Direction of angular velocity

1. Nov 21, 2015

Boomzxc

Im confused, is direction of angular velocity perpendicular to the plane of motion, or along the plane of motion??

From hyperphysics
- http://hyperphysics.phy-astr.gsu.edu/hbase/rotq.html
- https://www.dropbox.com/s/dpeisla93d6mv71/Screenshot_2015-11-22-09-51-40-1.png?dl=0

And wikipedia
https://www.dropbox.com/s/13g86di3prid46h/Screenshot_2015-11-22-09-37-03-1.png?dl=0

Please also provide explanation if possible!
Thank you!!!

2. Nov 21, 2015

nrqed

It is perpendicular to the plane of motion (following the right hand rule). This is chosen so that the torque is given by
$\vec{\tau} = I \vec{\alpha}$ and $\vec{\alpha} = \frac{d\vec{\omega}}{dt}$.

3. Nov 21, 2015

Mister T

The rotation is in the plane of motion, but the angular velocity vector is perpendicular to that plane.

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

4. Nov 21, 2015

ehild

It is the other way round: $\vec{v}=\vec{\omega} \times \vec{r}$

https://en.wikipedia.org/wiki/Angular_velocity

5. Nov 21, 2015

Mister T

Oh, yeah! Sorry. Where did that come from?

6. Nov 22, 2015

ehild

For the first, see it in a textbook, Landau's Mechanics, for example. For the second, go to the link https://en.wikipedia.org/wiki/Angular_velocity or expand the cross product
$\vec r \times \vec{v}=\vec r \times [\vec{\omega} \times \vec{r}]$
I meant, where did my mistake come from. As soon as you pointed it out I saw that of course it's dimensionally incorrect. I suppose I'm so used to dealing with $\vec{L}=\vec{r} \times \vec{p}$ and $\vec{\tau}=\vec{r} \times \vec{F}$ that it just came out of my brain that way.