Question

Three particles of masses $300g,800g$ and $700g$ are placed at positions $A(2,4),B(3,5)$ and $C(-3,-1)$ respectively. What is the moment of inertia of the system about the origin?

• A. $40.2{\text{kg-m}}^2$
• B. $42{\text{kg-m}}^2$
• C. $34{\text{kg-m}}^2$
• D. $30{\text{kg-m}}^2$

Answer 1

The correct option is A $40.2{\text{kg-m}}^2$

Let the particles be $A,B$ & $C$
Mass of $A=300g=0.3kg$
Mass of $B=800g=0.8kg$
Mass of $C=700g=0.7kg$

Position of $A=(2,4)$
Distance of $A$ from origin $r_A=\sqrt{(2-0{)}^2+(4-0{)}^2}=\sqrt{20}\text{m}$

Position of $B=(3,5)$
Distance of $B$ from origin $r_B=\sqrt{(3-0{)}^2+(5-0{)}^2}$
$=\sqrt{9+25}=\sqrt{34}\text{m}$

Position of $C=(-3,-1)$
Distance of $C$ from origin $r_c=\sqrt{(-3-0{)}^2+(-1-0{)}^2}=\sqrt{10}\text{m}$

Moment of inertia about the system
$I=I_A+I_B+I_c$

$=0.3\times (\sqrt{20}{)}^2+0.8\times (\sqrt{34}{)}^2+0.7\times (\sqrt{10}{)}^2$
$=40.2$
$\therefore$ Moment of inertia of the system about the origin is $40.2{\text{kg-m}}^2$
Hence the correct option is (a).