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Will a sphere spinning on its own axis come to rest given enough time, provided no other forces act upon it?

I know that if you have two spinning spheres in the depths of space and orbiting each other they'll eventually stop spinning -- they'll tidally lock each other. The internal friction of the shifting tidal bulge eventually overcomes the rotation.

But I don't see what forces would stop a sphere from rotating around its own axis, in absence of other forces. And yet, each point on the sphere (except points along the axis) is experiencing accelerated motion, which implies a force.

My intuition says that something must maintain the acceleration of each point, which implies that the acceleration causes internal stress. I know this to be true macroscopically (the sphere will have an equatorial bulge), but what forces maintain the acceleration of each particle, and will the stresses ever overcome the rotation of the sphere?

Clarification: I am not considering infinitely rigid spheres or point particles. I'm wondering about actual non-rigid macroscopic spheres, such as planets.

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When two orbitting bodies become tidally locked, they have not stopped spinning. They have simply joined their angular momentum. The time to complete one rotation is the same as the time to complete one orbit. –  Mike Dunlavey May 23 '13 at 18:19
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Your single spinning sphere with no external forces sees only static forces internally. It can't dissipate energy, and of course there is no other place for the angular momentum to go, so two separate laws of conservation say it will spin indefinitely. Note that in the tidally locked case, energy is dissipated, but total angular momentum of the two bodies together is still conserved. –  Olin Lathrop May 23 '13 at 19:16
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An isolated body that doesn't exchange any angular momentum with the outside universe will never stop spinning (by conservation of angular momentum). There is no way to absorb angular momentum within the body in internal degrees of freedom; the angular momentum must be transported away if you want to stop.

For example: if something is not rigid, you could have what's called "differential rotation", where maybe the crust of the planet (say) is spinning one way, but the core is spinning the opposite way. Ignore the problem of how to make this happen, just imagine it does. The core has one angular momentum, while the crust has another. But we just add them up, and they have to add to the total angular momentum of the planet initially; it won't be able to stop entirely. You can redistribute angular momentum, but you cannot dissipate total angular momentum within an isolated object, and you cannot make the total angular momentum go to zero without transferring it to the outside universe.

Tidal locking is just another way to redistribute angular momentum. Tidal locking happens because the angular momentum of (to take the most familiar example) the Earth's spin is transported to the angular momentum of the Moon's orbit. The way this happens is that little mound of matter that's raised on the Earth doesn't quite point toward the Moon because Earth's spin carries it a little. And this little bulge exerts a gravitational pull on the Moon, making it speed up (gaining angular momentum). Newton's Third Law tells us that an equal and opposite force is exerted on the Earth, causing its rotation to slow down (losing angular momentum).

To emphasize, that angular momentum gets transferred from a spin to an orbit. And this happens because the Moon is actually orbiting, and thus raising a tide. So, to answer your question in the comments of Olof's answer, tidal locking won't generally stop an object's rotation entirely, because the other object will always be orbiting it. Tidal locking could, in principle, eventually just make the Earth spin at just the right rate so that it rotates as often as the Moon orbits. (Though I think the numbers don't work out.) In fact, this is what happened to the Moon, which is why we only ever see one face.

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Conservation of angular momentum means that the sphere will continue to spin forever. In order to change the angular momentum you need to apply an external torque.

Note that this treats the sphere as a rigid body. If you consider just a small part of the sphere there are forces acting on it in such a way that the sphere remains undeformed. On a microscopic level these forces would come for example from bonds between the atoms in the small part and atoms in the rest of the sphere. These forces make that part accelerate so that it keeps up with the rotation. However, in the absence of external forces on the sphere there will be no change in the momentum of the center of mass of the sphere, or in the angular momentum.

If the sphere is not completely rigid, interactions between different parts of the system can redistribute the angular momentum internally in the sphere. But the total angular momentum of the system is still preserved.

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I don't think that's the case. If it were, then how does the earth tidally lock the moon? Can't the total angular momentum of the system be preserved in the particles of the sphere while the sphere itself stops spinning? You're absolutely correct for point-particle spheres (though it's not clear what it means for a point particle to spin), but I'm wondering about the internal stresses of macroscopic spheres, and whether there is a force resisting the rotation. –  So8res May 23 '13 at 16:49
    
@So8res: If the sphere is not considered a rigid body (which I guess is what you mean by "point-particle sphere?") the sphere can stop spinning, since internal degrees of freedom in the sphere can pick up the angular momentum. –  Olof May 23 '13 at 16:56
    
That's what I thought -- so the sphere can stop spinning, but does it? –  So8res May 23 '13 at 17:13
    
-1. This is not correct. You cannot absorb angular momentum into internal degrees of freedom. –  Mike May 23 '13 at 17:29
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@Mike: I agree that "the sphere stops spinning" was very badly phrased. The system as a hole will always have the same angular momentum. I've edited my answer. Anyway, the question has been changed to indicate that it is not about a rigid sphere, which is what originally assumed, so my answer is no longer relevant. –  Olof May 23 '13 at 18:31
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