Rolling Motion (Free Wheels) - Online Tutor, Practice Problems & Exam Prep

Rolling Motion (Free Wheels)

Alright. So, rolling motion, I also like to think of this as free wheels, is what we're going to talk about. And so far, we haven't talked about that yet. What we've seen is we've seen either a point mass moving around a circular path or we've seen rigid bodies moving around themselves. So, imagine sort of a cylinder that is free to rotate around its central axis. Something like this. Right? So think of this as a fixed wheel. Right? These are fixed wheels. They are fixed in place. Now, in some problems, we're going to have these rigid bodies or these shapes that are going to not only be rotating around themselves but they're also going to be moving sideways. So, they're both rotating, so they have omega because they're spinning and they're moving. And when I say moving, I mean, they're actually moving, moving sort of sideways. They're not fixed in the same place. We're going to think of these as free wheels. Okay? That's why I call this free wheel. And the best example, I think the most memorable example I can give you is actually rolls of toilet paper. So if you have a roll of toilet paper that's fixed in place as it normally is, this is a fixed axis. Okay? So here's the roll, and it can spin. It has ω but it doesn't move sideways. So it doesn't move sideways. I'm going to do it with no, which means v center of mass = 0. Okay? vcenter of mass is when you actually move sideways. If you spin in place, you're not actually moving. Now, so in this case, ω is not 0 but v center of mass is 0. Okay? The middle, center of mass in the middle, the middle of the cylinder doesn't move, it stays in place. Okay? Now, in the case of a free axis or a free wheel would be if you had a roll of toilet paper that is rolling sort of on the floor. Right? And it's doing two things here. So it's rolling, let's say this way. And if it's not only rolling this way but it's also moving. So if you combine this with this, you get this. Right? So it's sort of moving this way. So I can say that it has ω and it has a velocity of its center of mass is moving to the right. So ω is not 0, and the velocity of center of mass is not 0 either. What's special about these situations, the most important thing you need to know about these situations is that there's a relationship between these two numbers. K? And luckily, this relationship looks like something we've seen a lot of. So the velocity let's say you're spinning with ω here. The imagine that if your wheel spinning this way, then you're going in that direction. Right? There's a relation between these two. The velocity of the center of mass for a wheel of radius R is simply Rω. Now notice how we didn't use little r, so I'm going to write big R not little r. Because in this case, what we actually want is the actual radius of the the wheel or the disc and not a distance from the center. It's the actual radius of this thing. Okay? And this looks very similar to what we've seen. If you have a fixed axis. If you have a fixed axis like this, the velocity tangential at an edge here or here. These are tangential velocities. Right? These are let me make them blue. These are tangential velocities. These tangential velocities are rω, but we're not talking about a velocity of, a point at the the edge or any distance from the center. We're talking about the velocity of the middle of this thing because this thing moves sideways. K? 2 so this is the most important thing you need to know. 2 other things you need to know, is that the velocity there's a velocity at the top here, and there's a velocity at the bottom. So the velocity of the center mass is Rω. You should know that the velocity at the top is going to be twice the velocity at the center of mass. So it's 2Rω and the velocity at the bottom is 0 relative to the floor. K? Now your book, probably at some point, your book may derive these equations. How to arrive at them? Your professor may derive them. Here just for the sake of simplicity and time, I'm just going to give you these equations without deriving. Here's a really easy way to remember this. I'm going to draw this again here. Vtop velocity at the point at the top is 2Rω. Velocity in the middle is 1Rω, and velocity at the bottom is 0Rω. Okay? So 0, 1, 2, obviously this simplifies into Rω and this simplifies into 0. Okay. Those are the 3 velocities. Notice how this is different from this situation here. Here the velocity of a point at the top of a circle, of a cylinder, of a disk that spins around itself is rω and little r is the distance. Here, if you are a little edge at the top here, you are 2Rω because you're moving. The idea is that this Rω here, right, combines with this Rω to give you 2 of them. So I'll just mention that briefly, but those are the equations you need to know. Okay? Most of the time you need to know, the green one, you don't always need to know the yellow one. The green one is the most important one, but I'll give you the yellow ones just in case. Alright. Let's do a quick example. This is very simple. You just have to remember these 3 equations. Alright. So I have a wheel stripe of radius point, 30 centimeters. I actually made I actually meant to make this 30 centimeters or point 3. Sorry about that. So I'm going to say that it has a radius of point 3 meters. Okay? And it rolls without slipping along a flat surface with 10 meters per second. So it rolls without slipping. So if it rolls, it has a ω, and it rolls with 10 meters per second. So the wheel is actually moving. When I give you a velocity here, when I say that the wheel is 10, and give you the velocity of the center of mass of the wheel. Okay? So this velocity here, v center of mass = 10. Okay? Now there's something interesting here that we need to talk about. It says rolls without slipping. Roles without slipping is the condition for these three equations to work. These three equations are only true if you are rolling without slipping. Of a multiple-choice conceptual test. That this is the condition for rolling motion. Okay, rolling motion, the condition for rolling motion is that this is without slipping. That's a conceptual point there. Alright? So let's come back to this question real quick. So I want to know, a, what is the angular speed of the wheel? So angular speed of the wheel is simply ω. K? Now notice that I know v cm and I know R, and I'm looking for ω. Well, this is very straightforward. There's an equation that connects all 3 of them, and it is v cm = Rω. Therefore, ω = v cmR. Velocity = 10 divided by 0.3 ω therefore is 33.3 point 3 radians per second. Cool. Very straightforward. Part b, the speed of a point at the bottom of the wheel relative to the floor. This is just this v bottom right here, and if you know this conceptually, if you remember the equation, v bottom is just always 0, for a rolling wheel no matter what. Okay? So that's that's simple. ω is 33, and the bottom is 0. That's it for this one. Let's do the next example.

Speeds at points on a wheel

Alright. So here we have a car that accelerates from rest for 10 seconds. The initial velocity of the car is 0, and it takes 10 seconds accelerating. Its tires will experience 8 radians per second. Let's draw a little car here. The speed of the car, denoted as v, involves a basic car, and w deals with the wheels. If the car is moving in that direction, the wheel is spinning in the opposite direction. I'm giving you the acceleration of the wheel separately because this is linear, and I'm going to make a column here for angular. We have \( \alpha = 8 \). If the car is initially at rest, the initial angular speed \( \omega_{\text{initial}} \) is also 0. The tires have a radius of 0.4 meters. Can you see that? Yes, you can. We want to know the angular speed of the tires after 10 seconds. Part A: This looks like a motion problem, and it is. I have three motion variables here that are given, and I'm asking for one, and one of them is ignored. What’s ignored here is the number of rotations, \( \Delta \theta \), which I represent with a sad face. This indicates that I should be using the first equation for angular motion: \[ \omega_{\text{final}} = \omega_{\text{initial}} + \alpha t \] Given \( \omega_{\text{initial}} = 0 \) and \( \alpha = 8 \), calculate: \[ \omega_{\text{final}} = 0 + 8 \times 10 = 80 \] Therefore, \( \omega_{\text{final}} = 80 \) radians per second. Part B: We're being asked for the speeds at the top center and bottom of the tire. Since the tire is in rolling motion, or you can think of it as a free wheel, this means we can also use the three equations we just learned. For the top of the tire, \[ v_{\text{top}} = 2 \times r \times \omega = 2 \times 0.4 \times 80 = 64 \text{ m/s} \] The center of mass in the middle has a speed of, \[ v_{\text{cm}} = r \times \omega = 0.4 \times 80 = 32 \text{ m/s} \] And the speed at the bottom is always 0 m/s. So that's it for this question. We had to find \( \omega_{\text{final}} \), which is familiar material, and we have to find \( v_{\text{top}} \), \( v_{\text{cm}} \), and \( v_{\text{bottom}} \), and we got them. Let me know if you guys have any questions.

A long, light rope is wrapped around a cylinder of radius 40 cm, which is at rest on a flat surface, free to move. You pull horizontally on the rope, so it unwinds at the top of the cylinder, causing it to begin to roll without slipping. You keep pulling until the cylinder reaches 10 RPM. Calculate the speed of the rope at the instant the cylinder reaches 10 RPM.

• A. 0.21 m/s
• B. 0.42 m/s
• C. 0.84 m/s
• D. 1.68 m/s