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conservation of angular momentum (Definition)

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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Cross-references: vector, momentum, position vector, angular momentum
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This is version 1 of conservation of angular momentum, born on 2006-07-20.
Object id is 204, canonical name is ConservationOfAngularMomentum.
Accessed 1165 times total.

Classification:
Physics Classification: 45.50.-j (Dynamics and kinematics of a particle and a system of particles)
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