The Spinning Figure Skater. The outstretched hands and arms of a figure skater preparing for a spin can be considered a slender rod pivoting about an axis through its center (Fig. E10.43). When the skater's hands and arms are brought in and wrapped around his body to execute the spin, the hands and arms can be considered a thinwalled, hollow cylinder. His hands and arms have a combined mass of 8.0 kg. When outstretched, they span 1.8 m; when wrapped, they form a cylinder of radius 25 cm. The moment of inertia about the rotation axis of the remainder of his body is constant and equal to 0.40 kg•m2 . If his original angular speed is 0.40 rev/s, what is his final angular speed?

Question

A figure skater in a green outfit, arms outstretched on the left, and arms wrapped on the right, illustrating angular momentum.

Accepted Answer

Hey everyone welcome back in this problem. We have a researcher Okay, that's looking into rotational inertia. And so they build this central trunk. Okay? With rods that extend like arms and hands. Okay, When the rods are outstretched, we're going to treat them like thin rods and the rotating about a central axis when the rods are lowered, treat them like Collins hollow cylinders. Okay, Now, we're told some information about the mass and the size of the arms and the trunk. Okay. And what we wanna know is we want to know what is the speed when the rods are lowered? Okay, What is that total object speed when the rods are lowered? So let's go ahead and just draw a diagram of what is going on here. So we have this central trunk and it has some central axis. Okay? And it has two rods extending out, stretched from the central axis, pretend that those rods are the same. All right. And we know that it's rotating or can rotate. Alright. In this case we're told that the arms have a total mass of 10. kg and that each of them is 0.75 m long each. Or if they have a total mass of 10.2 kg, that means each of these is going to be 5.1 kg. Okay, 5.1 kg. And the length. So from the central axis all the way up to the outside of the rod 0.75 m. Alright. And we're told to treat these ones like thin rods. Okay, So when their outstretched these are thin rods. Alright, so we have an idea of what the object looks like. Now we're also told that the initial speed is 1. radiance per second. Okay, so the omega not the initial angular speed is 1. radiance for a second. Okay, great. So we know what's going on when the arms are outstretched. Let's draw a diagram when the arms go down. Ok? When the arms are lowered, so we still have our central trunk. It's the exact same. Still has that center axis. And now the arms are going to be outstretched in the middle like this and you can imagine them kind of on top of each other. We're told when they're like this, the radius of this cylinder is going to be cm or 0.3 meters. Again, it's rotating. We don't know anything about the speed of rotation. Okay, that's actually what we want to find. What is omega F. The speed when the rods are lowered. Alright. And we're told to treat this one like a hollow cylinder. Okay, now how can we find that speed that we're looking for? Well, let's recall conservation of angular momentum. Okay, we have no net external torques acting on this system. So we know we have conservation of angular momentum. Okay, so let's go down a little bit conservation of angular momentum tells us that we have L not equal to L. F. Okay. Where L is the angular momentum? So the initial angular momentum is equal to the final angular momentum. Now, when we're thinking about our angular momentum, we have two components. Okay, we have the angular momentum of the trunk itself and then we have the angular momentum of the arms. Okay, So we're gonna have L. Of the trunk. T not Okay, we're gonna have the angular momentum L. Of the arms initially. Okay. And that's gonna equal to the angular momentum of the trunk after it the arms have been lowered plus the angular momentum of the arms after they've been lowered. Alright now recall what is angular momentum? Well, it's I omega moment of inertia times angular velocity. Okay, so we're gonna have I of the trunk initially and the trunk stays the same. Okay. Before and after the arms lowered, the trunk is the same. So we're gonna take off the not notation there because it's going to be the same. So I have the trunk times omega naught plus omega not of the trump. Plus I have the arms initially times omega of the arms initially. Okay. And same thing on the right hand side. So this we have the moment of inertia of the trunk. Times the omega of the trump. Final A plus moment of inertia of the arms after they've been lowered times the speed of the arms after they've been lowered or the velocity. All right, so we see in this equation that we have all of these I moment of inertia. So let's go ahead and calculate those um before we continue with this equation. Ok, so let's start with I the moment of inertia of the trunk. Okay, well, if we go back up, we're actually told in the problem that the moment of inertia of the trunk is 8.1 kg meter squared K. And that its constant. So we're right to assume that it's the same before and after the arms are lowered. So this is going to be 8.1 kilogram meter squirt. Now let's move to the moment of inertia I of the arms when their outstretched before they're lowered. Well, if we want to consider both arms, what we can do is we can just say this this is two times the moment of inertia from one single arm initially. Now we're told to treat these as thin rods. So look at your table either in your textbook or that your professor provided the moment of inertia for a thin rod is given by one third Ml squared. Okay, so we're gonna have one third M L squared in this case the mass we're talking about one arm Is 5. kg. Okay, in the length? Well, that's the length of that entire rod 0.75 m. Okay, All squared. And working this out, we get a moment of inertia of 1. kg meters squared and doing the same thing for the moment of inertia of the arms after they've been lowered. Okay we're gonna have two times the moment of inertia of just one single arm. And in this case we're treating it as a hollow cylinder, moment of inertia of a hollow cylinder. Again, looking at your table is M. R. Squared. Well, in this case because they're hollow cylinders pointing kind of Lengthwise with the access the radius were given as .3. Okay, so mass is still 5.1 kg and our radius is 0.3 m squared. This gives a moment of inertia of 0.918 kg meters squared for the arms after they've been lowered. Alright, so now we have all of these moment of inertia. I and we can use those in our equation. So let's scroll down and work on our equation here. So we have the moment of inertia and we have all these omegas. Okay well let's think the trunk and the arms are moving together. They're all attached, it's one object moving together so the speed that omega, the angular speed or angular velocity, it's going to be the same for the trunk and the arms. Okay, so before the arms are lowered. So when the arms are outstretched you have one speed and after their lower do you have a different speed. Okay so omega not of the trunk is gonna equal the omega of the arms initial. Okay so we're just gonna call this omega and same thing after the arms are lowered. The speed of the trunk after they're lowered is going to equal the speed of the arms after their lowered. Okay, we're just going to call this omega F. Okay, This allows us to simplify our equation. So now we're just going to have and we can factor we're going to have omega not times the moment of inertia of the trunk plus the moment of inertia of the arms initially. Okay, that's gonna equal omega F times the moment of inertia of the trunk, plus the moment of inertia of the arms after they're lowered. Alright, so, we just simplified a little bit. Now we can go ahead and plug in the information we know. And again, omega F. This is what we're looking for. That final angular speed of the object. All right, So, omega not. We're given is 1.2 radiant per second. The moment of inertia of the trunk. We were given in the question is 8.1 kgm squared. The moment of inertia of the arms when their outstretched we found up here, 1. kilogram meter squared. On the right hand side. We have omega F. What we're looking for the moment of inertia of the trunk, which again, it's constant. It doesn't change when the arms are lowered. So this is 8.1 kg meters squared And the moment of inertia of the arms after they've been lowered, which we found above. To be 0.918 kgm squared. Okay. Alright, so that's great. We had information for every single one of those values except for omega F. Now we can just go ahead and solve for omega F. So on the left hand side, we're gonna have 1.2 radiance for a second, adding these values up. We're going to have 10 . kilogram meter squared. On the right hand side. Omega F times 9.18 kg meters squared. Multiplying these two values together on the left hand side, 1.2 radiance per second times 10.125 kg meter squared. We get 12.15 kg meter squared times radiance per second. And on the right hand side, omega F times 9.18 kg meters squared. Ok, dividing by 9.18 kg meters squared. The unit kilogram meters squared is going to cancel And we're gonna be left with the unit of radiance per second. So we have omega F equals 1.33 radiance for a second. Okay. And that is our final answer. So the final speed of the object after the arms have been lowered is 1. radiance for a second. Okay. And that's going to correspond with answer. B. Thanks everyone for watching. See you in the next video

Verified Solution