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Energy of Rolling Motion (Surface vs Air)

Patrick Ford
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Hey, guys. So you may remember that if we have a disc like object, like a cylinder or sphere, and it is moving on a surface while rolling around itself much like a toilet paper, if you throw it on the floor, would do, UM, that motion is called rolling Motions. Got a special name when you have an object on the floor like that, and what's special about it is that it has a both a linear velocity because it's moving the center of mass moves in a rotational velocity because it spins around itself. It's got two motions, so it has linear, kinetic energy and rotational kinetic energy. So let's look into that real quick, all right, so, as I just said, Well, like object rotating and moving around itself is called thes type of motion. Called really motion. You have a V, which is often referred to as the V of the center of Mass, and you have a an Omega because you're rolling around yourself and there are two types of motion here. What's also important about this is that there's a relationship between V. C, M and W, which is V. C M R W and the R is big Our radius. Okay. What that means is that those two variables v n w are linked. So if one grows, the other one has to grow by the same amounts. Now it's important to make a distinction that if you have an object that this Onley happens if you are on a surface. Okay, If you have an object that is rolling on air, these two variables v n w are not tied to each other. They're not tied. V. C m is not tied to W. Okay, so basically means that you cannot use this equation right here. The green equation. This Onley happens if you're rolling on the surface. And if you're rolling without slipping, lucky for you, all problems in physics, at least for you guys, is gonna be rolling without slipping. So you can just assume that to be the case. Okay, So to summarize, if you are rolling on the surface, this equation applies. If you are not rolling on the surface, this equation does not apply. So if you throw a ball and he rolls on the floor, that would apply. But let's say if you throw a baseball and it's spinning through the air and moving. You cannot say that V. C M equals R omega. That equation doesn't work. That's it. So let's do an example here. I have a solid sphere. This is the type of shape I have. Eso that's gonna tell me the moment of inertia and I of a solid sphere. I have it here. I of a solid sphere is solid. Sphere is to over five m r Square. And I'm told that the given that the masses to the radius is 0.3 and it rolls without slipping on a horizontal surface roll without slipping a horizontal surface means that this is called rolling motion. It means that the green equation is going to work. Okay, I'm gonna make the screen to match up with the other green up there with 10. This 10 is my the velocity off the center of mass. Okay, if I tell an object moves a 10 years per second, that's the velocity at the middle of the objects. The question here is let's calculate the linear rotational and total kinetic energy. Let's find the linear First we'll plug in the other one's linear energy is happened B squared. I got all these numbers. Half the masses of to the velocity is 10 squared. This will be 100 Jules. Okay. For kinetic rotational. It's gonna be half I Omega Square. I have I I is gonna be 2/5 m r squared and I don't have omega, but I can get Omega because I have a V. And these guys are related connected by this equation right here. Okay, so let's do that real quick, so v c m equals R omega. So? So omega is V c m. Divided by our V C. M is 10 are is 100.3 eso This guy here will be WB 33. Cool. All right. So I can put w over here. Notice that the two cancels with the two and then I'm gonna have 1/5. The masses to the radius is 0.3 squared and W, which is 10 over 100.3. Now, if you want what you could also do is instead of writing 33 here, I'm gonna actually gonna write this right. If you got a calculator, just put the 33. It's faster, but I'm gonna do 10 over 100.3 squared. And that's because if you notice, this cancels with this, okay? And then I'm left with two times 10 square, which is 200 divided by five. Make sure I'm doing this correctly, So we're invited by five. Yep. So this is 40 40 jewels, okay? And then for the total kinetic energy for total kinetic energy, we're gonna have kinetic linear plus kinetic rotational 100 plus 40 1 jewels. Alright, So that's linear rotational. And the total later, he's got two types of energy. So you had a linear with the rotational notice that they're not necessarily the same. Andi, remember that we can use this equation here because it's rolling on a surface. Cool. That's it for this one. Hopefully make sense. Let me know if you have any questions.