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Question

Three particles each of mass 1 kg are placed at the vertices of an equilateral triangle ABC of side 2 m. Find the moment of inertia of system about a line perpendicular to the plane of system and passing through middle point (D) of side BC of the triangle.



Correct Answer
A
5 kg-m2
Your Answer
B
3 kg-m2
Your Answer
C
2.5 kg-m2
Your Answer
D
1.1 kg-m2

Solution

The correct option is A 5 kg-m2
Given mass of particles, m=1 kg
Moment of inertia of body of mass (m)=mr2
Where r is the radius of rotation.
ΔABC is an equilateral triangle as all sides are equal.
In ΔABD;ABD=60
sin60=x2
x=2×32=3 m


Distance of masses from the axis of rotation is given by
AD=r1=3 m
BD=r2=1 m
CD=r3=1 m
Total moment of inertia of system (I) = I1+I2+I3
I=mr21+mr22+mr23
=1×(3)2+1(1)2+1(1)2
=5  kg-m2

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