Question

Three points masses, each of mass $m$ are placed at the corners of an equilateral triangle of side $l$, Then the moment of inertia of this system about an axis along one of the side of the triangle is $x$ times $m{l}^2$. The value of $x$ is

Answer 1

Moment of inertia of each of the point masses $\left(m\right)$ at B and C
about the side $BC=m(0{)}^2=0.$
Moment of inertia of point mass m at A about the side $BC=m(AD{)}^2$.
Now $AD=lsin{60}^{\circ }=l\frac{\sqrt{3}}{2}$(height of equilateral triangle)
$\therefore$ Moment of inertia of the system about side $BC$
$=0+0+m{\left(\frac{l\sqrt{3}}{2}\right)}^2=\frac{3m{l}^2}{4}$