Velocities in inertial and rotating frames of reference

[ryan88]

Hi,

I have a couple of questions about velocities in inertial and rotating frames of reference, related by the following equation:

\mathbf{v_i} \ \stackrel{\mathrm{def}}{=}\ \frac{d\mathbf{r}}{dt} = <br /> \left( \frac{d\mathbf{r}}{dt} \right)_{\mathrm{r}} + <br /> \boldsymbol\Omega \times \mathbf{r} = <br /> \mathbf{v}_{\mathrm{r}} + \boldsymbol\Omega \times \mathbf{r}

1. \mathbf{v_i} and \mathbf{v_r} both state which frame of reference they are measured in, however \mathbf{r} does not. Is this supposed to be in the inertial or rotating frame of reference?
2. If I use the equation to find the velocity in the rotating frame, does this mean that the value is represented in the rotating frame of reference? Or is it that the magnitude of that velocity is correct, but it still needs to be rotated to the rotating frame of reference?

Thanks,

Ryan

 

[ryan88]

Ok, after posting this I found the following post in the 'Similar Threads' section at the bottom of the page:

https://www.physicsforums.com/showpost.php?p=1650790&postcount=3

It gives a very unambiguous derivation of the equation that I posted earlier.

 

[ryan88]

After taking a closer look at the post I linked to, I have another question. I thought that the time derivative of a rotation matrix was given by:

\frac{\mathrm{d}R}{\mathrm{d}t} = \tilde{\omega}R

However, in his post, D H states:

\mathbf T&#039;_{R\to I} = \mathbf T_{R\to I}\mathbf X(\mathbf \omega)

Since matrix products are non commutative, doesn't this make the following incorrect?

\mathbf q&#039;_I = <br /> \mathbf T_{R\to I}(\mathbf X(\mathbf \omega)\mathbf q_R + \mathbf q&#039;_R) <br /> = <br /> \mathbf T_{R\to I}(\mathbf \omega\times\mathbf q_R + \mathbf q&#039;_R)