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# Rotation
2024-04-02 18:04:22 +02:00
Rotation is always viewed with respect to the axis of rotation, therefore in the following definitions the origin of the position is always implies to be the axis of rotation.
## Angular momentum
> *Definition 1*: the angular momentum $L$ of a point mass with position $\mathbf{r}$ and a momentum $\mathbf{p}$ is defined as
>
> $$
> \mathbf{L} = \mathbf{r} \times \mathbf{p},
> $$
>
> for all $\mathbf{r}$ and $\mathbf{p}$.
## Torque
> *Definition 2*: the torque $\mathbf{\Gamma}$ acting on a point mass with position $\mathbf{r}$ for a force $\mathbf{F}$ os defined as
>
> $$
> \mathbf{\Gamma} = \mathbf{r} \times \mathbf{F},
> $$
>
> for all $\mathbf{r}$ and $\mathbf{F}$.
The torque is related to the angular momentum by the following proposition.
> *Proposition 1*: let $\mathbf{L}: t \mapsto \mathbf{L}(t)$ be the angular momentum of a point mass, then it holds that
>
> $$
> \mathbf{L}'(t) = \mathbf{\Gamma}(t),
> $$
>
> for a constant $\mathbf{r}$ and all $t \in \mathbb{R}$ with $\mathbf{\Gamma}: t \mapsto \mathbf{\Gamma}(t)$ the torque acting on the point mass.
??? note "*Proof*:"
Let $\mathbf{L}: t \mapsto \mathbf{L}(t)$ be the angular momentum of a point mass and suppose $\mathbf{r}$ is constant, then
$$
\mathbf{L}'(t) \overset{\mathrm{def}} = d_t (\mathbf{r} \times \mathbf{p}(t)) = \mathbf{r} \times \mathbf{p}'(t),
$$
by [proposition](momentum.md) we have $\mathbf{p}'(t) = \mathbf{F}(t)$, therefore
$$
\mathbf{L}'(t) = \mathbf{r} \times \mathbf{F}(t) \overset{\mathrm{def}} = \mathbf{\Gamma}(t).
$$