Two balls of masses `m` and `2m` are attached to the ends of a light rod of length `L`. The rod rotates with an angular speed `omega` about an axis passing through the center of mass of system and perpendicular to the plane. Find the angular momentum of the system about the axis of rotation.

Explanation

Location of `c.m.`
`x_(c.m.)=(2mxx0+mxxL)(2m+m)=(L)/(3)`
The velocity of ball `A` w.r.t the center of mass `C` is
`v_(A)=omegax_(c.m.)=(omegaL)/(3)`
`v_(B)=omega(L-x_(c.m.))=(2omegaL)/(3)`
Angular momentum of a system about `C`
`L=2mv_(A)x_(c.m.)+mv_(B)(L-x_(c.m.))`
`=2mxx(omegaL)/(3)xx(L)/(3)+mxx(2omegaL)/(3)xx(2L)/(3)`
`=(2momega^(2)L)/(3)`
OR
`L=I_(c.m.)omega=[2mx_(c.m.)^(2)+m(L-x_(c.m.))^(2)] omega`
`=[2m((L)/(3))^(2)+m((2L)/(3))^(2)]omega=(2momega^(2)L)/(3)`
Note: For a particle moving in a circle is as follows.

Angular momentum of particle about `O`
`L_(0)=mvr=m(romega)r=mr^(2)omega`
OR
`L_(0)=I_(0)omega=mr^(2)omega`