Angular Velocity Direction: Perpendicular or Parallel to the Plane of Motion?

[Boomzxc]

Hi all, please help me here!
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!

 

[nrqed]

Hi all, please help me here!
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!

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 $\vec{\alpha} = \frac{d\vec{\omega}}{dt}$.

 

[Mister T]

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

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

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

 

[ehild]

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

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

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

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

 

[Mister T]

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

 

[ehild]

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

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.

 

[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.