Five particles of mass 2 kg are attached to the rim of a circular disc of radius 0.1 m and negligible mass. Moment of inertia of the system about the axis passing through the centre of the disc and perpendicular to its plane is
A
1kgm2
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B
0.1kgm2
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C
2kgm2
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D
0.2kgm2
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Solution
The correct option is B0.1kgm2 We will not consider the moment of inertia of ring because it doesn't have any mass. So, moment of inertia of five particle system I=5mr2=5×2×(0.1)2=0.1kg−m2.
Note: The masses are concentrated at fixed distance from the axis similar to that of ring. That is why you simply need to add all the individual contributions.