Resistive force exerted on a sphere (air resistance)

[alejo ortega]

Homework Statement

The following expression gives the resistive force exerted on a
sphere of radius r moving at speed v through air. It is valid over a
very wide range of speeds.
Rc= 3.1× 10-4rv + 0.87r2v2
where R is in N, r in m, and v in m/s. Consider water drops falling under their own weight and reaching a terminal speed.
(a) For what range of values of small r is the terminal speed
determined within 1% by the first term alone in the expression for
R(v)?
(b) For what range of values of Iarger r is the terminal speed
determined within 1% by the second term alone?
(c) Calculate the terminal speed of a raindrop of radius 2 mm. If there were no air resistance, from what height would it fail from rest before reaching this speed?

Relevant Equations

i don´t understand the question

i don´t understand the question

 

[jbriggs444]

Well, let's start by typesetting that equation and using some guesswork to fix the typos:$$R(v) = 3.1 \times 10^{-4}rv + 0.87r^2v^2$$Now you need to bring something to the table. Show us some effort. How would we determine terminal speed given this formula for resistive force? (i.e. "Calculate the terminal speed of a raindrop of radius 2 mm").

 

[haruspex]

Well, let's start by typesetting that equation and using some guesswork to fix the typos:$$R(v) = 3.1 \times 10^{-4}rv + 0.87r^2v^2$$Now you need to bring something to the table. Show us some effort. How would we determine terminal speed given this formula for resistive force? (i.e. "Calculate the terminal speed of a raindrop of radius 2 mm").

And with a few more corrections:
$$R(v) = 3.1 \times 10^{-4}rv ~\rm{Ns/m^2}+ 0.87\it r^2v^2 ~\rm{Ns^2/m^4}$$

 

[BvU]

Hello @alejo ortega ,

$\qquad$ !​

Relevant Equations:: i don´t understand the question

You'll have to agree that is not a very relevant equation !

Did you check the
https://www.physicsforums.com/threads/homework-help-guidelines-for-students-and-helpers.686781/

$\$

 

[alejo ortega]

And with a few more corrections:
$$R(v) = 3.1 \times 10^{-4}rv ~\rm{Ns/m^2}+ 0.87\it r^2v^2 ~\rm{Ns^2/m^4}$$

ok ,

Well, let's start by typesetting that equation and using some guesswork to fix the typos:$$R(v) = 3.1 \times 10^{-4}rv + 0.87r^2v^2$$Now you need to bring something to the table. Show us some effort. How would we determine terminal speed given this formula for resistive force? (i.e. "Calculate the terminal speed of a raindrop of radius 2 mm").

hello haruspex, this is a cuadratic ecuation

 

[alejo ortega]

ok ,

hello haruspex, this is a cuadratic ecuation

considering that if v is the terminal velocity, the resistive force is equal to weight

 

[alejo ortega]

considering that if v is the terminal velocity, the resistive force is equal to weight

I equaled the first term to 1% of mg
3.1× 10-4rv = mg/100
and solve to r :
r= (10^2) mg/3,1v

ok ,

hello haruspex, this is a cuadratic ecuation

 

[alejo ortega]

I equaled the first term to 1% of mg
3.1× 10-4rv = mg/100
and solve to r :
r= (10^2) mg/3,1v

but i believe that this proceeding its wrong :(

 

[haruspex]

I equaled the first term to 1% of mg

No, that’s a misunderstanding of the given information.
It is telling you that the first term gives you at least 99% of the value, with the quadratic term only contributing 1%.