Originally Posted by **Kiwi_Dave** So angular momentum of a particle is given by:
L = m (r x v)
and angular momentum is conserved in the absence of any forces.
I want to work in cartesian coordinates and I have a particle rotating around the z axis but not necissarily in the z=0 plane (i.e. dz/dt = 0). **How can the angular momentum be conserved when **
L/m = (**r** **x** **v**) = (y dz/dt - z dy/dt) **e**1 - (x dz/dt - z dx/dt) **e**2 + (x dy/dt
- y dx/dt) **e**3
L/m = (**r x v**) = - z dy/dt **e**1 + z dx/dt **e**2 + (x dy/dt - y dx/dt) **e**3
Ie the angular momentum has components in the x (e1) and y (e2) directions that are non-zero and have variable magnitude. |

"angular momentum is conserved

**in the absence of any forces**"

If the particle is rotating about the z-axis, there must be a radial

*centripetal force* acting on it.