Overflow from a Spinning Container A cylindrical container of height H and radius a is initially half-filled with a liquid. The cylinder is then spun steadily around its vertical axis Z-Z, as shown in Fig. E1.8. At what value of the angular velocity ω will the liquid just start to spill over the to

Question

Overflow from a Spinning Container

A cylindrical container of height [latex]H[/latex] and radius [latex]a[/latex] is initially half-filled with a liquid. The cylinder is then spun steadily around its vertical axis Z-Z, as shown in Fig. E1.8. At what value of the angular velocity [latex]\omega[/latex] will the liquid just start to spill over the top of the container? If [latex]H=1 \mathrm{ft}[/latex] and [latex]a=0.25 \mathrm{ft}[/latex], how many rpm (revolutions per minute) would be needed?

Accepted Answer

From Eqn. (1.47), the shape of the free surface is a parabola. Therefore, the air inside the rotating cylinder forms a paraboloid of revolution, whose volume is known from calculus to be exactly one-half of the volume of the "circumscribing cylinder," namely, the container. [latex]{ }^{8}[/latex] Hence, the liquid at the center reaches the bottom of the cylinder just as the liquid at the curved wall reaches the top of the cylinder. In Eqn. (1.47), therefore, set [latex]z=H[/latex] and [latex]r=a[/latex], giving the required angular velocity:

[latex]z=\frac{\omega^{2} r^{2}}{2 g}[/latex]. (1.47)

[latex]\omega=\sqrt{\frac{2 g H}{a^{2}}}[/latex]

For the stated values:

[latex]\omega=\sqrt{\frac{2 imes 32.2 imes 1}{0.25^{2}}}=32.1 \frac{\mathrm{rad}}{\mathrm{s}}, \quad N=\frac{\omega}{2 \pi}=\frac{32.1 imes 60}{2 \pi}=306.5 \mathrm{rpm}.[/latex]

[latex]^{8}[/latex] Proof can be accomplished as follows. First, note for the parabolic surface in Fig. E1.8(b), [latex]r=a[/latex] when [latex]z=H[/latex], so, from Eqn. (1.47), [latex]\omega^{2} / 2 g=H / a^{2}[/latex]. Thus, Eqn. (1.47) can be rewritten as:

[latex]z=\frac{\omega^{2} r^{2}}{2 g}[/latex]. (1.47)

[latex]z=H \frac{r^{2}}{a^{2}} .[/latex]

The volume of the paraboloid of air within the cylinder is therefore:

[latex]V=\int_{z=0}^{z=H} \pi r^{2} d z=\int_{z=0}^{z=H} \frac{\pi a^{2} z}{H} d z=\frac{1}{2} \pi a^{2} H,[/latex]

which is exactly one-half of the volume of the cylinder, [latex]\pi a^{2} H[/latex]. Since the container was initially just half filled, the liquid volume still accounts for the remaining half.

Question 1.8: Overflow from a Spinning Container A cylindrical container o...

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