Feynman's description of angular momentum

[stingray191]

I recently watched the Feynman Messenger Lectures on project tuva. I hope some of you have watched the second video lecture in the collection or will watch it in order to help me with my following question . In this lecture Feynman talks about angular momentum in terms of the time rate of change of the area swept out by the radial vector. That is, angular momentum is proportional to the areal velocity. I've always had trouble intuitively understanding angular momentum but feynman's geometrical description made sense to me and seemed simple. My question is this: Is interpreting angular momentum as being equal to the areal velocity times 2m valid in general? And if yes, then why is this way not taught in physics courses? Any answers would be greatly appreciated.

 

[Pinu7]

No, because if L, the magnitude of angular momentum, is proportional to the areal velocity, then so is 2L, 3L, and so forth.

 

[stingray191]

I meant that does the equation dA/dt=L/2m hold in general where da/dt is the areal velocity.

 

[Born2bwire]

"Leaning Towah!"

I haven't gone through all of his lectures yet but they are very entertaining. I never knew his accent was that strong though.

 

[Pinu7]

Consider a function r(t) (assume it is smooth or at least twice differentiable since we are considering a natural trajectory)

The area the function sweeps out from t to t+dt is $$dA=.5r(t)\times r(t+dt)$$ this is because he area will be an infinitesimal triangle with the sides as vectors r(t) and r(t+dt) and it is well known that the area of a triangle defined by two "side" vectors is one half times the cross product of the vectors. See the graph below I made( dA is the yellow region)
http://img8.imageshack.us/img8/614/areavelocity.jpg
Therefore, dA is equal to

$$\frac{dA}{dt}=.5\frac{r(t)\times r(t+dt)}{dt}=.5r(t)\times \frac{r(t+dt)}{dt}=.5r(t)\times v(t)$$

Obviously dA/dt (areal velocity) is proportional to angular momentum.