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Fluid Mechanics

  1. Dec 13, 2004 #1
    A thin spherical shell of mass 4 kg and diameter 0.2 m is filled with helium (density 0.18 kg/m3). It is then released from rest on the bottom of a pool of water that is 4 m deep. Determine the value of the upward acceleration of the shell.
     
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  3. Dec 13, 2004 #2

    Pyrrhus

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    what have you done???

    Apply newton's 2nd Law to the forces acting on the shell, buoyant force and the weight.
     
  4. Dec 13, 2004 #3
    I determined the weight to be 39.2 N. I used the formula B=ρgV = (.18)(9.8)(.004) = .007 and V=(4πr3)/3 = .004.

    Acceleration = Force/Mass = (.004-39.2)/4 = -9.799

    And this answer doesn't make sense to me, I think my problem is with the buoyant force. Thanks!
     
  5. Dec 13, 2004 #4

    Pyrrhus

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    You should have out for the buoyant force the density of the water, remember the buoyant force is equal to the weight of fluid displaced.
     
  6. Dec 13, 2004 #5
    Doing that I get (1)(9.8)(.004) = .0392...sorry I am not getting this at all
     
  7. Dec 13, 2004 #6

    Pyrrhus

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    You should have

    Applying Newton's 2nd Law

    [tex] B - m_{shell}g -m_{He}g = (m_{He} + m_{shell}) a [/tex]
     
  8. Dec 13, 2004 #7
    So that makes the upward acceleration -9.79?
     
  9. Dec 13, 2004 #8

    Pyrrhus

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    Check well, i get a = 0.46 m/s^2

    [tex] (\rho_{water}V - \rho_{He}V - m_{shell})g = (m_{shell} + \rho_{He}V)a [/tex]
     
  10. Dec 13, 2004 #9
    These are the values that I entered into the equation, I think the V (volume) might be incorrect, I am using .004 (V=(4Πr^3)/3)


    ((1∙.004)-(.18∙.004)-4)9.8=(4+.18∙.004)a = -9.79

    Thank you!
     
  11. Dec 13, 2004 #10

    Pyrrhus

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    I still get the same answer, i will write it to the simpler form

    [tex] a = g \frac{(\rho_{water} - \rho_{He}) \frac{\pi d^3}{6} - m_{shell}}{m_{shell} + \rho_{He} \frac{\pi d^3}{6}} [/tex]

    The density of water is 1000 kg/m^3!
     
  12. Dec 13, 2004 #11
    Yes, thank you. Also thank you very much for helping me out with this problem and the rest of the problems I have had tonight!!
     
  13. Dec 14, 2004 #12
    The upward force will be equal to the difference between the weight of the He plus the material of the shell, and that of the water the sphere displaces. But there will be a considerable drag force exerted against the sphere's upward travel by the water. The D.E. for this system will be: -m*x'' - k*x' + F = 0, where m is the mass of the sphere, m*x'' is the inertial force on the sphere (downward, hence negative), k is the coefficient of drag (the force is downward), and F is the buoyant force (upward). x' and x'' are the sphere's velocity and acceleration, respectively.
     
    Last edited: Dec 14, 2004
  14. Dec 14, 2004 #13
  15. Dec 14, 2004 #14

    Pyrrhus

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    This problem probably was meant for a non-viscous, stable, incompresible and irrotational fluid.
     
  16. Dec 14, 2004 #15
    The problem says it's hydrogen hydroxide, aka hydrogen monoxide...HOH...H2O...
     
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