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Direction of angular velocity

  1. Nov 21, 2015 #1
    Hi all, please help me here!
    Im confused, is direction of angular velocity perpendicular to the plane of motion, or along the plane of motion??

    From hyperphysics
    - http://hyperphysics.phy-astr.gsu.edu/hbase/rotq.html
    - https://www.dropbox.com/s/dpeisla93d6mv71/Screenshot_2015-11-22-09-51-40-1.png?dl=0

    And wikipedia
    https://www.dropbox.com/s/13g86di3prid46h/Screenshot_2015-11-22-09-37-03-1.png?dl=0

    Please also provide explanation if possible!
    Thank you!!!
     
  2. jcsd
  3. Nov 21, 2015 #2

    nrqed

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    It is perpendicular to the plane of motion (following the right hand rule). This is chosen so that the torque is given by
    ## \vec{\tau} = I \vec{\alpha} ## and ##\vec{\alpha} = \frac{d\vec{\omega}}{dt} ##.
     
  4. Nov 21, 2015 #3
    The rotation is in the plane of motion, but the angular velocity vector is perpendicular to that plane.

    ##\vec{\omega}=\vec{r} \times \vec{v}##
     
  5. Nov 21, 2015 #4

    ehild

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    It is the other way round: ##\vec{v}=\vec{\omega} \times \vec{r} ##

    See also
    https://en.wikipedia.org/wiki/Angular_velocity

    f79c5cb53b731791abb0dc6d12f63d94.png
     
  6. Nov 21, 2015 #5
    Oh, yeah! Sorry. Where did that come from?
     
  7. Nov 22, 2015 #6

    ehild

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    For the first, see it in a textbook, Landau's Mechanics, for example. For the second, go to the link https://en.wikipedia.org/wiki/Angular_velocity or expand the cross product
    ##\vec r \times \vec{v}=\vec r \times [\vec{\omega} \times \vec{r}]##
    Your formula is dimensionally incorrect.
     
    Last edited: Nov 22, 2015
  8. Nov 22, 2015 #7
    I meant, where did my mistake come from. As soon as you pointed it out I saw that of course it's dimensionally incorrect. I suppose I'm so used to dealing with ##\vec{L}=\vec{r} \times \vec{p}## and ##\vec{\tau}=\vec{r} \times \vec{F}## that it just came out of my brain that way.
     
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