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Fictitious force vectors

  1. Sep 13, 2013 #1
    1. The problem statement, all variables and given/known data

    Depending on the college or university, this may be a 2nd or 3rd year question. For me, it is a junior-level course.

    A train is heading due south through Edmonton at a speed of 300 m/s relative to the surface. Edmonton has a latitude of 53.5 degrees; the Earth has a radius of 6400 km. If the train has a mass of 1Gg (not sure, maybe a typo), what are:

    (a) the centrifugal force (magnitude and direction) of the train?
    (b) the coriolis force (mag. and dir.) of the train?
    (c) if there is a coriolis force on the train, how come it continues due south?

    2. Relevant equations

    I use star (*) to show the perspective of the non-inertial, Earthly reference frame.

    [itex]\mathbf{F_{\mathrm{cf}}}=-m\boldsymbol{\omega}\times(\boldsymbol{\omega}\times\mathrm{\mathbf{r}^{*}})[/itex]

    [itex]\mathbf{F_{\mathrm{co}}}=-2m\boldsymbol{\omega}\times\mathrm{\mathbf{v}^{*}}[/itex]

    3. The attempt at a solution

    For parts (a) and (b), I'm having a hard time determining my position and velocity vectors. I know to find the omega vector in terms of the non-inertial basis vectors given latitude and angular speed, but I don't really know what my position vector is since up until now, I've only done problems where something is rotating on a turntable and moving, but now i have to take into account that the Earth is curved.

    Once I've found a position vector r*(t), I can find velocity vector v*(t) = r'*(t). Then it's a matter of plugging it in and crossing the vectors.

    The problem is I'm not sure what r*(t) is. Would r*(t) = 300t i*? And then v*(t) = 300 i*? Remember i* is the basis position vector in the non-inertial frame (which is the Earth basically), i* is tangential to the surface of the Earth, and i* points "south". Are my guesses right (not likely), or is it much more complicated than this (quite likely)?

    For part (c), doesn't it just move south because the tracks create a third fictitious "m a" force that restrict its motion in one direction? An "m a" force is a fictitious force that is the result of acceleration of the non-inertial frame, so if I had a book on a car dashboard and stomped on the gas, the book experiences a fictitious "m a" force that pushes it off the dash (in my view).

    My biggest questions stem from parts (a) and (b).
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
    Last edited: Sep 13, 2013
  2. jcsd
  3. Sep 13, 2013 #2
    I want to think of this really simply and say is omega actually the angular velocity of the Earth's rotation, and the 300 m/s due south is used to find the velocity and position vectors (which are the ones I noted above in my attempt, which were very simple). Am I getting somewhere?
     
  4. Sep 13, 2013 #3

    SteamKing

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    You should know that 1 Gg = 1 Gigagram = 10^9 gram = 1000 tonnes
     
  5. Sep 13, 2013 #4
    Ok well that was trivial anyways...I know what a gigagram is, but if you saw the way my question has been formatted and typed out by my instructor (on word), he has (and often does) left symbols ambiguous. Anyways the information you provided is not terribly helpful to my question.
     
  6. Sep 13, 2013 #5

    ehild

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    Yes, omega is the angular velocity of Earth. The train travels due to South along the surface of the Earth. The position is Edmonton. Use polar angles or longitudes and latitudes as coordinates. What is the magnitude of the position vector r?

    ehild
     

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  7. Sep 16, 2013 #6
    |r| = 6400.

    the position vector is then r = 6400cos λ i + 6400sin λ j. But what confuses me is the i* j* k* vectors are all over the place. the problem is so easy as it's just a matter of plugging and playing but different frames of reference make these position vectors hard to find in the right frame.
     
  8. Sep 17, 2013 #7

    ehild

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    Centrifugal force and Coriolis force appear in the rotating frame of reference. The subject of these forces is the train now, they are exerted on the train and observed by somebody, standing on the ground, rotating together with the Earth. You can set up a coordinate system, appropriate for the observer on the Earth: with x axis to the East, y to the South and z axis up, for example. What are the components of omega in that system of coordinates?

    Or you can choose ω as the z axis, pointing towards the North Star. The train moves along a circle of radius ρ depending on the latitude, perpendicular to the axis of the Earth. You can choose one axis tangent to the circle (to East or West) and the third axis perpendicular to the other two. What angle does this axis enclose with the vertical?
    Answer giving the forces in terms of the bearings , North-South, East-West and with the angle they enclose with the vertical. (Vertical is parallel with the radius of Earth)Anyway, you have to give the direction of the forces with respects tp the Earth, in terms of East-West, North-South, up or down...


    ehild
     
    Last edited: Sep 17, 2013
  9. Sep 17, 2013 #8
    Let me define my non-inertial coordinate system, with basis vectors i*, j*, and k*. I want i* to point due south, j* to point due east, and k* to point perpendicular to the surface of the earth where the train is.

    Then we can express ω in terms of these basis vectors as follows:

    [itex]\boldsymbol{\omega} = \omega sin \lambda \textbf{k}^{*} - \omega cos \lambda \textbf{i}^{*}[/itex]

    And

    [itex]\textbf{r}^{*} = 6400 sin \lambda \textbf{k}^{*} + 6400 cos \lambda \textbf{i}^{*}[/itex]

    but when i go to find the Coriolis force, I'll need v*, which should be the time derivative of r* above. Am I totally lost? Feels like I am.
     
  10. Sep 17, 2013 #9

    ehild

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    No, v* is the velocity of the train, and it is given:
    ehild
     
  11. Sep 18, 2013 #10
    Yeah, i thought that originally.

    So v* = 300i*

    And what about the rest of my last post? am i in a good place, or not really? i'm thinking not really.
     
  12. Sep 18, 2013 #11

    ehild

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    r* is perpendicular to the surface of the Earth, so it is parallel to k*.

    The train travels to the South as its motion is constrained by the rails, as you correctly said. But it is not fictitious force. It is normal force from the rails.

    Without constraints, a moving body will deviate from the original velocity because of the Coriolis force. It has effect of motion of air and water of oceans. http://www.fjcollazo.com/documents/CoriolisEnvRpt.htm
     
    Last edited: Sep 18, 2013
  13. Sep 18, 2013 #12
    what's confusing is that if v* = 300i*, then my entire i* j* k* is moving with the train at all times, and thus r = 0 always. i don't see how r* = k*.
     
  14. Sep 19, 2013 #13

    ehild

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    The frame of reference is fixed to the Earth, it does not move with the train. You can put the origin into the centre of Earth, with the x and y axis parallel to the South and West directions in Edmonton. It is a rotating frame of reference then. You need to determine the magnitude and direction of the centrifugal force and those of the Coriolis force when the train is at some point in Edmonton, so r=6400 km.

    If you place the origin at the location on the train in Edmonton, so r=0, the frame of reference is both travelling along a circle (therefore accelerating towards the centre) and rotating. Then you have to take the fictitious force -ma into account where a is the centripetal acceleration on the latitudinal circle.
     

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