Angular velocity and those damn radians

[Nikitin]

I don't get radians. I understand it's a constant, ie: circumference/radius = 2pi because 2*r*pi equals circumference
BUt I don't get why it is so damn handy for measuring angles!

THe formula for angular velocity is \omega*r=(2pi/t)*r where omega=change in angle divided by time.

what the? 2pi/t equals \omega?? 2pi=6.28? Why would \omega*t*r equal 2pi*r aka the circumference of the circle? I must be misunderstanding this very badly... Though it is pretty late here :/

Please explain all this, slowly and clearly...

 

[256bits]

Well, there are 2pi radians in a whole circle.
And, the circumference, C, of the circle is of length 2pi*r.
So you automatically have a length by multiplying radains by the radius,
The radian is dimensionless.
Much easier than using degrees for angles.

secondly
If the circle, or more specificially, a spot on the circle has rotated 2pi radians in one second, then naturally omega=2 pi/t. If t changes then so does omega.

 

[Nikitin]

oops stupid me for some reason i wrote diameter when i meant circumference :facepalm:

thanks for the reply 256bits, i just read it and things are much more clear. but people, please keep them coming so i can get more information

 

[256bits]

You wrote circumference. I was just writing dowm my train of thought. ;facepalm undone.

 

[Nikitin]

well i edited the post afterwards :p

anyway i understand it all, thanks for your help. I just thought it was weird that multiplying an angle equivalent to 360 degrees with the radius of a circle would get you the circle's circumference. but once i stopped thinking in degrees i think it's easier to understand now =)

but thanks anyway

 

[HallsofIvy]

You are making a mistake when you say "I understand it's a constant". That would be like saying "inches" is a constant. "radians" are not numbers they are units of measurement.

 

[lyy1992]

Well, actually radian is another form of expressing angles, just like degrees and both type of expressions are convertable, just like converting km/h to m/s and vice versa.

Because of ω=change in angular displacement=angle traveled/ time, then ω=2Pi/t for standard expression.

The circumference you meant is actually the distance traveled by the rotating circle.
For ease, ω is look like velocity or speed, but is in angular motion, and that is why ω is called angular velocity.
ω*t = angular displacement,
ω*t*r= linear displacement

And the relationship between linear motion and angular motion is differed by r.

Hope this can give you the right concept.

 

[Nikitin]

yes, I understand it now on a superficial level...

but i still get this weird feeling when i multiply angular displacement with the radius and then i get the linear displacement...

From a purely algebraic standpoint this makes perfect sense but I just can't seem to picture this in my mind, like I can with almost all other concepts.

 

[BruceW]

I think everyone has trouble forming a picture in their minds of non-cartesian coordinate systems. For example, this is the equation for acceleration in the spherical polar coordinate system:
\vec{a} = ( \ddot{r} - r {\dot{\theta}}^2 - r {\dot{\phi}}^2 sin^2(\theta) ) \hat{r} + ( r \ddot{\theta} + 2 \dot{r} \dot{\theta} - r {\dot{\phi}}^2 sin(\theta) cos(\theta) ) \hat{\theta} + (r \ddot{\phi} sin(\theta) + 2 \dot{r} \dot{\phi} sin(\theta) + 2r \dot{\theta} \dot{\phi} cos(\theta) ) \hat{\phi}
I seriously doubt anyone has an intuitive picture in their heads for this equation.

Edit: The way I try to put myself at ease is by thinking: its just the cartesian system, but written using different variables. I can convert back to cartesian anytime I want, and the calculation can be done using either coordinate system.

 

[lyy1992]

yes, I understand it now on a superficial level...

but i still get this weird feeling when i multiply angular displacement with the radius and then i get the linear displacement...

From a purely algebraic standpoint this makes perfect sense but I just can't seem to picture this in my mind, like I can with almost all other concepts.

Then you will get used to it.