Energy of Rolling Motion - Online Tutor, Practice Problems & Exam Prep

Energy of Rolling Motion (Surface vs Air)

Hey, guys. So, you may remember that if we have a disk-like object like a cylinder or a sphere and it is moving on a surface while rolling around itself, much like a toilet paper would do if you throw it on the floor, that motion is called rolling motion. It's got a special name when you have an object on the floor like that. What's special about it is that it has both a linear velocity because the center of mass moves, and a rotational velocity because it spins around itself. It's got 2 motions, so it has linear kinetic energy and rotational kinetic energy. So let's look into that real quick. Alright. So, as I just said, a wheel-like object rotating and moving around itself is called, this type of motion is called rolling motion. You have a \(v\), which is often referred to as the \(v\) of the center of mass. And you have an \(\omega\) because you're rolling around yourself. And there are 2 types of motion here. What's also important about this is that there's a relationship between \(V_{CM}\), \(R\), and \(\omega\) represented by the equation \( V_{CM} = R \omega \), where \(R\) is the radius. Okay. What that means is that those two variables, \(v\) and \(\omega\), are linked. So if one grows, the other one has to grow by the same amount. Now it's important to make a distinction that this only happens if you are on a surface. Okay? If you have an object that is rolling on air, these two variables, \(v\) and \(\omega\), are not tied to each other. They are not connected. \(V_{CM}\) is not tied to \(\omega\). Okay? Basically, it means that you cannot use this equation here, the green equation. This only happens if you're rolling on a surface and if you're rolling without slipping. Lucky for you, all problems in physics, at least for you guys, are going to be rolling without slipping. So you can just assume that to be the case. Okay? So, to summarize, if you are rolling on a surface, this equation applies. If you are not rolling on a surface, this equation does not apply. So if you throw a ball and it rolls on the floor, that would apply. But let's say you throw a baseball and it's spinning through the air and moving, you cannot say that \(V_{CM} = R \omega\). That equation doesn't work. Cool? That's it. So let's do an example here. I have a solid sphere. This type of shape tells me the moment of inertia. The moment of inertia of a solid sphere is \( \frac{2}{5} MR^2 \). I'm given that the mass is 2, the radius is 0.3, and it rolls without slipping on a horizontal surface. Rolling without slipping on a horizontal surface means that this is called rolling motion, and it means that the green equation is going to work. Okay. I'm going to make this green to match up with the other green up there. This 10 is the velocity of the center of mass. Okay? If I tell you an object moves at 10 meters per second, that's the velocity at the middle of the object. And the question here is let's calculate the linear, rotational, and total kinetic energy. Okay? Let's find the linear first. We'll plug in the others. The linear energy is \( \frac{1}{2} mv^2 \). I've got all these numbers, half the mass is 2. The velocity is 10, \(10^2\), this will be 100 joules. For kinetic rotational, it's going to be \( \frac{1}{2} I \omega^2 \). I have \(I\). \(I\) is going to be \( \frac{2}{5} MR^2 \). And I don't have \(\omega\), but I can get \(\omega\) because I have a \(v\), and these guys are connected by this equation right here. Okay? So, let's do that real quick. So \(V_{CM} = R \omega\). So, \(\omega = \frac{V_{CM}}{R}\). \(V_{CM}\) is 10. \(R\) is 0.3. This guy here will be 33.33... Okay? Alright. So I can put \(\omega\) over here. Notice that the 2 cancels with the 2. And then I'm going to have \( \frac{1}{5} \), the mass is 2, the radius is 0.3 squared, and \(\omega\), which is \( \frac{10}{0.3} \). Now, if you want, what you could also do is, instead of writing \(33.33\)... here, I'm going to write this. Right? If you've got a calculator, just put the \(33.33\)... It's faster. But I'm going to do \( \frac{10}{0.3}\) squared. And that's because if you notice, this will cancel with this. Okay? And then I'm left with \(2 \times 10^2\), which is 200, divided by 5. Make sure I'm doing this correctly. So divided by 5. Yep. So this is 40 joules. Okay. And then for the total kinetic energy, for total kinetic energy, we're going to have kinetic linear plus kinetic rotational \(100 + 40\), 140 joules. Alright. So that's linear, rotational, and the total weight. It's got 2 types of energy. So you add up linear with the rotational. Notice that they're not necessarily the same. And remember that we can use this equation here because it's rolling on a surface. Cool? That's it for this one. Hopefully, it makes sense. Let me know if you have any questions.

A 150-g baseball, 3.85 cm in radius, leaves the pitcher's hand with 30 m/s horizontal and 20 rad/s clockwise. Calculate the ball's linear, rotational, and total kinetic energy.

• A. K_L = 6.75 × 10⁴ J ;K_R = 1.78 × 10⁵ J ;K_tot = 2.45 × 10⁵ J
• B. K_L = 67.5 J ;K_R = 1.16 J ;K_tot = 68.66 J
• C. K_L = 67.5 J ;K_R = 0.46 J ;K_tot = 68.96 J
• D. K_L = 67.5 J ;K_R = 0.018 J ;K_tot = 67.52 J

Ratio of energies of cylinder on surface

Hey, guys. So some of these rotational questions will ask you to find the ratio of one type of energy over the other. So I want to show you how to do one of these. So here we have a solid cylinder. A solid cylinder tells us that we're supposed to use I=12mr2 with mass m and radius r. It rolls without slipping on a horizontal surface. This is called rolling motion. And it means I can use this equation, vcm=rΩ. Because it's rolling like this. So vcm is tied to Ω. Alright. I want to know the ratio of its rotational kinetic to its total kinetic energy. So what you do is you follow what it's saying here and you set up a ratio like this. So it's saying rotational kinetic energy. So we're going to write Kr (rotational) at the top to total kinetic energy at the bottom. So the ratio of top to bottom, Ktotal, which is Kl (linear) plus Kr (rotational). And now what we're going to do is we're going to expand these equations as much as possible.

What I mean by expanding is, well, what does Kr stand for? Kr=12IΩ2. Kl is 12mv2 and Kr is 12IΩ2. And we're going to expand this as much as possible, meaning we're not going to stop there. We can replace I with this right here. I can also replace Ω with something else. The problem here is I have vs and Ωs. There are too many variables. Whenever you have a v and an Ω, you usually want to replace one into the other. So v=rΩ. And what we're going to do is whenever we have v or Ω, we want to get the Ω to become a v. Okay? So I'm going to write Ω=vr, and we're going to replace this here. The reason we do this so that we have fewer variables, so it's easier to solve this question.

Now before I start plugging stuff in, I want to warn you, you cannot cancel this with this. Right? That's not a thing. So don't get tempted to do that. What you can do is you can cancel the halves over here because they exist in all 3 of these guys here. Okay. So you can cancel the halves, and this simplifies a little bit. So we're going to do now is expand I. I=12mr2 and remember, we're going to rewrite Ω as vr, and then this whole thing is squared. Now we're going to do the same thing at the bottom. But before I do that, you might notice right away that this r cancels. Right? So that's another benefit of doing this thing here. Another benefit of doing this thing here is that it's going to cause the rs to cancel. Okay. So at the bottom, I have simply mv2+I, which is 12mr2, and then Ω, which is vr squared. And again, just like it did at the top, the rs cancel. Okay? The rs cancel. Let's clean this up a little bit and see what we end up with. I end up with 12mv2 divided by mv2+12mv2. And you may already see where this is going. There's an m in all three of these, and there's a v in all three of these. So everything goes away, and you end up with just some numbers left here. So you have 1 up here. And then this, there's a one here, right, that stays there, 1 + 2. So the mass The velocity doesn't matter. So all you have to do is do this thing here. Okay? There are two ways you can do this. If you like fractions, you can do with fractions. I'm going to do that first. So I'm going to rewrite this as a 2 over 2. And then I have 1 over 2 divided by I got a 2 at the bottom here. And then I can add up the tops the top here. So it's 2 + 1. So I have 1 over 2 divided by 3 over 2. I can cancel this 2, and then I end up with 1 over 3. If you don't like fractions, one thing you can do with this particular case is you can rewrite this like this, or half is point 5. This is a 1. This is a point 5. Right? This is better if you have a calculator. Point 5 divided by 1.5. And if you do this in the calculator, it's point 333, which is the same thing as this. Okay? Same. So that's it. The ratio is 1 third. And by the way, that ratio will change if you have a different I because this half here ends up showing up here and here. Or actually, that half ends up showing up here and here. Right? So if you have a different shape, this will be a different fraction and then your final list will be different. So the ratios change depending on what kind of shape you have. Alright? That's it for this one. Let me know if you have any questions.

A hollow sphere of mass M and radius R rolls without slipping on a horizontal surface with angular speed W. Calculate the ratio of its linear kinetic energy to its total kinetic energy.

• A. K_L/K_tot =0.15
• B. K_L/K_tot = 0.60
• C. K_L/K_tot = 0.71
• D. K_L/K_tot = 1.67