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.

Options

A. `(2)/(3) m omega L ^(2)`
B. `(1)/(3) omega ^(2) L m `
C. `(2)/(3) omega ^(2) L m`
D. `(1)/(3) L m`

Answer

A

Explanation

Location of CM,
`X _(CM ) = ( 2 m xx 0 + m xx L )/( 2m + m ) = (L)/(3)`
The velocity of ball, A w.r.t. the centre of mass C is
`V _(A) = omega X _(CM ) = ( omega L )/( 3)`
` V _(B) = omega ( L - L _(CM )) = ( 2omega L )/( 3)`
Thus, angular momentum of a system about C,
`L = 2 mv _(A) X _(CM ) + mv _(B) ( L- C _(CM ))`
`= 2m xx (omega L )/(3) xx (L)/(3) + m xx ( 2 omega L )/(3) xx (2 L )/(3) = (2 m omega L ^(2))/( 3)`