One of the biggest problems facing astronauts in space is the lack of gravity or the lack of apparent gravity. But luckily, we have a solution for this because one of the ideas that we have is we can build a massive space station with a circle of ring around it that rotates. So there's the central part here, and there's, like, a giant arm that goes across the space station, and this whole entire thing will rotate. And, basically, what you'll end up with is kind of like the effect that happens when you turn in a car. Right? As you turn in a car, the wall sort of pushes you toward the center. That's exactly what would happen here. If you were standing on the ring, you would be pushed toward the center. We want to do this to simulate gravity. So, basically, what we want is we want to simulate centripetal acceleration, which is \( g \), which is 9.8. What we have to do in this problem is figure out how fast the ring has to spin in terms of revolutions per minute or RPM so that this happens. Right? So let's go ahead and take a look here. We're told that the diameter of the spaceship is 500 meters, meaning, all the way across here, the diameter is 500. But we actually never use diameter in these types of problems. So instead, what I'm going to do is I'm going to use the radius. The radius of this is going to be exactly half of that, which is 250 meters. That's the number that I'm going to use. Alright? So how do we figure out the frequency or the RPMs? How do we actually get to this value here, which is RPMs? Remember that RPMs is sort of just like a modified version of frequency? It appears in this equation over here. So \( f \) is equal to \( \frac{1}{T} \). We don't know what the period is, but it's related to the RPM divided by 60. And this is really what we're going to sort of, this is what we're aiming for. Right? So in other words, I can rewrite this equation and see that RPMs is actually equal to the frequency times 60. Right? That's just directly from this relationship. So in other words, if I can get the frequency in this problem, I can figure RPMs because all I'm going to do is I'm just going to take whatever number that is and then multiply it by 60, and that's going to be my final answer. Alright? So then the question is, how do I actually get this frequency? That's really what I'm trying to solve for in this problem. So I'm going to go over here and do that. How do we actually solve for frequency? Remember, there's a lot of equations here for circular motion. A lot of them, unfortunately, include frequency. So let's try to see if we can sort of process an elimination of this. Right? We know we can't just directly use one over period because we don't have the orbital or the period of that rotation. Right? So we can't use \( f = \frac{1}{T} \). And we also know that frequency and period are related to the tangential velocity using these two equations over here. The problem is I don't have anything about the tangential velocity. I don't know how fast the points along the ring are moving around the circle, around that rotation. So, unfortunately, I can't use any of my tangential equations, my \( v_{tangential} \). However, remember that the tangential velocity when you divide it by \( r \) and square it, or square divided by \( r \), is equal to the centripetal acceleration. And we actually know something about this centripetal acceleration. We know that we want that to be 9.8, so we have a number for that. So instead of using the \( v^2/r \) relationship, remember that because we can use centripetal acceleration and actually sort of shortcut it to include period and frequency, we're actually going to use this relationship over here. You can actually go directly from acceleration to frequency. So we know that acentripetal is related to \( 4\pi^2r \times \text{frequency}^2 \). Alright? So notice how all of these numbers here I actually know. I have the centripetal acceleration. I want that to be 9.8, and I have what the radius is, which is 250, and the rest are just numbers. Right? So I can actually rearrange for this, and I could just divide by \( 4\pi^2r \). That cancels. This is going to be \( 4\pi^2r \). And then all I have to do is notice that this frequency is squared here, so I have to take the square roots. Alright? So I'm going to take the square roots of both sides. That's going to be \( \sqrt{\frac{a_c}{4\pi^2r}} \). Lots of variables going on here, so just make sure that you're plugging this in correctly. That's going to be frequency. And just go ahead and attach some numbers to this. This is going to be the square root of 9.8, and then I'm going to divide by \( 4\pi^2 \), and then the radius is at 500 or is that 250? Be very careful here. You're going to use the 250, not the 500. And so when you plug this into your calculator, do 9.8 divided by all of this stuff in parentheses and then square root that number. What you should get is you should get a frequency of 0.032 if you round it to the 2nd decimal place. Alright? So is this the answer? Remember, this is not the answer. This is frequency. This is that number that I'm going to plug into this formula over here and multiply by 60 so that I can get my RPMs. Alright? This is how many cycles it does per one second. So I'm going to have to take this number and multiply it by 60 to get how many revolutions we get per minute. And if you actually go ahead and do this, what you're going to see is that this is equal to this is equal to 1.92. So 1.92 is our final answer. So in other words, this space station has to rotate, you could also just round this down to like 1.9 if you wanted to. So, in other words, this space station has to rotate almost twice per minute in order to simulate gravity at 9.8 meters per second squared. Alright. So that's it for this one, folks. Let me know if you have any questions, and I'll see you in the next one.