Question

Four uniform rods of mass $m\text{kg}$ each form a rectangle as shown in figure. The rods have negligible area of cross-section. Find the position of center of mass of the rectangle made up of the four uniform rods.

• A. $(2,2)\text{m}$
• B. $(2,0)\text{m}$
• C. $(2,1)\text{m}$
• D. $(4,2)\text{m}$

Answer 1

The correct option is C $(2,1)\text{m}$


Given, mass of each rod $=m\text{kg}$
Co-ordinate of $COM$ of rod I, $(x_1,y_1)=(2,0)\text{m}$
Co-ordinate of $COM$ of rod II, $(x_2,y_2)=(0,1)\text{m}$
Co-ordinate of $COM$ of rod III, $(x_3,y_3)=(2,2)\text{m}$
Co-ordinate of $COM$ of rod IV, $(x_4,y_4)=(4,1)\text{m}$

Therefore, $x-$ co-ordinate of $COM$ of rectangle
$x_{COM}=\frac{m_1{x}_1+m_2{x}_2+m_3{x}_3+m_4{x}_4}{m_1+m_2+m_3+m_4}$
[$m_1=m_2=m_3=m_4=m$]
$=\frac{m\times 2+m\times 0+m\times 2+m\times 4}{m+m+m+m}$
$=\frac{8m}{4m}=2\text{m}$
and $y-$ co-ordinate of $COM$ of rectangle
$y_{COM}=\frac{m_1{y}_1+m_2{y}_2+m_3{y}_3+m_4{y}_4}{m_1+m_2+m_3+m_4}$
$=\frac{m\times 0+m\times 1+m\times 2+m\times 1}{m+m+m+m}$
$=\frac{4m}{4m}=1\text{m}$

The position of center of mass of the rectangle is $(x_{COM},y_{COM})=(2,1)\text{m}$