Time-dependent forces

0:00- [Instructor] Let's get some practice working with time-dependent forces. Here is the first question. Why don't you pause the video and give it a read and see if you can think about what's going on over here. All right, so we are dealing with a rocket-powered sled, all right?

0:15And as a result, its mass decreases as the fuel burns. So what's important is that throughout the motion, the mass is not a constant, okay? So now what's given is during the first four seconds of the burn, engineers measure the engine's thrust to be given by this particular expression, right?

0:36So that's the thrust force. It's changing with time. A and B are constants. And of course, t is in seconds. Our goal is to now find the change in the sled's momentum during the first four seconds. How do we do this? The big problem, I mean, there are two big problems.

0:56One is we have a time-dependent force. So force is changing with time, it's not a constant. And second, the mass is changing, like the mass itself is not a constant. I don't even have a function of mass over here. So how do we figure this out? Well, if I use F equals ma, I'm gonna be in trouble, because I don't know how the mass is changing.

1:14But we have a much more powerful theorem at our disposal. The impulse momentum theorem. It says that the impulse delivered to an object, which is basically the integral of the force over time, always equals change in momentum. Now, this is not a new law of physics, this is basically Newton's second law, but written in a different way.

1:36This is the integral version of the Newton's second law, which is called the impulse momentum theorem. And so what's important about this particular way of writing the Newton's second law is that, look, if I want to calculate the change in momentum, all I need to do is calculate the impulse.

1:50If the net force is a constant, then you can just pull it outside the integral, then we just have to integrate one with respect to t. That would be very simple. But if the net force itself is a function of time, then you really have to integrate that force with respect to time.

2:04And over here you can see the net force will be a function of time. So it'll be a great idea to pause the video and see if you can do this integral yourself and calculate the change in momentum. All right, so what we need to figure out is the change in momentum, that equals the impulse.

2:20And the impulse is basically the integral of the net force acting or on the object. Over here, since we're dealing in the horizontal, we are only considering the change in momentum in the horizontal. So we are only working with the net force in the horizontal.

2:34Since it's frictionless and we have no other forces in the horizontal, the thrust force itself is the net force. So a net force is this expression A t-squared plus B dt. And what are the bounds of this integral from time t1 to t2? Well, we are calculating the change in momentum for the first four seconds.

2:53So the time would be from zero to four, so zero to four seconds. And yeah, if we integrate this, we are done. So what's the integral of A t-squared plus B with respect to t? Well, the integral of A t-squared, it's just A t-cubed by three. And the integral of B with respect to t is just Bt.

3:17So you get plus Bt. And of course, don't forget the bounds. So we have zero to four and now we substitute the upper bound minus the lower bound. If you substitute the upper bound, you get a times four cubed, that's 64. So 64 A divide by three plus 4B. And when I substitute the lower bound, I get a zero here and I get a zero here as well.

3:40So this is it. This is our change in momentum, and we know the values of A and B, it's given us 200 and 500 in their appropriate units. So if I just plug this in to my calculator, which I'm sure you can do yourself, we will now find the change in momentum. And that happens to be about this much.

4:00Of course, usually in questions they could also ask you to round this off. We'll not worry about that. What is important for us is look at how complicated the situation is. The mass is changing, the force is vary with time. And yet using the impulse momentum theorem, we're able to figure out how much the momentum has changed in just a couple of steps, which is pretty awesome, right?

4:22But of course if we wanted to calculate what the change in velocity was from delta p, then, yes, we would now need to know how much mass was expelled. Anyways, onto the next question. Again, why don't you pause the video and see if you can try and solve this yourself first.

4:40All right, so in a crash test facility, a car moving with speed v0 is quickly brought to rest in time t to simulate a crash. So that's usually actually done. So instead of actually making your car crash into something where, you know, the car might get totaled and you may not be able to use it again, we instead quickly make it stop to kind of simulate, you know, what happens when a car would go into a crash, for example.

5:07It immediately stops, right? So that's what's happening over here. And the force that's transmitted to a test dummy of mass m, which is sitting inside the car, is modeled as given by this expression, which is, again, a time-dependent force and is a constant. Our goal is to find an expression for the peak breaking force, Fm, experienced by the dummy in terms of, you know, in terms of these variables.

5:36So how do we do this? Well, again, we can use the impulse momentum theorem. If you calculate the impulse and equate it to momentum, we'll be able to get an expression which involves Fm, which is what we need to figure out. And then we can rearrange and find an expression for Fm.

5:56So let's do this. So, on the left-hand side, we have the integral. What are the bounds of this integral? Well, it's given over here, right? t is the time it takes to bring this car to a stop. And so zero is the, you know, time t equal to zero. This force starts acting.

6:12So it's mentioned in the problem. So our bounds of the integral is gonna be from zero to capital T. What's the net force acting on this dummy? Well, this is the only force that's mentioned, so we're gonna assume there are no other forces in the horizontal, the net force in the horizontal, the net force on the dummy would be just this much.

6:29So I can take the negative sign outside and have Fm times sine of At dt. that equals a change in momentum. What's the change in momentum? Well, it's final momentum minus the initial momentum. The final momentum is zero because it goes to rest. Initial momentum.

6:45Well, we know the initial speed, so mv0, so that equals final momentum minus the initial momentum, zero minus mv0. So I can pull out the constant, Fm, which is what we need to figure out. So I can pull it out of the integral and that equals minus mv0. The negative and negative sign cancels.

7:04So we are left with on the left-hand side Fm times... What's the integral of sine At with respect to t? Well, that's gonna be cos At, it's negative cos At, negative cos At divide by A. So I get negative cos At divide by A, and, you know, the bounds are zero to capital T.

7:24And that equals mv0. And just to be clear, this minus and this minus sign canceled. So that's why I don't have those minus signs. And this negative sign comes because integral of sine with respect to time is negative cos, okay? So that's where the negative sign comes from.

7:41All right, so all we have to do now is substitute the bounds. So if I substitute the upper bound, I get cos of A capital T and minus the lower bound cos of zero is one. So cos of At minus one, but there's a negative sign, so it's one minus cos of At. So Fm divide by A times one minus cos of AT.

8:01That equals mv0, and we are done, we can now rearrange this. So if we rearrange, Fm equals A times mv0 divide by one minus cos of AT, and there we have it. That is the expression for the peak force acting on the dummy. Why would we need to calculate this in the real world, for example?

8:22Well, that's because we would want to learn what decides the peak force that's acting on, you know, a passenger traveling in the car. And then we can use that insights to design the car in such a way that we can reduce those parameters so that, you know, the cars are safer.

8:43The actual model for a force during a crash test would be very complicated and very different. But we are able to get a glimpse of, you know, how engineers do these things. So this is pretty awesome if you ask me, right? So that's the power of impulse momentum theorem.