Noninertial/fictitious forces (Movement on a rotating Earth)

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Homework Help Overview

The problem involves analyzing the difference in weight of an apple in two trains traveling on the equator at 50 km/h, considering the effects of fictitious forces due to the Earth's rotation. The subject area includes concepts from dynamics and non-inertial reference frames.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • The original poster attempts to calculate the apparent weight of an apple in two different frames of reference, incorporating Coriolis and centrifugal forces. Some participants question the definition of weight and how it is measured, while others clarify the need to focus on the difference in weights.

Discussion Status

Participants are actively discussing the calculations and definitions involved in determining the weight of the apple in each train. There is a focus on ensuring the correct application of forces and the implications of the rotating frame of reference. Some guidance has been offered regarding the measurement of weight and the relationship between apparent weight and normal force.

Contextual Notes

There is a mention of the need to calculate the difference in weights, as well as the assumption that weight is defined through gravitational force. The discussion also highlights the importance of understanding how scales measure weight through normal force.

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Homework Statement



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?


Homework 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


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

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