## Newton’s Spinning Bucket

A few weeks ago Andre Asiss gave a talk about spinning bucket and the possible implications to absolute space. Like Franklin Hu expressed at the meeting/conference, I too felt (and still do) that a simpler understanding exists.

Anyhow I sketched out a quick solution, and then blogged it here, and as things that are done too quickly my solution was not necessarily correct. I realized that the units did not match the next day and thought someone would jump all over it but nothing happened.

Since I do not like leaving messes (if I can) I have restructured my thoughts and also had a few email exchanges with Dr Andre Asiss who was very kind in his responses. Now the following is my solution and has nothing to do with Dr Asiss, whose solution remains a superior but more complex solution as given in Chapter 9 of his book on “Rotational Mechanics” (namely see Asiss’s equations 9.42 thru 9.45). The thing is that my solution shows the relationship between gravity and angular acceleration that may not be as obvious in Dr. Asiss’s somewhat more complex solution.

Reconsider the simple experiment that being the spinning bucket that was first devised by Newton and was more recently discussed by Andre Asiss^{1} in an open conference. Start with a bucket at rest and realize that the tensile layer remains stable and for the most part flat. Next we hang the bucket from a rope, twist the rope as much as we can and let the bucket freely spin due to tension within the rope.

At first the bucket will spin faster than the water within, but eventually the bucket and water will spin with the same angular velocity (: *v* is velocity, *r* is radius, *f* is frequency), at which point the tensile layer should be stable, resembling Fig 2.10.1.

From a pressure perspective we can say that the pressure for all molecules along the tensile layer is: *P*tensile = 1 atm. This applies to molecules along the tensile layer for both Fig 1 and Fig 2. Since the tensile layer is a line of equal pressure, that being 1 atm; can we not claim that the reason that molecule #2 is at a height (*h*) above the parabola’s vertex, is due to the increase in pressure associated with the centripetal forces? This certainly makes sense, which is to say that the pressure exerted by the added height (*h*) of the liquid along the *y*-axis must equal the pressure increase due to centripetal forces associated with the rotation.

We can expand the above and say; normally a centripetal force along the *x*–*z* plane will cause the force to be only along that *x*–*z* plane. If it was not for the bucket’s walls the liquid would spill radially outwards along the* x*–*z* plane and continually so, if there was no gravity. However, for the liquid in the spinning bucket, the *x*–*z* plane is limited by bucket’s walls. Therefore the centripetal forces results in a pressure increase along the *x*–*z* plane, which must equal the pressure increases along the *y*-axis.

Realizing that pressure is force over area and then considering a unit of surface area allows us to consider changes in forces, which is really due to pressure. Consider molecule 1, which is elevated above molecule 6 by height (*h*). The weight of that unit column is:

weight=ρgh 1)

Where ρ is the liquid’s density, g is the gravitational force and h is the height as shown. The force exerted by this unit column downwards along the negative *y*-axis should equally be:

F=ρgh 2)

Next consider the forces in the *x*–*z* plane. The centripetal acceleration (*a _{c}*) is:

a=v²/r=ω²r 3)

What is the centripetal force of a unit volume of liquid? Consider a unit column in the *x*–*z* plane has a mass as defined by the liquid’s density multiplied by radius (r). However for any rotating body the centripetal forces are measured at the center of mass hence instead of *r* we should use *r*/2, when equating these forces. Therefore for an angular velocity (*w*) and radius (*r*) the centripetal force due to centripetal acceleration for a unit column is :

F=aρ(r/2)=(ω²/2r)(ρr)=ρω²/2 4)

Equating the forces of the unit column along the *y *axis to *x*–*z* plane:

ρgh=ρω²/2 5)

Dividing both sides by density (ρ) gives:

gh=ω²/2=ar/2 6)

It is an interesting result as the densities in question drop out as; 2*h* vs *r *becomes inversely proportional to; *g* vs *a _{c}*. This is a very interesting result as it also shows why gravity is nothing more than a form of acceleration, which has indirect bearings on subjects that so many in this group are highly involved in i.e. relativity. Of course to me, I have trouble getting the same sense of absolute space arguments that Dr. Asiss asserts. Perhaps it is because I remain the simpleton.

Hopefully others will have improved insights that you may convey

One should further note that if we took a stir stick on a high speed drill, and inserted it into a bucket of water, then the vortex that we witness will resemble Newton’s spinning bucket. More complex/complete mathematical analysis of the spinning bucket can be found^{1,2,3}. Of these solutions you will find that Asiss’s is the most complex, Munson’s and the “Bucket argument” are a more complex version of my overly simplified understanding.

- A. Asiss “Relational Mechanics and implication of Mach’s Principle with Weber’s Gravitational Force” Aperion Montreal 2014

- B.R. Munson, T.H. Okiishi, and W.W. Huebsch “Fundamentals of Fluid mechanics, 6
^{th}edition., Hoboken NJ John Wiley & Sons 2010

See: http://demonstrations.wolfram.com/PressureOfARotatingFluid/

- “Bucket Argument” (Feb 2017): internet: https://en.wikipedia.org/wiki/Bucket_argument

Thanks you again for reading what I wrote

Kent

Off topic: I also remain confused as I see that the other week’s blogs I thought that were no responses, but seemingly they were. Strangely I received no email notifications of responses. Am I doing something wrong?