TITLE: How to calculate velocity vector from scalar angular velocity and position vector in 2D?
QUESTION [1 upvotes]: I would like to know, if I have an angular velocity as scalar, how can I calculate the velocity vector.
I know that the product of angular velocity and the length of the distance gives the speed, but I would like to know vector of that.
I tried solving this problem many times, but honestly, I dont even know what I did, Im complete beginner, and couldnt really find a solution, or answer.

REPLY [1 votes]: I assume you want to find the linear velocity if you know the angular speed. You need to realize that all the points on a rotating rigid body have the same angular speed and angular velocity but different linear speeds and velocities.  A point on a rigid body that describes a circle of twice the radius as another point will have twice the linear speed because it covers twice as much distance in the same amount of time.  The linear speed $v$ is related to the angular speed $\omega$ by $v=\omega r$.
The corresponding vector equation is the cross product $\vec v=\vec \omega \times \vec r$. The direction of the velocity vector is perpendicular to the plane defined by $\vec \omega$ and $\vec r$ using the right hand rule.  Put the thumb of your right hand perpendicular to the plane of rotation and match the circulation of your four fingers to the circulation of the rotating body. The velocity vector at a point is tangent to the circle at that point.
Example
The plane of rotation is $xy$ and have position vector $\vec r$ in that plane.  The angular velocity is perpendicular to the plane.
$\vec \omega=\{0,0,\omega\}$ and $\vec r=\{r \cos\theta,r\sin\theta,0\}$.  Then $$\vec v=\vec \omega\times \vec r=\{0,0,\omega\}\times \{r \cos\theta,r\sin\theta,0\}=\omega r\{-\sin\theta,\cos\theta,0\}.$$Both the velocity and position vectors are 2D.
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