Question

Four particles of masses $1\text{g},2\text{g},3\text{g}$ and $4\text{g}$ are kept at points $(0,0),(0,3\text{cm}),(3\text{cm},3\text{cm}),(3\text{cm},0\text{cm})$ respectively. Find the moment of inertia of the system of particles (in ${\text{g-cm}}^2$) about $x$ axis.

• A. $45{\text{g-cm}}^2$
• B. $4.5{\text{g-cm}}^2$
• C. $81{\text{g-cm}}^2$
• D. $8.1{\text{g-cm}}^2$

Answer 1

The correct option is A $45{\text{g-cm}}^2$

Given:
$m_1=1\text{g},m_2=2\text{g},m_3=3\text{g},m_4=4\text{g}$
Co-ordinates of mass $m_1$$(x_1,y_1)=(0,0)$
Co-ordinates of mass $m_2$$(x_2,y_2)=(0,3\text{cm})$
Co-ordinates of mass $m_3$$(x_3,y_3)=(3\text{cm},3\text{cm})$
Co-ordinates of mass $m_4$$(x_4,y_4)=(3\text{cm},0)$

Therefore, moment of inertia of the system about $x-$ axis
$I=m_1{r}_1^2+m_2{r}_2^2+m_3{r}_3^2+m_4{r}_4^2$
(where $r_1,r_2,r_3$ and $r_4$ are the perpendicular distances from $x$-axis)

$\Rightarrow I=m_1{y}_1^2+m_2{y}_2^2+m_3{y}_3^2+m_4{y}_4^2$
$I=1(0{)}^2+2(3{)}^2+3(3{)}^2+4(0{)}^2$
$\therefore I=45{\text{g-cm}}^2$