JIPMER2005PhysicsRotational Motion
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
Options
- A1 ~kg ~m ^2
- B0.1 ~kg ~m ^2
- C2 ~kg ~m ^2
- D0.2 ~kg ~m ^2
Correct answer
B. 0.1 ~kg ~m ^2
Step-by-step solution
The moment of inertia of the 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 therefore, MI of the system = MI of particle =5 2 (0.1)^2=0.1 ~kg ~m ^2