https://youtu.be/3b7ncc1O1cA
PHYS 1101: Lecture Twenty-One, Part Four
The last topic for this lecture is the idea of rolling motion. It really ties together these distance quantities that are happening, which are the distance and the motion along the circle when something is rotating and this rotational motion. It really connects it together. This is an example of a wheel rolling across the ground. That’s what we mean by rolling motion. Let me show you an animation to see a very important connection between rolling motion and the motion of this wheel as a whole that you observe. Here’s my wheel in this animation, and it’s going to roll to the right. It’s not slipping as it rolls.
So notice that it starts out with this line pointing right here at the edge of this pause button. I want to let this roll through one revolution and as it rolls forward in one revolution we need to think about the connection between the distance that it travels and this circular motion that has happened. Let’s watch it and I will show you what I mean. I’m going to let it go around one time. Oops. There! Perfect. I stopped it perfectly. Let me grab a snapshot of that and let’s do a little analysis of it.
Okay, there’s the picture. And now I’ve added a line here to remind us of where it started. So, at the start of the problem when that wheel was here, that black line, a scratch on the wheel, if you want to think of it, was pointed straight down. As it rolled to the right, if it didn’t slip, I wanted to point out that as it rolled this surface of the wheel came into contract with the ground as this rolled forward and we let it roll forward one complete cycle. By the time it ended that black line was back to pointing straight down.
If you think carefully about it, in the process of rolling one through one revolution, if this surface of the wheel always stayed in contact with the ground as it rolled forward, then the distance around this wheel has to be the distance along this line. This has to be 2πr, the circumference of the wheel.
If you don’t believe it, imagine you have a sticky line of tape all the way around this wheel. As you roll it out, the tape sticks to the ground. When you roll this one revolution and it moves to the right, you will stick to the ground one layer of this tape that was wrapped around this wheel, it really is true.
Well, let’s think about the rotational characteristics of this wheel. I know, for example, that the tangential velocity of this wheel is equal to the circumference divided by the period. This we learned back in chapter five. I also just argued to you that in the time it takes for this wheel to complete one cycle, it has moved forward this distance of 2πr, and in how long? How long did it take to go around once? Well, it’s the same value of the period. So I am going to write here, to the side, the distance to the right analysis shows that the center of this wheel, if I just followed the general motion of this wheel, that it moved forward the distance of 2πr.
Also, in that same time, the period, the time it takes for the wheel to go through one cycle, we’ll notice that these things are the same. The tangential speed at the edge of the wheel has to be equal to the speed of the center of the wheel, just as you watch that wheel move to the right and translate along. Very important connection that at first isn’t obvious. So let me highlight then this has to equal that because the right sides are equal.
That’s said, or summarized, here in this red box. Let me emphasis, then, that this Vcenter, what I mean by that is, for example that this is the speed of the bike, the speed of the car, it’s the speed of the wheel, the center of the wheel as it moves to the right, that distance over time is equivalent to the tangential speed of the edge of the wheel, that distance over time.
Okay, we can use that important connection in a lot of problems that have to do with rolling motion to connect some of these variables. Let’s do a final example for this lecture. I’ve got a motorcycle that accelerates uniformly from rest and reaches a linear speed of 22 meters per second in a time of 9 seconds. That to me is giving me the information that vcenter is equal to 22 meters per second. I right away can draw the conclusion that that’s vT; that’s the tangential velocity at the edge of my wheel. That’s what the rolling motion analysis immediately tells me.
The radius of the tire is .28. What’s the magnitude of the angular acceleration of each wheel? Here’s my wheel. It’s accelerating.
I’m going to sketch my acceleration in this direction. Just because that will be my nice, positive convention. So I start out with an initial tangential velocity of some value. Let me do here my time, position, and initial tangential velocity. At some later point I have a time, position, and tangential velocity and I have an aT tangential acceleration in between.
What are the knowns that I have? Summarized above, I know that vT is equal to 22 meters per second. I know that my final time is nine seconds. I know the radius of the wheel is .28. What do I want? I need the angular acceleration. Notice that’s not the tangential acceleration, but the angular. So I want Alpha. But I have all this information that’s about the kinematics along the circular path.
From this combination, I can determine what the tangential acceleration is, and then do I know the relationship that connects aT to Alpha? Yeah, I do. I know that aT is equal to r times Alpha. So if I go after aT first, which I am inclined to do because of this kinematic information I have about this kinematics along the circle. Then I can use this equation to get the thing I want: Alpha.
I forgot one more kinematic variable that I need. It’s always helpful; I should be more disciplined about going through my list here to be sure I’m not leaving anything out. The motorcycle does accelerate uniformly from rest up to this speed. If it starts at rest then I know my initial tangential velocity is zero. All of these are zero at this initial instant. And then my final time is 9 seconds. I don’t know about the final path along the wheel. The final tangential speed I do know. It’s 22.
Okay, so let’s take that approach. Can we get aT from the things that we know? And you know what? That equation one is looking really good again. vT is equal to vT0 plus aT times the time. Again, I know a lot of these variables so I can go after aT. This one is zero. aT is going to be just vT divided by time. The final velocity dived by time. That’s 22 meters per second divided by 9 seconds, is 2.44 meters per second squared.
aT now, this relationship can be used to get Alpha. Little algebra rearranging here. Alpha is equal to aT over r. So Alpha is 2.44 meters per seconds2 divided by 0.28 meters. That’s 8.73 radians per second2.
Comment on the units here. The meters cancel. So my equation looks like I’ve got no units in the numerator, divided by seconds2. And that’s what it should like for Alpha in is units. I know then to put back my placeholder of radians per second2. That I have radians. If I had been left with some other unit here another meter or kilogram, something would have been wrong. No units per seconds2, that’s what I am after. That’s a good sign. So there would be the answer to that problem.
So perhaps the important connection to make in this problem was to recognize that it involves rolling motion and connecting the linear speed that’s the motion along the line for this motorcycle as it moves, say, horizontally to the right. Its speedometer reads 22 meters per second, and to recognize that that means that the wheels are going around with a tangential speed that’s also 22 meters per second. That was a critical connection to make in order for us to realize we had a lot of information about this rotational motion of the wheel to figure out this angular acceleration of the wheel or the tire.
Okay, so that’s how the rolling motion information can help you out. That brings us to the end of lecture 21.