Question

A rigid body consists of a $3\text{kg}$ mass connected to a $4\text{kg}$ mass by a massless rod. The $3\text{kg}$ mass is located at $\overrightarrow{r_1}=(3\hat{i}+2\hat{j})\text{m}$ and the $4\text{kg}$ mass at $\overrightarrow{r_2}=(8\hat{i}+7\hat{j})\text{m}$. Find the length of the rod and the co-ordinates of the centre of mass

• A. $5\text{m},(\frac{41}{7},\frac{34}{7})\text{m}$
• B. $5\sqrt{2}\text{m},(\frac{41}{7},\frac{34}{7})\text{m}$
• C. $5\text{m}$, $(6.5,0)\text{m}$
• D. $5\sqrt{2}\text{m},(5,0)\text{m}$

Answer 1

The correct option is B $5\sqrt{2}\text{m},(\frac{41}{7},\frac{34}{7})\text{m}$


Let $COM$ of system = $(x,y)$
Length of rod $=\sqrt{(8-3{)}^2+(7-2{)}^2}$
$=\sqrt{5^2+5^2}=5\sqrt{2}\text{m}$

$x$-cordinate of centre of mass.
$x=\frac{m_1{x}_1+m_2{x}_2}{m_1+m_2}=\frac{3\left(3\right)+4\left(8\right)}{3+4}$
$=\frac{9+32}{7}=\frac{41}{7}\text{m}$
$y$ - coordinate of centre of mass.
$y=\frac{m_1{y}_1+m_2{y}_2}{m_1+m_2}=\frac{3\times 2+4\times 7}{3+4}$
$=\frac{6+28}{7}=\frac{34}{7}\text{m}$

So, $COM\Rightarrow (\frac{41}{7},\frac{34}{7})\text{m}$