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Wikipedia has a example of a equatorial train in its article on Coriolis Force:

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

In this example, a frictionless train is located on the equator, and can either stay still, travel at earth rate east (with earth rotation), or travel at earth rate west (against earth rotation). The article describes the forces applied in the inertial frame, the fixed-earth rotating frame, or the fixed-train rotating frame. The perceived weight of the train should be equal between the frames of reference, but different between each of the train's velocity options.

I'm trying to draw out the forces of each of the 9 cases (3 velocities * 3 possible reference frames). I've got every case making sense except one: the east traveling train from the fixed-earth rotating frame. I'm not all that skilled at Newtonian dynamics, so I would appreciate the help pointing out my error.

Here is the results of the cases thusfar (Ft is the reactionary force of the track, positive is off-earth, negative into the ground):

Still train, inertial space: Ft = m*(Ag - Acpl) = m*(Ag-(v^2)/r)

Still train, earth space: Ft = m*(Ag - Acfg) = m*(Ag-(v^2)/r)

Still train, train space: Ft = m*(Ag - Acfg) = m*(Ag-(v^2)/r)

West-moving train, inertial space: Ft = m*Ag

West-moving train, earth space: Ft = m*(Ag - Acpl - Acfg + Acor) = m*Ag

West-moving train, train space: Ft = m*Ag

East-moving train, inertial space: Ft = m*(Ag - 2*Acpl) = m*(Ag-2*(v^2)/r)

East-moving train, earth space: Ft = m*(Ag - Acpl - Acfg - Acor) = m*(Ag - 4*(v^2)/r)

East-moving train, train space: Ft = m*(Ag - 2*Acfg) = m*(Ag-2*(v^2)/r)

Where:

r = radius of earth at equator (assumes perfect ellipsoid)

v = linear velocity of earth rate at equator (assumes perfect ellipsoid)

Ag = absolute acceleration due to gravity as seen by an object not rotating with earth

Acpl = absolute acceleration due to centripetal force for an object rotating at earth rate on the equator. I'm subtracting this from gravity, as the acceleration is realized in the velocity changes to move in a circle, thus it's not felt by the track. = (v^2)/r

Acfg = absolute acceleration due to centrifugal force for an object on the equator when the reference frame is rotating at earth rate. This acceleration always points off-earth. = (v^2)/r

Acor = absolute acceleration due to coriolis force for an object rotating at earth rate on the equator in a reference frame rotating at earth rate. This acceleration points down when moving against the reference frame rotation, and up when moving with it. = 2*(v^2)/r

As you can see, everything lines up like it should except for the east-moving train in the earth-fixed rotating reference frame. Clearly I missed something or am thinking about this wrong. Please help! :)

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    – Danu
    Feb 16, 2016 at 16:09
  • $\begingroup$ @Danu How is this a homework question? $\endgroup$
    – Yashas
    Feb 16, 2016 at 16:17
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    – Danu
    Feb 16, 2016 at 16:20
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    $\begingroup$ Ha, thanks - well this isn't a check my homework question. I've been out of school way too long for that! I'm simply trying to understand coriolis force, and running into a roadblock. I'm sorry it appears I haven't put forth effort, I actually ran through the problem on paper and derived out the accelerations myself from the cross products. I guess I'm more asking for someone to "correct my understanding" than to "check my work", because I'm clearly not getting it. $\endgroup$
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    Feb 16, 2016 at 16:53
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    $\begingroup$ Note that we have a different definition of "homework" here, which you can see on this Meta post about homework-like questions (hint: it doesn't have to be school-related work for it to be homework here). $\endgroup$
    – Kyle Kanos
    Feb 17, 2016 at 11:24

1 Answer 1

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Should have seen this earlier...

Centripetal and centrifugal acceleration increase with the square of velocity, so moving twice the speed produces four times the acceleration. In this case, I was correct on the east-moving train in earth-fixed case, but incorrect in the other two reference frames. All frames should produce Ft = m*(Ag - 4*(v^2)/r)

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  • $\begingroup$ I think this was a perfectly good question. In fact, I've been mulling over the same Wikipedia article and the same discrepancy with a 2 in one frame and a 4 in another ... Of course, doubling the speed/frequency makes the force fourfold. I'm really angry for not noticing that. :/ $\endgroup$
    – BoLe
    Mar 27, 2016 at 20:29
  • $\begingroup$ BTW the reactionary force here, or what the scale would show, is about 100 * (1 - 0.35 * 4) = 98.6 % of the weight at the poles. $\endgroup$
    – BoLe
    Mar 27, 2016 at 20:35

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