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Anderson Video - Torque of Bullet Into Door

Professor Anderson
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 >> Hello class, Professor Anderson here. Welcome to another edition of the Learning Glass Lectures on physics. Let's start today's conversation with something that you might find rather relevant to your homework and this is the problem of dealing with the bullet going into the door. So, in this problem we have the following scenario. We have a door, here's the hinge on the door and we're going to fire a bullet into the door okay. It's going to stick in the door and the door will swing open and the bullet is lodged in the door. So, this is the before picture and this is the after picture. And what we want to figure out is how fast does that door swing open. What is this omega final. Remember omega is angular speed, how fast is that door rotating in radians per second and we'll give you some parameters here. This bullet has mass m, it's coming in at v, the mass of the door is capital M, the width of the door we will call a. And let's give you some real numbers because we're going to plug these in at the end. So, the data that we have is the following: mass of the bullet is 10 grams. In SI units that is of course 0.01 kilograms. The speed of the bullet is 400 meters per second. The mass of the door capital M is 13 kilograms. And the width of the door is 1.1 meters. And now we want to figure out how to solve this problem for omega f. All right, this is indicating the hinge, that's our axis of rotation. So, let me ask you guys, how do I attack this problem, what principle might I employ here to solve this problem? Anybody have a thought? Andy, grab the mic and let's have a chat. You're like I am never sitting in front again. That's okay, I'll pick on the people in the back too. So, Andy, how might I approach this problem? How do I attack it, what do you think? >> I really don't know. >> And it's okay to have no idea. >> Yeah, I'm not really sure. >> Okay, so let's think about the principles that we do understand already okay. We understand principles like conservation of energy right. Is energy conserved in this case or specifically, is kinetic energy conserved in this case? When the bullet hits the door and sticks in the door what do you think? >> Yes. >> Okay, what do you guys think, anybody else think that the kinetic energy is conserved? What sort of collision are we dealing with, is it elastic or inelastic? >> Inelastic. >> It's inelastic. Anytime things stick together right that is inelastic and is kinetic energy conserved in the inelastic collision? >> No. >> No, it's not conserved. Why, because you had to bend wood out of the way, you had to push wood fibers out of the way to get that bullet to stick in the door and that cost energy okay. So, kinetic energy is not conserved. What about momentum? One of the principles that we talked about was conservation of momentum and there were two types, there's linear momentum and then there is angular momentum and one of those is conserved here. Andy, which one do you think is conserved in this case? >> Angular momentum. >> Exactly right, conservation of angular momentum. So, what is angular momentum? For some reason we write it with an L again, we've run out of letters, okay. We're down to this. We can say whatever angular momentum is in the initial case has to be there in the final case and this is what conservation of angular momentum says. So initially, the only thing moving at all is the bullet. If the bullet does -- well let's change the question, does that bullet have angular momentum? It does about this axis okay. Even though the bullet is moving in a line you might think it's got a linear momentum which clearly it does. It also has angular momentum because we're looking about this point right here. And if I caught it with the door we know what would happen it would swing around okay. So, it certainly has angular momentum and we know what angular momentum is. Angular momentum is r across p, that's how we write it with the cross product. Or we write it mvr sine phi. Mass times speed times the radius times the angle between r and p. So, for the bullet it's moving to the right, the radius is just the door itself, that angle between them is in fact 90 degrees. So, we get Li equals mv, in this case the radius is just a, the length of the door. The angle between them is 90 degrees, but the sine of 90 degrees is 1 and so we just get mva. What about the right side of this equation? Angular momentum finally. Well now we have an object that consists of the door and bullet in it and it's rotating. We know that angular momentum is equal to i times omega. But i consists of two things, the door plus the bullet and it's rotating at omega final. And now we need to figure out some of these moments of inertia, we can set it all equal and we can solve for omega final, so let's do that. All right, one thing we need to know is the moment of inertia of the door. The door is a slab of wood and it's rotating about its hinge. And you can calculate the moment of inertia by doing that integral or you can look it up in the book because they've already done those integrals and what they tell you is that the moment of inertia of the door is 1/3 ma squared where a is the width of the door, how far from the hinge is the edge of the door. We also know that the momentum, the angular momentum of the bullet, it looks like a particle at the end of a string right, it's just a mass at the end of this door. And a particle at the end of a string has moment of inertia of ma squared. And so now we can put all this stuff together all right. We have L final equals L initial. L final we just said was i door plus i bullet times omega final is going to equal L initial which is mva. I can now solve this for omega final and I get mva divided by i door plus i bullet and now I can put all that stuff by i door plus i bullet and now I can put all that stuff in there for those i's. We have mva divided by i door which we said was 1/3 capital M a squared plus i of the bullet which is little m a squared. And we can simplify this a little bit, we can cross out an a on the top and one of those a squared on the bottom and so it becomes this. And now you have all those numbers okay. And if you take those numbers and plug them in and you want to doublecheck what I got you should end up with omega final is 0.84 and the units of course, are radians per second. This is how fast that door is going to rotate, not that fast, but it's a reasonable number. Whenever you get answers like this in terms of all these variables and again, I urge you to keep your variables all the way to the end. Look at the answer and see if it makes sense okay. What it says is if I shoot that bullet even faster, the door is going to swing open faster. That makes sense to us right? V is on the top, if v goes up then omega f goes up okay. It also says that if the door is a lot bigger it's more massive it's not going to open as fast because capital M is in the denominator and that makes sense too. And then some of these other things like the width of the door, the relationship of little m to big M you can look at. But in those limits the answer makes sense and so we say this has got to be the right answer. All right, questions about that one? Everybody mostly okay with that one? You don't have to be 100%, but if you're, you know, 75% okay with that one I'll accept that.