conservation of angular momentum

If a particle 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,

$\displaystyle \frac{d\mathbf{L}}{dt}$ $\displaystyle =$ $\displaystyle \frac{d}{dt}\left( \mathbf{r}\times\mathbf{p}\right)$  
  $\displaystyle =$ $\displaystyle \left( \frac{d\mathbf{r}}{dt}\times\mathbf{p}\right) + \left(\mathbf{r}\times\frac{d\mathbf{p}}{dt} \right).$  

Consider the term, $ d\mathbf{r}/dt\times\mathbf{p}$ . Since $ \mathbf{p} = md\mathbf{r}/dt$ , it follows that

$\displaystyle \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,

$\displaystyle \frac{d\mathbf{L}}{dt} = \left(\mathbf{r}\times\frac{d\mathbf{p}}{dt} \right) = \mathbf{N}.$

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.



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