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Question

Three particles each of mass 2 kg are placed at the corners of an equilateral triangle of side 6 m. What is the moment of inertia of the system about an axis passing through the height of the triangle on the same plane as the particles?

A
12 kgm2
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B
36 kgm2
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C
48 kgm2
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D
40 kgm2
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Solution

The correct option is B 36 kgm2

Let the three particles be A,B & C. AD is the height of the triangle and the axis of rotation.
The height of an equilateral triangle passes through the mid-point of the base.
BD=DC=3 m
Moment of inertia of the system
=MI=mA×0+mB×BD2+mC×DC2
MI=0+2×32+2×32=36 kg m2
The moment of inertia of the system is 36 kg m2.

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