# Velocity in rotating frame projected onto axis in inertial frame

I want to look at the projection of the velocity of a particle in the rotating frame onto an axis in the inertial frame as a function of time. For example, I am calculating

\begin{equation} \frac{d\boldsymbol{r}}{dt}=\frac{d\boldsymbol{r}'}{dt}+\boldsymbol{\Omega}\times\boldsymbol{r}' \end{equation}

Where $\boldsymbol{r}$ is the position vector of the particle, the un-primed is the inertial frame and the prime the rotating. $\Omega$ is the angular velocity of the rotating frame with respect to the inertial, and can also be a function of time. Then I would like to look at the $z$-axis component of $\frac{d\boldsymbol{r}}{dt}$ which is $\frac{dz}{dt}=\frac{d\boldsymbol{r}}{dt}\bullet\hat{\boldsymbol{k}}$. Then from above I have

\begin{equation} \frac{dz}{dt}=\left(\frac{d\boldsymbol{r}'}{dt}+\boldsymbol{\Omega}\times\boldsymbol{r}'\right)\bullet\hat{\boldsymbol{k}} \end{equation}

Which now means I need to look at the dot products of $\hat{\boldsymbol{i}}'\bullet\hat{\boldsymbol{k}}\quad, \hat{\boldsymbol{j}}'\bullet\hat{\boldsymbol{k}}\quad$ and $\hat{\boldsymbol{k}}'\bullet\hat{\boldsymbol{k}}$. I am confused about how to proceed with doing this?

• – Frobenius Sep 4 '18 at 11:28

To define the orientation of the rotating frame you need a 3×3 orthonormal rotation matrix $\mathbf{R}$. This will give you the local to the inertial frame transformation of vectors using a matrix-vector product.

$$\mathbf{c} = \mathbf{R}\, \mathbf{c}'$$

Now the columns of $\mathbf{R}$ are exactly the dot product you described, or the components of the local coordinates expressed in the inertial frame

$$\mathbf{R} = \left[ \begin{array}{c|c|c} \boldsymbol{\hat{\imath}}' & \boldsymbol{\hat{\jmath}}' & \boldsymbol{\hat{k}}' \end{array} \right] = \left[ \begin{array}{c|c|c} \boldsymbol{\hat{\imath}}' \cdot \boldsymbol{\hat{\imath}} & \boldsymbol{\hat{\jmath}}' \cdot \boldsymbol{\hat{\imath}} & \boldsymbol{\hat{k}}'\cdot \boldsymbol{\hat{\imath}} \\ \boldsymbol{\hat{\imath}}' \cdot \boldsymbol{\hat{\jmath}} & \boldsymbol{\hat{\jmath}}' \cdot \boldsymbol{\hat{\jmath}} & \boldsymbol{\hat{k}}'\cdot \boldsymbol{\hat{\jmath}} \\ \boldsymbol{\hat{\imath}}' \cdot \boldsymbol{\hat{k}} & \boldsymbol{\hat{\jmath}}' \cdot \boldsymbol{\hat{k}} & \boldsymbol{\hat{k}}'\cdot \boldsymbol{\hat{k}} \end{array} \right] = \left[ \matrix{ \hat{\imath}_x' & \hat{\jmath}_x' & \hat{k}_x' \\ \hat{\imath}_y' & \hat{\jmath}_y' & \hat{k}_y' \\ \hat{\imath}_z' & \hat{\jmath}_z' & \hat{k}_z' } \right]$$

Now specifically, the kinematics of a rotating frame are derived from the transformation of coordinates:

$$\mathbf{r} = \mathbf{r}_0 + \mathbf{R}\, \mathbf{r}'$$

Here $\mathbf{r}_0$ is the origin of the rotating frame in inertial coordinates, $\mathbf{R}$ the 3×3 orientation matrix, $\mathbf{r}'$ the position vector in the rotating frame, and $\mathbf{r}$ the position vector in the inertial frame.

Take the derivative of the above to get

$$\mathbf{v} = \mathbf{v}_0 + \boldsymbol{\Omega} \times \mathbf{R}\, \mathbf{r}' + \mathbf{R}\, \frac{\partial \mathbf{r}'}{\partial t}$$

This hinges on the property that the rotation matrix represents 3 unit length vectors and hence $\dot{\mathbf{R}} = \boldsymbol{\Omega} \times \mathbf{R}$.

• I'm still a bit confused. Is your primed the local or inertial frame? I'm also still unsure how to use this information to obtain the velocity of $\dot{z}$(the z-component of velocity in the inertial frame). Do I need to calculate the components of $\boldsymbol{R}$? – jamie1989 Sep 5 '18 at 14:32
• The z-direction of velocity is the 3rd component of $\mathbf{v}$. Yes, you need to define $\mathbf{R}$ based on axis-angle, euler-angles or quaternions. The primed unit vectors $\mathbf{\hat{\imath}}'$ etc contain the components of the rotating coordinate system expressed in the inertial system. – ja72 Sep 5 '18 at 17:00
• Ah, so i've been calculating the velocity in the inertial frame by just plugging in $\boldsymbol{r}'$ without the $\boldsymbol{R}$ in yours, which I can now see is wrong? What would you say is the easiest way to go about obtaining $\boldsymbol{R}$ axis-angle, euler-angles or quaternions? I'd like to keep $\boldsymbol{\Omega}$ time dependent in general does this change $\boldsymbol{v}$ at all? – jamie1989 Sep 5 '18 at 17:43
• If $\mathbf{\Omega}(t)$ is a function of time, then $\mathbf{v}$ isn't affected, but $\dot{\mathbf{v}}$ is. If you have a single fixed rotation axis, then use the axis angle method. If you have a free body, then euler-angles is a good start. It is best to work with a common inertial reference frame early on in order to avoid confusions. That is why I am starting with inertial position $\mathbf{r}$ first and deriving everything from there using time derivatives. – ja72 Sep 5 '18 at 17:48
• The rotation axis $\hat{\boldsymbol{n}}$ where $\boldsymbol{\Omega}=\Omega\hat{\boldsymbol{n}}$ is not fixed, it is a function of time so I'm looking at Euler angles. It doesn't seem easy to obtain these angles from a time dependent rotation axis, am I over complicating this problem or is it just not that trivial? – jamie1989 Sep 5 '18 at 19:12