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.

Find the components along the x, y, z axes of the angular momentum l o

1.	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.
Answer» Solution :If position vector of a particle is `vecr` and LINEAR momentum `VECP` than angular momentum `vecl=vecrxxvecp` but `vecr=xhati+yhatj+zhatk` and `vecp=p_(x)HATI+p_(y)hatj+p_(z)hatk` are the respectively COMPONENTS of `vecr and vecp` `vecl=l_(x)hati+l_(y)hatj+l_(z)hatk` `:.vecl=[xhati+yhatj+zhatk]xx[p_(x)hati+p_(y)hatj+p_(z)hatk]` `l_(x)hati+l_(y)hatj+l_(z)hatk=\|(hati,hatj,hatk),(x,y,z),(p_(x),p_(y),p_(z))\|` `=hati(yp_(z)-zp_(y))+J(zp_(x)-xp_(z))+hatk(xp_(y)-yp_(x))` From this `l_(x)=yp_(z)-zp_(y)` `l_(y)=zp_(x)-xp_(z)` and `l_(z)=xp_(y)-yp_(x)` The particle moves in the XY plane hence `Z=0 and p_(z)=0` `:.vecl=0hati+0hatj+(xp_(y)-yp_(x))hatk` `:.vecl=(xp_(y)-yp_(x))hatk` Therefore, when the particle is confined to move in the XY-plane, the direction of angular momentum `(vecl)` is along the Z-direction.