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Say I have the angular velocity vector of a body as a function of time. How can I determine the axis of rotation/location of the body?

we have the equation:

$\frac{d\vec{r}}{dt}=\vec{\omega}(t)\times \vec{r}$

can we have somthing like:

$\frac{d\vec{\omega}}{dt}=\vec{r'}(t)\times \vec{\omega}$

where $\vec{r'}$ points towards $\vec{r}$ or equals to it?

EDIT: I now ask about axis of rotatiob ot about the location of the body. If im not mistaken there are sitations where the angular velocity does not point in the direction of the axis of rotation.

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  • $\begingroup$ The angular velocity vector itself points along the axis of rotation... $\endgroup$ – lemon Jun 19 '16 at 7:41
  • $\begingroup$ @lemon i've edited my question $\endgroup$ – proton Jun 19 '16 at 7:58
  • $\begingroup$ To your last sentence: that's only if the centre of mass of the rotating object is not at the origin of your coordinate system. $\endgroup$ – lemon Jun 19 '16 at 8:03
  • $\begingroup$ @proton your forumla is wrong. $\endgroup$ – Yashas Jun 19 '16 at 8:32
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Your formula itself is wrong.

$$ \frac{d\vec{r}}{dt} = \vec{v} = \omega \times \vec{r}$$

Angular Velocity ($\omega$) is an axial vector whereas velocity is a polar vector.

Axial vectors are vectors whose directions oriented in a non-intuitive way. For example, when you rotate a disc by a certain angle ($d\phi$), the $d\phi$ vector lies along the axis of rotation.

A polar vector is a vector whose direction makes sense to a layman. For example, the direction of the velocity vector is in the direction of motion of the body.

Lets begin with rotation in a plane. You have the intuition for polar vectors, so if you know the direction of the velocity vector and the radius vector, you can use the right-hand rule to find the direction of the angular velocity vector.

$$\vec{v} = \omega \times \vec{r}$$

If you do it correctly, you will find the angular velocity vector to be oriented along the axis of rotation.

For cases where you have the rotation in more than one dimension. You can split the angular velocity vector into two mutually perpendicular components. These components are not along the axis of rotation.

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  • $\begingroup$ first, i thank you for the answer. second, given $\vec{r}$ and its derivative i can find $\vec{\omega}$. is there a way to find $\vec{r}$ givven $\vec{\omega}$? $\endgroup$ – proton Jun 19 '16 at 8:39
  • $\begingroup$ For uniform circular motion, you will have to integrate $\vec{\omega}$ to obtain $\theta$ as a function of time using which you can obtain the position vector. In non-uniform cases, you will almost never come across such situations. $\endgroup$ – Yashas Jun 19 '16 at 8:44
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You need two things. The angular velocity vector $\vec{\omega}$ at some time frame, and the linear velocity vector at some reference point $\vec{v}$.

The axis of rotation is located at a point relative to the reference point

$$ \vec{r} = \frac{\vec{\omega} \times \vec{v}}{ \| \vec{\omega} \|^2} \tag{1}$$

You can apply this at every instance.

See related post1 post2 and post3.

The proof is easy but it requires the use of the vector triple product identity $a \times (b \times c) = b (ac) - c (ab)$. Substitude above $\vec{v} = \vec{r} \times \vec{\omega}$ and carry out simplifications.

Simple example

The center of mass of an object is moving with $\vec{v} = \pmatrix{5 & -1 &0}$ and the rotational vector is $\vec{\omega} = \pmatrix{0 & 0 & 2}$ at some instant.

The point on the axis of rotation closest to the center of mass is located at $$\vec{r} = \frac{\pmatrix{0&0&2} \times \pmatrix{5&-1&0}}{ \| \pmatrix{0&0&2} \|^2 } = \frac{ \pmatrix{2 & 10 & 1} }{2^2} = \pmatrix{0.5 & 2.5 & 5}$$

To verify find the velocity at the center of mass with

$$ \vec{v} = \vec{\omega} \times (-\vec{r} ) = \vec{r} \times \vec{\omega} = \pmatrix{0.5 & 2.5 & 0} \times \pmatrix{0&0&2} = \pmatrix{5&-1&0} \;\checkmark $$

NOTE: That the axis of rotation is any point along the line, which includes $\vec{r}$ and all points parallel to $\vec{\omega}$.

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  • $\begingroup$ I added an example to show that my answer gives the correct answer. $\endgroup$ – ja72 Aug 10 at 16:45

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