Variation in pressure in a rotating tube?

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SUMMARY

The pressure variation inside a rotating tube filled with water is determined by the effective gravitational acceleration, which is influenced by the angular speed of the tube. In a rotating frame, the centripetal acceleration, expressed as ##\omega^2 r##, manifests as centrifugal acceleration pointing outward. The pressure at any point within the tube can be calculated using the formula ##p = \rho \omega^2 r^2##, where ##\rho## represents the density of water. At the center of the tube, the pressure is zero, and it increases with distance from the center.

PREREQUISITES
  • Understanding of angular velocity and its representation as ##\omega##.
  • Familiarity with the concepts of centripetal and centrifugal acceleration.
  • Basic knowledge of fluid mechanics, specifically pressure calculations.
  • Awareness of the properties of water, including its density denoted as ##\rho##.
NEXT STEPS
  • Research the effects of varying angular speeds on pressure distribution in rotating fluids.
  • Explore the implications of centrifugal force in engineering applications, such as centrifuges.
  • Study the relationship between pressure and density in different fluids beyond water.
  • Learn about the mathematical modeling of rotating systems in fluid dynamics.
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Students and professionals in physics, mechanical engineering, and fluid dynamics who are interested in the behavior of fluids in rotating systems.

Ravija
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What is the pressure experienced by any point at different locations inside a rotating tube filled with water?
The axis of rotation is through the center of the tube.
 
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Hi Ravija and welcome to PF.

Perhaps the easiest way to see this is in the rotating frame of the tube. Assuming that the rotation is in a horizontal plane, so that we don't have to worry about gravity, the centripetal acceleration points towards the center and is ##\omega^2 r## where ##r## is the distance from the center. In the reference frame of the rotating tube, this is seen as a centrifugal acceleration that has the same magnitude but points away from the center. The effective acceleration of gravity then is is ##g_{eff}=\omega^2 r## and becomes larger the more one moves away from the center. At the center the pressure is zero. At a point away from the center, the pressure is ##p = \rho g_{eff} r = \rho \omega^2 r^2##.

On edit: Here, ##\omega## is the angular speed of the tube and ##\rho## is the density of water.
 

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