Register to reply

Terminal Velocity

by sarmar
Tags: terminal, velocity
Share this thread:
sarmar
#1
Oct2-11, 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 Ám 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= 7Ám = 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
Phys.Org News Partner Science news on Phys.org
World's largest solar boat on Greek prehistoric mission
Google searches hold key to future market crashes
Mineral magic? Common mineral capable of making and breaking bonds
Doc Al
#2
Oct2-11, 12:45 PM
Mentor
Doc Al's Avatar
P: 41,310
Quote Quote by sarmar View Post
Vterm= sq.rt of 2mg/pA
Are you assuming a drag coefficient = 1?

A = cross section of the pollen grain
A= 7Ám = 0.000007m
7Ám is the diameter of the pollen grain.
sarmar
#3
Oct2-11, 01:14 PM
P: 8
Quote Quote by Doc Al View Post
Are you assuming a drag coefficient = 1?


7Ám is the diameter of the pollen grain.

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 ?

Doc Al
#4
Oct2-11, 01:20 PM
Mentor
Doc Al's Avatar
P: 41,310
Terminal Velocity

Quote Quote by sarmar View Post
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
That sounds OK.
I thought the diameter was the same as the cross-section.
No, A is the cross-sectional area.
Area of a circle? A=p*r^2
=pi*0.0000035^2
=3.85x10^-11 so,
A=cross-section=3.85x10^-11
Looks OK.

so then I just re-figure the mass ?
Why would you re-figure the mass?
sarmar
#5
Oct2-11, 02:17 PM
P: 8
Quote Quote by Doc Al View Post



Why would you re-figure the mass?

Well because I'm still not getting the right answer....
Doc Al
#6
Oct2-11, 02:45 PM
Mentor
Doc Al's Avatar
P: 41,310
Quote Quote by sarmar View Post
To find mass of pollen grain:
radius= 1/2diameter
= 1/2(0.000007)
= 0.0000035m
Volume= 4/3pi*r^2
That should be: 4/3pi*r^3.
sarmar
#7
Oct2-11, 04:15 PM
P: 8
Thank you!


Register to reply

Related Discussions
Terminal Velocity given Time at which velocity is .5Vt Introductory Physics Homework 6
Terminal Velocity Classical Physics 6
Terminal Velocity Introductory Physics Homework 7
Question about exhaust velocity and terminal velocity Special & General Relativity 5
Terminal Velocity Classical Physics 1