0
$\begingroup$

Suppose a system has angular velocity $\vec{\Omega} = \Omega_0\hat{z}$. Then according to Wikipedia, the velocity is given by $$ \vec{v} = \vec{\Omega}\times\vec{r} = \Omega_0r\sin\theta\hat{\phi} $$ where we used spherical coordinates. According to the same wikipedia page, we have that $$ \vec{\Omega} = \frac{\vec{r}\times\vec{v}}{r^2} = \frac{v}{r}\hat{r}\times\hat{v} = \Omega_0\sin\theta\hat{r}\times\hat{\phi} = -\Omega_0\sin\theta\hat{\theta} = \Omega_0\hat{z} - \Omega_0\cos\theta\hat{r} $$ Where am I wrong?

$\endgroup$
  • $\begingroup$ What do you mean where are you wrong? There is no question or statement to proof here... $\endgroup$ – QuIcKmAtHs Jan 16 '18 at 12:11
  • $\begingroup$ @XcoderX I assumed $\Omega = \Omega_0\hat{z}$ and got that $\Omega = \Omega_0\hat{z} - \Omega_0\cos\theta\hat{r}$ $\endgroup$ – JonTrav1 Jan 16 '18 at 12:14
  • $\begingroup$ You will only recover the components of $\vec{\Omega}$ that are perpendicular to $\vec{r}$. $\endgroup$ – ja72 Jan 16 '18 at 17:19
  • $\begingroup$ you are wrong that the first expression is for linear velocity and the second one is for rotational. Apples to oranges. Check also the cross product in spherical coordinates. $\endgroup$ – ja72 Jan 16 '18 at 17:45
0
$\begingroup$

Here is proof of the relationship $\vec{\omega} = \frac{\vec{r} \times \vec{v} }{ \| \vec{r} \|^2 }$.

The velocity of any point, located at $\vec{r}$, if the origin is at the rotation center is $\vec{v} = \vec{\omega} \times \vec{r}$ regardless of which coordinate represntation is used. Then

$$ \require{cancel} \vec{\omega} = \frac{\vec{r} \times \vec{v} }{ \| \vec{r} \|^2 } = \frac{\vec{r} \times (\vec{\omega} \times \vec{r}) }{ \| \vec{r} \|^2 } = \frac{\vec{\omega}(\vec{r} \cdot \vec{r}) - \vec{r}\cancel{(\vec{r}\cdot \vec{\omega})}}{\| \vec{r} \|^2} = \vec{\omega} $$

Use the triple vector product identity $a \times (b \times c) = b(a \cdot c) - c (a\cdot b)$ where $\cdot$ is the dot prodct, and $\times$ the cross product.

The fact that $(\vec{r} \cdot \vec{\omega}) = 0$ comes from the fact that only the perpendicular components of $\vec{\omega}$ to $\vec{r}$ contribute to linear velocity $v$.

The above is true regardless of the coordinate system used to describe it.

  • Example - $\vec{r} = \pmatrix{x \\y\\0}$, $\vec{\omega} = \pmatrix{0\\0\\ \Omega}$ and so $\vec{v} = \vec{\omega}\times\vec{r}= \pmatrix{-y \Omega \\ x \Omega \\ 0}$

$$ \vec{\omega} = \frac{ \pmatrix{x \\y\\0} \times \pmatrix{-y \Omega \\ x \Omega \\ 0}}{\| \pmatrix{x \\y\\0} \|^2} = \frac{ \pmatrix{0\\0\\ \Omega (x^2+y^2)}}{x^2+y^2} = \pmatrix{0\\0\\ \Omega} \; \checkmark$$

PS - Wikipedia isn't always correct. Always check derivations yourself.

$\endgroup$
  • $\begingroup$ Thank you for the comment. What happens if for example the rotation is of a solid sphere around an axis? Then $\vec{\omega} = \Omega\hat{z}$, and $r = (x,y,z)$ where $z$ is not necessarily zero. The velocity $\vec{v}$ will be the same, but when calculating $\vec{\omega}$ from the formula, we will get a value $\vec{\omega} = (-zx\Omega, -yz\Omega, (x^2+y^2)\Omega)\neq \Omega\hat{z}$ $\endgroup$ – JonTrav1 Jan 17 '18 at 8:13
-1
$\begingroup$

You should write it in the cylindrical coordinate(not spherical, because of rotation always around an axis).

$\endgroup$
  • $\begingroup$ Although helpful, this is more like a comment and not an answer. The OP expects to see the same results from different coordinate formulations, and none of that are shown here. $\endgroup$ – ja72 Jan 16 '18 at 16:40
  • $\begingroup$ I don't think $\Omega.r=0$ is correct in the spherical coordinate. because $\hat r$ is rotating in space and it can have an overlap with $\Omega$. but if you assume a cylindrical coordinate which its axis is along the $\Omega$ direction then the theory is consistent. $\endgroup$ – Rasoul-Ghadimi Jan 20 '18 at 13:05

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.