Problem Statement
A cylindrical tank of 1.5m in diameter and 3m in height contains water to a depth of 2.5m. Determine the speed of the tank so that 20% of the original volume is spilled out.
Solution
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Given Data:
Radius of the tank: \( r = 0.75m \)
Initial water depth: \( h = 2.5m \)
Height of the tank: \( H = 3m \) -
Water Depth After Spill:
Since 20% of the volume is spilled, the remaining depth is:
New depth = \( 0.8 \times 2.5 = 2m \)Rise in water level at the wall:
Rise = \( 3m – 2m = 1m \)Fall at the center:
\( z = 2m \) -
Equation of Parabolic Surface:
\( z = \frac{r^2 \omega^2}{2g} \)Substituting values:
\( 2 = \frac{(0.75)^2 \omega^2}{2g} \)Solving for \( \omega \):
\( \omega = 8.35 \, rad/s \)Converting to RPM:
\( N = \frac{60 \times 8.35}{2\pi} = 80 \, rpm \)
Explanation
When a cylindrical tank rotates, the water surface forms a parabolic shape due to centrifugal forces. The water moves outward, causing a rise at the tank’s walls and a fall at the center. The given problem states that 20% of the water volume is spilled, reducing the effective water level. Using the equation of the parabolic surface, we determine the required angular velocity.
Physical Meaning
The derived angular velocity represents the critical speed at which centrifugal forces balance the hydrostatic forces, creating a water surface depression at the center. If the tank rotates faster, more water will spill out. If it rotates slower, the depression at the center will be smaller.
