Editing Bucket argument (section)

===Newton's laws of motion===
The shape of the surface of a rotating liquid in a bucket can be determined using Newton's laws for the various forces on an element of the surface. For example, see Knudsen and Hjorth.<ref name=Knudsen2>{{cite book |title=Elements of Newtonian Mechanics |edition=3rd |author=Jens M. Knudsen, Poul G. Hjorth |page=143 |url=https://books.google.com/books?id=Urumwws_lWUC&dq=rotating+fluid+bucket+%22centrifugal+force%22&pg=PA143 |publisher=Springer |year=2000 |isbn=3-540-67652-X}}</ref> The analysis begins with the free body diagram in the co-rotating frame where the water appears stationary. The height of the water ''h'' = ''h''(''r'') is a function of the radial distance ''r'' from the axis of rotation '''Ω''', and the aim is to determine this function. An element of water volume on the surface is shown to be subject to three forces: the vertical force due to gravity '''F'''<sub>g</sub>, the horizontal, radially outward centrifugal force '''F'''<sub>Cfgl</sub>, and the force normal to the surface of the water '''F'''<sub>n</sub> due to the rest of the water surrounding the selected element of surface. The force due to surrounding water is known to be normal to the surface of the water because a liquid in equilibrium cannot support [[shear stress]]es.<ref name=Lerner>{{cite book |author=Lawrence S. Lerner |title=Physics for Scientists and Engineers |page=404 |url=https://books.google.com/books?id=eJhkD0LKtJEC&dq=shear+stress+%22pascal%27s+principle%22&pg=PA404 |isbn=0-86720-479-6 |publisher=Jones & Bartlett |year=1997}}</ref> To quote Anthony and Brackett:<ref name=Anthony>{{cite book |title=Elementary Text-book of Physics |author=William Arnold Anthony & Cyrus Fogg Brackett |page=[https://archive.org/details/elementarytextb02bracgoog/page/n141 127] |url=https://archive.org/details/elementarytextb02bracgoog |quote=pascal's law. |year=1884 |publisher=Wiley}}</ref>
{{Blockquote|The surface of a fluid of uniform density..., if at rest, is everywhere perpendicular to the lines of force; for if this were not so, the force at a point on the surface could be resolved into two components, one perpendicular and the other tangent to the surface. But from the nature of a fluid, the tangential force would set up a motion of the fluid, which is contrary to the statement that the fluid is at rest.|William Arnold Anthony & Cyrus Fogg Brackett: ''Elementary Text-book of Physics'', p. 127}}  Moreover, because the element of water does not move, the sum of all three forces must be zero. To sum to zero, the force of the water must point oppositely to the sum of the centrifugal and gravity forces, which means the surface of the water must adjust so its normal points in this direction. (A very similar problem is the design of a [[Centripetal force#Example: The banked turn|banked turn]], where the slope of the turn is set so a car will not slide off the road. The analogy in the case of rotating bucket is that the element of water surface will "slide" up or down the surface unless the normal to the surface aligns with the vector resultant formed by the [[vector addition]] '''F'''<sub>g</sub> + '''F'''<sub>Cfgl</sub>.)

As ''r'' increases, the centrifugal force increases according to the relation (the equations are written per unit mass):
:<math>F_{\mathrm{Cfgl}} = m \Omega^2 r \ , </math>
where ''Ω'' is the constant rate of rotation of the water. The gravitational force is unchanged at
:<math>F_{\mathrm{g}} = mg \ , </math>
where ''g'' is the [[Gravitational acceleration|acceleration due to gravity]]. These two forces add to make a resultant at an angle ''φ'' from the vertical given by
:<math>\tan \varphi =\frac{F_{\mathrm{Cfgl}}}{F_{\mathrm{g}}} = \frac {\Omega^2 r}{g} \ ,</math>
which clearly becomes larger as ''r'' increases. To ensure that this resultant is normal to the surface of the water, and therefore can be effectively nulled by the force of the water beneath, the normal to the surface must have the same angle, that is,
:<math>\tan \varphi = \frac{\mathrm{d}h}{\mathrm{d}r} \ , </math>
leading to the ordinary differential equation for the shape of the surface:
:<math>\frac{\mathrm{d}h}{\mathrm{d}r} = \frac {\Omega^2 r}{g} \ ,</math>
or, integrating:
:<math> h(r) =h(0) + \frac{1}{2g} \left( \Omega r \right)^2 \ ,</math>
where ''h''(0) is the height of the water at ''r'' = 0. In other words, the surface of the water is parabolic in its dependence upon the radius.