Question

Two masses `m_(1) and m_(2)` are placed at a distance r from each other. Find out the moment of inertia of system about an axis passing through their centre of mass.

Answer 1

We have, `m_(1)r_(1)=m_(2)r_(2) and r_(1)+r_(2)=r`
`rArr r_(1)=(m_(2)r)/(m_(1)+m_(2)),r_(2)=(m_(1)r)/(m_(1)+m_(2))`

Moment of inertia of the system about an axis passing through their centre of mass is
`I=m_(1)r_(1)^(2)+m_(2)r_(2)^(2)=m_(1)((m_(2)r)/(m_(1)+m_(2)))^(2)+m_(2)((m_(1)r)/(m_(1)+m_(2)))^(2)=((m_(1)m_(2))/(m_(1)+m_(2)))r^(2)`
Note : Here, `I = mur^(2)` where, `mu` is called reduced mass which is equal to `(m_(1)m_(2))/(m_(1)+m_(2))`.