#### Question

Find the components along the *x, y, z *axes of the angular momentum **l **of a particle, whose position vector is **r **with components *x*, *y*, *z *and momentum is **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.

#### Solution 1

*l*_{x} = *yp*_{z} – *zp*_{y}

*l*_{y }= z*p*_{x} – x*p*_{z}

*l*_{z} = *xp*_{y} –*yp*_{x}

Linear momentum of the particle,`vecp = p_x hati + p_y hatj + p_z hatk`

Position vector of the particle, `vecr = xhati + yhatj + zhatk`

Angular momentum, `hatl = hatr xx hatp`

=`(xhati + yhatj + zhatk) xx (p_x hati + p_y hatj + p_z hatk)`

`=|(hati,hatj,hatk),(x,y,z), (p_x, p_y,p_z)|`

`l_xhati + l_yhatj + l_z hatk = hati (yp_z - zp_y) - hatj(xp_z - zp_x) + hatk (xp_y - zp_x)`

Comparing the coefficients of `hati, hatj, hatk` we get:

`((l_x = yp_z - zp_y),(l_y = xp_z -zp_x),(l_z = xp_y - yp_x))}...(i)`

The particle moves in the *x*-*y *plane. Hence, the *z*-component of the position vector and linear momentum vector becomes zero, i.e.,

*z* = *p*_{z} = 0

Thus, equation (*i*) reduces to:

`((l_x=0),(l_y=0),(l_z=xp_y -yp_x))} `

Therefore, when the particle is confined to move in the *x*-*y *plane, the direction of angular momentum is along the *z*-direction.

#### Solution 2

We know that angular momentum `vecl` of a particle having position vector `vecr` and momentum `vecp` is given by

`vecl = vecr xx vecp`

But vecr = [xveci + yhatj + zveck], where x, y,z are the component of `vecr and vecp = [p_xveci + p_yhatj + p_zhatk]`

`:. vecl = vecr xx vecp = [x hati + yhatj + zhatk] xx [p_xhati + p_y hatj + p_z hatk]`

or `(l_xhati + l_yhatj + l_zhatk) `= `|(hati ,hatj ,hatk),(x,y,z), (p_x,p_y,p_z)|`

`=(yp_z - zp_y)hati + (zp_x - xp_z)hatj + (xp_y - yp_x) hatk`

From this relation we conclude that

`l_x = yp_z - zp_y, l_y = zp_x - xp_z, l_z = xp_y- yp_x`

if the given particle moves only in the x - y plane then z = 0 and 'p_z = 0' and hence

'vecl = (xp_y - yp_x)hatk` which is onl;y the z- component of hatl

it means that for a particle moving only in the x-y plane, the angular momentum has only the z-component