# How long will it take for rotating spherical container full of viscous liquid to stop rotating when there is no gravity?

Suppose there is a hollow ball of radius 1 meter. It is filled some viscous liquid. Then the ball is rotated at 1 radian per second. So the speed of any point on the ball's surface is 1 meter per second.

Also let us assume there is no gravity.

My understanding of viscosity is not too deep. So the next argument I am going to make may be faulty. From what I understand viscous forces make adjacent layers of liquid to move at the same speed. Suppose we keep applying external torque until every point in the liquid is moving at 1 meter per second. Now there will be no viscous force in the liquid and if we stop applying external torque this ball will go on rotating indefinitely.

But if the speed is same for every point in the liquid that must mean the angular speed is inversely proportional to distance from center. This means that water nearer to the center will have very high angular speed. This seems to be unrealistic.

So it seems that we cannot really achieve zero viscous force. There must be some limit on how high the angular speed can go. I think for some fixed torque we may be able to find a relationship between angular speed and distance from center.

Since there will always be a viscous force the ball will eventually stop rotating. My question is about how long it will be before the ball stops.

There seems to be two step to solve this problem. First find out out how the angular speed varies with distance at a steady state condition for some fixed torque and second from this find out how the angular speed of the ball will change with time when the external torque is removed.

I would love to analyze this problem further but I lack the mathematical background. My grasp of viscosity is elementary and most of the material I have found in the internet use complicated vector calculus to describe viscosity.

I would love it if you give an answer without using vector calculus. I would also love it if you just give some pointer to relevant reading material that are easy to follow.

• Just a little tidbit on what's wrong with trying to give it a constant angular momentum. The no slip condition will apply to viscous fluids where they interface with a wall. When you stop rotating the sphere, the fluid at the edge should stop with it, causing the angular rotation and straight line velocity to vary based on radius. – JMac Jan 20 '17 at 11:03
• I did not mean to say I want to stop rotating the ball. I wanted to say I would stop giving the torque. I will edit the question to make it more clear. Thanks for pointing it out. – Ibraheem Moosa Jan 20 '17 at 11:14
• The problem is difficult enough despite using vector calculus. In a lighter vein, your request to answer the question without using vector calculus is like asking one to win against nine-tailed fox (from Naruto) with hands and feet tied (not that you can win even if your hands and feet were untied). Perhaps someone can come up with an order-of-magnitude estimate. Also in solid body rotation angular speed, and not speed itself, is the same for all points in the fluid. – Deep Jan 20 '17 at 11:30
• I'm not sure if having the high angular velocity in the centre would really constitute a problem in the first place. I personally don't see how this system would slow down unless you consider more realistic factors like the surface friction slowing down the rotating sphere. If there's nowhere for the energy to dissapate it will not slow down according to Newtonian physics. – JMac Jan 20 '17 at 13:24
• I had a similar problem on a take home fluids exam, and it was just order-of-magnitude. Considering friction with the container was wrong by a lot, and the answer involved diffusion of angular momentum. – JEB Jun 24 '19 at 2:33

So the next argument I am going to make may be faulty. From what I understand viscous forces make adjacent layers of liquid to move at the same speed. Suppose we keep applying external torque until every point in the liquid is moving at 1 meter per second. Now there will be no viscous force in the liquid and if we stop applying external torque this ball will go on rotating indefinitely.

No, it isn't faulty. Given enough time all the layers will eventually move at the same angular velocity. It takes time because the viscous forces between the layers provide accelerating torque but even with acceleration taking place, it takes time for all layers to move at the same speed.

But if the speed is same for every point in the liquid that must mean the angular speed is inversely proportional to distance from center. This means that water nearer to the center will have very high angular speed. This seems to be unrealistic.

So it seems that we cannot really achieve zero viscous force. There must be some limit on how high the angular speed can go. I think for some fixed torque we may be able to find a relationship between angular speed and distance from center.

Here you're conflating things: angular velocity and tangential speed.

The angular velocity $$\omega$$ (in $$\mathrm{rad/s}$$) is independent of the distance from the center, at least when we've achieved equilibrium as described above.

But tangential speed $$v$$ does depend on distance from the center $$r$$

$$v=\omega r$$

Since there will always be a viscous force the ball will eventually stop rotating. My question is about how long it will be before the ball stops.

This is in contradiction with what you correctly wrote higher up:

Now there will be no viscous force in the liquid and if we stop applying external torque this ball will go on rotating indefinitely.

Without viscous forces in the liquid (at equilibrium) the sphere will keep rotating at constant $$\omega$$ indefinitely. The only way to change that, by Newton's second law, is to apply a braking torque to the shell. The layers of fluid will then continue to rotate but will experience angular deceleration due to viscous shear force. First the layer next to the shell will slow down, this will then affect the 'next' layer and so on.

The principle is the basis of a simple party trick in which participants are asked to try and spin a hard boild egg and a raw one (without knowing which is which). The raw egg is much harder to get to spin, due to viscous forces between the fluid's layers.

• I think it is worth mentioning that, even though the tangential velocity is varying with radius, this does not constitute a viscous deformation of the fluid, since deformations are only relevant over and above rigid body rotation. – Chet Miller Jan 20 '17 at 13:02
• Good point Chester. – Gert Jan 20 '17 at 13:05
• I thought viscous force between adjacent layers will be zero if both of the layers have same tangential velocity. – Ibraheem Moosa Jan 20 '17 at 13:07
• I did not quite understand "deformations are only relevant over and above rigid body rotation". Can you please elaborate. @ChesterMiller – Ibraheem Moosa Jan 20 '17 at 13:10
• If a fluid is experiencing a rigid body rotation, you don't expect any viscous stresses to be present. Newton's law of viscosity in 3D accounts for this by subtracting the local vorticity tensor from the velocity gradient tensor to yield the rate of deformation tensor. The viscous portion of the stress tensor is equal to twice the viscosity times the rate of deformation tensor. Look up the relationships for the viscous shear stresses in cylindrical coordinates. You will see that, for tangential velocity variations in the radial direction, only the derivative of angular velocity is importan – Chet Miller Jan 20 '17 at 13:17

The moment of inertia of a fluid body is difficult to calculate. Non-rigid mechanics means that we have to go to the integration of fluids from Navier Stokes terms. Look that up. The idea is to use the continuity equation ddt rho. The rho is the density. That density also appears in the force equation which is how you get a characteristic frequency for the rotation of the bulk fluid as a function of time. It should be inversely proportional to the mass of the object, the radius of the pointlike particle if in the Brownian regime, and it should have some dependence on the viscosity. A good way to go is to find the force from the viscosity term proportional to the velocity squared (for obvious reasons). That term is related to the collision time between particles, but in a bulk fluid, we approximate collisions from kinetic theory. Don’t worry about rarefied rotating fluids, those are tough to work out, use a particle model instead. If this is unclear, then use the divergence and time derivative in the continuity equation, plus all the forces acting on a test spherical section. The answer should show that inverse mass dependence is my bet.