Find the components along the `x,y,z` axes of the angular momentum `overset rarr(L)` of a particle, whose position vector is `overset rarr(r )` with components `x,y,z` and momentum is `overset rarr(p)` with components `p_(x), p_(y) and p_(z)`. Show that if the particle moves only in the `x-y` plane, the angular momentum has only a z-component.

Explanation

Angualr momentum of a particle is `vecL = vecr xx vecP`
Let, `vecr= xhati + yhatj + zhatk`, where (x,y,z) is the location of particle at some isntant and `vecp=p_(x)hati + p_(y)hatj + p_(z)hatk`, is the linear momentum of particle at any instant.
Then, angular momentum of particle is given by
`vecL = vecr xx vecp =|{:(hati,hatj, hatk),(x,y,z),(p_(x),p_(y),p_(z)):}|=hati(p_(z).y-p_(y).z) - hatj(p_(z).x-p_(x).z)+hati(p_(y).x-p_(x).y)`.........(i)
so, components of angular momentum along x,y and z-axis are
`L_(x) = p_(a)y-p_(y)z, L_(y) = p_(a)x - p_(x)z` and `L_(x)=p_(y)x-p_(x)y`..........(i)
So, components of angular momentum along x,y and z-axes are
`L_(x) = p_(x)Y - p_(y)z` and `L_(y)=p_(a)x - p_(a)z` and `L_(z) = p_(y)x -p_(x)y`, respectively Now, if particle is confined to x-y plane, then `vecr = xhati + yhatj rArr z=0` and `vecp =p_(x)hati + p_(y)hatj rArr p_(z)=0`
Substituting, in eq. (i), we get: `vecL = hatk(p_(y)x-p_(x)y)`
So, angular momentum only has a z-component.
As angular momentum is cross -product of `vecr` and `vecp`.
It is always perpendicular to plane containing `vecr` and `vecp`.