 So we're doing homework from lesson seven, orbital mechanics. We're doing example 3A. And now, circular force. Again, Aaron, don't forget the mass. Because the formula, the equation for acceleration, is so complicated, kids forget you've got to put an M in front of it for force. Which one do I want to use? Darn right. Oh, as it turns out, the little mass cancels. And one of the R's cancels. And in fact, I get orbital velocity is equal to the square root of 2. Not 2, Mr. Dewick. You're doing the wrong question. Good gosh. The square root of big G, big M over R, where M is not the mass of the Earth this time. It's going to be, we're talking about the mass of the moon, which I think is on your formula sheet. I have no idea what it is. Did you get this far? OK, because it says from the moon's center, that is a radius. R is going to be 2.5 times 10 to the sixth. By the way, once you have velocity, can you do b orbital kinetic energy? Yeah, kinetic energy is a half M v squared. In fact, what we noticed last day, it was a half M big G, big M over R. Because v squared was big G, big M over R. That's kind of a nice coincidence. I said, Dan, you don't need to memorize that, but it's kind of nearly cool. Orbital potential energy, that's going to be negative big G, big M over R. And the total orbit energy is going to be the potential plus the kinetic. And I think it's going to end up being half of the kinetic but positive if we do it right. We're doing the moon for starters. But it says it ends up orbiting 2,500 kilometers from the moon's center. What is the radius of the moon? Radius of the moon can't be times 10 to the eighth, because the Earth is 10 to the sixth. And it's not bigger than 1.74 times 10 to the sixth. Yeah, so it is above the moon. 2.5 times 10 to the sixth means it's orbiting above the moon somewhere. Be prepared. I mean, most of what we've done is with the Earth. So think about the questions that you've answered. An easy way for me to give you the same question, but different numbers, is to make up an imaginary planet and give you a different mass and a different planetary radius. So Earth, we've used quite a bit. Moon would be a good one. I'll move some of the stuff to the moon, same question. Or I think on your final exam, I think I'd put a planet do it with a radius of pi and a mass of E or something like that. If you're really alert, you might notice that. But what a great planet. Something like that. I think it's like 3.14 times 10 to the eighth or something like that. Well, it'd be a nerdly cool planet in my mind, but that's OK. Any others? So today is kind of wrapping everything up. I have a take-home quiz for you guys first, though. So if you take a look at the take-home quiz, the take-home quiz has no gravitation stuff on it. It's actually lessons 1, 2, and 3, where we looked at the kinematics of circular motion, horizontal, and vertical circles. I'll have a take-home quiz for you also on next class. And that's going to be on all the gravitation stuff. And what we're going to be doing today, lesson 8, is looking at some neat examples that combine energy and circular motion and other concepts, problems involving circular motion and other concepts. Example 1, roller coaster loops are a great example. Now, again, this is in our magic physics world, where everything works perfectly. In real life, roller coaster loops are not circular. If they made them circular, we'd black out. The G forces are tremendous. They're what's called a clothoid loop, or a clothoid loop, or a clothoid loop. I don't know how you pronounce it. I've never heard it pronounced. But it's a teardrop shape. If you look at a roller coaster loop from the side, it's not a true circle. It's a teardrop shape, which means the radius is constantly decreasing, which is why you don't black out when you go over them. But we're going to be in our magic physics world, or when we're going to have nice circular loops that we can analyze. And it says this. Pardon me? G forces. The blood wants to continue going in a straight line, but your body's being forced in a different direction. Same reasons why pilots would black out. So example one. At the top of a 12, it says centimeter. Oh, yes, we're using a little toy car. Radius loop. The 350 gram toy car experiences a force from the track of two newtons. Assuming that the car is released from rest and that the track is frictionless, find how high it was released from. And the range it travels after leaving the ramp. So I don't know about you. I don't know if they're still made. When I was a little kid, I played with the little matchbox cars, and we bought the plastic race car track set that you fit together with the little purple or whatever. You stuck them together with these little things. They're orange. You got the little loop kit with it as well. Some of you're nodding your head. Some of you have seen this stuff. Talking about that. What do they want us to find? It's release height. Is there a change in height between here and here? I think the best way to solve this is going to be energy. I'm going to start out solving this. And also because this isn't a nice straight track. It's a curvy track with loops and stuff. I'd like to use a scalar rather than a kinematics forces projectile approach. I'm going to start out by going kinetic energy initial and potential energy initial equals kinetic energy final plus potential energy final. Now we need to talk about what I mean as initial and final. I'm going to talk about this right here as location A and right here as location B. And I'm going to say location A is initial, location B is final. I'm not talking about the top of the ramp right now. And the reason is they gave me information at the top of the loop. So I'm going to use that. Are any of these zero? Does say starts from rest. Am I on the ground right here? Nope. Am I on the ground right here? Nope. Am I stopped right there? No. Yes. So none of these are zero. Now I'm going to oh, they want me to find the release height h, which I think is going to be initial height. So I'm going to just expand this. mgh initial equals a half mv initial squared plus mgh, sorry, not initial, final squared and final. mgh initial equals a half mv final squared plus mgh final. E and I do notice they didn't need to tell me how much the car weighed, which is kind of nice. And I guess to get the h by itself, I'll eventually just end up dividing by g. I'll come back to that. Let's walk down this. Do I know the final height? Well, if the height, if the loop has a radius of 0.12 meters, 12 centimeters, how far from here to here? 12. How far from here to the top? 12. How far from here to the bottom? What's the total height? 0.24, 24 centimeters. So I know that. Do I know g? Yep. Do I know v at the top? Well, I don't know it. So I'm going to write a little note to myself here. Need v final. Ah, but here's the key. What path is this car traveling in? Circle. I'm going to solve this. I'll just flip her there for a second. OK. Stay alive, bulb. Just last me till the end of the year, please. I've got you budgeted for next year, but not for this year. It's cheaper to buy a new projector, $600. I'm going to solve this using circular motion. So I'm going to make a little note here at top of loop. What are the forces acting on this car at the top of the loop? Get the obvious ones. Get the obvious ones. Gravity. And there is a normal force. Am I on the inside, or am I on the outside of the loop? Inside, so which way is the normal force? Also downwards. I wrote mv for some reason, mg. Let's try that again, Mr. Dewick. Don't try and do five steps ahead, Mr. Dewick. Just do this right. What path is this traveling in? A circle. Where must my net force be toward the? They're both toward the center. They're both winners. My equation's going to look like this. mg plus fn equals, what am I right here that I wanted to? Yeah, fc, absolutely. Except which fc am I going to use, the one with the period or the one with the v in it? I'm going to get this. mg plus fn equals mv squared over r. If I want to get the v by itself, I think it's going to be times everything by r. You know what, this is yucky and complicated. I'm just going to give up and start sticking in some numbers, I think. The mass, OK, now this time the masses don't cancel. Says the masses, 350 grams is 0.35. That's a heavy car. 0.35, that's about a half pound. That's a heavy car. 9.8 plus, what's the normal force? 2. By the way, what if I wanted to figure out, or what if I told you that it just barely made it over the loop? What would the normal force be if it just barely made it through? 0, so we can later on, we'll calculate minimum height required for safety. 2 equals 0.35 over 0.24 v squared. 0.35 times 9.8 plus 2. I get, oh, that's for an r, not a height. You're right. Thank you, Irwin. Sloppy me. It says r, which is 0.12. I agree. The left side, though, does work to 5.43, I think. Does it not? I hope. So we get this. 5.43 equals 0.35 v squared all over 0.12. How would I get the v by itself? What would I do with the 0.35? Divide, what would I do with 0.12? Times, and how to get rid of squared. I'm running out of room on a test, I'd show that. But since we're running out of room, I'm going to go straight to my calculator here. So you guys said I would times by 0.12, divide by 0.35, and then square root that. And I get a velocity of 1.364. And since that's not my final answer, I'll carry an extra sig figure 2. So I'll go v equals 1.364. And that's v final. And now I'll plug that back into there by 0.12. Yeah, you can go that, divide it by that, and then move it over. Yeah, let's see if we can get h initial then. So h initial is going to be a half, that's 0.5, 1.364 squared. I'll use the number I still got on my calculator, plus 9.8 times. What's the final height? That we said was 0.24, two radiuses, two radii. And to get the h by itself over here, yes, Taylor, divide by 9. I divide by g, right, 9.8. OK, 0.5 times that answer squared plus 9.8 times 0.24, enter, and then divide that by 9.8. Minimum height, 0.335 meters, 33 and a half centimeters, 0.335 meters. Now part b is nerdily cool, but we're not going to do it in the interests of time, although I kind of like it. Part b says, find the range. To do that, you would have to find the speed at the top. You could do that repeating what we've done here with energies, but instead, you're trying to again find v final, but you're not worried about the loop. Your h initial would be 0.335. You could get the speed right at the very, very tippy top of this. It's your final height is 6 centimeters, and you would have an angle of 30 degrees, and we'd have a lovely projectile question. But in the interests of time, I'm going to pass. Yeah. No, otherwise, I'd do it. OK. Example 2. A satellite in orbit around Earth, is that right? So you know what we should do, though? I guess, Aaron, to be clear for people that are away. So I'm going to go, nice question, but pass part b, right? The math nerd wants to do it, but I want to get you guys going on the homework. Example 2. Satellite in orbit around the Earth is at radius big R and speed little v. Satellite is hit by a small rock, and it slows down somewhat. What happens to the radius of its orbit? And it says, assume the satellite begins and ends in a circular orbit. I think what this is asking is, how are your orbital speed and your radius related? You say that again. How are your orbital speed and your radius related? And the fact that I said orbital speed and radius in the same sentence says to me that if I want to somehow prove what's going on, I'm going to start out by going, I don't know, we're going to convince ourselves. Gravity never changes. It's always big G, big M, little m, all over R squared. Which FC do I want to use, the one with the period or the one with the velocity? We're trying to figure out how radius and speed relate. I'm going to use the m v squared over R. M cancels. One of my R's cancels. And I want to get the R by itself. I want to ask myself, when I get the R by itself, I'm going to say, OK, if the v gets bigger, will my R be bigger or smaller? So if I get the R by itself, let's see. It's on the bottom. It would move on the top over here. In fact, I think I get this. R equals big G, big M, over v squared. Is that right? Is the v on the top or on the bottom? Bottom. So as v gets bigger, what happens to the orbital radius? It gets smaller. In fact, we said to orbit around the Earth's surface, we did that question. You'd have to be going really fast, because your orbital radius would be as small as it could be. As v gets smaller, your orbital radius gets bigger. So if you're going to end up in a stable orbit, because you got into a collision and slowed down, you'd have to end up further away from the Earth. And you'd need a very nice coincidental meteorite that struck this object at just the right angle to send it further away from the Earth and magically give you just the right orbital. This is a theoretical question, in other words. But yeah, as it turns out, final orbit would have a radius smaller than r. It says, explain your answer. I would write this, but I didn't want to write this down here. I wrote my equations up here. And then I would simply say this. As v decreases, r decreases, except Troy on the test. I probably write the word decreases and decreases instead of down arrow and down arrow. Sorry. As v decreases, r increases. That's what I said. Let me make that arrow a little clearer so you can read it properly. Sorry. Sorry. Isn't it ii? Slump. As this number gets smaller, this number gets bigger because we're dividing by a smaller number. And in fact, if this is twice as small, four times as big, I think, because it's an inverse square thing that we've talked about as well. What that means is the further a satellite is away, the less kinetic energy it has to have. Oh, but the more potential energy you have to burn to get it there. So somewhere, there's a magic, good, best, best efficiency zone where you're getting your best bang for your buck. Little technical comment, according to the author, the final orbit is actually not quite a circle. It's an ellipse that passes through the collision point and then moves closer to the Earth on the far side of the orbit. In other words, don't write this down, but it would look much more like that where you're a little closer here and a little further there. Next page. Aaron, do you remember you asked me a question a little while ago on the test? I like this question. I like this question. And specifically, the concept that I like is that part there. Barely makes it around. Now, they want us to find the normal force, so we're going to have to do a good free body diagram here. The problem is in our free body diagram, I think what we have to start out doing is we need to figure out this initial height so that we know how fast he's traveling down there or she's traveling down there. And I'll convince you of that in just a second. So we're going to start out. It says, find the normal force at the bottom. So I'm going to say this. At bottom, what are the forces acting on this roller coaster? Get the obvious ones. Gravity down. What else? Normal force up. Why did I draw normal force so much bigger? What path is this roller coaster tracing out? A circle. Where must the net force be then toward the center? Normal force has to be winning. Until he levels out, normal force has to be winning. So we would have this. Winner minus loser equals mv squared over r equals the net force towards the center. Do I know the mass? I think they said 50 kilograms. We're going to ignore the weight of the roller coaster. We'll just look at the weight of the passenger. Do I know g? Sure, 9.8. Do I know the mass? Yep. Do I know the radius? 4 meters. What I don't know is the speed at the bottom. I don't think it would be 0 because that would mean you'd come to a stop. I don't think you'd come to a stop suddenly. You keep going. Do you not? I don't think the speed is 0 at the top because if it was 0 at the top, you'd come crashing down and die. But here's what we're going to have to do, I think. We're going to have to, because we don't know v at the bottom, we're going to have to now go to some energy here. We're going to have to go to top of loop. At the top of the loop, what are the forces acting on this person? Get the obvious ones. mg. Now there is a normal force down, except because I said just barely makes it around, what do I want the normal force to be? That's what I want to be 0, not the velocity. I want them to feel like they're just about to fall, but they're still safe. So I'm going to say at the top, mg equals mv squared over r. v squared at the top is grr. grr. Oh, mass cancels. Now how does that help me? So we looked at the bottom and we said, really, we need to find out how fast he's traveling at the bottom. Well, to find out how fast he's traveling at the bottom, if I know how high he is here, all the potential is going to at the bottom be totally kinetic. So over here where I wrote at the bottom, put a little comma if you have room off to the right, we're also going to say this. Potential energy initial is all going to be kinetic energy final. Mgh initial is going to be a half mv final squared. And yes, Evan, once again, yay, mass cancels. For me to find the velocity at the bottom, which I can then plug in here, I think, I need to find my initial height. I need to find my initial height. How can I? Sorry? Well, if I divide by g, I still don't know v, though. If I know this, I got that. I got this. So what I need to find is this. Well, let's look at the top of the loop. Let's call this here location A at the top of the hill. And let's call this here location C. I'll call the bottom location B, B for bottom, so that gives me an easy way to remember this. Seems to me we can say this. The kinetic energy at A and the potential energy at A has to equal the kinetic energy at C plus the potential energy at C. However much energy you got at the top of the hill, that's how much you got at the top of the loop. Now, I think we are starting from rest, so I can safely say that. But the rest I'm going to expand. This is going to be mghA equals 1.5 mvC squared plus mghC. Yay, mass cancels. We're going to get this. ghA, initial height, is equal to 1.5. Now, this is the velocity at the top of the circle, which I don't know, except I have an expression for the velocity at the top of the circle. What is v squared the same as? Grr. So I'm going to write 1.5 grr, OK? And if you have a bit of patience, something nerdily cool, and it is only nerdily cool, is going to happen. Plus, now look up, please, all of you. This is height C. But from here to here, how many radiuses, how many radii is that? So instead of writing hC, as I notice I have an r right here, I'm going to write g times 2r. What happens to the g's? So this is kind of unusual. Now, what this is telling you, Brandon, is if you could somehow transport, using like a Star Trek teleporter, this roller coaster to the moon, you wouldn't have to change any of the heights. It would work just fine. Or if you could beam it to Mars, or Jupiter, or wherever, anywhere in the universe, this roller coaster would work just fine no matter what planet you were on. The gravitational field, as it turns out, completely cancels, which is kind of nice. Oh, so I now have this. The height to start out with is 0.5 1⁄2r plus 2r. I think those are like terms, aren't they? What's 1⁄2r plus 2r? 2.5r. And this is what I wanted to show you guys. When you're designing any roller coaster, and it has a loop, the minimum safe height is 2.5 times the radius. If you build the roller coaster hill, 2.5 times the radius of your bigger loop, your biggest loop, they'll make it through just barely if we ignore friction. So of course, you build a huge safety margin in. But I love, Evan, the fact that starting with this mess, we get such a clean answer as I've ever given you all year long. I mean, even gravity canceled. I don't think that's ever happened before. So the math nerd within me, the physics nerd within me, says, it's kind of cool, actually. And I'm willing to bet most amusement park ride designers know this. Now, this is for a circular loop. I don't know what it is for a clothoid teardrop shaped loop. I suspect it's different, probably considerably less. But this will just barely get you around. Now, what the heck were we trying to find? We were trying to find how fast you were going at the bottom. And we said, I think we can solve this using energies. Using energies. And once we find how fast we're going, we can find the normal force over here. So are you ready? I'm going to go like this. G times 2.5 r equals 1 half v final squared. 1 half on this side is the same as what over here. 2 g 2.5 r equals v squared. Right, Brandon? Because you're carefully writing this down. I don't know where you were, but I don't think you were. By the way, Brandon, what's 2 times 2.5, folks? Careful. 2 times 2.5, not 10. 5. Can I write this then? I think v squared equals 5 gr. What does this question ask us to find the normal force at the bottom? We started out by writing the normal force equation. Troy, how would I get the normal force by itself? OK. So now we're going to take this over here. We're going to get this. Normal force equals mv squared over r plus mg. But what's v squared the same as? 5 gr. The normal force this person is going to feel is going to be m 5 gr. We took care of the squared because we plugged in an expression for v squared over r plus mg. Holy smokes, what happens to the r's? And in fact, here's what I have. 5 mg's plus 1 mg. This person's going to feel six times their weight at the bottom. That's also, I like this question, and it's a nice, clean answer, Erin, kind of like, wow. I expected yucky decimals and junk and holy smokes. As it turns out, final answer, the normal force is going to be 6 mg, which is going to be six times, was it 50? Is that what I said the mass was? Six times 50 times 9.8. They're going to feel 2,940 newtons. Now, before you all freak out, I said I like this question. Well, I don't. I do and I don't. I like this concept here. And I like this concept to get the 2.5 r. In fact, you're going to find on your take home quiz, there's a question where I say to you, a roller coaster just barely makes it through how high it is the first hill have to be. I think that's a fair game question. The rest of this asking you to extend it to the normal force and find all that, I did because it was nearly cool. I don't know if I, in fact, I know I would not feel comfortable asking that. Not on a written section and even as a multiple choice, that's a little tough in my mind. Yeah. How many genes can a human body handle? It depends on the circumstance. Most of you could handle for a short while. First of all, it depends on whether it's positive genes or negative genes. Negative genes is when the blood feels like it's being forced away from your head. That's when you black out. Positive genes is like when you hang upside down and you feel like the blood's getting forced towards your head. We don't deal with that as well in terms of physical issues, but we won't black out. It's just there you're more likely to burst a blood vessel in your brain and get an aneurysm and die. The answer to your question, there was a test pilot in the 50s who, the US military, to test their ejection seats. How do you test ejection seats on the ground? Out in Nevada, they have a rocket sled track where they would basically strap a guy to the front of this and slam this rocket sled into a cargo net. And he was hitting, I believe, around 200 or 300 Gs, but he was also, his blood vessels in his eyes would pop and he'd have bruises everywhere. Like it was, they wanted to find out what someone could survive and recover from in a few weeks. And the answer seems to be in certain situations and those were straight on G horses, not positive or negative. We're pretty robust. I suspect this here, you'd be very, very sore. I know the original roller coasters that they built with loops back in the 1910s and 1920s, they built them with circular loops, not knowing any better. And I think the Coney Island one, they had a permanent ambulance station there. And it was a regular thing. You would go through the loop to loop and you would get nosebleeds. That was the big one. I got nosebleed, I didn't bleed at all. Oh, you know, that kind of thing. And it wasn't until Six Flags in California, I think, in late 60s, early 70s, they were the ones who realized, don't do a circular loop. So they were the first ones to do a successful and safe roller coaster loop. Yeah, you either do what's called a clothoid loop or you do a corkscrew. In a corkscrew, what you're doing is you're keeping your head as close to, or your heart really, as close to the center of the radius as possible. So your head and your heart are not undergoing very much circular force because our are small. Your body is rotating around, but you don't really care if your feet aren't getting the blood that they would like. They don't seem to be complaining too much. So you've seen the Playland, the Playland one has a corkscrew, right? Does it have a, yeah, does it have a regular loop too? I don't think it does. No. How many of you've been on California screaming at Disney? Okay, so that's a clothoid loop. And if you ever go there, if you're a physics nerd like me, try entering it sometime with, instead of having your head pushed back, like you're supposed to for safety, try leaning forward just before you get there and in the loop, try and straighten up. You will not be able to. I mean, you can really try, you will not be able to. The G-forces won't let you, okay? So, oh, and raise your right hand. All of you, all of you right now, everybody raise your right hand, repeat after me. I promise not to sue Mr. Dewick. Thank you. I just had horrible images of, I'm gonna try standing up. No. Okay. Homework, number one. Homework, number one. Number four. Number five, please notice in number five. I think we're talking about the earth in number five, it doesn't say, but we're talking about the earth. And please notice they gave you an altitude, not a radius, and it's in kilometers. Six is also good, find the gravitational force. Seven is also good, find the strength of the gravitational field. And number eight, this is where you're finding the minimum speed that it has to have at the top of the loop and the height from which it must be released. So that you can try as well. A nine B, that's all lovely review, but I'm gonna pass there. And we're officially done circular motion. What else are you working on today? Cause it's a shorter lesson. Take home quiz, and you have the great big circular motion review thingy. That answer key is online, and that will be due the day of the test. And your test is not gonna be this week, it's gonna be probably towards the middle end of next week. I think I said right now, I'm looking at Wednesday. Yes, if you happen to have it remembered and you use it, I'll clue in. You've already, you've had me long enough to know I despise in my mind memorizing things that I can derive in two seconds. But that one was about a 10 second derivation. I might, I guess. It's a nice, no, 2.5 times the radius. Okay, fair enough. I mean, certainly for number eight, you can use 2.5, but I would practice deriving it on number eight just to see if you can. It's gonna be 2.5 times the radius. Oh, 10 radius, 2.5 times 10, 25 meters. Oh, there's the answer. I don't want you to just go like that, right? Okay, let me hit stop. Oh, nope, let me hit stop.