13. Rotational Inertia & Energy

Conservation of Energy in Rolling Motion

# Sphere on rough and smooth hills

Patrick Ford

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Hey, guys, let's check out this example of conservation of energy in rolling Motion. Now it's special about this example. Is that gonna have an object that's gonna roll down a hill and go up the other? But in the first hill, there's going to be static friction. Therefore, there is angular acceleration Alfa, which means you go from no speed to rolling faster and faster. On the second hill. However, there is going to be no static friction. So there is no angular acceleration, Alfa, which means that you don't slow down. So here you spin faster and faster. But here, you're going to go up and your rotation is going to stay constant because there's nothing to slow it down. Okay, now your velocity increases, your V grows, and then your V goes down, right slows down. So the V acts as you would expect, but the Omega gets faster, but then does not get slower as you go up the hill. So let's check it out. Let's draw this real quick. I'm gonna draw two hills, Um, like this and we're gonna call this initial point here a okay, and then the initial velocity here will be zero vehicle zero. Um, and it has some sort of initial heights here, which were the problem Calls it, um h one. Okay. And I wanna know how far up he gets over here on this side. We're gonna call this H two, and I wanna know what maximum heights does it obtain on the second hill in terms of H one. Okay, so it's gonna be something like this, some multiple of each one, and that's what we want to know What it ISS. So it says it's a solid sphere. So the moment of inertia is by table. Look up to over five m r. Squared has a massive Emiratis are This is a literal solution. We're not gonna have actual numbers here. Um, it's initially at rest. We got that. They're on top of a rough hill of height. H one says the sphere rolls, rolls down the rough hill, then rides in a smooth, horizontal surface. So this is rough, Onda. What? That means that explains it here. The first hill has enough friction to cause the sphere toe roll without slipping. Okay, roll without slipping means no kinetic friction. But there is going to be some static friction which causes the object to to accelerate toe, have in an angular acceleration to roll faster and faster. For this piece, it is smooth, which means there is no angular acceleration, which means whatever velocity, whatever omega you have here, let's call this point B. Whatever. Omega, you have a point B will be the same omega. Then you have at point c okay, because there's no change in angular velocity in that piece. For the third interval here we have a way have a long, smooth hill. Smooth hill means that there is no friction. Therefore, there is no angular acceleration. Okay, again, just like the horizontal surface. So there is Alfa here, but there is no Alfa here and here. What that means is that imagine a block, the block speeds up, moves to the constant speed and then slows down on the way up. The sphere will do that, except so it gets faster, same speed and then slows down and stops. But the rotation is going to be different. It's going to accelerate on the way down because it has an Alfa. Then it spins at a constant rate and then as it goes up the second incline because there is no static friction. It's a smooth hill. Um, then there is going to be no Alfa. What it means is that it's going to go up, but it's gonna keep rolling at the same rate. Get to the top stop right vehicle zero. But it's still rolling, and then it starts going down. Okay, so it's kind of weird. It goes up, but it doesn't really stop rolling. Which you would usually expect is that it goes up and slows down and stops. And then it comes back down spinning faster. On this case. It's gonna keep rolling, go up, keep coming down. That's because there's no friction. So that special. That's what's special about this question. What we're gonna do is we're gonna write an energy equation from A to B, and then we're gonna right on energy equation from, um, see too deep. There's nothing. There's nothing special happening from PTC. In fact, BNC are really the same thing. Nothing happens there. So let's write the first one, which is the energy equation from A to B K, a. U. A. Work non conservative between A and B is K B um, you be There is no kinetic energy at the top of the first Hill because it's initially arrest. There's potential energy because it has a height. There is no work non conservative. Remember that. Even though, uh, work non conservative is you. And friction, even though there is friction, static friction does the work of zero in these problems, there's no work at all. There is Connecticut. The end because we have some velocity, but be is on the floor. It's the lowest point, so potential energy is zero. So what we have here is simply potential energy going into kinetic energy. Okay, Potential energy is M G h a R. In this case, I guess we're calling this H one. So let me do that. G h one, um, equal kinetic energy. Now, what types of kinetic energy does this ball have? Well, it's not only falling, but it rolls and speeds up, so it has both types of energy, so I'm gonna write half M v b squared plus half I omega B squared. OK, One object has two motions, so we have to kinetic energies. Cool. Now what we're gonna do is, as usual, we're gonna expand I and we're gonna rewrites Omega. Why do we rewrite omega? Because I have a V in Omega. Whenever you have the two velocities, you wanna get rid of one so that you only have one. So the omega will become a visa instead of having viene omega. I have V and V. That's better. Um, this is rolling motion, which means I can write that the velocity of the center of mass Remember, this velocity of center mass is linked with the rotation by this equation. So v C m is our omega. Therefore, Omega is V over the radius. So I can use this to replace Omega. I'm gonna replace Omega with V over R. Okay, let's rewrite this plug in the eye and then keep going. So MGH one equals half M v b squared, plus half, and then I have I. I is the moment of inertia of the sphere solid sphere, which is to over five to over five m r square. Now look what happens. A bunch of stuff is gonna cancel these two cancers. With this to the are cancers With er, there's EMS here. They There's one M in every term and they all refer to the same mask so we can cancel those as well. I have a two down here and I have a five down here. So to get rid of those guys, I have to multiply this by two. And by 500 words, multiply the whole thing by 10. Let me make a little space here. So it's not messy. When I multiplied the holding by 10 I get 10 g h one 10 times half this five V b squared 10 times 1/5. It's too v b squared, and that's it. So this adds up to seven V B square and then I have VB is, um 10 g h one divided by seven. Take the square root of both sides. Okay, 10/7. That's what we get for That's, um now. So we got that cool. So that's V B. I found the velocity here. So now what I'm gonna do is I'm gonna write an equation from I'm gonna move over here, and I'm gonna write conservation of energy equation from B to C to show you how this looks like You can think of this problem is really two questions that are merged together. And I want to show you separately what these parts look like. Andi, I'm sorry. It's actually from see too deep. Okay, so see if to these going up So k c you see work non conservative, K d you deep. All right, so is there kinetic Canada Point? See, the answer is yes. Sees the bottom of the hill. What kind of energy? Well, not only do you have velocity, but you also were rolling and you got to the bottom rolling and you kept rolling. So you have both. You have half m v squared plus half. Um, I Omega Square. And I wanna make the point that this velocity here at point C is really the same as the velocity of point B. Okay, so Omega Psi is the same as Omega B. And V C is the same as VB because you want a smooth surface. Nothing happened there. It's as if you basically just went immediately up another hill. Okay, so I'm gonna put a VB here, and I'm gonna put in omega being here. Um what about potentially? There's There's no potential energy at the bottom of the hill There's no work non conservative because there's no work done by you. And there's actually no friction at all. What about kinetic Energy? Point D. This is the trickiest part of this whole question is to realize that even though you are highest points even though you are at the highest points, you do have rotational energy, your linear stops because the object does this and stops right, Then it starts come back down. But it kept It keeps rolling as it goes up. Okay, so you still have this. This is the most important part of this whole question. This entire question basically exists for this one purpose. Okay? And there is potential energy that point. I'm gonna write m g the heights at point. Do you recall that H two. Okay. And by the way, H two is our target variable. So we're gonna simplify a bunch of stuff, and we're gonna be ableto we're gonna be left with h two. Okay, so what I gotta do here? We gotta expand this guy, and we gotta expand this guy and we gotta rewrite omega. We have viene omega. So we're going to rewrite um, omega and two v by writing Omega equals V over R, which we can do because it's a rolling motion problem. Half M V B squared plus half This is the Omegas. It's to over five m r. I'm sorry. This is the I the moment of inertia. So it's to over five m r squared omega. We talked about how we can rewrite this into vb over are so omega b is vb over our square. That's the left side. On the right side, it can expand the rotational energy here, which is going to be half half. Um, I I is to over five m r squared Omega squared. OK, I'm just doing this in one shot. Um, que rotation is half I omega squared. I'm replacing the I. It's the eye right there. And I'm replacing W with v B over r squared plus m g h to now. I didn't really have to do this. Um, this one last step here, I'll tell you what I mean, But I wanted to do this to make it painfully clear. Notice that this thing here is exactly the same as this thing here. It's exactly the same, and it's because this rotational kinetic energy at point B is the same is the rotational kinetic energy at point D. Okay, let me go up to the drawing real quick. So you're rolling at B, you're rolling, and this rolling never slows down. So whatever energy you have here, you have it here at sea, and then you go up and you have it here at D. Okay, so what you could have done is you could have canceled these guys. Um, right here. Okay, I'm gonna not cut it out there because it's gonna be messy. I'm gonna cut it over here, but again, you could have canceled those too, because those energies are the same. They don't change. So really, all you have is this and this I can cancel the masses and I'm solving for H two h two equals V b squared, divided by two g. Noticed that the G went down there. Ok, now I do have an expression for V b so I can plug it in. VB is right here, so it's gonna be one over to G and then the B squared. Notice that DBS inside of a square root. So 10 g h 1/7, which means that the square root cancels with the two. And then I'm left with one over to G times 10 G H. 1/7, and this will cancel nicely. This cancels with this and then I have 10. Divided by 14. 10 divided by 14 H one and 10. Divided by 14 is 140.71 H one. Okay, so we're done here. The final answer is that H two is 0.71 h one. Now. One question that were asked here is part of the bonus question is why are these difference? Right? So conservation of energy usually would dictate that you started the height and you come back to the same height. Well, the reason they're different is because of the energies, right? All of the potential energy went into kinetic at the bottom, right? So there's a full energy conversion. But on the way up on Lee, some of the kinetic energy went back into potential because some of it is still in spinning. So here you are, very high with no spinning. Here you are a little bit lower because you're spinning so energy spinning costs energy. Right? So basically what happened is you went from initial potential, but no initial, No in their show, Rotational to a potential at the end and rotational at the end. Okay, so what happens is some of this went into here and some of its went into here because your initial height got to split into height and rotation. You don't get as much height, right? So if we did this in terms of energy is, let's say, 100 jewels got split where, like, 70 Jules went here and then 30. Jules went here, right? That's why you see ah, lower heights. Because you spend some of that heights energy into rotation energy. So you didn't recover as much of your heights. Okay, so that's it for this one. Hopefully, make sense. Let me know if you have any questions.

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