Rotating systems

In terms of the problem you were given
[imath]\dot{r} = \dot{R} R^T r[/imath]

To simplify things let's take motion in the xy plane so we have
[imath]R = \left ( \begin{matrix} cos( \theta ) & - sin( \theta ) & 0 \\ sin( \theta ) & cos( \theta ) & 0 \\ 0 & 0 & 1 \end{matrix} \right )[/imath]

And we get
[imath]\dot{R} R^T = \left ( \begin{matrix} 0 & - \dot{ \theta } & 0 \\ \dot{ \theta } & 0 & 0 \\ 0 & 0 & 0 \end{matrix} \right )[/imath]

So
[imath]\dot{r} = \left ( \begin{matrix} 0 & - \dot{ \theta } & 0 \\ \dot{ \theta } & 0 & 0 \\ 0 & 0 & 0 \end{matrix} \right ) r = \left ( \begin{matrix} 0 & - \dot{ \theta } & 0 \\ \dot{ \theta } & 0 & 0 \\ 0 & 0 & 0 \end{matrix} \right ) \left ( \begin{matrix} x \\ y \\ 0 \end{matrix} \right ) = \left ( \begin{matrix} 0 & -y & 0 \\ x & 0 & 0 \\ 0 & 0 & 0 \end{matrix} \right ) \dot{ \theta }[/imath]

Notice that [imath]\omega[/imath] here will be in the z direction, so notice that
[imath]r \times \omega = \left ( \begin{matrix} x \\ y \\ 0 \end{matrix} \right ) \times \left ( \begin{matrix} 0 \\ 0 \\ \dot{ \theta } \end{matrix} \right ) = \left ( \begin{matrix} 0 & -y & 0 \\ x & 0 & 0 \\ 0 & 0 & 0 \end{matrix} \right ) \dot{ \theta } = \dot{R} R^T r = \dot{r}[/imath]

or [imath]v = r \times \omega[/imath].

I don't know why the author didn't finish the job properly. All he did was draw a parallel to the cross product and say "Here's an example and voila!" It was all there but for some reason he simply didn't pick the right example.

-Dan
 
topsquark said:
Like I said, you did what the problem as the author said. But the units are all wrong. [imath] [xyz]^T[/imath] cannot be an angular speed because it has the wrong units. This is not a derivation of [imath]\dot{r} = \omega \times r[/imath].

Loosely speaking: Let there be an axis of rotation and take polar coordinates r, [imath]\phi[/imath] in the plane perpendicular to the rotation axis. let there be an angle of [imath]\theta[/imath] between v and r. (v is the total velocity, all we want is the tangential component.)

Then [math]\vec{\omega } = \omega \hat{k} = \dfrac{d \phi }{dt} \hat{k} = \dfrac{v ~ sin( \theta )}{r} \hat{k} = \dfrac{v r ~ sin( \theta )}{r^2} \hat{k} = \dfrac{ \textbf{v} \times \textbf{r} }{r^2}[/math]
and from there we get [imath]\omega = v_t \times r[/imath]

-Dan
Click to expand...
Is \(\displaystyle \omega = v_t \times r \) correct?

One more question, \(\displaystyle \angle \phi\) lies between \(\displaystyle \overrightarrow{\omega} and \overrightarrow{r}\) as given in the following picture? doesn't it?

1655142375663.png
 
Last edited:
Dhamnekar Winod said:
Is \(\displaystyle \omega = v_t \times r \) correct?

One more question, \(\displaystyle \angle \phi\) lies between \(\displaystyle \overrightarrow{\omega} and \overrightarrow{r}\) as given in the following picture? doesn't it?

View attachment 33053
Click to expand...
No! It is not correct. I wrote that at 1AM and didn't check what I was saying. After all those times I told you your equation didn't have the right units I didn't check to make sure mine did! Everything I stated was correct except for that last line, which I corrected with the post I made this morning. [imath]\omega = v \times r[/imath] implies the unit equation [imath]1/s = m/s \cdot m = m^2/s[/imath], which is distinctly untrue.

In my derivations I skipped out of spherical polar coordinates and went to 2D polar coordinates. You can do this problem for an arbitrary 3D motion but I think it's easier to work with the derivatives of the unit vectors rather than the transformation directly.

-Dan
 
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