Incompressible fluid in horizontal test tube

In summary, the object in the fluid would be subject to the same external forces as the object without the fluid. The displaced mass of fluid would be ρV. The object would move along the tube as a result of the external forces.
  • #1
hvthvt
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Homework Statement



An incompressible fluid with density ρ is in a horizontal test tube of inner cross-sectional area A. The test tube spins in a horizontal circle in an ultracentrifuge at an angular speed ω. Gravitational forces are negligible. Consider a volume element of the fluid of area A and thickness dr' at a distance r' from the rotation axis. The pressure on its inner surface is p and outer surface is p+dp.
(a) Show that dp=ρω2r'dr'.
(b) If the surface of the fluid is at a radius ro where the pressure is po, show that the pressure p at a distance r≥ro is p=po+ρω2(r2-ro2)/2.
(c) An object of volume V and density ρob has its centre of mass at a distance Rcmob from the axis. Show that the net horizontal force on the object is ρVω2Rcm, where Rcm is the distance from the axis to the center of mass of the displaced fluid.
(d) Explain why the object will move inward if ρRcm>ρobRcmob and outward if ρRcm<ρobRcmob.
(e) For small objects of uniform density, Rcm = Rcmob. What happens to a mixture of small objects of this kind with different densities in an unltracentrifuge?

Homework Equations



ω=v/r
A= π*r^2
F=ma
p=F/A

The Attempt at a Solution



I have really tried everything I could but i simply don't understand the way how to tackle these problems. I would really like to know how these answers come about. Can anybody please help me with discussing this??
I would really appreciate it!
 
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  • #2
That small volume element is in circular motion. What is its acceleration? What are the forces acting on it?
 
  • #3
a=ω^2r is the acceleration. The force acting on it comes from the pressure on its inner surface i guess? p=F/A with F=ma so substituing the above equation for acceleration would give: p= m*ω^2*r/A
I don't know how to deal with the area A. I need to show what dp is, so I don't need the entire area A. But how can I tackle this?
 
  • #4
hvthvt said:
a=ω^2r is the acceleration. The force acting on it comes from the pressure on its inner surface i guess? p=F/A with F=ma so substituing the above equation for acceleration would give: p= m*ω^2*r/A
I don't know how to deal with the area A. I need to show what dp is, so I don't need the entire area A. But how can I tackle this?

Consider a disk element thickness dr. What is its mass?
 
  • #5
Hmm.. Its volume times its density? The volume should be A*dr I guess?
 
  • #6
But there's no A in the equation which I should get. I'm so lost!
 
  • #7
I solved a! Finally. But what about b? Should i integrate the answer from a?
 
  • #8
Integrate the left hand side of (a) from ##p_0## to ##p##, and the right hand side from ##r_0## to ##r##.
 
  • #9
How can I show that the net horizontal force is ρVω2R?
F=ma= and a = ω^2r but how can I show that it is 2R? And can anybody help me with the explanations for questions d and e ?
 
  • #10
hvthvt said:
How can I show that the net horizontal force is ρVω2R?
F=ma= and a = ω^2r but how can I show that it is 2R? And can anybody help me with the explanations for questions d and e ?
In part c, you have to recognize that you've created artificial gravity (in the horizontal direction). The acceleration you calculated is the artificial gravitational acceleration. You need to use Archimedes principle. The displaced mass of fluid is ρV. Parts d and e can be explained also by artificial gravity arguments.

Chet
 
  • #11
hvthvt said:
How can I show that the net horizontal force is ρVω2R?
Consider the situation without the object in the fluid. The region that the object occupies would instead be occupied by the same volume of the fluid, in exactly the same location.
Would that body of fluid be subject to the same external forces as the object?
Would that body of fluid move along the tube as a result of those forces?
 

What is an incompressible fluid?

An incompressible fluid is a fluid that has a constant density regardless of changes in pressure. This means that the volume of the fluid remains constant even when pressure is applied to it.

What is a horizontal test tube?

A horizontal test tube is a cylindrical tube used for conducting experiments on fluids. It is placed horizontally and is used to hold incompressible fluids for testing.

Why is it important to use a horizontal test tube for studying incompressible fluids?

Using a horizontal test tube allows for better observation and analysis of the behavior of incompressible fluids. It also minimizes the effects of gravity on the fluid, making the results more accurate.

What types of experiments can be performed with an incompressible fluid in a horizontal test tube?

Incompressible fluids in horizontal test tubes can be used to study properties such as viscosity, density, and pressure. They can also be used to observe the movement and behavior of fluids under different conditions.

How do scientists measure the density of an incompressible fluid in a horizontal test tube?

Scientists can measure the density of an incompressible fluid in a horizontal test tube by using a hydrometer or a densitometer. These instruments measure the density of the fluid by comparing it to the density of water.

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