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This is originally a problem in programming, but since almost no one on Stackoverflow know how to solve this I went here instead; http://stackoverflow.com/questions/23003612/javascript-angular-velocity-by-vector-2d

I want to convert X and Y velocities to angular velocity, this is the formula I am currently using to calculate the initial velocity by the x and y values and then turn it into angular velocity for my circle object:

Av = Sqrt(Vx^2 + Vy^2) / R

Angularvelocity = Squareroot of (Velocity x^2 + Velocity y^2) / Circle's radius

This is how it simulates in my programming: http://jsfiddle.net/yzb9P/2/ (Click to change the balls position)

Now since a square root can't be negative, this won't work when the ball is supposed to rotate anti-clockwise. So, I need a signed version of the initial velocity that also can be negative, how do I calculate that?

I've heard about that the Wedge product is working for this, and I've read many articles about it too, but I still don't understand how to use it, please help!

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Hi Murplx, and welcome to scicomp! On the stack exchange network, we strongly discourage cross-posting the exact same question. We advise that you delete all duplicate posts and only keep one post in the forum most relevant to your question. –  Paul Apr 12 at 21:18
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1 Answer 1

A cross product will tell you both the magnitude and the sign of your angular velocity. In general, angular velocity is defined by a vector as

$$\vec{\omega} = \dfrac{\vec{r}\times\vec{v}}{|\vec{r}|^2}$$

where $\vec{\omega}$ is the angular velocity, $\vec{r}$ is the vector from the center of rotation to the point under consideration, and $\vec{v}$ is the velocity vector.

In two dimensions this becomes

$$\omega = \dfrac{r_x v_y - r_y v_x}{r_x^2+r_y^2}$$

where a positive value corresponds to counter-clockwise rotation and negative means clockwise.

Edit:
After looking at your JSfiddle some more, it looks like your implementation has the more serious issue of giving rotation even when the ball hits straight down on a flat surface. If you have non-zero velocity, you're computing a non-zero angular velocity without regard to the vector directions involved. The cross product will eliminate this effect. I believe this is what you want: http://jsfiddle.net/yzb9P/7/

Edit 2:
If you want your ball to roll more realistically, there is another bug in your posted code. Try this version, the ball rolls down a slope without stopping unnaturally. I only edited the update() function. http://jsfiddle.net/yzb9P/8/

Edite 3:
A final update with improved implementation of elasticity for the ball: http://jsfiddle.net/yzb9P/11/

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THANK YOU VERY MUCH!!! Just one last question, goto line 188, and change the elasticy (0.8) to 0.1, why does the ball go through the line? –  Murplyx Apr 12 at 5:52
    
I do not have that problem so I have no idea what the issue might be. It still runs fine for me even if ball.e=1. I will say that setting 0<ball.e<1 adds some damping to the system which slows the ball down and makes things more stable. Also, this code does some "physics-like" things by using a very simple time stepping scheme and without properly solving the equations involved with elastic collisions. This may look ok most of the time, but will likely fail in certain cases. –  Doug Lipinski Apr 12 at 15:14
    
jsfiddle.net/yzb9P/9 This is how I want it to be, but with corrected and adjustable elasticy, the concept of elasticy in jsfiddle.net/yzb9P/8 is wrong, because the ball goes right through if ball.e = 0 –  Murplyx Apr 12 at 15:21
    
Is there anyway to correctly then calculate the elastic collision with ball.e? –  Murplyx Apr 12 at 15:23
    
Oh, I see. I misread your previous comment and used 1 instead of 0.1. This is related to the stability and accuracy of your time stepping method. How to fix that may be a much bigger question than I can answer here. Let me see what's really going on. –  Doug Lipinski Apr 12 at 15:35
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