# Recalculating angular velocity from velocity

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?

• What do you mean where are you wrong? There is no question or statement to proof here... – QuIcKmAtHs Jan 16 '18 at 12:11
• @XcoderX I assumed $\Omega = \Omega_0\hat{z}$ and got that $\Omega = \Omega_0\hat{z} - \Omega_0\cos\theta\hat{r}$ – JonTrav1 Jan 16 '18 at 12:14
• You will only recover the components of $\vec{\Omega}$ that are perpendicular to $\vec{r}$. – ja72 Jan 16 '18 at 17:19
• 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. – ja72 Jan 16 '18 at 17:45

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.

• 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 $\vec{\omega} = (-zx\Omega, -yz\Omega, (x^2+y^2)\Omega)\neq \Omega\hat{z}$ – JonTrav1 Jan 17 '18 at 8:13

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

• 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. – ja72 Jan 16 '18 at 16:40
• 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. – Rasoul-Ghadimi Jan 20 '18 at 13:05