Vector Product and Angular Velocity

TL;DR

The vector product (or cross product) of two vectors gives a third vector perpendicular to both, crucial for understanding rotational effects. Angular velocity describes how fast an object rotates and the direction of its axis of rotation. The vector product helps relate linear and angular quantities, like position and velocity in rotational motion.

1. The Mental Model

Imagine pushing a door open. The force you apply and your distance from the hinges work together to create a turning effect (torque). The vector product helps us precisely describe this turning effect, where both the magnitude and direction matter.

2. The Core Material

The vector product (or cross product) is an operation between two vectors, say $\mathbf{A}$ and $\mathbf{B}$, that results in a new vector $\mathbf{C}$. This new vector $\mathbf{C}$ has a magnitude and a direction.

Its magnitude is given by:
$|\mathbf{C}| = |\mathbf{A}| |\mathbf{B}| \sin(\theta)$
where $\theta$ is the angle between $\mathbf{A}$ and $\mathbf{B}$.

Its direction is perpendicular to both $\mathbf{A}$ and $\mathbf{B}$, determined by the right-hand rule. If you point the fingers of your right hand in the direction of $\mathbf{A}$ and curl them towards $\mathbf{B}$, your thumb points in the direction of $\mathbf{C}$.

In Cartesian coordinates, if $\mathbf{A} = A_x\mathbf{i} + A_y\mathbf{j} + A_z\mathbf{k}$ and $\mathbf{B} = B_x\mathbf{i} + B_y\mathbf{j} + B_z\mathbf{k}$, then:
$\mathbf{A} \times \mathbf{B} = (A_yB_z - A_zB_y)\mathbf{i} + (A_zB_x - A_xB_z)\mathbf{j} + (A_xB_y - A_yB_x)\mathbf{k}$

Angular Velocity ($\boldsymbol{\omega}$)

Photo by Bimal Chhetry on Pexels

Angular velocity is a vector quantity that describes how fast an object is rotating and about which axis it's rotating. Its magnitude is the angular speed (radians per second, rad/s), and its direction is along the axis of rotation, again given by the right-hand rule. If you curl the fingers of your right hand in the direction of rotation, your thumb points in the direction of the angular velocity vector.

Relating Linear and Angular Velocity

Photo by Joel Zar on Pexels

For a particle rotating about an axis, its linear velocity $\mathbf{v}$ at any point is related to its angular velocity $\boldsymbol{\omega}$ and its position vector $\mathbf{r}$ (from the axis of rotation to the particle) by the vector product:
$\mathbf{v} = \boldsymbol{\omega} \times \mathbf{r}$

This equation is fundamental. It tells us that the linear velocity vector $\mathbf{v}$ is perpendicular to both the angular velocity vector $\boldsymbol{\omega}$ (which is along the axis of rotation) and the position vector $\mathbf{r}$.

graph TD
    A["Angular Velocity (ω)"] --> B["Axis of Rotation"];
    C["Position Vector (r)"] --> D["Particle Location"];
    A --> E{"Vector Product (ω x r)"};
    C --> E;
    E --> F["Linear Velocity (v)"];
    F --> G["Perpendicular to ω and r"];

3. Worked Example

Let's say a particle is rotating about the z-axis with an angular velocity $\boldsymbol{\omega} = 2\mathbf{k}$ rad/s (meaning 2 rad/s in the positive z-direction). The particle is located at a position $\mathbf{r} = 3\mathbf{i} + 4\mathbf{j}$ meters in the xy-plane. What is its linear velocity $\mathbf{v}$?

Using the formula $\mathbf{v} = \boldsymbol{\omega} \times \mathbf{r}$:
$\mathbf{v} = (2\mathbf{k}) \times (3\mathbf{i} + 4\mathbf{j})$

We can expand this using the distributive property and the cross product rules for unit vectors ($\mathbf{k} \times \mathbf{i} = \mathbf{j}$ and $\mathbf{k} \times \mathbf{j} = -\mathbf{i}$):
$\mathbf{v} = (2\mathbf{k} \times 3\mathbf{i}) + (2\mathbf{k} \times 4\mathbf{j})$
$\mathbf{v} = 6(\mathbf{k} \times \mathbf{i}) + 8(\mathbf{k} \times \mathbf{j})$
$\mathbf{v} = 6\mathbf{j} + 8(-\mathbf{i})$
$\mathbf{v} = -8\mathbf{i} + 6\mathbf{j}$ m/s

So, the linear velocity of the particle is $-8\mathbf{i} + 6\mathbf{j}$ m/s. This means the particle is moving with a velocity that has components in the negative x-direction and positive y-direction, consistent with counter-clockwise rotation about the z-axis.

4. Key Takeaways

• The vector product $\mathbf{A} \times \mathbf{B}$ yields a vector perpendicular to both $\mathbf{A}$ and $\mathbf{B}$.
• The magnitude of the vector product is $|\mathbf{A}| |\mathbf{B}| \sin(\theta)$, representing the area of the parallelogram formed by $\mathbf{A}$ and $\mathbf{B}$.
• The direction of the vector product is found using the right-hand rule.
• Angular velocity $\boldsymbol{\omega}$ is a vector describing both rotational speed and the orientation of the rotation axis.
• The right-hand rule also determines the direction of $\boldsymbol{\omega}$ from the direction of rotation.
• The linear velocity $\mathbf{v}$ of a point in a rotating system is given by $\mathbf{v} = \boldsymbol{\omega} \times \mathbf{r}$.
• This relationship is crucial for connecting rotational motion to the motion of individual particles.

Common Mistakes to Avoid:
- Confusing dot product with vector product; they are very different operations.
- Incorrectly applying the right-hand rule, leading to the wrong direction for the resultant vector.
- Forgetting that the order matters in a vector product: $\mathbf{A} \times \mathbf{B} \neq \mathbf{B} \times \mathbf{A}$ (instead, $\mathbf{A} \times \mathbf{B} = -(\mathbf{B} \times \mathbf{A})$).
- Using scalar angular speed instead of the vector angular velocity when calculating linear velocity.

5. Now Try It

A satellite is orbiting the Earth with an angular velocity $\boldsymbol{\omega} = 0.001\mathbf{j}$ rad/s (along the positive y-axis). If a sensor on the satellite is located at a position $\mathbf{r} = 5\mathbf{i} + 2\mathbf{k}$ km relative to the satellite's center of mass, calculate the linear velocity of the sensor due to the satellite's rotation. What does the resulting vector tell you about the sensor's motion? You should find a linear velocity with components in the x and z directions.