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Angular velocity - spherical coordinates

  1. Dec 5, 2005 #1
    how do you express angular velocity in spherical coordinates?
    like the earth rotates with constant speed, so the direction of the angular velocity vector is out the north pole.
    if it was spherical coordinates , how do you specify that direction?
    i know that z = r cos theta so
    [tex] \hat{k} = r cos \theta \hat{e_{\theta}} [/tex] ????
    the problem is:
    for a particle moving in a horizontal plane on the surface of the earth show that the magnitude of the coriolis force is independent of the direction of motion of the particle.
    so the question i asked isn't the HW problem and a direct answer would be much appreciated.
    also, this is sort of a dumb question, but does the [itex] \theta , \phi [/itex] plane count as a plane? if it doesn't, why wouldn't it?
    Last edited: Dec 5, 2005
  2. jcsd
  3. Dec 5, 2005 #2
    z hat, whats a zhat?

    [tex] z= \rho cos(\phi) [/tex]
    Last edited: Dec 5, 2005
  4. Dec 5, 2005 #3


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    Generally, you would express any vector in spherical coordinates in the form of

    [tex]\vec v = v_r \hat e_r + v_{\theta} \hat e_{\theta} + v_{\phi} \hat e_{\phi}[/tex]
  5. Dec 5, 2005 #4
    i know how a velocity is expressed in spherical co-ordinates.
    i just don't know how to say the direction of the angular velocity vector.
    magnitude of the coriolis force =
    [tex] F_{cor} = | 2 m ( \vec{\omega} \times \vec{v'} ) | [/tex]
    where [tex] \vec{v'} [/tex] is the velocity in the non-inertial frame.
    wht's rho? am i using the wrong notation for this too?
    [tex] ( \hat{e_r} , \hat{e_{\theta}} , \hat{e_{\phi}} ) [/tex] is i think the most common way to express it, but it's not so type friendly. I generally use:
    [tex] ( \hat{r} , \hat{\theta} , \hat{\phi} ) [/tex] for spherical coordinates, but that's besides the point.
    now that the confusion to nomenclature has been cleared up:

    i need to know the angular velocity of the earth in spherical coordinates please.
    is it just [tex] \omega \hat{k} = \omega r cos \theta \hat{e_{\theta}} [/tex]
    Last edited: Dec 5, 2005
  6. Dec 5, 2005 #5
    my rho is your r.
  7. Dec 5, 2005 #6
    from your post the #2 i am inferring that i am correct in my assumption that

    [tex] \omega \hat{k} = \omega r cos \theta \hat{e_{\theta}} [/tex]
  8. Dec 5, 2005 #7


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    For the Coriolis force, [itex]\omega[/itex] is in the [itex]\hat z[/itex] direction. It may be helpful to write that as [itex]\hat z = -\sin \theta \hat e_{\theta} + \cos \theta \hat e_r[/itex]. Then if you write the velocity using spherical unit vectors you can evaluate the cross product noting that [itex]\hat e_r \times \hat e_{\theta} = \hat e_{\phi}[/itex] etc.
  9. Dec 5, 2005 #8
    thank you.
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