Physics: Understanding Rolling Motion

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Picture a bicycle cruising past you. We'd like to describe the physics of this situation. It's a little bit different than situations we've seen before. It's a complicated object with rolling parts and moving parts, but your first reaction when you see a problem with a complicated object is you say, "Let's just pretend all the mass is concentrated at the center of mass and let's just deal with it as a point object." Sometimes that's okay, but in this case, we could get into trouble. For instance, if we're dealing with kinetic energy, you remember that when you had just a stationary wheel, you can't think about kinetic energy just in terms of the kinetic energy of the center of mass. You have to worry about the rotational kinetic energy.
So what's the story with the bike? It's a more complicated problem than we've seen before, because it has both rotational motion and translational motion put together. All of the rotational motion story that we've considered so far has involved a fixed axle. I take a bike wheel and I mount it and I spin it. It doesn't have to be a bike wheel, but it's some fixed rigid object that is rotating around about a fixed axis. And everything we've talked about was that simple case. And now, we're trying to say, "What happens if the center of that object, that rotating object, is also, at the same time, moving?"
So here's my little picture of a bicycle. Let's try to analyze and understand what's going on. The bicycle is moving across with velocity v. That's the velocity of the bike. The wheels are spinning and they've got some angular velocity, , and the very first thing that we surely have to think about is how are these related. After all, if you're going fast, the wheels are going to spin faster. But the connection may or may not be obvious to you. After all, if you just have a fixed wheel and you ask, "What's the velocity?" there's no single answer. Over here the velocity is up, over here the velocity is down. It's different everywhere. And it's different at different radii. So the connection between these two just requires a little bit of thought. The easiest way that I know of to connect the two is to imagine yourself on the bicycle. So if you're on the bicycle and you look down, maybe you kind of look at the wheel, what do you see? You see just a spinning wheel that's center of mass of the wheel is motionless, with respect to you. You're in the reference frame of the hub of the wheel.
So let's draw a picture of what you see when you're sitting on the bicycle. You are looking down and you see a spinning wheel. And let's imagine looking at a little speck of dust or paint at the very outside edge of the wheel. So that's going around in a circle, and let me look at it at the very bottom. At the very bottom, that little speck is moving backwards. When I'm looking down, I see the bottom of the wheel moving backward with respect to me, and the top of the wheel is moving forward. And the velocity, the magnitude of the velocity, the speed of that point is just given by R. This is now just ordinary rotational motion with a fixed center. So this is our old formula, v equals R, and that's correct, it's fine. And now, think about the motion of this dot compared to the motion of the ground right underneath it. As a bicycle wheel is rolling, it's not skidding. Here's my bicycle where and its' cruising along. Here's my spot, this orange piece of tape, and take a look at it. At the moment that it touches the ground, it's in contact with the ground. It's not skidding and scraping, it's momentarily moving exactly the same as the ground below it. So if the bicyclist sees this spot on the wheel moving in this direction with a certain v, you must also see the ground moving in the same direction with the same v, because the bottom of the wheel and the ground are connected, they're' doing the same thing. So, as far as the bicyclist is concerned, the bottom of the wheel and the ground are moving backward with a velocity magnitude R. And so what does it look like from the perspective of the ground? If the bicyclist thinks the ground is moving this way with v equals to R, then the ground must think the bicyclist is moving the other way with v equals R. And that's the answer. The velocity of the bike, with respect to the ground, is just R.
A lot of work and a lot of thinking to get a really simple formula, and you say, "I know that formula, I've used it before." But, no, it's a different formula. This formula refers to the speed of a spot on a wheel. It depends on exactly what radius you're talking about. What is R here? It's the radius of the bike wheel. So this formula is really kind of different. It looks the same. This formula is telling you about a linear or translational velocity of the bicycle or, if you like, of the center of mass of the system as it moves along. So it's familiar-looking and easy to believe.
What's the acceleration of the bicycle? The bicycle is moving in a straight line, so the acceleration is very simple, it's just dvdt. So the acceleration of the bike is just the derivative, with respect to time, of this equation. The radius of the bike wheel is not changing with time, so it's just R times the dt, or . So another familiar-looking formula, just bear in mind that it's got a little bit of different information content than it did before. This is the linear acceleration of the bike, related to the angular acceleration of the wheel.
When you look at acceleration of spots on a rotating wheel, they have two components. This is this tangential component, R, that's what we're interested in here. When you have a rotating solid object, there is another component to the acceleration of spots, and that's the centripetal component. Centripetal acceleration has another formula, it's pointing in a different direction and it's completely irrelevant here. It's useful to think about that. The bicycle is moving in a straight line. It's true that spots on the wheel are moving in circles and are accelerating centripetally as well as tangentially, but the bicycle - this is the whole story, it's just R.
This idea that the bottom of the wheel is in contact instantaneously with the ground is an important one. It's kind of weird. Let's just think about it a little bit more. If I'm in the reference frame of the bicycle, if I'm sitting in a frame where the hub is at rest - this is the motion of the wheel, the bottom is moving backward, the top is moving forward. What I'm interested in is not the motion of the wheel with respect to the bicyclist, but the motion with respect to the ground. So how do you move from one reference frame to another? The velocity of the bike wheel, with respect to the bicyclist, plus the velocity of the bicyclist with respect to the ground, that's what we want. So we want to add these arrows to the velocity of the bike as a whole, with respect to the ground. If I draw that, it's kind of a funky picture, but the whole bike is moving, so every point is moving with velocity equal to the velocity of the bike. This is sort of a funny picture, but I'm adding this translational motion to the rotational motion in the previous picture. I'm adding velocity vectors. They add like vectors, so, for instance, the arrows at the top add up and the arrows at the bottom cancel out.
So here's the grand final picture. This is the velocity of various spots on the wheel, with respect to the ground. The top of the wheel is moving very rapidly. The hub, of course, hooks to the rest of the bike, it's just moving with v center of mass. And the ground spot, the bottom spot, is at rest instantaneously. If it wasn't at rest - think about tire tracks. As the tire rolls along, it leaves behind a nice image of the tread. And if there was not instantaneous rest at the bottom, if the bike wheel was scraping, you wouldn't see tread marks, you'd see scrape marks. So that's sort of physical evidence that this weird fact is really true.
There's a consequence of this weird fact; the bottom piece of rubber on the tire is momentarily at rest, with respect to the ground. Let's think about forces for a second. When you're pedaling your bike hard and you're trying to speed the bike up, you're accelerating. F equals ma, there must be a net force on the bicycle. So where's that net force being applied? You think, "Oh, I'm the one who's applying forces. I'm the one who's speeding up the bicycle." Well, you should be careful. Internal forces do not cause acceleration of a system. It has to be an external force, so I'm going to argue that it's really an external force that has to be accelerating the bike. What's that external force? There's a little piece of rubber that is instantaneously in contact with the ground. The only force here that's relevant is friction, and it's not kinetic friction. This thing isn't scraping, it's static friction. This point is momentarily at rest. What's the direction of the frictional force? Well, if you're trying to speed up the bicycle, the bicycle wheel is pushing backward against the ground, so, by Newton's Third Law, the ground must be pushing forward on the bicycle. It's funky. The force of static friction is pushing forward on this point and that's the net external force - you have to include both wheels, of course, which is pushing the bicycle forward. You might have thought that friction was always slowing you down, but, in this case, static friction is what's speeding you up. There might be other frictional forces, like air drag, that's slowing you down, but that's a different story.
Static friction doesn't do any work. If you have a point of contact to do work, you need force times distance and there's no relative motion here. So who's doing the work to create the extra kinetic energy as the bicycle speeds up? That's you. You're doing the work, but it's the friction force that's applying the force, static friction.
One last comment - I began this story by asking questions about kinetic energy, or at least thinking about kinetic energy as a source of worry with the bicycle, because we've got both rotating motion and translating motion. Let me focus my attention just on a single wheel, like a ball rolling down a hill or rolling across the ground. What's its kinetic energy? Well, I'm not going to prove this to you, but the kinetic energy is just exactly what you'd think it should be. It's mv^2, where m is the mass of your object and v center of mass is the translational velocity. That's the translational kinetic energy, but you have to add to that I - that's I around the center of mass, times ^2. So there's two pieces the kinetic energy; the translational piece from the bulk motion, and there's this rotational energy I^2, and that just adds in. It's nice and simple. and v are related, so that's good and that makes this formula end up being even easier to work with. And I've argued to you that the external force, the important one, is static friction doing no work, so the kinetic energy is not going to change with time, at least not due to friction. It might change in time if you're rolling down a hill, but we can use conservation of energy. That'll be a powerful technique.
So if you have a bicycle and you look at this formula, you have to use the total mass, the mass of the bike plus the person plus the wheels, it's the total mass. And you have to add in the rotational kinetic energy of both wheels and this formula is going to tell you what you need to know when you're really trying to solve problems involving just not pure rotation, not just pure translation, but rolling motion.
The Physics of Extended Objects
Rolling
Understanding Rolling Motion Page [3 of 3]