Coriolis Force and Centrifugal Force

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Homework Statement:
Occasionally one hears the rumor that railroad tracks wear out on one side. The reason for this is the Coriolis force. Check this claim using an ICE with mass m = 400 t, which is traveling north at v = 300 km / h at the 50th degree of latitude (northern latitude).
a) Calculate the normal force and the Coriolis force. Which splint is more worn out?
b) What radius would a curve have to have if the centrifugal force should correspond to the Coriolis force with the same mass and speed?
Relevant Equations:
Equations for Coriolis force and Centrifugal Force
Hello!

So for a) I have done the following

m = 400t = 400000 kg
v = 300 km/h = 83,3 m/s
##\alpha## = 50 (degrees)

Now this is the formula for the Coriollis effect

$$ F = 2 \cdot m (v × \omega) $$

So in a book I found in the library regarding physics I've found a similar example that used this formula. (In the example the weight was given the velocity as well as the angle through "50 degrees of the lattitude" phrase. But there was no explanation nor steps to how this formula came to be. Which means either it is something really easy that expects the reader to see it, or it is commonly used when dealing with corriolis effect in the northern lattitude? Can anyone please clarify this it would be much appreciated.

$$ F = 2 \cdot m \cdot v \cdot \omega \cdot sin(\alpha) $$

Now I've calculated the corriolis effect using the second formula and I've gotten

##F = 3712,23 N##

For the normal force I'm not sure by what is exactly meant by normal force I assumed that it is this,lead by the examples and problems I've solved in the lectures prior.
$$ F = m \cdot g $$

F = 3924000 N

So according to my calculations the normal force is much greater than the corriolis force. Now under a we are sussposed to give this answer

"The rails are used more in direction (choose East West South North) by a factor of (choose 1/10 1/100 1/1000 1/10000) weight.

Now this is the part I'm uncertain. How can I determine in which direction the rail is used more. I'd go as far as saying if the coriollis force was greater than the normal force I'd say the rails are used more in the direction north because the ICE is moving north.If I'm not mistaken the normal force should be at an 90 degree angle,in regards to the point or object that the force is acting to.But in which direction does the normal force opperate, in regards to north west south? Also would anyone be able to confirm if my calculations are right.

Thank you in advance!
 

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  • #2
haruspex
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How can I determine in which direction the rail is used more.
Which way does the Earth’s rotation vector point?
According to the cross product formula you used and the right-hand rule, in which compass direction is the force?
Wrt wear on the tracks, it is unclear exactly what is to be considered. Is it the wear on the side of the rail from contact with the flange, or wear on the top surface?
If on the side, bear in mind which side of a train wheel has the flange. If on top, you would also need info re the ratio of the height of the train's mass centre to the distance between the rails.
I searched for references to the phenomenon, but none that I found clarify it.
 
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  • #3
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Okay so for the wear on the tracks the professor said it is irrelevant if it is on the side or on top. We're simply suspossed to decide in which direction the track are being used more based on the values of the forces (coriollis und normal force). Hopefully this clarifys it a bit

Now the rotation vector of the earth should point east I think. Since the earth is rotating in that direction I'd make sence if pointed that way.

Now in which compass direction is the force according to the hand rule and crossproduct formula. If I recall the right hand rule right the force should be pointing north? So the coriollis force is pointing north?
 
  • #4
haruspex
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the rotation vector of the earth should point east I think.
A rotation vector points along the axis of rotation, so the choice is N or S. In principle you should use the right hand rule to decide which, then use it again to take the cross product.
In practice, though, it is simpler to imagine the train approaching the North pole and consider which way its EW momentum has to adjust.
the professor said it is irrelevant if it is on the side or on top.
True. Since the flanges are on the inside edge, the same rail gets the greater wear on both surfaces. The magnitude of the extra force between wheel and rail is different, but you are not asked for that.
 
  • #5
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Okay so I do not understand what "EW momentum" is I've tried googling it but nothing came up. Now if the rotation vector points along the axis and the direction can only be N and S, if I would to use the right hand rule the direction should be North. Now here " then use it again to take the cross product. " I assume you are meaning the direction of the force that I calculated using the cross product (Coriollis force). So here the thumb should be pointing in the direction of the force, which is also North. I hope that is what you meant.
 
  • #6
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Okay so for the wear on the tracks the professor said it is irrelevant if it is on the side or on top. We're simply suspossed to decide in which direction the track are being used more based on the values of the forces (coriollis und normal force). Hopefully this clarifys it a bit

Now the rotation vector of the earth should point east I think.Since the earth is rotating in that direction I'd make sence if pointed that way.
No. You are confusing the rotation vector of the entire earth with the velocity vector at that point. The "rotation vector" is in line with the axis of rotation. It is not the same as the velocity vector at a point some distance from the rotation axis.
Now in which compass direction is the force according to the hand rule and crossproduct formula. If I recall the right hand rule right the force should be pointing north? So the coriollis force is pointing north?
You are over-complicating this particular part of the problem. Consider the velocity vector of an earth-stationary object at the initial location of the train and compare it with the velocity vector of an earth-stationary object at the final location of the train. Those are different due to the change in distance from the earth's axis of rotation. What can you conclude from that?
 
  • #7
jbriggs444
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Okay so I do not understand what "EW momentum" is
East-West momentum. As the train moves due north, its east-west momentum [as seen from an inertial frame] changes.

Now if the rotation vector points along the axis and the direction can only be N and S, if I would to use the right hand rule the direction should be North.
The direction of the rotation vector should be north, yes.

Now here " then use it again to take the cross product. " I assume you are meaning the direction of the force that I calculated using the cross product (Coriollis force). So here the thumb should be pointing in the direction of the force, which is also North. I hope that is what you meant.
Personally, I have little patience for the "right hand rule" for the cross product. But the idea is that you point your thumb along the axis of rotation, your fingers in the direction of the velocity. The palm of your hand is then in the direction of the resulting apparent force. But that is either right or wrong. With no useful mnemonic to remember it by, you end up chasing east-west momentum and working it out by hand.

[Note that in the temperate latitudes, both thumb and fingers are pointed in somewhat the same direction thumb up toward the North star and fingers horizontally toward the North pole]

A quick trip to google says that that the above is indeed 180 degrees off. Your palm (or middle finger) points in the direction of the cross product but there is a minus sign, so this is the opposite of the direction of the Coriolis force.
 
  • #8
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Okay this was quite a lot of information, so forgive If I simply missed something but the normal force should be acting to the north and the coriollis force in the opposit direction of where the train is headed so south?
 
  • #9
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The eastward velocity of points on the earth's surface decreases for points farther from the equator (farther north in the northern hemisphere). What force can you conclude from that? It is good to work on getting the vectors, right-hand rule, etc. figured out, but your end result must agree with that simple fact.
 
  • #10
jbriggs444
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Okay this was quite a lot of information, so forgive If I simply missed something but the normal force should be acting to the north and the coriollis force in the opposit direction of where the train is headed so south?
Normal force [strictly speaking, "the normal component of a contact force"] always acts at right angles to a surface. That is the definition of the normal force. Once you pick out a surface, the direction of the normal force is fixed.

No, that guess is wrong.
 
  • #11
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Okay well, Im really not getting this with the directions and forces, I just cannot wrap my head around what east and west have to do with this. If we have a normal force that is always acting at a right angle on a surfce, how can the track be used in any direction for that matter? Also the earth rotation vector is pointing north what does that actually mean in regards to this problem? How can the track be used in east or west direction?
 
  • #12
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It is a very simple, step-by-step, problem.
1) A train closer to the North Pole is closer to the axis of the earth's rotation.
2) Things closer to the center of a rotation are moving slower due to the rotation. Do you know the equation for the velocity of a point at distance, ##R##, from the axis of rotation at the rotational rate of ##\Theta##? In this case, the rotation is around the earth's axis and the motion of a point on the earth's surface is from west to east.
3) Moving slower in the west-to-east direction implies that when the train goes north there is a westward force causing it to slow down (F=mA) Where do you think that force is coming from?
 
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  • #13
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It is a very simple, step-by-step, problem.
1) A train closer to the North Pole is closer to the axis of the earth's rotation.
2) Things closer to the center of a rotation are moving slower due to the rotation. Do you know the equation for the velocity of a point at distance, ##R##, from the axis of rotation at the rotational rate of ##\Theta##? In this case, the rotation is around the earth's axis and the motion of a point on the earth's surface is from west to east.
3) Moving slower in the west-to-east direction implies that when the train goes north there is a westward force causing it to slow down (F=mA) Where do you think that force is coming from?
Well if it is a westward force it should be coming from the west?
 
  • #14
jbriggs444
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Well if it is a westward force it should be coming from the west?
A westward force is one that points west. Not one that comes from the west.

[We name winds for the direction they come from, but little else follows that convention]
 
  • #15
jbriggs444
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If we have a normal force that is always acting at a right angle on a surface, how can the track be used in any direction for that matter?
You understand that trains can roll along tracks, right?
 
  • #16
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You understand that trains can roll along tracks, right?
I do but I dont grasp the concept of normal force having an influence on that. And for the upper question, the force is coming from the east than?
 
  • #17
haruspex
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I do but I dont grasp the concept of normal force having an influence on that. And for the upper question, the force is coming from the east than?
Viewed from a non-rotating frame, the train has its northward velocity, but also an eastward velocity caused by the rotation of the Earth. As the train travels North, the local eastward velocity of the Earth reduces, so the rails must exert a force to reduce the train's eastward velocity to match.
 
  • #18
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Viewed from a non-rotating frame, the train has its northward velocity, but also an eastward velocity caused by the rotation of the Earth. As the train travels North, the local eastward velocity of the Earth reduces, so the rails must exert a force to reduce the train's eastward velocity to match.
Okay I'll see I'll give this some more thought
 
  • #19
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Okay to give a final answer, the direction is East. We were given an explanation in class and now I actually understand what the deal was.And my calculations for the radius were also correct. Thank you for your help!
 
  • #20
Frodo
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1. Imagine a train starts travelling north from the equator on a line which runs to the North Pole.

2. Viewed from outside earth, the train has an eastward velocity due to the earth's rotation. At the equator, this velocity is 24,860 miles/24 hours, or 1,035 miles/hour.

3. Let the train continue until it reaches the North Pole. Viewed from outside earth, the train now has zero eastward velocity.

As the train has lost the eastwards velocity it had at the beginning, the train must have been accelerated towards the west during its journey.

The only way the train can get accelerated to the west is for a force to be applied towards the west. Hence the rails must have pushed the train westwards.

Therefore the inside of the east rails will get more worn.

This makes many assumptions including a perfectly flat railway track. In reality I doubt there is much visible effect.

In the same way, a shell fired northwards from a battleship sitting on the equator appears to the battleship observer to veer to the right.

Seen from outside earth, the shell travels in a straight line.

So Coriolis Force was invented as a virtual or fictitious force to explain to an observer sitting in a rotating frame (the earth) why the shell veers to the right. If Newton's Laws are correct, the only way it can veer is if it experiences a force. Hence the Coriolis force is magicked up so that the motion can be explained by an observer sitting in a rotating frame.

Of course, for an observer outside earth, who is not rotating, there is no need to magic up this fictitious force. Newton's Laws work fine for him and the shell travels in a straight line just as Newton's Laws require.

A virtual or fictitious force is a force which disappears if you make a suitable change of reference frame.
 
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  • #21
Frodo
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Homework Statement:: Occasionally one hears the rumor that railroad tracks wear out on one side. The reason for this is the Coriolis force.
That does rather assume that trains run in only one direction.

Whereas a train running north from the equator to the North Pole needs to be decelerated to lose its eastward equatorial velocity of 1.035 miles/hour, a train travelling south from North Pole to the equator needs to be accelerated eastwards to gain 1,035 miles/hour.

Both sides of the rails should therefore be similarly worn. :cool:
 
  • #22
haruspex
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That does rather assume that trains run in only one direction.
Which is true in many places since there would be separate tracks for northbound and southbound.
 

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