# Terminal Velocity

1. ### sarmar

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:
= 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
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

### Staff: Mentor

Are you assuming a drag coefficient = 1?

7µm is the diameter of the pollen grain.

3. ### sarmar

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 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 ?

### Staff: Mentor

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

Why would you re-figure the mass?

5. ### sarmar

8

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

### Staff: Mentor

That should be: 4/3pi*r^3.

8
Thank you!