Free Fall with Air Resistance Drag Force

hi everybody physics ninja here today what I want to do is I want to look at a freefall example again I'm gonna add air resistance and I'm gonna model the air resistance using this formula for this resists the force or this drag force if you look in our common text books it's often written like this one-half a Rho is the density of the fluid and in this case will be the density of air a is the cross-sectional area of the object so kind of the bigger it is the more drag force there is CD is a drag coefficient alright some number associated into the geometry and also V squared here now that's kind of the important part about this problem V is the speed of the object and that direction of that force is always opposite of the motion so what I want to do now is I want to try to calculate what the terminal velocity is of this object that's going to fall and that has this air resistance acting on it I previously did a video where my drag force was only proportional to the speed I'll put the link in the description for that video I'm gonna follow similar steps now the question when would you use the equation for drag force that's proportional to the speed versus speed squared this kind of depends on something called a Reynolds number that's kind of a topic for a separate video and its own discussion kind of depends on how fast the object is moving through the fluid so in this case now we're going to consider the V squared term for the drag force and let's see how we find this terminal velocity again if you liked the video give it a thumbs up consider subscribing to my channel and if you have any questions just leave them in the comment section alright let's get started first thing we want to do is just set up an equation of motion for this problem here I'm gonna take down to be the positive direction all right so now we're just gonna add up all the forces right Newton's second law says you sum up all the forces this is what we call the net force and that has to be equal to mass times the acceleration of the object so I'm going to take down to be the positive direction so that means I have an mg and again here I'm gonna have minus this drag force FD which is given by this expression over here that has to be equal to Ma alright now we could substitute with this drag force however just to simplify the notation a little bit what I'm gonna do is I'm gonna group together all these terms and I'm gonna call this the letter K okay so just to simplify the notation and K it will just be a constant for any given object here so K will be one-half the density of the fluid cross-sectional area and the drag coefficient alright now if you substitute back in my expression over here what you end up getting here is mg a minus K now I'm gonna keep the velocity out there because that term is gonna be important like this and that they're equals to Ma alright this is our equation of motion that we really want to solve for we're gonna go to the next page we're gonna have to do a little bit of calculus but even from this expression here actually you could see what's going to happen initially if I just drop the object if I drop the object at time equals to zero that means that the velocity would be equal to zero so that means there would be no drag force initially right so the acceleration initially would simply be equal to little G but what happens now is as it starts to speed up this term here starts getting bigger right and it'll keep getting bigger which means that this drag force gets bigger and it will keep getting bigger up until it's actually equal to the weight right once the drag force is equal to the weight actually we're gonna reach terminal velocity and at that point my acceleration is going to be zero once I've reached terminal velocity so let's go to the next page and look at terminal velocity and calculate the velocity at any moment in time when this object is falling all right so this is what we have our equation of motion where K was defined like this and let's just write velocity or terminal velocity terminal is when there's no acceleration so if you set a equals to zero over here you can simply rearrange this right we're gonna get zero has to be equal to the weight acting down and our resistive force acting up so rearrange this expression now you should get that the terminal velocity let's just write it as V subscript T for terminal that is going to be equal to the square root of this guy mg and divided by K and now you can go ahead and substitute this expression so we have now our expression for terminal velocity so we have mg if I divide by this K constant the twos gonna end up here in the numerator and I'm gonna have density a and my drag coefficient so a couple things that kind of makes sense from this expression is if area is big that would tend to make my terminal velocity small right that kind of makes sense also if this drag coefficient is big and that it will also tend to make the terminal velocity small right so those kind of intuitively I think kind of makes sense we can go ahead and substitute values in here just to get kind of an idea and I'll do that kind of at the end I'll consider somebody skydiving and we'll calculate what the terminal velocity is from this expression but for now I'm going to go back to my equation one over here which is my equation of motion and now my goal will be to find what is V as a function of T so this is kind of after a very very long time right this is what's gonna happen but what how does V vary with time if time is not very long so now we have to do a little bit of a calculus so that's gonna take a couple steps for us to get to this expression of the velocity as a function of time starting now from my equation of motion again if you just simply Express the acceleration as the rate of change of the velocity with respect to time you're left with this now what I'm gonna do here this is not required but I want to just kind of get things in a kind of a neater form so let's just divide through by the mass you do that this term here will simplify alright you'll only be left with little G and this second term here will be K divided by M V squared and again K is just this constant that's associated to a whole bunch of other terms one more thing I'm gonna do here I'm gonna factor out a couple terms here and you'll see why in a minute it's just because I want to get this into a nice kind of integral form so let me just switch the order here first and then the last thing I'm just gonna factor out this K over m term just so I have the V squared term by itself and if you do that now while clearly this other term if i factor out a negative sign I have to have another negative sign here and here this here has to switch this here has to be M over K so if I do multiply through again things are going to cancel out now this looks more complicated however you're gonna see why I did that in a minute now what I want to do is I want to group all the terms that have V on one side so this is what I'm gonna do so I have DV over here what I'm gonna do here is I'm gonna have a fraction because I'm gonna take this whole term that's in the bracket because that term there has a V and we're gonna bring it downstairs over here V squared minus again just the number here G M over K these are simply just the constant factor over here and if I do that now I have to bring my DT on the other side and I'm still left with this minus K over m so it looks like this minus K over m and this is DT all right this is a really important equation here this is kind of my equation that I want to solve for it's pretty straightforward to solve for this now what we're going to do is we're simply gonna integrate on both sides of this and that's going to allow us to get an expression for V as a function of T we have to do still a little bit of math so let's go on the next page and do some calculus here but it's pretty straightforward so to evaluate this function over here what you end up doing is we're simply going to integrate on both sides we're gonna integrate the right hand side from zero to T and during that time period the speed is gonna go from zero all the way to some final value or some value at time T all right all we have to do now is just evaluate both of these integrals what I'm gonna do here is I'm going to integrate I'm gonna introduce some new notation here just to simplify this I don't want to have to write these terms down all the time so what I'm gonna call B squared will be all of this term here in that red box G times a mass divided by that other constant K all right now if we rewrite this this here becomes zero to V and you'll see why I pick the value B squared in just the second over here V squared my is b-squared right this kind of has a nice symmetric form to it and this here's just keeps the same minus K over M DT alright well start with the left hand side this integral here is something you can look up if you look this up in an integration table integral table what you're gonna find here is that this is kind of a well-known value and the value of this integral here if you look it up looks something like this it's going to be minus 1 over B now it's an inverse tan hyperbolic and the argument here will simply be B sorry V divided by B so if you'll go ahead now and substitute our value for B what you're gonna end up getting here is going to be minus now the square root is going to be introduced because I only have be here and I've defined B squared as this term over here I can flip that fraction over here this is my inverse hyperbolic tan function and again this whole term here will simply be V divided by B so you can simply write it like this GM over K all right so this is really the left hand side of this integral now again this has to be evaluated between the limits of V final here and 0 okay the nice thing about this inverse hyperbolic tangent is if you substitute 0 in this expression you end up getting 0 so evaluating it at those limits you simply are back to my expression over here ok we now have to look at the right-hand side of this expression this one's super easy because negative K over m is simply a constant term you take that out and now you're simply integrating over DT so this whole thing simply becomes something that reduces to this alright we're not quite done with the math now really our goal was to get to what is V as a function of time what is this function that was really the goal of doing the whole thing and V is a function of time is to tucked away inside over here so we have to do is isolate V as a function of time and that's actually pretty straightforward because I have an inverse function here so it's kind of a little bit easy to isolate that but let's go ahead and the next page do a little bit of math to get to our final expression first thing I'm going to do here is we have negative signs on both sides and cancel those and now what I want to do is just bring this factor in the front of this inverse hyperbolic functions bring that on the other side okay if you bring that on the other side what you're gonna be left with here is again this inverse tan hyperbolic let's keep everything in here just the same GM over K all right now we have to group together the terms so I have kind of square root is gonna definitely be a square root term over here and the time I'm just gonna leave it the way it is you could see here I'm gonna have a square root of M in the numerator and simply M in the denominator so that means I'm left with square root of M and the denominator other terms here I'm gonna have square root of K because I have square root of K on the left hand side and simply K on the right hand side and the last bit now you'll see that little Geel end up coming upstairs over here all right and all you have to do now is really simply on both sides you have to evaluate the tan hyperbolic and what that's going to do is it's going to eliminate this hyperbolic function right it's going to eliminate this whole term that means that the left hand side will simply reduce to this GM over K and on the right hand side now I'm gonna have this hyperbolic function it's not the inverse anymore of this entire term over here so square root of I'm going to switch the order here just for preference and multiplied by time all right and we have one last step to do is simply bring this square root term over here actually this term here should look familiar to you all right this term here was exactly how we define this terminal velocity so if I just kind of rename it again terminal velocity and I bring it up Aires let's go ahead and just separate this what I'm left with now is that my V is a function of T is going to be equal to the terminal velocity and then multiplied by this whole term tangent hyperbolic of square root of G K over m multiplied by time if you use our definition of this terminal velocity again G M over K you can simplify this term here in the bracket of the hyperbolic tangent function and our final expression kind of looks it has a nice kind of compact form terminal velocity you calculate tangent hyperbolic now the argument here again I'm introducing VT go ahead and do this for yourself but you should end up getting something like this instead of having root of G if I introduce the variable VT is my terminal velocity it ends up simplifying to something like this and this is kind of a nice simple form so the last bit I have to do now is I could take this one step further I guess I could kind of just eliminate our variable K which I previously did but this is it let's go have a look now at the functional form of this hyperbolic tangent see what it looks like alright now I want to consider just this silly example over here let's calculate first our terminal velocity of the skydiver here and well I just put in some numbers they're somewhat realistic maybe not for this girl over here but let's do a mass of 75 kilograms little G is roughly 9.8 if I'm not too far away from the surface of the earth density of the air say approximately one kilogram per meter cube the cross-sectional area again this will depend on how you're gonna jump right if you're kind of bodies elongated like this and let's just say you're about a foot wide and maybe two meters tall so kind of this is the cross-section and I'll give it a drag coefficient of 0.5 or you go ahead and you punch in the numbers over here what you're gonna end up getting here is approximately of 50 meters per second you know that may not tell you a whole bunch if you convert that now if you convert this to miles per hour that should give you something that's a brown 112 miles per hour okay that's kind of the terminal velocity that you know a body that's kind of spread out like this with these types of parameters which are kind of realistic gives you approximately it's something a little bit bigger than 100 miles an hour now let's go ahead and make the plot now just to see how the velocity looks like as a function of time and I also want to compare to the case where there is no drag right if there's no drag then you're simply in freefall so you know that in that case the velocity as a function of time should always increase and if I release from rest then this is it right it's just the function that increases linearly with time so let's go ahead and compare our case with air resistance versus the case with no air resistance okay are no drag all right so just rearrange us a little bit so again I have my expression with air resistance or with drag and this is what the function looks like so the red line and this plot here starts at zero and then it increases and then you see over here it kind of bends over alright and it approaches if I go out in time it's kind of might be a little bit hard to see this let me just go out and write some of these values here this is kind of 18 this is 10 seconds and really what I'm approaching over here is as time goes on I'm approaching 50 meters per second right here all right this value over here is 50 along this line and you don't have to go very far before you reach this 112 miles per hour you're looking at about 15 seconds and then you've hit that value the blue line on the other cases where we just have just regular old freefall so that case would just continue to increase and you see that the slope here the slope is simply equal to little G that's the slope of the line the acceleration of the line you notice that in the first kind of two seconds they're almost one on top of each other oh maybe definitely for the first one second there's not a lot of difference between the two and then there's some separation here because the object starts to slow down that drag force gets bigger over here and remember over on this part over here the acceleration is simply zero all right the slope of that line gives you the acceleration and which is a flat value over here okay so here's the comparison between drag versus no drag showing you the plot again I just kind of plotted both of those using the values that I calculated for this particular example all right thanks for watching folks

Loading