Noninertial/fictitious forces (Movement on a rotating Earth)

[jasker]

1. The problem statement, all variables and given/known data

Two trains are traveling east and west on the equator at 50km/h. What is the difference in weight of an apple (mass = 200g) in each train? In which train is it lighter?


2. Relevant equations

x denotes vector (cross) product, bold denotes a vector quantity. Subscripts rot and in denote the rotating and inertial frames of reference respectively. Ω is the angular velocity of rotating earth. r is the position vector to a point on the equator (magnitude here equal to Re, the radius of earth)

The apparent force due to motion on a rotating earth is:
Frot = marot
= main - m[2Ω x vrot + Ω x (Ω x r)]
= F + Ffict

The fictitious force Ffict consists of two terms, the Coriolis force and the Centrifugal force:
Coriolis force F = -2mΩ x vrot = -2mΩvrot
Centrifugal force F = -mΩ x (Ω x r) = -mΩ^2 * Re


3. The attempt at a solution

On the train traveling east, the Coriolis force points toward the axis of rotation of the earth, increasing the weight of the apple:
Frot = mg + 2mΩvrot - (mΩ^2 * Re)

On the train traveling west, the Coriolis force points outward from the axis of rotation of the earth, decreasing the weight of the apple:
Frot = mg - 2mΩvrot - (mΩ^2 * Re)

At this point, I am stuck. Is what I have correct so far? How do I calculate the weight of the apple given the two equations above? Do I just plug in these values?
m = 0.2kg
Ω = 2π rad / 86400 sec = 7.27*10^-5 rad/s
Re = 6378.14 km @ equator
vrot = 50 km / h = 13.89 m/s

Thank you!

[Mindscrape]

Two comments
1) The problem wants the difference in weights.
2) How is "weight" normally measured?

[jasker]

Well I would assume that it would be using the gravitational definition of weight, weight being the force exerted on a body by gravity, so weight would be equal to F = mg, and would be expressed in newtons (kg * m/s2)

So the weight of the apple at rest would be F = main = mg = 0.2kg * 9.8m/s = 1.96N

On the train traveling east, the weight would be:
Frot = mg + 2mΩvrot - (mΩ^2 * Re) = 1.96N + 2(0.4kg)(7.27*10-5 rad/s)(13.89 m/s) - (0.2kg)(7.27*10-5 rad/s)2(6378140m) = 1.95406578 N

On the train traveling west, the weight would be:
Frot = mg - 2mΩvrot - (mΩ^2 * Re) = 1.96N - 2(0.4kg)(7.27*10-5 rad/s)(13.89 m/s) - (0.2kg)(7.27*10-5 rad/s)2(6378140m) = 1.95245009 N

Thus it would be lighter on the train traveling west. Does that all look right?

[Mindscrape]

Typically scales measure the normal force, by seeing how much a spring compresses. So really you should have Frot-N=0, but Frot=N, so you're good.