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Coriolis force question

OK,
I'm back from raquetball, but way too tired to think about this now.

I think that CurtC is wrong, but I'm not sure. I think we both agree that the plumb bob doesn't point to the center of the earth because of the centrifugal force produced by the earth's rotation.

I don't think this effect has anything to with the fact that the earth is not a perfect sphere.

If the earth was a perfect sphere the plumb bob would still not point to the center of the earth because of the earth's rotation and there would be a small difference in the angle of a wall which was built using geometric techniques to set a 90 degree angle from a wall which was built using levels or plumb bob's.

Does the fact that the earth is a geoide (not quite an ellipsesoid I think) make it so what we think of as level is tilted slightly so that it ends up being exactly perpendicular to the plumb bob? I don't know. I think this might be what CurtC is saying and I'm not sure he's wrong about that. I need to think more about this to understand what CurtC is saying and what the facts are.

As to what Hamish was saying about the Ceptimus/davefoc algorythm violating conservation of angular momentum, I thought about this some and don't think it does but I can't focus enough right now to get my thoughts out.
 
The dropped coin actually falls in the trajectory of an elipse, with one focus at the earth's centre. Over the height of the building this elipse is indistinguishable from the parabola that davefoc and I are calculating.

Dave - the buliding doesn't 'lean' relative to the local inertial reference frame. It is correct that it should line up with a plumb bob, as the same forces act on the building as on the line.

Once the coin is in free fall it will follow an elipse and the plane of the elipse will intersect the centre of the earth, so the coin will land north of the plumb bob, and can't be dropped down the south side of the building, as davefoc already stated.
 
Ok, I don't think the difference is insignificant, I think it accounts for the 2.2 inch discrepency.


As to what Hamish was saying about the Ceptimus/davefoc algorythm violating conservation of angular momentum, I thought about this some and don't think it does but I can't focus enough right now to get my thoughts out.
Click to expand...

Yes it does, and it's quite easy to show.

Angular momentum, L = mr<sup>2</sup>w where m is the mass, r is the distance from the centre of rotation and w is angular velocity.

The tangential velocity,v, is rw


If you assume that v is constant over the fall then, at the top, where the velocity is v<sub>1</sub>

[/i]L=m(r<sub>0</sub>+h)<sup>2</sup>w=m(r<sub>0</sub>+h)v<sub>1</sub>
[/i]

Where h is the height of the building and r<sub>0</sub> is the distance to the centre of roatation at the ground. At the bottom, where the tangential velocity of the penny is still v<sub>1</sub>

L=mr<sub>0</sub>v<sub>1</sub>

Which isn't the same as it was at the top, therefore conservation of angular momentum is violated.

I'm prepared to be proven wrong. If someone can show that the effect I'm describing really is negligible then I'll accept the 7.33 inch figure. I'll repeat that it is entirely possible that I've made a blunder but I don't think so at the moment.
 
The angular momentum is v<sub>t</sub><sup>2</sup> / r.

Where v<sub>t</sub> is the tangential component of the velocity, and r is the distance of the tangent from the point of rotation. I think you are confusing the radial and tangential components of the velocity.
 
davefoc wrote:
Does the fact that the earth is a geoide (not quite an ellipsesoid I think) make it so what we think of as level is tilted slightly so that it ends up being exactly perpendicular to the plumb bob?
Click to expand...
Yes.

Think of it this way - what we consider "level" is the plane that the surface of a body of water makes at that point. Think about an Earth with no land mass, just all ocean, and try to imagine it being perfectly spherical. At that point at 40.75 degrees latitude, the plumb bob would not be exactly perpindicular to the surface of the water. But since it's water and free-flowing, the water that's more "uphill" according to the plumb bob (North) would flow to the downhill side (South), eventually causing the Earth to bulge out at the Equator. Just like it does now. That shape will be an ellipsoid. I think the "geoid" term you use means "the shape of the Earth," which is very close to a mathematical ellipsoid.

So our definitions of "plumb" and "level" are always exactly perpindicular to each other, and are a combination of the effects of the gravitational pull toward the center of the Earth, and the centrifugal force outward.
 
ceptimus said:
The angular momentum is v<sub>t</sub><sup>2</sup> / r.

Where v<sub>t</sub> is the tangential component of the velocity, and r is the distance of the tangent from the point of rotation. I think you are confusing the radial and tangential components of the velocity.
Click to expand...

Go and check in a textbook. I think you're getting confused with angular momentum and centripetal acceleration.
 
CurtC wrote:
That shape will be an ellipsoid. I think the "geoid" term you use means "the shape of the Earth," which is very close to a mathematical ellipsoid.
Click to expand...
I usually hear of the shape of the Earth described as an oblate spheroid. Is there a mathematical difference between an ellipsoid and an oblate spheroid?
 
Hamish,
What you said about the angular momentum is what I thought you were going to say.

My thought was that you needed to take into account the angular momentum of the whole coin/earth system. I now think this was wrong.

I am not agreeing that you are right here, only that I am not sure. But I need to go, so I can't think about this right now.

As to what CurtC is saying about the shape of the earth making it so that the plumb bob hangs exactly perpendicular to the plane of the local ground, I think that's wrong. I think that plumb bob might hang perpendicular to the local ground though because the level we use to determine the local ground is level is caused to be tilted by the same mechanism that tilted the plumb bob. I think this would be true if the earth was a perfect sphere also though. I am going to have to think about this some more though. Maybe what CurtC is saying that the shape of the earth was formed because of centrifugal force from the rotation and per force it turns out to be just the right shape to make the local ground level. Interesting. Has the shape of the earth changed over time as the earth's rotation has slowed down? Maybe too small an effect to be able to detect.

It is interesting that something dropped will not follow the path of the plumb bob, but instead drops toward the center of the earth. I think CurtC agrees with this.
 
I think the way to get a pretty exact answer is:

Use a non-rotating intertial reference frame with origin at earth centre, and one axis (z) along the earth's spin axis.

1. Calculate the elipse the coin traces out.

2. Find where it intersects with the earth surface

3. Allow for the rotation of the earth during the fall.

I'm not sure my math is up to the task, but I'll maybe give it a go later. I expect the eastward offset (relative to a plumb line) to be about 7.33 inches, and a northward offset of about the same - I'm not sure - but I do expect it to be north rather than south.
 
davefoc wrote:
Maybe what CurtC is saying that the shape of the earth was formed because of centrifugal force from the rotation and per force it turns out to be just the right shape to make the local ground level. Interesting. Has the shape of the earth changed over time as the earth's rotation has slowed down?
Click to expand...
Yes, that's what I'm saying. The Earth is the shape it is, exactly because of the centrifugal force. It's how we use "sea level" as the definition of level, and the seas are free to flow as a result of the centrifugal force.

And I don't think a dropped object will fall towards the center of the Earth - it will hit just East of the plumb bob, and very slightly South. Ceptimus, since the ellipse that the dropped object will follow is at its Northernmost peak at the moment it's dropped (imagine an ellipse slanted at 40.75 degrees, with the "pointy" end and the northernmost point at the same location), it can go only South from there. After only nine seconds, it won't have gone very far South, but only a slight amount. There's no way it will hit North of the plumb bob.

Edited to add: wait, I retract that. The ellipse will have the center of the Earth as one of its foci, so maybe will be slightly North. Hmmm. Good luck with that ellipse calculation.
 
What's all this talk about centrifugal force?
I thought there was no such thing!

BillyJoe
 
Is there a good reason why my approach was wrong? I don't see why you have to go on calculating elipses using linear dynamics when the problem can be much simplified using rotational dynamics without making any simplifying assumptions (except for all the ones about air resistance and the like). Ok, you have a nasty integral to do but I was able to do that without too much trouble. If I have made a serious mistake, can anybody determine where I made it?

To recap, I'm standing firm on my 12.4 cm (or 4.88 inches).
 
Hamish,
I may be missing something here, but it looks to me if one takes into account the angular momentum issue the effect is to increase the eastward displacement slightly.

LT = m * (R+H)*VT
LS = m * R * VS

where
R = radius of earth at ESB latitude
H = height of drop
VT = velocity at top
VS = velocity at surface

assuming angular momentum conserved

m * (R+H)*VT = m * R * VS

VS/VT = 1.000060

This says that the coin is going slightly faster horizontally at the surface than it was at the top. This looks like the skater issue to me except that instead of pulling in her arms to spin faster the earth is pulling in the coin, but the earth doesn't pull it in very far compared to its radius so the increase in speed is small.
 
I don't think that's right davefoc. The path we calculated is for a parabola, whereas the real path, with angular momentum conserved, is an elipse. The math for the two curves is very much the same (if the one elipse focus is moved an infinite distance away the two are the same) but my feeling is that the 'arms' of the elipse will be inside those of the parabola.

I've not actually done the math yet, so I reserve the right to be completely wrong on this. :) My feeling is that the coin will fall slightly short of our calculated distance (7.33) but only by an infinitesimal margin. Hamish's calculation is way different, so at least one of us must be wrong.
 
CurtC,
I think we agree that the coin is displaced eastward. The only issue on that score is how much.

But I think the coin is also going to drop significantly north of a locally vertical line and I think you don't, but it sounds like you are equivocating a bit there.

A thought experiment:
At the instant the coin is released gravity is turned off. Which way is it going. I think it is going exactly tangent to the circle that it has been following around the earth. That is it is going exactly east with no downward deflection.

Now gravity is turned back on. What happens? Gravity sucks it towards the center of the earth. Are there any other forces acting on it, excluding much smaller lunar and solar gravitational forces, no. So the coin is just pulled toward the center of the earth which at the ESB latitude is somewhat north of a direction pointed to by a plumb bob.
 
The path we calculated is for a parabola, whereas the real path, with angular momentum conserved, is an elipse. The math for the two curves is very much the same (if the one elipse focus is moved an infinite distance away the two are the same) but my feeling is that the 'arms' of the elipse will be inside those of the parabola.
Click to expand...

My mind is reeling. I need to go. I don't quite get what you're saying, but I'm going to think about it. It seems like the coin isn't going to follow a parabola because the earth is curved and gravity is not completely parallel through its path. I'll take your word for it that as a result the coin is now following the path of an ellipse instead. That seems like a very small effect here, though. But it seems like the nature of that effect is to increase slightly the velocity of the coin parallel to the surface. I don't see how it could slow it down.
 
Just got the ellipse issue. I guess I can take old Isaac's word on that one. The coin is essentially orbiting the earth around the center of gravity. It just it runs into the surface before it completes a revolution.
 
davefoc said:

assuming angular momentum conserved

m * (R+H)*VT = m * R * VS

VS/VT = 1.000060

This says that the coin is going slightly faster horizontally at the surface than it was at the top. This looks like the skater issue to me except that instead of pulling in her arms to spin faster the earth is pulling in the coin, but the earth doesn't pull it in very far compared to its radius so the increase in speed is small.
Click to expand...

This is rather misleading, since the figures that matter are the tangential velocities relative to the ground. If you work out the relative velocities and take the same ratio, you get

VS/VT = 2.00006

The 0.00006 isn't that significant, but the fact that relative to the Earth's surface the tangential velocity of the coin at the surface is twice that at the top is quite relevant.

I think.:confused:

Trying to translate all the rotational dynamics into linear dynamics is a very tough prospect at the moment. I find it all so much simpler to start with spherical polar co-ordinates and do the whole problem in a co-ordinate set where the acceleration and velocity really are perpendicular.

I'll try another argument. Think about the motion of the penny relative to various points on the tower. If you use your original model you get very weird results since points on the tower must move relative to each other and the tower ends up slightly skewed (7.33 inches over it's height). Any vertical taken in this co-ordinate frame cannot remain a vertical over the course of the experiment.

It's very hard to convince myself, let alone others, that the errors in using a cartesian frame really do ammount to a significant difference but I'm fairly confident of my result.

As you may have guessed from my incoherent rambling, I'm quite tired. I'll try and think this through some more tomorrow.
 
After some research, I've decided that my math isn't good enough (or perhaps I'm just too lazy) to work out ellipses in three dimensions and their intersections with oblate spheroids. I might try a computer simulation of the problem though - the math is simpler there.
 
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