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Calculate the terminal velocity for a pollen grain

  1. Oct 2, 2011 #1
    1. The problem statement, all variables and given/known data

    Calculate the terminal velocity for a pollen grain falling through the air using the drag force equation. Assume the pollen grain has a diameter of 7 µm and a density of 0.3 g/cm3.


    2. Relevant equations

    Vterm= sq.rt of 2mg/pA
    Volume= 4/3pi*r^2
    Density = m/v

    3. The attempt at a solution

    I am given the answer but need to show how to get there.
    Here is what I have, can someone point out where exactly I am going wrong?
    Thank you!!

    Vterm= sq.rt of 2mg/pA

    p= air density of 1.3kg and

    A = cross section of the pollen grain
    A= 7µm = 0.000007m

    To find mass of pollen grain:
    radius= 1/2diameter
    = 1/2(0.000007)
    = 0.0000035m
    Volume= 4/3pi*r^2
    = 4/3pi(0.0000035)^2
    = 5.13x10^-11 m^3
    Density = m/v
    m=DV
    =0.3g/m^2 (5.13x10^-11 m^3)
    = 1.54 x10^-11g
    to kg = 1.54 x10^-14 kg

    So, Vterm= sq.rt of 2mg/pA
    Vterm = sq.rt. of 2(1.54 x10^-14 kg)(9.8m/s^2) / (1.3kg/m^2)(0.000007m)
    =0.0018m/s

    But I know this is wrong since I am given the final answer, which is 0.145m/s
    I've worked it out so many times and I'm stuck. Please help.
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Oct 2, 2011 #2

    Doc Al

    User Avatar

    Staff: Mentor

    Are you assuming a drag coefficient = 1?

    7µm is the diameter of the pollen grain.
     
  4. Oct 2, 2011 #3

    I am not given any information for drag coefficient. So I suppose I was assuming 1. I found online the drag coefficient of a spherical object is approx. 0.5

    I thought the diameter was the same as the cross-section.
    Area of a circle? A=p*r^2
    =pi*0.0000035^2
    =3.85x10^-11 so,
    A=cross-section=3.85x10^-11

    so then I just re-figure the mass ?
     
  5. Oct 2, 2011 #4

    Doc Al

    User Avatar

    Staff: Mentor

    That sounds OK.
    No, A is the cross-sectional area.
    Looks OK.

    Why would you re-figure the mass?
     
  6. Oct 2, 2011 #5

    Well because I'm still not getting the right answer....
     
  7. Oct 2, 2011 #6

    Doc Al

    User Avatar

    Staff: Mentor

    That should be: 4/3pi*r^3.
     
  8. Oct 2, 2011 #7
    Thank you!
     
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