Transcript for:
Comprehensive AP Physics 1 Exam Review

All right, this is going to be a review video of all the topics on the AP Physics 1 exam. Starting with the kinematic formulas. Here they are. You get three kinematic formulas. These three formulas relate five different kinematic variables. What are those variables? Well, one of them is time. You see here and here and here. This is the time it took. It's really like a delta t. Time it took to go from some initial position to some final position. That's what we mean by time. So effectively, we're considering t at the beginning to be zero. We start our clock right when the motion begins. But we don't necessarily do that with the position. The initial position might not necessarily be zero. Maybe this starts at like X equals 50 or something. But whatever that is, that's your initial position. X is your final position. Why don't they write final? I don't know. You just got to know that X without a zero on it means final position. And you might be like, final position? Does that mean like the very end of the process? No, not necessarily. Uh if your if your guy got thrown up over here and he's free falling up and he comes back down, you might be like, "Oh, Xfinal, that's got to be down here because that's when the problem ends." No, doesn't have to be. You can choose that as Xfinal, but you could also choose this as Xfal. It's like, why would you do that? Well, maybe you want to know the time it took to go the whole way. Well, if I want to know this entire time, then I'd probably want to choose the end of my analysis to be down here. That'd be Xfinal. It'd probably be Yfal. You know, vertical is Y. You'd probably call it Yfal. Maybe I didn't want to find that. Maybe I wanted to find, hey, what's the time it took to get to the top? In which case, I might be like, well, then I want to choose the end of my analysis to be up here. If I choose this location as yal or xfal, then the time that I would solve for in my formula that I would plug in, either one would be the time it took just to get to the top. So, depending on what you're looking for, choose xfal to be wherever is convenient for you to solve for the thing you want to solve for. It does not have to be like, you know, when the universe ends, where is this object? It doesn't have to be the end of the problem. It could be wherever you want. And if you choose whatever point you choose is xfal, there's going to be a final velocity. This v right here is the final velocity. Again, it doesn't mean like you know when the universe ends, what's the final velocity? It means at the point of analysis that you're looking at, what's the velocity at that xfal? Um, and you've got a v initial over here. This is again similarly the initial velocity it had at the beginning of your analysis. And then you've got an acceleration. The acceleration right here is the most important variable. I'll tell you why in a minute. This acceleration uh you know points right if it's speeding an object up moving to the right but it could point left as well. It could point left on an object moving to the right if it's slowing it down. So you might be like oh it's negative if it's slowing it down. Well yeah it can be negative if it's slowing it down but it doesn't have to be. Consider this guy free falling. The acceleration due to freef fall is acceleration due to gravity 9.8. But on the AP exam you're going to use -10. That's very nice of them. This makes the numbers come out nice. You're going to use -10 for your acceleration due to gravity. is constant on the way up, at the top, and on the way down. So, acceleration is -10 all the way to the top. You might be like, wait, at the top, isn't the acceleration zero? No, the final velocity is zero at the top. But the acceleration is not zero at the top. It's -10 the entire way on the way up and on the way down. And so, look at what happened. On the way up, I was slowing down and I had negative acceleration. And that makes sense. Negative acceleration, slowing down. But on the way down, I was speeding up and I had negative acceleration cuz it was negative the whole way. How could I be speeding up and have negative acceleration? Well, it's cuz my velocity was negative and my acceleration was negative on the way down. And if you ever had that case, if the acceleration and velocity ever point in the same direction, you're going to be speeding up even if they're both negative. So, you got to be careful. They'll try to trick you. They'll be like, "What's the acceleration at the top? You want to be zero." But it's really vital is zero up there. And this is this is good. It's like secret. There's like secret code words you got to watch out for. If the question asked you, hey, what's the maximum height that the object reached? You're like, max height. Where should I choose my my xfal or my yfal? You'd pick it up here, right? Because you know v is zero at the top. So, choose that as your xfal. What would your time correspond to? Your time will correspond to the time that it took to get up to here. That' be the max height. Uh time it took to get to the max height. What if you wanted to find total time it took all the way back down? Well, you don't have to do it again. Like, gravity is symmetric. The time it took to get up to some height is equal to the time it took to get back down to that same height. If you get to the same height, this is symmetric. You just multiply by two. Multiply that time it took to get to the top. That's how much total time it took to go all the way up and all the way back down. Similarly, let's say you shot this thing up. It's symmetric in this way, too. You shot this thing up at 30 m/s. That was your initial velocity. Well, if you come back down to the same position, you're going to be coming back down symmetrically with -30. It's just going to have the opposite sign, but again, oop, not squared. It's going to be negative of whatever your initial velocity was for freef fall, which is kind of handy. Good things to keep in mind. Why did I say acceleration is the most important variable here? Well, it lets you know that these things could be negative. Velocity could be negative. Displacement could be negative. How could this displacement, this is delta x here, xfal minus x initial. In fact, some people like moving this x initial over here that makes it delta x. Kind of cleans it up. This is the displacement that could be negative, too. Can go to the left, could go down. If you end up to the left or down of where you started, it's going to be negative displacement. So, acceleration highlights that your values can be negative and how they're related to each other. But it's more important because these formulas are only true only true if the acceleration is constant. And that's true for freef fall. That's why it's nice. Okay, for freefall, these formulas always work because the acceleration is always 9.8 or -10. Where else might they be constant? Anytime you have a constant force. So if your force is constant, right, F equals ma. If your force is constant, your acceleration is probably going to be constant. That is to say if your net force is constant. So but if your acceleration is not constant, maybe you got a mass on a spring, it's got changing acceleration. These don't work at all. You can only use these for cases where there is constant acceleration. What if there's no acceleration? What if your acceleration is zero? Uh well then V initial equals Z final. And it's like, oh yeah, duh. That's what it means for acceleration to be zero. You just maintain your velocity. Uh down here, what if your acceleration was zero? Well, yeah. Then v final squar equals v initial squar because you didn't change your velocity. That what that's what it means to have no acceleration. What about the middle one here? What if acceleration is zero? Well, that means delta x, the displacement is rate time time. And you're like, oh, wait, that sounds familiar. That that's kind of like distance equals rate time time. But here it's it's really displacement is rate times time. So if you want to use like distance is rate times time, you could do that if the acceleration is zero. You don't even really have to use the kinematic formulas. But it's good to remember this one. This one's not given. Distance is rate times time isn't given on the formula sheet, but it's good to memorize because there's a secret fourth kinematic formula that they don't give you. Why don't they give you it? I don't know why they don't. Maybe they figure you could figure it out from this. It's this. Let's say you were like, you know what? I love this formula. Displacement equals rate time time. I'm going to use it even when there is acceleration that's not zero. But what velocity do I plug in? Like I can't plug in my initial velocity because it didn't have that the whole time. I can't plug in my final velocity. It didn't have that the whole time. I'm accelerating. You might be clever. You might be like, "Oh, dude. You know what? I'm going to use the average of the velocities. I'm going to take V initial plus V final and divide by two to give me the average velocity and I'm going to multiply that by T. Is that going to work?" Yeah. Turns out that does work. You could just take like distance or displacement here equals average rate times time. And if acceleration's constant, this does work. This is the fourth secret kinematic formula. You might be like, when would I need to use this secret kinematic formula? And for that matter, when would I how do I know to use any of these? There's like a million of these kinematic formulas. How do I know which one to use? Well, look at this one up here is missing acceleration. Acceleration is nowhere to be found in it. So that means if you were given a problem where some reason they gave you the delta x, they gave you val. They gave you v initial and they wanted you to find time. Well, this is the formula to use because look, if I don't know acceleration, and I don't want to find acceleration, then I don't want acceleration to be any part of my formula whatsoever. So, it's like if acceleration was completely left out in your problem, it wasn't given to you and you weren't asked for it. Use the formula that's missing acceleration. You look at the rest of these. What's this one missing right here? This one's going to be missing delta x, the displacement. So, if you aren't given the displacement and you don't want to know the displacement, then use that formula. This one right here is missing vfal. So, if you weren't given vfal and you weren't asked for vfinal, use that one. And this last one down here is missing time. So, if for some reason you weren't given the time and they aren't asking you to solve for the time, then you'd use this formula down there. And that's a little recap of all the kinematic formulas. All right. Well, what if you have a 2D kinematics problem, i.e. the object's moving through two dimensions at the same time, like x and y. Well, that's why the AP puts these little x subscripts on here. The way you'll see these formulas on the formula sheet is with these little x's. What that's meaning is that you could use these for any direction you want, but only one direction. So the motion in each direction is independent. That means you could do VFAL in X, V initial in X, A in the X direction, but you can't do like VFAL in X, V initial in Y, A in the Z direction. It's all going to be for one direction. So in the vertical direction, you could put V final in the Y, V initial on the Y, A and the Y. But you can't be mixing them up. You got to pick one direction at a time. But the motion in each direction is independent. What do I mean by that? What I mean is that the acceleration vertically. Now, I wish they would have wrote y here. The acceleration vertically, if you're going to do these for the vertical direction, the acceleration vertical direction is always -10. So easy. These are all just going to be -10 for freely flying objects. But if you ever plug in -10, you've committed. You've committed to the y direction. That means this has to be the final velocity in the y direction, and this has to be the initial velocity in the y direction. And these are all going to be for the y direction. What are you going to do for the x direction? Well, the acceleration in the x direction for a projectile is zero. The gravity is not pulling right or left. So, whatever velocity the object has in the x direction, like let's say it was 40. Well, then it's going to stay 40, stay 40, stay 40 the entire trip. The velocity in the x direction never changes. So, you don't even need to be using these kinematic formulas. What do you do if the acceleration is zero? you're just going to be like, well, distance in the x direction equals rate in the x direction times time. So, it's kind of unfortunate they chose to wrote this write this as x. It's really y. Like, the only reason you'd ever be using these in the 2D kinematics is for the y direction. So, let's try a problem. Show you what I mean. Let's take this one down here. Let's say you started 45 m off the ground. You fire this thing horizontally with 40 m/s. Okay? So, it stays 40 the whole way. Let's say we want to know how far did it make it before it strikes the ground. Well, that's distance in x. So over here, I want distance and x. Well, velocity and x, I know what it is, and it stays 40 m/s the whole time. But I don't know what the time is. So I'm have to use the vertical information to get time. So I use one of these formulas here to get the time. Let's see. We have delta y. That's what this is right here. This is deltay, but technically with a negative sign. You got to be careful. Delta y is going to be -45 because it moved downward. So delta y we have. We always know acceleration is -10. We also know V initial. You might be like, "Oh yeah, V initial is 40." Not in the Y direction. It isn't. V initial. Oh, I'll color code it here. Pink here. V initial in the Y direction is zero. People are like, zero, but it's moving. Yeah, I know, but it's not moving vertically. This has no vertical motion in the y direction. So, we got delta y, v initial y, a. We want to find the time. What got left out? What got left out was vf final. So I come up here and I go, all right, I want the formula with no V final. That has VFAL. Okay, I don't want to use that one. This one has VFAL. So I'm just going to use this formula right here. This one has no VFAL in it. So I'm going to say delta deltay right here is going to be -45. That's deltay equals V initial is 0 * T. So I don't have to include that 12 -10 is the acceleration* T^2. And if I solve that for t, I get that the time that this took was 3 seconds to get all the way down to the bottom. So now I can take that 3 seconds. I can plug it into here and 3 seconds and I get that, oh well then this object must have moved 120 m to the right. That's how I can solve that one. But notice I never plugged any x variables into my y equation and I never plugged any y variables into my x equation. You just got to keep them separate when you're solving these. What if you started with a situation where you did have vertical velocity? Well, then you got to break it into components. You got to say, well, okay, the initial vertical velocity right here, you got to do trigonometry. This would be, you know, hypotenuse 40 time s of 30° and that's going to actually add up to 20 m/s in the vertical direction. That's a V initial in the y direction. What would v initial in the x direction be? It'd be this piece right here. You break this up. You say V in the X, I'm not going to put initial because it just stays that way the whole time. V and X is going to be 40 * cossine of 30. And if you plug that in, 40 cosine 30 is like 34.6 m/s. So this x component of velocity stays the same the entire way. That means at the top when there's only x component, guess what? I'm moving 34.6 m/s for my total speed cuz that's all I got right there. So I maintain that the whole way. my vertical velocity keeps changing. And when I'm plugging into these formulas, I only plug in vertical variables into the vertical equation. Horizontal is in the horizontal equation. You could also get the speed. The speed is this total hypotenuse. Let's say I knew these two numbers. I knew the vertical was 40. I knew the horizontal was this. The total speed of an object would be the square root of v in the x^2 plus v in the y^ 2 cuz that's how the Pythagorean theorem works. So if I plugged in vx 34.6 6 34.6^ squar + 20^ squar and I add that all up. Guess what? I get 40 when I take a square root because that's how that is all related. Okay, so we just showed that vertically this initial using tri trigonometry this initial velocity v initial in the y direction which was 40 sin 30 was 20 m/s. And we showed that 40 cosine 30 using trigonometry, this component right here, this v in the x direction was 34.6 m/s. Uh what if we wanted to graph these variables? Well, graphing acceleration is actually the easiest for projectiles cuz what's a in the x direction? It's zero. The accel acceleration zero because the velocity in the x direction just stays constant the whole way. So ax is zero. What's that graph? It's just zero. We love that one. A in the Y isn't even much harder. If you wanted to graph A in the Y direction, A in the Y direction is -10 the entire way. On the way up and on the way down -10. Well, then that means acceleration in the y is -10. What's the graph of that? It's just, you know, we can call this -10. So, graphing acceleration is actually the easiest. What's the next easiest? Probably velocity. How about velocity for the x? I said this is constant. Remember, this is 34.6 the entire path, even at the top. 34.6 6 all the way down. So you know whatever wherever 34.6 is that's what Vx would be and it be constant. Vy is going to be a little more interesting. It starts off positive points upward. Then it gets to zero then it becomes negative. So vy is going to look something like this. And then where should it cross? Let's try to find out how long would it take to get all the way to the top here. Well I have 20 m/s of velocity. What this is saying is acceleration is subtracting 10 m/s every second. That's what meters/s squared means. Gravity subtracts from my vertical velocity 10 m/s every second. So after 1 second, it gets to, you know, a velocity of 10 upward. And after 2 seconds, I subtract another 10 and I get to zero. So this is going to take exactly 2 seconds to get up. You don't have to do it that way. You could use a kinematic formula to figure it out, but it's good to have a little intuition. You know, gravity subtracting 10. So subtract 10 from 20, I get 10. Subtract 10 again from 10, I get zero. Zero is when I reach maximum height. So this would take 2 seconds. So if I wanted to plot this out, I'd say vy starts somewhere here and then it like goes through here. It goes through two and it looks like this. Something like this. Because at max height, I have zero velocity. So I started with a positive velocity. I got zero velocity. And now over here I have a negative vertical velocity. That's why I've got a negative vertical velocity over here. So, vy is just a straight, you know, diagonal line with a slope of exactly the slope of a velocity graph is the acceleration. So, the slope of this line would be -10 if you were looking for it. All right. How about horizontal position? Well, this just increases at a constant rate. My horizontal position just increases at a constant rate. What does that look like? It just looks like this. The straight line this way with a slope of whatever my velocity in the x direction is. That'd be the slope of this line because the slope of your x graph is your velocity graph. What would my vertical position look like? Like position in the y direction. Well, I started at zero. I got to a high y value and then I got to a low yvalue. So, it just kind of looks like the way you would think this thing would look. It just peaks out at two and it comes back down. It's symmetric. Remember, it takes just as much time to go up as it does come back down. So, this is going to take a total of two to get up, two to get down. and it is curved kind of like the path it takes through space except this is a path through time. So this is what it looks like to graph some 2D motion variables. All right, this is the main force formula you get on the formula sheet. It's Newton's second law. A lot of people prefer to write it like net force= ma instead of a is net force over m. But no matter how you write it, this is true. What are some forces you might plug in up here? Because this is the net force. You might have many forces up here. You got to add them like vectors. So if they point up, you know, make positive, down, negative, or whichever way you want to call positive. You got to be consistent. One of the forces you plug in here is good old force of gravity. We love the force of gravity because it's always just m * g. You're not given this formula, but hopefully you used it enough that you've kind of memorized it. Uh g is the magnitude. What we mean by the symbolic letter g is the magnitude of the acceleration due to gravity, which is pos10 m/s squared because we took the magnitude. So this formula gives you the magnitude of the force of gravity. You always got to know which way it goes, but that's easy. It always goes straight down. So the force on a 6 kilogram mass would be 6. 6 * 10 would be 60. And the units for force is Newtons. What if you're in an elevator? Doesn't matter. Mg 60 newtons straight down. You could also, another name for this is the weight. Mg is also the weight of an object. People say the mass. No, the weight. The weight is m * g. The mass is m. So weight is mg. What if you're on a ramp? Well, it doesn't matter. Force of gravity still straight down. M * G. It doesn't matter where you're at. Okay, that's the force of gravity. What are some other forces you might plug in up here? Well, there's the normal force. That's the force exerted by a surface. So, this floor, so normal forces can only push. So, this floor is below the 6 kg mass. So, it pushes up on the 6 kilogram mass. And if this object's just sitting here with no acceleration, well, then the net force has to be zero. So that means the forces have to be balanced. What does that mean? Well, the upward force has to equal the downward force. So if my downward force is 60, then my upper force also has to be 60 because it has to be balanced. So there's no formula for the normal force like gravity has a formula. You just have to figure it out based on, you know, Newton's second law. Let's look at an example where you have to do a more complicated case. Let's say your uh elevator is accelerating. Let's say it's accelerating upward with 4 m/s squared of acceleration. Now, how do you figure out the normal force? Well, first of all, you can say that it's got to be bigger. Like this upward force has to win. If you're accelerating upward, then that net force upward has to be bigger. You know, you have to have net force upward. So, the upward force has to be bigger. How do you figure it out? You use Newton's second law right here. You say acceleration, plug it in is four. Fine. Equals the normal force, which we want to find. You're like, I don't know it. I know you want to find it. So, just plug in a like a symbol here, fn. You're going to find the normal force minus because gravity goes down. You got to treat these like vectors. minus the force of gravity is 60 divided by the mass. The mass here is 6. If you solve this for FN, you're going to get that FN is 24 plus 60 is going to be 84 newtons of normal force. So the normal force here was 84 Newtons. That's how you can find that. What if you didn't have an acceleration up of four? What if you just had a velocity constant velocity of four? This messes people up sometimes. They're like, "Oh, I still put four right here." But no, this wouldn't be four. Now, if you have a constant velocity of four, this would be zero. Well, that means this FN has to just be 60. You would just have an FN of 60. It would just match what gravity is. You'd be like, why? I'm moving upward. Shouldn't my upward force be bigger? No. Only if you're accelerating upward would your upward force be bigger. If you're moving up with a constant velocity, your acceleration zero, your forces are balanced. So, be careful there. Your force would just have to be 60 here. Um, what about the normal force on this ramp here? This trips people out. They're like, "Ah, dang, dude. How do I figure out the normal force on a ramp? Well, you got to break up this gravity into the component that goes perpendicular to the ramp like that. Part of gravity goes that way and part of gravity goes this way parallel to the ramp. What are these? Well, you could prove it that if that's 30, that's 90. This is some weirdo phi angle. Um, you could just see right off the bat that like this would be 90. It turns out this right here is going to be the 30. It's not clear, but if you just trigonometric, you know, identity your way around, that's going to be 30° right there. Um, if you want to figure out what the normal force is, it's going to balance this component of gravity that points straight into the ramp. This perpendicular component, what's that piece? It's the adjacent piece to this 30. So that means this normal force on a ramp is mg force of gravity this hypotenuse times cosine of 30. And this component right here, this component of gravity that tries to push the thing down the ramp, this component of gravity, this component of gravity that's like parallel to the ramp is going to be mg. That's the opposite of this 30 side. So it' be hypotenuse time s of 30. People mix this up all the time. People go through fits trying to figure out whether it's cosine or s but the little trick to remember is like I mean it's gonna sound dumb but like if you try to push something into the ramp you know like this way which is like this perpendicular amount normal force you push it in the ramp it's like it makes a harsh sound like like a cosine makes a sound but if this gravity is trying to make it slide down the ramp right this component here tries to make it slide down the ramp it's going to slide it's going to sound like sign it's telling you just listen to it it's telling you which one to is sin theta and cosine theta. So try to remember those. Um you like wait I don't even get it. What do I do here? Well that just means mg is 60 cuz full 60. 60 * cosine 30 would be what your normal force is. And your force of gravity in the parallel direction would be 60 * sin 30. But sin 30 is just a half. So this would be 30 newtons going down the ramp. And what if you wanted to figure out the acceleration of this object parallel to the ramp? So now you got to imagine this ways and that ways. You always got to pick one direction. Like you got to do acceleration in the x, net forces in the x or acceleration in the y, net forces in the y. But on a ramp, you're going to do acceleration parallel equals net force parallel. So I don't need to worry about the normal force here. I'm going say the net I'm going say the acceleration of the force down the ramp parallel to the ramp is 30 newtons divided by the mass of 6 gives me an acceleration of 5 m/s squared and more generally I mean it's just mg sin theta you know divided by m just means that the acceleration down a ramp is just g sin theta whatever that theta is why would you have to know the normal force well because maybe there's friction we'll get to that in a minute but but what does this system thing mean like it says system right Like what the heck does that mean? Why why are they writing system right here? Cuz sometimes you want to treat the whole system as a single mass. Like this one right here. Let's do this one right here. This one's simple. Whole system is a single mass. Well, like the 20 is applied to it. But let's say they they kind of move together. They don't like slide against each other. I got 20 newtons on this whole system. I don't want to treat it individually. I just want to say the acceleration of my entire system is the external force on that system. So I don't worry about any forces internally like four applied to six or six applied to four. That's all internal. That can't affect the acceleration of the system. Only external forces can. So I got this 20 newtons pulling to the right. And that's it. If there's no friction on the ground, I got 20 newtons divided by the total mass. This whole thing has a mass of 10. So my acceleration of that system would be 2 m/s squared. Now you could ask about internal forces if you wanted. You could be like, well, okay, what's the force that the six exerts on the two? Clearly the six has to be dragging the four along with it. Or sorry, the six exerts on the four. The six is exerting a force on the four forward. How do I figure out what that force is? Well, now just analyze that single box. So now you can say acceleration is net force over mass just for the four though. So we're going to just put the mass of four. We know the acceleration is two. The whole thing is accelerating with two is the net force. The only force on it forward. Notice we don't worry about vertical forces. We're just dealing with acceleration in the x. So if you're doing acceleration in the x, net force in the x, the only force that way is the force, you know, exerted by the six on the four. And so you plug that there, you divide by the mass of four, you get that that force from the six on the four is 8 newtons. Now let's say you wanted to know the force on the six by the four. So the force exerted by the four on the six that'd be equal and opposite. You don't have to do it in the formula. It's just 8 newtons backward. So whatever two objects exert on each other are always equal and opposite. even if they're different masses. That's Newton's third law. And the last force is tension. Another force you might plug in here is tension. It also doesn't have a formula just like normal force. You have to just use Newton's second law to find it. So tension is the force exerted by a rope. It pulls. It always pulls. So it's always going to pull on the object. Tension. The force T. You could write it as T. You could write it as FT. This force always pulls. How do you figure it out? Well, for this example down here, you'd again want to treat it as one big system. These are forced to move with the same acceleration. So you're going to treat it like one big mass. Anytime these boxes are going to have to move with the same acceleration, you can treat them like one big mass. So I'll say that the acceleration of that system would be the net force external. It's only external forces that matter. These tensions are internal. Look, the four pulls on the six, the six pulls on the four. That's internal. They just negate each other. And if there's no friction on the table, then the only the only external force making this thing go is the force of gravity on this 4 kg mass. So 4 * 10 is 40 newtons straight down. I'm going to treat the direction of motion as positive minus nothing. There's no other forces slowing it down or speeding it up that are external. Divided by the mass, which mass the total mass. I'm treating this as one big mass. So I treat it as 10 kg and I get the acceleration of 4 m/s squared. Now I can find the tension. If I wanted to, I could say, okay, let's analyze just the 6 kg mass. Okay, well it's acceleration just like everything else in the system is four. So four is the acceleration equals there's only one force like gravity and normal go this way but that's that's not in the direction that I've got this acceleration. This is the acceleration in the x direction. That's which way this four goes in the x direction for the six. So I need to only look at forces in the x direction. So all I got is tension t. That's it. If there's no friction that's all I got. T over 6. That tells me that the tension in this rope is 24 newtons. That's the tension here. 24 newtons. It's also the tension right here. 24 newtons. So these are some force problems that you might see. Another force that has a formula that's given to you on the formula sheet is the frictional force. Now there's really two types of friction. Static friction, kinetic friction. Static friction is when the two surfaces are gripping, not slipping against each other. And kinetic friction is when the two surfaces are slipping. So friction tries to prevent slipping. And static is so good at it that it stops the two surfaces from slipping. and kinetic just tries to slow down that slipping that's occurring. Now, you might think, oh, static means not moving, kinetic means moving, and that's not necessarily the case. Often that's the case, but not always. So, imagine this blue line here is a blue rubber sheet. And instead of pulling on the 100 Newton mass, I pull on the blue rubber sheet, so it's all grippy. Well, that makes this mass go with it. And now it's moving. And it's not slipping, though. It's moving because it's not slipping. So, if the ground starts moving, the ground's like, "Hey, don't slip on me." That means I need you to move. But if I make you move the 100 Newton mass and the ground doesn't move, now you're going to have kinetic friction. Or another case, imagine you got a tire. If you're using a car under normal operating conditions, like this is going to be static friction right here. How do I know that? Cuz there's no slipping. If there's slipping, there's going to be squealing and smoke and screech marks. You know, you be Fast and Furious out here. But if there's none of that, this is just using grippy. That's why you want these tires to be grippy. Grippy static friction to push backward on the ground. It pushes forward on you. it propels you forward. So, it can be the case that you have static friction and moving. But always, if you're gripping and not slipping, that's static friction. If you're slipping, that's kinetic friction. So, how do you use this formula? Well, there's this crazy mu here. This is the letter mu if you've never seen it. It's the coefficient of static friction or the coefficient of kinetic friction. Just if you want to find the static force, plug in mu s. If you want to find kinetic, plug in mu k. So, let's do that for s. Let's say we plugged in mu s. So we're going to find the static friction force less than or equal to mus would be 0.5 and it is always true that mu s should be bigger than or equal to the coefficient of kinetic friction.5 times the normal force is going to be 100 newtons how do I know that well gravity here this is the weight of the box 100 newtons means the weight the mg is 100 newtons down this balanced 100 newtons up so that normal force is also 100 newtons so I get my static frictional force is less than or equal to 50 newtons what does that mean does it The static frictional force is 50. No. Let's say I pulled with 20 newtons on this rope. Well, it's not like friction is going to be like boom, 50 newtons backward and it flies backward. No. Friction only does what it has to. So static friction is lazy. It's like you pull with 20. Fine. I'll pull with 20. Okay. So what if I pulled with like 30 newtons forward? Well, friction's like, "No, I don't want I don't want you to move. I'm going to prevent you from slipping. So I'm pull back with 30." You're like, "Okay, fine, buddy. I'm going to pull with like 49 newtons." Well, it's like, "Yeah, I could do that. I could do this all day. 49 newtons backwards. So normal force only balances what you do. It's not just always equal to this number. This is just the maximum amount it could possibly be. But once you exceed that, okay, so once you're like 60 newtons, well now it's like, oh damn, dude. I can't do that. I can only go up to 50. Now you start sliding, it turns into kinetic friction. So let's find the kinetic friction. You do the same thing, but you plug in 0.25. So if I plug in 0.25 here, 0.25 of 100 means the kinetic friction is going to be 25 newtons. It's always got to be less than or equal to the maximum static frictional force. So you just got this 25 newtons backward of frictional force trying to slow you down. What if I pull what if I pull with like a 100 newtons? Well, kinetic friction is like, look, dude, I'm pulling with 25. That's all I got. I'm like, okay, I'll pull with a,000 newtons. It's like, dude, 25 is all I got. So unlike static friction that can like mimic you and mock you, whatever you do, it just tries to do it the exact same thing. Kinetic friction is just the value. So when you're using this formula for kinetic friction, it's just always equal to that value. We're using it for static friction. What you're really getting here is the maximum possible static friction. So typically acceleration speeds you up or slows you down, but it's centrial acceleration, which they give you this formula here on the formula sheet. Centrial acceleration neither speeds you up or slows you down, and yet it's still acceleration. So you might be like, wait, how can that be? If I got acceleration in the direction of motion, then it's going to speed me up. And if I got acceleration opposite the direction of motion, it's going to slow me down. Well, centrial is like, I'm going to do neither. I'm going to exert my acceleration perpendicular to the direction of motion. And this this is changing your velocity because that's what acceleration has to do. But it's not speeding you up or slowing you down. It's just changing your velocity's direction. So I was moving that way, then my velocity is that way, then that way. So my magnitude of velocity is remaining the same, but it's changing its direction. And that's a change. Acceleration is a change in velocity. So you're changing the velocity and that means you need a net force to do it. Like you might be like, well, that doesn't count. That's not real acceleration. Yeah, it is. You plug you plug this acceleration straight into F= ma, you solve it just like anything else. But the forces now are centripal forces. That just means any force that happens to point inward or outward from the circle. Uh, in other words, not tangential. So you're not going to put any forces this way. Those are the forces that would speed you up or slow you down. you're going to worry about forces that are perpendicular to your motion. So, imagine you got this uh you know roller coaster loop-de-doop. What's a force? What's one force on this cart at the bottom? Well, you'd have mg down and you'd have a normal force up. Which one's got to be bigger? Well, you got to have an acceleration into the circle for something that's doing centrial acceleration. So, this normal force is going to have to be bigger. How would you figure out how much bigger? Well, you just plug into this formula, right? You see like inward is positive. We're always going to treat inward as positive because that's which way the centrial acceleration points. So I'd have normal force positive minus mg would equal mv^2 / r. That's how you would solve for what the normal force is. You just solve for it. What if I'm up here? Well, now my gravity is still down. Mg is down. But this time my normal force is also down because this this surface is above my cart and surfaces can only push. So how would I plug these in? Now you might be like, "Oh, they're both negative." No, they're both pointing into the center of the circle. Now, these would both be positive. And you would solve for normal force. What if I'm what if I actually do have this car right here? So, let's say I'm driving around in a circle. What force could be causing this car? Now, this is a bird's eye view. So, this car is not driving on the wall. You're watching it from a drone aerial view. You're like, what force could be doing this? Well, it's the it's the, you know, the tires, that static frictional force exerted on the tires pushing inward. So in that case, the force that you would plug into this side over here would be the force of static friction. Let's say you got a ball rotating in a circle. It's going in this vertical circle at 12 m/s. It's a 2 kg ball. So this is a going in a circle. You know, it's a vertical circle. Uh just consider the case down here. So you'd have 20, we'll put some numbers in here. 20 newtons of gravitational force. You'd have a certain amount of tension upward. So plug it into this formula. You'd be like tension minus 20 would equal m. So 2 * v^2 v ^2 would be 12^2 over r would be the radius of 4. If you plug all that in and you solve for tension at the bottom you get like 92 newtons. If you solve all that, what about at the top? Would tension be bigger or smaller? Tension's going to be smaller cuz look now down here gravity was working against you. It's like tension's like, "Dang, dude. I got to make all this force inward and you're not even helping me. You're hurting me. You're pushing out of the circle." But now gravity up here is like, "Okay, fine. I'll exert my force into the circle." Tension's like, "Woo, thanks, man. Now we're both working together." Nei, you know, teamwork makes the dream work up here. So forces inward are both going to be positive. They're pointing in. So I'd have t + 20 = mv^2 over r is 4. And if I solve for tension up there, I get tension's only 52 newtons. So not as big. What if I'm at this like corner angle right here? Well, now gravity's still straight down 20 newtons, but tension points inward. Tension's always inward. Now tension's like, dang, dude. I got to do all of it. But at least gravity is not working against me out of the circle. So, if we were to do this one, it would just be the only force, the only net force centrially would be tension would equal m is 2. V ^2 would be 12^2 over 4. And at that point, it'd be like 144. * 2 would be 288 / 4. Uhoh, got to do math here. That'd be like 72 Newtons. So, not as much force as tension had to do at the bottom. A little more than it had to do at the top, but still, you know, at least gravity wasn't hurting you or helping you. It was just insignificant. You don't plug gravity at all into this formula since it wasn't a centrial force at all. It was just a tangential force. So, here's a formula that's given to you for the force of gravity. Now you might be like force of gravity. We already had a formula for force of gravity. It was just FG equals mg. And yeah, that's true and that is right. But this is only useful if you know what G is. Now on the surface of the Earth, we know G is 10 m/s squared. But let's say this was like the surface of Jupiter or or Jupiter pulling on some moon. Like I don't know what the G is over here at the moon caused by Jupiter. Well then you got to go back to first principles and use this formula here. So this says that every two masses in the universe pull on each other equal and opposite based on this formula right here. So earth pulls on moon the exact amount that the moon pulls on earth and Jupiter pulls on its moon the amount that the moon pulls on Jupiter and the earth pulls on the sun the amount the sun pulls on earth. Every two forces between every two gravitationally interacting objects are equal and opposite even if they're not the same size. Counterintuitive. So what do you do with this formula? Well look at objects orbit. Celestial objects orbit. They orbit in a circle. So this is a centrial problem. You're like, "Oh, damn." Okay. Let's say I wanted to know the speed. Speed. How fast is M2 going in the circle? Well, you'd plug this into here. You'd get that big G mass one, mass 2 over R2. Now, keep in mind R is the center to center distance from the center of one of the masses to the center of the other. That's what we mean by R. Take M1 M2 over R2 equals M. Which M is this? It's the moving mass, the M2. is the mass that's doing the accelerating. M2 speed of mass 2 ^ squ over again center to center distance. Now we're assuming mass one is so massive that it just kind of sits here and mass 2 goes around it which is usually how things work out in space. So we can simplify now m2 cancels m2. The mass of m2 doesn't matter. No matter what the mass was out here it'd be moving with the same speed. It's kind of interesting. And then we cancel one of the rs and we get that v of a gravitationally orbiting object would be big g. the mass of the orbited mass, you know, the one that didn't cancel over the center to center distance, not squared. The square got cancelled, so just over r. So, it's a good way to find the V of a moving object. But let's say you did want to find G. You're like, dude, I want to know what the G is on Jupiter. Like, what is the G at the surface of Jupiter? Well, G is the acceleration due to gravity. You could also call it the gravitational field, but it's the magnitude of the acceleration due to gravity. Acceleration, what that means is take the force that Jupiter exerts, set it equal to m a. You want to find this A. That's really what G is. Well, so let's imagine put a little mass here. You could have big G. This would be the mass of Jupiter times the little mass that you put your little test mass, right? You put a little test mass m right here near the surface divided by this would effectively be the radius of Jupiter squared because you want to know what's the G at the surface. That'd be the radius of Jupiter because center to center distance. Do you have to add the little bit of the mass here? Nah. Radius of Jupiter is gonna be so big it dominates. But theoretically, you could add that little radius there equals that little test mass that's going to accelerate times a. Well, these masses cancel. Remember, everything falls at the same rate. That's because these masses cancel. And that means the A, which is really G, the G on Jupiter is going to be big G, mass of Jupiter / radius of Jupiter squared. So, it looks just like this formula here, but it only has one of the masses instead of two of the masses if you wanted to solve for it. And then sometimes the AP makes you deal with densities instead of masses. That's fine. If you wanted to get this formula in terms of density of Jupiter, if density of Jupiter equals the mass of Jupiter times the volume of Jupiter, well, that means the mass of Jupiter is just the density density. By the way, this is row row density of Jupiter time volume. Sorry, volume of this is Jupiter of Jupiter. Well, what's the volume of Jupiter? Jupiter is a sphere. You're like, "Dude, I don't know what the volume of a sphere is." But they give it to you on the formula sheet. It's 4/3 pi r cubed, and that's the radius of Jupiter. So, you could plug this whole thing in. You're like, I don't know where to plug it. This is the mass. Mass is row v. Plug the mass into the mass and you'd get a new formula. It'd be big G. Okay, it'd be row of Jupiter 4/3 pi radius of Jupiter. I know it's gnarly, but you should be able to do it. Cubed. All right. all of that divided by radius of Jupiter squared. And look at you can cancel some stuff out. You could be like, well then two of these radius of Jupiter's cancel with two of those and I get this formula for G in terms of density that is big G. It's 4/3 pi density time radius. So you just end up with one density time radius. So if I like triple the density of the planet and triple the radius, well triple time triple gives me 9 times the gravitational field that I would have had before. So if you were to like triple the density of Earth, triple the radius of Earth, you actually get 9 times the G. G would be like 90 instead of 10. So there's a ton of formulas on the equation sheet for work and energy. And one of them is the formula to find the work done. But you don't always have to use this formula to find the work done if you know what work means. Work means how much energy you give or take away from something. So say I took away 50 jewels of kinetic energy. I slowed something down. Well, then I did negative 50 jewels of work because I took energy. Well, say I gave something 100 jewels of gravitational potential energy by raising it up. Well, then I did positive 100 jewels of work on it because I gave it 100 jewels of energy. Similarly, let's say I compress a spring and give it 8,000 jewels of energy. Well, I gave it energy, so positive and I gave it 8,000. So, the work I did was positive 8,000. Notice we didn't have to use force at all over here. But it is true that you could also find it an al alternative way. If you know the force on an object that was doing the work and you're exerting some force in some direction, and you know how far that object went, the distance through which it went, then you can multiply those two and get how much work was done. But this parallel sign here is because it's only the parallel component of the force that matters. So forces in the direction of motion do positive work. If I had a force back this way, it'd be doing negative work. That's what this F cosine theta means. Theta here is between F, the force doing the work, and the direction it moved. Well, if I'm moving forward and my force was backward, that's 180°. Cosine of 180 is negative 1. And that's just why you get a negative F * D. And conversely, if I had a force that was perpendicular like this, well, now none of it's in the direction of motion and therefore it does no work whatsoever. Cosine of 90 is zero and this perpendicular force does zero work whatsoever. Now, one other thing to keep in mind is that if the AP ever asks, hey, what's the net work done? This is a special keyword, net work done. This is the total work done. In other words, all of the F*s DS. This is the total work done on an object. Well, this is specifically always the change of kinetic energy. This is just a fancy way of asking what's the change in kinetic energy. So, an individual amount of work might change your spring energy. It might change your gravitational energy. But the network done on an object, the total network done on an object is going to be the change in kinetic energy of that object. So, if you ever hear network, just replace it in your mind with change in kinetic energy. That's usually easier to find. Okay. What if what if you don't want to deal with work at all? Like, I don't like dealing with work. Why I just put everything in my system? Because it turns out the total initial energy of a system plus any work that was done on it, right? Cuz the work is just the change in energy and this is going to be external work has to equal the final energy of a system. Um, but I don't want to deal with work. So you know what I do? I make it so there's no external work. How do you do that? Put everything in your system. Just put everything in your system and this is zero. There's no external work if there's nothing external to your system. So let's include the block and the spring and the earth and everything down here. And that way I know E initial equals E final. In other words, energy is conserved. So let's try this problem out here. Let's say this 5 kg kilogram mass flies down here and it gets stopped by the spring fully. And I want to know, well, how much did the spring compress when it fully stopped it? All right, we'll do conservation energy. We include everything. What kind of energy does this mass have? It has kinetic energy because it's moving. If you're moving, you got kinetic energy. So I got 1/2 * 5 * 10^ 2 plus. It's got gravitational potential energy. You could ignore the deltas if you want and just find the values of the gravitational potential energy is m is 5, g is 10. This isn't the speed of 10, but g is 10. And the height here would be two equals. So that was it. There's no other energies here. I don't do the spring energy because it's not compressed yet. So my initial, you know, my initial energies is the state of the system right here. My final is when it's fully compressed. I add up all the energies in the system. Final would be I got no kinetic now because I'm not moving. I got stopped. stopped as code for not moving. So, no kinetic energy, but I do have gravitational potential energy. I have 5 * 10 * 6 plus I've got a 12 * 100 * x^2. That's my spring energy. And that's it. That's all I got. So, if I add this all up, let's see. 12 5 * 10^ 2 is 250. And the units of energy are jewels. That's how much kinetic energy I had to start. uh the gravitational that I had to start was plus 100 Jew of gravitational equals I ended with 300 jewels of gravitational. So gravitational potential energy went up because this mass went upward and then I have plus a 50 * x^2 and if you solve this for x you get x is exactly 1 meter. So that's how much it would have been compressed if you solve this for X. But a lot of times the AP likes asking conceptual questions like, hey, the change in energy of just the mass, what was the change in energy of just the block, right? The 5 kilogram mass. Well, that's going to be the final energy of the mass minus the initial energy of the mass. But notice it just says block, not block Earth. It's got So that's all that's only kinetic energy. A block alone can only have kinetic energy. Block and earth could have kinetic end potential due to gravity but block alone is just kinetic energy. So how much kinetic energy did it end with? None. How much kinetic energy did it start with? 250. So what was the change in energy of the system consisting of just the block? Negative -250. It lost energy because it lost its kinetic energy and that's all it could have. Well, let's ask a new question. For this whole process, what was the change in energy consisting of the block and the earth? Well, block and earth can have kinetic and gravitational potential. So now I got to be like, well, my block on earth started with 350 and it ended with 300. The spring's not part of the system here. So I my final is 300 minus my initial is 350. Uh this system also lost energy. It lost 50 jewels. It didn't lose as much, but it lost 50 jewels. All right, one more. Let's say I asked you or the AP asked you, hey, what's the change in energy of the system consisting of just the spring? Well, the spring started with no energy, but it ended with, let's see, 50 * 1^ 2AR would be 50 Jew of gra of spring potential energy. So, it ended with 50. It started with none. The change in energy of the spring system alone would be 50 jewels of energy. It gained 50. This is not a coincidence. Look, our system of block and Earth, which was everything else besides the spring, lost 50. Our spring gained 50. That's because energy is a zero sum game. If if one system loses it, the other one has to gain it. something has to gain it and you don't create or destroy energy. That's why energy is conserved. Uh there's one more formula that's not given. It's the thermal energy. So if you want to know how much thermal energy was generated, let's say there was like friction down here. You know what I mean? You want to know how much thermal energy was generated? Well, it's going to be, you know, negative whatever the work done was cuz thermal energy is going to gain some energy by doing work, but it's going to do negative energy or negative work to get that energy to it. If you do negative work, you gain the energy. How much does it gain? It gains the force of friction times the distance that it moves. So it looks just like this formula here. So they don't give you the formula for the thermal energy. But you can find it. Force of friction times distance. What's the force of friction? Well, you could write it as mu k * fn * d. And then that would be a formula you could plug in. It always goes on the final side because thermal energy only shows up after something happened. So you could do plus fkd over here. Or if you wanted to treat it as an external work done, you could move it to this side and you could write it as negative FKD on the left side. But you would not write it as negative FKD on the right side or positive FKD on the left side. But this is how you do it. And sometime if they just ask you for thermal energy, find the thermal energy, just write just write plus thermal energy and solve for it. You don't always have to write mu kfn d. If they want you to find D, great. And you know, plug that in or something like but if they just want you to find Ether thermal, sometimes you can find it just by plugging it in as Ether thermal and solving for that variable that you called Ether thermal. So just like there's a general formula for the universal gravitational force, there's also a more general universal formula for the gravitational potential energy. So instead of using mg delta y or mgh, if you got planets, you know, you don't know what little g is, again, you got to use this formula here. Um, so say you got two masses separated. The formula is this. The gravitational potential energy in jewels is negative. This negative freaks people out. I'll tell you why it's there in a minute. Negative big G. Same old big G is the force formula. One of the masses times the other mass divided by the radius between them. The center to center distance between them. Not squared though, just center to center distance between them. And this gives you the gravitational potential energy. Why is it negative? Well, that's just cuz we choose to say that when they're infinitely far apart, if you made this distance between them infinitely far apart, we call that zero potential energy. Why? I mean, it's just a convention. It's just we could have picked anywhere. We picked we picked infinity because it was the only like unique choice. And that means that the potential energy is always negative. Does that matter? Not really. Just plug it in. It'll it'll be fine. Does it make sense? Well, yeah, cuz look, if I if I move these farther apart, they should gain potential energy. Like, if I raise something off the Earth, it should gain potential energy. That's still true. If I make this r get bigger, then this quantity becomes less negative. Think about it. You're dividing by a bigger number. So, instead of like -10, it becomes like5. That's still increasing the potential energy. So, it's still true that as things move farther away, they increase potential energy. As they move closer together, they decrease potential energy. All of that's still true. So, don't let the negative sign freak you out too much. Um, what kind of problem might you solve here? You might be like, well, I want to find the escape velocity of an object. What does that mean? It means, you know, we say what goes up must come down. But that's not really true. What goes up must come down, unless you shoot it fast enough. In other words, if you shoot it at the escape velocity at this speed, it goes up and it just never comes back down. It goes off forever. That's i.e., it escapes. So, how can we find this escape velocity? We're going to use conservation of energy. So I'll put everything in my system. The mass of the moon, the mass of the little probe that I send out. Um, and we'll just say that E initial equals E final. Okay. So what kind of energy does this have initially? It has gravitational potential energy. Has negative big G mass of the moon. Mass of the little probe that I'm sending out over the center to center distance they start is the radius of the moon. Do you have to add in the radius of the probe? Yeah, but it's so small compared to the radius of the moon. It doesn't matter. Oh, and I messed up. Look, it's not squared. Don't forget. You get so used to writing square for the force. You want to square it there. Don't do it. That's the gravitational potential energy plus the kinetic energy. So, it starts with some kinetic energy. How much? Well, 1/2 mass of the probe times this escape velocity. Like what I'm solving for here is the escape velocity squared equals well, I want it to get out to infinity. Once it gets, you know, it's to escape means to go forever, never come back. So it eventually gets to r is infinity. But if I divide by infinity here, 1 over infinity is zero. So I have no gravitational potential energy to end with because r is infinity plus and I want it to just get there. I don't want it to get there with a bunch of extra speed. So I want it to get there with a velocity of zero. I just want it to get out there. So it's got no kinetic at the end if I just want the minimum speed that I should send this. So now I can start simplifying. Look, I can cancel MP. Divide on both sides by MP. I could solve this for V escape velocity and what I get is the square root of 2 big G mass of the moon or whatever planet it is that you're on divided by the radius of that moon or whatever the radius of the planet you're on. So this is a example problem of solving gravitational potential energy to find the escape velocity. What other problem might show up? Well, you might have one like this over here. You got three objects, a mass 2 m, a mass m and a mass 3 mm. They're all separated by L. You want to find the total gravitational potential energy. Well, you just got to pair them off, right? There's only two masses in here. One, two. You can't be throwing three masses in here. So, just pair them off. I'll do the M and the 2 M first. So, I got negative big G M * 2 M would be 2 * M^ 2 M * 2 M divided by distance between them. L. Don't write it as 3 M. You're not adding the masses. You're multiplying the masses. So, M * 2 M 2 M^ 2. Then I have another one minus. We'll do M and 3 M. So minus big g m * 3 m is 3 m^ 2 divided by the distance between them is l. And there's only one more pair. It's the 2 m and 3 m. So I have minus cuz it's minus big g. 2 m * 3 m would be 6 m^ 2 / l. I add all these up. I get 6 + 2 is 8 + 3 is 11. So I get negative big g 11 m^2 l gives me my total gravitational potential energy in this system. So there's two formulas for power on the AP formula sheet. You get this top one and the bottom one. The top one's usually more useful and this gives you the definition of what we mean by power. So power is the amount of work done per time which works the change in energy. So it's how much energy you're giving something per time. That means it has units of jewels per second and we give that the name watt. So watts is the unit of energy. Um so what is it? How do you even do this? Let's say you want to know the power that this car develops over 4 seconds from getting to z from 0 to 30 m/s. Well, just ask yourself, okay, what kind of energy did it gain? Well, gained kinetic energy. How much? It gained 12 m * v ^ 2 30 squared. Well, how long did it take it? You divide by the time that took 4 seconds. Okay, easy, easy. It's not that hard. You do all that, you get 11,250 watts. Okay, what about this one here? You take a 3 kilogram mass. You raise it up four meters in two seconds. Okay. Well, how much energy did it get? What what kind of energy did it get? It got gravitational potential energy. How much did it get? It got mg * h was four. That's how much it gained. Just find the energy it gained. Divide by how long it took and you solve that and you get that this was oh that was 60 watts. So that's one way to get the power. This is the average power because we took it over a long period of time. You might you also get the in formula for instantaneous power. So this is the power at a given moment in time. How do you get from this top formula to this bottom one? Well, watch. Work done is force times distance divided by the time. Okay. Well, this is also equal to force times distance per time. Well, distance per time, that's just the velocity. So force times velocity gives you the instantaneous power at that moment. What does that mean? Let's say you're exerting 100 newtons at a given moment and this object's moving like 4 m/s at that moment. Well, if I want to know how much how much power am I exerting per time at that moment, you know, how much how much work per time at that moment, I just plug in my force. 100 newtons, plug in my velocity, 4 m/s, and I get 400 watts of power is the instantaneous power at that moment. Maybe it gets bigger or smaller after that, right? Maybe this four goes up to five or six. If it maintains four, then what this means is that every second I'm delivering 400 jewels of energy into this block. So kinetic energy was not a vector. In other words, kinetic energy going right or down or up or left. That's all the same kinetic energy. It's always positive. But momentum is a vector. That's why they got these little vector symbols up here, these arrows. So the definition of momentum, which is P, represented by the letter P, is mass time velocity. It's a vector. So momentum to the right is different from momentum to the left is different from momentum up and down. You mainly have to keep track of the signs. Leftward momentum is going to be negative. Downward momentum is going to be negative. What changes the momentum? Well, we could just see here. Let's say I ask what's the change in momentum per time? Well, that'd be the mass times the change in velocity per time. So here I have change in momentum per time. Change of velocity per time. That's change in velocity per time. That's just the acceleration. So ma, but we know what ma is. It's the net force. So net forces change the momentum. Or I could rearrange this if I just multiply by delta t. Force time delta t would be delta p, which gets you this formula right here. Force time delta t equals delta p. What does this mean? Means if you exert some external force, this would be some external force on the system. Exert it for a certain amount of time, you multiply that, that tells you how much momentum the system will gain in a particular direction, which could mean lose. Like if you had momentum to the right and your force was to the left, your your gain could be 20 newtons to or 20 Newton seconds to the left, which would mean you lose 20 units of momentum to the right. Uh we give this a name. The change of momentum or the force times time is called the impulse. So J is referred to as the impulse. Why' they pick J for impulse when there's no J and impulse? I don't know. I call it the gym impulse to keep track of that. So J is F delta T delta P. If you have no net external force, then there's no change of momentum. That means you could have conservation of momentum. Just like energy, if there's no work done, energy is conserved. If there's no net external force, there could be internal forces like these masses can hit each other. But if there's nothing external to the system, then the momentum will be conserved. What does that look like? Let's try it right here. So if these two objects hit and stick together and we consider them both part of our system, well then P initial is going to equal P final total. So let's see P initial MV. So I have 4 * 6. 4 * 6 plus 2 * now you got to be careful -3 equals they stick together. So 4 + 2 is a mass of 6 * V. I'm just doing MV plus MV. If you solve this for V, you get a VFAL of positive3 meters/ second. So afterward, they stick together and they're moving to the right at 3 meters/s. That's called an totally inelastic collision. When they stick together, um they're going to lose kinetic energy. They don't lose any momentum in that collision because there was no external forces, but they do lose kinetic energy. It turns to like thermal energy. If it's an elastic collision, that means kinetic energy is conserved. So let's just check. Is this collision elastic? Well, whether it's elastic or inelastic, if there's no external forces, momentum will be conserved. So, let's find out. So, momentum initial equals momentum final. That's true for elastic or inelastic. So, again, 4 * 6 mv plus 2 * -3 mv equ= it says afterward the 2 kg mass moves 9 m/s. So 2 * pos 9 plus 4 * unknown vf final. Um but if you add up this whole side here you get this whole side is 18 = 18 + 4 vfal. The only way this can be true is if vfal is zero. So forward isn't even moving afterward. The vfal of the four is going to be zero afterward. Was this an elastic collision? You have to check if kinetic energy was conserved. Though beforehand you had so kinetic energy initial was 12 4 * 6^ 2 plus 12 cuz kinetic energy is 1/2 heavy squar 2 * 3^ 2. Notice I don't care about the negative sign. I'm going to square it anyway. So if I add all that up I get 81 jewels of kinetic energy before the collision. How about afterward? Well, it's only the 2 kg mass moving afterward. So I'll have 1/2 m. It's going 9 afterward. 9 m squared. Don't forget to square it. Kinetic energy equals 81 jewels. Look at that. And there was no energy for the 4 kg mass. These are the same. Therefore, it's an elastic collision. Speaking of elastic collisions, there's some special rules you should know. If you know it's an elastic collision, this doesn't work for any elastic, but if you know it's an elastic collision, if a mass comes in and hits a mass of an equal size, so 4 kg hits 4 kg, this first mass just stops during the collision. This 4 kg would stay put. And the next one goes out with the exact same speed. They just swap velocities. So equal masses in an elastic collision will just swap velocities like a Newton's cradle. This one stops. That one goes right out. What if the other mass is bigger than the first one in an elastic collision? Well, then this 4 kg mass would bounce backwards and this 8 kg mass has to bounce forwards. Um, and then what if the big mass hits the little mass in an elastic collision? Well, the 4 kg mass obviously goes this way, but afterward the 8 kg mass will also go that way. This is helpful because in a multiple choice, maybe like they only give you one option for little mass hitting off a big mass that bounces backward and you know, oh shoot, well then that's got to be the only possible option. Or maybe they give you a situation where equal mass hits equal mass. You don't even have to do conservation of momentum here. You know this one stops and that one goes right out with the same speed. So this can save you some time. Also, another good thing to know is let's say it was a totally inelastic collision and they stick together. If equal mass hits an equal mass at rest, since you just double the mass and the momentum masses stay the same, these are just going to have to go out with half of the velocity that the incoming mass came in with. So, if equal mass hits equal mass, they stick together and the one of them was at rest, you just have this velocity and that's their velocity heading outward. So the center of mass is a location within some mass distribution. So say you had a 6 kg mass and a 2 kg mass. If you found the location of the center of mass, that would tell you where the system would balance. In other words, if I just connect this 6 kg to the 2 kg with a very light rod that has no mass itself. The location of the center mass would tell you exactly where it balances. So wherever this would balance at the pivot or fulcrum sometimes it's called is where the center of mass would be. And it has to be proportionally closer to the larger mass. So since this 6 kg mass is three times bigger than this 2 kg mass, the center mass has to be three times closer. So if this is 16 m and this is 4 m, 4 m, 4 m, 4 m, the center mass would have to be right here because that way I'm 4 m from the six and I'm 12 m from the two and 12 is three times bigger than four and so the six is three times closer to the center of mass. So let's prove it though. But let's say you didn't have numbers that were that nice. You wanted to prove the center of mass was right here. There's a formula. You get this on the formula sheet. It's x center of mass is the sum of all the m * x's over the sum of all the m's. What does that mean? It means pick all the masses. So, and choose a place to measure from. You might be like, where do I measure x from? Well, you get to pick. So, I'll just pick the left hand side. I like picking the leftest most mass as x= 0. All of our x's are going to be measured from there. So, if you use this formula, it's like, okay, let's add them all up. 6 * Well, where's the 6? It's at zero. It's at x= 0. Plus 2 * how far is it? It's 16 away from x= 0. So 16 / the total mass would be 6 + 2 is 8. And if you multiply that all out, look at what you get. 4 m, which is exactly what we found. 4 m from the point we were measuring x= 0 from, which is exactly there. You could have chose this side. If you did that, you would have got like 12 or -12, which tells you, oh, it's 12 from there. So, you could choose x equals 0 wherever you want. And if you're consistent, it'll all work out. Uh, how about velocity center mass? Since the center mass is a location, it could move right here. If they're at rest, it wouldn't move. But if these guys are moving around, well, this center of mass could change its location. It's not just going to stay put while all the masses head away. So, say you had a 6 kg mass heading 3 m/s to the right and a 2 kg mass heading 1 m/s to the left. You wanted to know what's the velocity of that center mass? I know right now it's like right here. But where would the center mass go? Like what would the velocity of that be? Is it going to be heading rightward, leftward? Let's find out. So we're going to do this. We're going to say take the sum of it looks just like this formula. Instead of mx, it's going to be mv and you divide by the total mass. So let's do it. 6 * 3. So 6 * pos3 plus 2 * -1 because this is a velocity of negative 1 goes leftward over the total mass of 8 gives you a center mass velocity of pos2 m/s. So this this location of the center mass would be moving this way at 2 m/s. And note something very important um this this sum of the mvs well look at mv is the momentum that means this numerator is just the total momentum of the system. So if you take the total momentum of the system divided by the total mass, you get the velocity of the center of mass. And think about it, the only thing that can change the total momentum of the system is an external force. Remember external force times time gives you the change in momentum of the system, the impulse. If you have no external forces, then there will be no change no change in the velocity of the center of mass. So if you can only change momentum, total momentum by external forces, and you have none of those, then you're not going to be able to change the velocity of the center of mass. What does that mean? It means after these collide, they come in, right, and they collide. Those are internal forces. This center of mass is going to keep chugging right here, keep on trucking at exactly 2 m/s. after these collide and head out because that was an internal collision. Only some third mass, right, causing some external force could cause this center of mass to go faster or slower or deviate from its path. Only external forces can affect the motion of the center of mass. So the angular kinematic formulas look just like the regular kinematic formulas except they just have all of their angular doppelgangers in here. So time is still time. That's the only thing that doesn't change. But instead of like original position, initial position, you have initial angle. So the angle, the number of radians you are from say zero. And then theta final would be the number of radians you are from zero. Um you've got omega which instead of change in position over change in time. Omega is defined to be change in theta over change in time. Uh so you could have the final omega, omega is the angular velocity. It's measured in radians per second. So instead of meters/s, omega is measured in radians per second. You could have your initial angular velocity also in radians per second. And then maybe you're speeding up or slowing down, which means maybe you have an alpha. Alpha is the angular acceleration just like acceleration's change in velocity per time. Alpha is the change in angular velocity per time. And so you again three formulas. You got five variables. There's a still a secret formula. Remember the secret formula? Distance is rate times time. Instead of distance, now it's just going to be delta theta. Instead of just rate, you have to take the average rate. Omega plus omega initial over 2 times the time. So displacement, angular displacement is going to be average angular velocity times time. And then similarly to before, if alpha is zero, if you're spinning at a constant angular velocity, you can actually just do theta is the angular rate times time because it'll only be one. Um, you got to be careful. Sometimes they might tell you revolutions instead of radians. Maybe they say this thing goes through two revolutions. Well, you got to be like, well, two revolutions. There's two pi radians per two revolutions. So, I'd have to multiply 2 pi radians per revolution to turn this into radians. You want this in radians. Radians/s radians/s squar. Uh, again, you need three of these variables to solve for a fourth. And you'd pick the formula that's missing the thing that you're not given and don't want to find. Typically, you treat clockwise as negative and counterclockwise as positive. That bothers people. They're like, "Wait, but the clocks go clockwise." Yeah, but think about it. In math class, you treated this way as positive. And look at which way that is. That's counterclockwise. I mean, you can pick whichever direction you want. If it says to pick counterclockwise as positive, then you have to treat that as positive. But if it doesn't say that, you can typically just choose as a convention whichever direction counterclockwise or clockwise that you want to pick as positive or negative. So let's say you found some of these angular variables using kinematic formulas or something and you wanted to convert them into like normal variables, right? Instead of angular displacement, just get displacement or omega, turn omega, angular velocity into regular velocity. How do you do that? Well, these formulas are given to you on the formula sheet. And let me just show you where they come from. They come from the fact that this first one's really just talking about arc length. If you want to know how much arc length did something go through, like what is this? We usually call that s in math class. Turns out you can find that by just doing r the radius of that point times delta theta. How much angle in radians right from here to here. How much angle and radians did this thing did this thing go through? And this is really the definition of a radian. I'm embarrassed to say how far I got in math class without realizing one radian means. So if we make this one, right? One radian. One radian is the point. If I make that one, it's the point where this arc length here s is exactly equal to this radius here. That's the magical point. It only happens at one angle. And I'm like, dang, really? That's arc length of this equals that? Yeah, that's the angle when you have one radian. That's what we mean by radian. And that's why these formulas work. But over here it says X center of mass. Like what is this talking about? It doesn't say arc length. Well, if you're rolling without slipping. So you got a wheel and the edge is painted. As it rolls, it goes through a certain arc length. And that arc length is going to leave an arclength line of paint on the ground. But think about it. That also means the center of this object is going to move forward by the exact same amount that this paint was left on the ground. Think about it if you don't believe me. But that means the center of mass is moving the same amount as the arc length. And so that's why they write it like this. The center mass of an object that's rolling without slipping. It's important. This has to be rolling without slipping. The distance this center moves forward is just equal to the arc length. And we know the arc length is r * delta theta. What about these other formulas? Well, this is the velocity of some point. If I know the omega, how do I figure out the velocity? It turns out you just do r * omega. Why do you do that? Look, we could divide both sides by delta t. Divide both sides by delta t. Delta x over delta t is the definition of velocity. Change in position over change in time. Well, change in theta over change in time is the definition of omega, the angular velocity. So that's why this is true. I affectionately know this as V is row omega. Row omega is how you get the V. So if you know the omega, you can get the V by just multiplying R by omega and that gets you the V. And it's also the V of the center mass. So just like up here, if you're rolling without slipping, then the center of this object is moving with a V center of mass that is equal to R * omega. So that's also true. Uh, what about a is r alpha, affectionately known as alpha? Well, I'll do the same thing. Take this, take the change in v over the change in t. Well, I'll do the same thing to the right side. Change in omega over change in t. But change in v, velocity over change in time, that's the definition of acceleration. And change in omega over change in t is the definition of alpha. Alpha is the angular acceleration. So if you know the angular acceleration alpha and you you want to know what the regular acceleration is, well then you could just do r * alpha affectionately known as alpha. You might be like hold up what's this t doing here? Well that's tangential acceleration. That's because if you remember we already had an acceleration of something going in a circle. It was centrial acceleration. If you're going in a circle you've got centrial acceleration. It's v ^2 over r. That's separate. This centrial acceleration is just keeping you going in the circle. It's not speeding you up or slowing you down, but this tangential acceleration right here, this does speed you up or slow you down. This is going to be if it's in the direction of motion, this tangential acceleration, it's going to speed you up. If it's opposite the direction of motion, it's going to slow you down. But centripal just changes your direction. And you can have both. I mean, you could have centrial and tangential, then your total acceleration would point like in here somewhere. And then just like before, if you're rolling without slipping, the tangential acceleration r alpha alpha is telling you also the acceleration of the center of mass is found by doing r * alpha if you're rolling without slipping. So just like forces cause acceleration, torques are the thing that causes angular acceleration. How do you find the torque? You use this formula right here typically. But even with this formula, people are bad at this. It's like people have very little intuition about torque. So you got to be very careful. Step one, you draw the force because forces cause the torque. Over here I already have it drawn. We'll do the easy mode. Got 30 newtons straight down. Great. Next, you draw the R. R is from the axis to where that particular force is applied. And this can be different for each force. So you draw it from the axis to where the force is. A little arrow. This is R. And the R for this 30 Newton torque is going to be 3 m because this if this whole thing is six half of it if we're assuming this is the middle it's going to be three. Then you have to identify what is the angle between R and F. So this angle in here in RF sin theta is the angle between the R and the F. Well the angle between R and F here is 90°. So that makes it easy because S of 90 is going to be one. So you can figure out the torque from this 30 Newton force is going to be R which is 3 * F which is 30 time s of 90. S of 90 is just one. So that maxes out. If you want to max out your torque, make sure your angle is 90 there. And you get 90 units which are Newton meters of torque caused by this 30 Newton force. Okay, let's try a harder one. Say you want to know the torque from the 60 newtons. Now, that's all at an angle and stuff. Doesn't matter. Same process. Draw the force if you haven't already. We've drawn it right there. Now, draw the R for that particular force. It goes from the axis to where the force is applied. It's going to be this way. That's the R. And the R again is going to be 3 m. It goes from the center to here, but it goes leftward. All right. Now, identify the angle between R and F. You might be like, uh, 90? No. Uh, 60. No. Look, we got to go from from R to the F. That's this whole angle here. So what is that? Well, this is 90 plus a 60. This whole thing would be 150°. So you got to be very careful. So this torque from the 60 would be 3 m * 60 newtons * s of 150°. But luckily s of 150 is the same as s of 30. So 3 * 60 sine of 30. uh this also gives you 90 newton meters. Now if you just draw this carefully. So if you draw your r extended all the way out, draw your f all the way out. It turns out the sign of this angle, this angle, this angle or that angle will all give you the same value. So signs of supplementary angles are the same. So you could pick this one or that one or that one or that one. You just you just can't draw some artificial line here and pick that one. You know what I mean? So you have to pick this one or that one. and I could extend it through or this one or that one. All of those would work, but you got to pick a correct one. So, you get 90 Newton meters. Now, you might be like, "Oh, so these must cancel, right? These torqus cancel cuz look, one force was down, one force was up, but that's not how torques work." You should think of torques. It's helpful to think of torques as being clockwise or counterclockwise. And this 30 is trying to rotate it clockwise. Well, what's this 60 doing? It is also trying to rotate it clockwise. These are working together. Our net torque as it stands, our net torque would be 90 + 90 180 Newton m in the clockwise direction. So, this thing's going to start spinning like crazy. Let's say we didn't want it to spin. Well, you better exert some sort of counterclockwise torque because the only way to get this to stop spinning in the clockwise direction is to exert some counterclockwise torque. Let's do it at the end here. Let's say I want to exert some force right here. I want to know how big should it be to prevent this thing from angularly accelerating. Well, effectively what I'm saying is that I want the clockwise torqus to equal the counterclockwise torqus. You just want this to be balanced. We know what the clockwise torques were. That was 180. Okay. Equals. Well, let's find the torque from this guy. Remember, step one, draw the force. I just did that. Step two, draw the R. R goes from the axis to where the force is. Again, it's 3 m. So, it's going to be 3 m time the f that we don't know times s of what's the angle going to be? Well, let's just give ourselves the best shot of doing this. So, 90°. So, sine of 90 gives us one. So, if I solve this for f, I see that I got to exert 60 newtons over here at the edge to counteract all the torque from the other direction. All right, let's try one more over here. One where the forces aren't already drawn. This is where people get really messed up. So, you got a beam of 5 kg and it can rotate around this pivot that's like hanging here. This is a side view. So this is going to rotate down from gravity and you got this rope holding it up at 30° and the the beam is 4 m. So first things first, draw all your forces. So let's draw a force here. We got force of gravity straight down. So there's a force of gravity and it's a if it's a 5 kg mass of a rod that's going to be 50 Newton. So gravity acts at the center of mass. Then you got a tension over here. Tension is going to point uh this way as it always pulls. So tension is going to be something like that. And then you're also going to have like forces from the beam over here. It's probably going to try to hold it up. Probably going to try to push it out. Turns out it won't matter. Here's why. These are exerted at the pivot point. In other words, R is going to be zero for those forces. So if I try to draw R from the pivot to where they're applied, they're applied at the pivot. So they apply no torque. So if something has an R of zero, it applies no torque. All right, let's try to figure out how big is this tension in order to hold up this whole system right here. So, we're basically going to do torque clockwise equals torque counterclockwise. So, is this gravity trying to torque this thing clockwise or counterclockwise? Clockwise. It's going to be going that way. How much torque? Well, we drew the force. Step one. Now, we draw the R. R goes from the axis to where the force is applied. If this whole thing is four, gravity's at the center. That R is 2. So we're going to have 2 * a force of 50 time s of 90 gives us 1. So that's our torque clockwise equals this torque from the tension is going to be going counterclockwise. It better be or this thing would never balance. That's the only other torque involved. So you draw R. You know we had draw the force. Now we draw R. R goes from the axis all the way to where the force is. Well that's 4 m. So I have to do 4 m time the tension time s of the angle between r and f which is t that really is 30° now. So that is the 30° and if I solve this sine of 30 is a half. So I end up getting that the tension here is actually going to be 50 newtons in order to solve this. So long story short draw your r your f draw your theta. A a force could exert no torque if the R is zero. A force could exert no torque if the F is zero, obviously. And a force could exert no torque if the angle is either zero or 180. Right? That makes sense. If I push straight toward the pivot, my angle here is 180 between my force and my R. Well, s of 180, this isn't going to get this thing to move. You can't just push toward the hinge. You also can't pull away from the hinge in order to get it to spin. That's not going to get this thing to spin. So forces can exert torqus but they don't have to. So just like mass measures the degree to which a system will resist acceleration, rotational inertia represented by i measures the degree to which a system will resist angular acceleration. So the bigger this I rotational inertia, the harder it will to get something to start spinning and angularly accelerating. How do you find this eye? Well, it depends on what system you have. But if you have a system of point masses that are all orbiting around some pivot, you could just add up all the mr squares. And so in other words, add these all up. You'll have a 1 kg mass at 6 m away. So 1 * 6^ 2. And then you'll have a 3 kg mass. So plus 3 kg mass at 4 m away from the axis. And then you'll have a 2 kg mass. So plus I'm making them all plus. Just like mass can't be negative, rotational inertia can't be negative. So 2 + or 2 * 3^ squar and if you add all that up, you get likeund 102 kilogram m squared of the units. So this measures the degree to which something's going to resist angular acceleration. If you make it more massive, more rotational inertia. If you push the mass farther away, more rotational inertia. And the and that matters more. So the if you could double how far away a mass is from the axis, you get four times the rotational inertia. If you double the mass, you just get two times the rotational inertia. But these are masses that are orbiting, you all at a specific location. All the masses there. What if you have an object that's spinning around its center? Like this 1 kg mass wasn't spinning around its center. It was orbiting around an axis. If you do have this case where objects are spinning around the center, there's just specific formulas that you don't have to memorize, but they'd be given to you. So the rotational inertia of a cylinder it turns out is 12 m r^ 2 mass of the cylinder time radius of the cylinder squar. If you have a sphere rotating around its center like spinning right these are spinning around their center rotational energy of a sphere spinning around its center is two fifths it turns out m r 2 and then if you have a hoop the hoop is the special case since all the mass is at the edge of the hoop it's equivalent to orbiting this is the case where spinning is the same as orbiting and you just get a rotational inertia of mass of the hoop time r^ 2 so equal masses all thing equal all things being equal equal masses equal radi the hoop would have the biggest rotational inertia. The cylinder would have the next biggest and the sphere would have the least biggest. What if you had a rod rotating about its center? It turns out the formula for that formula for rotational inertia of a rod about its center is 1112th ml 2. L is the total length of the system, not the radius. So in other words, you do 12th. The mass is 5. The total length is 6. So if you do this, if you do 112th * 5 * 6^ 2, you get 15, I believe 15 kilogram me squared. Okay, fine. That's the rotational energy of a rod about it center. What if you wanted to know if you rotate it here? So say you want to start changing it. If I change the axis to over here, start spinning the whole thing around to end. Now, what would the rotational inertia be? Well, there's a formula for it. It's over here. It's called the parallel axis theorem. The new rotational inertia I prime i.e the I about this end will equal the rotational inertia about the center of mass which we just found. It was 15. So I take that 15 and I add to it the mass of the whole object times how far I moved it. This d this d right here represents how far did you move the axis? I moved it this way and I moved it 3 m. So, if I want to figure out the new rotational inertia about the edge, I could do rotational inertia about the center of mass, which I just found is 15. 15 plus the total mass is five. And the distance I moved it was three. And you add this, it's going to get bigger. The rotational inertia about the center mass is always the smallest you can get. If you move it away from the center mass, it'll always get bigger. And in this case, it gets bigger. It becomes like 60 kg m squared. So, now we know how to find the net torque and we know how to find the rotational inertia. We could put it all together into the equivalent of Newton's second law for angular, which is instead of a is net force over mass, we have alpha is net torque over i. So we already figured out, remember we did this one a minute ago. We figured out, oh, you add up all the torque, you get a total of 180 Newton meters of net torque because these are working together. This exerted 90 newton meters. This exerted 90 Newton meters. You divide by the I. And we just figure that out. The I for a beam, you know, 1112th 112th ML squared. We just figured that out. We got that number. It was what was it? 15 15 kilogram me squared. So if you divide these out, you're going to get that this this thing is going to angularly accelerate with an alpha of 12 radians. The units of alpha are radians/s squared. Just like a is me/s squared. Alpha alpha is radians/s squared. So 12 radians per second squared in the clockwise direction. Let's try a harder one. We got to draw the forces. Okay, this thing, say this thing's hanging down. If I let go, it's going to start swinging down like this. So, first we got to draw the forces. What's the force? Well, the force of gravity and it acts at the center and it goes straight down and it's a 5 kg mass. So, same beam, but I'm just hanging it from the edge. Now, it's going to fly down. So, we got 50 newtons exerted right here. Okay, let's try to figure it out. We want to find the torque. So, we draw the force. Then, we draw the R. R goes from the axis to where the force is. So, this is 3 m that ways. So our torque is going to be 3 * 50 times the sign of the angle between R and F. So it's this angle right here, which remember if we just draw R, we draw F, it's any of these angles. It's that one, that one, that one, or that one. So I could I could extend this and say it's the same as this. So the sign of supplementary angles are the same. But look at this angle right here. This really is the same as this angle right here. That is the angle between vertical and the rod. between vertical and the rod. So this is 30. In other words, you can make this sign of 150. Either way, sine of 150 or s of 30 gives you a half. Then you divide by the i. So what's the i of a beam rotating about its end. We just did this. Remember it was you figure the rotational inertia about the center. You add md^2. We just did it for the 5 kg mass at 6 m and we got 60 as the total I for that system. And if we divide the two, we're going to get that thing rotates at 1.25 25 radians/ second squared. So if something's moving, it's got kinetic energy. And if something's spinning around its center, it's got rotational kinetic energy. This is the formula you get on the formula sheet. Looks just like the formula for regular kinetic energy 12 mv ^2, which each with each variable replaced by its angular counterpart. So instead of mass, you put the angular version of mass, which is i. And instead of V, you put the angular version of V, which is omega, and you get the formula for rotational kinetic energy. This works with almost every formula. So remember the formula for work done. This is just a good rule to remember. Work done was F delta X. Well, let's just see if we replace force with its angular counterpart torque and we replace delta X with its angular counterpart delta theta, you get a new formula for work. And this one is also true and given to you on the formula sheet. So, it's good to know that you can usually translate regular formulas into their angular counterparts by just replacing the regular variables with their angular counterparts. But let's try to calculate it. Let's try to figure out what the rotational kinetic energy is of the cylinder rotating around a circle. So, we'll use this formula. We'll say it's 1/2* I. Yeah, like I don't know what I is. You wouldn't have to memorize it. The I of a cylinder would be given. It turns out it's 12 m r^ 2. So 1/2 * 3 * the radius here would be 2^ 2. But be careful. Notice there's a half just from the formula, but there's another fraction from the i. So often there will be an extra fraction here. Don't forget you'll have two fractions here. Don't leave one off on accident. And then times omega squar. The omega says it's 4 radians/s squared. And if you calculate all that, I'm pretty sure you get an k of 48 jewels and it's in regular jewels. It's not like some special units. 48 jewels of rotational kinetic energy. What if you got a mass that's orbiting? like this mass isn't spinning around its center, but it is sort of like, you know, rotationally doing something at an omega. How would you do this? Well, you could figure out either way. You could do 1 fi omega squ or 12 mv^2 because this thing does have a velocity in a certain direction. The center is moving. But let's just do it this way first. We'll use this formula first. We'll say equ= 12 m. What's the m of a point mass orbiting? That was mr 2. So it would be 3 * 2^ 2 and then time omega^ 2 which would be 4^ 2. So 12 I omega^ squar and I'm pretty sure with this one you get 96 jewels of energy. Uh kind of makes sense like 3 kg rotating around its center is going to have less rotational kinetic energy than 3 kg that's actually pushed out 2 mters away. If all the masses at the edge you get more rotational inertia. So this is going to have a more rotational kinetic energy. What if you were like dude I don't want to do 1/2x squ Why can't I just do 1/2 mv^2 here? Like this has a certain speed v. Why can't I just do one? You can't. You could do 1/2 mv square here. 1/2. The m is 3. The v though you got to be careful. What's v? Well, this is omega. This is r. If you remember v is row omega. So v is going to be r which is 2 * 4 and that gives us 8. So it' be time squared. And this also gives you 96 jewels. So you could do it either way. What you can't do is add these up. You can't be like, "Oo, I got 96 jewels from rotational with this guy and a 96 jewels from translational kinetic energy." And no, you have one or the other. How would you ever could you ever have both? Yes, you could. You could have both. Let's say the center of your object is moving with some velocity and you're rotating around your center. Well, now you've got rotation around the center, rotational kinetic energy, and the center of the object's moving. got translational kinetic energy, which is the name we give for this normal 1/2 m squ. Now, okay, so how would you figure out a problem like this? Well, let's say you wanted to know you start this thing at rest up here. You start a cylinder at rest up at the top of a ramp, height h, and you want to know what's the speed that's going to have all the way at the bottom. What what center of mass speed and what angular velocity for that matter? Well, the easiest way to do it is energy. If you ever have something start up high, go down low, and you want to know speed, energy is going to be the easiest way to do it. We'll do conservation of energy. We'll put everything in our system. E initial equals E final. So, what kind of energy do we start with? MGH. You're like, I don't know the M. That's okay. You've got to have faith. It'll cancel out. So, MGH. We don't know the M. We don't even know the R. You might be like, I can't even do this. You just got to you got to have faith. You got to go for mgh. This is going to turn into well, it's spinning. So, we've got 12 I and the I of a cylinder we just said is 12 m r 2. I don't know r. I know it'll cancel out. Watch. Something magical is going to happen. Omega^ 2. But down here we saw omega is time r is v. That means omega is v / r if we divide. So it means over here I could write I could write the omega here. I got 12 i. Omega is going to be v / r squared. So like that. Plus I've also got regular translational kinetic energy. So I'm going to have 12 m regular v ^2. And I can cancel a bunch out. This is why we didn't have to know the mass. Turns out the mass did not matter, which is kind of interesting. Really massive cylinder, you know, very small mass cylinder, they'll they'll tie at the bottom. It doesn't matter what the mass is. It's kind of interesting. Also, look at this. This R squ is going to cancel out with this. Or if I move these inside, if I square both of these and get rid of that, that R squ cancels with this R squ. Turns out the radius doesn't matter either. Really high radius cylinder versus really small radius cylinder doesn't matter. Radius, mass don't matter. So they're both going to tie at the bottom, which is kind of interesting. So what do we got left? We got g * h equals we've got a half of a half, which is a 4th of v^2. This did get squared plus a half of v^2. That's going to be 34s v ^2. And then if we solve this for v, we're going to get the square<unk> of 4/3 gh. That's what v is right here. That's what the v is. How would I get the omega? Well, V is omega or omega is V over R. So, if I just take this and I divide by R, omega is V over R. I could just take this divide by R and that would give me what my omega is. Do I need to know the R for that? Yes. So, now you would have to know what the R is. It'd have to give you what the R is to solve for the omega. But you wouldn't need that R to solve for what the V is, which is kind of interesting. So, just like something can have translational kinetic energy or rotational kinetic energy, something can have regular momentum or angular momentum. So the formula for regular momentum is P= MV. So then the formula for angular momentum which we represent with L is replace M with I replace V with omega and you get the formula for angular momentum. So this is going to be for an object like spinning around its center like this cylinder right here. here. So if we want to find the L of this cylinder, we can say the angular momentum would be the I, which for a cylinder again is 12 m, which the mass here is 5 R 2 * 2^ 2 that's the I* omega would be 6 and you get an L here of 60 and the units are going to be kilogram m^ squar/s. That'd be how many units of angular momentum this cylinder has. But that's not really all that interesting. Usually you have a collision of some sort. So if you ever have a collision where things are rotating, you know, that's going to be a key indicator that you're going to use conservation of angular momentum. When will angular momentum be conserved? Well, just like the formula for momentum says that you'll get a change in momentum if there's a force times time. Well, if you replace all these with their angular counterparts, you realize you'll get a change in angular momentum if there's a torque time. But that means if there's no external torque. So if the external torque is zero then you'll have no change in angular momentum that means angular momentum will be conserved if there's no external torque. So if you include everything in your system so that anything interacting with each other or in your system then there's no external torque. So for instance let's say we drop right on the edge of this while it's spinning. We drop just a 5 kilogram mass right on the edge. Okay let's do conservation of momentum angular momentum. Figure out how fast will this thing be spinning afterward. Obviously, it's going to be going slower. We've increased the I that means we decreased the omega. Like a ice skater bringing her arms in decreases the I increases omega. Well, we've increased the I and that means we've decreased the omega. So, we're going to do L initial equals L final. We know what L initial was. We just found it. This thing's spinning with 60 units of angular momentum. How much L do we have finally? I omega, but one big I, like one big total I. This thing stuck together. So it' be 12 * m is 5 * you know r 2 is going to be * 2^ 2 m r^ 2 is the formula for the rotational inertia of a cylinder plus we got this point mass on the edge. So a point mass orbiting all at a certain radius is going to be mr 2. So 5 * 2 ^ 2 and then times omega final. And if you solve this for omega final, you get omega final is 2 radians per second is how fast it's going to be spinning. So basically we tripled the I and we got 1/3 of the omega. So we got 1/3 of the omega because we increased the i by a factor of three. Um what if you got a point particle moving in a straight line? A lot of people think oh well then it can't have angular momentum because it's moving in a straight line but that's not true. It can in fact this is the formula to figure out how much angular momentum it has. This is given to you on the formula sheet. You're like how wait how can this point particle move in a straight line have angular momentum? Well think about it. If you say there's no angular momentum beforehand in this entire system, but then after this hits and sticks, this whole thing starts spinning. Well, afterward there's angular momentum, well then there must have been some beforehand because angular momentum would be conserved if we include everything in our system. So point particles moving in a straight line can have angular momentum. They have this much here. So you could do r, which is, you know, you could use this formula. R would be the full eight. You know how far you are from your axis. Your axis is over here. That's your axis. How far away you are, eight. So we do this. We could say has 8 * m is 2 * v is 6 is the velocity times the angle angle between r. r would be this. That would be r and the v which would be that. So if this is 60 and that's 90 then this up here would be 30. So we'd have to do times sin 30. But sin 30 is just a half. So 1/2 you get you actually can cut this r by four. get 4 m * 2 kg * 6 m/s. And you might be like, wait a minute. Wait, mvr, this r right here, this is the four. This is how close it'll ever get to this axis. You might wonder, can I always do that and then avoid the sign? Yes, if you want to, you could do m time the v times I'm going to do it as capital r. This is the closest the object will ever get. If you draw a straight line straight down and you ask, "Hey, what's the closest I'll ever get to my axis, that's the this r here." So you could do it like this. In fact, I like rewriting it Mr. V because then I can remember it as Mr. V. You find the angular momentum of a particle. Well, that's Mr. V. You go ask Mr. V about that. So let's try it here. Let's say this thing hits and sticks and this whole thing starts rotating and I want to know, well, how fast is this thing going to rotate afterward if this hits and sticks? So we'll do conservation of angular momentum. We'll put everything in our system. We'll say that L initial equals L final. L initial is just this amount right here, right? It's 4 * 2 * 6. So that's 8 * 6 is 48 units of angular momentum to start with. To end with, we have one big I. This whole thing is going to be spinning. So I'll have M R 2 because the point mass at the end here is M R 2 for its I. So we're doing I * omega. So 2 * 4^ 2 is 2 * 16. Uh, plus we have the eye of the rod about its end. And the eye of a rod about its end, you don't have to memorize it, but the eye of a rod about the end is 1/3 m L 2. So, time 16 times omega final. And if you add this all up, you get that, oh shoot, omega is going to just be 1 radian per second. So, didn't exactly give it a lot of angular momentum or sorry, angular velocity, but it did give it some, which means it had to have some to start. So, the angular momentum here was conserved. There's just as much angular momentum to end with as there was to start with. That means there had to be some to start with. So a point mass can have angular momentum, but you could just always find it using Mr. V. So springs will exert a force if you stretch or compress them away from their natural length. And the amount that they'll exert a force is given by Hook's law right here. This is on the formula sheet. Says the amount of force the spring is going to exert is negative the spring constant K. So the spring constant K is a number that tells you how many newtons of force the spring will exert per meter you stretch it. So if I pulled this out one meter, it would exert five newtons. And if I pulled it out 2 m, it will exert 10 newtons and so on. And then if you multiply that spring constant by the amount you've stretched it, delta x from its unstretched length, you'll get the spring force. At least if you have an ideal spring or sometimes called a linear spring given by this formula right here. You might be like, what's this negative sign all about? Well, think about it. If my displacement is positive, i.e. I pull this thing to the right, right? I've displaced it from equilibrium. So this delta x is not the length of the spring. It's how far you've stretched it from the unstretched length of the spring. So this would be 3 m right here. If I stretch it out with positive displacement, my force is going to be left. It's trying to bring me back to my natural length of the spring. So my force is negative. That's why this negative is here. It's negative of the direction you stretch the spring. And if I compress the spring with a negative displacement, well then negative times that negative displacement will give me a positive force cuz again the spring is trying to bring me back to its natural length. Similarly, if you hang a mass from it, say I hang a 2 kg mass from this spring. Well, its natural length is right here. That's where it wants to be. So if you stretch it out, the spring's like, uh-uh, no, I want to go back up there. I'm going to go up here. This is where my spring force is going to be. And I know that the kx upward, at least the magnitude of it upward. Usually just it's easiest to just find the magnitude of the force and then treat the sign according to which way it goes. So my spring force would be upward kx. My gravity would be downward. If this is 2 kg then I would have 20 newtons downward. That just means kx has to equal mg. In other words k which is 5 * x which is what I want to know. I want to know how far did this thing stretch right here equals 20. So this thing must have stretched 4 m. It's hanging down 4 m below its unstretched length right here. This would be 4 m right here. All right. So, what's this got to do with simple harmonic motion? Well, if you stretch this thing out and you let go, I mean, because of this, you know, opposing forces, the spring's going to oscillate. It's going to go back and forth and back and forth with a consistent period or frequency. The period is how long it takes the number of seconds. Periods and seconds. How number of seconds it takes to go through a full cycle. So, all the way back and forth. That'd be the time for one period. And if you take one over that number, you'd get the frequency, which means frequency is also equal to one over the period. These are related in this way. Frequency would have units of one over seconds. That's just in hertz. So the frequency is the number of times it does this in 1 second. It's the number of cycles in 1 second. And the period is the number of seconds in one cycle. So these are just inversely related. You could always turn one into the other. Say you wanted to find the period of a mass on a spring. How would you get it? you get a formula for it. So this is the formula for the period of a mass on a spring. You take 2 pi * root mass over k. So if we wanted to find it for this mass over here, it would be 2 pi * the square root. The mass on it is 2. So if I increase the mass, I get more period. It's more sluggish to movement. It's got more inertia. That makes sense. And if I increase the k value, I get less period. That means there's more force. More force means, you know, it's going to be moving faster. So it doesn't take as long to go through a full period. So that kind of makes sense. So I'd get 2 pi <unk>2 over5 would be the time it takes to go through a full cycle. And if I took one over that, I'd get the frequency if I wanted it. Similarly, you can get the period of a mass on a string like a pendulum. So it's just a pendulum. Uh turns out this does the mass of this bob here, this pendulum bob does not affect the period of a pendulum. The only thing that affects the period of a pendulum, the time it takes to go forward and back is the length of the pendulum and the g. Now we're assuming all the mass is down here at the end. So g is the gravity on Earth. So if I wanted to find the period of this mass on a string just rotating back and forth, I'd take 2 pi <unk>3 / 10. G on Earth is 10. If you increase the length, it takes longer to go through a full cycle. That makes sense. The rotational inertia has increased. So it's harder to rotate. Takes longer. And if you increase the G, again, just like K, you've increased the force. So this thing's going to move faster. It doesn't take as long to go through one full cycle. But let's say you wanted to find the speed. Let's say you're like, you know what? I want to know say I pull, you know, this hangs down at four. I pull it down an additional three. So this point right here, by the way, this is called the equilibrium position. This point here is when the forces are all nice and balanced, right? So 20 newtons of force up from the spring, 20 newtons of force down from gravity. It's at equilibrium. But if I displace it from equilibrium, an extra 3 m, well, now it's out of equilibrium. Down here, the force of gravity is too big. Or sorry, the spring force is too big. It's going to like shoot it back up. Gravity just stays 20. So the spring force like get back and it pulls up right here. Then we get to equilibrium, but guess what? It's moving with some speed here. So it's not just going to stop there. It's got inertia. It's got some velocity at equilibrium. It's going to shoot past that point. And now the spring's got too little force. Now gravity wins. Gravity pulls it down, slows it down, comes back down here again. You get to equilibrium, but now you got velocity the other way. And this thing oscillates forever. What if you want to figure out what this speed is? Well, there's a trick. You might be like, okay, well then I got to do like some, you know, there is a formula. There's a 1/2. Remember 12 kx^2 is the formula for the spring potential energy. But this x in here is how much the spring's been stretched. So you might be like, "Dang, dude. I got to put like 4 + 3 as the amount of stretch." And that's true. That would tell you how much spring potential energy there is. But there's a trick to find the velocity. You can pretend. So two wrongs, it turns out, makes a right here. You can pretend like all of your x's in this formula and this formula right here are measured from equilibrium and ignore the potential energy due to gravity entirely. And it turns out those two completely offset and you get the right answer. What am I talking about? Watch. Let's say we choose this as our initial point and this is our final point for energy conservation. I'll do E initial equals E final. Well, down here I'm going to say what do I got? I got spring potential energy. We got 12 k x^2 x should be 7. You know, by all rights, it should be seven, but I'm going to pretend like it's three. I'm going to measure all my x's from this equilibrium point. It's like, okay, that's wrong. But watch, there's going to be another wrong. Equals. It doesn't have any uh gravitational potential here, and it doesn't have any kinetic energy. Equals. I'm going to say has kinetic energy at the midpoint here. So, where I want to find the speed of the object. So, 1/2, mass is 2 unknown v ^2. Now, by all rights, I should have plus mgh because look, it raised up in the air. It raised up by three. But I'm going to pretend like that's not there either. These are the two wrongs. You neglect gravity entirely. And you pretend like all x's are measured from equilibrium. Those completely offset. And now when you solve for v, if you do this, you get like 4.7 m/s. That's exactly the right answer. You can do this too for like accelerations. If you just neglect the force and potential energy due to gravity and you just pretend like there's the spring force and spring potential energy there measured from equilibrium, those two effects offset entirely and those two wrongs will make a right. But you got to make both wrongs. You can't include gravity and not do this or do this and include gravity. You got to not include gravity and measure x from the equilibrium position to get these answers right. So say you wanted to graph the motion of the same scenario we had before. Well, first thing to do is probably figure out what the period is. The period for the formula is 2 pi roo<unk> mass, which is 2 over k, which is five. And it turns out you get about 4 seconds. So down here, we should go through a full cycle in 4 seconds. But where should I start? Well, if we call this y equals 0, this equilibrium point, if we call this y equals 0, then I started 3 m below it. So if I start my clock down here, when I let go, I'm going start down here at -3 m. And then I should go through a full cycle in four seconds. That means I go all the way up and all the way back down in four seconds. So back down here I should back down to four negative three at 4 seconds and then at 8 seconds I go through another full cycle. So it's going to look something like see if we can draw well. Go something like that and then like that and then like that and then like that and so that you have from trough to trough here would be exactly 4 seconds. That'd be the period. What if you wanted to write an equation down for this? Well, they give you these on the formula sheet. They're like, "Hey, you can use this one or you can use that one." So, which one would be best here? Well, I've got this looks more like cosine, but looks like negative cosine. So, I'd be like, well, shoot. My best equation down here, you know, you mess around with it how you need to. y is a function of time equals and I'm going to do y instead of x here. It doesn't matter. Vertical or horizontal is amplitude. My amplitude is three, but I started off down here. So, my amplitude is 3 m, but I need a negative cosine. Negative cosine starts down here. So negative cosine. So just so you remember if you forgot it here. Negative cosine starts down here. Negative cosine. Positive cosine starts up here. Positive cosine. Positive sign starts at zero and goes up. And then negative sign starts at zero and goes down. So that's negative sign. So these are all the different options you could do. But I had negative cosine time 2 pi. Okay fine. 2 pi times the frequency. You're like, "Dang, dude. I don't know the frequency." But you do. Period's four. So the frequency is one over the period. Frequency is one over the period. Period is one over the frequency. So I can just write 2 pi / so times 1 over 4. So 2 pi over 4 * t. You're like, what do I plug in for t? Nothing. That that is the variable. So just to be clear, what this function's giving you is you I can now plug in any time I want. So plug in 2 seconds or 8 seconds or 6 seconds. If I plugged in 6 seconds, this I plug in my six for t, this would spit out a number. I could tell you right now the number would spit out. it would spit out exactly positive3 because I know I'm going to be at exactly positive3 when that occurs. But I could theoretically plug in anything like pi for t. I could plug in 2. I could plug in anything I want now and this would give me the height of that mass as a function of time. So you just got to use the right cosine sign or negative cosine negative co negative sign. Use the right amplitude. Plug in your period. So another way to write this if you don't like thinking about it like this, you could do a sin 2 pi over the period because one over the frequency is the period time t and you get your function for the harmonic oscillator. So pressure is defined to be the force per area, the amount of force exerted divided by the area over which that force is exerted. So it's got units of newtons per meter squared because area is in meters squared. That's important. If they give you centimeters or something, you got to convert that thing to meters and square it. But we give this a name. Newtons per meter squar is named Pascals, named after blaze Pascal scientist. Okay. So, let's try to find out the pressure right here. Like imagine at the bottom of this thing, right? You're underneath it or something. And it's like exerting a pressure or it's exerting on the pressure on the the floor that it's sitting on. We want to know how much pressure is there in that zone right there. Well, we use this formula. The force was, you know, the force of gravity straight down. 3 * 10. So 30 newtons of gravitational force straight down divided by the area. Well, if this is a cube with sides 5 cm, I can't do 5 * 5. Got to convert this. So I got to do 0.05 squared. It's got to be in centimeters. And then I square it. And what I get out of this, if you do that calculation, you end up getting 12,000 Pascals, which sounds like a lot, but I mean, it's not necessarily. Let me show you it. Let's say you're underwater. This water can exert pressure, too, right? So, the weight of this mass exerted pressure on the surface below it. Well, if you're underwater, the weight of that water above you that has gravity. You might not think it does, but it does. That has a weight. That weight can push down on you. and it pushes down on you with a certain amount of pressure which you get from this. So this is the gauge pressure. It's effectively the force per area but just caused by a fluid. You know if you're submerged in a fluid to a depth H. This gives you the pressure that you would measure on like a meter. That's why it says gauge pressure that you would measure on like a measuring device underneath you. So let's try to figure out it here. What would I get? I get the pressure is row. Well for a water. Okay. So this is one thing you should know. So for water the density is 1,000. It's nice. 1,00 kg per meter cubed. So this has to be the density of the fluid. This is g on the planet you're on, which for us would be 10. So we say the density down or the density down here is 1,000 times the g is 10 time h was if I'm 5 m below the water, I get 50,000 pascals. So even more than a 3 kg mass, I get 50,000 pascals cuz I had more, you know, more weight above me from this water pascals if I'm underwater. Is that a lot? I mean, not even necessarily. Think about it. Like, you're sitting on Earth, right? You're like standing on Earth. There's a huge column of air above you. That air is a fluid. It has weight. It's weighing down on you. How much is it weighing down on you? Um, 100,000 Pascals. This is one with five zeros. So, the normal ambient just air pressure just pushing on us all the times. And it pushes in from all all sections. It doesn't just push down. It doesn't make this can of beans like fly straight down to the bottom. It tries to crush it from all directions. It's trying to crush you, but you've always been born and known this pressure. You've you've never not known this pressure, so it doesn't feel weird to you cuz you that's all you know. But yeah, 100,000 is the air pressure. Sometimes it's called P or PTM. And so what is this formula here? You get both these formulas. Well, this is up here is the total pressure. So this would be the total pressure exerted inside of a fluid. You take that gauge pressure due to the weight of the fluid, you know, the water that you're in. But then if you add in that air pressure from above, normally we don't do that. If you pump up your bike tires, the number you're getting is the gauge pressure. It's not adding 100,000, right? When your bike tires are flat, technically they're at 100,000 Pascals, but that's not what we mean when we talk about gauge pressure. So gauge pressure is just the extra that's not in addition to the air pressure. But if you add in that air pressure, if you add in 100,000, then you get the total pressure. Sometimes it's called the absolute pressure. This would be like the entire pressure in there if you're interested in finding that in Pascals. Speaking of Pascal, he came up with a principle that says that an increase in the pressure anywhere in an enclosed system, an increase in a pressure here, increases the pressure everywhere by that same amount. That just gets distributed everywhere in the system. How would we use that? Well, let's say I come down and push down over here with some F1 and I increase the pressure here. Well, that increases the pressure over here. And the pressure at this point has to be equal to the pressure at that point. How do I know that? Well, I know that the pressure at equal depths in a fluid that's not moving. Fluid at equal depths, the pressure at these equal depths has to be the same. And just to hammer that point home, say I got different shaped containers here, but they're all full of water and they're all the same height. If I ask you, how's the pressure in this shape at the bottom compared to the pressure here compared to the pressure here? Pressure is row GH. Row GH. At the same depth in a non-moving fluid, the pressure's got to be the same. You might be like, "Wait, shouldn't this one have more pressure? There's like more weight of water above it." Well, the pressure at the bottom is the same. The force will be bigger. Remember, pressure is force per area. If you solve for force, so don't conflate force and pressure. Force is related to pressure. Pressure times area. So, this has more area at the bottom. That means there's more force at the bottom, but it doesn't mean there's more pressure. the force per area at this point and at this point and at this point and at this point the force per area are all the same at equal depths. So over here I can say like well at equal depths here uh the for the pressure has to be the same. So that means pressure is f per area. So f_sub_1 over a1 has to equal f_sub_2 over a2. Those are the pressure here and the pressure there. So if I solve for f_sub_2 let's say I want to know what f_sub_2 is. If I increase f_sub_1, f_sub_2 is just going to be f_sub_1 * the ratio of a2 to a1. Well, that means well, if I make a2 really big, the area over here really big, and the area right here really small, I could exert a really small force f1, maybe 10 newtons. But if a2 is a thousand times bigger than a1, I'll get like 10,000 newtons over here. This is how you make a hydraulic lift. You could lift really heavy objects. Now, you have to push down a long way. might have to push down for like 10 meters here to get like 1 centimeter out over here, but it's still an advantage. You could turn a little force into a big force. And then we can use this principle of, you know, pressures the same at equal depths for a non-moving fluid to figure out this problem. Let's say you have a an oil of unknown density. Like I don't know what the density of this oil is. I want to figure it out. And it sits here in this configuration. This is water and this is oil. They don't mix. I want to figure out what's the density of this oil. We'll pick a point that's convenient. At this point, let's say I say at these points, the pressures have to be equal because they're at equal depths. Equal depths. Okay? So, I say the pressure at this point would be row of water * g * 4 cm. So, 0.04 equals the row of the oil. Row of the oil time g its depth is 8 cm. So, 08. And I cancel my g's. I know the density of water is a th00and. So if I solve for this density of the oil, I end up getting 500 kg per meter cubed is the density of the oil. And look, that makes sense. Look, the water was getting held up with, you know, the pressure down here was getting created by 4 cm of water. And we know the density of water is 1,000. Well, the pressure down here was getting created by 8 cm of oil. Okay, if the oil's column is two times bigger, then its density must be two times smaller to create the same pressure at this point. And then you can create a barometer out of this. This might not look like a barometer, but the idea is let's say this is open to the air right here. This is open to the air, but inside of here it's sealed and there's a vacuum. So there's no air in here. No atmospheric pressure. We got atmospheric pressure out here pushing down. Remember the PN the 10 the 5th. So that's 10 5th. So the pressure right here from this column has to be equal to the pressure right here that's open to the air. These are at equal depths. So I say 10 5th which is the pressure right here equals the pressure from this column of water be row gh the density of water is 1,000 g is 10 h is what we want to find out how high would this column be this would have a height of 10 m this is why you don't want to make a barometer out of water a water barometer would be 10 meters tall usually make it out of mercury mercury has a much bigger density a much smaller height you don't need a barometer that's 10 m tall but in other words like this column of water is getting held up just by the pressure, right? That pressure that we don't notice, that pressure from air pressure from the atmosphere, that's what's holding up this 10 m tall column. That's how it works when you drink with a straw, too. You vacate the air right here. And the what's happening is the air pressure from the atmosphere is pushing the water up the straw. You're not pulling the water. You're eliminating the force downward on the water right here. And the air pressure is pushing this water back up the straw. So because the pressure increases as you increase in depth below a fluid that means if you got a submerged object or even a partially submerged object the pressure is greater at the bottom than it is at the top because the lower you go the deeper you go the more the pressure but that means there should be a net force upward and there is tries to push you out of the fluid. It's called the buoyant force. The buoyant force is that force that makes you float. So what is this is the formula you get. What are these variables? Well row is density but density of what? Not the density of the object submerged. This has to be density of the fluid that's getting displaced. So if you're in water, it'd be the water. If you're in air, it'd be air. And this V similarly isn't necessarily the volume of the object. Could be, but this is specifically the volume of the displaced fluid. So if you're submerged in water, it'd be the volume of that water that you've displaced. What does that mean? Well, think about it. Let's say you put an object in a container and the container is completely full to the top with water. You you let this object float in here, it's going to push some of that water out. So, some of that water's out here. This would be the volume. We're talking about the volume of that displaced water. But think about it. These together, row, density of the fluid time volume of the fluid you've displaced. That's mass. Mass of what? Mass of the fluid displaced. But look about it. We got row * v. This right here is mass. Mass time g. That means this whole formula buoyant force is equal to mass of the fluid displaced time g. That's just the weight. So in other words, if you weighed this fluid displaced, you put on a scale, you get a weight, an FG of this fluid that you've displaced, that is exactly equal to the buoyant force. So a conceptual way to think about what's the buoyant force. It's the weight of the fluid that you've displaced. So what does that mean? Well, let's try a problem over here. Let's say you got a raft. It's 12 kg and the density of the raft is 200 kg per meter cubed and it's floating in water and maybe the side of it's like 5 centimeters. Let's want to figure out like, okay, how much volume of this fluid is displaced, right? It's floating and you got some amount of fluid here getting displaced by this raft. It would have got pushed over the side if this pool was, you know, perfectly full to the top. Well, here's what you cannot do. You can't be like, well, I got a density here. Density is 200. Density is mass per volume. And so, if I take 200, which is the density of the raft, equals the mass of the raft over V, and I solve for V, I'll get a volume. In fact, I'll get 0.06 m cubed. That is not the volume displaced. That's the volume of the entire raft. I want to know the volume of the fluid displace. So, I have to use this over here. I have to say that well, there's going to be some gravity on this object. It's going to be 12 * 10 is 120 Newtons of force downward. I'm going have some buoyant force upward. This buoyant force is going to be row. V of the fluid displaced time g. I just set these equal if it's floating. So, row of the fluid is a,000. I got a,000 for my density of water. That's the density of the fluid displaced, not the density of the object. And then I got to multiply by V of the displaced fluid. That's what I want to solve for. I don't know how much fluid is getting displaced. Time G equals M, which is 12 * G. That's why I got 120. I can cancel the G's if I want. Helps to write this out symbolically. If I solve this for V, I get 12 over a,000. I get 012 m cubed. That's how much fluid is actually getting displaced. That's how much water got displaced by this. It's only the part that's underwater that's getting displacing that fluid. What if you want a harder problem? What if you say, "Okay, well, now I want to put an unknown mass on top that does make this fully get submerged. How much mass should this be? If I want this raft, the same raft to get fully submerged." Well, now now the raft is fully submerged. You know, the volume of the water displaced, it's the entire object. So if an object's ever fully underwater, then the volume of the fluid displaced is equal to the volume of that entire object. You're displacing all of your volume worth in fluid now. So now I can say, well, okay, so I got gravity down. I got buoyant force up. Buoyant force up is going to be 1,000 the density of water time V * G. But the V now really is 006. I'm fully submerged. The volume displaced is my full volume of this raft, which is 006. We found it right here. Time G. Well, that upward buoyant force up has to be balanced by the forces of gravity down, which would be m, the unknown m * g. Plus, well, this raft is 12 kg. So, 12 kg * g. Again, I can cancel out the g's. If you solve this for m, you get that the amount of mass you can put on there to fully submerge. This would be 48 kg. So, fluids can be gases or liquids, but liquids are kind of a special fluid because they're incompressible. Why does that matter? Well, if I try to pump a certain amount of volume into this pipe of a liquid, that same volume better come out the other side in the same amount of time or otherwise you'd be like compactifying it into this region. And that's impossible because it's incompressible. So, the volume flow rate, it's called, is going to stay the same everywhere in a fluid or a moving liquid. So, the volume flow rate here and here and here has to be the same because you can't be compressing the liquid's volume. And so, what does that mean? Like, where's this where's this formula come from? This is the continuity equation. You get this on the formula sheet. This is really the volume per time. You might not realize it. You're like, "This ain't volume per time. That's area times velocity." Well, look, think about it. Volume per time. What is volume? Well, if I take this cross-sectional area here, that A, and multiply by a certain length, so like L1. Well, then I get the volume of this little cylinder. You know what I mean? So, this this volume one I can think of as I'll do volume 2 over T2. volume one I can think of as area 1 times length 1 / T1 and then similarly up here I imagine this little area imagine a length L2 so up there volume 2 would be area 2 * length 2 gives me the volume of this cylinder up here divided by T2 well look 1 / T1 that's just how fast the water is going that much distance over this much time l1 over T1's V1 and A1's just A1 so this really the volume per time. This is giving you A1 V1 is the volume per time, the volume flow rate. In other words, at 0.1 and A2 V2 really is the volume per time flowing at 0 2 where these areas are the these cross-sectional areas. So let's say you wanted to figure out, okay, fine. If I'm going 6 m/s of this region down here, then how fast is the liquid moving up here where it's only got a cross-sectional area radius of 2 cm. So let's use this formula. We'll say A1, we'll pick one down here. Well, the area cross-sectional area of a circle is pi * r. And you got to use meters here if you want to get this thing right. 0.04 squared. P<unk> r² is the area of a circle. So 004 2 for the radius squar * v1 would be * 6 equ= m/s equals well up here we'd have p<unk> * 002 cuz the radius is only 2 cm. So 002 ^2 * unknown v and if you cancel cancel cancel solve this for v you get not surprisingly 24 m/s. Why is that not surprising? Well up here it's got to be going faster. I've reduced the area. If you decrease the area the speed of the water comes out faster. That's why you connect a little hose you know nozzle on your hose and the speed goes way up because your area went down. And look if I cut my radius by two or by half, you know, I get half the radius. Well, you square that radius to get the area. That means I have a fourth of the area. Well, if you've if you've cut the area by 1/4, and this equation's got to remain the same, then you need four times the speed in order to make this work out. And I started with six. That means this had to be 24 at the top. Okay, so pressure equals row GH is a great way to find the pressure in a fluid that's at rest. This only works if you're at rest. If you're moving, if your fluid's flowing and every liquid's flowing, then this is not going to give you an accurate pressure. You got to use something fancier. The thing you got to use is Bernoli's equation. This monster right here. This is on the formula sheet. How do you use Bernoli's equation? You pick two points. You pick 0.1. You know, we'll pick 0.1 down here at this bottom of the pipe. You pick 2 up here at the top of the pipe. And you say that all of these things added up at 0.1 have to equal all of them at 0 2. Does that sound familiar? Yeah. Sounds like conservation of energy because this is derived from conservation of energy. I'm not going to show you the derivation, but this is where you get it. So, what does this stuff mean? Well, P1 would be the pressure down here at 0.1. Ro G Y1. This is like ro g h but instead of h being the depth y is the height. So you know this is like if I call this h equals 0 or y equals 0 then the y at 0.1 is just going to be zero plus 12 row v1^2 the speed of the fluid down there. Row is the density of the fluid that is flowing equals pressure up here at 2 plus row g y2. So that would be 8. Now y2 has a height of 8 plus 12 row v2. So say I wanted to know all right what is the pressure down here like what exactly is the pressure down here at this point at P1 and let's say over here this is open to the atmosphere the water's just flowing out into the open atmosphere well let's use Bernie's equation to figure out what is P1 so I'd use this I'd say that P1 plus we just said this is zero zero because we're setting the bottom as H equals 0 or Y equals 0 plus 12 density of the water let's say we have water here so density of a th00and * V1 would be 6^ squared. So that's my V1^2 equals P2. Well, what's the P out here? If it's open to the atmosphere, I can either say it's zero and then I'd be using gauge pressure. I'd say the gauge pressure out here is zero. So I can say it's atmospheric pressure 10 the 5ifth, but then I'd be solving down for P1 as the total pressure. So let's try to solve for P1 as the gauge pressure. What we're finding here is the gauge pressure. And I know that cuz I'm setting the the pressure out here is zero, which means the gauge pressure is zero. That means when I solve for this P over here, I get the absolute. If I plugged in 10 to the 5if 5th here, well then I'd be solving for the total pressure down here at P1. You can do it either way. Just got to be consistent. So plus row is a,000. G is 10. Y2. Well, this is all the way up here at 8.2 is 8 m above where we called y equals 0 plus 12,000. And you're like, well, wait, they didn't give us the velocity here. But remember, we solved for it. And this is important. I don't care if this is 8 meters. It didn't matter. I don't care what's going on. The volume flow rate has to be respected. The volume per time has to be constant throughout this material. And so remember, we solved for it with a 4 cm 6 m/s speed. This was coming out at 24 m/s. We solved for that a second ago. You have 1/4 of the area. You have to have four times the speed. That just has to be the case. So I know this is 24. And it's important that regardless of what this height is, like this has to be 24. volume flow rate has to be respected. So now I got my formula. I just solve for P1. This is going to be the gauge pressure at P1. And if you do this, if you solve for P1, you end up getting what? 350,000 Pascals is what the pressure has to be down here to maintain the situation. If I plugged in 10 5th or I could just do it right now. I could just convert. I could just say, well then P total, right? I could just add 10 to the 5th to this. So if I want the total pressure down here, so the total pressure would be add a 10 to the 5th and you get 450,000 Pascal. So that's that's how you can use Berni's equation. What if you've got a moving fluid, but there's no significant difference in the height. In other words, if you had Bernoli's equation, you'd have a row G Y1 and you'd have a row G Y2, but they're the same on both sides. You could just cancel them. What happens then? Well, you get a simplified Bernoli's equation. And this is often referred to as Bernoli's principle. And think about it. What it says is that if you increase the speed like this equation's got to be satisfied. If you increase the speed at one spot, the pressure over there's got to decrease so that this thing is equal. Or conversely, you decrease the speed. You got to increase the pressure. What does this mean? Well, think about it over here. We got this flowing liquid right here. I com, you know, I make the area get smaller. That means this has to come out with a much higher velocity. If the velocity over here increases at V2, then the pressure has to decrease. I'd have less pressure at 0 2 than I do at 0.1. This is counterintuitive. People think, wait, more pressure, more velocity, right? No. Bernoli's principle says more velocity, less pressure in a continuously moving fluid. Like, what does that mean? Well, imagine I had this case over here. Right here, the the water is moving fastest. So, that means it would have the least amount of pressure. It would only be able to hold up maybe a small column. Over here, I've got the next fastest. It would be able to hold up a slightly higher column. And then over here, this would be able to hold up the highest column because the speed is the smallest. So where the speed is the smallest, you get the most pressure. And where the speed is the biggest, you get the least pressure. Is there any real life applications? Yeah. Imagine you blow air over a piece of paper. Well, the velocity's gone up. We've increased the velocity. That means we've decreased the pressure. Inversely related. If we've decreased the pressure above the paper, the pressure below the paper stayed the same. So this paper would fly up. Like you'd get this paper to move up into the stream of moving air. It's counterintuitive. Or say you like blow with a straw between two cans. You should try two empty soda cans. Blow with a straw straight between them. Well, you got fast moving air here. Fastmoving air means low pressure air. But that means the pressure from the sides here stay the same. They get pushed together. You might think they get pushed apart, but these cans actually get pushed together because by increasing the velocity, you've decreased the pressure between them. They get pushed apart. You can hold up a ping-pong p ball in a flowing, you know, hair dryer stream of air. The pressure in this region is smaller because the velocity is faster. Smaller pressure there, but larger pressure outside keeps it from flying out. It's weird. Looks like a magic trick. It just stays in that stream of air. And your roof get blown off if you got, think about it, really high winds out here. Well, really high winds means the pressure is very small, but the pressure inside your house could still be very big because that's, you know, normal atmospheric pressure. and roofs could get blown off or something from this Berni's principle, which is the idea that as the speed of a fluid goes up, the pressure is going to go down. Okay, last one. It's called Torichelli's theorem. It's just an application of Berni's equation. So, bro really just solved one problem with Berni's equation and got a theorem named after himself. I don't know, he must had a really good PR agent or something. So, what he did is this. He considered this problem. Let's say I got a container that's open at the top. This doesn't have a lid on it. This is open to the air, right? So there's just open air up here and it's so atmospheric pressure at the top. But you cut a little hole in the side and this causes water to flow out and it flows out at a certain speed. Let's say you want to know, well shoot, what speed does that water flow out? And that's that's cool to know. And you got to assume this hole is really small. I'll show you why in a minute. Well, let's say the, you know, the height below the top of the container is five and the height below the bottom of the container is two. And you want to use Bernoli's equation to figure it out. Let's do it. You got to choose two points. So if you want to use Bernoli's equation, pick two points. I'm going to pick this point up here at the top. I'm going to pick this point right here. You know, you can consider it right before it flies out or basically right after just right at the hole there. Basically 0 2. And we'll do Bernoli's equation. Okay. 0.1. What's the pressure up here? Well, it's zero. If I'm doing gauge pressure because it's open to the atmosphere. You could put atmospheric pressure if you want. Doesn't matter, but it's going to be zero or 10th. Either way works out. Plus row is the density of the water. And I'll we'll do it symbolically first. So, row density of the water time gives the height. Now, what what should y1 be? Oh, I'm going to call this h equals 0. So, this this region right here is h equals 0. So, this would be 5 m. It would be this height here above. I'm going to call this h. So, I'll call this h. The height above the hole to the top of the container plus one half row v1. Well, what's the speed of the water up here? If this hole is small, if we've cut a really small hole, the volume flow rate is small, and if this area up here is much bigger than the area down there, the rate at which this is decreasing is so small, we're going to say it's insignificant. Is it zero? No. Is it close enough to zero to ignore? Well, according to Torchelli in the AP exam, yes, it is. So, we're going to say that's zero equals P2. What's the pressure right here? Well, this is also open to the atmosphere. Look, like this hole is open to the atmosphere. That means it has zero for a gauge pressure. Um, you could you could do 10 to the 5ifth both of these, but it doesn't matter. They would both cancel either way. Plus row g y2. What's y2 right here? Well, it's zero. We said that this is h equals zero. So that's 0 plus 12 row density of the water* v at the whole squared. And look it, we just get we just get row g= 12 row v ^2. That should look really familiar. That looks like mgh= 12 mv². And because the rows cancel, it wouldn't have mattered. It is equivalent to mgh= 12 mv ^2. So this is the theorem. It says that the speed that this water is going to be coming out of this hole <unk>2 gh is exactly the speed you would have got if you would have just taken a little piece of water right here dropped it from a height h and checked hey when it got back down to this point how fast is it going that way? Well it's going to be coming out of this hole the exact same speed that it would have dropped this way. That's not a coincidence. Bernoli's equation just comes from conservation of energy. So it's not really that surprising that that this is going to be the answer. But this is Torchelli's theorem again. Bro must have had bro got so happy he solved one problem. He got a theorem named after him. He just says that you know the speed that a water's being coming out of the hole in the side is going to be 2g. How deep you are. So in other words, this would be root 2 * g is 10 * the h is 5. This is the depth below the top would be five and you get this is going to be 10 m per second. Well, if you made it this far, congratulations. That's basically every single topic you might see on the AP Physics 1 exam. I hope this helped. I wish you the best of luck.