Problem Statement
A cylindrical vessel of radius 0.5 m is rotated at an angular velocity of 20 rad/s. Due to rotation, the free surface deforms and a volume of air appears above the original level. By equating the volume of air above the line AB to the volume of a corresponding paraboloid, determine the uncovered area at the top of the vessel.
Solution
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Given:
Vessel Radius, \( r = 0.5\,\text{m} \)
Angular velocity, \( \omega = 20\,\text{rad/s} \)
Acceleration due to gravity, \( g = 9.81\,\text{m/s}^2 \) -
Step 1: Compute \( z_3 \)
\( z_3 = \frac{r^2 \omega^2}{2g} = \frac{(0.5)^2 \times 20^2}{2 \times 9.81} \approx 5.1\,\text{m} \)
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Step 2: Express \( z_1 \) and \( z_2 \) in terms of \( r_1 \) and \( r_2 \)
\( z_1 = \frac{r_1^2 \omega^2}{2g} = 20.38\,r_1^2 \) (a)
\( z_2 = \frac{r_2^2 \omega^2}{2g} = 20.38\,r_2^2 \) (b)
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Step 3: Relate \( z_1 \) and \( z_2 \)
It is given that \( z_2 = z_1 + 1.5 \). Writing this using the expressions from (a) and (b):
\( 20.38\,r_1^2 + 2 = 20.38\,r_2^2 \) (c)Rearranging, we have:
\( 20.38\,r_2^2 – 20.38\,r_1^2 = 2 \) (d)Dividing both sides by 20.38:
\( r_2^2 – r_1^2 = 0.098 \) (e) -
Step 4: Volume Relationship
The volume of air above AB (of the original surface) is equal to the volume of the paraboloid (POQ-MON). That is,
\( \pi \times (0.5)^2 \times 0.5 = \frac{1}{2} \pi \Bigl( r_2^2 z_2 – r_1^2 z_1 \Bigr) \) (f)Substituting \( z_1 = 20.38\,r_1^2 \) and \( z_2 = 20.38\,r_2^2 \) into (f) and canceling common factors gives:
\( 0.25 = 20.38 \Bigl( r_2^4 – r_1^4 \Bigr) \) (g)Thus,
\( r_2^4 – r_1^4 = 0.0122 \) -
Step 5: Solve for \( r_1 \)
Note that \( r_2^4 – r_1^4 \) can be factored as:
\( (r_2^2 – r_1^2)(r_2^2 + r_1^2) = 0.0122 \)From (e), we have \( r_2^2 – r_1^2 = 0.098 \). Therefore,
\( 0.098\, (r_2^2 + r_1^2) = 0.0122 \)Solving for \( r_2^2 + r_1^2 \):
\( r_2^2 + r_1^2 = 0.1244 \)Now, with the system:
\( r_2^2 – r_1^2 = 0.098 \) and \( r_2^2 + r_1^2 = 0.1244 \)Adding the two equations:
\( 2r_2^2 = 0.2224 \quad\Longrightarrow\quad r_2^2 = 0.1112 \)Subtracting the first from the second:
\( 2r_1^2 = 0.1244 – 0.098 = 0.0264 \quad\Longrightarrow\quad r_1^2 = 0.0132 \)Hence,
\( r_1 = \sqrt{0.0132} \approx 0.115\,\text{m} \) -
Step 6: Determine the Uncovered Area
The uncovered area is given by the area corresponding to \( r_1 \):
Area \( = \pi r_1^2 = \pi \times (0.115)^2 \approx 0.0415\,\text{m}^2 \)
Explanation
In a rotating cylindrical vessel, centrifugal forces cause the free surface of the liquid to form a paraboloid. The vertical displacements at different radial positions are given by \( z = \frac{r^2 \omega^2}{2g} \). By relating the heights at two radii (\( r_1 \) and \( r_2 \)) and equating the volume of the air above the original level to the volume of the formed paraboloid, we obtain equations that allow solving for \( r_1 \). This value is then used to calculate the area of the region where the liquid has receded.
Physical Meaning
This analysis illustrates how the rotation of a vessel deforms the free surface of a liquid. The uncovered area represents the region from which the liquid has receded due to centrifugal forces. Understanding this phenomenon is important in the design and operation of rotating containers, as it affects both fluid distribution and pressure.
