It is proposed that future space stations create an artificial gravity by rotating. Suppose a space station is constructed as a 1000-m-diameter cylinder that rotates about its axis. The inside surface is the deck of the space station. What rotation period will provide 'normal' gravity?

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Welcome back everybody. We are trying to replicate the force of gravity on a cylindrical space station here, we're told that the diameter of our space station is 1200 m, meaning that the radius Since it's just the diameter divided by two is equal to 600 m. And we are tasked with finding what the period of rotation should be. In order to replicate the force of gravity. Let's draw in some things over here to get a better understanding of this. Say you are an individual who is in the space station. So normally, right, say you were just standing on earth, you would have a force of gravity going downward and a normal force going upward. This is the same type of thing except what is causing this balance right here is a rotation of the space station and therefore a centripetal acceleration acting towards the center of the circle. So let's first make observations about our little guy here. We have that the sum of all forces in this up and down direction is equal to mass times acceleration. On the little guy we have that the normal force minus the force due to gravity is just equal to zero because he's just standing there, he's not going up, he's not going down. This means that the normal force is equal to the force due to gravity. But now let's take into the fact that we're trying to replicate this force of gravity using rotation. So let's have the sum of all centripetal forces is equal to mass times that centripetal acceleration. In this case, that centripetal force that is acting on our little guy here is going to be equal to our normal force. We have that are normal force is equal to our mass times are centripetal acceleration. And we already established that our normal force is equal to the force due to gravity on earth times mass times centripetal acceleration. And if we divide by mass on both sides, we can clearly see that we have that our force due to gravity is equal to our centripetal acceleration. This is great. But what we need to figure out remember is we need to figure out the period required to essentially guarantee this balance right here also related to radius. So here are my thoughts on this. Let me scroll down just a little bit here. When I see centripetal acceleration, I know that this is equal to a tangential velocity divided by our radius of rotation. We also know that are tangential velocity is equal to the radius of rotation times are angular velocity. And we know that our angular velocity is equal to two pi divided by R period. I know that's a lot of formulas I threw out you, but let's just substitute in one at a time and you'll see where I'm going with this. So we have that our centripetal acceleration is equal to our force due to gravity and we just established here or sorry. We also know that our centripetal acceleration is equal to when we substitute this value into here. We get R squared. All right, that looks like an end. We get R squared times omega squared divided by ar these RS will cancel out leaving us with our omega squared. Now let's go ahead and then plug in this right here and you'll see that we get our force or sorry, our acceleration due to gravity is equal to our radius times two pi over T squared. So here's what I'm going to do. We need to isolate this T term in order to be able to solve for it. So let's manipulate it one at a time here. I'm gonna divide both sides by R and you'll see that this R and this are canceled out. So I'm gonna go ahead and remove them from our equation here. I will then take the square root of both sides because that gets rid of this power and it gets rid of these parentheses. Now we have this value on the left and this value on the right here. What I'm gonna do is I'm going to now multiply both sides by, let's see here. I want to multiply it by E over the square root of G over. Are we going to do the same thing on this side here? It'll be a little squeezed, but it's T over the square root of G over our, these terms are going to cancel out. And this term on the right hand side is going to cancel out. And what this ultimately boils down to then is that we get that our orbital period is equal to two pi divided by the square root of our acceleration due to gravity on earth divided by our radius. We have finally found a formula that works relating our radius to our period. So let's go ahead and plug in our values here. We have on top two pi divided by the square root of 9.81 divided by our radius of 600. And when you plug all of this into our calculator, we get a period of 49.1 seconds to replicate the acceleration due to gravity in the space station corresponding to our final answer. Choice of a. Thank you all so much for watching. Hope this video helped. We will see you all in the next one.

It is proposed that future space stations create an artificial gravity by rotating. Suppose a space station is constructed as a 1000-m-diameter cylinder that rotates about its axis. The inside surface is the deck of the space station. What rotation period will provide 'normal' gravity?

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Key Concepts

Centripetal Acceleration

Gravitational Force

Rotational Dynamics