Free Fall with Air Resistance (Drag Force) (video 2 of 2)
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- Опубликовано: 7 фев 2025
- Physics Ninja looks at a problem of air resistance during free fall. The terminal velocity is calculated and the velocity of the object as a function of time. In this case the drag force is proportional to the magnitude of the velocity squared (appropriate in cases with a large Reynolds number). I previously looked at the case where the air resistance was linear with speed. • Free Fall with Air Res...
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Thank you so much for this! I've really been struggling with this problem, but now I understand :)
thanks professor. I've made a program to perform a numerical integration and it was fun. As the speed approaches 50m/s, the DF gets closer and closer to mg. As a result, the acceleration gets closer and closer to 0, contributing very little to the change in velocity. And this goes on and on till the end of times :D
Awesome!
Great video, it was interesting to see how the final function ended up to be hyperbolic tan. Just one question How do you know whether to use tanh inverse and coth inverse since both of them have derivative 1/(1-x^2)? Does it depend on initial velocity?
hey professor, Great Vid! the two videos really helped me, however i have a quick question, Ive spent time trying to understand how you converted v(t)=Vt tanh(√gk/m) to v(t)=Vt tanh (gt/Vt). I am doing this for a report so it would be really appreciated if you can help me! thank you in advance
I would like to see the integral for a known distance of fall to get the time of fall and not a known time of fall.
Thank you for providing this detailed video. I am however unsure about a few things. I oriented the free-body diagram to make gravity negative and drag force positive. Using an integral table, The integral of [dv/(v^2-b^2)] = [(1/2b)*ln|(v-b)/(v+b)| + C.]
you used [dv/(v^2 + b^2)]= [(1/b)*tan^-1(v/b))]
Please explain how you get to using the hyperbolic trig inverse function (tanh^-1) instead of the common trig inverse (tan^-1 or arctan).
The parameter of a hyperbolic function is a hyperbola, functions are used to model sinusoidal shapes on a hyperbola rather than a circle for trig functions. Hyperbolic functions are equal to their own second derivative because they are like "half exponentials" it takes two derivatives to complete the cycle.
Wouldn't I need to integrate again to complete the problem? (Other worked examples integrated from 0 to t' and from 0 to v'(t) and if this reasoning is true, then I would need to integrate again).
ruclips.net/video/j0pL486fDH0/видео.html
I remember this gave me a headache when I first made the video. I’ll have to redo and get back to you.
He used the inverse hyperbolic tangent because the derivative of tanh^-1x is 1/1-x^2, the usual tan^-1(x) derivative results in 1/1+x^2, there are other ways to do it but the inverse hyperbolic tangent is the most effective and easiest.
thnx for the explanation it solved my problem :D
Hello my question is this,
Taking the air resistance into account,the acceleration of a free falling body can be described approximately by a(v)=g-av^2.
Here g is the gravity acceleration and a a constant.
Determine the velocity v(t) of the body that is released from rest.
Thank you!
Watch the video! It’s the same case treated here. My drag force has more terms but just group them together and call that a. To have an equation for acceleration just divide through (my net force expression) by the mass.
Physics Ninja thank you!
11:59 What happens here? I don't understand how you move that to the other side.
I have a problem that consider tho phases.In theFirst phase,soldier jump from a plane an fall.After some time he opens his parachute.When I try to find integral constant I encounter with an error because my initial value is higher than the limit value for second phase.How can I fix that.Thanks
i see your point.. you will not be integrating from 0m/s to some v_final. You need to integrate from v_0 to v_f. The differential equation is the same but the limits are different. Your curve will be on exponential decay to the terminal velocity.
@@PhysicsNinja I can imagine it decreases exponentially but I can not find the integral constant using tanh(constant is inside the tanh).Is there different equation for these kind of cases.
@@fevzidenizyenigun1754 sometimes the drag force is simply -bv(t) and everything gets lumped into the constant b. If you know the terminal speed you can calculate b.
If you substitute the given numbers, it turns out to be 70 m/s, not 50 m/s. What happened?
I agree.
I'm trying to build the least aerodynamic object possible. For instance would it be a flat plate, a flat plate that's concave, a flat plate that's concave with a bigger flat plate behind it, a flat plate with a parachute behind it, etc etc
What about a cube, but the rear half of the cube is half a perfect sphere?
Why the 1/2? Why not just define the drag coefficient as (drag coefficient)/2?
This is one of the standard definitions of the drag force. It’s not unique. Sometimes the 1/2 is absorbed into the drag coefficient but sometimes it’s not.
when is try plooting the v(t) into desmos, i just get a straight line
You forgot to add the exponent
i did it differently but still got the same graph
the finale answer is,
sqrt(-sin(2*tan^-1(e^k/m*t*b)))^2 *b+b) b=gm/k, k = p*a*d/2
wtf
If you set v(t) as terminal velocity. You get tanh^-1(1) which is apparently infinite?
It's takes an infinite amount of time to reach the terminal velocity. It's kinda like an exponential decay - which doesn't ever reach 0 but after a long time it approaches 0. In this case we approach the terminal velocity without ever reaching it.
HOW THE DRAG FORCE APPLIES IN PROJECTILE MOTION?
Truc Lam Nguyen tougher problem. Check book by Marion and Thornton. I could make a video but not till next week.
Yes, it is. I think the drag force then relates to the angle, velocity,...I got messed with it. I hope you make that kind of video
Please calculate this question
For the constant Cd... how do i calculate this? I am trying to run a simulation but don't know what this value is.
www.grc.nasa.gov/www/k-12/airplane/dragsphere.html
@@PhysicsNinja I am now slightly more confused. I can determine the value of Cd for a sphere by calculating the Reynold's number of the sphere. However, the Reynold's number depends on the velocity of the object. Therefore, doesn't this mean that the reynold's number of a falling object changes as it moves? And if this is the case, this would mean that Cd would also change over time, meaning that the terminal velocity changes? This doesn't make any sense to me...
@@cameronmaddock9942 the Reynolds number is basically constant over a large range of velocities. I would use the value from the approximate terminal velocity. You can also see from the graph that the value is 0.5 for several orders of magnitude of Re numbers.
@@cameronmaddock9942 If you want to include the velocity dependency on the drag coefficient (through the Reynolds number), you would need an empirical equation for drag coeff versus Re. These are available in the literature. Unfortunately, to include this detail in the free fall model would require numerical integration of the differential equation.
when would the Euler's method be used then?
For the integral of 1/(v^2-b^2) you used at 9:30, I can't seem to find any integral table online that has this exact integral shown. Is there a source you could possibly provide? I'm trying to write a paper :))
Henry Trinh you may find something in terms of logs but then use some relationships with between tanh and logs for conversions. I agree it’s not obvious but it’s the most common way the solution is written. You may find the integrate for 1/ v^2 + b^2 then switch over to tanh.
@@PhysicsNinja i actually have found the integral in an integral table i have however it is not the way you have it, mine is = --(1/b)arccoth (x/b) = (1/2b) *ln ({x-b}/{x+b}) soooo... what happened there?? i'm confused
Math is cool man
can u calculate the terminal velocity on mars?
yes, but it will take alot longer to reach terminal velocity as the atmosphere is very thin compared to Earth's.
Another way I've done this is with partial fraction decomposition. It's a little tedious and the equation looks complicated.
How do you get time?
solve for t
@@mariovelez578 how exactly? :)
Thanks, u really helped someone! What program you used to plot the final graph?
Gabriel Luiz looks like it’s plotted using Desmos. (www.desmos.com)
@@PhysicsNinja Thanks!!!
Thanks for the interesting video. I have one question though.
At around 9:30, shouldn't the integral be log|(x-b)/(x+b)|/2b since the integrand is 1/(v^2-b^2) and not 1/(v^2+b^2) ?
Good question! There is a relationship between tanh(-1) and logs. I’m pretty sure both are equivalent.
@@PhysicsNinja I see, thanks!
im in grade 11…