# Particle dynamics

## Linear momentum

The linear momentum of a particle is a **vector** magnitude
given in an inertial reference frame:

$$\mathbf{p} = m \mathbf{v}$$

![momentum](figs/particulaMomLineal.png)


:::{card} Second Newton's Law
$$ \mathbf{F} = \dot{ \mathbf{p} }  $$
:::


:::{card} Principle of momentum conservation
$$ \mathbf{F} = 0  \rightarrow  \mathbf{p} = constant $$
:::


## Angular momentum of a particle

:::{card} Definition
The angular momentum of a particle with respect to a point ($O$) is:

$$ \mathbf{L} = \mathbf{r} \times \mathbf{p} $$
:::

![angular momentum](figs/particulaMomAngular.png)


:::{card} Rate of change of the angular momentum
$$  \mathbf{M}_{O} = \frac{d}{dt}\left( \mathbf{r} \times \dot{\mathbf{p}} \right) = \dot{\mathbf{L}} $$
:::


:::{card} Principle of angular momentum conservation
$$ \mathbf{M} = 0  \rightarrow  \mathbf{L} = constant$$
:::


## Equations of motion for a particle

For a single particle, we have a system of differential
equations:

$$
    \begin{eqnarray*}
       \mathbf{F} &=& \dot{ \mathbf{p} } \quad\quad \text{Newton (1643-1727)} \\
       \mathbf{M}_{O} &=& \dot{\mathbf{L}} \quad\quad \text{Euler (1707-1783)}
    \end{eqnarray*}
$$

Note these amounts to **six** equations for 3D point-like 
particles. However, they are **redundant**, and solving just
three of them is enough. 

Picking the first subset (Newton's second law):
Since $\dot{ \mathbf{p} } = \frac{d}{dt}( m \mathbf{v} ) = \frac{d}{dt}( m \dot{\mathbf{r}} ) = m \ddot{\mathbf{r}} $:

$$  m \ddot{\mathbf{r}} = \mathbf{F} $$

Which is a system of 3 ordinary differential equations (ODE):


$$
\begin{eqnarray*}
    \ddot{x}(t) = \frac{1}{m} F_x(t) \\
    \ddot{y}(t) = \frac{1}{m} F_y(t) \\
    \ddot{z}(t) = \frac{1}{m} F_z(t)
\end{eqnarray*}
$$

This ODE system can then be solved analytically (ideal cases only), or, more realistically, using **numerical integrators**: Euler, Trapezoidal, etc.