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Coriolis force

  1. Dec 13, 2006 #1
    I need some help,...

    On the surface of the earth a cartesian coordinate system _E will be installed at the latitude ϕ. The axes of the coordinate system are aligned as following:

    _x1-axis: points up
    _x2-axis: points north
    _x3-axis: points east

    The underscores in front of the variables are a reminder that these variables relate to the non-inertalsystem _E.

    The exercise I should solve asks the question:
    How does the coriolis force depend on the latitude?

    In the book I read, the coriolis force was deduced to the following formula:

    _Fc = -2*m*(w x _r')
    (two times the mass multiplied with the cross product of the angular velocity with the derivative of the position vektor, where the position vector is relating to the _E-system)

    Now my problem:
    Isn't there something missing in this formula? Because as I understand the coriolis force, the distance of the motion to the axis of rotation is of importance when calculating the coriolis force. But in this formula above, the position vector relates to the _E-system, so the distance to the axis of rotation is nowhere to be found in the formula.

    The result as it reads in the book:
    _Fc = -2*m*w*(_x3'*cos(ϕ) - _x2'*sin(ϕ), _x1'*sin(ϕ), -x1'*cos(ϕ))
    _r = (_x1, _x2, _x3)

    Can somebody explain to how I must understand this exercise. I'm pretty sure the book is correct. What am I missing? I really would appreciate any help...

    sorry, missed the right subforum for this thread
    Last edited: Dec 13, 2006
  2. jcsd
  3. Dec 13, 2006 #2


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    Your understanding is not correct. The equation is. It does not matter how far you are from the axis of rotation. What matters is the angle between ω and v, which does happen to depend on the distance from the axis on the surface of the earth for most directions of velocity.
  4. Dec 13, 2006 #3
    As far as I can remember that formula is correct, it assumes that the position is measured as a latitude (see polar co-ordinates) and that the velocity has no upward component
  5. Dec 14, 2006 #4
    Thanks for the answers!

    So the coriolis force does not depend on the distance to the axis of rotation. But I do not really undestand it, why does it depend on the distance to the axis on the surface of the earth? What does that mean exactly?

    I have almost exactly the same problem with the centrifugal force, for which the formula
    _Fz = -m*(w x (w x _r))
    was deduced(in relation to the exercise in my first post).
    Now here again there is no distance to the axis of rotation. How can this be explained?

    Sorry if these questions are kind of silly, but I just want to understand it correctly.
  6. Dec 14, 2006 #5


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    For many problems of interest v is parallel to the surface of the earth, so the angle between v and ω is different at different lattitudes for all directions of v except for east and west.

    The centripetal force does depend on the distance from the axis. It is in the ω x r product. ω and r have constant magnitude, but the angle between them depends on lattitude. Figure out what the angle dependence is and you will find a product of factors that equals the distance from the axis.
  7. Dec 14, 2006 #6
    Thanks. Before I think about your advice more carefully, just two short questions:
    1-Why should r have a constant magnitude? It depends on the time, doesen't it?
    2-When you talk about the position vector r, you talk about it in relation to the _E-system, the system installed on the surface of the earth?
  8. Dec 14, 2006 #7


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    I should have been more explicit in saying that I was comparing to cross product at different points on the surface of the earth. As long as you are on the surface, r has constant magnitude, or very nearly so on the scale of the earth's radius. That does not mean the vector r is constant, only that its length is fixed.

    I do not think the r is what you think it is. It is the positon vector from the center of the earth to the particle. It can be expressed in different coordinate systems, but it is the same vector. Take a look at the first 8 pages of this

  9. Dec 15, 2006 #8
    Therein is the problem. What exactly is the vector r or _r? How can you express the same position vector in two different coordinate systems when their origins to not coincide?

    I read the link you gave me, and its more or less similar to the theorie within the book I read and where the exercise is from, but its not as confusing!

    I should have been more carefully when describing my problem! Let's see if I can do it a little better!

    The equation of motion in my book is:

    m*_r'' = F - r0'' - m*w x (w x _r) - m*(w' x _r) - 2*m*(w x _r')

    F = some real force (like gravity)
    _Fz = - m*w x (w x _r) = centrifugal force
    _Fc = - 2*m*(w x _r') = coriolis force
    _Fd = - m*(w' x _r) = some other apparent force which I dont know, but is of no relevance if w=const

    r0'' = the position vector from the origin of a fixed E coordinate system (e.g. at the center of the earth) to the accelerated _E-System at the surface of the earth.

    Now I would think that the coriolis force (or centrifugal force) would need the vector r0 in someway to be correct. Because I think, that the vector r is not the same as the vector _r. r = r0 + _r. Hence the coriolis force should be:

    _Fc = - 2*m*(w x (r0 + _r)')

    But in my book the formula isn't used that way, and thats exactly what's confusing me... I hope you understand what Im trying to say!
  10. Dec 16, 2006 #9


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    If I understand correctly, you (or your text) want to move the origin of the rotating coordinate system to the surface of the earth. In doing so you have written the position vector from the earth's center to the position of a particle as the sum of two vectors. That is OK. In the centripetal force term, the vector will be the sum of two vectors. Your equation is missing an m that I have added in red

    m*_r'' = F - m*r0'' - m*ω x (ω x _r) - m*(ω' x _r) - 2*m*(ω x _r')

    In the constant ω case, the first two terms after F in this equation are in fact

    - m*ω x ω x( r0 + _r)


    ω x ω x r0 is the centripetal acceleration r0'' (the only acceleration for constant ω; and here the derivative is wrt the inertial reference frame) of the origin of the rotating coordinate system.

    In the Coriolis term that you have written

    _Fc = - 2*m*(ω x (r0 + _r)')

    the derivative wrt time is the derivative in the rotating coordinate system, where r0 is a constant. So this reduces to the term in the book equation. The term with the ω' only comes in if the rotation is accelerating.

    If you go through the derivation in the link I posted, being careful to distinguish between the derviatives wrt time in the different frames and write r as r0 + _r, the book's equation will come out for the constatant ω case. If ω is not constant, then the r0'' term includes the contribution from ω' x r0 which is the tangential acceleration of the origin on the rotating coordinate system.
  11. Dec 17, 2006 #10

    I think I got this part! If I'm calculating the centripetal acceleration form the viewpoint of the _E-System, I need to inlcude the r0 vector in my calculations.
    The part I have marked in the quote, what does that mean? I am also not sure if Im understanding you correctly when you use the term 'wrt time', is this a shortform for 'derivative with respect to time'? Other than that, the centripetal force chapter can be closed, thanks!

    Is it not just the magnitude of the r0 vector that is constant? I have a problem to distinguish constant in the rotating system, and constant in the fixed system. But it would makes sense, because the _E-System rotates on a constant latitude and therefore does not experience any coriolis force.
  12. Dec 17, 2006 #11


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    wrt is shorthand for with respect to. Sorry for not defining it.

    I don't understand what you mean by your last sentence, but the result does not depend on that. The vector r0 is constant in magnitude, but not in direction in the inertial frame. A change in r0 in that frame is tangent to the constant-latitude circle on the earth's surface. From the geometry you can see that the rate of change of that vector in the inertial frame is

    (dr0/dt)_inertial = ω x r0

    In the rotating frame, with the origin fized at the surface of the earth the vector r0 is drawn from the center of the earth at coordinates (0, 0, -r0) to the origin. This is constant in time in the rotating frame. Furthermore, the basis for all of this calculation is that the difference between time derivatives in the two frames for any vector A is

    (dA/dt)_inertial = (dA/dt)_rotating + ω x A

    Substituting r0 for A and using the know derivative of r0 in the inertial frame you have

    (dr0/dt)_inertial = (dr0/dt)_rotating + ω x r0 = (dr0/dt)_rotating + (dr0/dt)_inertial

    0 = (dr0/dt)_rotating

    Note that if A = ω (or any vector parallel to ω)

    (dω/dt)_inertial = (dω/dt)_rotating + ω x ω = (dω/dt)_rotating + 0

    (dω/dt)_inertial = (dω/dt)_rotating
    Last edited: Dec 17, 2006
  13. Dec 17, 2006 #12
    I think I finally understood it and I got no question unanswered!
    Thanks for the time you took to help me, I appreciate it a lot.
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