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Physics Question on System of Particles & Rotational Motion

Five particles of mass 2 kg2\,kg are attached to the rim of a circular disc of radius 0.1 m0.1\,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

1 kg m21\,kg\,m^{2}

B

0.1 kg m20.1\,kg\,m^{2}

C

2 kg m22\,kg\,m^{2}

D

0.2 kg m2 0.2\,kg\,m^{2}

Answer

0.1 kg m20.1\,kg\,m^{2}

Explanation

Solution

The moment of inertia of the given system that contains 55 particles each of mass =2 kg=2\, kg on the rim of circular disc of radius 0.1m0.1 m and of negligible mass is given by
=MI= MI of disc +MI+ MI of particle
Since the mass of the disc is negligible therefore, MIMI of the system =MI= MI of particle
=5×2×(0.1)2=0.1 kg m2=5 \times 2 \times(0.1)^{2}=0.1 \,kg \,m ^{2}