Question

Four point masses, each of value $m$ are placed at the corners of square $ABCD$ of side $l$. The moment of inertia of this system about an axis passing through $A$ and parallel to $BD$ is

• A. $4m{l}^2$
• B. $2m{l}^2$
• C. $3m{l}^2$
• D. $m{l}^2$

Answer 1

The correct option is C $3m{l}^2$


From diagram,
Distance of mass at $A$ from axis
$r_1=0\text{m}$
$r_2=l\sin {45}^{\circ }=\frac{l}{\sqrt{2}}$
simillarly,
$r_3=l\sin {45}^{\circ }=\frac{l}{\sqrt{2}}$

$r_4=\sqrt{(AB{)}^2+(BC{)}^2}=\sqrt{(l{)}^2+(l{)}^2}=l\sqrt{2}$

Total moment of inertia of system about given axis of rotation = Sum of moment of inertia of individual particle about given axis of rotation.

$\left(I\right)=I_1+I_2+I_3+I_4$

$=m{r}_1^2+m{r}_2^2+m{r}_3^2+m{r}_4^2$

$=m(0{)}^2+m{\left(\frac{l}{\sqrt{2}}\right)}^2+m{\left(\frac{l}{\sqrt{2}}\right)}^2+m(\sqrt{2}l{)}^2$

$=\frac{m{l}^2}{2}+\frac{m{l}^2}{2}+2m{l}^2$

$=3m{l}^2$