Question

Three particles each of mass $1\text{kg}$ are placed at the vertices of an equilateral triangle $ABC$ of side $2\text{m}$. Find the moment of inertia of system about a line perpendicular to the plane of system and passing through middle point ($D$) of side $BC$ of the triangle.

• A. $5{\text{kg-m}}^2$
• B. $3{\text{kg-m}}^2$
• C. $2.5{\text{kg-m}}^2$
• D. $1.1{\text{kg-m}}^2$

Answer 1

The correct option is A $5{\text{kg-m}}^2$

Given mass of particles, $m=1kg$
Moment of inertia of body of mass $\left(m\right)=m{r}^2$
Where $r$ is the radius of rotation.
$\Delta ABC$ is an equilateral triangle as all sides are equal.
$\therefore$ In $\Delta ABD;\angle ABD={60}^{\circ }$
$\Rightarrow \sin {60}^{\circ }=\frac{x}{2}$
$\Rightarrow x=2\times \frac{\sqrt{3}}{2}=\sqrt{3}\text{m}$


Distance of masses from the axis of rotation is given by
$AD=r_1=\sqrt{3}\text{m}$
$BD=r_2=1\text{m}$
$CD=r_3=1\text{m}$
Total moment of inertia of system ($I$) = $I_1+I_2+I_3$
$I=m{r}_1^2+m{r}_2^2+m{r}_3^2$
$=1\times (\sqrt{3}{)}^2+1(1{)}^2+1(1{)}^2$
$=5{\text{kg-m}}^2$