Cinématique du Point Matériel en 3D et Cas Avancés

TL;DR

You'll learn how to describe an object's motion in full 3D space using vectors for position, velocity, and acceleration. We'll explore different coordinate systems to simplify problems and tackle relative motion between moving frames. This topic builds on your 2D understanding, adding depth and complexity for more realistic scenarios.

1. The Mental Model

Imagine tracking a fly in your room. Its position changes in all three dimensions, and its speed and direction are constantly shifting. Cinématique du point matériel en 3D is simply the mathematical toolkit to precisely describe that fly's journey, even when the room itself is moving.

2. The Core Material

When moving from 2D to 3D kinematics, you're essentially adding a z-component to all your vector quantities. Everything you learned about position, velocity, and acceleration in 2D still applies, but now in three orthogonal directions.

### Position, Vitesse, Accélération en 3D

In a Cartesian coordinate system (your standard x, y, z axes), the position vector $\vec{r}(t)$ of a point P at time $t$ is:
$\vec{r}(t) = x(t)\vec{i} + y(t)\vec{j} + z(t)\vec{k}$

Where $\vec{i}$, $\vec{j}$, $\vec{k}$ are the unit vectors along the x, y, and z axes, respectively.

The velocity vector $\vec{v}(t)$ is the first derivative of the position vector with respect to time:
$\vec{v}(t) = \frac{d\vec{r}}{dt} = \dot{x}(t)\vec{i} + \dot{y}(t)\vec{j} + \dot{z}(t)\vec{k}$
Here, $\dot{x}(t)$ means $\frac{dx}{dt}$, and so on. The magnitude of the velocity vector is the speed: $||\vec{v}|| = \sqrt{\dot{x}^2 + \dot{y}^2 + \dot{z}^2}$.

The acceleration vector $\vec{a}(t)$ is the first derivative of the velocity vector (or second derivative of position) with respect to time:
$\vec{a}(t) = \frac{d\vec{v}}{dt} = \ddot{x}(t)\vec{i} + \ddot{y}(t)\vec{j} + \ddot{z}(t)\vec{k}$
Again, $\ddot{x}(t)$ means $\frac{d^2x}{dt^2}$.

### Coordonnées Spécifiques (Cylindriques et Sphériques)

Sometimes, the motion itself or the geometry of the problem makes Cartesian coordinates messy. That's when cylindrical or spherical coordinates become your best friends.

Coordonnées Cylindriques ($\rho, \phi, z$)

Think of these as polar coordinates in the xy plane, with an added z component.
- $\rho$: radial distance from the z-axis (like r in polar).
- $\phi$: azimuthal angle (angle around the z-axis).
- $z$: height along the z-axis (same as Cartesian z).

The unit vectors are $\vec{e}\rho$, $\vec{e}\phi$, $\vec{e}z$. They depend on $\phi$.
$\vec{v} = \dot{\rho}\vec{e}\rho + \rho\dot{\phi}\vec{e}\phi + \dot{z}\vec{e}_z$
$\vec{a} = (\ddot{\rho} - \rho\dot{\phi}^2)\vec{e}\rho + (\rho\ddot{\phi} + 2\dot{\rho}\dot{\phi})\vec{e}_\phi + \ddot{z}\vec{e}_z$

Coordonnées Sphériques ($r, \theta, \phi$)

These are great for objects moving around a central point, like satellites orbiting Earth.
- $r$: radial distance from the origin.
- $\theta$: polar angle (angle from the positive z-axis).
- $\phi$: azimuthal angle (angle around the z-axis, same as cylindrical).

The unit vectors are $\vec{e}r$, $\vec{e}\theta$, $\vec{e}\phi$. They depend on both $\theta$ and $\phi$.
The expressions for velocity and acceleration are more complex here. It's usually better to derive them or look them up when you need them, but here's the velocity:
$\vec{v} = \dot{r}\vec{e}_r + r\dot{\theta}\vec{e}\theta + r\sin\theta\dot{\phi}\vec{e}_\phi$

The key insight is that choosing the "right" coordinate system can drastically simplify your calculations, often by making some components zero or constant.

### Mouvement Relatif dans des Repères en Mouvement

This is about observing motion from a non-inertial (accelerating) reference frame. Imagine you're on a spinning merry-go-round, trying to describe the path of a ball thrown from its center.

Lets denote quantities in a fixed inertial frame as $R_0$ and in a moving frame as $R$.
The absolute position $\vec{r}{P/R_0}$ of a point P is equal to the position of the origin of the moving frame $O'$ relative to the inertial frame $\vec{r}{O'/R_0}$ plus the position of P relative to the moving frame $\vec{r}{P/R}$:
$\vec{r}{P/R_0} = \vec{r}{O'/R_0} + \vec{r}{P/R}$

Taking derivatives gives us the composition of velocities formula:
$\vec{v}{P/R_0} = \vec{v}{O'/R_0} + \vec{v}{P/R} + \vec{\omega} \wedge \vec{r}{P/R}$
Where:
- $\vec{v}{P/R_0}$ is the absolute velocity of P.
- $\vec{v}{O'/R_0}$ is the velocity of the origin of the moving frame.
- $\vec{v}{P/R}$ is the relative velocity of P as seen from the moving frame.
- $\vec{\omega}$ is the angular velocity of the moving frame relative to the inertial frame.
- $\vec{\omega} \wedge \vec{r}{P/R}$ is the velocity due to the rotation of the moving frame (often called the entrainment velocity component due to rotation).

This leads to the composition of accelerations formula:
$\vec{a}{P/R_0} = \vec{a}{O'/R_0} + \vec{a}{P/R} + 2\vec{\omega} \wedge \vec{v}{P/R} + \vec{\dot{\omega}} \wedge \vec{r}{P/R} + \vec{\omega} \wedge (\vec{\omega} \wedge \vec{r}{P/R})$
Here:
- $\vec{a}{P/R_0}$ is the absolute acceleration.
- $\vec{a}{O'/R_0}$ is the acceleration of the moving frame's origin.
- $\vec{a}{P/R}$ is the relative acceleration.
- $2\vec{\omega} \wedge \vec{v}{P/R}$ is the Coriolis acceleration.
- $\vec{\dot{\omega}} \wedge \vec{r}{P/R}$ is the angular acceleration component.
- $\vec{\omega} \wedge (\vec{\omega} \wedge \vec{r}{P/R})$ is the centripetal acceleration.

This looks like a lot, but each term has a distinct physical meaning.

graph TD
    A["Description du Mouvement en 3D"] --> B["Coordonnées Cartésiennes"]
    A --> C["Coordonnées Spécifiques"]
    C --> C1["Cylindriques ($\rho, \phi, z$)"]
    C --> C2["Sphériques ($r, \theta, \phi$)"]
    B --> B1["Position $\\vec{r}(t)$"]
    B1 --> B2["Vitesse $\\vec{v}(t)$"]
    B2 --> B3["Accélération $\\vec{a}(t)$"]
    A --> D["Mouvement Relatif (Repères en Mouvement)"]
    D --> D1["Composition des Vitesses"]
    D --> D2["Composition des Accélérations"]
    D1 --> D1a["$\\vec{v}_{Absolue} = \\vec{v}_{Entraînement} + \\vec{v}_{Relative}$"]
    D2 --> D2a["$\\vec{a}_{Absolue} = \\vec{a}_{Entraînement} + \\vec{a}_{Coriolis} + \\vec{a}_{Relative}$"]
    style D1a fill:#fff,stroke:#333,stroke-width:2px;
    style D2a fill:#fff,stroke:#333,stroke-width:2px;

3. Worked Example

Let's find the position, velocity, and acceleration of a particle whose movement is described by:
$x(t) = 3\cos(t)$
$y(t) = 3\sin(t)$
$z(t) = 4t$

1. Position Vector:
   $\vec{r}(t) = 3\cos(t)\vec{i} + 3\sin(t)\vec{j} + 4t\vec{k}$
   This describes a helix winding up the z-axis.

2. Velocity Vector:
   Differentiate each component with respect to $t$:
   $\dot{x}(t) = \frac{d}{dt}(3\cos(t)) = -3\sin(t)$
   $\dot{y}(t) = \frac{d}{dt}(3\sin(t)) = 3\cos(t)$
   $\dot{z}(t) = \frac{d}{dt}(4t) = 4$

   So, $\vec{v}(t) = -3\sin(t)\vec{i} + 3\cos(t)\vec{j} + 4\vec{k}$

   The speed $||\vec{v}(t)||$ is:
   $\sqrt{(-3\sin(t))^2 + (3\cos(t))^2 + 4^2} = \sqrt{9\sin^2(t) + 9\cos^2(t) + 16}$
   $= \sqrt{9(\sin^2(t) + \cos^2(t)) + 16} = \sqrt{9(1) + 16} = \sqrt{25} = 5$
   The particle moves at a constant speed of 5 units/time.