I like the way you clearly explain how the mathematics of the physical phenomena involved. Thank You very much Sir. A lot of videos leave out the STEPS.
It's been years since I took Diff Eq. and needed a refresher that clearly explained why you were doing things. Every other video made huge leaps and I got lost each time. Thank you so much!
bro i didn't even know how integrals rly work before this vid (my physics class is moving faster than my calc class). thx so much for this vid- I got it now!
Hmm... isn't drag force not linearly proportional to v, but rather proportional to v^2? The video was still hugely useful though in helping me get started on deriving it since I wasn't sure where to start. Thank you!
For fluids like air, it's actually quadratic air drag that applies, not linear! You can check out this video ( ua-cam.com/video/WcSkGQsFKME/v-deo.html ) for projectile motion in 1 dimension with air drag, and this video ( ua-cam.com/video/CjNWvScg4T0/v-deo.html ) 2 dimensional projectile motion with air drag!
Whether the drag force can be treated as solely linear or quadratic depends on the speed of the object. Both the linear and quadratic terms are always present. In the low velocity limit, the linear term dominates. This makes the quadratic term negligible. As the velocity increases, the quadratic term becomes more relevant. At large enough speeds the linear term is the one that can be neglected, which is what yields the model your referring to.
I find that your method of solving this differential equation kind of messy. In the third line: mg- kv = m(dv/dt) is already a linear 1st order differential equation. Just rearrange to: dv/dt + (k/m)v = mg/k; calculate the integrating factor: I = e^integral(k/m).dt; then multiply both sides by I and carry out a few steps of calculation to get the same solution.
@@phandinhthanh2295 So before the substitution I am reading it as take the integral with respect to v. I get you have to replace dv with du, but why must you take the derivative if it’s already derived to v, since acceleration is dv/dt. 1/u is equivalent to the pre-subisution, which is already derived. Like I get what he does, I just don’t get why he does it. Like isn’t it already equivalent. I am seeing dv is more of a notation than a variable idk if I am getting that bit wrong(do we not know the derivative of velocity already, like isn’t the derivative of velocity dv/dt = (mg-kv)/m Thanks for help
@@MarkSmith-vo1vn I'm not sure I understand your question. Everytime I make a substitution when I solve a integral, I always take a derivative of the new variable wrt the old variable to get the relationship between the differential(the du, dv-part) of the new and the old variable. B/c you have to change not only the variable in the integrand(the function under the integral sign) but also in the differential(dv-> du). Perhaps, he wants to simplify the integral making it easier for us to follow the steps.
How did you still retain = mg/k? From your method, we divide both sides by m. Gives us g - (k/m)v = dv/dt. Then we add (k/m)v to both sides We now only have dv/dt + (k/m)v = g. Can you explain if there was something else you did?
In air, yes! You can check out this video ( ua-cam.com/video/WcSkGQsFKME/v-deo.html ) for projectile motion in 1 dimension with (quadratic) air drag, and this video ( ua-cam.com/video/CjNWvScg4T0/v-deo.html ) 2 dimensional projectile motion with (quadratic) air drag!
A constant that involves the density of the fluid(air) it is moving through, the shape of the object and the area of the object that is in contact with the fluid (air)
Yes! And that's where the constant K comes in. It is proportional to the area pf the object you're studying. Hreater area, greater K, so that mg/K denominator gets smaller. And don't forget the negative exponent of the exponential function. If it gets vreater,than again the result is gonna be smaller (smaller velocity, in this case).
I like the way you clearly explain how the mathematics of the physical phenomena involved. Thank You very much Sir. A lot of videos leave out the STEPS.
at 2:31 "But since this is a physics problem, we need to put bounds on our integration" I am fucking DYING LAUGHING AT THIS LOLOLOLOL
Thank you this was absolutely brilliant!!!
Thank you... And... I'll also recommend this Chanel to my friends who are facing difficulties in understanding Physics
It's been years since I took Diff Eq. and needed a refresher that clearly explained why you were doing things. Every other video made huge leaps and I got lost each time. Thank you so much!
Sir - this video is outstanding. You should make more!
Great explanation for low Reynolds numbers
bro i didn't even know how integrals rly work before this vid (my physics class is moving faster than my calc class). thx so much for this vid- I got it now!
Would it work if I used V = dS/dt, and solved for a known distance (S) to get the time of fall?
your handwriting is so nice! also what kind of notebook is that (im kind of a stationery freak)
Hmm... isn't drag force not linearly proportional to v, but rather proportional to v^2? The video was still hugely useful though in helping me get started on deriving it since I wasn't sure where to start. Thank you!
For fluids like air, it's actually quadratic air drag that applies, not linear! You can check out this video ( ua-cam.com/video/WcSkGQsFKME/v-deo.html ) for projectile motion in 1 dimension with air drag, and this video ( ua-cam.com/video/CjNWvScg4T0/v-deo.html ) 2 dimensional projectile motion with air drag!
Whether the drag force can be treated as solely linear or quadratic depends on the speed of the object. Both the linear and quadratic terms are always present. In the low velocity limit, the linear term dominates. This makes the quadratic term negligible. As the velocity increases, the quadratic term becomes more relevant. At large enough speeds the linear term is the one that can be neglected, which is what yields the model your referring to.
Outstanding explanation
I like your method ❤️
It may be correct for friction force of sphere body in the liquid (stokes formula)...
I like so much the mathematics specially the calculation
Amazing video
We need more videos like this!!!!!!!!
thanks man , you did that i always wanted.
Fantastic. Just what I needed! Thank you.
doesn't viscosity affect drag as well, why isnt it a factor in this equation
Absoulutley great job
Shouldn't be -kv?
I find that your method of solving this differential equation kind of messy. In the third line: mg- kv = m(dv/dt) is already a linear 1st order differential equation. Just rearrange to: dv/dt + (k/m)v = mg/k; calculate the integrating factor: I = e^integral(k/m).dt; then multiply both sides by I and carry out a few steps of calculation to get the same solution.
Could you help me understand why we take the derivative of u with respect to v. Like I get we have to get it to u, but what’s the point of doing it.
@@MarkSmith-vo1vn It's a substitution. it's for changing the variable from v to u so he can make use of the property integral{du/u} = ln|u| + c
@@phandinhthanh2295 So before the substitution I am reading it as take the integral with respect to v. I get you have to replace dv with du, but why must you take the derivative if it’s already derived to v, since acceleration is dv/dt. 1/u is equivalent to the pre-subisution, which is already derived. Like I get what he does, I just don’t get why he does it. Like isn’t it already equivalent. I am seeing dv is more of a notation than a variable idk if I am getting that bit wrong(do we not know the derivative of velocity already, like isn’t the derivative of velocity dv/dt = (mg-kv)/m Thanks for help
@@MarkSmith-vo1vn I'm not sure I understand your question. Everytime I make a substitution when I solve a integral, I always take a derivative of the new variable wrt the old variable to get the relationship between the differential(the du, dv-part) of the new and the old variable. B/c you have to change not only the variable in the integrand(the function under the integral sign) but also in the differential(dv-> du).
Perhaps, he wants to simplify the integral making it easier for us to follow the steps.
How did you still retain = mg/k?
From your method,
we divide both sides by m.
Gives us g - (k/m)v = dv/dt.
Then we add (k/m)v to both sides
We now only have
dv/dt + (k/m)v = g.
Can you explain if there was something else you did?
Your are a life saver :')
Thank you! Helpful
Thank you
its brilliant ...! D =bv ^2 please solve this.... urgent.
Yup got that drag test in 43 minutes 🥶
Very good and easy to understand and thanks
Great explanation!
Is it apply to the case when sphere is drop in water
Wow just wow
Excellent video
shouldn't drag force be proportional to v^2
That's at velocities much greater than the critical velocity. Watch the 8.01x lecture on Drag and Resistive forces and you'll see what I mean :)
In air, yes!
You can check out this video ( ua-cam.com/video/WcSkGQsFKME/v-deo.html ) for projectile motion in 1 dimension with (quadratic) air drag, and this video ( ua-cam.com/video/CjNWvScg4T0/v-deo.html ) 2 dimensional projectile motion with (quadratic) air drag!
When the velocity is very high, then its equal go v^2
What is the value of k?
Its a constant
A constant that involves the density of the fluid(air) it is moving through, the shape of the object and the area of the object that is in contact with the fluid (air)
Intergrate v-t I wanna c the formula
nice and superb
Actually helpful
doesn't surface area affect the equation ?
Yes! And that's where the constant K comes in. It is proportional to the area pf the object you're studying. Hreater area, greater K, so that mg/K denominator gets smaller. And don't forget the negative exponent of the exponential function. If it gets vreater,than again the result is gonna be smaller (smaller velocity, in this case).
Drag force is non-linear and is a square function of the velocity. So this is not correct.
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