Alright, let's get started. Let's get started. Okay, so welcome everybody to the last class, which is the seventh day C. So remember, we've covered all the material. And on Tuesday I gave an overview of everything we've learned since the midterm.

And I asked you to send me problems or general physics questions. And I would solve the problems in class as long as I got them before last night. So I got a few problems. I also got a request at like five o'clock this morning. Sorry. I didn't have a chance to do that one this morning.

But I have a bunch of problems here to solve, some of which were quite tricky, some of which were pretty easy. I also have the practice final. Has anybody looked at the practice final yet? Yeah? So I can solve the problems in the practice final. I can solve the requested problems by strong feelings.

Requested problems? Alright. Final is too easy? I haven't made the solutions yet, but I will. Practice final look easy or hard or what? Alright, good. Well, I hope that bodes well.

Alright. So first problem I was asked to solve is 5.118. Okay? And this one should strike close to home. It says a physics major is working to pay his or her college tuition by performing in a traveling carnival.

He rides a motorcycle inside a hollow transparent plastic sphere with a vertical circle of radius 13 meters. That's quite a big sphere, right?

Radius 13 meters. Wow. This thing is like 70 feet high. Have you seen a 70 foot wide plastic sphere with a physics major riding a motorcycle inside? Somebody Google video search that and see if you can find it.

Okay. The physics major has mass 70 kilos and the motorcycle has mass of 40 kilos. What minimum speed must he have at the top of the circle if the tires of the motorcycle are not to lose contact with the sphere?

So he's up here at the top of the circle. Let's say he's riding a chopper. Okay. So what are the forces on this guy at the top? Here's our free body diagram. Gravity and normal force, right?

Gravity is this way and the normal force is also down. So now we're asked what minimum speed must he have at the top of the circle if the tires are not to lose contact with the sphere?

What's the approach? Right. So let's translate this requirement that the motorcycle is not going to lose contact with the sphere. Just think that it moves in uniform circular motion, right?

Because if it's going to make it up here and come back down, this is a lot like the vertical loop that we had with the pilot pilot, right? In order to move in a circle it has to have uniform circular motion. So the force needs to be mv squared over r, right? And what we're looking for in the end is the velocity, right?

It is the velocity. So, okay, so what does that tell us then, right? This is mv squared over r equals n plus mg.

How does that help us? That means that all the force at that point will be zero. That's right. We're looking for the minimum velocity, right?

The normal force can be any amount depending on how fast it's going. So let's set this one to zero and say mv squared over r is just equal to mg, right? So then the v's cancel, v squared equals gr, v equals square root of gr, which is 11.3 meters per second.

Okay, this will translate 11.3 meters per second into something we actually understand. How fast is that in miles per hour? What was the conversion unit used there?

11.3 pretty fast? Yeah, I like that, approximately. Or magnitude pretty fast. Okay, 11.3 meters per second, let's see. It's pretty close to the same as 100 meters per second. Yeah, the same bolt as 10 meters per second, right? 10 meters. So that's pretty fast, but not that fast for a motorcycle, right?

Well, that's good. We should keep our physics major safe. And the mass is canceled, right? Oh wait, and then there's another part to this problem. At the bottom of the circle, the speed is twice the value calculated in part a. What is the magnitude of the normal force exerted on the motorcycle by the sphere at this point?

Okay, so at the bottom of the motors, at the bottom of the loop, we have mg down, the normal force up, right? And we have the same condition, uniform circular motion,

and we call this VB, and we'll just use the, let's do this, 2V squared over R, right? Same mass, same radius, twice the velocity is equal to N plus mg, right?

Nope, it's equal to, of course, it's inward, so N minus mg, right? Well, this one here, so this is 4 times mv squared over R, right?

And mv squared over R is equal to mg. So this is 4 mg, which is equal to N minus mg, so N is equal to 5 mg. All right, so I expressed the force in terms of the velocity required, and I was told already that you needed twice the velocity of what you had before,

which means 4 times the force, because the force goes like mv squared, right? I already knew what the force was last time, which is mg. So I just said, this is the force from last time, just replace that with mg. Now, 4 mg on this side. Now, last time, the normal force on the top of the loop, the normal force and gravity help each other provide the centripetal force.

Here, they're opposing each other, you need to do more, you need to have a stronger force at the bottom than you do at the top, right? Anyway, it's not quite uniform circular motion though, it's because you have different velocities. So we're considering this, so you might object to that and say,

how could we have uniform circular motion? Well, you're right, we don't. We have momentary uniform circular motion, right? You're moving in a certain circle at one point, and another circle at the top, these different velocities. Okay, all right, that was one problem requested.

I'm doing these in the order of the problems. Next one is 573. This is a real headache. Okay, here you have two blocks sitting on top of each other,

block A, block B, and a force this way. Block A weighs 2.4 newtons, so 2.4 newtons, block B weighs 3.6 newtons. The coefficient of kinetic friction between all surfaces is 0.3.

Find the magnitude of the horizontal force necessary to drag block B to the left at constant speed if A rests on B and moves with it. Okay, so the question is, what is the force necessary on block B so that it moves at constant speed?

We've got sort of a kinetic friction. This is a coefficient of kinetic friction, yes. So, what is the key phrase here? Constant speed means constant speed means no acceleration. No acceleration means net force is zero, right? So, what forces do we have on B?

Let's draw a three-body diagram for B, right? We've got, somebody said normal force, so that's up from the ground, right? So, we have a weight down. What else? We have the dragging force. Find the normal force from A on what?

There's the weight of A, that's true. So, this is the weight B, and there's also the weight of A, right? This is the dragging force, and what else? Friction, right?

Okay, so we're interested in the motion of B, but A is sitting on top of B. So, the easiest way to handle this is to just think of B and A together, right? You can just think of them together, or you can just focus on B and account for A's weight as a force on B, which is really the same thing as thinking of A and B together, and this being a single weight, see what I mean?

Either path, as long as you're thinking about it consistently, can get you to the right answer. Okay, so basically, we need to set, and we want this guy to balance this guy, right? So, if we can figure out the force of friction, we can figure out what this force is, because I have to have the same magnitude to give us constant velocity.

So, the whole game is, what is the force of friction? The force of friction, of course, frictional force is the normal force times the coefficient, which we're given. What's the normal force? Well, the normal force here is going to be weight A plus weight B.

The normal force is just going to balance the weight of the two blocks, okay? So, that gives, so this plus this is 6 newtons times 0.3. So, this is 1.8 newtons, okay?

That's part A. Part B says, let's make this a little nastier, and part B says, the A is held at rest, okay?

So, it's a very similar setup, except now, and I'll drop, I'll redraw the diagram so we don't get confused. Now, B is here, like before, A is here like before, except there's a string attaching A to the wall.

Oh, man, all right? So, A is held at rest. Now, the same question. Find the magnitude of the horizontal force F, necessarily drag block B to the left at constant speed, okay? It's the kind of problem only a physics person would do, because it never happens in real life. Okay, so now we're going to think about the free body diagrams for A and for B, right?

So, first let's think about the diagram for B. Okay, I remember we had this force again, all right? So, let's start off with the forces we had before, all right? We have the dragging force. We have frictional force, all right, F.

We have the normal force. We have the weight, B, right? What else do we have? What's new?

Friction between these two blocks, right? Now, which direction is that friction going to go? To the left or to the right? Acting on block B, block B is going to move to the left. Friction always closes the motion. So, this one is going to be, call this one F, B, okay, to the right.

And there's also, there's this force down from the weight of A, right? Okay, now let's think about what happens to block A.

Block A has what forces on it? We've got its weight, right? There's a normal force up from block B, all right? This is, call this NA, all right?

Friction. Okay, in which direction is friction going to go? Y to the left? Because it's reversed. Okay, in which direction? Okay, sure, so A is moving to the right relative to B, right? The rope is pulling it to the right, so friction is opposing that.

So, this one we call F, A, and then, of course, this tension, all right? Okay, so what do we do here? Well, we know something about this one, right? This one is just going to be equal to WA, right?

It was driven in balance because that guy's not going anywhere, right?

Why would it be the same magnitude? Okay, so what principle are you applying there?

The principle of universal loving particles. Let's not get down that road. But you're on an interesting track. Why would those two forces, are they the same, and if so, why? Well, I would say that R, because the normal force on B is equivalent to the weight of A, and then the normal force on A.

The normal force on B? Yeah, I'm sorry, no? The normal force between the ground and the normal force basically The normal force here, the normal force which gives us this normal force

Yeah, not the one that gives us the friction between the ground and B is the same as the weight of A, and the normal force on A is the same as the weight of A and it said that the friction of kinetic friction between all substances The same normal force and the same coefficient of friction

means these are the same, right? So we'll just write that like that. So let's see what we can figure out. We know what NA is, right?

NA is just the weight of A, which is given 2.4 N That's why we wrote NA NB, what's NB going to be? Weight A plus weight B

This is NB here

NB is weight A plus weight B

So now the frictional force FA is equal to FB is mu times weight of A So that's 0.72 N

Let's keep track of what we know We know this one of course, that gives us this one and we were able to use that to figure out this one, which is the same as this one A is going to be held constant force

A is not going anywhere So that means we know this one as well Everything is the same FB is T And what were we looking for again? This one And we know F F hasn't changed since last time

F is mu times WA plus WB Which is the same as it was before 1.8 N So now this guy has to balance both of these two So it's just going to be the sum of these two guys

So F minus F minus FB is equal to 0 Which means F is equal to F plus FB Which is 2.52

So this problem is a little nasty But in the end just put down the free body diagram Be very careful about counting for all the forces So we need to do that careful work No tricky concepts here

Next one is 675 A small block with a mass of 0.09 kg Which is 0.09 kg He's attached to a cord

There's a block and a table There's a hole in the middle Attached to a rope which goes through the whole thing This small block with a mass of 0.09 kg

It's attached to a cord It's happening through a hole in a frictionless horizontal surface The block is originally revolving at a distance of 0.4 meters from the hole

With a speed of 0.7 meters per second The cord is then pulled from below Shortening the radius of the circle at which the block revolves to 0.1 meters

At this new distance the speed of the block is observed to be V2 equals 2.8 meters per second Question What is the tension in the cord in the original situation when the block has V equals 0.7 meters?

What's the tension in the cord in the final situation when the block has 2.8 meters per second? How much work was done by the person who pulled the cord? So what's going on with this block? What forces are at play on the block? We draw the block We've got what? Gravity

Normal force And tension So this thing has a net force on it But it's moving in a circle So we have uniform circular motion Which means this force has to be equal to what? mv squared over r

And the question we're being asked is What is the tension in the cord? Well to know the tension we need to know the mass Check We need to know the velocity Check We need to know the radius Check So part A is just plug in So just plug in mv squared over r We have all the numbers You get 7.1 Newtons

For the initial case And Oh sorry, 0.1, this is the second case Just plug in and you get 0.11 Newtons here

And 7.1 Newtons here Okay? So that's the tension in the rope in both configurations The next part of the question says

How much work was done by the person who pulled on the cord? How do we calculate work? Work done Force times distance, right? So how do we do that? Let's see Can I just do force times distance as a constant force?

No, so I need to do an integral, right? So now I need to express the force as a function of distance, right? Okay, this could work Is there an easier way? Conservation of energy Conservation of energy, how do I do that?

1.5 pi omega squared plus Well, is there any potential energy? No potential energy, but there's an energy added by pulling it So the initial kinetic energy plus the energy added should be the final kinetic energy

I think this is called the work energy theorem or something But in the end you can derive it from conservation of energy, right? You can say u1 plus k1 plus work other equals u2 plus k2

And if this drops out, then the work is just k2 minus k1 So another thing you don't have to memorize, but just give it a fancy name for some reason So we can just calculate the difference in the kinetic energy How do we calculate the kinetic energy?

It's just v squared, right? Which we have So this is the easiest way to calculate it What about this one? How do we do this one again? Well, we need to know the force as a function of the radius, right?

That's just over v squared Yeah, but isn't the radius of the velocity also going to change with the radius? We know it is, right? It's not the same for different radii So that's going to be a bit of a tricky interval to set up So much better to just use this approach

There's nothing wrong with the other approach This will work This will get a headache Okay, next problem is 786

786 says a particle moves along the x-axis while acted on by a single conservative force parallel to the x-axis The force corresponds to the potential energy function graphed in this figure So it comes down, negative, up, a little less negative, and up again

This is point A, C, point 0, point 0 here

Okay, a particle moves along the x-axis, acted on this force Here's the potential energy function Okay, that's what we're looking at Question A, what is the direction of the force on the particle when it's at point A?

So what's the relationship between potential and force? It's the negative derivative, right? So what you can do is take the derivative of this and invert it Or you can just ask yourself, if I put a little ball right here, which way would it roll?

It would roll down, thank you, yes Useless but always correct answer It would roll this way, right? So the force is going to be towards positive x Let's check that Well, what's the derivative here?

The derivative is negative, right? So we take the negative derivative, that's positive, so the force is positive So the force here is greater than 0 in the x direction What about at point B, it says? Well, at point B, the derivative is positive, so the force is going to be negative Also, if we put a ball here, it would roll towards the negative value, so it's negative force

C, at what value of x is the kinetic energy of the particle a maximum? How do we figure out where the kinetic energy is a maximum? Well, say, kinetic energy plus potential energy is some constant, right?

Then the kinetic energy is a maximum and potential energy is a minimum, right? It's going to be here Give it a fixed amount of energy, you get the most kinetic energy and you minimize the potential energy What is the largest value of x reached by the particle during its motion? If it's released from A Let's see, it's tricky, how do we figure that out?

To know how it's going to move The question is, what is the largest value of x reached by the particle during its motion? It starts out here Two point something, yeah

Well, what you do is you say, this is, you can never have more You can never be any place on the graph that requires it to have more potential energy than the total energy of the particle Does that require a negative kinetic energy? That means you can imagine it's like a Tupperware lid here, it can never break through So if it starts here, it's going to go down here, be very fast, come back up here, be slow

We still have non-zero kinetic energy, meaning that it's moving It's going to get over this hump, come down here When it comes up here, it's going to run out of kinetic energy, i.e. stop So, kinetic energy equals zero means velocity equals zero, that's when it's turning around So this is the maximum point it can reach

Two point something What value or values of x correspond to points of stable equilibrium? Stable equilibrium means no net force, right? No net force means the derivative of this thing is zero So that's here, here, and here

It's asking for stable equilibrium, which means the second derivative is going to be positive, right? So this is stable equilibrium because if you move away from it, you force back towards it You move this way, you move that way, you move this way, you go that way This is an unstable equilibrium because as soon as you move away from it, you're pushed further away from it

Alright, so this one's stable, stable, unstable Can I think about that?

Next one, seven point four nine A fifteen kilogram stone slides down a snow-covered hill Got some weird shaped hill

Got stone This is point B, this is point A

Okay, fifteen kilogram stone Fifteen kilogram This is twenty meters high This distance is fifteen meters Slides down a snow-covered hill leaving point A with a velocity of ten meters per second There is no friction on the hill between points A and B

But there is friction on the level ground at the bottom of the hill Between B and the wall After entering the horizontal region, the stone travels a hundred meters And then This is a hundred meters

And then runs into a very long light spring with force constant two meters per meter The coefficients of kinetic and static friction between the stone and the ground are So kinetic friction is zero point two Static friction is zero point eight

What is the speed of the stone when it reaches point B, first of all? Okay, so complicated situation But first thing we're asked is The stone starts with ten meters per second Rolls down and creates a hill at point B What is the velocity? So we've seen this problem a million times in a million configurations, right?

This is potential energy turned into kinetic energy Nothing complicated about that So we solve with energy So K1 plus U1 equals K2 plus U2 K1 is one half mv0 squared

This is the initial velocity Initial potential energy is mgh Where h is given at twenty meters Final kinetic energy is mv squared plus zero This is the bottom

Anyway, remember we're interested in the difference in the height of the potential energy So here we have everything we need, right? Mass cancels V squared is equal to V0 squared plus

2gh So mv is 22.2 meters per second Okay, that part was easy Next it says How far will the stone compress the spring?

So now we have a stone The fact that it's come down the hill and whatever happened on the hill is irrelevant now, right? We have a stone moving horizontally across a friction-ish Friction-ful? Friction-y? Fictional?

It sounds like fictional Fiction-ly? Friction-ly? I believe I've never wondered what that word was for Okay A surface with friction until it hits a spring Alright, and the question is How far will the stone compress the spring?

So, how do we approach this one? Minus work done by friction equals Right?

No, it's not rolling It's sliding The one time my perfect sphere confuses me, right? I'll try to be more sloppy with my drawing Yeah, so kinetic energy First potential energy was turned into kinetic energy

Now kinetic energy is being turned into energy from friction And energy of the spring Okay, but it's a little bit tricky because once it hits the spring It's still losing energy to friction, right? So we can't just say, we can't do it in two steps We can't say, how much does it lose due to friction And then just turn the rest of that into spring

We have to solve it simultaneously So we have to be a little careful about how we express this work due to friction So if we define x here as the placement of the spring Relative to its equilibrium point Then we need to write this As 100 plus x Times force Times the force of friction

And we know how heavy this thing is, yeah So the force of friction Is just going to be mu k times m g Because the normal force is just going to be m g, right? So we have One half m v

This is supposed to be squared but not two m v squared Where we know this velocity, right? From here Minus 100 plus x Times Mu k m g Is equal to one half k x squared Okay, so this is our critical equation

And you see that x comes in in two different places So we know m, we know v squared We don't know x, it's what we're interested in We know this, we know this, we know this We know everything but x We have a quadratic equation in x So that would just come down to the math part And to solve it you get

x is equal to 16.4 meters Whoa, that's pretty far I guess why they said this is a Very long, light spring Right, or how did this rock get going at 10 meters per second already, right?

Somebody should know This is dangerous playing with rocks Okay, so the only This is an energy problem, right? The only tricky part here Was making sure you took into account the fact that there's additional Friction after it hits the spring Okay, the next part of the problem Will the stone move again After it has been stopped by the spring?

So it's using kinetic friction Right, and then it gets stopped And now the question is Is the force of the spring strong enough to move it To overcome the static friction Once it has stopped How do we figure that out? Well, let's think about the free body diagram of this thing, right?

It's going to be a force through the spring here Force of friction, static, right? So we just need to compare the magnitude Of these two guys, right? And figure out which one is larger So what does the force do to the spring? Well, it's just going to be one half kx squared Right? That's where it comes to a stop

I'm sorry, it's going to be Kx, of course The work is one half kx squared So the force due to the spring Is 32.8 Newtons Okay How much static friction can you have? Well, this is just mu s times m g

Right? The block is 15 kilograms We know m, we know g We know mu s Mu s is a pretty large number, 0.8 So this gives 118 Newtons So you can have up to 118 Newtons of static friction Right? Whereas the spring is only providing 32.8 Newtons

Remember, static friction is a funny thing There's a maximum value For this You can have up to zero, right? If you have a down to zero You have something resting on a surface with friction And there's no force on the sideways And there's going to be no static friction It only resists the force As you crank up a horizontal force

The force of friction ramps up So it gets to some maximum After which it can no longer keep it stationary And then it's overcome So just because If you just did this calculation naively You might think, oh the force of friction Is going to move this guy Because the force of friction is stronger

Than the force of the spring Obviously that's not the case The force of friction is less than or equal to This maximum value Does that make sense everybody?

Next one, 8.97 Jack and Jill Are standing on a crate Jack and Jill are standing on a crate

At rest on a frictionless horizontal surface Of a frozen pond So there's a crate on the pond We've got Jack And Jill Jack has massed 75 kilograms

Jill has massed 45 kilograms And the crate Has massed 15 kilograms They remembered that they must Fetch a pail of water

So each Jumps horizontally From the top of the crate Alright I'm not sure how that's Going to get in their water Just after each Just after each Just after each one jumps

That person is moving away From the crate with a speed of 4 meters per second relative to The crate Everything starts out at rest But if you're going to jump sideways this way The crate is going to go that way right?

And they were given the velocity relative To the crate So the question is What is the final speed of the crate If both Jack and Jill jump simultaneously In the same direction? And it says hint Use an inertial coordinate system Attached to the ground

Alright So a coordinate system attached to the ground We're given the velocity Relative to the crate Alright so So in the coordinate system of the ground The crate is going this way And we're asked

For the final speed of the crate so we'll call that V Jack and Jill are flying through the air At least I'm just going to have to draw their stick figures Okay Jack and Jill are flying through the air So what is their velocity? If this is V Then what is their velocity? On the system of Relative to the ground

The crate is V with respect to the ground So Jack and Jill are going to be V plus four Let's make them going positive You're confusing otherwise

So they jump positive Okay So this velocity is Negative V And their velocity is four

With respect to the crate So it's going to make their velocity what? Their velocity relative to the crate is four meters per second So their velocity is then what?

Four minus V right? So their velocity is Four minus V This velocity is minus V Okay so the question is And we've given all the masses And the question is What is the final speed of the crate? What kind of problem does this sound like?

Momentum right? Definitely momentum problem And we can say There's no external forces This whole thing is supposed to be on top of a frictionless icy pond Which sounds like momentum already And so we can say that Initial momentum is equal to Initial momentum is equal to Final momentum

What is the initial momentum? Zero because nothing is moving So we just need to write the final momentum In terms of the thing we want to figure out So we figured out We know what the velocity The mass of the crate is The mass of the crate times its velocity Plus We can treat these guys as

Oh Jack and Jill Times their velocity Treat them as the same Logic in the air if they have the same velocity And then we have An equation that just has v

And so we can solve this for v because we know All the masses Ok good And so you solve for that and you get V equals 3.56

Ok But this problem is not done What is the final speed of the crate if Jack jumps first? Then a few seconds later Jill jumps in the same direction

You're doing two steps right? First Jack jumps off You figure out what the speed of everybody is Using conservation momentum Then Jill jumps off You figure out what the speed of everything is Giving conservation momentum The question is what is the final speed of the crate Again if Jack jumps first Ok

Now we have an intermediate quantity So we call this v prime After Jack jumps Now I guess it's assumed that No matter what's going on You move 4 meters per second relative to the crate When jumping Ok

So now the crate The crate is 15 right? Ok so when Jack jumps off You can think of Jill as just part of the crate Or the crate is part of Jill And so p initial P final Zero Mass of the crate plus the mass of Jill

Times negative v prime This is the mass of the crate After Jack jumps Plus M Jack Times 4

4 minus v prime Alright So this is 60 kilograms here This from Jack is 75 kilograms

Ok so we can figure out what v prime is Again we have an equation with just v prime And we get v prime equals 2.22 meters per second Ok Now We have to solve the problem with Jill jumping Off this moving crate right?

So now We have P initial Equals P final P initial is not zero anymore Cause after Jack jumps the crate is moving With Jill on it right? So we have 60 kilograms Times v prime

This is the initial Configuration And the final configuration Is Mass of the crate Times Minus v Which is what we are looking for Plus The mass of Jack

Right Jack is going at 4 minus v prime And that hasn't changed Plus The mass of Jill Who is going at 4 minus v

We don't need to include Jack in this configuration do we? We already accounted for him This is the momentum of Jill plus the crate If we added Jack in it would all be zero again

This is Jill plus the crate After Jack jumps So we just need to account for Jill and the crate We are doing a system now Of just Jill and the crate Because We are trying to figure out the velocity of the crate is We just call that Minus v And we know every time somebody jumps they get

4 meters per second relative to the velocity of the crate So we just did this in two steps This encapsulates Jack jumping off Now Jill jumps off We have another equation where we only have v as an unknown and we already know v prime Right? So this gives us

v is equal to 5.22 meters per second Ok? So here we got 4 meters per second If they jump individually 5 meters per second if they jump one at a time Why is that? Each jump they have to make sure the

velocity relative to the crate Is 4 meters per second That's right It's a bit of an artificial situation Each person does the same jump Regardless of what's happening So you compare whether the crate is Speeding up the same amount The constant condition is that You have 4 meters per second velocity

Relative to the crate Which is different depending on the crate etc. Ok, the next part of the problem says What if Jill jumps first? But it's just really an exercise in the same sort of game Ok, that was 897

We did 749 already Ok? Next one is 10 We did this one already 1095 1095

1095 says the 500 gram bird is flying horizontally At 2.25 meters per second Not paying much attention But suddenly it flies into a stationary vertical bar To blink

Alright, here's our bird Ok? Velocity is 2.25 meters per second Mass is 0.5

kilograms It hits the bar It flies into a stationary vertical bar hitting at 25 centimeters below the top Ok, so this is 0.25 meters The whole bar is 0.75 meters

and has a mass of 1.5 kilograms and it's hinged at the base That's sort of unlikely This bar is standing at perfect equilibrium through the waiting for a bird Ok, nobody ever said these physics problems were realistic

The collision stuns the bird so that it just drops to the ground afterwards But soon recovers to fly happily away That's why I didn't make that part Ok, so What is the angular velocity of the bar just after it's hit by the bird and just as it reaches the ground Ok, so the bird

knocks into the bar gets it moving and the question is What is the angular velocity of the ball? Alright, so we're looking for omega bar How do we figure out how fast this thing spins? What is it that the bird gives the bar?

Angular momentum, right? This thing is going to rotate But does the bird have angular momentum? No It has what? It has linear momentum Does that mean it doesn't have angular momentum? Everything with linear momentum has angular momentum relative to some axis Just like the moon

is going in a straight line but it has angular momentum relative to the center of the Earth Ok, so How do we do conservation of angular momentum? Well, we say L initial equals L final and then we figure out how to write angular momentum in terms of things we want to know So for the bird, how do we write the angular momentum?

Well, it's just m v r, right? I think the bird is a particle We know its mass We know its velocity We know the radius and the pivot point This is 0.5 meters, right? So we know how much angular momentum it carries

relative to that pivot point How do we express the final angular momentum of this ball? I omega, right? Why I omega? Sure We want an expression to use omega Right? Because that's what we're looking for Also, this thing is not a particle

So you can't just write it as m v r Since you integrate over it and turn out a constant angular velocity but the radii are different So that's what this does for you This encapsulates something about the shape of the ball I never remember what that is but it says here that it's

one third m l squared times omega, right? And so we know the mass of the bird We know its initial velocity We know this distance We know the length of the bar, the mass of the bar So we can calculate what omega is Right? Omega equals

two radians per second But then it says what is the angular velocity of the bar as it reaches the ground Why is it not going to be the same as what the bird is? Gravity, right? So how do we incorporate gravity to speed something up?

Well that sounds like an energy problem usually You're turning potential energy into kinetic energy So you're speeding something up by transforming the potential into kinetic energy But here we have rotation, right? We have rotational kinetic energy and we have potential energy So let's think about how to write that

mu one plus k one is equal to mu two plus k two There's no friction so we don't need to worry about that Right? So first of all how do we write the kinetic energy? One half i omega squared

We know omega, we know i so this term is fine This guy here we can write as one half i omega final squared That's success because we've introduced an equation of quantity we want to know So hopefully this is going to tell us something about what we're trying to figure out Now we just need to figure out the potential energy What's the difference in the potential energy between when the bird hits it

and when it hits the ground? Potential energy of a complex object and it's changing its direction also Center of mass, right? Anytime you have potential energy of a complicated object Just think about the motion of its center of mass The fact that it spins is irrelevant There's no change in potential energy

due to motion relative to the center of mass Only a change in the center of mass will change its potential energy Alright, imagine for example Where is the center of mass in this thing? Well, halfway up by symmetry Imagine if this thing rotated about this point Would there be any change to potential energy? No, exactly the same

In both cases it's as if there's a point mass At the center of mass because the center of mass hasn't changed But when it hits the ground Where is the center of mass? It's at the ground, right? So we'll call this one zero And this one we'll write as mg and then This has to be the mass of the bar

Which we're given And then we need the height of the center of mass Which is L over 2 So the potential energy stored in its initial configuration Is just the height of its center of mass relative to the height of its center of mass

Okay, that's all you need for that one Omega squared, omega squared Then you just solve for omega Omega f is 6.58 radians per second So it speeds up a lot

What's this about that? Wait, so the whole bar bends and hits the ground It doesn't bend, it's rigid It flops around that hinge So it's fixed in place And the center of mass also moves horizontally, right? But that doesn't change the potential energy

Okay Next one 10.39

Find the magnitude of the angular momentum of the second hand on a clock About an axis through the center of the clock's face Alright, so Here's the clock Take the second hand The clock has a length The clock hand is a length of 15 cm

and a mass of 6 g

The clock hand, yeah I'm going to take the second hand to be a slender rod rotating with constant angular velocity about one n Find the magnitude of angular momentum So This is a very similar problem we just solved in terms of angular momentum

We expressed it as i omega Here again we have a rod rotating around one end So this is going to be the same one third ml squared times omega And we're going to have to find the magnitude of angular momentum

We know this mass, we know this length, so we just need to know omega. So what's omega? Sorry? Two pi over sixty, right? Omega is literally radians per second, right?

How many radians does a second hand go through in one minute, right? Well, it's two pi radians in sixty seconds, right? A second hand. So, we have everything we need then, right? We just plug that in. One third ml squared.

So this turns out to be zero point one radians per second. And so the answer is four point seven one times ten to the negative six kilograms meters squared per second. So this is really just a unit problem.

And once you recognize that you should calculate the angular momentum in terms of i omega, it's just a question of converting the information to give you the information you need. Okay. Next one is thirteen point seven nine.

This one is three dots. A five thousand kilogram spacecraft five thousand kilograms is in circular orbit two thousand kilometers above the surface of Mars

two thousand kilometers above Mars. How much work must the spacecraft engines perform to move the spacecraft to a circular orbit that is four thousand kilometers above the surface? So I'll call this H zero. H one is four thousand kilometers.

Okay? And the question is how much work is done by the engines? Conservation of energy. Okay, so

k one plus u one plus work other is k two plus u two. Right? And now let's think about what this the question is asking how much work needs to be done by the engines? How much work must the spacecraft engines perform?

So the only thing that's going to happen is this thing is going to change its position, right? This is the only outside work done. Just this guy. It's going to move through gravitational potential that will be accounted for here. It will change its kinetic energy that will be accounted for here. So basically we just need to solve for w. Now the question really says solve for w.

So as long as we can express the kinetic energy and the potential energy in terms of the information we have then we can solve for the work of it, right? So first of all what about the potential energy? Well this is minus big G which is universal times the mass of Mars, right?

The mass of the earth. The mass of the spacecraft plus r, right? Now r here is going to be the radius of Mars plus our height, right? The radius of Mars plus height

is distance from the center of Mars, right? Remember we're given these two heights. That's why I've expressed it that way because we're given these values relative to the surface of Mars. We can just include that here. We can call this height 4,000 kilometers plus rn and it will be fine. Alright, so we have an expression for u1.

The same expression works for u2, right? That's no big deal. What about the kinetic energy? How do we calculate the kinetic energy? It's a particle, right? It's not spinning. So the kinetic energy is simple.

It's one half mv squared, right? How do we calculate the velocity? We're given the mass. How do we calculate the velocity?

We need to know the velocity of a radius of an orbit around Mars at a certain radius. So the condition of an orbit is circular motion and that circular motion will be provided by gravity, right? So

GMm over r squared, right? So v squared is equal to g, so these little m's cancel, right? Mm over r,

right? So v is equal to square root of g, mass of m over r, which is the radius from the center of Mars, right? And so this, again, we can express in terms of the quantities we know, right? And so this is determined now just by the height.

So let's try this. So one half mv squared is GMm over r, m plus h, okay? So that leaves us with this equation right over here.

So the initial kinetic energy is one half m GM over r, m plus h, zero, right? Initial potential energy is minus

GMm over r plus h, zero plus work from the engines is equal to one half m GMm over r, m, s, h, one

minus G m, m, m over r plus h, one. And now we have an expression for w in terms of the radii, so the radius of Mars we just look up. The mass of Mars we look up. Everything else in this equation

we know, so we can just solve. So the key to this problem was understanding it's an energy problem and that the kinetic energy will also change. One was expressing the kinetic energy in terms of the height.

To do that you have to think about the orbit and the velocity of the orbit. Okay, those are all the problems that I got at midnight last night. And I should say that only two people sent me any problems. And one of those was a single problem. So almost the entire class

today was dial a problem by a single person. That person could have been you if you sent me any requests but you relied on your major answer. Okay, so anything else? Random physics questions? Do we go to the lab on Monday?

Do we go to the lab on Monday? There's a lab final on Monday. The final is 224 The final is 4-6 Do we still go to the lab? There is no class next week. Do not go to class after the semester is over.

Why do lights flicker? Why do lights flicker? Because the other ones have electricity going through them and they're just giving off the light. So that one sometimes uses electricity and sometimes it doesn't. Okay, well how do these lights work? How do these lights work?

So how does a light work in general? Well, there's two kinds of lights, right? Incandescent lights, how does that work? That's heat, that's resistance. It has electricity through a wire with resistance, it glows, gives off light. That's a very nice kind of light. But these lights are not incandescent lights.

These are fluorescent lights. How does a fluorescent light work? It's filled with gas and then it's ionized by an electric potential, so it's a plasma. It's a little tube of plasma created in the gas. And it's alternating current. So it's on, off, on, off all the time.

So that's why fluorescent lights, even when they don't look like they're flickering, need a headache because they actually are flickering just really really fast faster than you can tell. So sometimes they go wonky because it requires the perfect conductance of the electric current for it to be switching. Any of these things go wrong and the frequency of the plasma coming on and off

gets off from the AC and then it doesn't work very well. And that's why when you first flip on fluorescent lights they sputter a little bit before they get going and then they get in sync. Okay, other random questions? What's going to be on the lab final? What's going to be on the lab final? Appropriate lab questions.

Just like, it's going to be multiple choice because you're bringing a scantron. It's going to be a bunch of problems you should know how to solve. I wish I knew more questions. Yes? I will, yes after you finish it.

Calculators are allowed just like on the midterm. You get a cheat sheet in the final, yes but I will also provide all the equations I think you'll need to know. You can bring one piece of paper covered on one side with anything you want that's not pornographic or otherwise

illegal. Or human skin. Okay, thank you everybody.