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Conservation of Energy in Rolling Motion

Patrick Ford
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Hey, guys. So in this video, we're gonna talk about conservation of energy in rolling motion problems. Now rolling motion. If you remember, it's a special kind of rotation problem where we have an object that not only spins around itself, but it also moves side lease so similar to if you have a toilet paper on the wall, it's rolling on a fixed axis, right? That's not really motion really emotions if you get the toilet paper and you throw it on the floor and it's going to roll and it's going to rotate and move on the floor so it rolls on the floor. So let's check out how that stuff works. All right, so remember, if on object moves while rotating, this is called rolling motion on Git. Does this on a surface without slipping Weaken? Say, let me draw that real quick. Usually sure like this V. C. M. And there is an Omega. At the same time, we say that the velocity in the middle here equals to our Omega, where R is the radius of that wheel shape Well, like objects. Okay, so this is an extra equation that we get to use all right now, remember that in order for an object to start rotating eso for it to start rotating, in other words, to go from omega zero to an omega of not zero, or to rotate even faster in both of these cases we have Alfa. We have an acceleration. There needs to be static friction. Okay, you have to have static friction in order to roll. Friction. Static. Okay, now the role of static friction enroll emotion what it's doing. It's essentially converting some of your velocity into omega. So think about it this way. If you have this guy and there's no friction, it's going to move on the surface. It's going to move in a service like this. Notice that I'm not rolling in. What friction does is get some of this V here and starts to turn it into rotation. Right? If this was completely frictionless ice and you threw, a diskette wouldn't roll. It will just go like this, but friction is what causes it to roll at the same time. So it's taking some V and changing into omega now, technically, what it's doing, Um, it's getting some linear kinetic energy and converting it into rotational kinetic energy. Okay, now, that being said, friction static does that without dissipating any energy because you're converting from kinetic kinetic so it stays within mechanical energy. So we're gonna say that even though there is static friction, the work done by static friction is zero. Okay, the work done by static friction is zero. Alright, So very briefly here to summarize if you have no acceleration if you have a no acceleration Um, I'm sorry if you do have an acceleration If acceleration is not zero, there has to be a static friction. But the work done by static friction is zero. Okay, The phrase that I want to remember here is that you need FFS in order to have Alfa Okay, Need fs in order to have Alfa in these kinds of problems. Now, the word without the term without slipping means that there is going to be no kinetic friction and a vast majority of rotation problems we're going to be You're gonna have some acceleration, but it is going to roll without slipping. So what that means is that you have static friction, but you have no kinetic friction. Um but even though you have static friction. It doesn't do any work. So when you write the work equation or when you write the conservation of energy equation, the work done by static friction is zero. Who Let's get started. Let's do an example here. So we have a solid cylinder. Solid cylinder is the shape it tells me that I'm supposed to use. I equals half M r square the masses m the radius is our So this is a literal solution. We're gonna solve this with letters were going to arrive in equation here. Um, instead of actually getting numbers, it is released from rest. Okay, so the masses and the radius is our It's released from rest Soviet Initially go zero from the top of the inclined plane of length L blah, blah, blah. Let's draw this here. So you've got a solid cylinder all the way at the top here, Um, this plane has a length of L, and it makes an angle of data with the horizontal. The cylinder rolls without slipping rolls without slipping means that there is no kinetic friction. No kinetic friction is that there's no rubbing off the cylinder on the surface, it just rolls and I want to derive an expression for the linear and angular speed at the bottom of the plane. So when it's here, I wanna know what is the final and what is Omega Final and we're gonna use conservation of energy. This is very similar to when we solved when we use conservation of energy to find the final velocity of a block at the bottom. The big difference. Now it's obviously not block. It's rotating, so we have rotational energy as well. Okay, so it's similar to this. Alright, so kinetic Initial plus potential initial plus work non conservative equals kinetic final plus potential final. All right, there's no kinetic energy beginning because it's not moving. Initially, it starts from rest. I do have a potential energy because I have a height. So someone to write MGI h initial Plus working on conservative to works you You're not doing anything you're just watching on. Then there's the work than by friction. And I wanna be very explicit here that even though there is static friction, even though there is static friction, the work done by static friction is zero. Okay, so there is basically no work. Um, there's nothing here there is kinetic Canada at the end. Now what type of kinetic energy that we have at the end? Well, what's special about really motion? One of the things that special role emotion is that the object is rolling around itself and moving at the same time. Right? So it's sort of doing this s o. I have the one object has two types of motion. So it has two types of kinetic energy. So I'm gonna write half M V squared Final plus half I Omega Square Final. There's no potential energy at the end because you are on the grounds. Okay, so what we're gonna do is we're gonna expand I and we're gonna rewrite omega. The reason we're gonna rewrite Omega is because we have V n Omega. And remember, whenever we have viene omega, what we always want to do is instead of having two variables Wien omega, we wanna rewrite omega. So we have V and V, which is the same variable. So we're gonna change omega into V, and we're able to do this because enrolling motion. We have this extra equation right here. Okay. So I can use that equation too. Replace it so v equals R Omega. Therefore, Omega is V over or so here instead of omega, I'm gonna write V over R. Okay, and then I'm going to plug in. I hear I is half m r squared, okay? And I'm just gonna go ahead. I know. I'm writing backwards here. Sorry about that. I'm gonna go ahead and write this whole thing out. Okay? So if you get here, this is the most important part. As long as you can get here. The rest is just a lot of cutting. We're gonna cancel out a bunch of stuff, so notice that they are square cancers with this Are this happens all the almost all the time, right? That they are is cancel in the conservation of energy equation. So look out for that notice that I have Mm mm. Every one of these three terms as an m. They all refer to the same object because I only have one object so I can cancel the masses as well. And then I'm left with I'm left with some fractions here. I'm gonna multiply this whole thing. My four, because I really don't like fractions. So, um, multiplying so this becomes four G H. Initial four times have becomes to the final and four times a quarter is just one. So this becomes V Final Square. Okay? And obviously these two combined to be three v final square, and I am almost ready to plug it in. V final will be four g h I This three is gonna go down there, and then I take the square root of it. There's one more thing I have to do. Um, which is noticed that I'm not giving h right. The problem doesn't give us a check. The problem gives us l and feta, but I hope you remember that H equals l sign of data. Can you see that? As you can. So I'm going to rewrite this as four g l sign of data divided by three. Okay, now, So this is the final answer. I just wanna make one quick point here. I wanna notice how I wanna actually show you. This guy here. Notice what this looks like. This looks very similar. This looks very similar to what this final velocity here for a block would look like. You might remember that the final velocity after a non object drops the height of H Um, irrespective of whether it's straight down or at an angle is V Final is the square root of two g h and I want to show I want to point out that this is very similar. But instead of two g h I have 4/3 g h right. And that's because the form the form off the solution, um, in rotation problems is similar or the same. The form is similar or the same too linear to the equivalent linear motion problem. Okay, What I mean by that is that you should expect that this final answer here will look like this. The difference is, but it has a different coefficients has a different coefficient. Okay, In fact, the coefficient in rotation in rotation, the coefficient is lower. Then it would be in linear. When you get out of here, it's lower than it would be in linear. What does that mean? Well, to, um, this is 4/3, which is 1.33 in this number has to be lower than a to the reason why I'm making this point is so that you can feel a little bit more comfortable if you remember that this is how it works, which you probably should. When you're solving the question like this, you can look at it and say, Hey, this looks kind of like what it would look like and then your motion, I'm probably right. The other way this is helpful is you can check to make sure whatever quite fish and you've got is less than what you would have gotten, Um, what you would have gotten in linear motion. So if it's less than two in this particular case, then you're good to go. If you've got something that's like a 2.5, let's say you got five over to GH five over. Choose 2.5. That's more than two. So now you know that you've done something wrong. The reason why it's lower the coefficient slower is because you are moving slower, right? It's a lower coefficient because you're moving at a slower pace, and that's because if you're rolling while, um, if you're rolling while coming down the hill, you have two types of energies. Therefore, you move overall in the slower you would have a lower fee going down. Okay, so that's it. That's how these problems work. I hope this makes sense. Let me know if you have any questions.