Free Fall with Air Resistance Terminal Velocity

hi everybody physics ninja here I'm back after the summer break today I want to look at a problem that involves free fall except I'm not going to add a little bit of air resistance to make it a little bit harder so what we're gonna do we're gonna start off with the model for air resistance that I'm gonna use we'll do a free body diagram I'll then apply Newton's second law to that diagram write down an equation of motion we're gonna look at the solution in kind of the short time limit and what happens after a very long time and at the end we're gonna do some calculus and actually get the complete solution of the velocity as a function of time and make sure everything makes sense so again with all my videos if you liked it give it a thumbs up and also if you like my channel please consider subscribing if you have any questions about the video or anything else feel free to send me an email or leave it in the comments section I'll get back to you alright let's get started alright so first thing we're gonna do is we're gonna consider an object this object is just gonna be a ball for example and the ball has a certain mass therefore there is a weight acting downright the earth is pulling down on this and the magnitude of that force is M times little G all right now in addition to this what I'm gonna do here is I'm also going to add a resistive force when this goes down and interacts with the air around it there's a lot of different ways you can model resistive forces in this case here I'm gonna take the simplest possible resistive force and I'm gonna call it B multiplied by the velocity of the object now in some cases if you look in the textbooks typically you might see a negative sign the negative sign typically only tells you the direction of this force and clearly the direction is always opposite of the velocity so that's why they kind of put that there but we know an object that is going to just release from rest is going to fall down therefore we've already taken the direction into consideration in this diagram so what I'm gonna do is I'm just gonna eliminate this there is no need for it we have the direction of this force and that is the magnitude of that force so have a look at a couple limits right initially what do we have initially if I just drop the object if we drop it drop that ball that means that the velocity at times 0 has to be 0 so if I look at this equation if I substitute 0 and this equation means that my resistive force at least initially has to be 0 and then what happens well as I start getting faster right as I start accelerating downward this force here is actually going to get bigger and it's going to continue to get bigger until this resistive force is equal to the weight so let's see so this is initially what we have after a long time let's write that down after a long time what we're gonna have is both forces will balance so we're gonna have that the resistive force is going to be equal to the weight so we're going to have that be multiplied by V that's the magnitude of our resistive force and this is really going to be my final speed I can never get faster than that this is what they call this terminal velocity and if I rearrange my equation now an expression for my terminal velocity is going to be given by this simply the weight of that object mg and I bring the be downstairs over here divided by B this here's the expression for the terminal velocity for this type of resistive force so initially I'm gonna have 0 if I just drop the object and eventually my final speed should be this one down over here simply the weight divided by that constant one more thing I want to look at are what are the dimensions of this constant again if we just look at the dimensions of this resistive force this guy we know we're in Newtons let's use a square bracket to denote the units and that has to be in units of kilograms meters per second squared that means the units of this side must also be in kilograms meters per second squared so let's have a look so I'm gonna have the units of B multiplied by the units of velocity which are meters per second so at the end what I can do is I can just simplify this a little bit I can cancel out a meter I can cancel out one of the second and what I'm left with right here are the units of this constant over here this resistive constant which are going to be measured in kilograms per second all right now let's go to the next page we'll write down Newton's second law and we'll talk about some limiting cases as well right what I want to do now is I want to apply Newton's second law to this problem we have forces two forces acting on the system what I'm going to do is I'm going to assume that the down Direction is going to be my positive direction and I need a coordinate system when I write down in second law a noon second law remember says that the net force which simply equals to the sum of all the forces acting on that object have to be equal to the mass of the object multiplied by the acceleration of that object now this term over here the sum of the forces must include two terms the weight and this resistive force and both of these forces are in opposite directions so we've got to be a little bit careful let me start off with the law with this side over here MA that doesn't change and now I'll add up all the forces so I have the force acting down which is the weight mg and I have the resistive force acting up minus B times V all right this is it this is Newton's second law says mass times acceleration equals the sum of the forces and these are in opposite directions so again if you'll have a look at the beginning let's look at those limiting cases again so we're gonna have the short time short T and long T remember what we said initially initially if we just drop the object that means that this initial velocity have to has to be equal to zero so that means that this entire term here goes away and that means our acceleration is simply going to be the acceleration due to gravity and that's just what we have right so in the short time limit this is going to behave very very similar to your standard freefall problem in the long time limit what happens again this term over here starts getting bigger and bigger as time goes on and eventually it's equal to the weight and when that happens the acceleration goes 0 it goes to zero and this is when we have our terminal velocity right the thing doesn't just keep accelerating the velocity doesn't increase forever and ever we eventually reach a terminal velocity and once we do that that's said there's no more acceleration in that case so these are the two limiting cases that our final equation has to give us in order to make sense with the physics all right so let's go back now and what we're gonna do is we're gonna actually use a little bit of calculus in order to solve this equation it's pretty straightforward there's just a couple steps that you got to be careful of and I'll highlight those for you all right so we're gonna do here we're gonna start by just rewriting this left-hand side this is M the acceleration I'm gonna write it in terms of the velocity this is DV over DT and I'll just rearrange these terms over here - be V Plus the weight mg alright next term actually I'm gonna make the marker a little bit thinner here I'm gonna divide through by the mass I can divide through all the terms by the mass that'll simplify things a little bit these terms will cancel out both of those will cancel out rewriting that expression now as DV over DT equals the minus b v over m and plus the acceleration due to gravity alright we're just about done now rewrite this term just to make the integration a little bit easier it's kind of useful actually just to factor out this term over here this is minus B over m I'll factor that out and here I'm gonna be left with the velocity and now if you got to be a little bit careful here there's gonna be a negative sign because when these multiply they have to give me this positive in this term here becomes the weight divided by B and again you need to have these two terms in order to cancel the term in front of the bracket here all right now we're just about done I'm gonna go over here what I'm gonna do is the term in the bracket I'm going to bring it in the denominator and I'm gonna bring the time over here in the numerator so at the end it looks something like this and this makes the integration much simpler alright let's write it out minus the weight divided by B all right that term is there and then over here I'm left with minus B over m multiplied by time alright this is my equation that I need to solve and the way I solve this now is I'm going to integrate on both sides I'm gonna integrate from zero all the way to the time that I'm evaluating let's call that T Prime and here I'm going to assume that at time zero that my velocity is zero and that's true if I drop it if I had some initial velocity there at time zero then this term over here the bottom of the integral would be v-0 and my final one will be V Prime all right the right-hand side of this is pretty straightforward because the term in the front is simply a constant so all of this here is simply minus B divided by M multiplied by T prime again no need for a constant here because I have the limits of integration that take care of everything here and now this term here you have to be a little bit careful so this guy becomes the natural log of u substitute V Prime in here minus mg divided by B and again now you substitute zero but you still have that term there natural log of this whole thing minus mg divided by B I'm gonna use some of the properties here of the natural log if I take the difference between 2 natural log terms you end up taking the ratio of both of them so it looks something like this natural log now I'm gonna have a huge bracket over here this is V prime minus mg divided by B all of this is over the second term which is negative and don't let that scare you mg divided by B just be careful and there that has to be equal to minus B over m T Prime all right this looks like some complicated math however what we have to do now is what we want to do is we want to isolate in this expression of the velocity term and the way you do that is by raising both sides using an exponential function that'll eliminate this natural log so let's go on the next page and finish this off alright as I said what we're gonna do now is simply raise both sides to the exponential that'll simplify this greatly so on this side we're gonna eliminate this natural log we're gonna have mg over B we're still left with this entire term here same thing minus mg over B and now the right hand side of this expression becomes exponential of minus B over m multiplied by the time which I've called T prime so all we have to do now is simply eliminate or isolate V in this expression so we'll start off by bringing this guy upstairs and then we have to bring this whole term over on this side and notice they're the exact same term so at this point what I'm do is to just write the final answer that V prime which is a function of T Prime at the end you could just drop the prime values they're not really needed this is going to be equal to mg divided by B and now there's kind of a big bracket here because there's two terms and it's one minus exponential of the whole term that we have over there all right there you have it folks that is our final expression not bad all right only a few kind of mathematical steps and actually we get this result which is kind of rich in physics one thing we could do is we could say well what happens when T prime equals to zero so that means that we're looking for the velocity of that object at time zero well if you substitute 0 exponential of 0 gives me 1 so what you're gonna have is mg be bracket 1 minus this exponential term of 0 which also gives me 1 so you get 0 how about in a very long time what if time is much much bigger then actually this terminal here a few let's just say that time is very very big we'll look at what does that mean in just a minute but again if the time is very very big this term over here the second term will go to 0 because it's exponential of minus something that's going to be very very big so that term goes to zero the second term in this bracket goes to zero and that means that the velocity as a function of time is simply left with this first term and again this is my terminal velocity now if you remember this term B over m or the units of B rather all right we could take a one step further here actually the other thing I want to look at is look at this term over here this B over m I just want to make sure that something here is clear this term over here that appears in the exponential again the entire units of the terming the exponential have to be dimensionless so if I have time here that means that the characteristic time here we can call it tau is M over B and this here is going to have units of seconds because it has to have units of seconds because it multiplies times this guy over here okay and the whole term here must be dimensionless if you take an exponential of any function so this is kind of the characteristic time so although in the previous case I said that when T was tending toward infinity or T was very very long it really only has to be long with respect to this characteristic time so if this characteristic time is 5 well my time might only have to go to 20 it doesn't have to go all the way to infinity all right there's one more limit I want to take the other limit is what happens in the short time instead of at T equals to zero what happens when T is kind of small it's nonzero but kind of in the short time limit so in short time limit and again now we have a reference point so in short time limit that's when T is much less than this characteristic time which is equal to M over B what you can do is you can expand this exponential term alright and that limit over here that means that this exponential term of mine is B over m multiplied by T Prime in the short time limit this term is approximately 1 minus B over m T Prime so that means that your velocity in that time limit all you have to do now is just substitute that in for that exponential term is 1 minus instead of exponential what I'm gonna do is I'm gonna replace it by the approximate value of the exponential which is this term which is 1 minus B over m T Prime now if you carry that out what's gonna happen well this one - this one that one's going to cancel and at the end this negative is going to multiply with this one so that'll turn into a positive and then you're gonna be left with this M it's gonna cancel with that one this B is going to cancel with that one and at the end my final result in the short time limit is simply equal to little G the gravitational judex the gravitational acceleration multiplied by the time actually this is exactly what you have in the case of no air resistance all right so let's just write that down so same as no air resistance alright so in the final group page what I want to do now is I want to just give a graphical representation of what this result looks like all right so let's go ahead and plot this result down so here we have velocity or my giant expression over here what I'm gonna do now is well remember our two time limits that we looked at at time zero we had this result over here at time zero we had that the velocity V prime was equal to zero and the other thing we had that in the very long time limit so that's as I'm going down over here on along this axis what you end up getting is that the velocity tends to ward a constant value and that constant value again was simply given by this leading term over here this was the terminal velocity VT in the long time limit which was equal to the weight divided by that resistive constant B now the rest of this function looks like this again it's an exponential term and this term goes away and the curve kind of looks like this try to make it as smooth as possible alright looks something like that for a sketch and that's pretty good the other thing that you can look at is what happens over here in the short time limit right in the short time limit that's when T was small remember if this was a case without any air resistance what you would find would be that the velocity as a function of time would simply be equal to little G multiplied by the time it would be linear in time and that's kind of what you have over here all right this is kind of a short time limit which we find that the velocity is little G times T Prime alright folks there you have it folks that's kind of looking at a simple model of air resistance and what happens to the velocity of the object as a function of time ok I hope you liked the video hope you learned something let me know if you have any questions

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