Suppose that a cylindrical container of radius r and height L is filled with a liquid with volume V , and rotated along the y-axis with constant angular speed ω. This makes the liquid rotate, and eventually at the same angular speed as the container. The surface of the liquid becomes convex as the centrifugal force on the liquid increases with the distance from the axis of the container. The surface of the liquid is a paraboloid of revolution generated by rotating the parabola y = h + ω2x2/2g around the y-axis, where g is gravitational acceleration and h is shown below. (You can take g=32ft/s2 or 9.8m/s2). Express h as a function of ω. (2) At what angular speed ω will the surface of the liquid touch the bottom? At what speed will it spill over the top? (3) Suppose the radius of the container is 2 ft, the height is 7 ft, and the container and liquid are rotating at the same constant angular speed ω. The surface of the liquid is 5 ft below the top of the tank at the central axis and 4 ft below the top of the tank 1 ft out from the central axis. (It might help to draw a picture.) (a) Determine the angular speed of the container and the volume of the fluid. (b) How far below the top of the tank is the liquid at the wall of the container?

Algebra & Trigonometry with Analytic Geometry

13th Edition

ISBN:9781133382119

Author:Swokowski

Publisher:Swokowski

Chapter11: Topics From Analytic Geometry

Section: Chapter Questions

Problem 18T

See similar textbooks

Related questions

Concept explainers

Question

Suppose that a cylindrical container of radius r and height L is filled with a liquid with volume V , and rotated along the y-axis with constant angular speed ω. This makes the liquid rotate, and eventually at the same angular speed as the container. The surface of the liquid becomes convex as the centrifugal force on the liquid increases with the distance from the axis of the container. The surface of the liquid is a paraboloid of revolution generated by rotating the parabola

y = h + ω2x2/2g

around the y-axis, where g is gravitational acceleration and h is shown below. (You can take g=32ft/s² or 9.8m/s²).

Express h as a function of ω.
(2) At what angular speed ω will the surface of the liquid touch the bottom? At what speed will it spill over the top?
(3) Suppose the radius of the container is 2 ft, the height is 7 ft, and the container and liquid are rotating at the same constant angular speed ω. The surface of the liquid is 5 ft below the top of the tank at the central axis and 4 ft below the top of the tank 1 ft out from the central axis. (It might help to draw a picture.)

(a) Determine the angular speed of the container and the volume of the fluid.

(b) How far below the top of the tank is the liquid at the wall of the container?

Expert Solution

Step 1

Consider the given:

Express volume in terms of “h”, “r” and “w”.

$\begin{matrix} V = \int_{0}^{r} 2 π x y d x \\ V = \int_{0}^{r} 2 π x [h + \frac{ω^{2} x^{2}}{2 g}] d x \\ V = 2 π \int_{0}^{r} (h x + \frac{ω^{2} x^{3}}{2 g}) d x \\ V = 2 π {[h \frac{x^{2}}{2} + \frac{ω^{2} x^{4}}{8 g}]}_{0}^{r} \\ V = 2 π [h \frac{r^{2}}{2} + \frac{ω^{2} r^{4}}{8 g}] \\ \frac{V}{π} = h r^{2} + \frac{ω^{2} r^{4}}{4 g} \\ h r^{2} = \frac{V}{π} - \frac{ω^{2} r^{4}}{4 g} \\ h = \frac{4 V g - ω^{2} π r^{4}}{4 g π r^{2}} \end{matrix}$

Trending now

This is a popular solution!

Step by step

Solved in 3 steps

SEE SOLUTION Check out a sample Q&A here

Knowledge Booster

$Application of Integration$

Learn more about

Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, calculus and related others by exploring similar questions and additional content below.