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Coriolis and centrifugal forces

  1. Dec 20, 2013 #1
    1. The problem statement, all variables and given/known data
    There is a fixed coordinate system and a rotating one. the origins don't coincide.
    The situation is like a stone falling from a tower on earth, where the rotating system has it's origin at the base of the tower.
    Are the coriolis and centrifugal forces dependent on the choice of the coordinate system?
    According to the equation-yes, because r is measured in the rotating system (is it?), but my intuition says the forces are constant.
    If i choose a closer coordinate system to the moving object then the imaginary forces, the coriolis and centrifugal, are smaller, no?

    2. Relevant equations
    The "F" notation is for the fixed coordinate system and "R" for the rotating one.
    [tex]\vec{a}_R=\vec{a}_F-2\vec{\omega}\times\vec{V}_R-\vec{\omega}\times\left( \vec{\omega}\times \vec{r} \right)[/tex]
     

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  2. jcsd
  3. Dec 20, 2013 #2

    tiny-tim

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    Hi Karol! :smile:
    Yes of course.

    r and v are measured in the rotating frame.

    The coriolis force and the centrifugal force are creations of the coordinate system … of course they will usually be different for different systems … they are fictitious corrections that enable the observer to regards newotn's laws as applying.
    Why?? :confused:
     
  4. Dec 21, 2013 #3
    If so, which coordinate system do i choose if i want to know how far from the base of the tower the stone will fall?
    If i choose at the center of the earth i will get an other answer than if i choose it at the base of the tower
     
  5. Dec 21, 2013 #4

    tiny-tim

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    Hi Karol! :smile:
    You're ignoring a more fundamental fictitious force: the fictitious force due to the translational acceleration of the observer (in this case, the centripetal acceleration of the tower round the earth).

    This is not negligible (in fact, it's huge!).

    So use the centre of the earth. :wink:

    (of course, even if we use the centre of the earth, there's still the centripetal acceleration round the sun … but that's rω2, which is about 2300/3652, or about 2% … and of the sun round the galaxy, which is even smaller)
     
  6. Dec 21, 2013 #5
    I know the tower has centrifugal acceleration, but it doesn't affect the falling stone, or should i add this acceleration to the acceleration i get for the stone for the frame at the base of the tower? is it true that i will get 2 different answers for the distance the stone is falling from the base of the tower?
     
  7. Dec 21, 2013 #6

    tiny-tim

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    the tower's frame of reference (ie with the origin at the tower) has a fictitious force due to the translational acceleration that it has in the centre-of-earth frame

    since it's far larger than anything else in the calculation, i don't see any point in treating it as a correction … you should not use the tower-origin rotating frame of reference at all
     
  8. Dec 21, 2013 #7
    Yes, i know it's far larger because the radius of the earth is big in comparison with the tower, and i have understood your proposal to use the center of the earth as the origin.
    The fact is i read in some book a solved example to the tower problem in which they used the tower frame.
    I want to understand, in principle, what should i do with these problems. should i add the forces the tower frame feels to the stone or not?
     
  9. Dec 21, 2013 #8

    vanhees71

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    The forces on the tower frame is just the centrifugal force, and this is already included in the gravitational acceleration [itex]\vec{g}[/itex], which can be set constant for this problem. The EoM. for the free fall from the tower thus reads
    [tex]\ddot{\vec{x}}=-\vec{g}-2 \vec{\omega} \times \dot{\vec{x}}-\vec{\omega} \times (\vec{\omega} \times \vec{x}).[/tex]
    where [itex]\vec{x}[/itex] is the position vector in the reference frame at the tower and [itex]\vec{\omega}[/itex] the constant angular velocity wrt. to this same frame, i.e.,
    [tex]\vec{\omega}=\omega \begin{pmatrix}
    -\cos \beta \\ 0 \\sin \beta
    \end{pmatrix},[/tex]
    where [itex]\beta \in [-\pi/2,\pi/2][/itex] is the latitude where the tower is located on Earth.

    Now you can also neglect the centrifugal acceleration in the EoM, because [itex]\omega=2 \pi/\text{d}[/itex] is very small at the time scale the free fall is happening. The remaining equation, including the Coriolis acceleration only, is then pretty straight-forward to solve!
     
  10. Dec 21, 2013 #9
    I don't understand that. centrifugal force isn't the acceleration g, and how is it included. and what do i do with it concerning the stone, if at all, and if i don't want to neglect the centrifugal force.
    Excuse me, but what EoM means?
     
  11. Dec 25, 2013 #10
    I feel c certain centrifugal force on the earth. if the imaginary forces are coordinate system dependent, according to which coordinate system do i feel them? it must be with the origin at the center, and not on the surface, and this is the only one the can be used for the force has a certain magnitude, no?
     
  12. Dec 26, 2013 #11

    PeroK

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    The constant force of gravity that we feel at a point on the Earth's surface is, in fact, a combination of gravity and the centrifugal pseudo-force caused by the Earth spinning. So, the centrifugal force is effectively bundled up with gravity and there is no need to calculate it separately.

    Apart from at the poles and equator, this combined force doesn't point at the centre of the Earth. So, things that are falling vertically are not actually falling directly towards the centre of the Earth. But, because the force is effectively constant (except over large distances) we don't notice this or normally have to take this into account.

    See the note on "apparent gravity":

    http://en.wikipedia.org/wiki/Gravity_of_Earth

    If the Earth stopped spinning, then you'd find that things you thought were "vertical" were no longer vertical.
     
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