Question

Three particles each of mass m gram, are situated at the vertices of an equilateral triangle ABC of the side $1$cm. The moment of inertia of the system (as shown in figure) about a line AX perpendicular to AB and in the plane of ABC in gram-$c{m}^2$ units will be :

• A. $\frac{3}{2}m{l}^2$
• B. $\frac{3}{4}m{l}^2$
• C. $2m{l}^2$
• D. $\frac{5}{4}m{l}^2$

Answer 1

The correct option is D $\frac{5}{4}m{l}^2$

Perpendicular distance of mass at A from axis AX $r_1=0$

Perpendicular distance of mass at B from axis AX $r_2=\frac{l}{2}$

Perpendicular distance of mass at C from axis AX $r_3=l$

Moment of inertia of the system $I=I_1+I_2+I_3=m_1{r}_1^2+m_2{r}_2^2+m_3{r}_3^2$
$I=0+m{\left(\frac{l}{2}\right)}^2+m{l}^2$
$I=\frac{m{l}^2}{4}+m{l}^2$
$I=\frac{5}{4}m{l}^2$.