OK, then let's get started with 7C, right? And so remember last week we talked about how we covered all the material that I wanted to cover all the way through gravitation. Today I'm going to go over most of that material just to give you a reminder and overview of what's

going to be covered. Can we close down the chit chat? What's going to be covered on the final of the material? I know there should be a surprise to you. I posted a practice final exam today, which you can look at. Just like for the practice midterm, it's the same length as the real final will be.

It's approximately the same difficulty as the real final. The distribution of problems, easy problems, hard problems, topics, is very similar to the final. So if you found the practice midterm to be a good predictor for what the midterm was like, you'll feel the same way about the final

and if not, then again, you'll feel the same way about the final. So I encourage you to study for it the same way I encourage you to study for the practice midterm. Don't use it as a study guide. Use it as an assessment for how prepared you are. Study, study, study, study, study. Sit down and take it like a final exam. Give yourself an hour for two hours.

Take it, see how you do. One of the most important things about doing well on an exam is practicing. And I don't mean just looking at problems. I mean practicing, taking the exam, time management, not panicking, et cetera, et cetera. So the more of my kind of exams you take, the more comfortable you'll be and the better you'll do.

So here's an opportunity for you to take an exam that I run. I'm going to post the solutions on Thursday and we can go over them in class on Thursday if people have questions. But I don't think that's going to take all of Thursday and so if you recall last week, I invited you, send me any problem from the book that you want solved or, and nobody sent me any problems,

like exactly zero problems, which means you guys spent the whole weekend studying and found no problems in the book you couldn't solve, for which I'm very proud of you guys. So what I want you to do next is choose any physics problem. Ask a physics question, something you can't figure out, whatever, a topic.

I'm interested in blah, blah, blah, I don't know what. I'll present something about it on Thursday, okay? Dial a lecture. So if you have a problem you want to see solved or you have a physics question, let me know. As long as you let me know before Wednesday evening, I'll prepare it for Thursday. Otherwise, you'll just do the practice final

and a bunch of problems I pick out from the book, none of which would be problems that are actually on the final. But I'm writing the final from the book. If you choose a problem from the book which is on the final and ask me to do it on Thursday, I will, okay? All right, questions about that?

And remember the 7C final is different from the 7LC final, because these are two different classes. It's sort of strange, I know, but LC is different from 7C. So this class is the final December 13th, 1030 to 1230 in this room, okay? And it'll work just like the midterm.

Please bring an acceptable calculator. I don't wanna have to walk around and take away people's calculators and makes me feel like a jerk, okay? You've had plenty of warning by now. Yes, I suggest you have a calculator. It's not that difficult a burden, I believe.

I encourage you to bring a calculator. You can pass the exam without a calculator. You could pass the exam by putting the pencil in your teeth, but I wouldn't recommend it. Other questions? So just to clarify, do you have a final in the lab class and then also a lecture title? That's right, because they are different classes officially.

Is it from 46th on Monday? I think it's from 46th on Monday. If you have a conflict with that time, send me an email. Okay? Is the 7LC final gonna be held in the lab room? Yes, it's in MSTB, whatever.

Yeah, I sent the information an email. Again, check your emails. I thought you guys were online all the time. You get emails in your brain, whatever. Yeah, didn't see it, huh? I noticed you said you provide an equation sheet. Are you gonna clarify all the equations or are we gonna need to derive some from the equations up there?

I'll provide all the equations you need. There will be some math required on the exam calendar. Any other questions? Okay, so here are all the topics that we've covered since the midterm.

We talked about conservative forces. We talked about those energy diagrams, potential energy. Momentum and impulse. We have collisions, which is fun. And then we did the whole rotation and the angle of momentum topic. And then finally, gravitation. So this is everything on top of the midterm. And I should say the final is cumulative in a way that physics is cumulative.

Everything you need for the second half of this class, you need the first half of the class for the second half of the class. The final will focus on problems and things we learned in the second half, but they're still gonna require you to know things from the first half. The first half seems simple and basic in comparison to what we did in the second half, right? You still need to know kinematics and energy

and that kind of stuff for the second half. But that's why today's review basically just covers the second half. So conservative forces. Remember, conservative forces are forces where the potential energy depends only on your position, not on how you got there, right? Independent of what your path was.

So the classical example is gravity. And here you have a runner who starts in some position. You got the low road, you got the high road. If she goes up to the high road, she spends a lot of internal energy creating positive potential energy, but then loses it when she comes back down. Similarly, if she takes a low road,

she gains kinetic energy, loses potential energy, but then gains potential energy back when she gets you. And this is because the potential energy is just dependent on height, right? It's M-G-H. So it doesn't matter, there's no history there, right? And it's linear. So if you go up, you gain, and when you come back down, you lose in exactly the same proportion as you gained.

Right, so there's two reasons there. One is that it only depends on the current situation. There's no derivatives or history there. And the other is that it's linear. Okay, but there are, of course, non-conservative forces like friction, right? If you have a choice to move something and you're sliding across the floor, so there's interface between the object and the floor,

and there's rubbing there, then you're doing work proportional to the distance traveled, right? The force applied times the distance traveled. So if you take a longer path to move this photon, right, so you have to move it around the table, it's gonna take more work than if you took the direct path, as if the table wasn't there.

And if you zigzag around and took it to 7-Eleven and came back, then it's gonna take a lot more work. Right, so here, the energy does not depend just on the position. So like by moving out of the way and then coming back, you've lost and then regained the position, regained that energy that you lost. The energy spent here is gone.

It's turned into internal energy of the two objects that have rubbed. It depends on the distance, not on the final position. So here, history is explicitly included in this D term. And we talked about a nice way to represent forces and to think about where they come from in terms of the potential.

And the relationship between the force and the potential is this one, and the force is the negative derivative of the potential with respect to distance. And I find this gives you a nice intuitive handle on how systems should work. If you can write down the potential function, so for example, a spring, has potential function one-half kx squared. That's the kinetic energy stored in a spring, right?

You can use your gravitational intuition, which you have because you exist in a world with macroscopic gravity, and think, what would happen if I put a ball in here? You could put the ball at the energy level of the object you're considering, and think about its motion gravitationally. So if you put a ball here, it would roll down, pass through here with a lot of velocity, come back up, and oscillate, which is exactly what happens if you pull on a string,

which is a moral equivalent of giving the ball energy. You put some energy into a ball, into an object attached to a spring, it's gonna oscillate in exactly the same way. All right, so you can get a sense for what's gonna happen to a system just by looking at the potential energy diagram, right? And if you can get the potential energy diagram,

then you can derive the forces. You don't have to, you know, look, just stare at the problem and derive them from your own brain. You can just take the derivative. It's really nice to have a mechanistic manner to derive the forces. And here, of course, the force is just negative kx, as we know from the spring, right? And gravity, if you think about what would happen here,

you put a ball on this surface, and this is just gonna roll down to negative infinity because gravity always just pulls you down. And the energy, of course, is the sum of potential and the kinetic. And here, it's just an elaboration on that.

If you think about an object tracked inside this well, it starts out with maximum potential energy and no kinetic energy, right? And then that turns into kinetic energy here as the potential energy goes to zero, right? This is its maximum kinetic energy, but the total energy is constant, right? Just switching between kinetic and potential energy.

So it goes from potential energy to kinetic energy, back to potential energy, right? And it's trapped, essentially, by this well. We never get out of here because being here would require more energy than it has, right? It has this much kinetic energy to start here, so it can never be up here. It would require having more energy than is available.

So it's sort of like a potential well. Later on, you guys will learn about these things for quantum mechanics, like an electron in an infinite potential well. So all solutions to these equations. And this is very generalizable if you have some arbitrary potential function. So here, for example, if you view this blue curve,

you can think about what the motion of different objects would be if they were trapped at a certain energy level, right? And remember, the extrema here are the cases where they have no kinetic energy and a lot of potential energy. This is too high up from the point of my finger, so I'll use the mouse, right? If you have an object with this much energy,

then at this point it's gonna have no kinetic energy because it's all potential, right? But it's still trapped by this bump here. It can move freely here, right? It can exist anywhere here because then it has some non-zero kinetic energy and non-zero potential energy. But to get over this bump would require more potential energy than is available in the system as kinetic energy, right?

It would require essentially negative kinetic energy here, which is, of course, not possible. And an object with a smaller amount of energy would get trapped in one of these wells. And it has a very complicated force structure, but in every case, you can just derive it by taking the negative derivatives. And I encourage you,

because there are example problems like this on the practice final, to do this in two steps because it's very easy to make a mistake, but take the derivative, then invert the derivative. Taking the derivative and inverting it, just make a mistake about half the time. Then we can get counterintuitive. So doing two steps, both of them are pretty easy, okay?

The topic of momentum is one that was very important. And first we identified momentum in its intuitive definition, which is mass times velocity, which we know. But then we connected it to acceleration.

Because remember acceleration is the derivative of velocity. So if p is equal to mv, then the derivative of momentum is mass times acceleration. And since mass times acceleration is nothing but force, that let us connect force to momentum. And when I first learned this,

I thought this was a really intuitive way to think about force. It's something that changes the momentum of an object, right? It's a much more direct definition than just mass times acceleration because different masses receive different accelerations for the same force, but it's very clear what a force is. And it changes the momentum. And whether that's in terms of mass or velocity,

that depends on the details of the object and its mass. But regardless of its mass, this is true. And also allows you to trivially derive the concept of conservation of momentum, right? There are no external forces on an object. No force means no change in this momentum. So its momentum is gonna be conserved, right?

And then there's the definition of impulse. Impulse is this change in momentum, right? So for example, we got this guy throwing a ball and there's a few quantities you can think about. You can think about the kinetic energy the ball has gained. So he's throwing, he's got a certain force. He applies that force over a distance. That's the change in kinetic energy, right?

This is also the question of, for how long was he applying the force? And that determines the impulse because that determines the change of momentum. He's applying the force over certain times. That changes the momentum. So delta P is the impulse. It's the change in the mind.

And this is important because it's harder to change momentum quickly than it is to change momentum slowly, right? If you, for example, a racket, like a tennis ball hitting a racket, it requires a very large force to change the direction of the ball, change the momentum of the ball, basically inversely, the momentum in a very small amount of time.

Time that the racket is in contact with the ball is a very small amount of time. So it's a large force that's required to change the momentum of the ball, which of course is the job of the racket, right? That's why you hit the ball with the racket. Okay. And then we thought about systems, an object with no external forces

and it doesn't have any change in its momentum. But sometimes we're interested in something that's happening inside an object. So we made this arbitrary definition. We call something a system when we think about the external forces on that system. And hopefully we constructed our system so there are no external forces so we can apply constraints like conservation momentum. And then we just need to solve problems

of what happens inside the force, inside the system. So here's the classical example, our two skaters on a rink. There's no friction due to the ice and there's no force on these skaters externally. They can do complex stuff internally, pulling each other, you know, dip each other, throw each other. But there's no change in their net momentum.

There's gonna be a very powerful constraint you can use to solve problems. Okay, and this helps you solve problems like the recoil of a rifle, right? Rifle, there's no external force on the rifle when you pull the trigger. And so the final momentum is gonna be the same as the initial momentum.

But of course the bullets moving very quickly that direction carries some momentum that way, requiring the gun to go that way, the opposite direction in order to have total net momentum be unchanged. And the reason, of course, that the bullet goes much faster than the gun is just because of the mass of the rifle is much larger than the mass of the bullet.

All right, they carry the same momentum, which is why the recoil is powerful. Remember, we saw that movie of all those people who underestimated recoil and hooked themselves in the head or dropped their pants or whatever, okay? And this brings us right to the topic of a collision. We thought, well, what happens when two things collide, right?

And one of the simplest cases is two sliders on a trap, so you have no friction. You know the initial velocities, right? You know the initial masses. If you're given one of the final velocities, then how do you figure out what the other velocity is? All right, we can use conservation momentum. The initial momentum is equal to the final momentum.

You know the initial momentum because you know the masses and velocities to start with. You know what the final momentum is. It has to be the same value, and you can decompose it into things you're interested in. So you know these masses, of course. You're given one of these velocities, and you can just solve for the other. So conservation momentum is very helpful if you're given one of those final states, right?

But what if you're not? Well, in that case, you're gonna apply another constraint, or you can add the momentum constraint, but if you don't have VA or VB in the final case, you have two equations in one, sorry, two unknowns in one equation, can't solve that without more information. So if you know that the collision is elastic

because they've bumped, you can imagine there's a little spring here. In fact, in this diagram, you can draw these little bumpers to tell you that it's elastic, then conservation of kinetic energy can be applied, and then you have two constraints and two equations, right? But there are other kinds of collisions, of course.

We have elastic collisions where the kinetic energy is stored and the potential energy of these little springs between them, right? That's the kind of thing that we like to think about. It's convenient, it's simple, it's clean, but there are also inelastic collisions where these things stick together, energy is spent forming this new object. And so after the collision,

you can't say that all the kinetic energy is restored, right? And we thought about some example, some thought experiments where we could prove, remember this, we thought, can you prove that two objects stuck together have smaller kinetic energy than the initial state or that kinetic energy is not conserved?

Anybody remember the simple proof of that concept? We've got a small object, it's a super big object, it doesn't matter. Small object hits a super big object and? Like they stick to each other, they won't move. Okay, and so what do you conclude?

Okay, I don't quite see how that follows. Well, I remember there was like something mathematical, both examples, just that velocity and energy of the equation is very good, and the momentum of the equation is linear,

it's the power of one. So whatever the equation is, this is in the, or this is in the momentum part, it's going to decrease by zero, zero and square. Okay, yeah, I'm sure there's a mathematical

way to extract that. But just think about the rest frame of this final object, right? Transfer it to a frame, some inertial frame, where this thing is at rest. What's the kinetic energy in that frame? Zero, right? In that same frame, what is the kinetic energy of the initial state?

It's non-zero, right? Because you have two objects moving relative to each other. So in that frame, I mean, in the inelastic case, there's a frame where the final state has no kinetic energy. But the initial state obviously has greater than zero kinetic energy. So that proves that kinetic energy is lost.

You can't do the same thing in the elastic case, right? In the elastic case, there is no frame in which there's zero kinetic energy, because you have two objects moving. There's no frame you can choose in which both objects have zero velocity. Okay, and so to summarize, in both cases, elastic and elastic momentum is conserved, right?

Kinetic energy is conserved in elastic collisions, and in elastic collisions, it's reduced, right? Not conserved doesn't mean you can gain kinetic energy, it just means it's reduced. But the total energy is always conserved. That kinetic energy goes somewhere, internal energy or heat or something, right? Okay, every time we've talked about compact objects

and aggregates of objects with different positions, we thought about the case of lots of particles, all the different positions, so R1, R2, R3. How do you treat the motion of a big, complex object? We thought, well, let's think about where the center of mass of this thing would be. If we define RCM to be the center of mass,

it's a mass weighted average, right? Where particles that have more mass get larger weight than its average, see? If we were just averaging, it would be R1 plus R2 plus whatever, divided by the number of particles, right? This is a weighted average, where each term gets a different contribution depending on its fractional weight, M1 over all the masses, M2 over all the masses.

Those are the weights. And similarly, we can define the velocity just by taking the derivative of this thing, right? Derivative of R1 is V1, derivative of R2 is V2. And then we can define this, this is the total mass, and the total momentum is the total mass

times the velocity of the center of mass, which is just the sum of all these guys, right? Because VCN is the sum of the momentum of the individual objects divided by the total mass. So therefore, the total momentum is just the sum of the momentum of each of the particles, which makes sense. And what are you supposed to take away from this?

This is just a construct, this center of mass construct, just shows you the intuitive derivation of it. Why it makes sense for it to be a mass weighted average, and why the velocity should also be a mass weighted average. And then we calculate the, where the center of mass is for simple objects,

and by symmetry, it has to be in the middle, right? And for objects with symmetry, it needs to be in the geometric center. And that means if you're flat, if you're donut, even if there is no mass, at the center of mass, it's still at the center of your bi-symmetry, and you don't actually mean stuff at the center of mass.

And then because the center of mass is just a sum of parts, if you have a complicated object, you can calculate the center of mass of the sub-objects, and then have it together, right? Because it is just a sum. So you can always consider that you're breaking up each of these center of mass calculations into their sum of the particles inside there.

And so you can generalize it to break it up sort of recursively. The center of mass of a complete system is the mass weighted average of the center of mass of the individual of the subsystems, which are in turn, of course, calculated by adding up the pieces inside. But you don't want to have to do this integral over the shape of a oxygen, of a water molecule,

but you can do the individual spheres and add them together, okay? And the whole concept here, the whole reason we focus on center of mass is to take a hard problem, which is when the whole system is moving, and there's internal motion, and break that up into two problems we can solve. Often in physics we come to a problem, we don't know how to solve it, so we break it into two pieces,

and hopefully we can solve those pieces separately. So for example, I remember I tried to draw this on the board and looked carefully, but if you have a motion like this, and you think, put a dot on here, and understand its motion, you see it's quite complicated, right? It doesn't just fall straight down. The center of mass is of course falling straight down, but if you look at just the edge,

right, the one edge of the wrench, then it swings around in a complicated motion, which would be difficult to derive, unless you break it into two pieces. Falling motion in the center of mass, which you know how to do, it's just free falling objects, and then rotation with respect to the center of mass, which is also simple, because it's just circular motion. All right, so you can combine those two things

to get a fairly simple motion, a fairly simple understanding of complex motion. The same thing applies to something like this. You shoot a rocket up in the air, it explodes in the middle of the air, its center of mass is not gonna change from a parabolic motion. Why? Because there's no external force. It's got an external kick when you launch it here, right?

But then it's lost its drive or whatever, it's just a projectile, and there's no external forces on it. There's an internal force, I'm trying to some sort of internal fuel through an explosion, it breaks into two pieces, velocity relative to the center of mass, right? So all you need to do to calculate the trajectory of this one or that one,

is calculate the trajectory of the center of mass, which is the projectile motion, and add to it the motion of this object with respect to the center of mass. And in a problem like this, you'll be given enough information to do that. They'll say, each projectile has one meter per second of velocity with respect to the center of mass, or they'll tell you the force of the explosion

from which you can calculate these things. And so it's always two simple kinds of motion added together that give you a more complex motion. It'd be very difficult however to in one step derive this function, which has some sort of a kink in it here. Okay, and we took all the lessons that we learned of motion in linear motion,

translation of motion, applied it to rotational motion. And we tried to use all the same ideas because we didn't wanna get confused in our heads and have a whole different system for thinking about rotation. We tried to solve every problem with the same small set of tools. In the end, that's sort of our philosophy of physics, right? Build the smallest number of tools with which you can solve any problem.

So it wouldn't make sense to develop a whole new set of equations and tools just for rotation, but it really is very similar to translation of motion. And so of course the basic quantity is just the coordinate. Rather than having x, now we have this coordinate of an angle, which is always measured with respect to the x-axis, right? And we use units of radians, which are sort of bizarre but also sort of natural.

I mean, the definition of a radian is the angle at which the arc length s is the same length as the radius r. So that's sort of cool, but it's just the definition. All right, and then we expanded that to velocity. Linear velocity is of course dx, dt.

So here we have d theta, dt. And the only wrinkle in our extrapolation from linear from translational motion to rotational motion is this question of the direction of the velocity, right? In translational motion, we have, you know, we know the direction we're going, so we can just say velocity points in that direction. In rotational motion, we're spinning either clockwise or counterclockwise

as the vector can't be moving, right? We need to have a stationary vector to describe the magnitude and the direction. So we take the length of the vector to be the magnitude, and then we put it along the axis of rotation because that defines where the rotation is, right, around the axis of rotation. And then we have enough freedom to say up or down

that gives us the positive or negative sign for rotating one way or the other. And it's totally arbitrary. We could have said up was clockwise or up was counterclockwise, but most people in the world are right-handed, and so it's natural for most people to just use the right hand and say motion in this direction is up.

All right, that's the one big distinction between linear, translational, and rotational motion. And the same story applies for angular acceleration. It's just the derivative of angular velocity and the same game can be applied for the direction of angular acceleration, right? If your angular acceleration's in the same direction as your angular velocity, you're speeding up. If you're in opposite directions, you're slowing down.

So as long as you have the same standard for acceleration and for velocity, you can use your intuition there. And all of the equations we had formed were kinematics in translational cases, giving math to the same equations, I love this, to the same equations in rotation

except with the different variables, right? Every time you have x, you put a theta. Every time you have v, you put an omega. Every time you have an a, you put an alpha. And so all of your intuition about how this stuff works and all your memorizing of these equations can be applied trivially. In fact, in my opinion, this is really the only equation for kinematics

you really have to remember, right? Because this one is a simple case of that one, right? This is a special case, right? Can be derived from here. So this is really the fundamental equation for kinematics. And the one for rotation is essentially identical just for different coordinates.

Okay, and then we took on more complex topics like rotational energy, right? And here, I think in the example in class, it's something like a rotating z drug, a spit or something. But here, we just have a bunch of particles. Each one has kinetic energy of its motion, one half mv squared. And the kinetic energy of an object,

which is of course made of particles, is just some of the kinetic energy of all those objects. But we replace v with r omega, right? We wanna think about it in terms of angular velocity and radius. And then we grouped all the mr squareds together and said, we can write the kinetic energy of rotation as one half i omega squared

if we define this new concept i, which is mr squared summed over all the objects, right? It's called the moment of inertia. And it's our attempt to extend translational motion to kinetic, to rotational motion that we built this concept of basically rotational mass or moment of inertia, right?

And remember, we saw the video of the girls with the juice boxes doing that experiment. It's harder to rotate something if the masses are far away from the axis of rotation, right? It has a larger moment of inertia. It means it's gonna contain more energy, which means it takes more work. It feels harder to get something spinning

if it has a larger moment of inertia. And the moment of inertia goes like r squared. So as you increase r, it grows very rapidly. And it depends of course on the axis you've chosen, right? Here we have this crazy object. If you're rotating around axis b, then mostly you're dominated by the position and the mass of object x. If you're rotating around an axis

that points through object a, then it's gonna be dominated by masses b and c and their distance from the axis. So you can't say this object has a certain moment of inertia and be done with it. It has a certain moment of inertia around the given axis. These things are simple to calculate for particles, right? It's just mr squared or sum.

But if you have an object which is continuous, we think of it as a continuous piece of mass and then you're gonna need to do an integral because every piece of mass is a different radius. So you can't trivially add this stuff up. Smoothly varying functions, you need to weigh together and you do pieces of the integral and it's the integral of radius squared. This is the mass turns into the dead state, right?

You have this volume integral, which can be tricky and complicated. So for most standard cases, it's just been done for you. And I'm never gonna ask you to do this integral in some sort of crazy time crunch situation like a final exam. So if you need this on a problem, of course,

I'm gonna provide it. Some of you seem to memorize them, and some of them are easy like this one. All of the mass is at distance r, so the interval is just mr squared. But the key thing to understand is the further away from the axis of rotation that the mass is distributed, the larger the moment of inertia, okay?

And remember we did that problem where the balls are rolling down the plane and it's different arrangements as a solid sphere and a thin sphere and a hollow cylinder and a full cylinder. We thought about which one's gonna go fastest and dependent all on the relative moments of inertia, okay? And then angular forces. Here we've introduced the concept of torque.

And torque is very similar to force in that it causes an angular acceleration the way force causes a translational acceleration. But the key is the direction of the force matters and the position where the force is applied matters relative to the axis of rotation, right? So in this case, Fc is not gonna do anything to rotate that nut

because it's pointed towards the axis of rotation. Fa and Fb are gonna have more luck because they're pointed perpendicular to this axis of rotation, to the lever arm. And force b is gonna be more effective than force a because it's further away, right? The longer the lever arm, the more effective the force is gonna be in producing torque, right?

So if you wanna think about rotating in some crazy shaped object like a and you apply different forces to it, you have to think about the magnitude of the force and the length of the lever arm, right? And the way you figure out the lever arm is you draw a line of action through the force

and you take the lever arm as the perpendicular line from the line of action to the rotation, right? So if you're rotating around, for example, point O. Point O here, F3 has no lever arm because the line of action goes right through the force, right through the rotation point, and that makes sense. If you're pulling on the axis, you're not gonna spin this thing.

You have to pull off axis in order to get any spin, right? F1 and F2 are gonna get it to spin in different directions, right? Which makes sense also because one's spinning on the top, one's on the bottom, and they have different lever arms. So in a simple case like this, when the lever arm and the line of action are perpendicular,

then the torque is just the magnitude of the force times the lever arm. But in more complex situations, you need to evaluate, take into account also the angle between the force and the distance between where the force is applied in the axis of rotation. And so then we have the vector form, R cross F, where R and F are vectors, right?

When we used cross product, we reviewed cross product to remember that A cross B is out of the plane, is normal to the plane created by A and B. Because remember two vectors define a plane and the cross product is perpendicular to that plane. And A cross B is not the same as B cross A, right?

And so if you wanna calculate torque, there's a few ways you can do it and all get the same answer. So here you have a lever, literally, and somebody pulling on it. You define a force straight up in this case, right? So one thing you can do is say, well, I'm gonna project the line of action down here

until I can make a perpendicular line to the axis of rotation. This is my lever arm here. It's R sine phi, right? It's not the full R. The full R is the length of this arm. The lever arm here is R sine phi. And I can just multiply the force times the lever arm. So I get FR sine phi.

Or I can say, I'm just gonna take the component of the force that's gonna do me any good, right? Not the radial component, which is not gonna turn the thing, which is the tangential component, which is F sine phi. Multiply that by R and I get RF sine phi. Same answer. Or I can say, I don't care about all that. I'm just gonna take the equation I know,

which is R cross F, and I'll get RF times sine of the angle between, because that is the cross product equation after all. And you get RF sine phi. So, you know, math works. It's all self-consistent. There's lots of ways to think about how to solve this problem. They all give you the same answer. All right, either you're taking the fraction, the component of the force that's doing anything useful,

or you're projecting the line of action. Okay, and this helps us, if we understand rotation, it helps us solve complicated problems like throwing a baton and spinning at the same time, because we can break it down into two kinds of motion we already know. One is projectile motion of an object,

and here we're just thinking about the center of mass. So while it is rotating, we don't draw it as rotating, because we're only thinking about the motion of the center of mass, plus simple rotation. And that lets us describe a rather complex motion of one object, one end of its constant. And it has kinetic energy both in its translation and in its rotation.

And then we talked about rolling. And this is, this can be quite counterintuitive if you don't break it down. If you have a wheel which is rolling without slipping, we define rolling without slipping to mean that the wheel has no velocity with respect to the ground, the point that it's touching. That doesn't mean the wheel is not moving.

It means that at any moment, the point that's touching the ground is not slipping with respect to the ground. Velocity is zero with respect to the ground here. And the reason the wheel moves is not because the point that's touching the ground is moving, but rather different points then come into contact with the ground. It's this point, then it's this point, then it's this point, then it's this point.

And so you think about that as motion of the center of mass, where every part of the wheel is the same velocity. One thing moving, motion of the center of mass, plus the rotation about the center of mass, the way to get rolling without slipping is for this velocity to be exactly countering this velocity.

Which means, of course, at the top, you're gonna get two times the velocity of the center of mass. And so that gives you the wheel instantaneously at rest. And this only happens if the tangential velocity is equal to the center of mass velocity, which means vcn equals r omega. That's the critical thing.

If you're doing a problem which is where you get rolling without slipping, it allows you to connect the velocity of the center of mass to the tangential velocity of the edge of the wheel. Because you know the relationship between them in order to have it roll without slipping, which usually solves your problem because it reduces one degree of freedom in the problem,

one unknown rule, gives you an extra equation you can use to handle the problem. But of course, this requires friction. You can't roll without slipping on a frictionless surface. If you try to ride a bike on a pure sheet of ice, the wheels are just gonna spin. There's gonna be no connection

between the motion of your center of mass and the angular velocity of the wheels. And then rotational work is a natural extension. Remember, linear work is just force times distance, so rotational work is just gonna be tau times theta, our new rotational force, our rotational coordinate, exactly the same equation.

So here, for example, she's applying a tangential force. The angle through which she applies it determines the angular work that she has done, the rotational work that she has done. And the last concept in the rotation universe was this one of linear momentum versus angular momentum.

Linear momentum of course is just MB. Angular momentum is gonna be linear momentum times R. It depends on the axis about which you're rotating and how far you are from that axis, which can also be written as I omega. We have the same definition of the direction

of angular momentum that we use for angular velocity. So if you're rotating a certain direction, the angular momentum points the direction predicted by your thumb. Okay, and then we say, well, what about crazy-shaped objects? You know, you have some weird blob, and I think this is just something I got by typing weird blob into Google research,

and this book came out. And if you wanted to calculate the rotational, the angular momentum of this object, you'd say, well, I'm just gonna calculate it by summing over all the tiny little particles inside this weird blob, right? But what is the angular momentum of a bunch of weird particles where each one has MR squared omega, right?

And then you could just take the MR, every single particle has the same rotational velocity. So you can use a single omega, take that out of the sum, and then recognize that the rest of it is just the moment of inertia, just MR squared. So the total angular momentum of any shaped object

can be described by I omega, where I encapsulates the shape of the object, right? Any weirdness in the shape of the object is described by the moment of inertia. So if you're given the moment of inertia, or you can calculate it, then you can describe it in terms of simple motion, as long as it all has the same angular velocity, right? If it's a soft object, so it's not rigid,

then it's gonna be much more complicated, but we're usually dealing with the motion of rigid objects. And just like in the case of translational motion, we thought about how force is connected to a change in momentum. Here, torque, which is I alpha, right, moment of inertia times acceleration, this is the same way F is MA,

can be connected to change in angular momentum, which means no torque, no change in angular momentum. So no external torque means no change in angular momentum, conservation of angular momentum, right? If there's no external torque, tau is equal to zero, dL dt is equal to zero, that means that L is constant.

So angular momentum is constant if there's no external force, okay? And then the general definition of equilibrium was not only translational, that is that there's no net force in the object, so there's no linear acceleration, but also there's no net torque, because you can imagine an object with no net force on it,

but where something is being spun, and in that case, it's not in equilibrium. So no torque and no force. And remember, we did that problem with the marbles in the jar. We figured out what all the forces in the marbles were, we needed not only to require to the agreement of the forces, but also that there was no rotational total net torque around any axis,

which gave us one of the constraints we needed, okay? And then the last topic was gravity, and gravity really is a mysterious topic, and Professor Mircea lectured to you guys about this. You know, gravity's fascinating because on one hand, it's super weak, right? It's the weakest of all the forces. On the other hand, it's the one that dominates

the large-scale structure of the universe, right? Do you guys know what the largest structures known in the universe are? So obviously you know about the solar system, right? Solar system's part of the galaxy. Galaxies tend to, galaxies are obviously very far apart. The distance between stars in a galaxy is much smaller than the distance between galaxies.

But the galaxies have gravitational influence on each other, right? So even though our sun is, of course, rotating around the center of the galaxy, galaxies cluster together, and it turns out there's these galactic clusters, right? A bunch of galaxies closer to each other than they are to something else that have a constant gravitational, sorry, that have a self-consistent gravitational interaction.

They're like rotating around the center of mass of the cluster of galaxies. Bigger than that, there are clusters of galactic clusters. They call these super clusters, okay? So these, I mean, this scale's already mind-boggling, right? So these are super clusters of galaxies, clusters of galactic clusters, each of which contain galaxies, which contains zillions of stars, okay?

Then they started looking, and they noticed that these super clusters tend to line up in sheets, okay? So if you look in one direction, the super cluster seems to go on forever. If you look in another direction, you run out of super clusters pretty quick, beyond which are enormous voids. So the super, the galactic super clusters are in sheets, and if you keep looking

further and further out, and this is what Hubble was really good at, you notice that the galactic super clusters form bubbles. So there are these voids in which there is nothing. It's just incredible voids, which are the centers of these bubbles, the surface of which are galactic super clusters, which of course contain clusters of galaxies,

which contain galaxies. So this is the largest scale structure that we've ever discovered in the universe, right? And the thought is that these bubbles, these are like the initial foam from the initial states of the, quantum states of the universe. We had random fluctuations from when the universe was like the size of a dot. You get these little fluctuations, which then get expanded to a ginormous scale

by inflation and cosmic evolution. And so you look at those bubbles, and you're looking at the initial first, second two years of the universe. So it's really, it's amazing stuff. It just goes to show you what you can learn by just looking, right? You have no idea what you're gonna find. You just look, and you'll find crazy stuff.

Anyway, gravity is what dominates all this, right? If you didn't have those initial fluctuations, imagine the universe started out as just a point, and you have an explosion, and there's just gravity after that, right? Then the universe is gonna be spherically symmetric, right? And that doesn't lead to a very interesting universe. It's hard to have life if you don't have,

inhomogeneity of density of matter, right? And so all of the interesting structure of the universe has to come from somewhere, and it has to come randomly. So it comes from these quantum fluctuations in the initial first few moments of the universe. That seeded all the inhomogeneity that we have now,

which determines basically everything about the structure of the universe. And after that, gravity just takes over. Okay, so we do know a lot about gravity. We know how to write in terms of the force equation, right, because it's proportional to the masses and to this constant G. We don't know why G seems to be so small,

but through experiments, we've determined the structure of this force. You know, we can measure the gravitational force of different radii. They've done this from distances down to like now less than a centimeter up to, you know, obviously cosmic scales. They can look at rotations of galaxies and interactions between super clusters of galaxies. Testing it at a really small scale is very difficult

because the masses are so small. Obviously, if you want to test two small things at a small scale, you need small masses, physically small masses, which means low values in the mass, which means it's very difficult to test. For a while, people thought that perhaps there were additional spatial dimensions, like spatial dimensions that were on the scale

of like a centimeter that we just couldn't tell if they were there because we couldn't probe gravity on that scale. But then some guys at the University of Washington set out to disprove it and they built some very, very sensitive experiments and showed that gravitational force follows this rule even down to a millimeter scale, which is crazy that over like 20 orders of magnitude,

this equation holds. Okay, and so any two objects with mass can be attracted to each other, right? Here's the equation. And the force is always attracted because gravity, because mass is a little bit positive. And the nice thing about gravity is that you can do the same trick we usually do,

which is take a complex object and summarize it in terms of a point particle. You have some weirdly shaped object. You want to think about the gravitational impact of it. You say, where's the center of mass? I'm gonna put a mass, a particle with that same mass at the center of mass of that particle and then I don't need to worry about the shape of the object. As long as I'm outside the object. If you're inside the object, it's more complicated, right?

But as long as you're outside an object, you can replace any object with a point particle at its center of mass with a force of the same mass and you get all the right results. So when we think about how the earth goes around the sun, we don't need to think about the gravitational effect of every piece of the sun, every piece of the earth. We just think about two point particles.

The sun and the earth. Did anybody here see Comet Ison? Yeah. Anybody see, you actually saw with your eyes? No. No? Nobody went and looked at it? Pretty crazy cosmic event though. Okay. And remember that gravitational force

is of course a vector, it has a direction and so you can just add the vectors like you can with anything else and you can get confused easily with gravitational problems in multiple dimensions, but just break them down into X and Y. Remember we did that earlier in the year. Every time you have a two dimensional problem, turn to one dimensional problems.

Okay. And of course, Earth's gravity, you just say, you know, one of these ends is the mass of the earth, the other one is the mass of your test object and so if you're at the radius of the earth, then this force of course is just MG. That's where the solution comes from, right? Comes from the evaluation of this function and the mass of the earth. And of course it's gonna fall off very rapidly

as you go away from the earth, right? This is the surface of the earth and as you get further and further away, right? What happens below here? Does this keep shooting up towards the crazy values? Decreases. Decreases? How do you know it decreases? This is directly proportional to the radius.

It's proportional to the radius. Just what happens at zero? Zero. Zero, right? If you're in the center of the earth, okay, you're cooked like a... There's no gravitational force on you, right? Because all the gravitational forces cancel. There's no net gravitational force on you, okay? And gravitational work is just motion

through the potential, the gravitational potential, right? So it's a force. You apply the force through a certain distance and you calculate the work. It's just an integral of FDR where F happens to be this gravitational function. So no special knowledge needed to calculate gravitational work, just your knowledge of gravitational force, the dependence of it,

and already your knowledge of how work is calculated. Just plug that in, right? Okay, what about light, right? I told you that gravity attracts anything with mass, right? GMM over R squared. What about how much mass do photons have?

Zero, right? Does gravity attract photons? Yes. Yes? So how do you reconcile all those ideas? Gravity works on things with mass. Photons have no mass. Gravity works on photons. Gravity works on energy. Gravity works on energy, exactly. So our simple understanding of gravity

and Newtonian gravity is it works on mass, right? But general relativity tells us it actually works on energy. So energy is what bends space-time and makes things attract each other, gravitation and energy. None of this, of course, is what you need to know for the final, just for your own dedication as physics majors. But that's why gravity bends light,

because it's because light has energy and gravity actually works on energy density, okay? Okay, so that's the whole final review. I have a couple of examples here that are pretty basic, but do you have questions about what's gonna be covered or anything else?

The other comment I wanna make is evaluations. The evaluations, I think, are open now on triple E. Please go in and fill it out. You know it's anonymous. It's very important for me to hear about what your experience was like, how to improve it, feel free to vent, whatever.

These things do get taken into account when I go up for being promoted. They read the student evaluations, and if somebody wrote anything particularly nasty, that kind of stuff comes up. Anyway, so you have some power, you have some influence. But also, you can influence the way I will teach this class next year. So if you have ideas, something you really liked,

tell me if something you really didn't like, tell me if this, whatever, something about the class that you were too embarrassed to tell me that really bothered you, please fill out the evaluation. Also, if you really love the class, please fill out the evaluation. I don't think they're closed.

Anyway, if you guys filled out the evaluations, thank you. If you didn't, please go fill out the evaluation. Okay? Okay. Anything else? All right, well, I can go through this problem, but it looks like you guys are antsy to get going. Should I go through it? Yeah, all right.

Okay, so here's an example problem. We have a large crate with mass m on the floor. We have a coefficient of friction between the crate and the floor, which is mu. So we have static friction and kinetic friction, okay? A woman pushes downward and then angle theta below the horizontal on the crate.

What is the magnitude of the force? What magnitude of force is required to keep the crate moving at constant velocity? Okay, so constant velocity means no acceleration. No acceleration means no net force, all right? So here we're gonna have to deal with kinetic friction, right? So first here's our free-body diagram. We have mg down force.

We have the normal force up. We have frictional force resisting the motion. And we have this woman, for some reason, pushing down on the subject. And so in y, we have down, we have gravity down. We have the normal force up. And we have the component of her pushing down. So the normal force is gonna be whatever is necessary

to avoid crushing the floor. So it's gonna counteract her pushing down and the force of gravity. Once we have the normal force, we can figure out what the force of friction is. Friction is, of course, just mu times n. And in the x direction, we have the cosine theta component of her pushing and the frictional force.

These two things have to balance because we have no net force, which is why we have constant velocity. So we know f cosine theta. We can take the little f and replace it with mu k n, where n is the normal force that we solve for using the y component. And then we just plug in and we have everything we need to calculate what f is.

So that was the question was, what magnitude of force is required? Then it says if the static friction is greater than some critical value, the woman cannot start the crate moving no matter how hard she pushes. Calculate this critical value under x.

Why is that? If the static friction is large enough, she can't get it moving no matter what. No matter how hard she pushes. Remember that the direction she's pushing is fixed.

Now we're just talking about the magnitude. So the harder she pushes, the harder she's pushing down as well as sideways, meaning that the normal force is gonna grow, which means the frictional force is gonna grow, right? So for some value of the frictional force, the down compensates more than the sideways

and she can't push it. So we still have the same arrangement, mg f sine theta equals n for the y component. Now the x component, we have mu s, right? Because now we have static friction. So, but the equation looks very similar. We can solve this equation, but there are some configurations for which f is infinite, right?

And the configuration is when this denominator here becomes zero. So based on what mu s is, if this condition is zero, then there's just no way for her to push hard enough to get this thing moving. Because the harder she pushes, the more frictional force she's inducing, which counteracts the force she's applying.

Okay? Here's another one. Blocks A, B, and C are placed as in by ropes and negligible mass.

Placed in, as in the figure, I guess. Both A and B, we lost some of the, all right, we're gonna have to guess where the missing text is. It's like it is last week in November. Okay, so this kinetic friction, right? Block C defends with constant

velocity, I'm thinking. Constant move, constant smell. Block C descends with constant velocity. Draw two separate block free body diagrams, okay? So there is free body diagram for A has its weight and the force, tension and force.

Free body diagram for B, there's two tensions. There's friction, there's weight, and there's a normal force, okay? Find the tension in the ropes connecting something with B, B, B, C. Does anybody happen to have their book with them?

5.45? Oh wait, here it is. Okay, block, block C descends with constant velocity. Draw two separate free body diagrams

for the force acting in A and B. Okay, we'll do that. Find the tension in the rope connecting blocks A and B. A, find the tension in the rope connecting blocks A and B. Okay, so there is no net force on A, right? Because the whole thing is moving

with constant velocity, right? So that means that all these forces have to balance. There's only two forces on A, there's the tension and the friction. So the tension here has to equal the friction on A. That's what I'm saying. T equals mu A. Find the tension in the rope connecting blocks A and B. So this one has to be equal to this,

and we know what the frictional coefficient is, and we know the weight of A, 25 newtons, so we can calculate this force, this tension here. Next, it asks us, what is the weight of block C?

Okay, so we're gonna need to use the fact that C is in equilibrium also, right? Okay, C has tension up and weight down, so the weight of C is the same as the tension on this rope, right? We have to find the weight of C, which is really another way of saying, what is the tension on this rope? And so the tension on the rope

is really what we're looking for, and to solve that, we can use B, right? Because here we have tension one, which we already solved from looking at A. We have weight, we have normal force, we have tension two, and we have that friction here, but what we really wanna find out is tension two, and so we can just look at this component, right?

Because the normal force and the weight are gonna cancel each other out, and so T1 plus FB equals T2, in this case, and we know T1, and we know the friction on D, right? And so we can solve, and so that one's easy to solve.

Yeah, I was just thinking that. The N2 is gonna balance part of the weight, right? The other, there is some point of the weight that's along the surface. Yeah, I think there's a missing term, but.

angle. So it should be T1 plus FB plus WB sine theta. Sine. Yeah, because if the angle is zero, it should be zero. Thanks. I'll give you some extra credits for noticing

that. Okay. Here's another one. Angle for minimum course. This one says a box of weight W is pulled at constant speed along a level 4 by a force at an angle theta above the horizontal this time. The coefficient of kinetic friction is mu K. The terms of

mu K and W calculate F. Okay, so calculate F, I guess, so that it's pulled at constant speed. And again, this is just draw the two by a diagram. Right, here's the forces. Balance the forces in X and Y. If we do Y, then we can derive the normal force because it's got the constant balancing weight and it's diluted a little bit by this guy

pulling up. Right, the normal force is smaller than it would be without this force because this is pulling it up, essentially reducing the weight. The normal force is W minus the concentration from here, so there's a reduced weight. And that's the normal force which

helps us get the frictional force because friction is mu times N. Okay, I've just plugged in N and that's going to balance F cosine theta. Right, and we know they balance because it's constant speed which means no acceleration, which means no force, so which means the net is zero. And so then here again, this looks very similar to the previous problem where you're

pushing down. Of course, the angles are different, the signs are different. You can calculate what the force is, you know, or have constant velocity. And then it says for certain values, calculate F for theta ranging from zero to 90 increments of 10 degrees. Okay, so we have

the function, we just evaluate it for different values of theta. It looks like this. For larger values of theta, the force required gets larger. Does that make sense? Does that make sense? If you're pushing horizontally, then you need less force. Theta equals

zero required less force than theta equals 90 degrees, for example. Why is that? Because you're what, sorry? Right, but shouldn't it be the opposite then? Because you're pulling

and you're applying a force this way, you're lessening the frictional force by lessening the weight of the object, right? So there are two effects here which is why it's a complicated function. On one hand, you are lessening the frictional force but you're

also applying less force horizontally which is what you need to get the motion to go. So the second part of the problem says, from the general expression part A, calculate the value of theta for which the value of F is a minimum. Okay, so it says essentially at what angle is the force required for constant speed a minimum? So it's a minimization

problem. You take the expression, take the derivative with respect to theta, find the value for which the second derivative is positive, because we want a minimum, right? And you can solve this, you can differentiate this equation, set it equal to zero and you

get this expression, right? Because it has the value of theta for which F is a minimum, so theta is arctangent of the kinetic friction, right? Okay, those are all the examples I have for today. So remember, for Thursday, we will go over the practice final and

I'll solve any problems you guys send me or describe any kind of topic you request by Wednesday evening. So we have a quiz tomorrow. Are the quizzes ever announced in advance? No. Well at least you should write your own conclusion then. Okay, any other questions?

Okay, see you guys on Thursday.