Three particles, each of mass m grams situated at the vertices of an equilateral triangle ABC of side λ 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-cm 2 units will be :
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Moment of inertia of the system about AX is given by
M Ι = m A r 2A + m B r 2B + m C r 2C
M Ι = m(0) 2 + m( λ ) 2 + m( λ sin30°) 2
= m λ 2 + 
= 
m λ 2
Alternative : Moment of inertia of a system about a line OC perpendicular to AB, in the plane of ABC is
Ι CO = m× 0 + m × 
+ m × 
∴ Ι CO = 
According to parallel-axis theorem
Ι AX = Ι CO + Mx 2
where x = distance of AX from CO, M = total mass of system
Ι AX = 
I AX = 
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