Finding pressure with given radius

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SUMMARY

The discussion focuses on calculating pressure within a test tube filled with water being spun in an ultracentrifuge at a constant angular velocity (ω). The derived formula for pressure at any radius r is P = 1/2 (ρ)(ω)^2 + P(atm), where ρ represents the density of water. The specific case presented involves an angular velocity of 3.8 x 10^5 rad/s and a radius of 10 cm, questioning the pressure at r = 13 cm. The distinction between gauge pressure and atmospheric pressure is also addressed, clarifying that they are not the same.

PREREQUISITES
  • Understanding of centripetal acceleration and its relation to angular velocity (ω)
  • Knowledge of pressure equations, specifically P = P(atm) + (ρ)g(h)
  • Familiarity with the concepts of gauge pressure versus atmospheric pressure
  • Basic principles of fluid dynamics and density calculations
NEXT STEPS
  • Research the derivation of pressure formulas in rotating fluids
  • Learn about the effects of angular velocity on fluid pressure in ultracentrifuges
  • Explore the differences between gauge pressure and absolute pressure in practical applications
  • Examine real-world applications of pressure calculations in laboratory settings
USEFUL FOR

Students in physics or engineering, particularly those studying fluid dynamics, as well as professionals working with centrifugation processes and pressure measurement techniques.

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Homework Statement


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A Test tube filled with water i being spun around in an ultracentrifuge with constant angular velocity, w (w = omega). The test tube i slying along a radius and revolving in a horizontal plane. The free surface of the water is at radius r(0) (with smaller parenthesis) as shown

a) show that the pressure at any raidus r within the tube is given by: P=1/2 (row)(omega)^2+P(atm)

b)suppose that w=3.8x10^5 rad/s and r(0)=10cm, what is pressure at r=13
Does it matter whether it's gauge or atmospheric pressure?

question, is it asking whether gauge pressure and actually pressure are the same?

Homework Equations


centripetal acceleration =v^2/r = r(omega)^2
Pressure= P(atm) + (row)*g*h
where row is density, g is gravity (or acceleration in this case), and h is dept
density = (row) = m/V = mass over volume



The Attempt at a Solution


took simple approach
a= r(omega)^2

and used this "a" (acceleration) as"g" (gravity)
and r-r(0) (since that's the dept of watr) as h
so using Pressure= P(atm) + (row)*g*h
P= P(atm) + (density of water)*r*(omega)^2*(r-r(0))
in symbols :P
P= P(atm)+p*r*w^2(r-r(0))
but this doesn't match the answer ;)
 
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This is how I would start, does this work?
 

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