II$ = $(0,1)$
$(x_3,y_3)=$ co-ordinate of $COM$ of rod $III$ = $(1,2)$
$x$ - co-ordinate of COM of system $=\frac{m_1{x}_1+m_2{x}_2+m_3{x}_3}{m_1+m_2+m_3}$
$=\frac{m\left(1\right)+m\left(0\right)+m\left(1\right)}{m+m+m}=\frac{2}{3}\text{m}$
$y$ - co ordinate of COM of system $=\frac{m_1{y}_1+m_2{y}_2+m_3{y}_3}{m_1+m_2+m_3}$
$=\frac{m\left(0\right)+m\left(1\right)+m\left(2\right)}{m+m+m}$
$=\frac{3m}{3m}=1\text{m}$
$\therefore (x_{com},y_{com})=(\frac{2}{3}\text{m},1\text{m})$