PlanetPhysics/Conservation of Angular Momentum

If a \htmladdnormallink{particle {http://planetphysics.us/encyclopedia/Particle.html} is subject to no torque then the angular momentum is conserved}

The angular momentum, $$\mathbf{L}$$ of a particle with position vector, $$\mathbf{r}$$, and total linear momentum, $$\mathbf{p}$$ is given by $$\mathbf{L} = \mathbf{r}\times\mathbf{p}$$. If some force, $$\mathbf{F}$$, acts on that particles, then the torque is defined similarily as $$\mathbf{N} = \mathbf{r}\times\mathbf{F} = \mathbf{r}\times d\mathbf{p}/dt$$.

Taking the time derivative of the angular momentum equation, $$\begin{matrix} \frac{d\mathbf{L}}{dt} & = & \frac{d}{dt}\left( \mathbf{r}\times\mathbf{p}\right)\\ & = & \left( \frac{d\mathbf{r}}{dt}\times\mathbf{p}\right) + \left(\mathbf{r}\times\frac{d\mathbf{p}}{dt} \right). \end{matrix}$$ Consider the term, $$d\mathbf{r}/dt\times\mathbf{p}$$. Since $$\mathbf{p} = md\mathbf{r}/dt$$, it follows that $$\frac{d\mathbf{r}}{dt}\times\mathbf{p} = m\left(\frac{d\mathbf{r}}{dt}\times\frac{d\mathbf{r}}{dt}\right). $$ But, given an arbitrary vector, $$\mathbf{A}$$, $$\mathbf{A}\times\mathbf{A}=\mathbf{0}$$ (the zero vector), so the expression for the time derivative of the angular momentum becomes, $$\frac{d\mathbf{L}}{dt} = \left(\mathbf{r}\times\frac{d\mathbf{p}}{dt} \right) = \mathbf{N}.$$ Writing the above simplistically as $$d\mathbf{L}/dt = \mathbf{N}$$ is is clear that when the torque is zero, then the angular momentum is constant in time; it is conserved.