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Terminal Velocity 
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#1
Oct211, 12:23 PM

P: 8

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
Oct211, 12:45 PM

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#3
Oct211, 01:14 PM

P: 8

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 crosssection. Area of a circle? A=p*r^2 =pi*0.0000035^2 =3.85x10^11 so, A=crosssection=3.85x10^11 so then I just refigure the mass ? 


#4
Oct211, 01:20 PM

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P: 41,300

Terminal Velocity



#5
Oct211, 02:17 PM

P: 8

Well because I'm still not getting the right answer.... 


#7
Oct211, 04:15 PM

P: 8

Thank you!



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