Vector Product and Angular Velocity

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Vector Product and Angular Velocity

TL;DR

You'll learn how the cross product $\vec{a} \times \vec{b}$ encodes rotation, why angular velocity $\vec{\omega}$ is a vector pointing along the rotation axis, and how $\vec{v} = \vec{\omega} \times \vec{r}$ gives velocity for any point in a rotating body. You'll also see why this relation is the backbone of rigid-body mechanics.

1. The Mental Model

A cross product takes two directions and spits out a third one perpendicular to both — it measures "how much turning" exists between them. Angular velocity is just that turning rate packaged as a vector: its direction tells you the axis, its length tells you the speed. Once you accept $\vec{\omega}$ as a vector, the velocity of any point on a spinning object falls out of a single cross-product formula. Rotation is fundamentally a vector quantity, not just a number of degrees per second.

2. The Core Material

2.1 The Vector (Cross) Product — Mechanics, Not Just Definition

For two vectors $\vec{a}$ and $\vec{b}$, the cross product is defined as

$$\vec{a} \times \vec{b} = |\vec{a}||\vec{b}|\sin\theta \, \hat{n}$$

where $\theta$ is the angle between them and $\hat{n}$ is the unit vector perpendicular to both, chosen by the right-hand rule: curl your right-hand fingers from $\vec{a}$ toward $\vec{b}$, and your thumb points along $\hat{n}$.

In Cartesian components, if $\vec{a} = (a_x, a_y, a_z)$ and $\vec{b} = (b_x, b_y, b_z)$:

$$\vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ a_x & a_y & a_z \\ b_x & b_y & b_z \end{vmatrix} = (a_yb_z - a_zb_y)\hat{i} - (a_xb_z - a_zb_x)\hat{j} + (a_xb_y - a_yb_x)\hat{k}$$

Three properties you must internalize, because they get used constantly in rigid-body problems:

1. Anti-commutativity: $\vec{a} \times \vec{b} = -\vec{b} \times \vec{a}$. Order matters — swap the vectors and the answer flips sign.
2. Zero for parallel vectors: if $\vec{a} \parallel \vec{b}$, then $\sin\theta = 0$, so $\vec{a} \times \vec{b} = \vec{0}$. This is why $\vec{\omega} \times \vec{r}$ vanishes for points on the rotation axis.
3. Magnitude = area: $|\vec{a} \times \vec{b}|$ equals the area of the parallelogram spanned by $\vec{a}$ and $\vec{b}$. This geometric fact is exactly why angular momentum $\vec{L} = \vec{r} \times \vec{p}$ and torque $\vec{\tau} = \vec{r} \times \vec{F}$ use this operation — they both care about the "lever arm" component perpendicular to $\vec{r}$.

2.2 Angular Velocity as a Vector

For a particle undergoing pure rotation about a fixed axis, we normally write $v = \omega r$ for speed, treating $\omega$ as a scalar. But this hides information: it doesn't tell you which way the rotation axis points, or which way the particle is actually moving (its velocity is tangential, not radial).

The fix is to define $\vec{\omega}$ as a vector:

• Direction: along the rotation axis, given by the right-hand rule (curl fingers in the direction of rotation, thumb gives $\vec{\omega}$).
• Magnitude: $|\vec{\omega}| = d\theta/dt$, the rate of change of the rotation angle.

With this definition, the velocity of any point in a rigid body rotating about a fixed axis, where $\vec{r}$ is the position vector from any point on the axis to the particle, is:

$$\vec{v} = \vec{\omega} \times \vec{r}$$

Why does this work? Decompose $\vec{r}$ into a component $\vec{r}_\parallel$ along the axis and $\vec{r}_\perp$ perpendicular to it. Since $\vec{\omega}$ is along the axis, $\vec{\omega} \times \vec{r}_\parallel = 0$ (parallel vectors), so only the perpendicular part matters:

$$\vec{v} = \vec{\omega} \times \vec{r}_\perp$$

This has magnitude $|\vec{\omega}||\vec{r}_\perp|\sin(90°) = \omega r_\perp$ — exactly the familiar $v = \omega r$ formula, but now $r_\perp$ is correctly the perpendicular distance from the axis (not the full distance to some arbitrary origin), and the direction comes out automatically tangential to the circle of rotation, which is exactly what a rotating particle's velocity should be.

This matters for rigid bodies with many particles: every particle in the body shares the same $\vec{\omega}$ (that's what "rigid" means — no relative rotation between parts), but each has its own $\vec{r}$ and hence its own $\vec{v}$. This single vector equation replaces having to track a different speed and direction for every particle separately.

flowchart TD
    A["Rigid body rotates about fixed axis"] --> B["Angular velocity vector ω<br/>direction: along axis (right-hand rule)<br/>magnitude: dθ/dt"]
    B --> C["Pick any particle at position r<br/>from a point on the axis"]
    C --> D["Decompose r into r_parallel (along axis)<br/>and r_perp (perpendicular to axis)"]
    D --> E["ω × r_parallel = 0<br/>(parallel vectors)"]
    D --> F["ω × r_perp ≠ 0<br/>tangential, magnitude ω·r_perp"]
    E --> G["v = ω × r<br/>(one formula, every particle)"]
    F --> G

2.3 Differentiating Rotating Vectors — Where $\vec{\omega}$ Comes From Naturally

There's a deeper reason $\vec{\omega} \times \vec{r}$ shows up everywhere: it's literally the time-derivative of any vector of fixed length that's rotating. Suppose $\vec{r}(t)$ has constant magnitude $r$ (a particle tracing a circle, or a unit vector attached to a spinning body). Then:

$$\frac{d\vec{r}}{dt} = \vec{\omega} \times \vec{r}$$

This is a general kinematic identity, not something you have to re-derive case by case. It's why, when you later study angular momentum of a system of particles,

$$\vec{L} = \sum_i m_i \vec{r}_i \times \vec{v}_i = \sum_i m_i \vec{r}_i \times (\vec{\omega} \times \vec{r}_i)$$

you can substitute $\vec{v}_i = \vec{\omega} \times \vec{r}_i$ directly — the whole machinery of moments of inertia and the inertia tensor is built on this one substitution. Angular velocity isn't a side quantity you compute after the fact; it's the generator of velocity for every particle in a rotating system.

3. Worked Example

Problem: A rigid wheel spins about the z-axis with angular velocity $\vec{\omega} = 4\hat{k}$ rad/s. A point P on the rim is located at $\vec{r} = (0.3\hat{i} + 0.4\hat{j})$ m at a given instant. Find the velocity of P, and verify its magnitude using $v = \omega r$.

Step 1 — Set up the cross product.
$$\vec{v} = \vec{\omega} \times \vec{r} = (4\hat{k}) \times (0.3\hat{i} + 0.4\hat{j})$$

Step 2 — Use the basic unit-vector cross products: $\hat{k} \times \hat{i} = \hat{j}$, and $\hat{k} \times \hat{j} = -\hat{i}$.

$$\vec{v} = 4(0.3)(\hat{k} \times \hat{i}) + 4(0.4)(\hat{k} \times \hat{j}) = 1.2\hat{j} + 1.6(-\hat{i}) = -1.6\hat{i} + 1.2\hat{j} \text{ m/s}$$

Step 3 — Check the magnitude.
$$|\vec{v}| = \sqrt{(-1.6)^2 + (1.2)^2} = \sqrt{2.56 + 1.44} = \sqrt{4} = 2 \text{ m/s}$$

Compare with $v = \omega r$: here $r = |\vec{r}| = \sqrt{0.3^2 + 0.4^2} = \sqrt{0.09+0.16} = \sqrt{0.25} = 0.5$ m, so $v = 4 \times 0.5 = 2$ m/s. Matches.

Step 4 — Check perpendicularity (sanity check). $\vec{v} \cdot \vec{r} = (-1.6)(0.3) + (1.2)(0.4) = -0.48 + 0.48 = 0$. Good — velocity is tangential, exactly perpendicular to the radius, as it must be for circular motion. This confirms the cross product correctly produced a physically sensible result without you having to reason geometrically about angles at all — the algebra did it for you.

4. Production Pitfalls & Best Practices

4.1 Real-World Best Practices

• Always use a right-handed coordinate system ($\hat{i} \times \hat{j} = \hat{k}$, cyclic). Mixing left- and right-handed conventions silently flips every $\vec{\omega}$ direction you compute.
• Use the determinant formula for cross products, not memorized "diagonal" tricks — it generalizes cleanly to any component values and avoids sign slips.
• Measure $\vec{r}$ from a point ON the rotation axis, not from an arbitrary origin, when applying $\vec{v} = \vec{\omega} \times \vec{r}$. Using the wrong reference point adds a spurious translational velocity.
• Check perpendicularity as a sanity test: $\vec{v} \cdot \vec{r}_\perp$ should be zero for pure rotation. If it isn't, you've made an arithmetic error or there's translational motion mixed in.
• Decompose $\vec{r}$ into parallel/perpendicular components before plugging into formulas when the axis isn't a coordinate axis — this avoids errors from blindly applying $v = \omega r$ with the wrong radius.
• Remember $\vec{\omega}$ is the same for every point in a rigid body — don't recompute a different $\omega$ per particle; only $\vec{r}$ (and hence $\vec{v}$) changes.

4.2 Common Bugs & Anti-Patterns

Bug 1 — Using the wrong hand rule / sign:

# WRONG: treating cross product as commutative
v = r × ω   # computed as if same as ω × r

# FIXED: order matters, a × b = -b × a
v = ω × r   # (not r × ω, which gives -v)

Bug 2 — Using full distance instead of perpendicular distance from axis:

# WRONG: using distance from arbitrary origin, not axis
r_total = sqrt(x**2 + y**2 + z**2)
v = omega * r_total

# FIXED: use only the component perpendicular to the rotation axis
r_perp = sqrt(x**2 + y**2)   # if axis is z
v = omega * r_perp

Bug 3 — Treating $\omega$ as scalar when direction matters (e.g., combining two rotations):

# WRONG: adding angular speeds as plain numbers
omega_total = omega1 + omega2   # ignores different axes

import numpy as np
# FIXED: add as vectors, since each has its own axis direction
omega1 = np.array([0, 0, 3.0])      # rotation about z
omega2 = np.array([2.0, 0, 0])      # rotation about x
omega_total = omega1 + omega2       # vector sum, valid for angular velocity

Bug 4 — Forgetting that $\frac{d\vec{r}}{dt} = \vec{\omega} \times \vec{r}$ only holds for constant-length vectors:

# WRONG: applying this identity to a vector whose length is changing
# (e.g., a particle spiraling outward)
v = omega_cross_r_formula(r_changing_length)

# FIXED: add the radial stretching term explicitly
# v = (dr/dt) r_hat + ω × r   when |r| is not constant

4.3 When to Use What (Decision Matrix

5. Now Try It

Exercise: A rigid disc rotates about the x-axis with $\vec{\omega} = 5\hat{i}$ rad/s. A point Q on the disc is at $\vec{r} = (0.1\hat{i} + 0.2\hat{j} + 0.0\hat{k})$ m from a point on the axis.

1. Compute $\vec{v} = \vec{\omega} \times \vec{r}$ by hand using the determinant method.
2. Verify $|\vec{v}| = \omega r_\perp$, where $r_\perp$ is the perpendicular distance from the x-axis (not the full $|\vec{r}|$).
3. Check that $\vec{v} \cdot \vec{r} = 0$.
4. Now suppose a second rotation $\vec{\omega}_2 = 3\hat{j}$ rad/s is added simultaneously (the disc also spins about y). Find the new total $\vec{\omega}_{total}$ and recompute $\vec{v}$ for the same point Q.

Success looks like: you get $\vec{v} = (0, 0, 1)$ m/s for part 1, confirm $|\vec{v}| = 5 \times 0.2 = 1$ m/s in part 2, get exactly zero in part 3, and correctly add the vectors (not scalars) in part 4 to find $\vec{v}_{total} = \vec{\omega}_{total} \times \vec{r}$.

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