Is there some force of friction that acts on bodies due to rotation of Earth?. We all know earth is an non inertial frame. If there is some frictional force is there some way to prove it?. What I mean is if I keep an object on earths surface will it experience some frictional force because of earths rotation and is there some way to mathematically deduce an expression for it. (Please be simple)
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I suspect this question is rooted in the widespread misconception that some external force is needed to keep the Earth rotation. That is not the case. Angular momentum is a conserved quantity. A rotating object that is not subject to any external torques will rotate with a constant angular momentum. This is the rotational analog of Newton's first law. An external torque is needed to change the angular momentum of a rotating object.
That said, there is a good amount of friction on and in the Earth due to the Earth's rotation. One example of a frictional force due to the Earth's rotation are the winds. Tropical cyclones (hurricanes and typhoons) are an extreme example of winds caused in part by the Earth's rotation. The seasonal changes in the distribution of the major winds over the course of a year result in small but observable changes in the Earth's rotation rate. See this answer regarding whether storms can make the day longer or shorter.
The atmosphere is a part of the Earth as whole. Changes within the Earth such as those described above don't change the Earth's total angular momentum a bit. That requires an external torque. Such a torque does exist. The ocean tides raised by the Sun and Moon result in friction at the ocean floor, and this very gradually slows the Earth's rotation rate. The evidence of this slow down is recorded in some banded rock formations.
David Hammen
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what if you place something on earths surface, will it be right to say friction force acts on it even if the object is at rest. – Einstein'sLover Nov 03 '14 at 07:11
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@AmritanshuSrivastava - To some extent. Consider a hockey puck at rest on a frozen lake. The hockey puck stays fixed even when a breeze blows across the lake. (This may not be the case if that breeze turns into a 100 kph wind.) However, friction is not needed with no breeze and no variations in the Earth's rotation rate. All that is needed is the normal force. – David Hammen Nov 03 '14 at 13:19
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-1 because there has to be a small friction force to keep bodies rotating along a latitude line when they are not on the equator – André Chalella Sep 25 '21 at 00:26
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I learned that the effect I described is canceled because of the oblate spheroid shape of the Earth. I'd like to retract my downvote, but it's locked until somebody edits the answer. – André Chalella Sep 25 '21 at 19:33
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No, there is no frictional force opposing centrifugal force. The shape of the earth is an oblate spheroid such that the surface of the earth, where the earth is flat (for instance: the surface of the ocean) is at right angles to the sum of the centrifugal force and the gravitational force. Thus the normal force and gravity provide all the necessary forces to maintain axial centripetal acceleration, that is, to keep the object stationary with respect to Earth's surface.
But in reality, weight is the only reason why the earth has its sphere-like shape at all. The natural structure of randomly slammed together rocky solids is a lumpy potato shape, as we can tell by looking at asteroids. What happens if we make an inclined plane out of a liquid? Obviously the liquid puddles down so that the surface of the liquid is normal to the local weight vector. What happens if we make a planet heavy enough that it's drawn into a sphere by its weight, like a liquid? The surface puddles down so that the surface is normal to the local weight vector, which is to say, the earth is a little bit thicker across the middle than it is from pole to pole: an oblate spheroid, not a sphere.
André Chalella's answer identifies a true statement:
Where the surface of the Earth is parallel to the surface of a sphere concentric with Earth, the sum of gravity, centrifugal force, and the normal force is small but nonzero and in the direction tangent to the surface and in the direction of the equator. If the object is not moving, there must be an equal and opposite force of static friction.
But he then imports a false statement: "the Earth can be approximated as a sphere concentric with Earth for this purpose", which leads to a false result: "a stationary object where the earth is flat detects such a frictional force". Since the very force which we are trying to calculate is the reason the Earth isn't a perfect sphere, we can't assume that the Earth is a perfect sphere when trying to calculate it.
g s
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So you're saying this is why planets become oblate spheroids, and now what I described doesn't happen on Earth. I'm glad to learn that. But I did some math with reference ellipsoid WGS-84 and found that the sum is still significantly nonzero. Did I pick an inappropriate ellipsoid? I'd also be more than happy with a trusty source that I could understand, but I've been unable to find one. Thanks for all the effort! – André Chalella Sep 25 '21 at 19:19
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I found sources! Had to look up "equatorial budge" instead of "Earth oblate spheroid" etc. I'm glad to have learned that the thing I thought of is the source of a major phenomenon. Do you think I should remove my answer? I'd also like to cancel my downvote in the other answer, but it's locked now until an edit is made. – André Chalella Sep 25 '21 at 19:31
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My personal thought would be to make a note explaining the mistake at the top of the mistaken answer. I don't know if there's a recommended policy. – g s Sep 25 '21 at 21:01
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I edited my answer. Feel free to suggest more modifications or other courses of action. Once again, thank you very much for your time and effort! – André Chalella Sep 26 '21 at 00:06
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Yes and no.
No, once the body is synced with Earth's rotation, there is no need for a friction force to do work to keep the body rotating. This is what the other answer meant, I think, and is probably what you wanted to know.
However, there may exist a friction force. It is small and does not produce any work, because it is perpendicular to the body's velocity.
This force is zero on the equator and is greatest in latitude 45º. It is zero at the poles too, because at the poles the (point) body doesn't travel in a circle (zero radius).
This force exists in all points of the surface where the surface is not locally tangent to an oblate spheroid which balances this force.
To understand this force, think of Earth as a simple sphere that spins along its North-South axis. Remember that Earth's rotation makes bodies all over the surface travel in a circular motion whose center is not the center of the Earth (except when the body is on the equator). That means that the resultant force on the body is a centripetal force pointing towards the North-South axis, but not to the center of the Earth. This force's direction is on a horizontal plane secant to Earth, containing the latitude line the body is on.
However, such resultant cannot arise from the "usual forces": weight and normal are on a straight line from Earth's center to the body. So there must be a tangential contact force to "fix" the direction of the resultant. That is the friction we're talking about.
If such friction did not exist, the body would travel along a great circle instead of a latitude line.
In my (simple) calculations the magnitude of the friction force is $m \omega^2 R \sin \theta \cos \theta$ ($R$ is Earth's radius and $\theta$ is latitude).
This force is the reason why Earth is not spheric, having become (approximately) an oblate spheroid (an ellipsoid) as mentioned earlier. Earth's radius is bigger on the equator than on the poles. Research equatorial bulge for more information.
I found this pic online that helps to understand the situation.
André Chalella
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I don't think this is right. You can sum $mr\omega^2 cos(\phi) \hat x - mgcos(\phi) \hat x - mgsin(\phi) \hat y + \vec F_n = \vec0$ without any need for friction. Normal force is under no obligation to be vertical, just opposite the sum of all forces between the supporting and the supported object. If we insist that $\vec F_n = -mg \hat r$, we don't end up with a missing friction term, we end up with the object drifting off into space because of its inertia. – g s Sep 25 '21 at 03:08
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1If we insisted that the Earth was exactly spherical (it's not, of course) then objects on its surface would want to move such that the Earth was closer to an oblate spheroid, so you'd end up with a friction force then, because towards the equator would be down-hill and we could treat the perfectly spherical earth as an inclined plane from high school physics. But since the Earth is an oblate spheroid, "downhill" has already been "filled in" by the matter uphill. – g s Sep 25 '21 at 03:20
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@gs sorry, I don't follow. What I mean is, the resultant is on a different direction than the weight, so there must be a tangential component in addition to the well known normal component. Maybe you can be more precise on your explanation? I cannot make a diagram due to my current circumstances, but maybe you can. – André Chalella Sep 25 '21 at 17:30
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I made an answer that responds to this. In the interest of making it a proper answer, rather than a jumped-up comment, I addressed it towards the OP, not to you. No disrespect is intended, I'm not trying to talk past you, just to meet the standards of the SE format. – g s Sep 25 '21 at 19:11