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This question came to me while I was in the pool last month. I took a basketball and I was making it spin on the surface of the water in a few different ways. When the ball rested on the surface of the water, a majority of the ball was above the surface of the water, indicating that the density of the ball is less than half of the density of water.

First I would spin the basketball so that the angular momentum vector was vertical, perpendicular to the water surface. Then I spun it so that the angular momentum vector was horizontal, parallel to the surface of the water. In each case, the spinning would slow down, although in the first case where the angular momentum vector was vertical the ball took longer to slow down.

Finally, I spun the basketball at an angle, so that the angular momentum vector was neither parallel nor perpendicular to the water surface. What I saw was consistent with what I saw earlier. The horizontal component of the angular momentum vector decayed more quickly than the vertical component, so that after maybe 10 seconds the rotation was essentially with a vertical axis.

So the question is why? Why does the horizontal component decay faster than the vertical component? I tried to think of it in terms of friction and normal forces, but there didn't seem to be a difference between the 2 cases.

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As a fluid mechanics guy, it seems clear to me that the resistance to spinning with a vertical axis should be much less than when the ball is spinning with a horizontal axis. Viscous friction is very different from ideal dry solid friction (where the shear stress is independent of relative velocity). In the case of viscous friction, the local drag shear stress on the object (ball) is dependent on the local velocity of the ball surface, relative to the (zero) velocity of the stationary fluid far from the ball. The velocities at the surface in each case are equal to the angular velocity times the distance from the axis of rotation. In the case of vertical rotation, the radius of rotation runs from zero at the vertical axis to the submerged contact radius (less than the ball radius) on the edge. In the case of vertical rotation, the radius of rotation is on the order of the ball radius at all locations. So the shear velocities (and surface shear stresses) in the vertical axis case will, on average, be much lower than those with a horizontal axis.

But, there is more to the story than this. The rate of change of angular velocity of the spinning ball is determined not by the local shear stress, but by the moment of the shear stress about the axis of rotation, integrated over the submerged surface. Since the radial moment arms in the vertical axis case are going to be significantly smaller than those for the horizontal axis case, the effect is magnified even more. The net results is that, for a given rotation rate, the drag torque with a vertical axis of rotation is going to be much less than the drag torque with a horizontal axis of rotation.

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