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I have a donut shaped sealed pipe attached to a fixed axis by a frictionless bearing , the pipe is filled with an incompressible liquid initially at rest.

Within the pipe I have a small pump which should normally cause a flow around the circuit clockwise.

As this (via the bearing) is an isolated system I understand that the angular momentum is conserved. L=0

By this understanding the internal force on the fluid is countered by the opposite force on the body/pump so it should begin to rotate counter clockwise so that Net L is still zero L=0

While momentum is conserved at L we now have two opposite rotations of the body and the fluid

Is this physically possible ?

Is the relationship complicated by the energy that is put into vorticity ?

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  • $\begingroup$ Yes. Momentum is conserved, including angular momentum. $\endgroup$ – Mike Dunlavey Oct 9 '16 at 13:52
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There are some features that need to be addressed.

  • The pump is affixed to the tube (should be obvious)
  • The pump is massless (it's position in the tube does not affect the center of mass of the tube)

If both true then the whole system is equivalent to a pair of cocentric counter-rotating flywheels with L=0. The tube/pump is one and the fluid being the other. They may have different angular speeds if they have different masses but their momentums should cancel due to Newton's third law (isolated system of two bodies).

Introducing friction in the flow does not change the result, only the angular speeds of these two bodies.

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Yes, this is physically possible. For simplicity, visualise a water hose, when the water is thrust out by the pump, an equal force acts on you. F = d(mv)/dt. So, since the pump is connected to the body, the force will act on the body in the opposite direction, and will this cause opposite rotations. Assuming no interaction(friction) between the water and the inner surface, and the water to be a non viscous fluid, there will be no vortex. As soon as you have viscosity into the consideration, you will be having a symmetric gradient, thus no vortex still, though there will be a possibility of turbulence near the pump, since the pump will try to keep the velocity constant along the radius(cross section) at the outlet, but the inlet will be having a gradual gradient.

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  • $\begingroup$ Thanks for your reply ! Can you please expand a bit on why there would be no vorticity .If I thought of this system like a rotating bucket then it would be a solid body rotation so would require Vorticity. This vorticity, since we started with none, must have been added by the pump ? $\endgroup$ – Tom Chester Oct 9 '16 at 10:11
  • $\begingroup$ Yes, you're right in saying about the vorticity in case of a bucket. That happens because there is no symmetry to cancel the effect of viscosity(lower and higher speeds). The toroid(or a doughnut) has water going through it, perpendicular to the cross sectional area, so the viscous forces are same along all radial directions (in the cross sectional area) and this symmetry cancels the reason for there to exist any vortex $\endgroup$ – Pranshu Malik Oct 9 '16 at 10:23
  • $\begingroup$ A bucket would be having a pump thrusting water parallel to the cross sectional area, and thus all velocities at a particular distance from the center is unique, and velocity changes at all radial distances, and thus you see a vortex at the center $\endgroup$ – Pranshu Malik Oct 9 '16 at 10:25

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