Transcript for:
AP Physics C Overview

We're about to review every single topic of AP Physics C mechanics. This is the only video you're going to need to study for your exam. We're going to go through kinematics, force and translational dynamics, work and power, linear momentum, torque and rotation, energy and momentum of rotating systems, and oscillations. Let's get started. Today we are going to be dealing with unit one of AP Physics C, that is kinematics. We've got several topics on our list for today. And we are going to make sure these are all perfectly aligned to College Board standards. So the first thing that we need to discuss is scalers and vectors. And the difference between these is that a scalar has just magnitude while a vector has magnitude and direction. So a scaler you could just represent as a dot whereas a vector would have to be an arrow with a direction pointing this way in this case, right? And the magnitude being essentially from here to here. A scalar is just a number. Um, but a vector has to be represented with this arrow. A couple of examples of vectors are displacement, velocity, and acceleration, which we'll get to in a second. The first thing that we need to deal with is vector addition. Vector addition can be represented by simply drawing the arrows and then adding them together, or by adding their components together. So, let's take a look at the graphical approach first. Let's say we have a set of axes. We have one vector going here and then another vector like this. And we're given that from here to here, let's say this is three, this is six. I know the scale isn't perfect. And this is two. And then this is three here on the x axis and y axis. Okay. How do we add vectors together? Well, it's very simple. We just add their x components and then we add their y components. So here the first x component is simply a two and the second x component is an additional one. Right? So we have 2 + 1 = 3 in our x direction and then in our y direction we have three here and then another three here. So we get 3 + 3 = 6. So that means our total vector or our resultant vector which essentially can be drawn as a line like this right from the original point to the end point. That's a resultant vector right there has a magnitude of three in the x direction and six in the y direction. If it also had it was in 3D there would also be a z direction but we're not going to deal with that for much of physics c especially the mechanics portion. Now if we have these labels x y and z there's another way that we can write this that's becoming increasingly popular in the AP physics mechanics exam and that is JK notation. So we have an I, a J with a hat and a K with a hat. Okay, we can also just say these as I hat, Jhat, and K hat. These are the exact same thing as X, Y, and Z. Okay, so how we would write this vector, this final vector here is we would say 3 I hat 6 J hat plus 0 K hat because there's no Z component, right? So it's just a zero. And that is our vector. We'll call the vector V. A lot of the time we will also have a little arrow over the name of a vector. Another way that we could write the magnitude of this vector as just one number is by using the Pythagorean theorem, right? Which takes the side here, we'll call this one b^2. We'll take this side here a squ. And then we'll try to find this length here c^2, right? So in this case we would have three here. So 3^2 + 6^2 right is equal to our resultant vector c ^2. We take the square root of both sides and then we would eventually get that the of 45 is c. Another way that we could write this is by indicating the angle. So if we have a vector let's say have a different vector that's right here. We could say and let's say this vector is 5 m long and this angle is 45 degrees. We could say that this is a vector that is 5 m long and it is 45° u south of west. And that's pretty much all we need to know for scalers and vectors. Now we're going to move on to real kinematics. Okay. AP Physics C uses the object model which indicates that when we have an object, let's say it's just like a box. We have a box, it is represented as a single point in space. So obviously when we have like a box, it's made up of billions and billions of molecules, but we're just going to assume that's like kind of like one big box. And generally we can just treat this as a point mass. Okay. When this object moves, we can represent its displacement. Its displacement as delta x and delta x is simply its initial x position. We'll call this x0 compared to its final exposed position. We'll call this x. So it's x - x0. Right? Because if we've already traveled this distance of x0, we simply needed to subtract x0 from our final placement. Right? So that is our displacement. Now what is the difference between displacement and distance traveled? So let's say I have a little dot and let's say it's a person and this person is going to kind of walk a little winding path throughout space and the person ends up right here. Displacement is how much the person has moved in total essentially. So that's if we could draw a vector from the first spot to the second spot. That is our displacement. That is the resultant vector. Remember I used that word before. Between the first place, let's call this I initial and the second place. Let's call this I final. Okay, that is displacement. Now, distance traveled is a scalar and a and distance traveled is essentially this whole long thing. So, let's say like this whole path here, this trail was like 20 m. But our displacement here is just this orange line. Let's say that's maybe 15 m. And this also has to have a direction. Let's say it's 15 m um 40 and it's 45° uh north from east. Okay, now as we've kind of discussed a little bit more of our vectors and things, we're going to move on to position, velocity, and acceleration. X represents position, right? A position function shows where an object is at any time. Velocity is the derivative of position, and it's basically just how fast you're going. And then acceleration shows essentially how fast you are speeding up or slowing down. Now you'll notice that each of these should have a little vector arrow on the top of them. Each of these is the derivative of the previous one. So acceleration is the derivative of velocity and velocity is the derivative of position. Acceleration therefore is also the double derivative of the position function. You'll find these equations as numbers four and five on your reference sheet because remember when we have the derivative let's say this velocity is the derivative of position right that means that position is the integral of velocity and that means that the that velocity is the integral of acceleration. There are a few more and we call them jerk which is essentially the uh derivative of acceleration and the next one is snap and then we have crackle and then we have pop. These should not be accessed on your AP exam which is kind of a fun thing to know. So now given this information we can write out the following equations. We can also say that x of t is equal to the integral of velocity which is equal to the double integral of acceleration. This also applies in graphical matters. So let's say I had a graph here. We have a basically just a quadratic function. This is x and this is t. If we took the velocity for that, if we took the derivative of, you know, let's just say this graph is x^2, then we'd obviously just get 2x. And we can just draw that as a straight line. And then we take the derivative again, we just get two, right? And this would be our acceleration graph. All right. And now we're going to get to equations 1 to three on your reference sheet. And these are the big three kinematics equations. No, this is not set for kinetic energy. These can be used for onedimensional motion analysis. In other words, you can only use these when you have a velocity and acceleration whatever in one direction. So I couldn't combine, for example, if I knew my velocity in the y direction and I knew my acceleration in the x direction, I have to use either just x direction or just y direction at one time. That does not mean that this equation cannot be completely switched to just be only for the y direction, right? By change of subscripts. This is just how it's written on your reference tables with all x's. One helpful way to way to write this second equation is just how we showed you with displacement earlier. If we have delta x is equal to vx kn * time + 12 ax t ^2 because again if we have x - x0 it's just equal to delta x. We're also going to briefly discuss what units you should be using for velocity, acceleration and position. Velocity is usually written with me/s. Acceleration is usually written with me/s squared. And of course, position is just meters. One other note is that often your acceleration value will be equal to little g because of the force of gravity, which is equal to approximately 9.8 m/s squared. On your exam, you can also use g= 10 m/s squared. But for laboratory experiments, 9.81 or 9.8165 or 9.8 eight will all give you better values um and better data. So, I just have a quick practice problem here for your big three kinematics equations. I would recommend taking a second to quickly solve this. Put your answer in the comments. Uh the pinned comment will have the answers to this problem. Okay, back to content. So, we're going to do this next topic, which is reference frames with a example question. That's the best way to teach this really quickly. Okay, so we have a car, a bird, and a UFO that are traveling along a flat earth in various vector directions and in various speeds. So we have a car, a bird and a UFO on a flat earth traveling at various speeds and in various directions. I want you to treat this direction as the positive direction. Okay, so previously when we had a set of axes, right, we used that graphical representation and we said that this was the positive direction, right? This was the positive direction as is generally considered the positive direction. Now we're going to consider this direction to be the positive direction. So that means that as we go left, we're going more positive. As we go down, we're going more positive. Now we can treat this essentially as a one-dimensional example because we're only going this way. We're not going this way at all. We're not going this way at all. Right? We're only going left and right. Okay? So let's take a look at the problem at this point. So, I'm going to give you a quick equation that you can use to solve these three problems, which I highly recommend you do. And that is the velocity of the object relative to a given reference frame is equal to the velocity of the object minus the velocity of the frame. And just given that equation, see if you can solve these three questions. Okay, now we're going to solve them together. So, what is the velocity of the bird relative to the car? All right, let's take a look at this. So we have this is the positive direction. So that means the car is going 60 60 m in the positive direction. Okay. However, the bird is going 20 m/s in the negative direction, right? So it's essentially - 20. Now we know that our reference frame is the car, right? Cuz we're going relative to the car. Okay. So our reference frame is going to be 60, right? And it's going to be negative because it's going to be minus 60 because this is positive. you know, we're just using this subtraction symbol that carries over, right? The velocity of the object though is -20, right? Because we're going 20 m/s, but we're going in the negative direction because we defined left to the positive direction. So, that means that, you know, -20 - 60 is going to be -80 m/s. So, we have a UFO, right? We're going relative to the UFO now. So, the velocity of the object is still -20, right? Still 20 m/s in the negative direction. But the velocity of the UFO is 7,000 m/s also in the negative direction. So it's -7,000. We're subtracting that value. And so what we get from that is we get -6980 m/ second, right? -6980 m/s. And of course, relative to the ground, the ground is supposed to just not be moving at all, right? So that would just be the reference frames velocity is zero. So we just get -20 m/s for that one. Now, what does this all mean? Well, that means that the bird relative to the car is essentially going backward even faster than the car is moving forward, right? Or relative to the UFO. Sure, they're still going the same direction, but it seems that the bird is almost going backward because of how slow it's going, right? Compared to the UFO. So the people in the UFO, right, if I had like a person who's or I guess an alien who's in the UFO, it seems to the alien as though the bird is actually going backward. Because as a person in the UFO, right, it seems as if you're standing still. Think about if you're in a car on the highway and you're going, you know, 70 mph, it seems as though the ground is almost going beneath you in the reverse direction, right? That's kind of what's happening here is it looks like the bird is actually going backwards because of how much faster the UFO is going in comparison to the bird. even though they're still going the same direction. Okay, one last important thing to know is that acceleration is constant and the same when is viewed from different reference frames. Okay, so if this UFO is accelerating, the bird notices that it's accelerating. I don't know how it would, but let's say it does, it's still going to notice it at the same rate as the car notices it. That's just an important detail to know. And that is going to conclude our discussion of reference frames. All right. Now, we're going to continue with the discussion of projectile motion. Projectile motion is a special case of 2D motion. And what makes it special, right, is that our acceleration in the y direction is always negative g or 9.8 m/s. And our acceleration in the x direction is always zero. Right? Now it is true that this g might be different. Let's say the question said, "Oh, you're on a different planet where g or where the acceleration due to gravity is 20 m/s." In that case, you would use that value. But generally in Earth scenarios, it's going to just be g. And there are two main scenarios that we always have to deal with when we're thinking about projectile motion. The first one is that we have an object that we are launching and it just kind of goes like this in a little arc path, right? We're just launching it from the ground. It's going to launch and it's going to end up somewhere at the same y value, the same height of zero or whatever this y value is. And it's going to land at the same height just in a different spot on the x- axis, right? For this scenario, we have a launching velocity, right? In kind of just like this diagonal direction or whatever this theta is here, it's just going to go in that direction. We're just going to call that vector v0. to find the velocity in the x direction specifically or the velocity in the y direction specifically which again we need to do to figure out how to use those big three kinematics equations because we can't just use it in any direction right we just simply take v 0 and we for the x direction we just use vx is equal to v kn * cos theta again that's just a theta here and similarly for the y direction we're just going to use v y is equal to v kn * sin sin theta. And so as we look at various paths along this arc, let's say the object is right here, we're going to have our vy up, our vy up, and then our vx to the right still. Okay, so now when we get to the peak of the parabola, right, this is at time t = 1/2 where t is basically just when it goes from here, but this is a t 0 equals z. And this is the final thing. This is when we're just halfway through, right? Because the parabola is symmetrical. There is only vx here and it's the same constant vx because again ax equals zero which means that vx has to be completely constant throughout the whole thing. Right? Right. So let's take a look at that first kinematics equation again. Right? We have vx is equal to vx plus a kn or plus axt. Well ax is zero. So that means that t doesn't matter. So vx is always equal to vx. Right? So we have the same vx but vy is changing because obviously we have our acceleration which in this case would be g. this here is going to just be zero, right? Cuz we're not going up at this point. We're not going down at this point. Um, so as we keep getting closer to this point, the velocity in the y direction vector is going to get smaller and smaller and smaller as soon as it approaches zero. And then it's going to get bigger, bigger, bigger, bigger, bigger as the um object keeps traveling down. So that's kind of our first little scenario here is that we have an object, we launch it and then it goes to um essentially the same y value. So the y displacement is equal to zero. Um but the x displacement is equal to some constant. So whatever this value is here. Now our second scenario here is let's call this scenario B. We have a our axis again here. We are launching although from although we are launching from an elevated position. So we're already at some height. Let's just call this height y. We're going to launch the projectile here with some velocity v kn. Um and there is no initial velocity in the y direction. But as soon as we start getting this kind of right we're just going to get this kind of trajectory and it's just going to kind of fall down towards the waters. Right? And so again, this is the same scenario. We have um the acceleration in the y direction is equal to g. The acceleration in the x direction is equal to zero. And it just kind of falls down to the earth. Now, in both of these scenarios, we're going to get the same little table that we can make. And I always recommend making this table for projectile motion problems. Let's say we have this little T-chart. We have the x and the y direction. We know that v in the x direction is equal to v cos theta. And again, this only works for um the zero for the kn for at t equals z for the y direction. It does work for all vxs because you do not have that acceleration in the in the x direction. But for v y we only know v y um based on this because v y changes throughout the projectile motion. Okay. Now ax is equal to zero. And we know that a y is usually equal to negative g, right? The thing that links these two is time. So time is always equal to time. And we use these, we basically can just plug all this stuff into various kinematics equations to kind of get what you need to know. And you can also use delta x, you can use deltay, um, whatever you need to do to solve those problems. So, just to finish off our video here, we have several practice problems that are equations that might be useful for you to have in your calculator for the AP exam. Again, just because you have Desmos does not mean that you cannot bring in a graphing calculator. Your graphing calculator will not be wiped. So, you can put basically whatever you want in it. These are just some practice problems that result in equations that might be useful. For example, if you want to find this vertical displacement launch with some velocity, right? If you have a multiple choice question where you don't have to show your work if you have this an equation for this in your calculator already, you can pretty much solve that equation like 10 15 seconds easily by just plugging in numbers, right? So definitely make sure that you take a stab at these problems and put them in your calculator. Again, answers will be in the pinned comment below. Unit two of AP Physics C force and translational dynamics. We're going to discuss center of mass, forces and free body diagrams, Newton's three laws and various types of forces. So the first thing that we're going to understand is center of mass. And there are three scenarios that we need to understand for this unit. Firstly, if we have a symmetrical mass distribution, it's very easy to determine center of mass. Let's say I just kind of have a box, right? A two-dimensional box. We are going to understand that our mass distribution that our COM is going to lie on the axes of symmetry. So let's say I have this as my y direction and then this as my x direction. If I wanted to find the center of mass in the y direction, I would simply kind of draw an axis wherever I wanted to just find the center in the y direction, find the center in the x direction, and that is my center of mass is right there in the middle. We could extend this to three dimensions, uh, whatever you really wanted to. But if we have a symmetrical mass distribution, we just know it's going to be halfway in between the end of the box and the start, right? The second thing that we want to know, number two, is if we have a system of masses. If we have a system of masses, there are two equations that we want to keep track of. The first one is that the center of mass in the x direction is equal to the sum of the masses* x over the sum of the masses. And that in the y direction, it's going to be equal to the sum of the masses* y over the sum of the masses. Again, let's take a quick example. If we have our set of axes here and we have one point that's right here at let's say this point is -3 0 and we have another point at let's say this is 6 3. Um obviously we need to assign some masses. Let's call this one 4 kg. Let's call this one 2 kg. Um let's see if we can find using these equations. Let's see if we can find the center of mass of this system. Okay. So what we're going to do to find firstly let's find the xcom. So if we want xcom we're going to get 4 * 6 right because of this guy right here 6 * 4 right here plus we have -3 and 2. So -3 * 2 over here over some of our masses which is obviously 4 + 2 and that is going to give us 3 and that is going to give us three. Now our sum in the y direction is going to be 3 * 4 + 0 * 2 over again 4 + 2 which is going to give us 2. So using this x coordinate and using this ycoordinate we can see that the center of mass of our system is at 3 comma 2. So using our three over here, our two over here, we can say that the center of mass is at x= 3 and y = 2. Okay. So our third scenario here is if we have a nonuniform density, right? It's not a symmetrical mass distribution. And if that's the case, then we can use the equation r is equal to the integral of the rs * dm over the integral of the dms. And you'll notice this is very very similar to this equation that we had up here. We've just replaced our summation signals, our sigas with integrals. Okay, so I'm just going to draw a quick onedimensional rod. We're going to assume that this direction is kind of just negligible and we're going to have the r direction going this way. Now in these scenarios, what we're going to have is we're going to have a lambda. And we can set lambda equal to m / l which is also equal to d mm / dr. Right? Now what we can do here is we can make a substitution. Right? We can set lambda is equal to dm dr. Therefore dr lambda is equal to dm. And then we can replace these dms with this substitution. So what we're going to get then is we're going to get r comm is equal to the integral of r lambda dr r over the integral of dr r lambda. In a lot of cases you may also get an additional equation for lambda. The problem may state that lambda is equal to something like beta r cubed. In that case you would simply substitute this in and solve the integral. And that's all you need to know for center of mass for AP physics C. Okay. And now we're going to discuss Newton's laws. The first of which states that an object at rest will stay at rest and that an object at a constant speed will stay at that speed. Both of these assuming that no forces act upon this object. So let's say I have an imaginary particle traveling through space and it's far away from any planets or black holes. So there's no gravitational forces that can be exerted on it. So it's kind of just coasting through space and let's say it's going 300 kilometers per hour, right? That particle is going to continue moving at 300 kilometers an hour through space unless something happens to it. Right? Unless a black hole sucks it in, unless a planet pulls it into its gravitational field, etc., etc., right? That particle will continue moving until something happens, until a force acts on it and causes acceleration. Conversely, if I have a particle in space that's just kind of sitting there at 0 kilometers per hour, it's also not going to move until something acts upon it, until a force acts upon it. All right, let's take a look at Newton's second law, which is very simple. The sum of the forces is equal to ma. Hopefully, at some point in your study of physics, you've already seen this equation. Um, that's pretty much all it is. Very simple. And the third one states that whenever we have a force B on A, it is equal to the negative force of A on B. So if I push somebody, there's an equal and opposite force exerted in the opposite direction. Right? An equal and opposite force. For every action, there is an equal and opposite reaction. Cannot stress that enough. Okay. So the first force that we're going to formally discuss in this video is FG. We know from our reference table that FG is equal to G M1 M2 / R2. FG always acts on the center of mass and it points towards the line connecting the COM of the two systems. So let's say I have a little imaginary planet and I have a little little person standing on top of it. The FG is going to point down towards this center of the mass of the planet right here. Right? And that is the FG of the earth acting on the person. Of course, every object exerts some gravitational force, but we know that in most cases it's negligible. Right? I in my chair am not exerting a great gravitational force on the earth. Another thing that we should note is that if we have ma right which is the sum of the forces from our Newton second law is equal to g m r 2. Again, we can just write this m1 m2 as little m big m if we want to. Then we can cross out little m and we can get that a is equal to g m / r squared when gravitational force is the only force acting on a person. This is also known as little g and it usually has a value of 9.8 m/s squared or 10 m/s squared. You can also use 9.8165 if you would like to. Um, whatever works. This is also often referred to as weight. In later units, we will deal a little bit more with gravitational force, but that's all you really need to know for now. The second force that we're going to keep track of is FN, which is known as the normal force or the apparent weight. The apparent weight, it is always perpendicular to surface that the force is exerted on and it's often equal to mg. Let's say I have a box on a plane that cannot be pushed through. All right, we have mg is always going down and then we have FN is actually pushing up, right? If we draw our little free body diagram here and in this case, the magnitude of FG is equal to the magnitude of FN. However, if I was to have a inclined surface like this one and the box was kind of on here, right? Then mg again is going to be straight down, right? However, FN will always point perpendicular. So, it's still going to be this way. In this case, we would have to say if we drew our free body diagram, kind of rotate it a little bit, get something like this, right? We could say that FN is equal to mg sin theta, right? because we've established that FG is equal to G MM / R2 which is also equal to M G. All right. So our next forces that we're going to study are static and kinetic friction. They use essentially the same equation which is that FF is equal to mufn. mufn is the coefficient of friction. It's commonly between zero and one but it can get up all the way to four if we have two very very very very friction frictiony surfaces. Right? Now there are two types of friction. We have static and kinetic friction. If we have static friction, it means the box is not moving or whatever object is not moving. If we have kinetic friction, that means it is moving. So let's say we have a box on a surface and it is being pulled by some sort of FT. That means that there must be if it's still at rest, there must be some force that is opposing it in the x direction as well. We're going to call that friction FF, right? And if it's at rest, it's called static because the box is static, right? It's just it's not moving. And what happens when we have a graph of FF, right? We see it increases, right? And then it decreases as soon as the maximum static friction is approached. And then we have kind of a lower result for kinetic, right? Once the FT becomes so great that it overcomes the force of friction, the box just kind of starts moving and we get a slightly lower friction. So again, kinetic is when the box is moving or whatever object is moving, but it's still exerting a force of friction to slow it down. Static is when the force of friction is so large that it's actually preventing the direction of motion, right? It's pre it's preventing FT from, you know, having an effect on the box and making it move. Another force that we should take a look at is the spring force. FS is equal to kx and is essentially what's called a restoring force. So let's say I have a box that is attached to a very poorly drawn string on a wall, right? If I stretch the box out this way, then the FS is going to exert some sort of force to try to bring the box back to its natural position, right? Before I stretched it out. That's a pretty simple one to deal with. The next thing that we're going to discuss are centrial forces. And we have discussed this briefly in the video before, but FC or centrial force is equal to MV^2 / R. MV ^2 / R, which is equal to M A sub C, which means that AC, if we cross out our M's, is equal to V^2 / R. Right? When we have an object that is in circular motion, we have an object that is in circular motion. We're going to have an AC pointing in a centrial force going in. Even if and we have our velocity going this way, we have a tangential acceleration going this way. Okay, so let's do an example. Let's say we have a circle and around that circle is a little car and we want to find the maximum speed of the car without the car slipping. So let's draw our FBD. And you know I haven't really explicitly mentioned this but FBD is essentially just draw the arrow that the force is pointing on and where it's pointing from. So here we're going to have FN going up. We're going to have FF kind of pointing in, right? We're kind of that's going to be our centrial force. And then down we have mg. So we know that our sum of the forces in the y direction is zero, right? Because the car is not floating up and it's not sinking into the ground either, right? So we know that f n is equal to mg, especially because they're also, you know, they are going in the same axis. There's no angle there. So that means that the sum of the forces in the y direction is equal to zero, which is equal to uh fn minus mg, right? Okay, so that means that FN as we've previously concluded but just confirming is equal to mg. We know that our sum in the x direction is simply equal to ff because both of these forces do not act in the x direction at all. Right? Which is equal to mv^2 / r which is equal to m a c which is also equal to mu mg because we can make this substitution right. It's not just mufn anymore. It's also mu mg as well. So now we can kind of use these two guys and put these two together. We can say that mv ^2 / r is equal to mu mg cancel our m's. We're going to get that v is equal to the root of g r mu. Okay. So the next thing that we're going to talk about is probably the most important part of forces and that is drag. This is like the one new thing that you're going to learn in physics C besides center of mass for forces and it's really really important. So here's the concept of drag. Let's say that I'm on the top of a really really really tall skyscraper. This is me. And I'm going to throw a ball. As this ball continues to kind of just go down the skyscraper. It's going to approach a time a point in time where it approaches its terminal velocity. What is terminal velocity? What is terminal velocity? Well, terminal velocity is when the ball stops accelerating, right? Like it should, and it reaches a constant velocity. So, let's say that velocity is like 50 m/s or whatever, right? That's pretty unrealistic. But let's say it's just starts moving at a constant velocity. It's not changing anymore. What does that mean? That means our forces must be equal. But what force is counteracting mg, right? In physics one, we kind of qualitatively understood air resistance as something that's going to slow something down. So, if I'm like a bird and I'm coasting through the air, air resistance is eventually going to make me slow down if I stop flapping my wings. Right? In AP Physics C, we're going to look a lot closer at air resistance. And air resistance is actually the force that's opposing the direction of motion here. It's actually opposing mg. What does that mean? It means that mg at terminal velocity is going to be equal to BV. Right? This is our sum of the forces equation. Right? Okay. So now that we've established that our sum of the forces, right, is equal to mg minus some drag force BV. We can set this equal to m a we can also set it equal to m dv dt. as we know from kinematics. And if you haven't taken physics C yet, you might not recognize this, but if you have taken physics C or if you are also taking a advanced mathematics course, you might be able to tell that we're entering a differential equation right now. So, we're obviously going to try to separate our variables, get our V's with our DV. And to do that, we're just going to do a little algebra first. And now we're going to make an interesting substitution. So we're going to say that instead of B over MG, we're going to say MG over B is equal to VT. Meaning that we now have 1 over VT, right? Minus V on one side, which is equal to B over M DT DV. We're going to then cross out our DVs and we're going to get a DV over here. So now we can make an integral right our friendly integral. So we're going to go from 0 to v of 1 / vt minus v d with respect to dv is equal to the integral of 0 to t of b / m t. Okay. So we're going to get just b over mt here. Pretty simple integral right. And then over here on this side, we're going to get ln of vt minus ln of vt minus v. Right? We can kind of transform this over here into ln of vt minus v over vt = b / m t. Raise everything to the e power. We're going to get 1 - v over vt equ= e to the b mt, which means that vt * 1 - eb mt= v. Then if you wanted to, you could substitute your uh mg over b back into vt. And so we've successfully solved our drag equation. Now on the AP, there might be some variations to this drag equation. And you might see something a tiny bit different. Probably nothing too crazy. Maybe we're throwing the ball up and then down. There's some initial velocity. Whatever. This is essentially what you want to know is you want to know how to take this sum of the forces equation, transform your A into a DVDT, separate your variables, etc. And that is the end of unit 2 of AP Physics C. Welcome to unit 3 of AP Physics C. That is work, energy, and power. In this unit, we're going to discuss several types of energy, power, and then we're also going to discuss stable versus unstable equilibriums. Okay. So, the first type of energy that we're going to go over is called kinetic energy. Kinetic energy is often abbreviated as K and it's equal to 1/2 mv^2 where m is the mass of the object and v is its velocity. Kinetic energy is known as the energy of motion. Meaning that when our velocity is zero, our K is also going to be zero because when we multiply this term, obviously we're just going to get a zero. The next type of energy that we're going to discuss is known as potential energy or potential gravitational energy. Potential gravitational energy is often abbreviated as UG u subg and it's equal to mgh or mg delta h depending on how you look at it. Let's explore the difference between h and delta h. Well, when we're comparing the difference in potential energy between two different points. Let's say we're at one point on the skyscraper and another point on the skyscraper right here. And we have our little h axis going this way, right? Our delta H is from here to here. But if we're looking at it from the ground, right? If we're looking at this relative to the ground, we might say that MGH is probably this whole thing, right? This is our MGH at this point. This is our different MGH at this point. Let's call this H2 and H1, right? However, our change in potential energy, our deltaU is essentially just whatever we would get by subtracting this guy and this guy. Gravitational potential energy is the energy of position, right? Because it depends on how far we are above some relative surface here. Now, if I changed it and I said I wanted to calculate our UG relative to some, you know, underground surface down here, then our H1 would be like this and our H2 would be like this. The last type of energy that we're going to discuss in this unit is known as spring potential energy. So this is kind of like gravitational energy in that we start with a u and it is another form of energy of position but it's got a subscript of s instead of g u s is equal to 12 kx^2. And you'll notice it's kind of a similar format to that 12 mv ^2. And why is that? Well, it's because when we integrate the spring force equation, we're actually going to get we get this x squar term, we get the half term. When we do the integral of some x dx, what we're going to get out of that is a 12 x^2. So spring potential energy is pretty easy to calculate. This x here is simply the distance between where we start the spring and then where we stretch it out. So let's say I draw a little diagram here. We have our little spring and a box. And I stretch out, we'll call this, we'll call this equilibrium position zero. We stretch it out to x. So the box is now kind of like right around here. Well, our spring potential energy is K, the spring constant, right? And then X squ is simply our distance here. So again, it's another energy of position, meaning that the position of your object is what determines how much spring potential energy you have. A few other notes that we need to make here. The first one is that work which is essentially the change in energy is equal to the integral from a to b of f do dx. And what does this dot mean? It means it's a dot product. And we just need to remember here that this dot product means that we may need to add a co- theta if we have a force that is not uh parallel to the direction of our x-axis here. Right? Another thing that you'll notice and this is called the work energy theorem is that work is equal to the change in kinetic energy. Right? Now it's also true that this is equal to the negative change in potential energy. There is an equation in your reference sheet that states that deltaU is equal to the negative integral from A to B of F of a conservative force dr. And what this CF here means again is a conservative force. So one example of a non-conservative force is air resistance. because when we have something that's whistling along in the air at a high speed, we may make some sound energy and that energy we basically just can't get back. Um, another thing is if we have heat energy, we often again cannot get that back. So, anything like a force that would create that heat energy would be a non-conservative force. But pretty much everything that you'll deal with in physics C will end up being a conservative force. Now, one other thing to note before we move on to power. If you have a question that asks you about mechanical energy, that means kinetic energy and potential energy. And finally, another important thing to note is that if we do not have a function for force, if we have a constant force across some interval, then we can simply write this as equal to F d, right? Just FD. And then we might also need to again throw in that cos theta, right? Because if we have a constant force, it's just going to be the same thing as if we integrated a non-constant force. If we integrate a constant, we're just going to get um that constant times d, right? And one final thing to note here is that conservation of energy is extremely important. energy is conserved and that is shown here in that our change in K plus our change in U is always equal to zero. Okay, we're now going to deal with power. So what is power? Power is the rate at which work changes. And as you might predict, power is equal to the derivative of work with respect to time. If we wanted average power, we would have average power is equal to work over time. And an interesting application of that is that we know that for a constant force, work is equal to FD, right? So we can work write this as FD over T, which is equal to F * D over T, which is equal to F * velocity. That's just an interesting application that could potentially show up on your exam. Um another important thing to note with that is that this is again a dotproduct. So we should be also writing this as fv cos theta. Now another thing to note here is that if we have power is equal to the derivative of work with respect to time that means that work is equal to the integral of power with respect to time as well. There's another interesting equation that's on your reference sheet that says that f is equal to the negative derivative of u with respect to x. An easy way to derive that is that we have work is equal to the integral of f dx which is equal to the change in kinetic energy minus the change in potential energy. Therefore the change in uh potential energy is equal to the negative integral of f with respect to dx. And therefore du dx is equal to f or f is equal to du dx. Just an interesting to thing to note. We should also briefly deal with the concepts of stable and unstable equilibrium. So what is stable equilibrium? A stable equilibrium is a place where the sum of the forces is equal to zero and U is at a local min. So let's say we have a little roller coaster track here and we have a ball that is kind of at the bottom of the track. There are no forces that are causing it to move up or to move down because it's at rest, right? And it's not moving. It's not going to move. And we are at a local minimum in the UG. Right there. U is going to be higher as we move this way. It's going to be higher as we move this way. Now, what's an unstable equilibrium? Unstable equilibrium, which I'm just going to kind of write on this part here, is a place where the sum of the forces equals zero and U is at a local max. So that would be something like this, right? though the sum of the forces is equal to zero. If I give this ball even a slight nudge, it's not going to just return to the bottom here like this one will. It's going to go this way or it's going to go this way, right? It's going to fall down the track. So, one thing that we discussed previously in this video was that energy is conserved unless work is being done. And we need to know that that means that EO is equal to EF. And that's like basically the best way to solve a lot of these energy problems. So let's say like I have a roller coaster here. I have a particle over here. It has some U. We'll just call it U K knot. And it has no kinetic energy, right? That means all of my energy here is just going to be equal to U KN. However, when it's at like right here on the track, it's going to have some different UG, right? It still has UG with respect to the bottom with respect to the bottom of the track, but it also has kinetic energies. And so I can set that equal to UF plus K. We'll just call it K because there's no K at the start, right? What you would do is you'd measure the height of here and here. You know the mass of the marble, hopefully the mass of the particle and you know G. So you'll be able to figure out these two terms and then using that you'll be able to find out what the velocity is, what the kinetic energy is at this point. So that's like a quick example of what you can do for energy. Now another thing you might want to know is how do you find the direction that work is being done? How do you know if work is a positive or a negative quantity? Well, when you have a positive quantity of work, you're essentially adding energy to the system. When you have a negative quantity of work, you're essentially taking energy away. AP Physics C unit 4 linear momentum. So, there's two main things that you need to learn about momentum. The first thing is that we represent it using the letter P and it's equal to MV. And the second one is that P is equal to PF for closed system. What that means is essentially that momentum stays conserved. It stays constant in a closed system. This is known as the law of conservation of momentum. So, how does that help us? Well, let's do an example problem to kind of figure that out. Let's say we have two carts. We have one 8 kg cart that is headed in the positive direction. We're going to find positive to be this way. We have another cart that is headed in the negative direction going 3 m/s. And that cart is going to weigh 18 kg. And then we also have let's make up a number for the velocity of this cart as of the starting position. Let's call it 6 m/ second. So after these two carts collide, which we see that they're going to collide because they're headed right towards each other. What is the velocity of the first cart if it's known that after their collision, this cart goes the other direction at 2 m/s? So, we can again use both of those principles we discussed before. P= PF and P= MV. So, we'll start off by saying PO equals PF. And we can then discover the PO's of each of these carts here. The P knots of each of these carts. So we'll have M. So 8 * 6 plus -3 * 18, right? Cuz we're going in the negative direction with this cart here is equal to, and we don't know the velocity of this cart after the collision. So we're just going to call it V * 8, right? 8 V plus we know this is going two in the other direction * 18. So we'll get 48 - 54 is equal to 8 V + 36. Meaning that -42 is equal to 8 V. Therefore -214 is equal to V or -5.25 m/s is equal to V. So that means that after this collision, we're going to have 5.25 m/s going the other direction. Let's take a look at another example in which the carts actually stick together after their collision. Let's say we have a cart of mass 12 m that is headed in the positive direction at a speed of 2 v kn. We have another cart that's headed at a speed of fif v knot with a mass of just m. We want to use the same principles after they stick together. What is the velocity of the total kind of conglomerate of these two stuck together? Right? So we're going to get 12 m * 2 v kn plus -16 v kn * m. Right? And this is going to be our P KN. And so after we've done that, we can write this as 24 V KN M - 16 V KN M is equal to V T * 13 M. We're just going to say VT is the total velocity of the carts after they have collided, right? And then we can just kind of cross out all the M's here. We're going to get that 8 v kn over 13 is equal to the total velocity of the carts after they collide. Another example that you might see is if you have two particles that are kind of flying at each other in space and they're not going in one dimension like those other particles were before, those other carts were before. How would you do a problem like this? Well, what you would want to do is you'd want to use your trig functions to figure out what the momentum in the x direction was for each particle, what the momentum for the y direction is for each particle. We know that momentum is conserved in every direction. It's conserved in both directions. So, what you do very simply here is you would find the velocity of both of them using the x direction, the y direction after their collision. And then you can use Pythagorean theorem to kind of put those two together. Right? So if I have some momentum px this direction, some momentum py in this direction, then I'm going to draw a resultant vector. That's kind of my like total sort of p. Now we're going to look at impulse. What is impulse? So impulse is equal to the change in momentum. So it's basically when we change the momentum of a system. Previously we looked at closed systems. Now we're going to look at systems where we have added something in the middle of the experiment or whatever. or we've changed the momentum of some object. Right? So, delta P is equal to J. J is our letter for impulse and it's equal to the integral of F with respect to time with respect to time. Okay, you notice earlier when we had the integral of f with respect to distance, that was work. In this case, we have a sort of a parallel here. we have a change in momentum versus a change in energy. Right? So this is just something to note really quickly. Obviously if we have an integral if we have f over here and t over here the area under the curve is going to be your impulse or your change in momentum. And that also does mean that if we have a constant force being exerted then that J is going to just be equal to FT or F delta T. Just going to take a look at Newton's second law in sort of momentum quote unquote form. So we have F is equal to M A. And we know from our drag equation and our kinematics knowledge that M A is equal to M DV DT. Meaning that D of MV is over DT is also equivalent to M A here right. Which is equal to DPDT. So our change in momentum over time is well our change in momentum with respect to time I should say is equal to our force. This is ma valid for when mass is constant and when it's changing. And the last thing that we're going to discuss in this unit is the difference between an elastic and an inelastic collision. So which one is which? So an inelastic collision happens when we have stuff that sticks together and slash or when k is lost. So kinetic energy is not conserved. If we have an elastic collision that's when two things are bouncing off of each other and that means that k is conserved. Remember that kinetic energy is not a vector. It's a scalar quantity. So, make sure you take that into consideration when you're kind of summing up the energies to see if a collision was elastic or inelastic. A collision that's perfectly elastic, which basically just can't happen in nature, means that all of the kinetic energy is conserved. When you're doing like a laboratory experiment or something like that, usually it's just, you know, a good proportion of that kinetic energy will be conserved. AP physics C unit 5 torque and rotational dynamics. In this video, we're going to get into kinematics, torque, and rotational inertia. The first thing that we need to know about rotational kinematics is that there is a way to relate rotational and linear kinematics. So, let's say that I have a circle, a little circle here, and it's rotating around this axis in the center, and we have a radius of r. So, let's say that we are rotating in this direction here, and that we've rotated some theta, some angle right here. How do I relate how far that we've linearly traveled? Right, this arclength distance right here and how far we've traveled in the angular way, right? The equation for that is very simple and it can be found in your reference sheet at this point. It's just theta is equal to x / r where this little arc length here is equal to x. You'll also see a diagram on your reference table that looks very very similar to this but just uses s instead of x. Similarly, we're going to relate angular velocity and angular acceleration to linear velocity and linear acceleration. So, we have omega is equal to v / r. And then we also have alpha is equal to a / r. And as you might guess, this is angular acceleration. We have here angular velocity and then this is angular position. If we have a delta here, it becomes angular displacement. Again, similar to normal kinematics, we have three big equations that we can use if the platform or the rotational object is in constant acceleration, constant angular acceleration, and those are the same. Essentially, we just swap out our variables. So, this is our first one here. This is our second one. And then our third one. And you'll notice these are all exactly the same as our big three kinematics equations. We've just replaced all of our x's with thetas, all of our v's with uh omegas, and all of our a's with alphas. A few important things to note about what I've just said. On any point on a rotating surface, right, if I rotate this entire surface some angle, right, every single point on this surface on this circle has traveled theta radians. Every single point has traveled the same theta radians. And every single point will also experience the same angular acceleration and the same angular velocity. What does not occur is that it experiences the same linear displacement or the same linear change of velocity, the same linear acceleration. Why is that? Well, if I'm a little bit closer, let's say I'm right here, this arc length here is obviously going to be a lot shorter than if I was at a larger r, travel the same theta, right? This arc length here is obviously a lot bigger. So you can see that our linear displacement is a lot larger when we're at a longer radius, when we're at a radius further from the center. All right, next up is our discussion of torque. Torque is pretty simple. It's defined by the Greek letter toao and it's equal to R cross F or RF sin theta. Now, how do I know what the theta is? How do I know what the R is? Well, let's say I have a little rod here and this is my pivot point. Our R is essentially the distance between where the force is applied. So, let's say the force is applied like this. And the angle is simply this little angle right here. So, basically torque is like the rotational version of force, right? We're just adding an R for our distance and we're adding a little sin theta in there. Now, we're going to talk about rotational inertia. Rotational inertia. So rotational inertia measures a rigid systems resistance to changes in rotation. It's related to the mass of a system and the distribution of the mass relative to the axis of rotation. All right, next up on our list is rotational inertia. Rotational inertia. Rotational inertia measures a rigid systems resistance to changes in rotation. It's related to the mass of the system and the distribution of that mass relative to the axis of rotation. The equation that you might first see on your reference sheet to relate uh rotational inertia with which by the way has the sign I is I is equal to the sum the M R 2. Now I mentioned earlier that the general form for I is going to be the sum of the M R 2. But we also have some specific shapes that we kind of want to keep track of for what this coefficient here will be. Right? We know that we have a collection of masses. We know they're arranged a specific way. Let's say like a cylinder or a rod. So there's going to be some multiple of MR squ that's attached to that. And we're going to do a quick derivation using integral calculus for a rod. However, you can look up some common rotational inertials. For example, for a hoop or for a mass that's attached at a further distance from the axis, whatever, for a sphere, anything. You can look those up, put them in your calculator or memorize them before the exam. You're going to find coefficients like a third or a half or a sixth or something like that. Just make sure you put them in your calculator. We're going to do this derivation just so you guys have an example, though. So, we have a rod right here. And our axis we're going to put right in the middle. Now, we're trying to use these little dms here, right? And we are going to have a total mass of just M. Before, we knew we knew that I is the sum of the M R 2. Now, we're going to say that I is the integral of the R squares with a little dm there. Right? Now, we kind of briefly touched on this in a previous um unit, but if we have this uniform density that we're assuming, this lambda is equal to the sum of the M's over the sum of the L's, which is equal to DM DL, then we can take DL M / L is equal to DM, right? Just doing some simple algebra here. Using this information, what we can do is kind of plug this guy in here, right? So, we have I is equal to R2 M / L DL. And I'm just going to change out our R here for an L just because R is what you'll see in the reference table. But if you use L in a problem, if that's what's given, it doesn't really matter. I could have just changed this DL to DR. It would not have mattered either way. But what I'm going to do now is I'm going to integrate this function. What are my bounds though? What am I integrating from? What am I doing? So I'm going to say that our total L, right, is just going to be this whole thing. Uh that should be generally implied in the problem. But what I'm integrating this function from is I'm integrating it from negative half of L, right? So I'm going to put L /2. And then over here I'm going to have positive L /2 positive L /2. Now we know from basic calculus that when we had this L /2 and L /2 what we can do pretty simply is just say that this is equal to 2 from 0 to L /2 L^2 m L D L. Now we also know that we can take out the constant here. So we have is equal to 2 m / L from 0 L over to 0 from 0 to L / 2 L^2 DL. This is a relatively simple integral. So we're going to get 2 M L we're going to get L cub over 3 from 0 to L / 2. And then you could simply solve this integral. What we're going to get here is we're going to get ml over 12 ml^2 over 12. And so we can see here the coefficient of our ml here is 112 for a rod when we're about the center. Now there's going to be a different coefficient if our axis was over here on the edge, right? Or if it was kind of like here in between because our mass is going to be a little bit differently distributed. It's not going to be equally from this thing. it's going to be, you know, there's going to be a lot over here in this case. Whatever it's going to be a different coefficient and that might be something that you could put in your calculator, but it's also something you could derive on the exam by simply just switching the bounds here and here. Now, the next thing that you might have noticed on your reference table if you're following along using that is that there's an equation under the integral equation for rotational inertia that looks like this. I prime is equal to IO M + M D^2. What does this mean? Well, this is known as the parallel axis theorem. And it states that if we want to find the I of some object, some system, whatever along a different axis other than from the center of mass axis, we can just add this MD² term to figure it out. So, we're just going to do another quick example. And like I said, we're going to find another coefficient for a similar problem as what we just did. So we found from the previous thing that if we find I around the center of mass, right? It's going to be 112 ml squar, right? This is obviously center of mass because it's a symmetrical object, right? We're just kind of in the center of both of these of both of these axes, right? But what if we want to find the new rotational inertia about this point P? Well, what we're going to do is we're going to add this term right here. So, we have 1112th ML^2 plus M. And then what's our D? Our D is the distance from when we had the center of mass to our new point. So, it's going to be L / 2^2, right? And what is this equal to? Well, when we factor it out, when we simplify, it's going to be equal to/3 ML 2. And that's just another example. we have a new coefficient because we move the point P. And that's how we do rotational inertia. That's how we do the parallel axis theorem. But what if we have a nonuniform rod? What if we have a rod with nonuniform density? So in the previous problems, we had a rod that has a standard density of just lambda is equal to m / l is equal to dm / dl. Right? What if we have a problem here where we have a rod with length L? We're trying to find rotational inertia about an axis that's over here. We have a mass of M. We have these little dms as per usual, right? We're just adding up all these little DMs. And we're given that the lambda that the density is equal to beta x cubed, right? Right? And that this is still equal to that dm dl or dm dx or whatever you want to call it. The first thing that we need to figure out is what is beta, right? We need to figure out what beta is going to be cuz we can't just have beta in the end answer. So let's try to find beta. So we know that dm dx I'm going to call x l, right? Same thing. They both mean position. dm dx. and we're going to set that equal to beta x cubed. If we put this x over here, so we get dm is equal to beta x cubed dx. And then take the integral of each of these from 0 to m over here and then 0 to our l over here. What I'm going to get is that m is equal to beta x 4 / 4. Right? Simple integral. Therefore 4 m / x 4th is equal to beta or I should say really this is l 4th right? The next step here is that once we know that lambda is equal to beta x cub which is equal to 4 m / l 4th * x cubed right because we this is just our beta. This is our x cubed over here. We can now use that equation that we had before. So I is equal to the integral of x^2 dm. And you'll notice I'm not using r square. I'm just using x instead of r. Um that's pretty common. It's just a little bit easier, right? We know that dm dx is equal to 4 m / l 4x cubed. Therefore, dm is equal to 4 m or l 4th * x cub dx. Right? Right? And that's getting a little bit long here. But now we can take this and do a pretty simple substitution. So I is equal to 4 m / L 4th the integral of X 5th DX. So of course once we integrate this and obviously we're going to go from 0 to L here we're going to get 4 m / L 4 * L 6 / 6 which is essentially equal to 2/3 m L^2 and that is our answer. We're going to finish up this unit with some quick notes on Newton's laws in rotational form. So, we can still write this sum of the torqus equals zero equations. We can still write sum of the torqus equals whatever equations. That's something that we can do with forces and it's something that we can do with torqus too. Another thing is that we have a if we have a constant angular velocity, it means that we have obviously no torque on the system. And the last thing is that you can still draw free body diagrams for torqus. It's just works the exact same way. And one final note as well that will be extremely useful for when you are solving problems. You should always use this in your reference table. The sum of the torqus is equal to I alpha is equal to the inertia times the angular acceleration of the system. AP Physics C unit 6 energy and momentum of rotating systems. We're going to go over rotational kinetic energy, torque and work, angular momentum and impulse, conservation of angular momentum, rolling systems, and the motion of orbiting satellites. We're also going to include a little bit of a bonus on gravitation that we need to finish up from forces earlier. Let's get started with rotational energy. So you may notice that on your reference sheet you have an equation for K wrote K rotational that is equal to 12 I omega^ 2 and you may also remember from previous units that we have K trans which is equal to 12 MV squared. So what is the difference between these two? Is there a difference? The answer is yes there is. So when we have a bike wheel for example, let's say you're on your bike, you know that the wheel is both spinning and moving this way, right? It's moving forward. When you're on your bike, when you're riding your bike, the whole bike is rolling, but it's also going forward, right? You have some rotational energy and you have some translational energy. And now we have rotational energy as the bike's wheel, you know, goes in revolutions. And we also have translational energy. This also occurs for when we have some sort of sphere, let's say rolling down an incline plane. We're going to start with mgh as the balls up here and we're going to end up with both rolling and translational energy at the bottom. So for example here let's say we have E kn is equal to EF the ball starts at rest and so we have MGH is equal to 12 MV^2 plus 12 I omega^ 2 and let's say that we're given that the I for a solid sphere is 2s 2 M R 2 then we can plug this into our equation here and we're going to get is equal to 12 mv^2 + 12 2 m r^ 2 * and then what are we going to use for w? Well, we can just use v / r. So, we're going to have v / r squar. And here we're going to have these r squars are going to cancel out. Here we're going to get plus 210 um + 2/10 mv^ 2. So, we're going to get 12 mv ^ 2 + 210 mv ^2. Obviously, that's just going to be 7/10 mv^2. So, mgh is equal to 7/10 mv^2. This could also be used to find the velocity, right? We would just cross out m from both sides and then solve for v. So, another note about rolling really quickly. If we have that wheel and we have some sort of translational kinetic energy, then that means that we have kind of this velocity, right? We have a velocity around the center of mass. But if we have a rotational thing, we're going to get some rotational going this way, some rotational going this way. Right? Because as we rotate along different points of the wheel, it's going to look like it's going this way. It's going to go like look like it's going this way. Right? At the center though, the velocity is zero. Right? You'll notice at the center of your bike, there's not really any velocity there. Right? There's no rotational velocity right there at the dead center. Now, when we add these two together, then we're going to get that rolling thing that we mentioned earlier, right? We're going to get that wheel that has some sort of angular velocity, right? Some sort of angular velocity, and we're also going to get that translational velocity moving forward. Another thing to note is that when we have an inclined plane, this is just another random note that we're just going to kind of put in here. If there is no FF on this plane, then instead of rolling, what the ball is going to do is just slide. Imagine if we put like a block here, right? The block obviously is not going to roll. But if there's no friction, the block just kind of slides down. The same thing is going to happen with the ball. There needs to be some sort of friction that pushes the ball so that it rotates or it will not do so. It will not rotate. Our next note relates work and torque. How much work do we need to rotate something? Well, we have work is equal to the integral of f. Well, we have work is equal to the integral of f dx, right? fd dx. And we know that f is equal to torque / r. So, torque / r dx. We can rewrite this as equal to torque * d x / r. What is dx / r? Well, it's equal to torque d theta. And this equation is also on your reference sheet in case you do forget it. Now, we can also use this equation, an interesting derivation, to kind of go back and figure out what our K rotational is, right? We know that torque is equal to I alpha. So we're going to get I alpha d theta and we're going to get I * d w dt d theta which is equal to i dw which is equal to i w ^2 / 2 which is equal to 12 i w ^2 obviously and of course our bounds here would just be something like 0 to omega Yeah, just an interesting proof there. And the last thing we're going to go over with regards to rotation stuff like that is angular momentum. Angular momentum. So let's say that we have F is equal to DPDT. And we know this from our previous unit from our momentum unit. RXF. So let's just write rxf is equal to rx dp dt meaning that torque is equal to d * rxp / dt which means that we have l is equal to rxp. What is l? L is our angular momentum. What does this often end up being? Well, it often ends up being L is equal to M V R sin theta, especially for a point mass. That's how we use that one. But we can also use that torque is equal to DL DT, right? Where L is again our angular momentum. Or we can use that change in L which is also known as angular angular impulse is equal to this integral of torque with respect to time. Additionally, we can see it for a larger object that L is equal to I W and you can also find that on your reference sheet. This is comparable to P= MV basically, right? Because we have the rotational equivalent of M is I and then the rotational equivalent here is just a V. The final thing to note here is that we can also use the same LO equals LF. L= LF. If there are no external torqus, we have conservation of momentum. Now we're moving back on to universal universal gravitation. So what is universal gravitation? Well, we remember from previous units that we have fg is equal to g m / r 2 where g is just a constant. It's 6.67 67 * 101 Newton m^ 2 over kg squar where we have two masses. Let's say we have m1 and m2 over here. And then r here is going to be the distance between the centers of each of these. Let's do a quick example question. Let's say we're trying to find the radius. So find the radius of a satellite a satellite in geocynchronous orbit. What does that mean? in geocynchronous orbit means that it goes around the Earth every 24 hours. So let's say we have a little miniature Earth right here and then we have a satellite that's somewhere around the Earth orbiting let's say it's just right here at this point. Well, we know that AC is equal to V^2 / R, right? So therefore if we have the sum of the forces is equal to m a which is equal to g m / r 2 then we can say that v^2 / r is equal to g m / r 2. Then we're going to cross out one of our rs over here. We're going to get v ^2 is equal to g m / r. So now we want to find v. We know that v is going to be the distance traveled per second or per unit of time. We know that our distance here is simply going to be 2 pi r. Right? We're just going to use a circumference formula. So it's going to be 2i r / t^ 2. So we're going to get 2^2 pi^2 r^2 over t ^2 is equal to g m / r. And at this point you might be thinking wow this looks pretty familiar. And you'll probably write if we have 4^2 r cubed = g m t^2 and we simply do a little bit more rearranging. And we're going to get 4^2 r cub over gm is equal to t^2. And I'm going to use a capital t because that's kind of conventional. Basically, this is just Kepler's third law. You do not need to know this or memorize this for the exam, but you should be able to derive it like how I just did. Another super super important thing to know that will come up in ENM is Newton's shell theorem. So, what is a shell? A shell is basically a hollow sphere. So, let me just draw this really quickly. A shell is just a hollow sphere. How does this affect FG? Well, here's the thing. If I have FG due to a uniform shell and I'm at the center, that is actually zero. If I'm on a mass outside the shell, if I'm over here, let's say I'm at point A, right? Then it's the same as a point particle. So if I have this shell here and this shell has some mass, let's call it 2 kg, then that's going to be we can treat this the same way using FG= GMM / R 2. We can use this as just calling this like a 2 kg mass centered at the center. But if I'm at the center of the shell, right? If I'm at the center of the shell or if I'm in the middle of the shell, if I'm anywhere inside of the shell, any mass inside the shell has no FG acting upon it. What if I have a solid sphere? What if I have a solid sphere? And we're going to treat this solid sphere as basically just being a bunch of shells nested in between each other. And you can kind of think about this as the Earth, right? The Earth has multiple shells in layers. Well, here's what's going to happen. The only shells that are going to affect you, the only shells that are going to affect you in terms of gravity or whatever are the ones that are like below you. So, let's say I'm like right here and I'm like in the center of the Earth. Basically, the only shells that are going to affect me are all of these ones here. These shells from here to here are just not going to affect me. So, I should only treat these shells as affecting me. And those shells can be treated as one point mass in the center. So let's say like the mass of every single one of these shells under me combines to like 400 kg. I can just treat that as a 400 kg mass in the center of that thing right there. And I can just use this value as my radius. The next thing that we're going to talk about is universal gravitational potential energy. Let's say that we have two masses. Mass over here and mass little m over here. We're going to try to find the work to get to infinity where ug infinity equals zero. So let's say that the work is equal to the integral of f do dr and that we're going to try to evaluate this integral from r some measurable r here some measurable r here to infinity. And we are going to integrate the gravitational force. Right? We're going to get gm mm / r 2 dr in here. Obviously, we can take out these constants. So, what we're going to get is we're going to get gm mm integral from r to infinity of 1 / r 2 dr. And what are we going to get from this result? Well, what we're going to get is we're going to get - g m over infinity plus g m / r. Now, when we have an infinity in this bottom term here, this is basically just going to go to zero, right? As we approach infinity, you know, we're just going to create such a massively large number at the bottom that whatever gm becomes is going to be insignificant. And so, this means that we need to find a negative here. There needs to be a negative here because the system gains this energy. System gains this energy. Thus, we can write the graph of U versus R as such. We have UG over here, R over here. It's going to look something like this. Let's kind of apply this idea really quickly to something called escape velocity where we're trying to find find the speed find the speed needed to reach infinity. So let's say I have the earth right here. I have a little mass right here and we're trying to get that velocity that we need to leave the earth. So let's say that E 0 equals EF. Always a good way to solve a problem. So - g m / r + 12 mv^2 is equal to 0 + 0. Right? Because both of these things are going to become zero. So that means that 12 v^2 is equal to g m / r. And therefore that the v escape is equal to of 2 gm / r. We could calculate the exact numbers by plugging in the radius and the mass of the Earth, but that would have to do for now. So here we can just kind of see that again that gravitational energy is a negative term and make sure you're keeping it as that negative term um when you're doing your calculations. Unit 7 of AP physics C is oscillations. So what is oscillations? What is this unit about? Well, we're going to focus on simple harmonic motion. And that means we're going to discuss physical pendula, simple pendula, and spring systems. So what is SHM? Well, SHM or simple harmonic motion is a cyclical a cyclical or periodic motion caused by a linear restoring force. A linear restoring force. And usually kind of like how we represented the drag force, it's equal to - k delta x. As I mentioned, there are three types of simple harmonic motion that we're going to be looking at. The first of these is a mass spring system. So let's draw a little spring right here. And we have our mass. Let's call this mass m on a surface. This is the first case that we're going to look at. The second case, the second case that we're going to look at is a simple pendulum. So this is if we have some little rope here. We're going to call this rope basically weightless or massless. It's going to be like negligible weight, negligible space taken up by this rope attached to something here, some little pivot point. And then we have a little mass on the end. And then our third case is what's called a physical pendula. Now a physical pendula is basically the same thing as our normal pendula except for it is a rigid system. So this is not a rope. This is maybe like a stick or a rod, right? So we have our same pivot point. We have our little rod here and then maybe we even have an additional mass on the end. The point is with a physical pendula, let me write that down. The physical pendula is that we have some sort of I we have some sort of rotational inertia going on. Whereas with our simple pendula over here, we do not have that I because this is essentially negligible here. Now, how long does it take for each of these to complete a cycle? I previously mentioned that we are dealing with cyclical or periodic motion. So, obviously the question comes up, you know, how long would that take? Well, there's different equations on your reference sheet for each one of these. And you'll notice it's essentially the bottom three on the right hand side of the mechanics side. So the period of a spring system is t subs is equal to 2 pi m / k. The period of a simple pendula is tp = 2i l / g. And the final one over here, our physical pendula t fizz is equal to 2 pi i over mg d. Another thing that we should note is that when we're relating frequency and period, t is equal to 1 / f. So 1 over frequency is equal to t. Also note that frequency has units of hertz and that period has units of seconds. Additionally, you'll see on your reference table that t is also equal to 2 pi over omega. So there's a few things to know here and we're going to focus on our spring system that we had before. So let me draw that spring system again really quickly. We have our little spring right here and it's attached to a mass of mass m. When I pull on this spring right here, when I increase the length of the spring, right, it's going to be stretched out. And you know from experience, just with your life experience with the world around you, that when you stretch out a spring, it wants to go back to where it used to be. And that's also kind of what we meant earlier by restoring force. The restoring force basically opposes the direction of displacement. So, if I were to compress the spring more, then the restoring force is going to make us go this way, right? If I compress a spring, it's going to basically shoot out back at me, right? But if I pull on the spring, if I pull it this way, then the restoring force is going to make us go back this way. So here's the thing. After I have extended the mass, so after I' after I've stretched out the spring here, something like over here, when the mass is basically where it would be without any forces before I stretched on it, this is called the equilibrium position. And this distance that I stretched it is the amplitude. We'll call this distance X. At the equilibrium position right here, the net force is equal to zero. Additionally, at this point here is when we have the maximum kinetic energy. Why is that? It's because all of our spring energy, all of our potential spring energy has dissipated, right? Because spring energy is equal to 12 kx^2. And here our x is just zero. So here we have all kinetic energy. Here we have all spring energy. Just an important thing to note. We can also translate this to our pendulum. When we have the pendulum all the way over here on this side, we're at our max gravitational energy. When we have it over here, we're at our max kinetic energy. Right? K max is over here. It's very simple in comparison to what we've dealt with with rotation and things like that. Now, you may be wondering, how can I find what x position this mass is at at different times when it's here or here? And for that we'll use the last equation on your reference sheet. That is x is equal to x max cosine of omega t + 5. All right. The x position is equal to the amplitude. This is our x max by the way. Time cosine of omega t + 5. Now this phi is just a constant. It represents a phase shift. If we don't have a phase shift, we can basically just cut off the equation right here. Right? It will just be cosine of omega t. You can also write this as cossine of 2 pi frequency t. Now we have something exciting a differential equation. So we saw earlier with our spring force that the sum of the forces was essentially just equal to that negative spring force right that restoring force that we experienced. And we called that equal to kx. So let's set that equal to m a. And while we're doing that, let's just set it to m * d times the second derivative of x. Right? And this should probably look familiar. Let's say this is - kx / m is equal to d^2 x over d t^2. Right? Just a derivative here. And now let's make a quick observation. We know that x of t as we described earlier is equal to a cosine of omega t. We could add the fs in there, but it's not really necessary. But if we have dx dt, that means it's equal to the derivative of this, right? So we're going to get omega a sin omega t. If I wanted to take the second derivative here, what am I going to get? Well, I'm going to get omega^ 2 a cos omega t. And so now that I have that information, I'm going to plug it into the equation that we just had. So going back up here for a second, we have this second derivative of the x of t function is negative and we're going to plug it in right here. I'm just going to shift sides here. Omega^ 2 a cos omega t and that is equal to - k / m * our x of t function which is a cosine omega t. And you'll notice very quickly, it seems as though we have some stuff that'll cancel out, right? This co of omega t term will cancel out. This one will too. And that's why I kind of said earlier if we had the fi in there, it doesn't really matter cuz it was just going to cancel anyway. Um, obviously I knew that and you didn't. So it's okay if that's fine. Um, the next thing that cancels out is our negative signs and our a's. And so what we're left with is that omega^2 is equal to k over m for our spring. Now, let's go back up to the equation we made over here. If we know this, then we can simply plug this back in. And what we're going to get is that we're going to get a is equal to omega^2 x. Right? That's just something we can put out there. Another thing that we can derive from this is that omega is equal to the of k / m. And this is our angular frequency. This is our angular frequency. A few other results that we can get is that vax our max velocity is equal to a * uh this here a * omega. Another thing we can get is a max is equal to a omega squar right because we're replacing our x with just this amplitude here. Um and our acceleration will matter and which direction it is. So we can either cancel the negative or keep it. Another thing that we might want to notice here is that changing the amplitude does not change the period. One last thing that we're going to discuss here is damped SHM. What is damped SHM? Well, when we have something like a spring that is sliding across a surface or a mass that is sliding across a surface, what's going to happen over time, right, is that our oscillations when we go this way and this way and this way and this way and this way and this way. You can see that our displacement is going to get smaller and smaller as we keep going. And that could be due to air resistance, it could be due to friction, whatever. In the real world, a lot of the time is what we're going to see is this damped SHM. I'm going to spell this out for us. This damped damped SHM. And what that means is basically just that the amplitude of the oscillations is going to get smaller as we go on. And so we saw normally we'd have a cosine graph or a sign graph, whatever, look something like this. When we have a damped SHM graph, it's going to just be like that. But over a longer period of time, what we're going to get is this kind of shape that gets smaller and smaller and smaller and smaller and smaller. We can see as the amplitude decreases as the time increases that the thing is on there. The next thing that we're going to discuss is energy in SHM. Let's calculate ET. So ET for a mass spring system is equal to our spring potential energy plus our kinetic energy. Right? We know that our kinetic energy is going to be equal to 12 mv^2 and that our spring potential energy is going to be 12 kx^2. So let's write that down. 12 kx^2 + 12 mv^ 2. And we know that our x term is just going to be equal to the a cosine of omega t. So we're going to have 12 k of a cosine omega t plus 12 m. And we know our angular velocity. We know our velocity equation, right? Right. It's going to be a shift over for more space here omega sin omega t^ squar. Where did we get this from? Well, we know that our dx of t up here obviously dx of t is just v of t. The next thing we're going to do is we're going to get 12 The next thing that we're going to do is we're going to get 12 k a^ 2 cossine^ 2 of wtus 12 m a^ 2 w^2 sin^ 2 of wt. We're going to factor some things out here. We're going to get 12 k a^ 2 of cossine wt - sin^ 2 wt. And when we just use our trig identities really quickly here, we're going to get that this whole number here is just equal to 1. So the energy of our the total energy of our mass spring system is just this guy right here. This 1/2 k a squ where a is our amplitude. And at this point, we're going to further discuss our pendulums. So, let's draw a simple pendulum right here where we have a length of L. We have a force of tension going this way. Let's call this our pivot point right here. We have mg going down. And then we also have some little angle right here. Right? Some little theta right here. This L, by the way, is drawn to the center of mass of this little guy right here. Now, mg is the reason why this pendulum is going to rotate. It's the reason why we're going to see a cycle of motion. Why? Because it's the force that is not along the R vector, right? This FT is going right along the R vector here. Meaning that if we took the cross productduct to find torque, we would just get zero. So, this FT is not really doing anything for us right now. Thus we can write our sum of the torqus equation as the torqus is equal to l mmg sin theta. We also know that because we're going in the negative direction because it's a restoring force is actually going to be a negative torque essentially. So we're going to get i alpha is equal to l mg sin theta. And this is where I introduce you to something called the small angle approximation. The small angle approximation says that for small angles sin theta is approximately equal to tangent of theta which is approximately equal to theta. And we can go into how this is derived with tailaylor series and things but you can just look this up on your own. If you have a small sin theta it's going to be roughly equal to theta or a small tangent theta it's going to be roughly equal to theta. So how can we apply this to our differential equation? Well we can basically just take out this sign term. So we're going to get I alpha is equal to L MG theta otherwise known as I d^2 theta over dt^2 is equal to l mg theta. Then we have I a W ^2 cossine of WT is equal to L MG a cos WT. And notice that when I say W, I really mean omega, but W is a little bit more familiar to us. Um, if you use a W as your omegas, it'll probably work as long as it looks something like an omega. As long as you're being consistent and you're not, you know, using W and omega as completely separate variables, it's okay. Just either use W or use omega. Call it W, call it omega. The technically right term though is omega. So once we've done this, we can see that as with our previous differential equations, we're going to have a bunch of stuff cancel. So there goes our I's, there goes our co wt. There's our a cos wt. So we're going to get i w ^2 is equal to l mg. What does this mean? It means that W ^2 is equal to L MG over I over I. So now what are we going to do? Well, we're going to go and we're going to go plug that into what we had before. So now that we know that we have this guy right here, we can rearrange this as alpha is equal to L MG over I theta. Right, we're our negative symbol over here. And then we're just going to plug this guy in. So we're going to get negative omega^ 2 theta. And you'll notice this is pretty similar to the one that we had with the spring. And that is all you need to know for simple harmonic motion.