Question

A system consists of particles of mass $4\text{kg}$ placed at $(4,0),6\text{kg}$ placed at $(0,-4)$ and $3\text{kg}$ placed at $(-3,4)$. How far from the origin must a particle of mass $10\text{kg}$ be placed so that the moment of inertia of the system about the origin becomes $300{\text{kgm}}^2$?

• A. $25\text{m}$
• B. $2.5\text{m}$
• C. $10\text{m}$
• D. $1\text{m}$

Answer 1

The correct option is B $2.5\text{m}$

Let the particles be $A,B,C$ and $D$
Mass of $A=m_A=4\text{kg}$
Position of $A=(4,0)$
Distance of $A$ from origin $=r_A=\sqrt{(4-0{)}^2+(0-0{)}^2}$
$\Rightarrow r_A=4\text{m}$
Mass of $B=m_B=6\text{kg}$
Position of $B=(0,-4)$
Distance of $B$ from origin $=r_B=\sqrt{(0-0{)}^2+(-4-0{)}^2}$
$\Rightarrow r_B=4\text{m}$
Mass of $C=3\text{kg}$
Position of $C=(-3,4)$
Distance of $C$ from origin $=\sqrt{(-3-0{)}^2+(4-0{)}^2}$
$\Rightarrow r_C=\sqrt{9+16}=\sqrt{25}=5\text{m}$
Mass of $D=10\text{kg}$
Moment of inertia of the system,
$I=m_A{r}_A^2+m_B{r}_B^2+m_C{r}_C^2+m_D{r}_D^2$
$\Rightarrow 300=4\times 4^2+6\times 4^2+3\times 5^2+10\times r_D^2$
$\Rightarrow 300=64+96+75+10r_D^2$
$\Rightarrow 10r_D^2=65\Rightarrow r_D^2=6.5$
$\Rightarrow r_D=\sqrt{6.5}\text{m}=2.5\text{m}$