Derivative of rotation matrix produces angular velocity vector

Question

I am trying to understand how and why the angular velocity vector is related to the derivative of a rotation matrix.

I understand that for a rotation matrix $R(t)$ one has $R(t)R(t)^T = I $, so $\dot{R}R^T + R\dot{R}^T = 0$ which shows that the matrix $R(t)\dot{R}^T = \hat{\Omega}$ is skew symmetric. I understand that then we can write $\hat{\Omega}$ in vector form $\omega$ such that we get for all vectors $v$, $\hat{\Omega}v = \omega \times v$.

But what else do we need to show that that $\omega$ is the angular velocity vector? Can we assert that $\omega$ will be unique?

Answer 1

The transformation matrix $~\mathbf R~$ transformed the components of a vector from body system (Index B) to inertial system (Index I)

$$(\mathbf u)_I=\mathbf R\,(\mathbf u)_B$$

take the time derivative you obtain the velocity components $~\mathbf v~$ in inertial system

$$(\mathbf v)_I=\mathbf{\dot{R}}\,(\mathbf u)_B$$

with

$$\mathbf{\dot{R}}=\,\mathbf R\,\left[ \begin {array}{ccc} 0&-\omega_{{z}}&\omega_{{y}} \\ \omega_{{z}}&0&-\omega_{{x}}\\ -\omega_{{y}}&\omega_{{x}}&0\end {array} \right]$$

$\Rightarrow$

$$(\mathbf v)_I=\mathbf R\,\left(\mathbf\omega\times\mathbf u\right)_B $$

or

$$\underbrace{\mathbf R^T\,(\mathbf v)_I}_{(\mathbf v)_B}=\left(\mathbf\omega\times\mathbf u\right)_B$$

thus you obtain

$$\mathbf v=\mathbf\omega\times\mathbf u$$

---

with : \begin{align*} &\mathbf{\dot{R}}=\mathbf{R}\,\mathbf{\tilde{\omega}}\\ &\Rightarrow\\ &\mathbf{\tilde{\omega}}=\mathbf{R}^T\,\mathbf{\dot{R}}\quad \text{and}\quad \mathbf{\tilde{\omega}}+\mathbf{\tilde{\omega}}^T=\mathbf{0}\\ &\Rightarrow\\ &\mathbf{R}^T\,\mathbf{\dot{R}}+\,\mathbf{\dot{R}}^T\,\mathbf{R}=\mathbf{0}\\ &\text{or}\\ &\mathbf{R}^T\,\mathbf{R}=\mathbf{I} \end{align*}

Answer 2

Take any basis vector $\hat{u}$ that is riding on a rotating coordinate frame and find as far as the components as measured by the inertial frame you have

$$ \frac{\rm d}{{\rm d}t} \hat{u} = \vec{\omega} \times \hat{u} \tag{1}$$

Now recognize that the rotation matrix $\mathbf{R}$ just has the three basis vectors of the body frame in its columns

$$ \mathbf{R} = \left[ \begin{array}{c|c|c} \hat{i} & \hat{j} & \hat{k} \end{array} \right] \tag{2}$$

and the relationship

$$ \frac{\rm d}{{\rm d}t} \mathbf{R} = \vec{\omega} \times \mathbf{R} \tag{3}$$

is just a shortcut for

$$ \begin{aligned} \frac{\rm d}{{\rm d}t} \hat{i} & = \vec{\omega} \times \hat{i} \\ \frac{\rm d}{{\rm d}t} \hat{j} & = \vec{\omega} \times \hat{j} \\ \frac{\rm d}{{\rm d}t} \hat{k} & = \vec{\omega} \times \hat{k} \\ \end{aligned} $$

So your question is really how do you prove (1)?

My favorite method is stating that under rotations the length of the basis vector must remain one $\| \hat{u} \| = 1$, or in derivative form

$$ \frac{\rm d}{{\rm d}t} \sqrt{ \hat{u} \cdot \hat{u} } = 0 $$ which quickly simplifies to $$ \hat{u} \cdot \frac{\rm d}{{\rm d}t} \hat{u} = 0 $$

which is interpreted as the derivative must be perpendicular to the direction (pretty intuitive) and that one (or only way) to ensure this is to use a cross product to define the derivative as the cross product is guaranteed to be perpendicular to both arguments.

$$ \hat{u} \cdot ( \vec{\omega} \times \hat{u}) = 0 $$

The above leads to

$$ \frac{\rm d}{{\rm d}t} \vec{r} = \vec{v} = \vec{\omega} \times \vec{r} $$ for $\vec{r}$ riding on the body and $\vec{\omega}$ is the angular velocity vector.