# Express Rotation of a body around a rotation axis in rotation about interial axes

I know that a body is rotating around its body axis (let us define it as $$y_{body}$$ with the angular velocity of $$\vec{\omega}(t)=0.05 rad/s$$. The rotation axis ($$y_{body}$$) is the interial y-axis which is rotated about 30° around the global x-axis and could be represented for example by the vector $$\begin{pmatrix}0 \\ cos(30°) \\ sin(30°) \end{pmatrix}$$ in the inertial coordinate system which is just defined as: $$x=\begin{pmatrix}1 & 0 & 0 \end{pmatrix}$$ $$y=\begin{pmatrix}0 & 1 & 0 \end{pmatrix}$$ $$z=\begin{pmatrix}0 & 0 & 1 \end{pmatrix}$$

Now I am looking for an angular velocity vector expressed in the inertial coordinate system which gives me the same rotation as in the body fixed coordinate system.

I have searched here in the forum and found something like this: Link: Applying angular velocity to a rotation matrix which could be quite useful.

So it would be: $$\vec \omega(t) \times \vec{r}(t) = W(t) \vec{r}(t)= W(t) = \begin{pmatrix} 0 & -\omega_z(t) & \omega_y(t) \\ \omega_z(t) & 0 & -\omega_x(t) \\ -\omega_y(t) & \omega_x(t) & 0 \\ \end{pmatrix} \vec{r}(t)$$

I think the vector $$\vec{r}(t)$$ should then just be the rotation vector of the body $$\vec{r}(t)=\begin{pmatrix}0 \\ cos(30°) \\ sin(30°) \end{pmatrix}$$

And in $$W(t)$$ all entries are 0 except of $$\omega_y(t)= 0.05 rad/s$$

So now it should be just multiplying those matrix and vector?

Is this correct or am I deeply confusing something :-D ?

I relly would appreciate your help!!

Thank you :-)

Each vector $$\boldsymbol{g}$$ that is riding on a rotating coordinate frame, whose basis vectors form the columns of 3×3 rotation matrix $$\mathbf{R}$$ is transformed from local to interial frame with

$$\boldsymbol{g}_0 = \mathbf{R}\, \boldsymbol{g} \tag{1}$$

this is a matrix vector calculation. As an example using $$\boldsymbol{g} = \pmatrix{g_x \\ g_y \\ g_z}$$ and a single rotation about the y-axis with angle $$\theta$$

$$\pmatrix{ g_x \cos \theta + g_z \sin \theta \\ g_y \\ -g_x \sin \theta + g_z \cos \theta} = \begin{bmatrix} \cos\ \theta & 0 & \sin \theta \\ 0 & 1 & 0 \\ -\sin \theta & 0 & \cos \theta \end{bmatrix} \pmatrix{g_x \\ g_y \\ g_z}$$

Now take the derivative of (1) to find the rate of change of $$\boldsymbol{g}_0$$

$$\tfrac{\rm d}{{\rm d}t} \boldsymbol{g}_0 = \left( \tfrac{\partial}{\partial t} \mathbf{R} \right) \boldsymbol{g} = \boldsymbol{\omega} \times \mathbf{R}\, \boldsymbol{g} \tag{2}$$

With the above example

$$\tfrac{\rm d}{{\rm d}t} \boldsymbol{g}_0 = \begin{bmatrix} 0 & -\omega_z & \omega_y \\ \omega_z & 0 & -\omega_x \\ -\omega_y & \omega_x & 0 \end{bmatrix} \begin{bmatrix} \cos\ \theta & 0 & \sin \theta \\ 0 & 1 & 0 \\ -\sin \theta & 0 & \cos \theta \end{bmatrix} \pmatrix{g_x \\ g_y \\ g_z}$$

which is a direct result of the rule of differentiation on a rotating reference frame

$$\boxed{ \tfrac{\rm d}{{\rm d}t} \boldsymbol{g}_0 = \tfrac{\partial }{\partial t} \boldsymbol{g}_0 + \boldsymbol{\omega} \times \boldsymbol{g}_0 } \tag{3}$$

In most cases $$\boldsymbol{g}_0$$ is fixed with time, so $$\tfrac{\partial }{\partial t}\boldsymbol{g}_0 = \boldsymbol{0}$$, as it only varies due to the rotation of the frame.

The next step is to find the angular velocity from a sequence of rotations. If $$\mathbf{R}_1 = \mathrm{rot}(\hat{j}, \theta)$$ and $$\mathbf{R}_2 = \mathrm{rot}(\hat{k}, \psi)$$ for example, the derivative of

$$\boldsymbol{g}_0 = \mathbf{R}_1 \mathbf{R}_2 \boldsymbol{g}$$

\begin{aligned} \boldsymbol{\omega} \times \boldsymbol{g}_0 & = \left( \tfrac{\partial}{\partial t} \mathbf{R}_1 \right) \mathbf{R}_2 \boldsymbol{g} + \mathbf{R}_1 \left( \tfrac{\partial}{\partial t} \mathbf{R}_2 \right) \boldsymbol{g} \\ & = \left( \hat{j} \dot{\theta} \times \mathbf{R}_1 \right) \mathbf{R}_2 \boldsymbol{g} + \mathbf{R}_1 \left( \hat{k} \dot{\psi} \times \mathbf{R}_2 \right) \boldsymbol{g} \\ & = \left( \hat{j}\dot \theta + \mathbf{R}_1 \hat{k} \dot \psi \right) \times \mathbf{R}_1 \mathbf{R}_2 \boldsymbol{g} \end{aligned}
Here $$\hat{i} = \pmatrix{1&0&0}$$, $$\hat{j} = \pmatrix{0&1&0}$$ and $$\hat{k} = \pmatrix{0&0&1}$$.
or $$\boldsymbol{\omega} = \hat{j}\dot \theta + \mathbf{R}_1 \hat{k} \dot \psi \tag{4}$$
So (4) gives us the kinematics of a 2DOF joint based on a rotation about y of $$\theta$$ followed by a rotation about z of $$\psi$$.