Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Coriolis Force

  1. Nov 24, 2008 #1
    When I studied "Earth Science" , I found a formula of Coriolis Force which is

    -2m[tex]\vec{}\omega[/tex]x[tex]\vec{}v[/tex]

    I am wondering how the formula is proven?


    I need a careful proof and diagram

    THX!
     
  2. jcsd
  3. Nov 24, 2008 #2
    The Coriolis Effect a result of using a rotating reference frame. I recommend getting comfortable with the Coriolis effect in 2d first. The site below is a nice site to get used to the polar coordinate basis and how to take derivatives using r, theta which are unit vectors that are dependent on time.

    Notice the equation for acceleration:
    [​IMG]

    the
    [itex]
    2\dot{r} \dot{ \theta }\bf{e_{ \theta}
    [/itex]
    is the Coriolis effect in 2d
    here is a nice video describing this:
     
    Last edited by a moderator: Sep 25, 2014
  4. Nov 24, 2008 #3
    I have known the effects of Coriolis force

    I want to see the "3-D" condition of Coriolis Force

    and the "critical" proof(by mathmatics and physic)......

    Thanks you for displaying the vedio!
     
    Last edited by a moderator: Sep 25, 2014
  5. Nov 24, 2008 #4
    hmm not sure what we are looking for by "critical proof", basically when you are using a different coordinate system these pseudo forces just show up, so being able to change the coordinate system seems to do it. Maybe I am misunderstanding what you are asking, dunno.
     
  6. Nov 24, 2008 #5

    LURCH

    User Avatar
    Science Advisor

    I never took a "proof-writing" math class, but I think that's what he's asking for (am I right abc?). Or perhaps a derivation of the equation?
     
  7. Nov 24, 2008 #6
    But this just requires changing coordinate systems from xyz to rho,theta,phi. If you prove that then the Coriolis effect is just a side effect of the proof. Am I missing something in my reasoning Lurch?
     
  8. Nov 24, 2008 #7

    D H

    User Avatar
    Staff Emeritus
    Science Advisor

    The Coriolis force is a pseudo force (aka inertial force, aka fictional force) that arises from wanting to express Newton's second law in a rotating frame. So, to understand how this term arises it is necessary to delve at least a little into rotating reference frames.

    I'm not going to prove this here, but

    [tex]
    \frac{d}{dt_I}(\boldsymbol{u}}) =
    \frac{d}{dt_R}(\boldsymbol{u}}) +
    \boldsymbol{\omega} \times \boldsymbol{u}
    [/tex]

    where [itex]\boldsymbol{u}[/itex] is any vector quantity and [itex]\boldsymbol{\omega}[/itex] is the angular velocity of the rotating frame with respect to some inertial frame. The subscripts I and R on the time derivative of [itex]\boldsymbol{u}[/itex] denote the time derivative of [itex]\boldsymbol{u}[/itex] as observed by someone fixed in the inertial and rotating frames, respectively.

    You can find a hand-waving proof of this conjecture in most sophomore/junior level classical physics texts. Good, solid proofs are hard to find. Some upper-level aerospace engineering texts do so. The result is an immediate consequence of the time derivative of the transformation matrix from the rotating to inertial frame (which takes a page or so to derive) or from the corresponding transformation quaternion (which takes a quarter page or so to derive).

    The above expression in operator form becomes

    [tex]
    \frac{d}{dt_I} =
    \frac{d}{dt_R} +
    \boldsymbol{\omega} \times
    [/tex]

    Using this expression, the second time derivative of the position vector from the perspective of inertial and rotating observers are related via

    [tex]
    \aligned
    \frac{d^2\boldsymbol{x}}{dt^2_I} &=
    \frac{d}{dt_I}\left(\frac{d\boldsymbol{x}}{dt_I}\right) \\
    &=
    \frac{d}{dt_R}\left(\frac{d\boldsymbol{x}}{dt_I}\right) +
    \boldsymbol{\omega} \times \left(\frac{d\boldsymbol{x}}{dt_I}\right) \\
    &=
    \frac{d}{dt_R}\left(\frac{d\boldsymbol{x}}{dt_R} +
    \boldsymbol{\omega} \times \boldsymbol{x} \right) +
    \boldsymbol{\omega} \times
    \left(\frac{d\boldsymbol{x}}{dt_R} +
    \boldsymbol{\omega} \times \boldsymbol{x}\right) \\
    &=
    \frac{d^2\boldsymbol{x}}{dt^2_R} +
    \frac{d\boldsymbol{\omega}}{dt} \times \boldsymbol{x} +
    2\,\boldsymbol{\omega} \times \frac{d\boldsymbol{x}}{dt_R} +
    \boldsymbol{\omega} \times (\boldsymbol{\omega} \times \boldsymbol{x})
    \endaligned
    [/tex]

    Newton's second law holds in an inertial frame, so the left-hand side is just Fext/m. Solving for the second derivative of the position vector as observed in the rotating frame yields for constant-mass objects

    [tex]
    m\,\frac{d^2\boldsymbol{x}}{dt^2_R} =
    \boldsymbol{F}_{ext} -
    \left(
    m\,\frac{d\boldsymbol{\omega}}{dt} \times \boldsymbol{x} +
    2\,m\,\boldsymbol{\omega} \times \frac{d\boldsymbol{x}}{dt_R} +
    m\,\boldsymbol{\omega} \times (\boldsymbol{\omega} \times \boldsymbol{x})
    \right)
    [/tex]

    The three terms proportional to mass are various fictitious forces (all fictitious forces are proportional to mass). As the earth's rotation rate is nearly constant, so the first fictitious force term can be safely ignored for earth-based observations. The second fictitious force term is the Coriolis force while the final term is centripetal force.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Coriolis Force
  1. Coriolis force (Replies: 12)

  2. Coriolis force (Replies: 0)

Loading...