A question on gravity and friction

Discussion in 'Physics & Math' started by DJ Erock, Aug 29, 2008.

  1. DJ Erock Resident Skeptic Registered Senior Member

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    The question begins with a sphere sitting on a flat surface. If these objects created little enough friction, would the rotation of the earth cause the sphere to move? If so, which way would it move? I would think that it is the opposite direction of the earth's rotation, but I'm not certain. Is the gravity of earth strong enough of a downward force to prevent the lateral forces of its spin?
     
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  3. temur man of no words Registered Senior Member

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    It will not move if the Earth is ideally shaped (which it is very close). If the Earth is a perfect ball (which it is not), then there will be a slight force pushing the body towards the equator and to the west.
     
    Last edited: Aug 29, 2008
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  5. BenTheMan Dr. of Physics, Prof. of Love Valued Senior Member

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    I changed the thread title fiction->friction.

    Either way, this is an interesting question. I'm not fully convinced that temur is correct. Care to qualify your answer? (Sorry if I'm being dense.)
     
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  7. Steve100 O͓͍̯̬̯̙͈̟̥̳̩͒̆̿ͬ̑̀̓̿͋ͬ ̙̳ͅ ̫̪̳͔O Valued Senior Member

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    I reckon that the ball would just stay still relative to an outside observer.
    As if the Earth were just spinning and sliding against it.

    I don't see any reason why the ball would move if there is no friction to drag the ball.

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    Last edited: Aug 29, 2008
  8. temur man of no words Registered Senior Member

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    I assumed the ball was at rest relative to the Earth surface in the beginning of the experiment. In the local non-inertial frame, there will be three forces at work: the gravitational attraction to the center of the Earth, the upward normal force from the surface, and the centrifugal force because of the rotation. In the ideally shaped Earth, these three forces exactly cancel (actually this is the definition of the “ideal shape”, I think it is called geoid but not sure). But if the Earth was spherical, then the gravitational attraction is colinear with the surface normal force, so the tangential component of the centrifugal force will not be canceled It will be in the direction of the equator (except of course on the poles). Now, when the ball starts moving towards the equator, there will be another force to account for: the Coriolis force. This force will be in the west direction.
     
  9. BenTheMan Dr. of Physics, Prof. of Love Valued Senior Member

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    Depending on the hemisphere, right?

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    So the ball rolls TOWARDS the equator, due to the centrifugal force?
     
  10. temur man of no words Registered Senior Member

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    The direction of the Coriolis force does not depend on the hemisphere.

    Yes.
     
  11. BenTheMan Dr. of Physics, Prof. of Love Valued Senior Member

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    Right. I just spend five minutes on wikipedia trying to convince myself otherwise, but realized that you were correct

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  12. temur man of no words Registered Senior Member

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    From the derivation of the formula I never really understood why there is Coriolis force, but when you think of a rotating horizontal disk, and you imagine you are walking from the middle to the edge of the disk (or in the opposite), you will find that since you have to be moving with the disk faster in order to stay on the radius, you will feel a force that pushes you in the direction opposite of the rotation. If you go towards the middle, you will feel a force that pushes you in the same direction as the rotation. Well I guess in the end it is just the same effect as if you jump on a moving walking way from the side.
     
  13. D H Some other guy Valued Senior Member

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    You need to do a better job of defining your terms and the initial conditions. Regarding the latter, temur assumed the ball started at rest with respect to the rotating Earth. This is the natural assumption. You apparently are assuming it starts at rest with respect to a non-rotating frame.

    Regarding the former, what do you mean by "perfectly flat"? I'll make a stab at it: A perfectly flat surface is a surface such that a ball placed at rest on the surface does not roll in any direction. In which case, the answer to your question is trivial: The ball doesn't move. Please don't get upset about me giving you a snooty answer. This snooty answer is very closely aligned with the concept of mean sea level.

    Depends on what you mean by "ideal shape". A somewhat simple model is the WGS84 reference ellipsoid. The geoid is a complex shape that represents a isopotential of the gravitational+centrifugal pseudo forces in the Earth-fixed (i.e., rotating) frame. The geoid is what I would call a flat surface -- except it is anything but flat.

    For more what-if games of this type, see this article: "Mean Sea Level, GPS, and the Geoid" by Witold Fraczek, ESRI Applications Prototype Lab. The what-if games, one of which is "what if the Earth was spherical" start on page 2 of the article.
     
  14. D H Some other guy Valued Senior Member

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    It follows by applying what some call the "transport theorem" twice. The transport theorem relates the time derivative of some vector quantity \(\mathbf q\) as seen by a pair of collocated observers denoted as A and B, one rotating with respect to another:

    \(\frac{d{\mathbf q}}{\left.dt\right|_A}\,=\, \frac{d{\mathbf q}}{\left.dt\right|_B} \,+\, {\mathbf{\omega}}_{_{_{A\to B}}} \,\times\,{\mathbf q}\)

    So, apply this to itself:

    \( \frac{d^2{\mathbf q}}{\left.dt^2\right|_A}\,=\, \frac{d}{\left.dt\right|_A}\left(\frac{d{\mathbf q}}{\left.dt\right|_A}\right)\,=\, \frac{d}{\left.dt\right|_A}\left(\frac{d{\mathbf q}}{\left.dt\right|_B} \,+\, {\mathbf{\omega}}_{_{_{A\to B}}} \,\times\,{\mathbf q}\right) \)

    \( =\frac{d}{\left.dt\right|_B} \left(\frac{d{\mathbf q}}{\left.dt\right|_B} \,+\, {\mathbf{\omega}}_{_{_{A\to B}}} \,\times\,{\mathbf q}\right) + {\mathbf{\omega}}_{_{_{A\to B}}} \,\times\, \left(\frac{d{\mathbf q}}{\left.dt\right|_B} \,+\, {\mathbf{\omega}}_{_{_{A\to B}}} \,\times\,{\mathbf q}\right) \)

    \( = \frac{d^2{\mathbf q}}{\left.dt^2\right|_B}\,+\, \frac{d{\mathbf{\omega}}_{_{_{A\to B}}}}{dt}\,\times\,{\mathbf q}\,+\, 2\,{\mathbf{\omega}}_{_{_{A\to B}}} \,\times\,\frac{d{\mathbf q}}{\left.dt\right|_B}\,+\, {\mathbf{\omega}}_{_{_{A\to B}}} \,\times\, ( {\mathbf{\omega}}_{_{_{A\to B}}} \,\times\,{\mathbf q}) \)

    Now denote frame A as the non-rotating ("inertial") frame I and frame B as the Earth-fixed frame R, ignore the tiny angular acceleration term, and take the second derivative of the position vector, and tada!

    \( \frac{d^2{\mathbf x}}{\left.dt^2\right|_I}\,=\, \frac{d^2{\mathbf q}}{\left.dt^2\right|_R}\,+\, 2\,{\mathbf{\omega}}_{_{\text{earth}}} \,\times\,\frac{d{\mathbf x}}{\left.dt\right|_R}\,+\, {\mathbf{\omega}}_{_{\text{earth}}} \,\times\, ( {\mathbf{\omega}}_{_{\text{earth}}} \,\times\,{\mathbf x}) \)

    Ignoring the angular acceleration is valid because the precession of the equinoxes has a period of 26,000 years, the lunisolar nutation has a period of 18.6 years, and the faster terms such as the Chandler wobble and polar motion are incredibly tiny in magnitude.
     
    Last edited: Aug 29, 2008
  15. temur man of no words Registered Senior Member

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    Thanks for the interesting link and explanation! What I meant was I can follow the mathematics line by line but just there was no “gut feeling” about Coriolis force. But now I have because of the rotating disk example above. However your derivation was a new way to me thanks for your time.
     
  16. Reiku Banned Banned

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    Simply to look at equations, can be about as fulfilling as looking at a peice of music. The notes denoted on a seperation of several octaves can look beautiful and neat, as much as a well described mathematical theorem can.
     
  17. Steve100 O͓͍̯̬̯̙͈̟̥̳̩͒̆̿ͬ̑̀̓̿͋ͬ ̙̳ͅ ̫̪̳͔O Valued Senior Member

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    I don't understand why the coriolis or centrifugal effects would make any difference when there is no friction.
     
  18. D H Some other guy Valued Senior Member

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    Friction has nothing to do with the Coriolis and centrifugal forces. The Coriolis and centrifugal forces are pseudo forces. They arise solely because the (a) the reference frame in which the equations of motion are expressed is not inertial and (b) we to use something akin to Newton's laws to describe the motion.
     
  19. Steve100 O͓͍̯̬̯̙͈̟̥̳̩͒̆̿ͬ̑̀̓̿͋ͬ ̙̳ͅ ̫̪̳͔O Valued Senior Member

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    Yes, but without friction the ball would not move from an outside observer's frame.
     
  20. D H Some other guy Valued Senior Member

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    You are forgetting about the normal force. The normal force is a constraint force. In the case that the normal component of the ball's inertial velocity relative to the sphere is zero, the normal force is whatever force is needed to make the normal component of the ball's inertial acceleration relative to the sphere be equal to zero.

    The OP has not yet specified what he means by "The question begins with a sphere sitting on a flat surface." As the OP raised the question with regard to the earth, most of us took this to mean that the sphere is initially at rest with respect to the Earth. I'll fill in some of these missing assumptions; the OP is more than welcome to change these assumptions (it's his problem after all).

    First consider the case of a rotating spherical Earth with the test sphere placed somewhere on the surface of the Earth and having zero inertial velocity with respect to the center of the Earth. Note that unless the test sphere is located at the north or south pole, the test sphere will have a significant velocity with respect to the point on the Earth directly underneath it. This situation is probably not what the OP was thinking about, but it is what you are thinking about. The Earth is spherical in this case, so the gravitational force is directed against the normal to the surface. The test sphere isn't moving inertially, so the normal force is simply the additive inverse of the gravitational force. When viewed from inertial space the test sphere does not move in this case.

    Now consider the case of a rotating spherical Earth with the test sphere placed somewhere on the surface of the Earth and having zero velocity with respect to the surface of the rotating Earth. The gravitational force is the same as above, but the normal force is not. Think of it this way: Suppose the Earth started spinning faster and faster. At some point in time, a loose rock resting on the surface at the equator will have achieved escape velocity and will simply fly away from the Earth. Just before the Earth reaches this critical rotation rate the rock is still held on the surface, but just barely. The Earth's rotation rate acts to reduce the normal force on an object at rest on the surface, making the test sphere will move toward the equator.
     
  21. Steve100 O͓͍̯̬̯̙͈̟̥̳̩͒̆̿ͬ̑̀̓̿͋ͬ ̙̳ͅ ̫̪̳͔O Valued Senior Member

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    But why would the rock speed up when the Earth did?
     
  22. D H Some other guy Valued Senior Member

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    I said what I did precisely because I want you the get out of the box into which you have trapped yourself. When someone says "Think about it this way," it means to look at the problem a bit differently. I specifically said "rock" rather than test sphere because it should be obvious that the rock will speed up as the planet rotates faster and faster. The goal was to make you realize that the normal force is a function of velocity. I thought that you might be able to see this if I could make you envision rocks flying off of the planet if it gradually spun up to a higher and higher rotation rate.

    Now suppose we have our a frictionless test sphere and a normal old rock of the same mass, located at the more-or-less the same location on the Earth, and both fixed with respect to the Earth. That the rock is subject to friction and the test sphere is not does not change the normal force. That both have a non-zero inertial velocity does change the normal force. The normal force is a function of velocity.
     
  23. Steve100 O͓͍̯̬̯̙͈̟̥̳̩͒̆̿ͬ̑̀̓̿͋ͬ ̙̳ͅ ̫̪̳͔O Valued Senior Member

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    Gotcha.

    I don't think the rock would actually take off though, because at the very moment it was traveling fast enough to take off, the force which enables it to do so would disappear.

    Much like a car with wings, I imagine.

    I hope.
     

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