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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 place 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 m20.2\, kg\, m ^{2}

Answer

0.1 kg m20.1\, kg\, m ^{2}

Explanation

Solution

The moment of inertia of the given system that contains 5 particles each of mass 2 kg2\, kg on the rim of circular disc of radius 0.1 m0.1\, m and of negligible mass is given by == MI of disc ++ MI of particles Since, the mass of the disc is negligible therefore, MI of the system = MI of particles =5×2×(0.1)2=0.1 kg m2=5 \times 2 \times(0.1)^{2}=0.1\, kg\, m ^{2}