An open cylindrical tank, 6 ft high and 3 ft in diameter, contains 4.50 ft of water. If the cylinder rotates about its geometric axis, (a) what constant angular velocity can be attained without spilling any water? (b) What is the pressure at the bottom of the tank at C and D (Fig. 5-6), when ω = 6.

Question

An open cylindrical tank, 6 ft high and 3 ft in diameter, contains 4.50 ft of water. If the cylinder rotates about its geometric axis, (a) what constant angular velocity can be attained without spilling any water? (b) What is the pressure at the bottom of the tank at C and D (Fig. 5-6), when ω = 6.00 rad/sec?

Accepted Answer

(a) Volume of paraboloid of revolution = [latex]\frac{1}{2} [/latex](volume circumscribed cylinder) [latex]=\frac{1}{2} \left[\frac{1}{4}\pi 3^2(1.50+y_{1}) ight] [/latex] .

If no liquid is spilled, this volume equals the volume above the original water level AA, or

[latex]\frac{1}{2} \left[\frac{1}{4}\pi 3^2(1.50+y_{1}) ight] =\frac{1}{4}\pi 3^2(1.50)[/latex]

and y₁ = 1.50 ft.

To generalize, the point in the axis of rotation drops by an amount equal to the rise of the liquid at the walls of the vessel.

From this information, the x and y coordinates of points B are respectively 1.50 and 3.00 ft from origin S. Then

[latex]y=\frac{\omega ^2}{2g}x^2 [/latex]

[latex]3.00=\frac{\omega ^2}{2 imes 32.2}(1.50) [/latex]

and ω = 9.27 rad/sec.

(b) For ω = 6.00 rad/sec,

[latex]y=\frac{\omega ^2}{2g}x^2 =\frac{(6.00)^2}{(2)(32.2)}(1.50)^2=1.26 [/latex] ft from S

Origin S drops [latex]\frac{1}{2}y=0.63 [/latex] ft, and S is now 4.50 - 0.63 = 3.87 ft from the bottom of the tank. At the walls of the tank the depth = 3.87 + 1.26 = 5.13 ft (or 4.50 + 0.63 = 5.13 ft).

At C, p_C = γh = 62.4 x 3.87 = 241 psf

At D, p_D = γh = 62.4 x 5.13 = 320 psf

Question 5.8: An open cylindrical tank, 6 ft high and 3 ft in diameter, co...

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