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Ch 13: Rotational Inertia & EnergyWorksheetSee all chapters

# Conservation of Energy with Rotation

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Sections
Conservation of Energy in Rolling Motion
Conservation of Energy with Rotation
Moment of Inertia via Integration
Energy of Rolling Motion
Moment of Inertia & Mass Distribution
Intro to Moment of Inertia
More Conservation of Energy Problems
Types of Motion & Energy
Parallel Axis Theorem
Intro to Rotational Kinetic Energy
Moment of Inertia of Systems
Torque with Kinematic Equations
Rotational Dynamics with Two Motions
Rotational Dynamics of Rolling Motion

Concept #1: Conservation of Energy with Rotation

Transcript

Example #1: Work to accelerate cylinder

Transcript

Hey guys! Let's check out this example here. We have a solid cylinder and we want to know how much work is needed to accelerate that cylinder. Solid cylinder means that I is going to be 1/2 MR^2 because that's the equation for the moment of inertia of a cylinder. Mass is 10 and radius is 2, I'm going to put these here. If you want to, you could already calculate the moment of inertia. I is 1/2 10R^2 and the moment of inertia is 20. We can already get that, 20 kilograms meters squared. It says it is mounted and free to rotate on a perpendicular axis through its center. Again, you have a cylinder which is the same thing as a disc and it has an axis. It's mounted on an axis that is perpendicular to it, so it looks like this and it's free to rotate about that axis. It just doesn't wobble like that, it rotates like this. Most of the time, you actually have this where the axis is horizontal, so it's on a wall. ItÕs something that's on a wall and you have the disc spinning like this. It says that cylinder is initially at rest so the cylinder spins around itself but it's initially at rest. _ initial = 0 and we want to know what is the work done to accelerate it from rest to 120 rpm. Remember, most of the time when you have rpm you're supposed to change that into w so that you can use it in the equation. Let's do that real quick just to get that out of the way. _ final is 2¹f or 2¹rpm/60. If you plug 120 here, you end up with 120/60 is 2. You end up with 4¹ radians per second. I'm going from 0 to 4¹ and I want know how much work does that take. Work is energy so hopefully you thought of using the conservation of energy equation. K initial + U initial + work non-conservative = K final plus U final. In the beginning, there's no kinetic energy because it's not spinning, it's not moving sideways. The potential energies cancel because the height of the cylinder doesn't change. It stays in place. Work non-conservative is the work done by you plus the work done by friction. There is no work done by friction, just the work done by you which is exactly what we're looking for. The kinetic energy which is only kinetic rotational, there's no linear it's not moving sideways, the center of mass of the disc stays in place so V = 0. The only type of kinetic energy we have is rotational, which is _ I_^2. We're looking for this, so all we got to do is plug in this number. Work is going to be _, I, we already calculated I over here. It was 20 and _ is 4¹^2. If you multiply all of this, you get that it is 1580. You get 1580 joules of energy and that's how much energy is needed to get the solid cylinder from rest all the way to a speed of 4¹ or 120 rpm. Very straightforward. Plug it into the energy equation because we're asked for work. Hope it makes sense. Let me know if you guys have any questions and let's keep going.

Practice: How much work is needed to stop a hollow sphere of mass 2 kg and radius 3 m that spins at 40 rad/s around an axis through its center?

Example #2: Which shape reaches bottom first?

Transcript

Hey guys! Let's check out this conservation of energy example. Here we have three objects of equal mass and equal radius but they have different shapes. Remember your shape is what determines what equation your moment of inertia has. It's usually something like a fraction MR^2, but the number in here depends in the shapes. If you have different shapes, youÕre gonna have different I equations. They're all released from rest at the same time from the top of an inclined plane. I'm going to have here a solid cylinder. Here is a hollow cylinder and here is a solid sphere. They're all from rest. They all have the same mass. They have the same mass, the same radius, they all start from rest and they all start from the top of the inclined plane. They're going to start from the same height as well. Everything is the same except the shapes and I want to know who reaches the bottom first if theyÕre released at the same time. This question will depend on your moments of inertia. What I want to remind you is that moment of inertia is a measurement of angular resistance, of rotational resistance. You can think that the greater my I, the heavier I am, the more resist rotation therefore I will get to the bottom last because I'm slower. More I, you can think of this as being heavier. It doesn't mean that I have more mass that's why I have heavier. I have more resistance therefore I am slower. A solid cylinder has a moment of inertia of _ MR^2. A hollow cylinder has a moment of inertia of MR^2 so you can think that there's a one in the front. A solid sphere has a moment of inertia of 2/5 MR^2. In this question, all we're doing is comparing these numbers because the M and R are the same. This is a little bit easier if you use decimals. This is 0.5, this is 1.0, and 2/5 is 0.4. You can see from here that this one is the lightest one because the coefficient number in front of the MR is the lowest. It's the lightest one therefore it is the fastest one, therefore it gets to the bottom first. The sequence is that the solid sphere is first. I'm going to write it like this. It gets to the bottom first. The second one is going to be the solid cylinder. The third one is going to be the hollow cylinder. There is a pattern here. There's a reason why the hollow cylinder is slower than the solid cylinder. Solid cylinder has very good mass distribution. The mass is very evenly distributed and remember, the more evenly distributed the mass, the lighter you are, the less I. Better mass distribution means lower I, which means you are lighter. A whole cylinder has all of its mass concentrated on the edge. It has worst mass distribution which means it has a higher I, which means it is heavier. It has a worse mass distribution, therefore it is heavier. A solid sphere is even more well-distributed than a solid disc. A solid disc has all the mass on a thin layer like this. A sphere has basically the most perfect mass distribution you can have that's why it has the most symmetrical one that's why it has the lowest of them all. The sphere is always fastest. That's it for this one. Let's keep going.

Example #3: Cylinders racing down: rolling vs. sliding

Transcript

Hey guys! Let's check out this conceptual conservation of energy question. Here we have two cylinders of equal mass and radius, so basically the same cylinder. TheyÕre released from rest from the top of two hills having the same height. It's the two very similar or so far identical situations. Let's say I have cylinder A on hill A and cylinder B on hill B. Everything is the same except that A rolls without slipping. That means that it's not only going to have a V, but it's also rolling so it has a _. Remember this is called rolling motion if you roll without slipping and it means that Vcm = R_, but B 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 going to see that. What this means is that it's moving this way with a V but it has no _. Basically it's falling, it's going down as a box would. You can think of this as a box going down because it doesn't roll. Before we answer the question, I want to talk about how that's even possible. Basically, the difference is that here you have just enough static friction to cause this thing to roll. Remember, you need static friction to 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. The question is who gets to the bottom with the fastest speed? Is VAf or VBf going to be the greater one? 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. Most people get this question, they 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 roll are faster than 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 are fast because we have wheels and cars. But in reality, the one that gets to the bottom with the greater speed is this one. Why is that? That's because of conservation of energy. Conservation of energy. K initial + U initial + Work non-conservative = K final + U final. In both of these situations, the kinetic energy in the beginning is 0. I have some potential energy. The work done by non-conservative forces is 0 as well. You don't do anything. You're just watching. In the first problem, there is in the first case, there is static friction but the work done by static friction is 0. It actually even though there is static friction, it doesn't do any work. There is some kinetic at the end and there is no potential at the end. All we have is that my potential energy in the beginning goes into kinetic final at the end. If they start from the same height and they have the same mass and everything, they have the same potential energy for both of them. Let's say that number is 100 Joules. In both problems, you start with 100 Joules and then you roll to the bottom. The difference is that in the first situation, for A the 100 joules is going to get split between kinetic linear and kinetic rotational because it has two types of energies. Let's make up some numbers here. Let's say that 80 goes here and 20. I'm sorry, I meant kinetic linear and kinetic rotational. Let's say 80 goes to linear, 20 goes to rotational. Totally making this up. For B, the entire amount, the entire 100 Joules is going to go into kinetic linear because there is no kinetic rotational, because it doesn't roll around itself. Notice how this guy ends up with a greater kinetic energy than A and that's irrespective of the division. No matter how these two numbers get balanced out, the 100 will always be bigger even if this was 10 and 90, 100 is still greater. One way to think about this, so the answer is that it's basically because this has one type of kinetic energy, 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 using up your potential energy and transforming into kinetic. 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 energy to get it to roll. Rolling costs you some energy therefore some of the energy that would have gone here is actually going here to cause you to roll. In fact the faster you roll, the more rotational energy you have so there's less energy that's left for your linear, for your V. Your W is stealing energy from your V, so you end up with a lower V. That's why. That's it. It's a big really popular question conceptually. Let me know if you have any questions. Let's keep going.

Practice: Two solid cylinders of same mass and radius roll on a horizontal surface just before going up an inclined plane. Cylinder A rolls without slipping, but cylinder B moves along a slippery path, so it moves without rotating at all times. At the bottom of the incline, both have the same speed at their center of mass. Which will go higher on the inclined plane? (Why?)