Okay, let's go ahead and get started. I'm going to leave this up because it's a cool picture of roller coasters and we'll come back to it.

Quiz? Very good. Most people did quite well. Notice there's a lot more green than there was on Monday. Just so you know, the next quiz will open today and will be due on Wednesday's lecture. Now, notice the biggest area of red. People are still trying to call something the centripetal force.

There is no centripetal force. Let me repeat that. There is no centripetal force. There is a centripetal acceleration, and if you happen to have a centripetal acceleration, there must be a net force coming from lots of sources that provides that centripetal acceleration.

But it will be usually a combination of many forces, and there is no force that is the centripetal force. So keep that in mind. Other than that, the quiz looked very good. Now, the other great thing from the quiz, and I'm going to put up this nice comment from it,

and the reason this is very useful, a number of people made comments like this. Notice how this is phrased.

Does friction counter component of gravity that pushes the block, or is it countering the force that is supplied? A number of people said this. Friction does not counter forces.

So that's going to be a source of confusion for you if you don't get rid of that idea. Friction does not oppose forces. Friction opposes attempted motion. So you want to go back, if you were thinking of it as friction opposing forces,

you need to go back particularly to that series I showed you on the website of the three blocks with friction. Version one, version two, version three. Because it really gets at this idea of friction opposes attempted motion.

Also the two blocks on top of each other. Additional podcast focuses on how do you identify the attempted motion that friction is going to oppose. And remember, it's tricky, because if I ask myself how do I walk forward,

you would think my attempted motion is moving forward, and so friction should be backwards. But the attempted motion is my foot pushing backwards on the floor. So the attempted motion is sliding backwards, so friction is forwards and actually makes me move.

Friction is moving you forward. So that, notice there's no force friction is opposing in that case. Because that's not what friction does. Friction opposes an attempted motion. Questions on that? I'm really good that this was asked on the quiz in exactly this way,

because it does highlight one of the challenges with friction. Now, the other thing just to keep in mind, if you go back and look at it, kinetic friction does show up whenever the object is moving.

Whether it's moving at constant velocity, or whether it's moving with a changing velocity, you use kinetic friction. Static friction is literally that, static, when it is not moving. And when it's just about to move, but it's not moving yet, and it's not accelerating yet, that's still static friction. Once it starts moving, that's kinetic friction.

And again, that sequence on the blocks on the ramp will hopefully clarify that. Questions? So today, we are doing circular motion, and we're going to look at what happens with,

actually, I don't want the arrow, centripetal acceleration with the magnitude v squared over r. And it's always in the direction, we're going to see what that means, radially inward.

And it could be an instantaneous acceleration if v is changing. But it can also be, and this is a little tricky, you've got to listen carefully, it's still an instantaneous acceleration with a constant magnitude if v, the speed, is constant.

But notice, since I'm moving in the circle, the direction is always changing.

So this is never constant acceleration, so I can never, a common error at some point is you see a circular motion, you compute an acceleration v squared over r, and then you plug it into the equations of constant acceleration. Does not work, because it is not constant acceleration. Because the direction, as I move in a circle, right, here's my spin thing in a circle, right,

as I move in a circle, the direction of radially inward is always changing. So the direction is always changing. Now, the other thing is, there are lots and lots of places where you get circles,

which is why I wanted to keep the roller coaster up. You should be able to see right here, these are vertical loop-de-loops, right? They're a circle, and they're vertical. So radially inward is down when I'm at the top, when I'm on the side it's parallel to the ground,

when I'm on the bottom radially inward is up. So you can see radially inward, as I said, always changes. This circle here is roughly horizontal, goes a little bit down. Again, radially inward is always changing, but it tends to always be horizontal. Maybe it's slightly tilted. So we can have circles fundamentally at any angle,

but the dominant ones are to have circles that are vertical and horizontal.

What problem do you see with those two pictures? If I didn't label them, which one is the vertical circle? You don't know. So you really have to be able to picture in your head whether I'm vertical or horizontal. Why does it matter whether I'm vertical or horizontal?

Gravity, right? If I do a free-body diagram in this case, I'm going to pick x and y like that. Maybe my object is here. I'm going to have my object, and I will have weight down. In this case, I have a normal force to the right, and that's x and y.

This object, I might still pick an x and a y, but that x and y isn't really going to be useful, so I don't want that one because this might be horizontal, but on the end of a string. So it might be spinning like this.

That's horizontal spinning. It's horizontal circle, but up and down for gravity still matters. So when I draw my picture of my free-body diagram, I'm going to have, say, an x and a y. I'm still going to have weight down.

I might have, I don't know, say, some tension in a string like that. That was my string making it go in a circle. But for both of these, in this case, at this point, I'm that way, so my centripetal acceleration is that way. At this point, I'm that way, so with this coordinate system,

I've picked the x so that my centripetal acceleration is a long x. As I move around, you know, over here, if I'm up there and I'm in red, I still have weight down, but now my normal force went down. Notice, in this case, if I moved over here,

and of course, my centripetal acceleration is now down. If I move here in this one, weight is still down, I've now changed my x. X is now that way.

And so my tension looks the same, and my centripetal acceleration looks the same. I can do that. I can move my coordinate system. For this one right here, I picked x that way. Well, actually, I guess I picked it the other way because I drew my arrow to the right.

And I drew my arrow to the right there. But the point being, you have to keep in mind whether you're moving your coordinate system, and if I'm moving in the plane like this, if I'm flat, usually y is still always up because I need gravity, and I move x around to line up with me

because all I care about is that radial direction that centripetal acceleration is. When I'm going vertical, I tend not to move my coordinate system because I'm going to need to use gravity to see where gravity is in the y direction, and x is just in that plane of my circle. And the centripetal acceleration will be where it is.

So this is probably the first most confusing thing about moving in a circle. Questions on this? Yes, they are.

That's why they have their acceleration. So we're going to do that in more detail. But notice, when I move from here to here, I've moved towards the center. I am now closer to the center than I was before.

Now, notice, right, acceleration is the change in direction of your velocity as well as the magnitude. It doesn't necessarily mean the direction you're moving in. And that's what the biggest pictorial confusion comes from. And we're going to do a lot with this.

If my velocity is that way, and my acceleration is that way, all that tells me is, in the next time instant, I'm going to be moving a little bit down. I've changed my velocity in the direction of my acceleration. It does not mean in the next instant I am moving radially inward. And since my velocity was that way,

in the time it took me to move down a little bit, I physically moved over a little bit. So moving over a little bit and down a little bit, over a little bit and down a little bit, and constantly doing that makes a circle. So it's the centripetal acceleration that is really why you're moving in a circle. So you know when you're moving in a circle, you have centripetal acceleration.

They go together. It's kind of an if and only if thing. And you'll see that in some of the examples we do. There was another question. Say that in here, we don't know anything yet about the relative magnitudes of these.

Notice in the vertical case at this point, they're at a right angle. And remember, any two things at a right angle are independent. So in this case, what we would see is the normal force being in the direction of the centripetal acceleration is completely responsible for it. And gravity has nothing to do with the centripetal acceleration.

It has to do with speeding you up as you go down the circle. When we do a loop-de-loop, your speed is changing as you go around. And we'll see that. Over here, notice what's happening in this picture. The tension now has two components. One in the direction of gravity, and one in the direction of my centripetal acceleration.

So depending on the relative magnitude of the component that's opposite gravity, that'll determine if I can stay flat in a circle. So if I can spin this flat in a circle, I have no acceleration vertically. Since I have no acceleration vertically, the component of tension has to cancel weight.

If I spin it, and it does this, right where it's falling and the circle's getting smaller, since it's falling, clearly weight was bigger than that component of tension because it accelerated downward as well as moving in a circle. Tricky enough?

So you just have to always pick two perpendicular directions. Think in terms of components with forces. You really have to. It's the best way to picture it. And then ask, what is the acceleration in each of those two directions? And often one will be zero and one will not be zero, but sometimes they'll both be non-zero. Excellent questions.

Any more? Yes. Be a little more specific about what you mean. You could in principle use polar coordinates, r and theta.

And any of the problems we do is not really needed because we're always going to be asking for you about the motion at one point. And since we're not going to be asking about it at two different points in the same problem, you can just pick an x and y coordinate for that motion and that's easiest to do. Since we don't move around,

what you're really doing is picking, if you're used to polar coordinates, r and theta. But since they're at a right angle and they're not changing, we can call them x and y. Other questions? So real quickly before we get into the fun demo,

two things. One is, what are some examples of the motion? We already saw one. We saw a roller coaster loop. That tends to be vertical. We saw spinning on a string or rope, which unfortunately can be vertical or horizontal.

That tends to be vertical, right? I can spin it in the circle like that, or I can spin it over my head like that. You got to draw it. There's, see, does anyone remember what a record player is? Yeah, there's records spinning around. If you happen to notice, your CDs do spin. So there's lots of examples of spinning disks

that things may or may not sit on. There's merry-go-rounds, right? You all remember those. Those rotate horizontally. And there's cars turning. And one thing to keep in mind, you will often have a banked curve.

This is something you need to learn to picture. So in this case, the car, this is from the side. From above, the car is going in part of a circle, but it's on a banked road. So the centripetal acceleration is still radially inward.

But now you might decide to pick one of our slanted coordinate systems, or you might decide, okay, I really want the acceleration to be along an axis, and you might pick a normal one. This is the one case on a banked curve where it's not always clear which acceleration you want to pick.

So these are things you want to picture. Questions on, and there's lots of others, but I just gave you a few here. The other thing you want to be aware of, and we said it here, right, with the thing going around the roller coaster, right?

If I drew the free-body diagram here, there is clearly an unbalanced force in both the x and the y, I'm sorry, mix them up, in both the x and the y direction.

This means I have an acceleration with both an x component and a y component. In this case, the x component is the radial centripetal acceleration, v squared over r. The y component is whatever it is coming from that force. And so the total acceleration

is the square root of these two squared, and it's going to be at some angle, right? We'll have a x, we'll have some a y, and the total acceleration will be at some angle theta, where as usual the tangent theta is a y over a x.

So there will be a few problems where you're asked for the total acceleration. And you have to just keep in mind that it might be speeding up. So we call this one the tangential acceleration

because it's tangent to the curve you're moving on. It's the one changing your speed. The radial acceleration is your centripetal acceleration. It corresponds to you changing your direction. Questions?

Now we have some fun. Oh yeah, sorry, question. So you were saying when it gets spun, it's being drawn to the center? No, it's being turned towards the center. No, you're actually being turned towards the center.

If you weren't being turned towards the center, you'd leave the ride. And we'll do that right now. Here's you. Here's Carnival Ride. It's kind of like that commercial, right? Your brain, your brain on drugs. There you are, you just got on the ride. The ride is going to be a vertical Carnival Ride.

We're going to spin you around in a vertical circle like this without a seatbelt. We're going to ask if you fall out or not. So our clicker question.

This one is phrased as water in the bucket. We're going to go to water in a minute. I don't like to start with water. I like to start because I've got to warm up. You know, it takes a lot of effort here. This is like SeaWorld, by the way. Those in the front three rows, you may get wet today. I'm going to spin it around, and the question is, is it going to fall out

no matter how fast I spin it? Will it stay in if I spin it fast enough or there's really nothing in the bucket and I'm cheating you? So most people think if I spin it fast enough, it stays in. So sure enough, it stays in. And as I slow down, you can hear it starting to fall out.

And we're going to come back to that. So now, the question is, will this work with water? I just realized I forgot to practice how to stop this demo.

Sorry. So, work with water.

Copy on that. That would have been bad to spill the water at the last minute. So it even works with water. So you would have been perfectly safe without your seatbelt. Try to get off the ride. Okay, so it works with water. And the real question is, why?

Right? What is happening here? This gets to the whole issue of moving in a circle. And what does it really mean? So if we look at the top of the swing, because that's really the most dangerous point where we expect to fall out,

we're completely upside down. And we were to ask for a free body diagram on ourselves. Now, there is kind of the sides to the bucket, so there could be some friction there that helps hold you on. But, you know, this was water. There is not really any friction between the water and the sides of the cup.

It pours out quite easily, as I demonstrated at the end, when I spilt it all over everything. So, water slides out, so there's no real friction there. But it is interacting with the bottom of the cup through our normal contact force, which can really only point down. And, of course, it has weight.

So those are our two forces on it. Now, in this case, I'm going to take up to be positive, which will give everything a minus sign, but I just want to do that because it's a common way to pick it. Now, we know at the top of the swing,

the bucket and the water is moving to the right. Why do we know that about the top of the swing? What happens at the top of our motion, by definition, in our description of motion? See, we have to actually remember kinematics and how we describe motion. What happens at the top of any motion?

The vertical velocity is zero, which means you're moving perfectly horizontally. So, if the total force equals zero, then our acceleration would equal zero, and we just keep moving horizontally.

So there would be no change in our motion, and we would fly straight that way, and that should be what happens. But there are forces. Now, the total force is greater than zero,

then our acceleration is greater than zero. Which way do we turn? Up or down, in this case? Well, notice which way I picked to be positive. Up, because up is positive. I just told you a total positive force,

which is a positive acceleration, so our curve would do something like that. We would curve up. Now, if, in this case, our total force is negative, now which way do we go? Down. Notice, the positive and negative isn't really the relevant thing. It's whether the force is down or the force is up,

which we've defined from our plus and minus sign for a coordinate. So if our force is down, we curve down. Now, you're used to things falling and curving down because there is almost always weight, and which direction is weight? Down. So nothing stays going straight, right?

It tends to always curve. There also is this question of the normal force, however, and what's going on there. Now, if what we want to do is move in a circle with a radius r, which is what I was doing with the bucket,

then I need an acceleration equal to v squared over r. That's what centripetal acceleration means. If I'm going to go in that particular... So this is a special case of a down curve.

And so what do we need to do? We look at it the following way. We write down what's going on, and I'm going to write f normal in red, and I'm going to currently make it just positive because I'm going to say we don't know which way it is yet. And then we know weight is down. And the question is, and that therefore... Whoops.

And I'm going to write in mg for weight. And we know that has to, if we want it to go in a circle, equal minus mv squared over r, and that will make us go in a circle. Now, this means if g equals v squared over r,

then fn equals zero to go in a circle, right? Because if g is going to equal that,

I can't have any fn. And this would be what we call just make it, because my normal force went to zero. Now, what happens then

if g is less than v squared over r, or I'm going faster? Anyone want to say what would happen then

if g is less than v squared over r? To achieve my circular motion, two things have to now happen. So I need to make my top equation an equality, and I'm telling you g is less than v squared over r.

I'm spinning faster than I was to just make it. What now happens? Has that some value, positive or negative? Well, look at my equation the way I wrote it. If this is too small and this is a negative number, what must this be? Negative to make it up, right? So at this point,

fn must be some negative number, which is good, because we know it can go down, so we're okay. So if I spin faster, then what's happening, notice spinning faster would mean a bigger acceleration,

so I need more force. So to achieve my spinning faster, I suddenly need to use my normal force, which is down and negative. If I go too slow, so if g is greater than v squared over r, which is going slower,

we want fn to be positive, which it cannot be unless I'm also attached. So if you have a seat belt, or if you're physically attached to the track, or if you're super-glued to your seat,

then your normal force can point up. If I'm physically attached, my normal force can point up, which is what I need to keep me up in the circle. So in this case, if my red line is the circle I want,

what's happening when I go really fast, gravity would only allow me to make a circle that big, and the normal force pulls me in and lets me achieve my circle, pushes me in. If I'm going too slow, gravity is trying to make a circle like that,

and I need a normal force to pull me up, because this is where the bucket is, because my hand is pulling the bucket there, and everything else falls out, unless I attach it. So those are the main concepts you want to have right now in your head.

So let's pause there for a moment. Think about this, look at it, tell me if you have any questions, and then we're going to do an example problem with it. There is a lot on this slide, yes.

So v squared over r, the whole thing is an acceleration. v squared is meters squared over seconds squared, so when I divide by a radius, I get meters per second squared. So what is something, if you look at this, and we're trying to explain it to someone, what is one thing you can take away from this? What happens to the normal force?

Does it have a set single value? Yes or no? No. The value of the normal force will change depending on the situation and what it needs to be. We already saw that with a ramp, right? When I put something flat, right now the normal force on me is equal to what?

My weight, mg. As soon as I'm on the side of a hill, gravity is down, normal force is perpendicular, it's less than that, it changes. Now, when I'm spinning around in circles, it gets even more complicated, because the normal force can adjust as needed.

Now, if I'm throwing in friction, particularly static friction, that gets even worse, right? Because the normal force can adjust as needed, static friction is less than or equal to the normal force, it can adjust as needed, so you've got two things that are adjusting, so you've got to pay attention to them.

So you can see where the problems might be. Questions. So now having said all of that, let's answer clicker question number two. Now that most of you had answered,

this is one of those rare cases that I would say it truly is a trick question. And we're done. It is D. And the reason it's D is it's inertia that is generating the normal force. And what do I mean by that?

Inertia is our tendency to keep moving in the same direction we've been moving. So at the top, I want to keep moving straight. The bucket, extra chain in there, the bucket's moving down, and I'm trying to move straight. And it's my desire to move straight that causes me to hit the bottom of the bucket and generate a normal force that moves me down.

And you notice, that's only needed when I'm going really fast, right? There was that special case where all of the centripetal acceleration was generated by the force of gravity itself. And so it really is inertia

that causes you to stay in the bucket. You want to keep moving straight, the bucket's turning, forces happen on you to make you also turn with the bucket. So at the subtle point, you know, it's not really one I would ask specifically on a test in this way, but it's important for you to realize when you look at any of these situations,

the objects want to keep moving straight so they're only going to turn if there's a force on them to generate that change in velocity. And that force can be from many, many different sources. And so you have to pay attention to all of them. So let's go to the problem. Like I said, these will be roller coasters,

they'll be loop-de-loops, they'll be ropes spinning around, they'll be things moving sideways on tables, they'll be merry-go-rounds, friction will sometimes do it, sometimes it's a normal force, sometimes it's just gravity. One thing please pay attention to, this is not really meant to trick you, it's just the way, it's the way problems are worded

for lots of different reasons. I am telling you in this case how high the loop is. What is the physical dimension that goes into the centripetal acceleration? Radius. What is the radius of a 16 meter high loop? Eight meters. So watch that. If you email me and say,

mastering physics just can't get this right, I get the answer wrong every time, I will probably not respond because you probably just need to divide by two somewhere. So keep that in mind. Again, look for the question, this is how fast are we going? And we just make it without falling.

So just make it, that's our key word. We're going to have a normal force of zero. So when we go to do this problem, we have our loop-the-loop, which is 16 meters high. We are at the top. We're going to pick down to be positive.

We're going to draw a free body diagram. There's only the weight. Normal force equals zero because we just make it. We know we have an acceleration of v squared over r in the downward direction because it's really inward. Notice, I do not put the acceleration,

this right here is not in my free body diagram. It's indicated separately, so I don't get confused. Having done that, I use f equals ma. So I'm going to have f weight equals m v squared over r.

The weight I know is mg. Notice, my masses cancel. I would not have needed to give you a mass in this problem. And v squared equals g times r, or v equals the square root of g times r.

And if I plug into that, I will get 8.9 meters per second. That is kind of the core of these centripetal acceleration problems, the most basic version. You identify your acceleration, you identify all your forces, and you're generally solving for something.

In this case, we're solving for v. Notice, as you go forward in this class, this is a great situation where there's lots of ways we may or may not want to know about v. Suppose your loop-de-loop was broken. Coming down the roller coaster, you go into loop-de-loop,

and it suddenly ends at the top. And I asked you, where do you land over here? What did I just do? When you leave the little loop, you're doing what type of motion?

Projectile motion. Your initial velocity is horizontal. You're in projectile motion. If you needed to know your initial velocity, you would find it from f equals ma. Because up until this point, just as you leave, you are doing circular motion. You know you just make it.

You know the acceleration right here was v squared over r. So you know now you're flying through the air at some speed 8.9 meters per second horizontally. You are no longer doing centripetal acceleration because you are no longer in a circle. You now only have gravity and its projectile motion. And you use everything we did from the first three weeks.

I've now connected two problems. And you hopefully don't land too far away and too hard. Questions on that? Now we look at the second part of the problem. By the way, I really encourage

most of the problems you do in this section, stop and think, okay, how could Professor Denton turn that into this plus a kinematics question? Just a hint. Think about that. Now the second part says, if we double that speed, now what's the normal force?

Notice turning it around. Now we know the speed, and we're going to be asked something about the forces. So now, we know we're going faster, so there has to be a normal force. So there's our free body diagram. We still take down to be positive.

And so now we have F normal plus mg is going to equal m v squared, let's call it 2, over r. And what we know is v2 equals twice the velocity we had before. Now you have a couple of choices at this point.

You could just plug in twice that velocity numerically. But as I said, I like to do these algebraically all the way to the end, and then plug in. So I would plug into here,

and this equals 4 times the mass times the velocity squared over r. And I know what this is, I just did it up here. If I take that equation, which I had used, I look over here and I see,

velocity squared over r is just g. And so now this equals 4mg. And now I have F normal plus mg equals 4mg,

which tells me F normal just equals 3mg. Now I do need to give you the mass, which I do in the problem, and when you plug it in, you get 1800 Newtons. So in this simple of a problem, it may not be clear

how that is really useful to keep it as a variable, but a lot of times with the information you're given, you may not have been given per se the initial velocity, and you may not have solved for it for some reason, but you would have had that equation and by keeping track and just doing 2v and pulling it out, you would have known to

substitute for g and get your mg. So there are times and places where it's really helpful to keep everything as a variable. Questions on this problem?

Yeah, questions? Sure. That was given as the 16 meters right there. Now, real quickly, so let's just review this so you have

kind of an overview going into the weekend as you got till Sunday to finish all this homework. Your focus is going to be F equals ma in an appropriate xy coordinate system. You need to pick this, and it may be tilted. You need to ask

what are all the interactions going on? Do I have weight? Do I have friction? What type? Do I have a normal force? Do I have any pushing or pulling? Do I have any ropes with tension? You have to ask what accelerations

do I have? Am I just about to do something so that that's telling me an acceleration is zero? Is it given as zero? Or some other constant? Is it circular? So that I know I have

v squared over r. And then you have to focus on what are you being asked to find so you know how to put all these pieces together. So you can see how this is much more complicated than kinematics in one way. The details, there are more details. In kinematics we just had basically constant acceleration

or zero acceleration which was a special case of constant acceleration. And we had our two components and that was it. So we used our two equations a lot. Here we still really only have one concept but we have way more details you have to keep track of. So that's what you want to be practicing so you get the idea

that it's all just one problem even though there's like 20.