Rotating systems

In terms of the problem you were given
r˙=R˙RTr\dot{r} = \dot{R} R^T r

To simplify things let's take motion in the xy plane so we have
R=(cos(θ)sin(θ)0sin(θ)cos(θ)0001)R = \left ( \begin{matrix} cos( \theta ) & - sin( \theta ) & 0 \\ sin( \theta ) & cos( \theta ) & 0 \\ 0 & 0 & 1 \end{matrix} \right )

And we get
R˙RT=(0θ˙0θ˙00000)\dot{R} R^T = \left ( \begin{matrix} 0 & - \dot{ \theta } & 0 \\ \dot{ \theta } & 0 & 0 \\ 0 & 0 & 0 \end{matrix} \right )

So
r˙=(0θ˙0θ˙00000)r=(0θ˙0θ˙00000)(xy0)=(0y0x00000)θ˙\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 }

Notice that ω\omega here will be in the z direction, so notice that
r×ω=(xy0)×(00θ˙)=(0y0x00000)θ˙=R˙RTr=r˙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}

or v=r×ωv = r \times \omega.

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

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

Then ω=ωk^=dϕdtk^=v sin(θ)rk^=vr sin(θ)r2k^=v×rr2\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}
and from there we get ω=vt×r\omega = v_t \times r

-Dan
Is ω=vt×r\displaystyle \omega = v_t \times r correct?

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

1655142375663.png
 
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Is ω=vt×r\displaystyle \omega = v_t \times r correct?

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

View attachment 33053
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. ω=v×r\omega = v \times r implies the unit equation 1/s=m/sm=m2/s1/s = m/s \cdot m = m^2/s, 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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