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Angular velocity and those damn radians

  1. Jul 26, 2011 #1
    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 [itex]\omega[/itex]*r=(2pi/t)*r where omega=change in angle divided by time.

    what the? 2pi/t equals [itex]\omega[/itex]?? 2pi=6.28? Why would [itex]\omega[/itex]*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...
     
    Last edited: Jul 26, 2011
  2. jcsd
  3. Jul 26, 2011 #2
    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.
     
  4. Jul 26, 2011 #3
    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
     
  5. Jul 26, 2011 #4
    You wrote circumference. I was just writing dowm my train of thought. ;facepalm undone.
     
  6. Jul 27, 2011 #5
    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
     
  7. Jul 27, 2011 #6

    HallsofIvy

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    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.
     
  8. Jul 27, 2011 #7
    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.
     
  9. Jul 27, 2011 #8
    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.
     
  10. Jul 27, 2011 #9

    BruceW

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    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:
    [tex] \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} [/tex]
    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.
     
    Last edited: Jul 27, 2011
  11. Jul 28, 2011 #10
    Then you will get used to it.
     
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