Solution Found!
An incompressible fluid with density ? is in a horizontal
Chapter 12, Problem 87P(choose chapter or problem)
Problem 87P
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’ a distance r’ from the rotation axis. The pressure on its inner surface is ρ and on its outer surface is p + dp. (a) Apply Newton’s second law to the volume element to show that dp = ρω2r’dr’. (b) If the surface of the fluid is at a radius r0 where the pressure is p0. show that the pressure p at a distance r ≥ r0 is p = ρ0 + ρω2(r2 – r02)/2. (c) An object of volume V and density ρob has its center 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 > ρob Rcmob 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 ultracentrifuge?
Questions & Answers
QUESTION:
Problem 87P
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’ a distance r’ from the rotation axis. The pressure on its inner surface is ρ and on its outer surface is p + dp. (a) Apply Newton’s second law to the volume element to show that dp = ρω2r’dr’. (b) If the surface of the fluid is at a radius r0 where the pressure is p0. show that the pressure p at a distance r ≥ r0 is p = ρ0 + ρω2(r2 – r02)/2. (c) An object of volume V and density ρob has its center 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 > ρob Rcmob 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 ultracentrifuge?
ANSWER:
Solution 87P
Step 1
In the figure the pressure at the left end of the volume element is and the pressure at the right side of the volume element is
. Hence the pressure difference is
Hence net force is
Now the mass of the volume element is
And the acceleration is
Hence, from the newton's second law the fore on the volume element is
Therefore we have
(proved).