Show that the angular momentum is equal to twice the product of mass and real velocity.

Explanation

Areal velocity : It is the time rate of change of area.
Let us consider a particle, rotating in XY plane about Z-axis.
Let `vecr` = position vector of particle, when it is at point P.
`vecr + Deltavecr` = position vector of particle when it is at point Q after a small interval `Deltat`. The displacement of the particle in small time `Deltat` is given by `vec(PQ) = vectau + Deltavecr - vecr = Deltavecr`
Let `DeltavecA` = the area vector swept by the position vector of particle in time `Deltat` , then `DeltavecA = 1/2 (vecr xxDeltavecr)`
Dividing both the sides by `Deltat` , we get `(DeltavecA)/(Deltat)=1/2(vecrxx(Deltavecr)/(Deltat))" "...(i)`
As `Deltat` is very small, so equation (i) can be written as
`{:(Lt),(Deltararr0):}(DeltavecA)/(Deltat)={:(Lt),(Deltararr0):}(1/2 vecrxx(Deltavecr)/(Deltat))`
or `(dvecA)/(dt)=1/2(vecrxx(dvecr)/(dt))=1/2 (vecrxxvecv)" "{ :. (dr)/(dt)=v}`
Multiplying both sides by mass (m), we get
`m(dvecA)/(dt)=1/2 (vecrxxmvecv)=1/2 (vecrxxvecp)`
`implies 2m (dvecA)/(dt) = vecrxxvecp=vecL " " { :. vecvxxvecp = vecL}`
`:. vecL = 2m (dvecA)/(dt)` Hence, angular momentum = `2 xx` mass `xx` areal velocity.