Question

Four particles each of mass $500\text{g}$ are placed at the four corners of a rectangle $ABCD$ with sides of length $5\text{m}$ and $10\text{m}$. What is the ratio of moment of inertia of the system when $AB$ is the axis to when $BC$ is the axis?

• A. $1:4$
• B. $2:1$
• C. $1:2$
• D. $4:1$

Answer 1

The correct option is D $4:1$


When $AB$ is the axis of rotation,
$r_D=AD=10\text{m},r_C=BC=10\text{m}$
$m_A=m_B=m_C=m_D=500\text{g}=0.5\text{kg}$

Moment of inertia of system of particles about AB,
$\Rightarrow I_{AB}=I_A+I_B+I_C+I_D$

$I_{AB}=m_A{r}_A^2+m_B{r}_B^2+m_C{r}_C^2+m_D{r}_D^2$
$I_{AB}=0.5\times 0+0.5\times 0+0.5\times {10}^2+0.5\times {10}^2$
$\therefore I_{AB}=100{\text{kg- m}}^2$


When $BC$ is the axis of rotation,
$r_A^{\prime }=AB=5\text{m},r_D^{\prime }=DC=5\text{m}$
Moment of inertia of system of particles about BC,
$\Rightarrow I_{BC}=I_A+I_B+I_C+I_D$

$\Rightarrow I_{BC}=m_A{r^{\prime }}_A^2+m_B{r^{\prime }}_B^2+m_C{r^{\prime }}_C^2+m_D{r^{\prime }}_D^2$

$I_{BC}=0.5\times 5^2+0.5\times 0+0.5\times 0+0.5\times 5^2$
$\therefore I_{BC}=12.5+12.5=25{\text{kg -m}}^2$

$\Rightarrow \frac{I_{AB}}{I_{BC}}=\frac{100}{25}=4:1$
Hence the correct option is (d).