Conservation of Energy in Rolling Motion - Video Tutorials & Practice Problems

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1

concept

Conservation of Energy in Rolling Motion

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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.

2

Problem

Problem

A solid sphere of mass M = 10 kg and radius R = 2 is rolling without slipping with speed V = 5 m/s on a flat surface when it reaches the bottom of an inclined plane that makes an angle of Θ = 37° with the horizontal. The plane has just enough friction to cause the sphere to roll without slipping while going up. What maximum height will the sphere attain? (Use g = 10 m/s^{2}.)

A

0.175 m

B

0.350 m

C

1.75 m

D

3.50 m

3

example

Sphere on rough and smooth hills

Video duration:

15m

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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.

4

Problem

Problem

You may remember that the lowest speed that an object may have at the top of a loop-the-loop of radius R, so that it completes the loop without falling, is √gR . Determine the lowest speed that a solid sphere must have at the bottom of a loop-the-loop, so that it reaches the top with enough speed to complete the loop. Assume the sphere rolls without slipping.

A

√(5gR)

B

√(13gR/5)

C

√(27gR/7)

D

√(10gR)

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