What density do I use to calculate buoyancy for a hollow cylinder float?

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Discussion Overview

The discussion revolves around calculating buoyancy for a hollow cylinder float, specifically addressing which density to use in the calculations—air density or water density—given that the cylinder is immersed in water with air trapped inside.

Discussion Character

  • Exploratory
  • Technical explanation
  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • One participant questions whether to use air density or the density of the cylinder for buoyancy calculations.
  • Another participant states that if the cylinder is immersed in water, the density of water should be used, providing a specific value for water density.
  • A participant references Archimedes' principle, emphasizing the importance of the weight of the displaced fluid.
  • One participant notes that the accuracy of the buoyancy calculation may depend on the depth of immersion and the compression of air at that depth, suggesting that the volume of air may be compressed by about 3% due to pressure.
  • Another participant agrees with the previous point but suggests that for most practical purposes, using the volume of the cylinder and the density of water would suffice, although they express doubt about whether the upthrust would be sufficient to keep the cylinder afloat.
  • A later reply mentions that the extent to which the cylinder submerges will depend on its mass, drawing a parallel to personal experience with buoyancy at depth.

Areas of Agreement / Disagreement

Participants express differing views on the necessity of accounting for air compression and the accuracy required for the buoyancy calculation. There is no consensus on whether the additional complexity of air compression is relevant for the problem at hand.

Contextual Notes

Participants highlight that the accuracy of the buoyancy calculation may vary based on the depth of immersion and the specific conditions of the cylinder and fluid. The discussion includes assumptions about the conditions under which the calculations are made, such as the depth of water and the density values used.

Who May Find This Useful

This discussion may be useful for individuals interested in fluid mechanics, engineering applications involving buoyancy, or those seeking to understand the complexities of calculating buoyancy for hollow objects in fluids.

mido
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I have a hollow cylinder float as seen attached, trying to work out the buoyancy. I work out the inner volume for the cylinder, but do I use air density or cylinder density ?
 

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What is the fluid it will be immersed in? If it is air, then you should use air density ρ=1.225kg/m3 at 15° to 20°C at sea level.

Buoyancy depends on the density of the fluid and the volume of the object,
FB=ρfluidVobjectg
 
As Archimedes is reputed to have said: "the weight of the displaced fluid"
 
it will be immersed in water, it is a hollow cylinder with air trapped in
 
The solution to this depends upon how accurate you need the answer. The air/water interface at the bottom is under pressure ρgh and the volume will be less than the cylinder volume, so the upthrust will be less. You may need to account for the volume of air being under water pressure of around 0.3m depth (about 1/30 atmospheric pressure) - which means that the original volume of air will be compressed additionally by about 3% (back of a fag packet calculation). Is that relevant for you? A more accurate answer can be obtained if you are bothered about greater accuracy than a percent.
Otherwise you can say the upthrust will be the volume of the cylinder times the density of water. (minus the actual weight of the material of the cylinder, of course.)
 
sophiecentaur said:
The solution to this depends upon how accurate you need the answer. The air/water interface at the bottom is under pressure ρgh and the volume will be less than the cylinder volume, so the upthrust will be less. You may need to account for the volume of air being under water pressure of around 0.3m depth (about 1/30 atmospheric pressure) - which means that the original volume of air will be compressed additionally by about 3% (back of a fag packet calculation). Is that relevant for you? A more accurate answer can be obtained if you are bothered about greater accuracy than a percent.
Otherwise you can say the upthrust will be the volume of the cylinder times the density of water. (minus the actual weight of the material of the cylinder, of course.)
that is correct, but i don't think he would want to go for THAT accuracy, involving the extra compression of air by 3%. He could simply obtain the volume, plug in the value of ρwater= 1000kg/m3 along with g=9.8m/s2 and voila.
ALTHOUGH yes, i don't think the upthrust would be enough to keep the cylinder afloat
 
It would all depend upon how far under the cylinder would go. on that vertical rod. and the mass of the cylinder. My lungs collapse enough to make me neutral at less than 3m depth.
AS with all Engineering matters, the Numbers Count.
 

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