Motion as a function of time: Resistive force example (part 1)

0:00- A tiny drop of water falls vertically from rest through air. It experiences a drag force of this particular form where B is a positive constant. Our goal is to find the drops, acceleration, velocity, and position as a function of time. So let's think about what's given, and we can directly start by trying to draw a free-body diagram.

0:23We're gonna draw a free-body diagram for a tiny drop of water that's falling vertically. So here's the tiny drop of water. What are the different forces that are acting on it? Well, there is the force of gravity and G acting on it, but along with that, there's also the drag force.

0:37What exactly is it? Whenever a particle moves through any fluid like liquids or gases, then these fluids will resist the flow of that particle and that resistive force is what we call the drag force. What's interesting is that that force depends on the velocity.

0:55Think about it. If the velocity is zero, then the drag force is zero. That makes sense because the particle is not moving. If it's not moving, then there is no resistance over there. But if it is moving, if there is some velocity, well the drag force depends on it.

1:10More the velocity, bigger the drag force. All right? Okay, that's on the magnitude. What about the direction? Well, first of all, notice that B is a positive constant. This number depends on the properties of the medium, which means that the direction of the drag force is in the opposite direction.

1:29There's a negative sign over here, opposite direction of the velocity. And again, that makes sense. That's what resisting means, right? So in order to understand the direction of the drag force, I need to know in what direction the particle is mowing. I need to know the direction of its velocity.

1:44Do I know the direction of the velocity of a raindrop? Yes. A raindrop will be mowing downwards. So since the raindrop is mowing downwards, the drag force will be upwards. That's its direction and its magnitude is going to be B times V. Notice I have not written the minus sign over here because we're only talking about the magnitude.

2:02We've taken care of the direction. Notice it's in the opposite direction of the velocity. Okay, these are the only two forces that are acting on it. So we are now ready to analyze the forces and calculate the net force and stuff. But before we do that, one last thing would be to choose our coordinate axis.

2:18So since the forces are only in the vertical, let's consider the downward direction as our positive direction. Let's call that as our Y direction. Let's completely our choice and let's consider the initial position at time equal to zero, the position of the raindrop.

2:34Let's call that as zero. With that, we need to now find the acceleration, the velocity, and position as a function of time. How do we do that? Well in this video we'll only focus on the velocity because we'll see, it's gonna take a while to get there. But how do we do this?

2:50Well, the first step is to calculate the net force and then equate it to Ma Newton second law. And then from there you can cal try and calculate the acceleration and from there we can try to calculate the velocity and from there we can try to calculate the position.

3:05So great idea to pause the video and see if you can calculate the net force and then equate it to MA. Alright, we have two forces. MG is positive because downwards is positive and then the drag force is upwards. So it's negative, so it's gonna be -B times V and again, the Y subscript to represent we are calling this the high direction and that should be equal to mass times the acceleration.

3:30So from here we can isolate the acceleration. We divide by mass on both sides, so we'll get G minus BV by M equals acceleration. And there we have it, we have found the expression for acceleration, right? Actually no, because if I asked you to calculate the acceleration at some time, say three seconds, can we do that?

3:49No, we can't. Even though we might have the values of GB and M, which are constants, we don't know what the velocity at T equals three seconds is. So this is not really our expression for an acceleration. This is not an acceleration purely as a function of time.

4:03There's also the velocity. So what do we do? Well, we need to either represent this entire expression purely in terms of acceleration or purely in terms of velocity. How do we do that? Well, we know the connection between acceleration and velocity. Acceleration is rate of change of velocity.

4:19So let's plug that in. So acceleration is just DV or DT and that should be equal to the right hand side. And look, now we have an expression that is purely in terms of velocity, but of course it's a differential equation. And solving differential equations requires using mathematical tricks and stuff.

4:37So we have to do some work over here. So now to get us to get some intuition behind how to solve differential equations, let's take a simpler version. Let's assume that this G, this constant in front was not there. Let's assume that this was our differential equation just to get some intuition.

4:53What would be the expression for VY? The question we should be asking ourself is what function, if we differentiate with respect to time, gives us the same function with some constant? Can you think of any function? If you're wondering exponential functions, you're absolutely right because that's the property of exponential function.

5:13Differentiating E to the power T with respect to time gives you E to the power T. But of course there's a constant over here. So what happens if you differentiate E to the power of some constant times T? Well, it'll be some constant times that same function, and this is exactly what we have.

5:28If that constant was -B over M then look, our velocity will be E to the power

5:37- B or MT. Because if you differentiate it, you get -B over M into velocity itself. Exactly solving or satisfying the differential equation. So even though this is not the actual differential equation we should be solving, it gives us some intuition that our velocity would be an exponential function, right?

5:57But of course that exponential function will not be of the same form because there is some constant over here. So the actual form would be quite different. And now we need to see how to solve for that. The method we're gonna use over here is separation of variables.

6:12What that means is we are gonna separate the two variables with velocity and time. So we'll try to keep all the velocity variable on the left side and the time variable on the right side. How do we do that? Well, we can first of all multiply by DT on both sides.

6:25So on the left hand side the DT cancels out, and on the right hand side we get this multiplied by DT. So we get DV equals this times DT and now we can divide both sides by this term. So we can DV divide by this term equals DT. So let's write that down over here.

6:41And look, we have separated our variables. We have velocity on one side and time on the other side, and once we have separated them, we can integrate on both sides. And what are the bounds of our integral? Well, our initial time is T equals zero zero and our final time is sometime T.

6:58So the bounds over here would be zero to T. And similarly, what will be the bounds over here? Since our variable is velocity, our bounds will be in terms of velocity. At time T equals zero what's the velocity? Well, it's given that it's the tiny drop of water falls vertically from rest.

7:15So initial velocity is zero and the final velocity, let's just call that as VY. And just to make sure we don't confuse our variable with the upper bound, let's use primes over here. Basically we are saying that VY is the upper bound and VY prime is a variable.

7:33DT is the upper bound and T prime is the variable. Okay, that's it. It doesn't change anything. Okay, so how do we integrate this expression? We'll do it nice and slowly. Let's first look at the left hand side. On the left side if you just ignore the constants, what you have is one over the function.

7:49So we should ask ourselves what function if you differentiate, you get one over the function? That's natural logarithm which means if you integrate this expression, we should get the logarithm, natural logarithm, natural log of this term. But of course if you were to differentiate this, you'll get one over this number times this constant would get pulled out.

8:12Remember when you're differentiating, that's a chain rule. So this points would get pulled out, but we don't have that constant over here. So to get rid of that, we need to multiply -M over B. Again, feel free to pause the video and differentiate this with respect to T to convince yourself you end up with this expression itself.

8:28Okay? So this is the left hand side, and of course we have to include the bounce from zero to VY and that equals what's the integral of one With respect to T prime? That's just T prime. So on the right hand side we just get T prime and the bounce would be zero to T.

8:44And again, it'll be a great idea to pause the video now and see if you can do the rest yourself. All right, let's first pull the constant out over here and then we'll substitute the upper limit. When we the upper limit, we'll get the natural log of G -BYMVY minus, we substitute the lower limit.

9:03And when you substitute the lower limit, this goes to zero. So you get minus natural log of G minus natural log of G. And there we have it, that's the left hand side. What's the right hand side? Well, if you substitute the upper limit you just get T minus zero.

9:16So you get just T, that equals T. And we've done the heavy lifting. Now we have to do more algebra over here to get to the final form. So let's do that. Okay, so what can we do? Well, we know the property of algorithms. Log A minus log B can written as log of a divided by B.

9:32So I can write this as the natural log of this divided by this. So let's do that. That should equal to T. So if I rearrange whatever is inside the log equals -B by M times T, and if I simplify inside the log function, I'll get one -B by mg VY. So wait, are we done over here?

9:54No, because our goal is to find the expression for velocity. How do we free it from inside the logarithms? Well, we can exponentiate it on both sides. Remember, E to the power natural log of something equals that something. So if we exponentiate it on both sides, we take E to the power on both sides.

10:10Then on the left hand side, E to the power natural log of this gives me just this. And on the right hand side we get E to the power this much. Okay, we are almost home. Let's do some more rearrangement and eventually we'll get an expression for the velocity. This expression looks scary, so let's try to get some intuition for it.

10:31And one way to do that is to try and plot a graph. So let's try to plot the velocity time graph. How do we think about it? Here's how I like to do it. First, let's think about at time T equals zero. If I plug in zero, you get E to the power zero and that's one.

10:46One minus one is zero. Zero times some positive number is zero. So at time T equals zero, our velocity is zero. And that makes sense because it starts from rest. At time T equals zero we knew that the velocity was zero. So this is quite comforting. So our graph should start from here, but now what happens when T increases?

11:06Well, the exponent increases, but there's a negative sign, which means the overall value starts reducing. And since it's an exponential function, it reduces very, very quickly. At large enough values of T this number would be so small I can completely neglect it, which means if I neglect this at very large values of T, I would just get MG by B times one, which means as time increases, the velocity approaches asymptotically MG or B, it approaches this.

11:40So it starts from zero and it starts approaching over here. So the graph looks like this, and sure enough, this is an exponential graph. So what's interesting is that there is a maximum value for this velocity. But how does that make in any sense? Well think about this, here is the raindrop.

11:58Okay? At time T equal to zero, the velocity is zero and the only force acting on it is gravity. So it's at accelerating down to nine point meters per second squared. It's velocity starts increasing. But as its velocity increases, the drag force starts increasing and as long as there's a downward net force, the velocity keeps increasing.

12:16So velocity keeps increasing, but eventually the drag force, if you wait long enough will be exactly equal to the force of gravity. It has to happen, right? And when that happens, the net force becomes zero. This means the stops accelerating, it doesn't stop moving, it already has gained some velocity because it no longer accelerates.

12:37That means the raindrop will continue to fall at that constant velocity. So notice it will have some maximum velocity when the drag force equals the force of gravity. And that maximum velocity is this value MG divided by B. And we give a name to that maximum velocity.

12:58We call it the terminal velocity. It's the maximum velocity an object achieves when it is moving through some fluid. So finally, look at our equation. We can write this as terminal velocity times this function. And so this is our final form for the velocity. So the expression for terminal velocity is MG divide by B.

13:23Does that make sense? Yeah, I mean for one you can directly calculate the expression for terminal velocity from the free-body diagram. At terminal the two forces are exactly equal. And so if you equate MG equals BV terminal B terminal equals mg divide by B. So you get the exact same expression even from the free body diagram, which is awesome.

13:42But we can also think about this more intuitively. This says that if the mass is higher, we have a higher terminal velocity. That makes sense, if there's more mass, then there's more force of gravity, which means the drop will have to accelerate for a longer time until the net force goes to zero, and therefore it would have reached a higher terminal velocity.

14:02So that makes sense. On the other hand, the value of B is in the denominator. Does that make sense? Well, yeah, if B is higher, then the drag force is higher for a given velocity, which means the net force becomes zero much quicker. Therefore the acceleration is for a smaller time.

14:18And so the maximum velocity, the terminal velocity achieved would be smaller. So it makes perfect sense that this is a denominator. And finally, to appreciate the idea of terminal velocity. If you consider actual raindrops, they travel a few kilometers before they hit the ground.

14:33Without the drag force they would be projectiles. And if projectiles were to accelerate for a kilometer, they would reach speeds of like 500 miles an hour, which would be very bad for us. But thankfully there is a drag force because of which they reach terminal velocity very quickly and that terminal velocity for raindrops happen to be somewhere around 20 to 30 miles an hour.

14:54We have drag force to thank for that.