Question

Three particles, each of mass $m$ grams situated at the vertices of an equilateral triangle $ABC $ of side $l\, cm$ (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 $A B$, in the plane of $A B C$ is

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