# Rotations and Angular Velocity

A rotation of a vector is a change which only alters the direction, not the length, of a vector. A rotation consists of a rotation axis and a rotation rate. By taking the rotation axis as a direction and the rotation rate as a length, we can write the rotation as a vector, known as the angular velocity vector $\vec{\omega}$. We use the right-hand rule to describe the direction of rotation. The units of $\vec\omega$ are $\rm rad/s$ or ${}^\circ/s$.

Rotation axis: $\hat\imath$ $\hat\jmath$ $\hat{k}$ $\hat\imath + \hat\jmath$ $\hat\imath + \hat\jmath + \hat{k}$

Angular velocity vector $\vec\omega$. The direction of $\vec\omega$ is the axis of rotation, while the magnitude is the speed of rotation (positive direction given by the right-hand rule).

The Greek letter ω (lowercase omega) is the last letter of the Greek alphabet, leading to expressions such as “from alpha to omega” meaning “from start to end”. Omega literally means O-mega, meaning O-large, as capital Omega (written Ω) developed from capital Omicron (written Ο) by breaking the circle and turning up the edges. Omicron is literally O-micron, meaning O-small, and it is the ancestor of the Latin letter O that we use today in English.

 Α α alpha Ι ι iota Ρ ρ rho Β β beta Κ κ kappa Σ σ sigma Γ γ gamma Λ λ lambda Τ τ tau Δ δ delta Μ μ mu Υ υ upsilon Ε ε epsilon Ν ν nu Φ φ phi Ζ ζ zeta Ξ ξ xi Χ χ chi Η η eta Ο ο omicron Ψ ψ psi Θ θ theta Π π pi Ω ω omega

The Greek alphabet, shown above, was the first true alphabet, meaning that it has letters representing phonemes (basic significant sounds) and includes vowels as well as consonants. The Greek alphabet was was derived from the earlier Phoenician alphabet, which was probably the original parent of all alphabets. This shows that the idea of an alphabet is so non-obvious that it has only ever been invented once, and then always copied after that.

## Vector derivatives and rotations

If a unit vector $\hat{a}$ is rotating, then the angular velocity vector $\vec{\omega}$ is defined so that:

Derivative of unit vectors.

$\dot{\hat{a}} = \vec{\omega} \times \hat{a}$

Derivative of general vectors.

\begin{aligned} \dot{\vec{a}} = \underbrace{\dot{a} \hat{a}}_{\operatorname{Proj}(\dot{\vec{a}}, \vec{a})} + \underbrace{\vec{\omega} \times \vec{a}}_{\operatorname{Comp}(\dot{\vec{a}}, \vec{a})} \end{aligned}

Using the same approach as #rvc-em we write $\vec{a} = a\hat{a}$ and differentiate this and use rkr-ew to find: \begin{aligned} \dot{\vec{a}} &= \frac{d}{dt} \big( a \hat{a} \big) \\ &= \dot{a} \hat{a} + a \dot{\hat{a}} \\ &= \dot{a} \hat{a} + a (\vec\omega \times \hat{a}) \\ &= \dot{a} \hat{a} + \vec\omega \times (a \hat{a}) \\ &= \dot{a} \hat{a} + \vec\omega \times \vec{a}. \end{aligned} Comparing this to #rvc-em shows that the two components are the projection and the complementary projection, respectively.

Derivative of constant-length vectors.

$\dot{\vec{a}} = \vec{\omega} \times \vec{a} \qquad\text{if \vec{a} is constant length}$

This can be seen from the fact that $\dot{a} = 0$ if $a$ is constant (a fixed length vector), substituted into #rkr-ed.

## Rotations in 2D

In 2D the angular velocity can be thought of as a scalar (positive for counter-clockwise, negative for clockwise). This scalar is just the out-of-plane component of the full angular velocity vector. We can draw the angular velocity as either a vector pointing out of the plane, or as a circle-arrow in the plane, which is simpler for 2D diagrams.

Show:

Comparison of the vector and scalar representations of $\vec\omega$ for 2D rotations.

In 2D the angular velocity scalar $\omega$ is simply the derivative of the rotation angle $\theta$ in the plane:

Magnitude $\omega$ is derivative of angle $\theta$ in 2D.

$\omega = \dot\theta$

Take $\hat{a}$ to be a unit vector rotating in the 2D $\hat\imath$–$\hat\jmath$ plane, making an angle of $\theta$ with the $x$-axis, as in Figure #rkr-f2. Then: $\hat{a} = \cos\theta \,\hat\imath + \sin\theta \,\hat\jmath.$ Differentiating this expression gives: $\dot{\hat{a}} = -\sin\theta \, \dot\theta \,\hat\imath + \cos\theta \,\dot\theta \,\hat\jmath.$ We now consider an angular velocity vector $\vec\omega$. Because the rotation is in the $\hat\imath$–$\hat\jmath$ plane, the angular velocity vector must be in the $\hat{k}$ direction. Thus $\vec\omega = \omega \hat{k}$. Now we can compute the derivative of $\hat{a}$ using #rkr-ew, giving: \begin{aligned} \dot{\hat{a}} &= \vec\omega \times \hat{a} \\ &= \omega\hat{k} \times \big( \cos\theta \,\hat\imath + \sin\theta \,\hat\jmath \big) \\ &= \omega \cos\theta \,(\hat{k} \times \hat\imath) + \omega \sin\theta \,(\hat{k} \times \hat\jmath) \\ &= \omega \cos\theta \,\hat\jmath + \omega \sin\theta \,(-\hat\imath) \\ &= - \omega \sin\theta \,\hat\imath + \omega \cos\theta \,\hat\jmath. \end{aligned} Comparing this expression to the earlier one for $\dot{\hat{a}}$ we see that $\omega = \dot\theta$.

The right-hand rule convention for angular velocities means that counter-clockwise rotations are positive, just like the usual angle direction convention.

Angular directions have long been considered to have magical or spiritual significance. In Britain the counterclockwise direction was once known as widdershins, and it was considered unlucky to travel around a church in a widdershins direction.

Interestingly, right-handed people tend to naturally draw circles in a counterclockwise direction, and clockwise drawing in right-handed children is an early warning sign for the later development of schizophrenia [Blau, 1977].

### References

• T. H. Blau. Torque and schizophrenic vulnerability. American Psychologist, 32(12):997–1005, 1977. DOI: 10.1037/0003-066X.32.12.997.

## Rotations and vector “positions”

The fact that vectors don't have positions means that vector rotations are independent of where vectors are drawn, just like for derivatives.

Show:

Rotational motion of vectors which are drawn moving about. Note that the drawn position does not affect the angular velocity $\omega$ or the derivative vectors.

## Properties of rotations

Rotations are rigid transformations, meaning that they keep constant all vector lengths and all relative vector angles. These facts are reflected in the following results, which all consider two vectors $\vec{a}$ and $\vec{b}$ that are rotating with angular velocity $\vec\omega$.

Derivative of rotating vector is orthogonal.

$\dot{\vec{a}} \cdot \vec{a} = 0$

Using #rkr-el and the scalar triple product formula #rvi-es gives: \begin{aligned} \vec{a} \cdot \dot{\vec{a}} &= \vec{a} \cdot \big( \vec{\omega} \times \vec{a} \big) \\ &= \vec{\omega} \cdot \big( \vec{a} \times \vec{a} \big) \\ &= 0. \end{aligned}

Angle $\theta$ between rotating vectors is constant.

$\theta = \cos^{-1}\left(\frac{\vec{b} \cdot \vec{a}}{b a}\right) = \text{constant}$

We first consider the dot product $\vec{a} \cdot \vec{b}$ and show that this is not changing with time. We do this by using the scalar triple product formula #rvi-es to find: \begin{aligned} \frac{d}{dt} \big( \vec{a} \cdot \vec{b} \big) &= \dot{\vec{a}} \cdot \vec{b} + \vec{a} \cdot \dot{\vec{b}} \\ &= (\vec{\omega} \times \vec{a}) \cdot \vec{b} + \vec{a} \cdot (\vec{\omega} \times \vec{b}) \\ &= \vec{b} \cdot (\vec{\omega} \times \vec{a}) + \vec{b} \cdot (\vec{a} \times \vec{\omega}) \\ &= \vec{b} \cdot (\vec{\omega} \times \vec{a}) - \vec{b} \cdot (\vec{\omega} \times \vec{a}) \\ &= 0. \end{aligned} Now $\vec{a} \cdot \vec{b}$ is constant and the lengths $a$ and $b$ are constant, so the angle $\theta$ between the vectors must be constant.

Rotating vectors parallel to $\vec\omega$ are constant.

$\dot{\vec{a}} = 0 \qquad \text{if \vec{a} is rotating and parallel to \vec\omega}$

From #rkr-el we know that the derivative is $\dot{\vec{a}} = \vec\omega \times \vec{a},$ but the cross product is zero for parallel vectors, so this the derivative is zero.

## Rodrigues’ rotation formula

Rodrigues’ rotation formula gives an explicit formula for a vector rotated by an angle about a given axis.

Rodrigues’ rotation formula for $\vec{a}$ rotated by $\theta$ about $\hat{b}$.

$\operatorname{Rot}(\vec{a}; \theta, \hat{b}) = \vec{a} \cos\theta + (\hat{b} \times \vec{a}) \sin\theta + \hat{b} (\hat{b} \cdot \vec{a}) (1 - \cos\theta)$

Assume $\vec{a}$ is not parallel to $\hat{b}$. Then let $\vec{v} = \hat{b} \times \vec{a}$ and $\vec{u} = \vec{v} \times \vec{b}$, so $\hat{u}, \hat{v}, \hat{b}$ is a right-handed orthonormal basis. Take $\phi$ to be the angle between $\vec{a}$ and $\hat{b}$. Then we do a rotation by $\theta$ in the $\hat{u}$-$\hat{v}$ plane:

\begin{aligned} \vec{a} &= a \sin\phi \,\hat{u} + a \cos\phi \,\hat{b} \\ \operatorname{Rot}(\vec{a};\theta,\hat{b}) &= a \cos\theta \sin\phi \,\hat{u} + a \sin\theta \sin\phi \,\hat{v} + a \cos\phi \,\hat{b}.\end{aligned}

Now we want to convert from the $\hat{u},\hat{v},\hat{b}$ basis to write the rotated result in terms of $\vec{a}, (\hat{b} \times \vec{a}), \hat{b}$. To do this, we need to work out what $\hat{u},\hat{v},\hat{b}$ are in terms of these other vectors.

\begin{aligned} \hat{v} &= \frac{\hat{b} \times \vec{a}}{\|\hat{b} \times \vec{a}\|} = \frac{\hat{b} \times \vec{a}}{a \sin\phi} \\ \hat{u} &= \frac{\vec{v} \times \hat{b}}{\|\vec{v} \times \vec{b}\|} = \frac{\vec{v} \times \hat{b}}{a \sin\phi} = \frac{(\hat{b} \times \vec{a}) \times \hat{b}}{a \sin\phi} = \frac{\hat{b} \times (\vec{a} \times \hat{b})}{a \sin\phi} \\ &= \frac{\vec{a} - (\hat{b} \cdot \vec{a}) \hat{b}}{a \sin\phi} = \frac{1}{a \sin\phi} \vec{a} - \frac{\hat{b} \cdot \vec{a}}{a \sin\phi} \hat{b}.\end{aligned}

Substituting these into the rotated vector expression above gives

\begin{aligned} \operatorname{Rot}(\vec{a};\theta,\hat{b}) &= a \cos\theta \sin\phi \, \left( \frac{1}{a \sin\phi} \vec{a} - \frac{\hat{b} \cdot \vec{a}}{a \sin\phi} \hat{b} \right) \\ & \qquad + a \sin\theta \sin\phi \, \left( \frac{\hat{b} \times \vec{a}}{a \sin\phi} \right) + a \cos\phi \,\hat{b} \\ &= \cos\theta \,\vec{a} - \cos\theta \,(\hat{b} \cdot \vec{a}) \,\hat{b} + \sin\theta \,(\hat{b} \times \vec{a}) + a\cos\theta \,\hat{b} \\ &= \cos\theta \,\vec{a} + (1 - \cos\theta) (\hat{b} \cdot \vec{a}) \,\hat{b} + \sin\theta \,(\hat{b} \times \vec{a}).\end{aligned}