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.

Explanation

When two objects of mass `m_(1)` and `m_(2)` 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)))^(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 =mu d^(2)`
In this problem, `mu = (m_(1)m_(2))/(m_(1)+m_(2))=(1xx2)/(1+2)=(2)/(3)kg`
`I=mu d^(2)=(2)/(3)(1.2)^(2)=0.96 kg m^(2)`