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
1 kg m2
0.1 kg m2
2 kg m2
0.2 kg m2
B.
0.1 kg m2
The moment of inertia of given system that contains 5 particles each of mass = 2 kg on the rim of circular disc of radius 0.1 m and of negligible mass is given by :
MI of disc + MI of particle
Since the mass of the disc is negligible
∴ MI of the system = MI of particle
= 5 × 2 × (0.1)2
= 0.1 Kg m2