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Ch 13: Rotational Inertia & EnergyWorksheetSee all chapters

# Conservation of Energy with Rotation

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Sections
Parallel Axis Theorem
Intro to Rotational Kinetic Energy
Moment of Inertia of Systems
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
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

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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?

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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

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