Hey, guys, let's check out this conservation of angular momentum. Example. So you stand on a stool that it's free to rotate about a perpendicular access to itself and through its center. So there's a stool, the stool. Let's get a little little still going on here. And this thing can rotate around self about an axis perpendicular to it through its center. So imagine that the stool is this and it rotates and you're on top of it. Perpendicular means makes a 90 degrees right, so this means that the stool rotates like this within access through its center. So just like a normal stool if you are on it. So you stand on the stool and it says it's suppose that your combined moment of inertia you in school is eight, um, and that it is the same whether you have your arms open or pressed about around your body. Now that's not true, obviously, because if you reduce your if you close your arms, remember, Moment of inertia is some version of M R squared. If you close your arms, your are is going to drop, which means your eye drops okay, but in this problem, just to make things simpler were saying that this change is negligible. Okay, so when you close, it's going to stay the same so initially. So that's not gonna happen. Um, initially with arms open or closed, your I will be, um, eight. So let me draw this here. You with your arms open, I equals eight. And then you with your arms closed over here. Something like this, right close against your body. Your I will be eight as well. And then it says here that you stand on the stool with arms wide open, holding 1 10 kg weight on each hand at a distance of centimeters from the central axis. Now, if you're on top of the stool, you're gonna rotate around yourself, so the central axis imagines a line through your body. Okay, so the central axis looks like this. Okay, so what we're actually going toe have is we're gonna have one draw this one more time. We're gonna have one last time. You holding weights with your arms wide open. So here, the way it's you're holding, they have massive 10. So the eye of the system is not going to be eight. Um the eye of the system is gonna be the i of you. Plus, these two weights both ways have the same mass. They are the same distance from the axis, which is points eight and points 8 m. And these guys are 10 kg each. Okay, so the moment of inertia of the system is you plus two times the moment of inertia off the little masses because they're gonna be the same moments of inertia. Okay, your moment of inertia is eight. But you'd have to calculate this piece here. Okay, so we wanna first. And that's what I wanna find out. First, we want to calculate the systems total moment of inertia about it's central axis. So I already have that. It's eight plus two. I i is the moment of inertia. Of the little objects. The objects are point masses because they're shapeless. Um, they have negligible size. The moment of inertia of a point mass, remember, Is m r squared. So this is gonna be too m r squared. The mass is 10 and the distances 100.8. And when you do all of this, you get that the eye of the system, the eye of the system is 20.8 20. kg times meter square. So this is part eight, Um, for part B. It says supposes, system spins with 60 rpm. We want to calculate the angular momentum that you have, So I calculated I and all for part B for part B. We wanna find what is l What is El of the system? If it has an rpm of 60 now, remember? L is I omega. I just found the I I just have to plug it in omega. I don't have omega. I have RPM. But I can replace right so I can write I And instead of Omega gonna write two pi r p m over 60 rpm is 60. So this whole thing just becomes one, and I end up with l equals 20.8 times two pi, and this is going to be 131. Okay, units is units are kilograms meters square over seconds. That's very straightforward. Just plug and play for part C. It says you bring the weights to your chest so that they lie on the axis of rotation. So the weights were out here, The axis of rotation like an imaginary line through the middle of your body. So you're putting the weights on the axis of rotation like this. Okay. And as calculate the systems new moment of inertia. So now what you have is not this picture. Instead, what you have is a situation where your arms are actually closed. So I'm gonna draw this one last time. So you have a situation where your arms are closed. So something like this. Okay, that doesn't look awesome. But that's we're gonna roll with. You're holding these weights like this. And remember, if you changed, um, if you change the way your arms air set up, it didn't affect your moment of inertia. But it will affect the moment of inertia of the weights because the moment of inertia of the weights depend on M r squared where r squared where r is the distance from the access. And that's changing. I used to have a distance of 0.8 from the axis and now the distance to the axis is zero. OK, zero. So the new moment of inertia I system I system new or final. I guess if this one is initial, is going to be Are you? Plus two i or eight plus two m r squared. But they are. This is a 10. The mass through our is becoming a zero. So this whole thing is gone and the Onley I you have is your I which is eight. Okay, that's it for that piece. And then the last part we're gonna we're gonna ask. Well, if we do this, how fast is it going to spend in terms of RPM? So what is my rpm final? So when my I initial waas when my I was 20.8, we had an rpm of 60. And then when my I final is eight instead of 28 what is the new our PM So I can use conservation of angular momentum to solve this ally, Um, l I equals I left. So else don't change. I can expand l into Iomega. Iomega initial final. Okay, now remember, um, here I have this in terms of omega, but I wanna I wanted in terms of rpm, remember, Omega is two pi r p m over 60. So you can just replace this here. I initial two pi r p m over 60 initial equals I Final two pi r p m final over 60. And then what you can do is you can cancel out the two pies and you can cancel out the sixties, and you're gonna be able to use this to solve for your final rpm. Okay, so let's just plugging it. RPM final is I initial, which was 20.8 times rpm initial, which is 60 divided by I final, which goes to the bottom over here, which is eight. So if you do this whole thing, you get 156 rpm, so your rpm is 156 okay? And that should make sense. You close your arms, you spin faster. Okay? Your your eye is dropping by a factor of roughly 2.5. So your rpm right, this is dropping. So your rpm is going to increase by a factor of of roughly 2.5 x. Okay, that's it for this one. Uh, this question. We had four steps, so I walked you through every single step, which you might see in a question like this is something much more straightforward where I give you this set up, and I just say, Hey, if the rpm initial is 60 what is the rpm final? Um, and then you're gonna have to figure out that you have to calculate my initial and then have to calculate I find here, I kinda held your hand through it, and I did in the multiple steps. But some questions you have to figure out, um, you have to do all of this stuff before you get to the final answer, and it may only ask you for the final answer, so you have to work your way there yourself. All right, so that's it for this one. I mean, if you have any questions and let's keep going.
You stand on a stool that is free to rotate about an axis perpendicular to itself and through its center. The stool’s moment of inertia around its central axis is 1.50 kg m2 . Suppose you can model your body as a vertical solid cylinder (height = 1.80 m, radius = 20 cm, mass = 80 kg) with two horizontal thin rods as your arms (each:length = 80 cm, mass = 3 kg) that rotate at their ends, about the same axis, as shown. Suppose that your arms’ contribution to the total moment of inertia is negligible if you have them pressed against your body, but significant if you have them wide open. If you initially spin at 5 rad/s with your arms against your body, how fast will you spin once you stretch them wide open? (Note:The system has 4 objects (stool + body + 2 arms), but initially only stool + body contribute to its moment of inertia)