# 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
$\vec r \times \vec{v}=\vec r \times [\vec{\omega} \times \vec{r}]$
Your formula is dimensionally incorrect.

Last edited: Nov 22, 2015
7. Nov 22, 2015

### Mister T

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.