Problem Statement
An open circular cylindrical pipe of radius R and height h is completely filled with water. The cylinder is rotated about its axis at an angular velocity ω. Determine the value of ω in terms of R and h such that the diameter of the exposed center portion is equal to the radius of the cylinder.
Solution
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Given:
Radius of the cylinder: \( R \)
Diameter of exposed center: \( R \)
Radius of exposed center: \( R/2 \) -
Equation of the Parabolic Surface:
The free surface follows a parabolic shape given by:
\( z = \frac{R^2 \omega^2}{2g} \) -
At the tank wall (Point A and B):
The water surface depression at the tank wall:
\( x + h = \frac{R^2 \omega^2}{2g} \) (Equation 1) -
At the exposed center (Point C and D):
The depression at the center where \( x = 0 \) and \( r_1 = R/2 \):
\( x = \frac{(R/2)^2 \omega^2}{2g} = \frac{R^2 \omega^2}{8g} \) (Equation 2) -
Solving for Angular Velocity \( \omega \):
Subtracting Equation 2 from Equation 1:
\( h = \frac{R^2 \omega^2}{2g} – \frac{R^2 \omega^2}{8g} \)Rearranging:
\( h = \frac{3 R^2 \omega^2}{8g} \)Solving for \( \omega \):
\( \omega = \sqrt{\frac{8gh}{3R^2}} \)
Explanation
As the cylinder rotates, the water surface forms a paraboloid. The central exposed portion occurs when the depression at the center equals the required value. By equating the derived expressions at the wall and center, the required angular velocity is determined.
