Conservation of Energy in Rolling Motion

Concept: Conservation of Energy in Rolling Motion

13m
Video Transcript

Hey guys! In this video we're going to talk about conservation of energy in rolling motion problems. 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 sideways. Similar to if you have a toilet paper on a wall it's rolling on a fixed axis, that's not rolling motion. Rolling motion is if you get the toilet paper and you throw it on the floor and it's going to roll and on itÕs going to rotate and move on the floor, so it rolls on the floor. Let's check out how that stuff works. Remember, if an object moves while rotating, this is called rolling motion, and it does this on a surface without slipping we can sayÉ Let me draw that real quick. Usually I show it like this Vcm and there is an w at the same time. We say that the velocity in the middle here equals to Rw where R is the radius of that wheel-like object. This is an extra equation that we get to use. Remember that in order for an object to start rotating, for it to start rotating in other words to go from w = 0 to an w ­ 0, or to rotate even faster. In both of these cases, we have alpha. We have an acceleration. There needs to be static friction. You have to have static friction in order to roll, fs. The role of static friction in rolling motion, what it's doing is it's essentially converting some of your velocity into w. Think about it this way. If you have this guy and there's no friction, it's going to move on a surface. It's going to move in a surface like this. Notice that I'm not rolling it. What friction does is gets some of this V here and starts to turn it into rotation. If this was completely frictionless ice and you do a disc, it wouldn't roll. It would just go like this. But friction is what causes it to roll at the same time. It's taking some V and changing into w. Technically, what it's doing, it's getting some linear kinetic energy and converting it into rotational kinetic energy. That being said, friction static does that without dissipating any energy because you're converting from kinetic to kinetic so it stays within mechanical energy. We're going to say that even though there is static friction, the work done by static friction is zero. The work done by static friction is zero. Very briefly here to summarize. If you have an acceleration, if acceleration is not zero, there has to be a static friction but the work done by static friction is zero. The phrase that I want you to remember here is that you need fs in order to have alpha in these kinds of problems. The term without slipping means that there's going to be no kinetic friction. A vast majority of rotation problems are you're going to have some acceleration, but it is going to roll without slipping. What that means is that you have static friction, but you have no kinetic friction. But even though you have static friction, it doesn't do any work. When you write the work equation or when you write the conservation of energy equation, the work done by static friction is zero. Let's get started. Let's do an example here. We have a solid cylinder. Solid cylinder is the shape. It tells me that I'm supposed to use I = _ MR^2. The mass is M, the radius is R. This is a literal solution. WeÕre gonna solve this with letters. We're going to derive an equation here instead of actually getting numbers. It is released from rest. So the mass is M, the radius is R, it's released from rest so V initial equals 0. From the top of an inclined plane of length L, let's draw this here. You've got a solid cylinder all the way at the top here. This plane has a length of L and it makes an angle of _ with the horizontal. The cylinder rolls without slipping. Rolls without slipping means that thereÕs no kinetic friction. There's no rubbing of the cylinder on the surface. It just rolls. I want to derive an expression for the linear and angular speed at the bottom of the plane. When it's here, I want to know what is V final and what is w final. We're going to 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. ItÕs rotating so we have rotational energy as well. It's similar to this. Kinetic initial plus potential initial plus work non-conservative equals kinetic final plus potential final. There's no kinetic energy at the beginning because it's not moving initially. It starts from rest. I do have a potential energy because I have a height, so IÕm going to write mgh initial plus work non-conservative. Two works Ð you, you're not doing anything you're just watching and then there's a work done by friction. I want to be very explicit here that even though there is static friction, the work done by static friction is zero. There is basically no work. There's nothing here. There is kinetic energy at the end. What type of kinetic energy do we have at the end? What's special about rolling motion, one of the things thatÕs special about rolling motion is that the object is rolling around itself and moving at the same time. It's sort of doing this. The one object has two types of motion so it has two types of kinetic energy. IÕm going to write _ mv^2 final + _ Iw^2 final. There's no potential energy at the end because you are on the grounds. What we're going to do is we're going to expand I and we're going to rewrite w. The reason we're going to rewrite w is because we have V and w. Remember, whenever we have V and w, what we always want to do is instead of having two variables, V and w, we want to rewrite w so we have V and V which is the same variable. We're going to change w into V. We're able to do this because in rolling motion, we have this extra equation right here. I can use that equation to replace it. V = Rw therefore w is V/R. Here instead of w, IÕm going to write V/R. Then IÕm going to plug in I here. I is _ MR^2. I'm just going to go ahead. I know IÕm writing backwards here, sorry about that. IÕm going to go ahead and write this whole thing out. 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 going to cancel out a bunch of stuff. Notice that the R^2 cancels with this R. This happens almost all the time that the R is canceled in the conservation of energy equation. Look out for that. Notice that I have M, M, M. Every one of these three terms has an M. They all refer to the same object because I only have one object so I can cancel the masses as well. Then I'm left with some fractions here. IÕm going to multiply this whole thing by 4 because I really don't like fractions. IÕm multiplying so this becomes 4gh initial. 4*1/2 becomes 2Vfinal. 4*1/4 is just 1, so this becomes Vfinal^2. Obviously these two combined to be 3Vfinal^2. I am almost ready to plug it in. Vfinal will be for 4ghi, 3 is going to go down there and then I take the square root of it. There's one more thing I have to do which is notice that I'm not given h. The problem doesn't give us h. The problem gives us l and _. But I hope you remember that's h = lsin_. Can you see that? IÕm going to rewrite this as for 4glsin_ / 3. This is the final answer. I just want to make one quick point here. I want to actually show you this guy here. Notice what this looks like. This looks very similar to what this final velocity here for a block would look like. You might remember that the final velocity after an object drops a height of H irrespective of whether it's straight down or at an angle is Vfinal is the square root of 2gh. I want to point out that this is very similar but instead of 2gh, I have 4/3gh and that's because the form of the solution in rotation problems is similar or the same to the equivalent linear motion problem. What I mean by that is that you should expect that this final answer here will look like this, but it has a different coefficient. In fact the coefficient in rotation is lower than it would be in linear. What does that mean? This is 4/3 which is 1.33. This number has to be lower than a 2. 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 ÒHey this looks kind of like what it would look like in linear motion. I'm probably right.Ó The other way this is helpful is you can check to make sure whatever coefficient you've got is less than what you would have gotten in linear motion. If it's less than 2 in this particular case, then you're good to go. If you've got something that's like a 2.5, letÕs say you've got 5/2gh, 5/2 is 2.5, that's more than 2. Now you know that you've done something wrong. The reason why the coefficient is lower is because you are moving slower. It's a lower coefficient because you're moving at a slower pace and that's because if you're rolling while coming down the hill, you have two types of energies therefore you move over all slower. You have a lower V going down. That's it. That's how these problems work. I hope this makes sense. Let me know if you have any questions.

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?

8m

Example: Sphere on rough and smooth hills

16m
Video Transcript

Hey guys! Let's check out this example of conservation of energy in rolling motion. What's special about this example is that youÕre going to have an object that's going to 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 alpha which means you go from no speed to rolling faster and faster. In the second hill however, there is going to be no static friction so there is no angular acceleration alpha, which means that you don't slow down. 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. Your velocity increases. Your V grows and then your V goes down, slows down, so the V acts as you would expect but the w gets faster but then does not get slower as you go up the hill. Let's check it out. Let's draw this real quick. I'm going to draw two hills like this and IÕm going to call this initial point here A. Then the initial velocity here will be 0, VA equals zero. It has some sort of initial height here which the problem calls it h1. I want to know how far up it gets over here on this side. We're going to call this h2. I want to know what maximum heights does it attain on the second hill in terms of h1. It's going to be something like this. Some multiple of h1 and that's what we want to know what it is. It says it's a solid sphere, so the moment of inertia is by table look up, 2/5MR^2. It has a mass of M, radius R. This is a literal solution. We're not going to have actual numbers here. It's initially at rest, we got that there. On top of a rough hill of height h1. It says the sphere rolls down the rough hill then rides on a smooth horizontal surface. This is rough and what that means, it explains it here, the first hill has enough friction to cause the sphere to roll without slipping. Roll without slipping means no kinetic friction, but there is going to be some static friction which causes the object to accelerate to have 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 w you have here, let's call this point B. Whatever w you have at point B will be the same w that you have at point C because there's no change in angular velocity in that piece. For the third interval here, we have a long smooth hill. Smooth hill means that there is no friction therefore there is no angular acceleration, again just like the horizontal surface. There is alpha here but there is no alpha here and here. What that means is that imagine a block, the box speeds up, it moves at a constant speed and then slows down on the way up. The sphere will do that, so it gets faster, same speed and then slows down and it stops but the rotation is going to be different. It's going to accelerate on the way down because it has an alpha then it spins at a constant rate and then as it goes up the second incline, because there is no static friction, itÕs a smooth hill, then there is going to be no alpha. What it means is that it's going to go up but it's going to keep rolling at the same rate, get to the top, stop. V = 0, but it's still rolling and then it starts going down. It's kind of weird. It goes up but it doesn't really stop rolling. What 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 going to keep rolling, go up, keep coming down and that's because there's no friction. That's what's special about this question. What we're going to do is we're going to write an energy equation from A to B and then we're going to write an energy equation from C to D. There's nothing there's nothing special happening from B to C. In fact B and C are really the same thing. Nothing happens there. Let's write the first one which is the energy equation from A to B. KA + UA + WNC between A and B is KB + UB. There is no kinetic energy at the top of the first hill because it's initially at rest. There's potential energy because it has a height. There's no work non-conservative. Remember that even though work non conservative is U and friction, even though there is friction, static friction does a work of zero in these problems. There's no work at all. There is kinetic at the end because we have some velocity, but B is on the floor. It's the lowest point so potential energy is zero. What we have here is simply potential energy going into kinetic energy. Potential energy is mghA or in this case I guess we're calling this h1. Let me do that, mgh1 equal kinetic energy. What types of kinetic energy does this ball have? It's not only falling but it rolls and speeds up so it has both types of energy. IÕm going to write _ mvb^2 + _ IwB^2. One object has two motions so we have to kinetic energies. What we're going to do as usual we're going to expand I and we're going to rewrite w. Why do we rewrite w? Because I have a V and an w. Whenever you have the two velocities, you want to get rid of one so that you only have one. The w will become a V, so instead of having V and w, I have V and V, that's better. 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. Vcm is Rw therefore w is V/R. I can use this to replace w. I'm going to replace w with V/R. Let's rewrite this, plug in the I and then keep going. mgh1 = _ mvB^2 + _É Then I have I. I is moment of inertia of the solid sphere which is 2/5. 2/5MR^2. Look what happens. A bunch of stuff is going to cancel. This 2 cancels with this 2. The R cancels with the R. There's Ms here. There's one M in every term and they all refer to the same mass so we can cancel those as well. I have a 2 down here and I have a 5 down here. To get rid of those guys, I have to multiply this by 2 and 5, in other words multiply the whole thing by 10. Let me make a little space here so it's not messy. When I multiply the whole thing by 10, I get 10gh1. 10*1/2 is 5vB^2. 10*1/5 is 2vB^2 and that's it. This adds up to 7vB^2 and then I have vB is 10gh1 / 7. Take the square root of both sides. That's what we get for that. We got that. That's vB. I found the velocity here. Now what I'm going to do is I'm going to write the conservation of energy equation from B to C to show you how this looks like. You can think of this problem as really two questions that are merged together and I want to show you separately what these parts look like. IÕm sorry it's actually from C to D. C to D is going up, so KC + UC + WNC = KD + UD. Is there a kinetic energy at Point C? The answer is yes. C is the bottom of the hill. What kind of energy? Not only do you have velocity, but you also we're rolling and you got to the bottom rolling and you kept rolling. You have both. You have _ mv^2 + _ Iw^2. I want to make the point that this velocity here at Point C is really the same as the velocity of point B. wC is the same as wB and vC 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. I'm going to put a vB here and I'm going to put an wB in here. What about potential energy? 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 at Point D? This is the trickiest part of this whole question is to realize that even though you are at the highest point, you do have rotational energy. Your linear stops because the object does this and stops then it starts coming back down but it keeps rolling as it goes up. You still have this. This is the most important part of this whole question. This entire question basically exists for this one purpose. There is potential energy at that point so IÕm gonna write mg, the height at Point D recalling that h2. By the way h is our target variable. We're going to simplify a bunch of stuff and we're going to be left with h2. What I've got to do here? WeÕve got to expand this guy and we got to rewrite w. We have V and w so we're going to rewrite w into V by writing w = V/R which we can do because it's a rolling motion problem. _ mvB^2 + _ this is the I, the moment of inertia so itÕs 2/5 MR^2. W we talked about how we can rewrite this into VB/ R, so wB is (VB/R)^2. That's the left side. On the right side, I can expand the rotational energy here which is going to be _ I. I is 2/5MR^2*w^2. I'm just doing this in one shot. K rotational is _ Iw^2. I'm replacing the I right there and I'm replacing w with (VB/R)^2 plus mgh2. I didn't really have to do this one last step here. IÕll 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 because this rotational kinetic energy at point B is the same as the rotational kinetic energy at Point D. Let me go up to the drawing real quick. You're rolling at B, youÕre rolling and this rolling never slows down. Whatever energy you have here, you have it here at C, and then you go up and you have it here at D. What you could have done is you could have canceled these guys here. I'm going to not cut it out there because it's going to be messy. I'm going to cut it over here. But again, you could have canceled those two because those energies are the same. They don't change. Really, all you have is this and this. I can cancel the masses and I'm solving for h2. h2 = VB^2 / 2g. Notice that the g went down there. I do have an expression for VB, so I can plug it in. VB is right here. It's going to be _ g and then VB^2. Notice that VB is inside of a square root, so 10gh1 / 7 which means that the square root cancels with the 2, and then I'm left with 1/2g(10gh1/7). This will cancel nicely. This cancels with this and then I have 10/14 h1/ 10/14 is 0.71h1. We're done here. The final answer is that h2 is 0.71h1. One question that weÕre asked here as part of the bonus question is why are these different? Conservation of energy usually would dictate that you start at a height and you come back to the same height. But the reason they're different is because of the energies. All of the potential energy went into kinetic at the bottom, so there's a full energy conversion. But on the way up, only some of the kinetic energy went back into potential because some of it is still in spinning. Here you are very high with no spinning, here you are a little bit lower because you're spinning. Spinning costs energy. Basically what happened is you went from an initial potential, but no initial rotational. Two, potential at the end and rotational at the end. What happens is some of this went into here and some of this went into here. Because your initial height got split into height and rotation, you don't get as much height. If we did this in terms of energies, let's say 100 Joules got split where 70 Joules went here and then 30 Joules went here. That's why you see a lower height because you spend some of that heights energy into rotation energy so you didn't recover as much of your height. That's it for this one. Hopefully it made sense. Let me know if you have any questions.

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 . Calculate 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.

10m

Conservation of Energy in Rolling Motion Additional Practice Problems

A solid cylinder with moment of inertia I = 1/2 MR2, a hollow cylinder with moment of inertia I = MR2 and a solid sphere with moment of inertia I = 2/5 MR2 all have a uniform density, the same mass and the same radius. They are placed at the top of an inclined plane and allowed to roll down the inclined plane without slipping. Rank them in order of total kinetic energy at the bottom of the incline, from higher to lower.

[a] sphere(1st), hollow cylinder(2nd), solid cylinder(3rd)

[b] sphere(1st), solid cylinder(2nd), hollow cylinder(3rd)

[c] hollow cylinder(1st), sphere(2nd), solid cylinder(3rd)

[d] hollow cylinder(1st), solid cylinder(2nd), sphere(3rd)

[e] they all have the same kinetic energy

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A solid disk is released from rest and rolls without slipping down an inclined plane that makes an angle of 25.0° with the horizontal. What is the speed of the disk after it has rolled 3.00 m, measured along the inclined plane?

A) 4.07 m/s

B) 6.29 m/s

C) 3.53 m/s

D) 5.71 m/s

E) 2.04 m/s

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A thin-walled hollow cylinder (I = MR2), with mass M = 3.00 kg and radius R = 0.200 m, is rolling without slipping at the bottom of a hill. At the bottom of the hill the center of mass of the cylinder has translational velocity 16.0 m/s. The cylinder then rolls without slipping to the top of a hill. The top of the hill is a vertical height of 6.00 m above the bottom of the hill. What is the translational velocity of the center of mass of the cylinder when the cylinder reaches the top of the hill?

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A thin-walled hollow sphere with mass 5.0 kg and radius 0.20 m is rolling without slipping at the base of an incline that slopes upward at 37° above the horizontal. At the base of the incline the translational speed of the center of mass of the sphere is v = 12.0 m/s. If the sphere rolls without slipping as it travels up the incline, what is the maximum vertical height that it reaches before it starts to roll back down?

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