Question

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

• A. $(3/4)m{l}^2$
• B. $2m{l}^2$
• C. $(5/4)m{l}^2$
• D. $(3/2)m{l}^2$

Answer 1

Answer: (C)

Moment of inertia of the system about $AX$ is given by

Moment of inertia
$=m_A{r}_A^2+m_B{r}_B^2+m_C{r}_C^2$
Moment of inertia
$=m(0{)}^2+m(l{)}^2+m{\left(l\sin 30^{\circ }\right)}^2$
$=m{l}^2+\frac{m{l}^2}{4}=\frac{5}{4}m{l}^2$
Alternative : Moment of inertia of a system about a line OC perpendicular to $AB$, in the plane of $ABC$ is

$I_{CO}=m\times 0+m\times {\left(\frac{l}{2}\right)}^2+m\times {\left(\frac{l}{2}\right)}^2$
$\therefore I_{CO}=\frac{m{l}^2}{4}+\frac{m{l}^2}{4}=\frac{m{l}^2}{2}$
According to parallel-axis theorem
$I_{AX}=I_{CO}+M{x}^2$
where $x=$ distance of $AX$ from $CO,M=$ total mass of system
$I_{AX}=\frac{m{l}^2}{2}+3m\times {\left(\frac{l}{2}\right)}^2$
$I_{AX}=\frac{m{l}^2}{2}+\frac{3m{l}^2}{4}=\frac{5}{4}m{l}^2$