port from mathematics-physics notes

docs/physics/classical-mechanics/newtonian-mechanics/rotation.md

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+# Rotation
+
+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
+
+    $$
+    \mathbf{L}'(t) = \mathbf{r} \times \mathbf{F}(t) \overset{\mathrm{def}} = \mathbf{\Gamma}(t). 
+    $$