Question

Three point masses, each of mass $m$ are placed at the corners of an equilateral triangle of side $l$. The moment of inertia of this system about an axis along one of the sides of the triangle is

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

Answer 1

The correct option is C $\frac{3}{4}m{l}^2$


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 $BC$
$=0+0+m{\left(\frac{l\sqrt{3}}{2}\right)}^2=\frac{3m{l}^2}{4}$