Rotating systems

[Win_odd Dhamnekar]

What are the meanings of highlighted expressions in the following image?



What is the meaning of highlighted expression and how the 3 × 3 square matrix is derived in the following image?





Is \(\displaystyle \dot{R}R^T\) equal to derivative of R times transpose of R? Is R(t) vector?

If any member knows the answers to these questions may reply.

 

[blamocur]

[imath]R(t)[/imath] is a 3x3 matrix corresponding to the rotation transform at time [imath]t[/imath]. .......................................edited
[imath]r(t)[/imath] is the result of the rotation of [imath]\rho[/imath] at time [imath]t[/imath]
Because rotation matrices are unitary we have [imath]R^{-1} = R^T[/imath], which means that [imath]\dot r = \dot R \rho = \dot R R^{-1} r = \dot R R^T r[/imath]
Since [imath]R R^T=I[/imath], i.e. constant we have [imath]\frac{d}{dt} (RR^T) = 0 = \dot R R^T + R \dot R^T[/imath].
Can you figure out the rest?

 

[blamocur]

[imath]r(t)[/imath] is a 3x3 matrix corresponding to the rotation transform at time [imath]t[/imath].
[imath]r(t)[/imath] is the result of the rotation of [imath]\rho[/imath] at time [imath]t[/imath]
Because rotation matrices are unitary we have [imath]R^{-1} = R^T[/imath], which means that [imath]\dot r = \dot R \rho = \dot R R^{-1} r = \dot R R^T r[/imath]
Since [imath]R R^T=I[/imath], i.e. constant we have [imath]\frac{d}{dt} (RR^T) = 0 = \dot R R^T + R \dot R^T[/imath].
Can you figure out the rest?

Sorry for the typo : [imath]R(t)[/imath] is a 3x3 matrix, not [imath]r(t)[/imath] ...............................fixed
Which way is the corner?

 

[topsquark]

Sorry for the typo : [imath]R(t)[/imath] is a 3x3 matrix, not [imath]r(t)[/imath]
Which way is the corner?

Just follow the sound of Mr. Khan's voice.

-Dan

 

[Deleted member 4993]

Just follow the sound of Mr. Khan's voice.

-Dan

Shhhhh - don't tell Steve - I ran away from the corner. It was dark and scary and I kept hearing "Khaaaaaaan".....

 

[Steven G]

You can't leave the corner until your time is up. Now you have to do double the original time. At least blamocur will be there to keep you safe.

 

[Win_odd Dhamnekar]

[imath]R(t)[/imath] is a 3x3 matrix corresponding to the rotation transform at time [imath]t[/imath]. .......................................edited
[imath]r(t)[/imath] is the result of the rotation of [imath]\rho[/imath] at time [imath]t[/imath]
Because rotation matrices are unitary we have [imath]R^{-1} = R^T[/imath], which means that [imath]\dot r = \dot R \rho = \dot R R^{-1} r = \dot R R^T r[/imath]
Since [imath]R R^T=I[/imath], i.e. constant we have [imath]\frac{d}{dt} (RR^T) = 0 = \dot R R^T + R \dot R^T[/imath].
Can you figure out the rest?

[math] \rho =\begin{bmatrix} x \\y \\z \end{bmatrix}, R(t) = Q(z,\alpha) = \begin{bmatrix} \cos{\alpha} & -\sin{\alpha} & 0 \\ \sin{\alpha} & \cos{\alpha} & 0 \\ 0 & 0 & 1\end{bmatrix}[/math]Now, [math]r(t) = R(t)\rho = \begin{bmatrix} -y\sin{\alpha} + x\cos{\alpha} \\ y\cos{\alpha} + x\sin{\alpha} \\ z \end{bmatrix}[/math]
At any instant,as observed in the fixed system,
\(\displaystyle \frac{dr}{dt} = \dot{R}\rho + R\dot{\rho}\) but the second term is zero since we assume \(\displaystyle \rho\)] to be constant, so, we have
\(\displaystyle \frac{dr}{dt} = \dot{R}R^T r = \begin{bmatrix} -y\cos{\alpha}- x\sin{\alpha} \\ -y\sin{\alpha} + x\cos{\alpha} \\ 0 \end{bmatrix}, \dot{R}R^T + R \dot{R^T} = 0 \)

\(\displaystyle \dot{R}R^T = \begin{bmatrix} 0 & -1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix}\) is anti-symmetric.

Now, how to prove \(\displaystyle (v)\dot{r}= \omega \times r\) using cross product operator \(\displaystyle \omega \times\) where \(\displaystyle \omega = [xyz]^T ?Note: \omega= \)Angular velocity of point P

 

[topsquark]

Now, how to prove \(\displaystyle (v)\dot{r}= \omega \times r\) using cross product operator \(\displaystyle \omega \times\) where \(\displaystyle \omega = [xyz]^T ?Note: \omega= \)Angular velocity of point P

I didn't check the rest but for this last part the units don't match. So you shouldn't be able to.

-Dan

 

[blamocur]

[math] \rho =\begin{bmatrix} x \\y \\z \end{bmatrix}, R(t) = Q(z,\alpha) = \begin{bmatrix} \cos{\alpha} & -\sin{\alpha} & 0 \\ \sin{\alpha} & \cos{\alpha} & 0 \\ 0 & 0 & 1\end{bmatrix}[/math]Now, [math]r(t) = R(t)\rho = \begin{bmatrix} -y\sin{\alpha} + x\cos{\alpha} \\ y\cos{\alpha} + x\sin{\alpha} \\ z \end{bmatrix}[/math]
At any instant,as observed in the fixed system,
\(\displaystyle \frac{dr}{dt} = \dot{R}\rho + R\dot{\rho}\) but the second term is zero since we assume \(\displaystyle \rho\)] to be constant, so, we have
\(\displaystyle \frac{dr}{dt} = \dot{R}R^T r = \begin{bmatrix} -y\cos{\alpha}- x\sin{\alpha} \\ -y\sin{\alpha} + x\cos{\alpha} \\ 0 \end{bmatrix}, \dot{R}R^T + R \dot{R^T} = 0 \)

\(\displaystyle \dot{R}R^T = \begin{bmatrix} 0 & -1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix}\) is anti-symmetric.

Now, how to prove \(\displaystyle (v)\dot{r}= \omega \times r\) using cross product operator \(\displaystyle \omega \times\) where \(\displaystyle \omega = [xyz]^T ?Note: \omega= \)Angular velocity of point P

Looks you are only considering one special case, i.e., rotation about z-axis - why?

 

[Win_odd Dhamnekar]

Looks you are only considering one special case, i.e., rotation about z-axis - why?

At the beginning the author says ' Consider a body which is rotating with the constant angular velocity \(\displaystyle \omega\) about some axis passing through the origin.'s'.
Author didn't say 'about all axis passing through the origin'. That's why I considered only one case.

 

[blamocur]

At the beginning the author says ' Consider a body which is rotating with the constant angular velocity \(\displaystyle \omega\) about some axis passing through the origin.'s'.
Author didn't say 'about all axis passing through the origin'. That's why I considered only one case.

I thought 'some' means 'arbitrary', and the following statements seem to confirm that. On the other hand, I just realized that you are not solving a problem but trying to digest some text from a book. Looking at special cases in this situation can actually be quite useful, so my question does not make sense.

 

[blamocur]

[math] \rho =\begin{bmatrix} x \\y \\z \end{bmatrix}, R(t) = Q(z,\alpha) = \begin{bmatrix} \cos{\alpha} & -\sin{\alpha} & 0 \\ \sin{\alpha} & \cos{\alpha} & 0 \\ 0 & 0 & 1\end{bmatrix}[/math]Now, [math]r(t) = R(t)\rho = \begin{bmatrix} -y\sin{\alpha} + x\cos{\alpha} \\ y\cos{\alpha} + x\sin{\alpha} \\ z \end{bmatrix}[/math]
At any instant,as observed in the fixed system,
\(\displaystyle \frac{dr}{dt} = \dot{R}\rho + R\dot{\rho}\) but the second term is zero since we assume \(\displaystyle \rho\)] to be constant, so, we have
\(\displaystyle \frac{dr}{dt} = \dot{R}R^T r = \begin{bmatrix} -y\cos{\alpha}- x\sin{\alpha} \\ -y\sin{\alpha} + x\cos{\alpha} \\ 0 \end{bmatrix}, \dot{R}R^T + R \dot{R^T} = 0 \)

\(\displaystyle \dot{R}R^T = \begin{bmatrix} 0 & -1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix}\) is anti-symmetric.

Now, how to prove \(\displaystyle (v)\dot{r}= \omega \times r\) using cross product operator \(\displaystyle \omega \times\) where \(\displaystyle \omega = [xyz]^T ?Note: \omega= \)Angular velocity of point P

What is [imath]\dot R[/imath] in your special case?

 

[Win_odd Dhamnekar]

What is [imath]\dot R[/imath] in your special case?

I got as author said \(\displaystyle (v)\dot{r}= \omega \times r \). So, I tagged this question as 'SOLVED'.

 

[topsquark]

I got as author said \(\displaystyle (v)\dot{r}= \omega \times r \). So, I tagged this question as 'SOLVED'.

I went back and scanned the whole thread. You correctly solved the problem, so far as I can see. But you should know that this is not a derivation of [imath]\dot{r} = \omega \times r[/imath]. The form [imath] [ xyz ]^T [/imath] is not an angular velocity. Whoever wrote the problem was using an analogy that shows the motion is obtained using the cross product, nothing more.

-Dan

 

[Win_odd Dhamnekar]

I went back and scanned the whole thread. You correctly solved the problem, so far as I can see. But you should know that this is not a derivation of [imath]\dot{r} = \omega \times r[/imath]. The form [imath] [ xyz ]^T [/imath] is not an angular velocity. Whoever wrote the problem was using an analogy that shows the motion is obtained using the cross product, nothing more.

-Dan

\(\displaystyle \omega= [0,0,1], \overrightarrow{r}\) have been given in #7.

\(\displaystyle \therefore [0,0,1] \times [-y\sin{\alpha}+ x \cos{\alpha}, y\cos{\alpha} + x\sin{\alpha},z ]= [-y\cos{\alpha} - x\sin{\alpha}, -y\sin{\alpha} + x\cos{\alpha}, 0] \)

So, we get \(\displaystyle \dot{r} =\omega \times r\) what author said/got.

 

[topsquark]

\(\displaystyle \omega= [0,0,1], \overrightarrow{r}\) have been given in #7.

\(\displaystyle \therefore [0,0,1] \times [-y\sin{\alpha}+ x \cos{\alpha}, y\cos{\alpha} + x\sin{\alpha},z ]= [-y\cos{\alpha} - x\sin{\alpha}, -y\sin{\alpha} + x\cos{\alpha}, 0] \)

So, we get \(\displaystyle \dot{r} =\omega \times r\) what author said/got.

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

 

[Win_odd Dhamnekar]

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

I don't understand how did you compute \(\displaystyle \frac{d\phi}{dt}= \frac{v \sin{\theta}}{r}\). When I glanced at your answer, what I understood is depicted pictorially below:



Now, would you explain me how did you compute \(\displaystyle \frac{d\phi}{dt}= \frac{v \sin{\theta}}{r} ?\)

 

[Win_odd Dhamnekar]

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

I agree that \(\displaystyle \frac{d\phi}{dt}\hat{k} =\frac{v \sin{\theta}}{r}\hat{k}\).

But finally, how did you compute from\(\displaystyle \frac{d\phi}{dt}\hat{k} =\frac{v \times r}{r^2}\hat{k}\) to \(\displaystyle \omega = v \times r\)

 

[Deleted member 4993]

I agree that \(\displaystyle \frac{d\phi}{dt}\hat{k} =\frac{v \sin{\theta}}{r}\hat{k}\).

But finally, how did you compute from\(\displaystyle \frac{d\phi}{dt}\hat{k} =\frac{v \times r}{r^2}\hat{k}\) to \(\displaystyle \omega = v \times r\)

\(\displaystyle ω=v×r\displaystyle \omega = v \times rω=v×r\)

That equation is INCORRECT - because dimensions do not balance. Look at response #8

 

[topsquark]

But finally, how did you compute from\(\displaystyle \frac{d\phi}{dt}\hat{k} =\frac{v \times r}{r^2}\hat{k}\) to \(\displaystyle \omega = v \times r\)

Yeah, this is why you should never do Physics at 1AM! Sorry about that.

You are looking for an expression for v. So
[imath]\omega = \dfrac{v \times r}{r^2}[/imath]

(I'm dropping the "t" in the v_t here. Note that v and r are at right angles.)

[imath]r \times \omega = r \times \dfrac{v \times r}{r^2} = \dfrac{1}{r^2} r \times (v \times r)[/imath]

[imath]r \times \omega = \dfrac{1}{r^2} ( ( r \cdot r ) v + (r \cdot v) r )[/imath]

Again, v and r are perpendicular so [imath]r \cdot v = 0[/imath] and [imath]r \cdot r = r^2[/imath], so we have
[imath]r \times \omega = \dfrac{1}{r^2} r^2 v = v[/imath]

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

-Dan