Question

Four uniform rods of different densities are connected to each other to form a square as shown in figure.


Linear mass density of $\text{A, B, C}$ and $\text{D}$ are given as $\lambda ,2\lambda ,2\lambda$ and $5\lambda$. Find out the distance of $COM$ of the square from the rod $\text{A}$ if length of each rod is $l$.

• A. $l$
• B. $\frac{3}{4}l$
• C. $0.55l$
• D. $0.45l$

Answer 1

The correct option is C $0.55l$

Let us consider this square on $X-Y$ plane as shown.


Consider $\text{O}$ to be the origin.
The distance from rod $\text{A}$ will be the $Y$ coordinate of $COM$ and position of $COM$ of rod $\text{A, B, C}$ and $\text{D}$ are respectively $(\frac{l}{2},0),(l,\frac{l}{2}),(\frac{l}{2},l)$ and $(0,\frac{l}{2})$.
Mass of the rods are $\lambda l,2l\lambda ,2\lambda l$ and $5\lambda l$ respectively.

Therefore, $Y-$ co-ordinate of $COM$ of system is
$Y_{COM}=\frac{y_1{m}_1+y_2{m}_2+y_3{m}_3+y_4{m}_4}{m_1+m_2+m_3+m_4}$
$Y_{COM}=\frac{0\times \lambda l+\frac{l}{2}.2\lambda l+l\times 2\lambda l+\frac{l}{2}\times 5\lambda l}{\lambda l+2\lambda l+2\lambda l+5\lambda l}$
$=\frac{\lambda l^2+2\lambda l^2+2.5\lambda l^2}{10\lambda l}$
$=\frac{5.5\lambda l^2}{10\lambda l}$
$=0.55l$
So, distance from rod $A$ will be $0.55l$.