Two objects of masses 1 kg and 2 kg separated by a distance of 1.2 m are rotating about their centre of mass. Find the moment of inertia of the system.

Two objects of masses 1 kg and 2 kg separated by a distance of 1.2 m a

1.	Two objects of masses 1 kg and 2 kg separated by a distance of 1.2 m are rotating about their centre of mass. Find the moment of inertia of the system.
Answer» Solution :When two objects of MASS m, and m, are SEPARATED by a distance .d., the distance of mass `m_(1)` from the centre of mass `=(m_(2)d)/(m_(1)+m_(2))` and the distance of mass `m_(2)` from the centre of mass `=(m_(1)d)/(m_(1)+m_(2))` Hence `I=m_(1)((m_(2)d)/(m_(1)+m_(2)))+m_(2)((m_(1)d)/(m_(1)+m_(2)))^(2)` On simplification `I=((m_(1)m_(2))/(m_(1)+m_(2)))d^(2)` Here `(m_(1)m_(2))/(m_(1)+m_(2))` is called reduced mass denoted by `mu`. Hence, for a system of two PARTICLES rotating about their centre of mass, moment of inertia `I=mud^(2)` In this PROBLEM, `mu=(m_(1)m_(2))/(m_(1)+m_(2))=(1xx2)/(1+2)=(2)/(3)kg` `I=mud^(2)=(2)/(3)(1.2)^(2)=0.96kgm^(2)`