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01 Rotating Liquid

Aim

To show that the surface of a rotating liquid forms a paraboloid

Subjects

Diagram

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Figure 1:.

Equipment

Presentation

The glass beaker is half filled with water. The beaker is submerged in a square reservoir (see Diagram). By means of the electric motor the glass is made rotating. Gradually the liquid climbs the wall of the beaker until it settles itself. The paraboloidic shape can be seen clearly. By means of a videocamera and projector, the paraboloid is projected on the blackboard. Chalk is used to draw the shape of the parabola on the blackboard. Now it is checked that the drawn shape is really paraboloidic by looking for the focal point (F) and course line(c). Our experience is that the positions of this point and line are found quickly by trial and error (until the distances of focal point and course line to the drawn line are equal: see Figure 2).

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Figure 2:.

Explanation

  1. In a rotating reference frame the liquid is in static equilibrium. In this reference frame the sum of the forces acting on the particles in the surface will be perpendicular to that surface. Two forces are acting on such a particle dm: gravity, F1=dmgF_{1}=d m g and the centrifugal force, F2=dmω2F_{2}=d m \omega^{2} r. Figure 3 shows: tan⁡α=dydx=ω2xg\tan \alpha=\frac{d y}{d x}=\frac{\omega^{2} x}{g} and from this y=12ω2x2g+cy=\frac{1}{2} \frac{\omega^{2} x^{2}}{g}+c. This is the formula of a parabola.

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Figure 3:.

The constant cc indicates the position of the lowest point of the rotating liquid. If the xx axis in Figure 2 is located in the surface of the liquid at ω=0\omega=0, then because of the conservation of mass and the assumed incompressibility of the water, one obtains: ∫0aydx=0\int_{0}^{a} y d x=0 After integration we find: c=−16ω2a2gc=-\frac{1}{6} \frac{\omega^{2} a^{2}}{g}

  1. Explaining can also be done from the point of view of hydrostatics (see Figure 4).

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Figure 4:.

Pressure in the liquid is a function of rr and zz. When rotating, the forcefield has two components: gravity, ∂p∂z=−ρg\frac{\partial p}{\partial z}=-\rho g and centrifugal, ∂p∂r=ρω2r\frac{\partial p}{\partial r}=\rho \omega^{2} r. So dp=−ρgdz+ρω2rdrd p=-\rho g d z+\rho \omega^{2} r d r. After integration: p=−ρgz+12ρω2r2+cp=-\rho g z+\frac{1}{2} \rho \omega^{2} r^{2}+c. So surfaces with equal pressure are determined by z=ω22gr2+const.z=\frac{\omega^{2}}{2 g} r^{2}+const., showing the parabolic relationship.

Remarks

Sources

1R10 Hooke’s Law
01 duplicate
2C20 Bernoulli Force
01 Magnus Effect (1)