# Particle dynamics

Before moving on to rigid bodies, we review the basic concepts of the dynamics of a particle (a point with mass $m$ and no dimensions).

## Linear momentum

:::{card} Definition
The **linear momentum** of a particle is the vector quantity, given in an inertial reference frame:

$$\vec{p} = m \vec{v}$$
:::

![Linear momentum](figs/particulaMomLineal.png)

:::{card} Newton's second law
$$ \vec{F} = \dot{\vec{p}} $$
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:::{card} Principle of conservation of linear momentum
$$ \vec{F} = \vec{0} \quad \rightarrow \quad \vec{p} = \text{constant} $$
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## Angular momentum

:::{card} Definition
The **angular momentum** of a particle about a point $O$ is:

$$ \vec{L}_O = \vec{r} \times \vec{p} $$

where $\vec{r}$ is the position vector of the particle from $O$.
:::

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

Cross-multiplying Newton's second law by $\vec{r}$:

$$ \underbrace{\vec{r} \times \vec{F}}_{\vec{M}_O} = \vec{r} \times \dot{\vec{p}} $$

On the other hand, differentiating the angular momentum with respect to time (with $O$ fixed):

$$ \frac{d\vec{L}_O}{dt} = \frac{d}{dt}\left( \vec{r} \times m\vec{v} \right) = \underbrace{\vec{v} \times m\vec{v}}_{=\vec{0}} + \vec{r} \times \dot{\vec{p}} = \vec{r} \times \dot{\vec{p}} $$

since the cross product of a vector with itself is zero. Comparing both expressions:

:::{card} Rate of change of the angular momentum
$$ \vec{M}_O = \vec{r} \times \dot{\vec{p}} = \dot{\vec{L}}_O $$
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:::{card} Principle of conservation of angular momentum
$$ \vec{M}_O = \vec{0} \quad \rightarrow \quad \vec{L}_O = \text{constant} $$
:::

A classic application of this principle is **spin stabilization of satellites**: Telstar I (1962) spun about its major axis at about 200 rpm to keep its orientation.

<img src="figs/telstar1.jpg" alt="Telstar I" width="260px">

## Equations of motion of a particle

For a particle we have the system of differential equations:

$$
\begin{aligned}
\vec{F} &= \dot{\vec{p}} &\qquad& \text{(Newton, 1643-1727)} \\
\vec{M}_O &= \dot{\vec{L}}_O &\qquad& \text{(Euler, 1707-1783)}
\end{aligned}
$$

In 3D these are six scalar equations, but for a point particle they are **redundant**: the second one follows from the first, as we have just seen. Newton's equation is therefore enough. Since $\dot{\vec{p}} = \frac{d}{dt}(m\vec{v}) = \frac{d}{dt}(m\dot{\vec{r}}) = m\ddot{\vec{r}}$ (with constant $m$):

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

which is a system of three ordinary differential equations (ODE):

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

This system can be solved analytically in simple cases or, in general, with **numerical integrators** (Euler's rule, trapezoidal rule, Runge-Kutta...).

:::{dropdown} Example: numerical integration with Euler's rule
To integrate $m\ddot{x} = F(t, x, \dot{x})$, it is written as a first-order system with state $(x, v)$ and advanced with a time step $\Delta t$:

$$
\begin{aligned}
v_{k+1} &= v_k + \frac{\Delta t}{m} F(t_k, x_k, v_k) \\
x_{k+1} &= x_k + \Delta t\, v_k
\end{aligned}
$$

For instance, for a mass on a spring ($F=-kx$), in MATLAB:

```matlab
m = 1; k = 10; dt = 1e-3; N = 5000;
x = zeros(1,N); v = zeros(1,N); x(1) = 0.1;
for i = 1:N-1
    v(i+1) = v(i) + dt/m * (-k*x(i));
    x(i+1) = x(i) + dt * v(i);
end
plot((0:N-1)*dt, x)
```

Euler's rule is the simplest but accumulates error (in this example the amplitude slowly grows); in practice, higher-order methods such as `ode45` are used.
:::