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]

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

 

[blamocur]

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

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

 

[topsquark]

Sorry for the typo : $R(t)$ is a 3x3 matrix, not $r(t)$
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]

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

$$\rho =\begin{bmatrix} x \\y \\z \end{bmatrix}, R(t) = Q(z,\alpha) = \begin{bmatrix} \cos{\alpha} & -\sin{\alpha} & 0 \\ \sin{\alpha} & \cos{\alpha} & 0 \\ 0 & 0 & 1\end{bmatrix}$$Now, $$r(t) = R(t)\rho = \begin{bmatrix} -y\sin{\alpha} + x\cos{\alpha} \\ y\cos{\alpha} + x\sin{\alpha} \\ z \end{bmatrix}$$
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]

$$\rho =\begin{bmatrix} x \\y \\z \end{bmatrix}, R(t) = Q(z,\alpha) = \begin{bmatrix} \cos{\alpha} & -\sin{\alpha} & 0 \\ \sin{\alpha} & \cos{\alpha} & 0 \\ 0 & 0 & 1\end{bmatrix}$$Now, $$r(t) = R(t)\rho = \begin{bmatrix} -y\sin{\alpha} + x\cos{\alpha} \\ y\cos{\alpha} + x\sin{\alpha} \\ z \end{bmatrix}$$
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]

$$\rho =\begin{bmatrix} x \\y \\z \end{bmatrix}, R(t) = Q(z,\alpha) = \begin{bmatrix} \cos{\alpha} & -\sin{\alpha} & 0 \\ \sin{\alpha} & \cos{\alpha} & 0 \\ 0 & 0 & 1\end{bmatrix}$$Now, $$r(t) = R(t)\rho = \begin{bmatrix} -y\sin{\alpha} + x\cos{\alpha} \\ y\cos{\alpha} + x\sin{\alpha} \\ z \end{bmatrix}$$
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 $\dot R$ in your special case?

 

[Win_odd Dhamnekar]

What is $\dot R$ 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 $\dot{r} = \omega \times r$. The form $[ xyz ]^T$ 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 $\dot{r} = \omega \times r$. The form $[ xyz ]^T$ 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. $[xyz]^T$ cannot be an angular speed because it has the wrong units. This is not a derivation of $\dot{r} = \omega \times r$.

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

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

-Dan

 

[Win_odd Dhamnekar]

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

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

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

-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. $[xyz]^T$ cannot be an angular speed because it has the wrong units. This is not a derivation of $\dot{r} = \omega \times r$.

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

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

-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
$\omega = \dfrac{v \times r}{r^2}$

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

$r \times \omega = r \times \dfrac{v \times r}{r^2} = \dfrac{1}{r^2} r \times (v \times r)$

$r \times \omega = \dfrac{1}{r^2} ( ( r \cdot r ) v + (r \cdot v) r )$

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

Therefore $v = r \times \omega$.

-Dan