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Drag force, bv or kv^2?

  • Thread starter iScience
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  • #1
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i have an object thrown up in the air with drag present.

i want to know how much time it takes to reach the peak ie when v=0

m[itex]\ddot{x}[/itex]=-mg-kv2

basically, i don't know which one to use for drag, i've seen bv being used before in problems, and i've also seen kv2 being used before in problems. which one do i use for this problem and when/how do i know when to use which one?

for this problem, if it's bv i can definitely do the problem. but if it's kv2, i don't know how to pursue the problem. for the above equation i gave, if it's kv2, then i can put acceleration in terms of t like dv/dt. but then when i move the RHS over to the right by dividing, i cannot do the integral because i do not have an extra v term in the numerator.
 

Answers and Replies

  • #2
jhae2.718
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In reality, drag is proportional to the square of velocity (##D = C_D \frac{1}{2}\rho v^2 S##), but is sometimes modeled as being proportional to the velocity so that you get a linear ODE instead of a nonlinear ODE.

If you choose to model drag as ##D \propto v^2##, the easiest way is probably to consider a numerical approach.
 
  • #3
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so in other words, it sounds like i can use bv OR kv^2 depending on whatever form helps me to solve the problem right? since b and k are just stuffed up constants?
 
  • #4
jhae2.718
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Well, no. The correct model is ##kv^2##; modeling the drag as ##bv## is a simplification that only works for small velocities. It's also generally used in introductory mechanics problems.

It sounds like you have some freedom with your model. What are you trying to accomplish?
 
  • #5
fzero
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i have an object thrown up in the air with drag present.

i want to know how much time it takes to reach the peak ie when v=0

m[itex]\ddot{x}[/itex]=-mg-kv2

basically, i don't know which one to use for drag, i've seen bv being used before in problems, and i've also seen kv2 being used before in problems. which one do i use for this problem and when/how do i know when to use which one?

for this problem, if it's bv i can definitely do the problem. but if it's kv2, i don't know how to pursue the problem. for the above equation i gave, if it's kv2, then i can put acceleration in terms of t like dv/dt. but then when i move the RHS over to the right by dividing, i cannot do the integral because i do not have an extra v term in the numerator.
You get an integral of the form

$$ \int_0^v \frac{dw}{1 - \kappa^2 w^2}, $$

which can be done by partial fractions:

$$ \frac{1}{1-\kappa^2 w^2} = \frac{1}{2} \left( \frac{1}{1+\kappa w} + \frac{1}{1-\kappa w} \right).$$

These partial fractions integrate as certain logs.
 
  • #6
465
4
Well, no. The correct model is ##kv^2##; modeling the drag as ##bv## is a simplification that only works for small velocities. It's also generally used in introductory mechanics problems.

It sounds like you have some freedom with your model. What are you trying to accomplish?
small velocities? actually i thought it was only proportional to v when the object itself had a small cross sectional area.
 
  • #7
19,954
4,108
i have an object thrown up in the air with drag present.

i want to know how much time it takes to reach the peak ie when v=0

m[itex]\ddot{x}[/itex]=-mg-kv2

basically, i don't know which one to use for drag, i've seen bv being used before in problems, and i've also seen kv2 being used before in problems. which one do i use for this problem and when/how do i know when to use which one?

for this problem, if it's bv i can definitely do the problem. but if it's kv2, i don't know how to pursue the problem. for the above equation i gave, if it's kv2, then i can put acceleration in terms of t like dv/dt. but then when i move the RHS over to the right by dividing, i cannot do the integral because i do not have an extra v term in the numerator.
There is a trick to integrating the force balance equation. Multiply both sides of the equation by dv/dt. You will end up with all three terms which are exact derivatives that can be integrated immediately.

Regarding the drag force on an object, in practice it is not proportional to either v or v2. The drag coefficient is a function of the velocity that needs to be measured. However, as implied by jhae2.718, in the limit of very low velocities, the drag coefficient is inversely proportional to the velocity. In this limit (laminar flow), the constant of proportionality can be measured or calculated using fluid mechanics.
 

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