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Anderson Video - Race Down Incline

Professor Anderson
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>> Let's do a race down an incline. And we'll do a race between a wheel rolling and a box sliding without friction. Okay; wheel rolling down the hill, no slip, versus a box sliding down the hill with no friction. And let's say that we start up here at height H. And let's pose the question, "Which one wins the race?" And let me ask you guys, what do you think, which one is going to win the race? Who has an idea? Jordan, what do you think, which one's going to win the race? >> (student speaking) The box. >> Why? >> (student speaking) Because you said so. >> Okay; [laughter] so you were actually listening to something? >> (student speaking) The picture. >> Okay. So it might seem like the box should win the race, but the key is that it's frictionless, right? And we know if things are frictionless, they move really easily. The fact that the wheel doesn't slip means that there's some static friction between the bottom of the wheel and the surface. And so you might say, "Well that's got to do something. It can't speed it up, it's got to slow it down." And you'd be right. But let's see if we can analyze this mathematically. And let's take a look at the idea of conservation of energy. When the wheel starts up there -- and we're going to say that this thing has mass M, radius R, and the box has mass M, what can we say about the initial energy? When it's up at the top, it's all potential energy, MGH. When it's at the bottom, it has center of mass motion, and it has rotational motion. And so have to add those two up, 1/2M VCM squared plus 1/2 I omega squared. But there's a relationship between VCM and omega if it's pure rolling. So we can put these terms together. We get 1/2M VCM squared, plus 1/2 I over R squared, which comes from this right here. And so we can lump it together, M plus I over R squared, all of that times VCM squared, and now you get the velocity of the box -- or sorry, the velocity of the wheel. How do we do it? Well, we have MGH. We're going to divide that whole thing by 1/2 M plus I over R squared. And then I'm going to take the square root of the whole thing. Okay; we just solved this last equation for VCM. And now we can simplify this a little bit more by the following. Half on the bottom means two on the top. So let's put a 2GH. I have an M in the top, and an M in the bottom. Let's divide everything by M. And if I do that, then this one becomes a 1 and this becomes I over MR squared. So this is the velocity of the wheel. What about for the box? The box at the bottom, the velocity of the box, that's not too bad, EI equals EF, MGH equals 1/2MV box squared. And so we can solve this very quickly for the velocity of the box, and we get the following. Square root of 2GH, divided by M. All right; I multiply it across by 2. I-- Oops. If we divide out the Ms, we don't have the M on the bottom, we just get square root of 2GH. Okay; that's the velocity of the box. So let's look at these two things. Is the velocity of the box bigger than the velocity of the center of mass? Jordan, what do you think? Is this answer bigger than the answer that we got for the wheel? >> (student speaking) Yes. >> Why? >> (student speaking) Well, because they're both square root 2GH, but you're dividing the velocity of the wheel by 1 plus I over MR squared. >> That's right. So if this is a positive number down here, which it is, I is always a positive number, M is always a positive number, R squared is, of course, always a positive number, then we get the box has a bigger speed than the wheel, and therefore it gets there first. It gets there first, and it's going faster when it gets to the bottom. All right? And now you can take this equation and you can use this for any object that's rolling down the hill; because we haven't said anything about I. So if it is a solid sphere, it has a particular I. If it's a hollow sphere, it has a different I. If it's a cylinder, it has yet a different I. And now you can tell which one is going to win the race, and which one is going to lose the race, just based on their moments of inertia. All right; everybody okay with that one?