13. Rotational Inertia & Energy

Conservation of Energy with Rotation

# Cylinders racing down:rolling vs. sliding

Patrick Ford

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Hey, guys, let's check out this conceptual conservation of energy question. So here we have two cylinders of equal mass and radius. So basically the same cylinder, um, the release from rest from the top of two hills having the same height. So it's the two very similar or so far identical situations. So let's say, have cylinder A on hill A and cylinder. Be put this over here on Hill B, and everything is the same, except that a rolls without slipping okay, rolls without slipping. So that means that it's not only gonna have a V, but it's also rolling. So it has a W. Remember, this is called rolling motion if you roll without slipping, and it means that the C M equals are omega. But be slides without rolling. Those are the two options you can have. It's either going to roll without slipping, or it's going to slip without rolling. You can have both. You could, but it's more complicated. You're not gonna see that. Alright. What this means is that it's moving this way with a V, but it has no Omega. Okay, basically, it's falling. It's going down as if as as a boxwood. So you can think of this as a box going down because it doesn't roll. Now, before we answer the question, I want to talk about how that's even possible. Well, basically, the difference is that here you have just enough static friction to cause this thing to roll. Remember, you need static friction. You need static friction toe have an angular acceleration when you have a rolling problem. Here, you have a situation where the hill is different for whatever reason, and there is no static friction, which is why it doesn't roll. Okay, now the question is, who gets to the bottom of the fastest speed, who has his VA f or V B f going to be the greater one. And I want you to take a guess or just sort of try to figure this out, think about which one do you think would get to the bottom first. Now most people get this question to guess this question wrong. Most people say that this one will get to the bottom first, and I suspect it has to do with the fact that you're going to think that objects that role are faster right then objects that slide. And that's probably because you associate sliding with some friction slowing you down or you associate rolling with wheels and wheels air fast because we have wheels and cars. But in reality, the one that gets to the bottom of the greater speed is this one greater speed. Okay, now, why is that? That's because of conservation of energy. So conservation of energy equation K initial plus you initial plus work non conservative equals K Final Plus you. Finally, in both of these situations, the kinetic energy, the beginning is zero. I have some potential energy. Okay, The work done by non conservative forces is zero as well. You don't do anything you're just watching now, in the first problem there is in the first case, there is static friction. But the work done by static friction is zero, right? So it actually, even though there is static friction, it doesn't do any work. There is some Connecticut, the end, and there's no potential at the end. So all we have is that my potential energy in the beginning goes into kinetic final at the end. So if they start from the same heights, they have the same, and they have the same mass and everything. They have the same potential energy for both of them. So let's say that number is 100 jewels in both problems. You start with the 100 jewels and then you rolled to the bottom. The difference is that in the first situation for a D 100 jewels is going to get split between kinetic linear and kinetic final. Okay, between kinetic linear kinetic final because it has two types of energies. Let's make up some numbers here. Let's say that 80 goes here in 20. I'm sorry, man. Kinetic, linear and kinetic rotational. Let's say 80 goes to linger. 20 goes to rotational, totally making this up for be the entire amounts. The entire 100 jewels is going to go into kinetic linear because there is no kinetic rotational because it doesn't roll around itself. So notice how this guy ends up with a greater kinetic energy than a and that's irrespective of the division, right? So no matter how this these two numbers get balanced outs, um, the 100 will always be bigger. Even if this was 10 and 90 100 still greater. Okay, so one way to think about this. So the answer is that it's basically because this has one type of energy, one type of kinetic, and this one has two types of kinetic. One way to think about this is to think about energy as money. It's expensive to get some sort of energy going on. You're you're using up your your potential energy and transforming into kinetic. Now, if you're falling and rolling, it takes a little bit of energy to get it to move. And it takes a little bit of manager to get it to roll. So rolling costs you some energy. Therefore, some of the energy that would have gone here is actually going here to cause you to roll. Okay. In fact, the faster you roll, the more energy, the more kinetic energy you have, And then the less linear energy, the more rotational energy to have s so there's less energy that's left for your linear for your V. Okay, so you're w is stealing energy from your V. So you end up with a lower V. That's why. Cool. That's it. It's a big, really popular question. Conceptually, let me know if you have any questions. Let's get going

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