Two astronauts, both 80 kg, are connected in space by a light cable. When they are 10 m apart, they spin about their center of mass with 6 rad/s. Calculate the new angular speed they'll have if they pull on the rope to reduce their distance to 5 m. You may treat them as point masses, and assume they continue to spin around their center of mass.

Hey, guys. Hope you tried this practice problem. Here we have 2 astronauts that are spinning around a, a common central point and we're going to reduce the distance between them and we wanna know how much faster they will spin as a result. Okay. So the astronauts have mass of 80, both of them. So I'm just gonna do this and say m equals 80 for both of these guys. They're connected by a light cable and when they are 10 meters apart, so the initial length of this cable here is 10 meters. I should write little l not to be confused now with, angular momentum big l. They so when this is happening, they're spinning around the center here with an Omega, with an Omega initial of 6. And I want to know what is the new angular speed? What is Omega final, if they that they will have, they pull the rope to reduce their distance to 5. So what is new omega? If the new distance is 5, what is the new omega? I hope you're thinking that the omega will, increase, that they will spin faster as a result of getting closer together to each other. Alright? So let's check this out. It says you may treat them as point masses and assume they'll continue to spin, around their center of mass right here. Okay? So this is conservation of angular momentum because I have I have objects that are spinning, a system that spins, and the radius was reducing which will cause a change in Omega. So l initial equals l final, I initial Omega initial equals I finalomeganfinal. I have, uh,omeganinitial. I'm looking foromeganfinal. I have enough information to calculate I initial and I final. And that's what we gotta do. So I'm gonna go off to the side here real quick and calculate I initial and I 5. So in the beginning, when they're spinning about around them around each other like this, their moment of inertia of the system, initial, is going to be the moment of inertia. Let's call this guy 1, this guy 2, I 1 initial plus I 2 initial. We're treating both of these guys as point masses. Point masses have moment of inertia of m r squared. So m 1 r 1 squared initial plus m 2 r 2, squared initial. Okay. And their masses are the same 8080. R is the distance. So if the total distance between them is 10 meters, the r is the distance from the axis of rotation to where the point is which in the beginning so our initial is 5. Okay. 5. Our initial is 5 because the axon rotation is right in the middle between the 10 meters. Alright. So that means I put a 5 here. Don't forget to square it. I put a 5 here. Square that guy. And if you do this whole thing here, if you do this whole thing here, you get 4,000. So the initial moment of inertia of the system is 4,000. K. Now at the end, what happens is because this length got reduced to 5, the distance between each of these guys in the points in the middle, right, it was 10. So each guy was 5 meters from the middle. Now it's gonna be 5. So each guy is gonna be 2.5 from the middle. So it's the same thing here. Isystem final is gonna be I 1+i2. It's the same thing as this except that instead of 5, we're going to have 2.5. Okay. So it's going to be 80 2.5 squared plus 80 2.5 squared. And if you do this, you get 1 1,000. Okay. Now we have these 2 i's, 4,000, 1,000, 4,000, 1,000. We're ready to roll. Okay. So Omega final will be I initial, which is 4,000 times Omega initial, which was 6 divided by ifinal, which is a 1000. You multiply this, you get 24 radians per second. 24 radians per second. Okay. Notice that your length went down by a factor of 2. This and then this went up 24 and then this went up by a factor of 4. And that's because the change in the length there affected your r and r is squared. So when you drop this by 2, you can think of this getting squared. So you end up with a decreasing 2, squared on one side caused an increase of 4 on the other side of the equation. So in this problem, you probably could have done this much faster by realizing this will just give you this. But it is a little bit tricky because it's the length and the r vector, is the is half of the length. So I wanted to do this the long way, the slow way, just to be more complete. Alright? So that's it for this one. Let me know if you have any questions and let's skip going.

Struggling with Physics with Algebra?

Watch next