Question

Five homogeneous bricks, each of length $L=9\text{m}$ are arranged as shown in the figure. Each of the bricks is displaced with respect to the next one in contact by $\frac{L}{6}$ as shown in the figure below. Find the $x-$ coordinate of the centre of mass relative to origin $\text{O}$.

• A. $2.1\text{m}$
• B. $5.4\text{m}$
• C. $4.5\text{m}$
• D. $6.3\text{m}$

Answer 1

The correct option is D $6.3\text{m}$

Given length of each brick is $L$
Let the mass of each brick $=m\text{kg}$


Initially all five bricks have centre of mass at horizontal distance of $\frac{L}{2}$ relative to origin $\text{O}$.


After displacement, the centre of mass of the bricks are given by:
$x-$ coordinate of centre of mass of brick $1,x_1=\frac{L}{6}+\frac{L}{6}+\frac{L}{2}=\frac{5L}{6}$
For brick $2,x_2=\frac{L}{6}+\frac{L}{2}=\frac{4L}{6}$
For brick $3,x_3=\frac{L}{2}$
For brick $4,x_4=\frac{L}{6}+\frac{L}{2}=\frac{4L}{6}$
For brick $5,x_5=\frac{L}{6}+\frac{L}{6}+\frac{L}{2}=\frac{5L}{6}$

$\therefore x-$ coordinate of COM of combined system
$x_{COM}=\frac{m{x}_1+m{x}_2+m{x}_3+m{x}_4+m{x}_5}{m+m+m+m+m}$
$=\frac{m\left(\frac{5L}{6}\right)+m\left(\frac{4L}{6}\right)+m\left(\frac{L}{2}\right)+m\left(\frac{4L}{6}\right)+m\left(\frac{5L}{6}\right)}{m+m+m+m+m}$
$=\frac{21mL}{6\times 5m}=\frac{21L}{30}$
But $L=9\text{m}$
$\therefore x_{COM}=\frac{21\times 9}{30}=6.3\text{m}$