Free Fall with Air Resistance - Terminal Velocity (video 1 of 2)
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- Опубліковано 25 сер 2024
- Physics Ninja looks at a problem of free fall where i include air resistance effects. The resistance is taken to be linear with velocity. We solve to obtain the velocity as a function of time and verify our result with some limiting cases and find an expression for the terminal velocity.
I appreciate the attempt to explain but without going step by step you just lose momentum. None of this really makes sense once you reach the integrals
Really appreciate these videos, thanks man!
The calculus manipulations were too complex for me, but the conclusion at the end of each calculation (initial, terminal, short term velocity situations) made sense with the calculation results. Perhaps you could create an appendix that puts in more steps so that those of us that like to play with numbers can have a chance of understanding the manipulations, even a little more. Thanks.
I think you tried to simplify the calculus parts, but it made it more confusing to someone that has gone through calculus.
Yeah I agree. I think you don't need to factor or make the integrals definite because you can solve for C at the end is a lot easier.
@@andrewkurtz1252 I think you have to factor to get v with dv so that you can take the integral on both sides.
As soon as I heard physics. Inja I subscribed nice job.
This video was super helpful! Thanks a lot
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!
Just subtite a with dv/dt and separate the integrals and solve both the intregals, your intial conditions will be your limits.
How would you determine a value for b? I’m doing an experiment where I have to find g from the free fall of steel balls but I also have to factor in drag and I have no idea how to determine the constant.
Great video. Really enjoyed it.
I didn't understand how you were supposed to integrate with an initial velocity
Just integrate from the initial velocity instead of from 0, 0 in this case is the initial velocity.
Thank you
Very useful. Thank you!
Only thing i dont understand is when you "factored" v
Did you intend to write dt instead of t at 09:15? Also shouldn't it be from 0 -> t instead of from 0 -> t' ?
Sam D yes should be dt but these rest looks good.
I think he's defined the upper limit as t' to differentiate it from the time function itself, just for clarity's sake
Great explanation. Simple but accurate. I have one question. Suppose we don't drop the ball but propel it through the air with a certain acceleration. Can we substitute g with this new acceleration and get the same results? Also, does this mean that the terminal velocity depends on the acceleration? So, with greater acceleration we can achieve greater terminal velocity? Thanks.
If instead of g we have an applied force F then the terminal velocity will depend on applied force. The terminal velocity will always be obtained when we balance forces ( in this case gravity and drag). Be careful when you say “terminal velocity depends on acceleration). Once terminal velocity is reached the acceleration goes to 0.
@@PhysicsNinja Yeah, I meant force. I always associate force with acceleration, thus the confusion. Thanks, for the reply. I just wanted to make sure the drag force can arise due to any kind of movement of the object and it's not somehow bound to the G force.
@@PhysicsNinja Yeah, I meant force. I always associate force with acceleration, thus the confusion. Thanks, for the reply. I just wanted to make sure the drag force can arise due to any kind of movement of the object and it's not somehow bound to the G force.
@@PhysicsNinja does that mean we can control terminal velocity, in horizontal direction also, if we control the horizontal force?
@@bandekarameya Yes! For projectile motion the drag force is always opposite of the instantaneous velocity so there are 2 components.
What if the body is a bubble under the water, 10 meters deep, for example. How will speed and acceleration be calculated, given that the size of the bubble will change and grow in each moment?
That’s an advanced problem. Gravity, buoyancy, drag, hmmm I’ll have to think about that one.
@@PhysicsNinja it's nice to make videos about it
What is b
Im here to understand what d and k stand for but here i got introduce to b. I feel like i wont able to grasp it if i dont understand what each variable stand for
Where did you find those?
How can I find out the approximated time in which Vt will be reached?
If we have the expression for v(t). Say you want to know how long it takes to reach 90% of vterm. Set v(t)=0.9*v_term and solve for t. Time will be the only unknown assuming you know b.
@@AndyMBBallSkills Thank you, I always tried to use limit to figure out, but it always diverged and I wasn't sure how I could do further.
(ooooomagod I'm gonna die)
Prof cancelled this class due to tech issues, and we still have a test on it and the previous chptr in 3 days.
🙄😳 wtf..... is happening.
Sorry but what if the body is moving down at t=0 ? Let's say at t=0, my v=8. How do I find out what is my v at t=10 ?
What I mean is, what if v0 is not 0. How do I find v(t) ?
@@leomadden7724 If you look when i integrate the equation for time i integrate from 0 to some time later t, my limits for velocity go from 0 to some final velocity. You need to redo the integration with new limits for velocity. For you case you are integrating from 8 (at time 0) to some final value v. After you get your expression you substitute t=10seconds
please can u tell me how the rule becomes if the body is cube not sphere.. i really need help
What would be the time of flight
what does b stands for
thanks what is time formula?
but the force is proportional to v squared not v
Marcos Pimentel check out my other video where I look at the c squared case. The drag forge dependence on velocity depends on the Reynolds numbers. For slow moving objects the drag forge depends on v and for faster moving it depends on v squared.
@@PhysicsNinja thanks
Can someone please explain -or tell me where to learn from- that how e^(-bt/m) became 1-(bt/m) in short time limit?
btw I found the derivative of the velocity formula with respect to time and evalueted it at t=0 and the slope is g at the beginning just as told but I am still not able to understand how e^(-bt/m) became 1-(bt/m) in short time limit.
That is the Taylor expansion for exp(-a) what a is small. The expansion is 1-a+a^2/2-....I only keep the first 2 terms of the expansion
@@PhysicsNinja thank you very much for your quick reply. You helped me quite a lot.
Mad lad
How can I find a value for b. I am dropping a basketball from different heights for my experiment and I don’t know how to find a value for b
b is the density of the air * the area of the object facing the direction of movement / 2 softschools.com/formulas/physics/air_resistance_formula/85/
@@realpowergaming1449 hey, I don’t understand why in the video Fr=bv whereas on the link you sent it is F=kv^2
Why is it squared in one and not in the other?
@@oscarc.2127 man can't do the integral of 1/(v^2-mg/b) dv looooool
Sir what is b
I wrote down the drag force as being proportional to the velocity of the object. In this case b is just the constant of proportionality. At the end of the day the force should have the proper units so b would need to be in units of [N/(m/s)]. It will depend on the geometry of the object and the density of the fluid. The exact expression can be found in books on fluid dynamics (i'm guessing). Depending on the speed of the object moving through the fluid sometimes the drag force is proportional to v^2 rather than just v. Again there is a constant term in from of the velocity term.
b is the fluid resistance constant....It depends on the density of air(ρ) , Area of falling body(A), and aerodynamic drag(C).
b=(ρCA)/2.
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