My question involves a modification of Newton's bucket experiment. If a sphere filled (say) one-third or one-half with water is rotated very very fast, will the water eventually spread out across and coat the entire interior surface of the sphere? If so, then does this mean that the sphere's inside is coated with water whose inertial force is everywhere at right angles to the surface? Might this offer a primitive mechanical model for the topological "gluing" of two ordinary 3-dimensional spheres, especially when the gluing is imagined according to the method of progressive longitudinal "slices"? (Unless it is better to imagine the two 3-d spheres “put through” each other, in keeping with the mechanical, force-oriented character of Newton’s experiment.) I'm trying to arrive at a historically plausible 18th-century version of 3-d "gluing" in order to visualize William Blake's Mundane Shell, which Bronowski dimly recognized as a 4-d sphere or torus (he wasn't sure which) in 1942.
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$\begingroup$ I think it would have to rotate infinitely fast for the water to coat the top. $\endgroup$– G. SmithCommented Jun 30, 2019 at 22:04
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1$\begingroup$ Hmmm, I wonder if that might fit with Blake's idea of Eternity as "expanding inward" at the vanishingly tiny center of a round(ish) physical object such as a globe, a flower, or a grain of sand.... $\endgroup$– Andrew CooperCommented Jun 30, 2019 at 22:51
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$\begingroup$ Duplicate: hsm.stackexchange.com/questions/9769/… $\endgroup$– safesphereCommented Jul 1, 2019 at 3:00
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1$\begingroup$ @AaronStevens The twins paradox on SE. $\endgroup$– safesphereCommented Jul 11, 2019 at 16:03
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1$\begingroup$ Sorry, safesphere and AaronStevens, the site won't let me change my profile for another 2-3 weeks, and I can't access Maxwell Cooper's profile. If the site administrator want to delete Maxwell, Andrew would be grateful. Seems odd that such Meta matters matter much on a physics site. $\endgroup$– Andrew CooperCommented Jul 12, 2019 at 15:06
1 Answer
Water will always take the shape of a surface with a constant potential (i.e a surface normal to the total force field). In a non rotating bucket (or sphere, or any container), the potential $U$ is only due to gravity and is given by: $$U(z)=gz+U_0$$ where $z$ is the height, and $U_0$ is some reference value. Thus, a surface with constant potential $U(z)=c$ is a horizontal plane with an elevation $z$ given by $$z=\dfrac{c-U_0}{g}.$$
In a rotating bucket, the total potential would be due to both gravity and centrifugal forces, and it is given by: $$U(z,r)=gz-\dfrac{1}{2}\Omega^2r^2+U_0,$$ where $\Omega$ is the angular velocity of the rotating container and $r$ is the radial distance from the axis. In this case, a surface of constant potential $U(z,r)=c$ is given by $$z=\dfrac{\Omega^2}{2g}r^2+\dfrac{c-U_0}{g},$$ which is an equation of a paraboloid. This analysis is independent from the shape of the container, thus the water surface will always be a paraboloid, whether it is in a cylindrical, cubical, conical, or a spherical bucket.
