Linear vs quadratic drag limiting displacement

[kurenaishu07]

Why does a surface with resistive force mkv^2 have displacement given by x = 1/k(ln(1+ukt)), hence divering, whereas a surface with resistive force mkv has displacement x = u/k((1-e^-kt)), and converges to x = u/k. Is this because the quadratic drag model is incorrect?

 

[liamkk112]

Why does a surface with resistive force mkv^2 have displacement given by x = 1/k(ln(1+ukt)), hence divering, whereas a surface with resistive force mkv has displacement x = u/k((1-e^-kt)), and converges to x = u/k. Is this because the quadratic drag model is incorrect?

yes. truthfully for a true quadratic drag model, you need to consider both linear and quadratic drag forces, the quadratic drag model is an approximation that works well for small velocities essentially. but as you can imagine this makes it much harder to solve (idek if there’s an analytic solution, i forget now, it’s in taylor’s classical mechanics textbook which i skimmed a while back)

 

[kurenaishu07]

Would i be required to know the correct quadratic drag model for hsc, or is it sufficient to know the one involving mkv^2, (carrotsticks textbook covers the correct drag model but has only 1 question on it).

 

[liamkk112]

Would i be required to know the correct quadratic drag model for hsc, or is it sufficient to know the one involving mkv^2, (carrotsticks textbook covers the correct drag model but has only 1 question on it).

nah definetly not, its out of syllabus

 

[kurenaishu07]

There has been 1 question in the hsc that requires the correct model in new syllabus (HSC 2022 4u)

Would this not imply that they can test it?

 

[liamkk112]

There has been 1 question in the hsc that requires the correct model in new syllabus (HSC 2022 4u)
View attachment 43639
Would this not imply that they can test it?

thats not the correct model (for quadratic drag)
the correct model would be both linear and quadratic in the same equation

 

[liamkk112]

this would be the "correct" model that much more accurately models the motion of a projectile experiencing a resistive drag force

 

[kurenaishu07]

Sorry let me reword, you cannot do this question using mkv^2, for this question the resistive force would be (mod(sdot))^2 X unit velocity vector, which simplifies into B which is the same as taylors quadratic drag equations -->

 

[kurenaishu07]

View attachment 43640
this would be the "correct" model that much more accurately models the motion of a projectile experiencing a resistive drag force

ohh i see what you mean

 

[liamkk112]

Sorry let me reword, you cannot do this question using mkv^2, for this question the resistive force would be (mod(sdot))^2 X unit velocity vector, which simplifies into B which is the same as taylors quadratic drag equations --> View attachment 43641

to clarify - both the quadratic drag and linear drag models, on their own, are firmly in syllabus.

however both combined, is not. quadratic drag appears to go to infinity, because it fails to consider other forces such as linear drag (and/or friction or other resistive forces) that would bring the object to rest; so when considering projectiles where friction isn't a thing, if you want an accurate model that considers quadratic drag, you have to include the linear term too

 

[kurenaishu07]

to clarify - both the quadratic drag and linear drag models, on their own, are firmly in syllabus.
View attachment 43642
however both combined, is not. quadratic drag appears to go to infinity, because it fails to consider other forces such as linear drag (and/or friction or other resistive forces) that would bring the object to rest; so when considering projectiles where friction isn't a thing, if you want an accurate model that considers quadratic drag, you have to include the linear term too

okay that makes alot more sense, thank you!!