So today's topic Angular momentum and work Okay, now you guys remember we've been doing angular rotations

We've been doing these angular forces called torque last time Right and we did that angular kinetic energy. For example, we said the kinetic energy of an object is One half and v squared of the center of mass plus one half I

Omega squared so here we have motion of the center of mass and motion relative to the center of mass Remember we constructed I just so that we could write this equation in this nice way where the rotational and the translational stuff looks similar Okay

Okay, and A couple quicker problems just to get your brains back in rotation mode Get your head spinning. So here we have a bowling ball rolling down a ramp and there is friction on this ramp The question is which of the following forces exists a torque on the bowling ball around its center

remember torque is is a force on the impact of the Lever arm Right and it's defined relative to an axis. So I've told you that there is friction but not told you whether or not

There's enough friction for the role of that something Okay Discuss with your neighbor and see you guys. You can't agree and vote again. So that didn't help

So what's the issue here? How do we figure out whether a force exerts a torque? The direction of the force with respect to the radius, okay It's essentially drawing the line of action of the force and figuring out whether there's a lever on. Yeah. Okay

So let's go through each of the forces. First of all, what are the forces on the on the on the bullet? Normal force weight and the frictional force are the forces on the ball, right? Where did these forces apply? Where does the weight fly? Where in the ball

They go through the axis of rotation, okay So the normal force definitely acts on one point right because it's one point of contact between the ball and the surface Right and that definitely points to the center of mass of the ball. So no torque right degree the normal force is no torque

Okay, what about gravity does gravity act on only one point in the ball? No, where does it act? Every tiny particle right but can we treat the ball as an effective particle of the center of mass? Sure because Gravity is outside the ball, right? We're interested in the ball as an object not the internal Fascinations of it. So gravity acts on basically the center of mass of the ball, which is also the rotation axis

So is there any torque from gravity? No, no, so then we're only left with one force, right? Friction right now a lot of you said D more than one of the above

Anybody still feel that's true? What was the confusion there Was it gravity Okay. So now the question is Is there enough information? Well, where does fiction apply at the bottom right at the point of contact between the ball and the ramp, right?

here's the Here's a free body diagram right gravity. You can draw from the center. The normal force is Here, but you can draw it in the center, right? Friction applies here. It's the only one that has a nonzero lever on

Do we get did I give you enough information to solve the problem? Yeah. No, this is no why not?

Very interesting. Okay, so she says it's not enough information depends on how much friction there is So let's think about that if there's no friction then what happens to the ball It slides right I totally agree but in this problem we said there is friction, right so that case is not relevant

If there's any nonzero friction, what's going to happen? Can it rotate and slide at the same time? Yeah, you can right rolling without slipping just means that there's a perfect link between the motion of its center of mass and its rotation So there's no sliding between the two surfaces

So it's possible for there to be less friction than you need to roll without slipping But still some friction and some friction is going to lead to some spinning even if it's not enough To make it roll without slipping So as long as there's any friction then it's going to be a frictional force and it'll be some torque tricky though

Questions about that okay, one tricky thing about the problems with rotation is that we We try to sometimes overthink them like where's the force get applied? Remember when you're thinking about the motion of the center of mass It doesn't matter where on the force where on the object the forces applied

You just think about it as motion in the center of mass Okay What about this one here Here we have a yo-yo on a horizontal surface Here there is enough friction for the yo-yo to roll without slipping. The string is pulled to the right

What happens to the yo-yo? Okay, so we have intuition here right, you know what happens if you pull on the yo-yo to the right it's going to move To the right, right or it could be D. The answer depends on the magnitude of the pulling force It's certainly not going to stay at rest or roll to the left Okay, so who wants to make an argument for why roll to the right?

Or against that The rest of the string is wrapped around it will begin to rotate

Roll okay, and what about friction?

I Mean, it's not sliding anywhere. It's not sliding right so but there you need friction to roll without slipping, right? So there is friction what direction is friction applying which directions are pretty much to the right to the right? Why is that because as you pull on the string the yo-yo is going to be rotating to the right if you?

Rotating this way, right? Yeah And if you look at a point At the end That point's gonna end up rotating and hitting the surface so that

So this you're saying the yo-yo so here the top of you is moving to the right the bottom the O's move to the Left so friction points to the right So what forces do we have on the area then pulling force which goes to the right? Frictional force which goes to the right Is it possible for the age of anything we move to the right no the net force on the yo-yo is to the right

Even though the two forces are applied to different places on the yo-yo, right? We're only concerned about the motion of the yo-yo the motion and seven master Yo-yo doesn't matter where on the yo-yo those are applied Or is it the top and the bottom doesn't matter both forces are to the right the yo-yo moves to the right Questions about that, right

So Say there was no friction for example

Okay, so say there's no friction what would happen when it's been it would slide right there's no friction there's no No, it would still be spinning because there's still four If the if the string was attached to the center, it wouldn't spin the train the string is not attached to the center So if there's enough friction between the string and the yo-yo right before assuming those are attached to each other

Then it would spin because it's w-torque. So it's been in place No, because there's still force in the center of mass. So the center of mass Yo-yo rolls to the right. Okay. All right. What about this case?

Here we have the same horizontal surface enough friction for the yo-yo to roll without slipping and now the string is pulled up What is the face of the video? All right, no 38 people

So what happens in this case what forces do we have? Mm-hmm. There's also weight in the normal force, right?

We ignored in the last problem because we're just thinking about in one dimension, but how does that play in here? So you're assuming that the force

Is the force is less than the weight doesn't leave the table, okay Forces So it's been in the same direction as last time

So what are the horizontal forces just friction So friction makes it roll to the right everybody agree with that there's nothing to balance friction, right?

Yeah, that's right, so in this case the right yo-yo again rolls to the right, okay questions about that Yeah, if you pull hard enough then Then you can lift off the ground and then there'd be no friction and they would just spin right?

Okay, let's do some examples From the book to get used to doing these calculations Okay, and then we'll talk about Angular work

Okay, here's a problem from the book 10.19 Says a 2.2 kilogram poop with diameter 1.2 meters rolls to the right without slipping and with the velocity of three radians per second

rolling this way and mass 2.2 kilograms Now the first part of question says how fast is the center moving?

Hey says what is the velocity of the center of mass? Okay, so we're told that it's rolling without slipping without slipping rolling without slipping remember means That the velocity of the center of mass is equal to angular velocity times the radius, right?

Now it's always true remember it's always true that the tangential velocity Is equal to the angle of velocity times the radius, right? This is the velocity of a point on the on the hoop but if Rolling without slipping then this is equal to this, right?

or So people get confused because they remember this This case here rolling without slipping is a different is a different equation right here. It's BCM not the tangent, okay?

These do not necessarily equal each other unless it's rolling without slipping which it is in this case Which makes this problem very easy because we're given the radius and we're given omega, right? so it's just 0.6 meters times 3 radians per second

1.8 meters per second Okay, then it says What is the kinetic energy Well, we know the kinetic energy is one half Mvcm squared plus

one half I Omega squared All right, we're given in the mass. We just calculated the velocity of the center of mass We can look up what I is for this case Is and r-squared, yeah, I check to look that one up right this one's very easy to figure out because remember

This is just the sum and I Squared over all particles right all the particles have the same radius and this becomes So that's simple right so we know that

We're given omega right and so we can calculate this very easily If you can manipulate it a little bit symbolically first you can say this one is one-half and R squared and Omega is V over R This quantity squared see the bars cancel, right and we get another one-half and V squared

So interestingly in this case the kinetic energy is MV squared So there's equal kinetic energy in the spinning and emotion But you don't have to necessarily do that when you have all the information already at this point

Okay, then it asks us What is the velocity of the loop as? Used by a person on the ground. What is the velocity of the highest point of the loop?

So up here So somebody's standing on the ground and measuring the velocity of a single point on the loop What's the velocity of that point to be why to be

So velocity the center of mass is VCM velocity of rotation, right Is

positive here And negative here with the same magnitude right which is what we need for rolling without slipping So the velocity at this point here is to VCM and Here it's zero right the velocity is a motion of the center mass plus the motion with respect to the center of mass

okay, so here it's Velocity at the top is to VCM and then it asks us predictively Find the velocities viewed by person on the ground of the lowest point velocity in the bottom Is zero now? I think this is a bit confusing, right?

Zero if you're looking at a rolling loop, it seems strange to say at any point of the loop has zero velocity All right And that's because the whole loop is moving. So how do we reconcile that intuitive? Understanding like if you would ask a person on the street who's not taking physics is any part of this loop motionless

If they know what are you an idiot, right? So then we do this calculation we get this answer I'm telling you it's correct, but how does it make sense of it?

Okay, explain it to me like I don't know any physics you mean intuitive explanation

Lost me already Where the corner is gonna stay at one point and then you change that like a hexagon Okay, but the point on the square is gonna move right the point at the bottom just like this

Like this, right every point is moving, but from the point where it gets to there until the other side Is Okay, if I watch this corner right it moves right it's velocities non-zero

Even though it starts off. All right, so it starts off the bottom. It's moving, right? Not at that instant, right so that's the key what we mean by the bottom changes every instant because the wheels rolling All right So if you think about it a person who's asked you ask this question is the wheel is there any part of the wheel? Motionless he's going to think about a specific part of the wheel and follow that wheel through its motion and say no everything is moving

But here we define something sort of funny which is the point which is touching the bottom Even if that point of the wheel is changing, right? the point here is that the reason this is confusing is what we mean by bottom here means the Part of the wheel which happens to be touching the ground at any moment and that meaning changes

Your explanation was a good one And I try to trick you by linking you to a specific part of the wheel which any part of the wheels of course moving You're right that the part which is at the bottom is never

Okay So anytime you get an answer physics problem think about it does this make sense, how would I explain this to somebody? That's generally possible until you get the quantum mechanics and then you just gotta Just Follow the math, okay So now let's think about rotational work

Attentional work. Okay, and for translational motion or linear motion. You're right work is force times distance All right, you have a force you apply to an object and moves through distance. Here's how much work you've done

in rotation We say work And the analog of force is now torque the analog distance is now theta Okay, so because we spent a lot of time building our analog basic objects and rotation to be similar to Translation than the higher level objects are sometimes very easy to define like this one. So this is angular work. It's

Torque applied to an object if that object is rotating then you're doing Angular work on it and that makes sense. You're giving it rotational kinetic energy, right? That's how you're doing work on it. Okay, and we can also think about angular momentum

Linear case momentum is and Right in the rotational case We can write I know man

Rotational inertia rotational velocity inertia times velocity This is also you can relate these two quantities By taking R and crossing it with P and that gives you the angular momentum

So why is this interesting? Well remember for forces? we said we said force is

Ma is dp dt, right? So No external force means Force equals zero means dp dt Equals zero conservation momentum, right in the linear case torque is I alpha, right? This is

Newton's law for rotation, it's just translation of all these terms into rotational terms And this is dl dt

The derivative of England momentum All right So just like in linear momentum force is the thing that changes the momentum of an object In rotation torque is the thing that changes the angular momentum of an object Okay, if you have a certain amount of angular momentum, you will keep that angular momentum unless somebody applies a torque key

So the earth is going to keep spinning until some massive You know celestial body hits it and applies an angular momentum Unless the body hits exactly the center of mass of the earth in which case it won't change the spin fortunately Okay, and so this means

conservation of momentum, right so dl dt equals zero if tau equals zero So, how does that help us? Well, let's think about what that means physically I could think about a simple case I We used conservation of momentum to solve interesting problems like collisions

Right we considered what happened to the momentum when two particles collided and have helped us figure out the kinematics of the outgoing particles Let's think about collisions with rotation Okay, and so a common collision with rotation is the clutch in your car Okay, and if that's in your car you have one wheel

All right, which is spinning With some velocity and then you have another one So one of these is connected to the engine and one of these is connected to the wheel This one a B

Okay, and this one has this thing has some moment of inertia I a I B and this one has some moment of inertia I a Okay, we know that we can calculate the car the angular momentum as I omega now if I say that okay So this is the initial state the final state is No question. The final state is that these two things touch each other

Right they come together So this is when you have the clutch pressed So the engine and the wheels are not connected to each other you release the clutch The two pieces come together and they have a single rotational speed, right?

Otherwise your clutch is grinding like crazy Okay So we can figure out for example What is that final rotational speed from knowing these initial conditions by insisting that the net momentum is conserved? that Initial angle momentum equals final angle of momentum. How do we do that? Well, we know the initial angle momentum. It's I a

W a plus I B WB right. What's the final angle momentum? Well, it's the moment of inertia What's the moment of inertia of this for these two gears together? I a plus I B. Why is that?

Why can't I just add them? Like why can I why do you combine two objects together the same axis? Why why is there moment of inertia the sum of the

two Right Moment of inertia is a sum So you can put two things together the moment of inertia is just the sum of the moments Right because you think about the calculation You can calculate the moment of inertia of some object by calculating the moment

of inertia of one piece and adding to the moment of inertia of another piece So if you know the moment of inertia of the sub parts, you know the moment of inertia of the system just like that That's what that's what tells you why this is a plus not minus or times or to you know To the second power or whatever

Here we put the final rotational speed. All right, and so we can solve for For this it's easy and it's sort of like a mass weighted average. See Except it's a moment of inertia weighted average of their of their angular velocities

So here's how we use conservation of angular momentum to help us solve collision problems with rotation Okay, now let's think about angular

kinetic energy I think about that I have a Quicker problem for you. Okay, here we have a figure skater And he's skating it starts out like this as he pulls his arms in he's gonna start going faster and faster We know what happens to his angular momentum. And what happens to his kinetic energy his angular kinetic energy

boom You guys totally figured this out All right argue with your neighbor try again Okay, we seem to coalesce into camps B and D

Okay, so B and D B L stays the same K increases and D L and K both stay the same. So it seems like most people agree That the angular angular momentum is unchanged. Why is that or does anybody disagree with that?

Why does the angle velocity increase? So so why does the angle of velocity increase? Why is he spinning faster because the angle of velocity

Okay, so you're saying angular momentum is constant why is that

Okay, so there's no external torque does everybody agree Okay, no external torque means angular momentum is constant, right Okay, and angular momentum initially is I initial omega initial Equals I final omega final, right?

Obviously, this is gonna change. Why does this change? Because he's moving his arms his arms now have different radius he hasn't changed his mass and so this has to change right and so the final Velocity is I initial over

final times Initial right? So you're saying he speeds up. Why does he speed up? You must speed up if this ratio is greater than one And so why is the initial? Moment of inertia larger than the final moment of inertia because this goes like M R squared and he's decreasing at R, right?

Okay, so we agree that this number here is greater than one. So the final Angular velocity is greater than the initial angle of velocity. Okay, so we all agree that L stays the same. What about kinetic energy?

I Sounds like you're talking about Okay Okay, so he's got some form of internal energy and you're saying he's turning that internal energy into kinetic energy

Why do you think kinetic energy goes up you just explain to me where the energy could come from to increase his kinetic energy But how do we know this kinetic energy increases? If it does anybody else

What's that Angular velocity increases. So what? so Kinetic energy, however has two components. All right, not just omega not just omega. There's I also right and

I is going down. It's omega squared, right? Okay, so think about this Right. This part is constant, but you have this part also

So if angular velocity goes up and angular momentum is constant Kinetic energy has to increase What was your argument

Is

There's no need to translate this back to tangential terms, right? It's hard though because he's a complicated system, right? He doesn't have a single R for his position is a complicated me to describe system described by the moment inertia So angular momentum is conserved and kinetic energy increases and then the next question was asking was how can the kinetic energy increase?

Right and the answer is given to us already by Jeffrey, which is he's got energy in his muscles He's doing work on himself by pulling his arms in it's not trivial Right, if you're spinning around to pull your arms in takes energy

You're giving yourself more kinetic energy than you do this and that takes some work It's interesting. You can make yourself spin faster without pushing on anything. I Saw this anybody here watch

Marvel's Okay, so There's a scene the scene in that show where

There's a ceiling and there's some door with a hatch And one of these you have to spin it and it's like I don't know some ridiculous mode distance off the ground and super person jumps up Grabs this thing and without touching anything just the strength of her abs or whatever turns that Is that possible?

No, you're not anchoring yourself on anything. Yes. Yes. Why? I think it's possible because she has momentum in a direction. She grabbed on say she's okay So she starts right under it say for example

She's hanging from it and she goes, you know Is that the equivalent of pulling yourself up by your bootstraps or is it physically possible

All Right, so if you turn your legs one way and everybody the other way then what's gonna happen just right back, right? And then you can let go Okay, okay so

Let's get back to reality. So angular momentum is interesting So angular rhythm is interesting and important it also affects things not just like, you know What super agents can do in a crazy universe but also what happens in our actual universe and conservation of angular momentum is one of the critical Concepts that determines the structure of our universe. So for example, why is Jupiter spinning?

Magic, great. Why is Jupiter spinning torque? Why is Jupiter spinning Does Jupiter have to be spinning No, why not why not? Well, so why is it spinning? Why are most of the planets spinning?

Well, it's certainly true that angular momentum will be conserved in the universe, right? My guess would be that when like They started farming his planets like everything didn't collide at the center of mass. So it gave it towards when it collided

So how is the solar system form? You have a big ball of gas right or dust and that coalesces gravitationally pulls itself together What are the chances that some massive universe or like some massive scale ball of gas has exactly zero angular momentum

Seems to be very unlikely, right? So if it has any angular momentum to begin with it's gonna have any momentum afterwards no matter how much rearranging it does Right things coalesce gravitationally You're reducing the are by a lot Those things pull in just like the figure skater pulling his arms in Gravity does work to pull this stuff in it makes a small amount of angular momentum of a huge blob of gas

Into fairly rapidly spinning planets. So that's why the planets are spinning the solar system is spinning The Sun is spinning around the center of the galaxy. So initial angular momentum from the initial state You know of these balls of gas So angular momentum important stuff on a sort of galactic scale

Okay. All right. Let's do some more examples. Are there questions about that?

Well, there's planets all do orbit on one plane except one of them which one Uranus it's off kilter, right and this planet spin all at different rates Right, and it's been in different axes. Like the earth is tilted slightly

You know why that is? Well, it comes from collisions example, you know The tilt of the earth is likely due to a collision of the earth and some massive body which may have also formed the moon a bajillion years ago Any time you absorb a large impact change anything with them then slightly sideways

Yeah, I think I don't remember anymore Neptune that's spinning basically horizontally

And that's possible due to some impact, right? massive Impact a long time ago, you know has to be a long time ago because Neptune began a sphere Any massive impact is going to disrupt the spherical shape for a while until it coalesces back into a sphere So you don't learn a lot about the history the universe by looking the history of our full system by looking at the rotation

Of all the objects and for example, you know the moon there's a size of the moon Which is almost always dark inside which is facing the Sun, right? It just so happens that the angular rotation of the moon exactly counteracts the motion of the moon

The motion of the moon around the Sun via the earth That's a perfect coincidence What's that? You say whoa The moon is another amazing coincidence, which is that the moon is the same size in our sky as the Sun Certainly doesn't have to be right. They're not like they're objects of the same size. It's just coincidence

This is the kind of coincidences you find in physics anything Well, there must be some real explanation for it can't just be a coincidence Sometimes there are just coincidences You look at an eclipse is a little corona for example, yeah, not exactly but pretty close I think it's within 1%

Okay, so let's do some problems using conservation of angular momentum, etc Here we have one from the book 1056 it says

Uniform holo disc has two pieces of thin light wire wrapped around this outer rim and is supported by the ceiling It looks like this the disc here and

We're given these two radii Our one is 30 centimeters our two is 50 centimeters Okay, suddenly one of the wires breaks and the remaining wire does not slip as the disc rolls down

Okay, so initial condition then this one falls and then it rolls down this wire without slipping So there's friction between the wire and this acts like a yo-yo after the wire is cut The remaining wire does not slip as the disc rolls down

Use energy conservation to find the speed of the center of this disc after it has fallen a distance Of 1.2 meters Okay, so as if it wasn't obvious it tell us to use conservation of energy, right? So kinetic energy plus potential energy is kinetic energy final plus potential energy final, right?

We can call the initial potential energy zero just to be convenient and What is the initial kinetic energy One

What is the initial kinetic energy one zero also, right Okay, so that was easy and then in the case so it's a look up what we Move in America is for this guy. It's one of the examples

That can't be right somebody look up at the somebody have their book

Oh That makes sense Always break the plus r2 squared. Yeah, because if our one goes to zero

If our one goes to zero it reverts to one half a month later Okay, so we know the moment of inertia, right? We're also told it moles without slipping. So VCM equals V tan Equals omega R, right and this is going to be omega R2 Right because it rolls around the outside radius obviously not the inside radius

Okay so here the kinetic energy is one half MV squared CM plus one half omega squared

Right plus M G Y, right? This is the potential energy It's just the mass gravity plus the distance it travels or I guess I should be H since I use the H up there And it depends if you define this as a positive or negative distance

Right because this obviously has to be a negative quantity because this is going to be positive I We have here in terms of the radii Omega we have in terms of VCM, right? Using omega is V CM over R 2

then we get one half and VCM squared plus one half and R 1 squared plus R 2 squared. This is I here I plugged in V CM over R 2 quantity squared

plus M G H Where I'm gonna call H, I guess I should call H here negative 1.2 meters and so this is equal to zero So I know M. I know M, I know R, I know R

I know R, I know M, I know G, I know H So I know everything in this except for M except for V squared which is what I'm looking for Thank you. Yes one half. Yeah

So it's another Classic case of potential energy turns into kinetic energy and then you connect the Rotational velocity to the motion of the center of mass by knowing that it rolls without slipping It looks like a complicated problem because they have this whole double wire and inner loop and outer loop The inner circle doesn't really matter at all

Except in the calculation of this which you can just look up So it's really just like a simple yo-yo problem made a tiny bit more confusing Questions? Okay. Let's do another one. Okay

So a large roll of paper with a radius 18 centimeters rests against the wall

R is 18 centimeters. This is a roll of paper like toilet paper and And it's held in place by a bracket attached to a rod through the center of the roll

So this is an axis of rotation and this is attached to the wall like this Okay The rod turns without friction in the brackets and a moment of inertia about the paper and rod The moment of inertia of the paper and rod about the axis is blah

They're 0.26 kilograms meters squared Okay, and so that's the moment of inertia around this axis here of the paper and the rod The other end of the bracket is attached

The rod turns without friction in the bracket moment of inertia of the paper and the rod about the axis Sorry, this moment of inertia is around this axis here The end of the bracket the other end of the bracket is attached by frictionless hinge To the wall so that it makes a 30 degree angle to the wall theta or theta is

30 degrees The weight of the bracket is negligible the coefficient of friction between the paper and the wall so there's friction here 0.25 Constant vertical force just gets coming around here is applied to the paper

Newton's and the paper unrolls All right, so we have the paper which spins around this axis and which is connected to a rod Connected to the wall which has a hinge at this angle We have friction between the paper and the wall and we have a force somebody's pulling on the paper

Okay, so sort of a realistic system Question is what is the magnitude of the force on the force that the rod exerts on the paper as it unrolls? Okay, so Let's have an understanding of what all the forces are

On this guy Think about the forces. Okay on the paper Okay, so what are the forces on the paper? gravity, right What else

Sorry, so the force from the rod This is the wrong Rod hinge here. Okay, and the rod goes along the rod, right? What else friction which is which direction friction goes down why is that?

All right, so the paper is being pulled down here so the motion is up here so friction is down here Force of friction. What else? Normal force sure which is this way, right and what else what's that?

The pulling force, yeah, which is here Is that everything we have one two, three, four five forces. Okay, so we're asked what is the force on the rod?

Or is due to the rod How do we figure this out? I think it's complicated forces applied everywhere things spinning Sorry, go ahead

Yes, I just heard you say the answer. I think you're sure now Yeah, think about the components This thing is not moving right the center of mass is not moving It's going to spin so the center of mass is not going to move Which means that the net forces on the center of mass are zero

Paper goes down Less paper sure. Yeah, there's an angle of the crease as you Paper, but here we can assume the papers next to the thing compared to the size But yeah, if you have an infinite demand for paper

But assuming that the center of mass of the roll paper doesn't move these forces all balance and this complicated looking problem It's actually just a simple equilibrium problem, right and so in X We have components So let's break up the force of the rod the force of the rod is the only one even that has an angle

everything else is pretty simple Force of the rod has a component like this and a component like this, right and in y this is F rod cosine theta and X this is sine theta

So the X equation we have F rod times sine theta is equal to the normal force right from the only two forces with X components and

in Y we have F rod Cosine theta is the only force that goes up Is equal to friction plus the weight plus the pulling force Now sometimes I find this confusing because it's strange to think of these forces balancing because they're being applied to different points

And they do have different torques and we haven't thought yet about whether the torques balance In fact, we know that torques don't because the paper is going to spin but we're just being asked about the motion of the center of mass or If we want to think about the motion of the center of mass to find the magnitude of one of the forces

Then we can ignore all that and just think about the motion of the center of mass which makes this problem much simpler Okay, and then we know of course that F is equal to mu K N. All right, we've put an N here and we know the weight We know we're given the pulling force. We know mu

We know theta. All right, we just want to know F rod. So there's enough information here to solve This system of equations to get the magnitude of the force of the rod Really the hardest part of this problem To do a little bit of algebra and it's the hardest part of this problem is figuring out what they're asking

To avoid all the trickiness. Let's make it clear diagram for yourself. Think about where all the forces are Think about what are they actually asking? We'll try to understand every single thing about the system from the beginning

And then they ask What is the magnitude of the angular acceleration of the roll

Find the torque line We're given I just given to us in the problem. They're at me asked to find alpha. So let's find the torque, right?

Okay. So what forces here are going to give us torque is the weight gonna give us or normal force force in the rod This force is gonna get the torque right and this force is gonna get the torque and the two will close each other

Right, they're going different directions so the torque Is equal to the pulling force minus the frictional force Times R right and so from there the angular acceleration

All right. No, we know this one. We're giving it. How can we find the frictional force? Find a frictional force from the UK n and we find the normal force normal force we find from here because we know the force in the rod f is equal to

UK n Is equal to UK that in here and you have the torque and then you just get the acceleration

All right, let's do another one

Okay Uniform marble rolls without slipping down a crazy path

This This is H from here here is 45 meters

From here here is 25 meters a marble starts here And The pit is 36 meters wide a uniform marble rolls without slipping out slipping

That's important right Down the path shown in this figure starting from rest. Hey find the minimum height H required for the marble to not fall in the pit

And so I think from here we can assume that it's Horizontal despite my drawing and that all of its kinetic energy is going to get turned into horizontal velocity Okay, so how do we solve this problem?

We need a strategy for several parts Okay, so you're thinking about how to turn this potential energy into kinetic energy, yeah, but we don't know this height, right?

So let's go backwards so first we do what Oh Okay, so from 45 to 25 right so it's gonna go like this

so Find the time of flight Right, assuming it's gonna have no initial y-velocity It's pretty easy to figure out how long it'll take from call from here to here right from that you can find The X Right, if you know how long you're gonna be in the air

You know how long you have to get 36 meters so you can figure out how fast you need to be going this direction, right? So from knowing the velocity here you can figure out to know if you Know how far this is you figure out the time of flight from the time of flight You figure out how fast you have to be going right?

So now we know the velocity we need here and how do we relate the velocity we need here? to this height Energy, right? It's potential energy in terms of the kinetic energy. Okay So then relate the X to kinetic energy to height, okay, so

It's gonna fall 20 meters, right so let's do one find the time of flight it's gonna fall 20 meters, right? one half gt squared All right, there's no initial y velocity so it's just falling from rest so I can calculate the time it's going to take to fall

Because I know everything already there 2.02 seconds Okay, so you have two seconds to fall 20 meters. Does that make sense? Well, you're accelerating in 10 meters per second So it sounds about right You have two seconds to fall 20 meters

So you have two seconds to get 36 meters across, right? So what velocity do you need to go 36 meters in two seconds, well, this is just distance equals rate times time, almost 18 meters per second.

So if you have a horizontal velocity of 18 meters per second here, you will make it across the pit. Everybody agree? OK, now the question is, what height do you need to get that horizontal velocity? OK, and here's where we use energy.

So we say kinetic energy initial plus potential energy initial is kinetic energy final plus potential energy final. So we call potential energy final zero. Kinetic energy initial is zero. So this is just mgh is the initial potential energy. And how do we write the kinetic energy?

Can I just do 1 half mv squared? Yes? You need to consider the rotation. I need to consider the rotation, right? If this thing was just a ball rolling with slipping, then all of its potential energy would turn into kinetic energy of its motion, right? But some of the potential energy gets sucked away into the spinning. So I have 1 half mv squared plus 1 half i omega squared.

And as usual, I have two possible speeds here, motion in the center of mass and rotation around the center of mass. But because I know it's rolling without slipping, I can link these two so I don't have two variables, right? v equals omega r.

And in this case, i is equal to 2 fifths mr squared because it's a solid sphere. OK, so now it looks like I have everything I need because I can replace this guy here.

So mv squared plus, so this becomes 1 fifths mr squared v over r quantity squared, right? And the r's cancel.

So this is mv squared times 1 half plus 1 fifth, right? And so now I can just, the masses cancel.

And so the height, was the question, was the height? Or the velocity, yeah, it's the height, right? Has the question, is v squared over g, right?

And so I know everything there because I already solved for v, right? I know what the velocity is so I can turn that into height. Second part of the question says,

the moment of inertia of the marble depends on its radius. That's certainly true. Explain why the answer to part A does not depend on the radius of the marble.

Notice how the radius canceled out. Why doesn't it depend on the radius of the marble? Because they cancel each other.

We already established the relationship that angular velocity is going, start tangential velocity is going to be the same as the velocity of center of mass. And since they're proportional, it's not going to matter.

It's our third base. Right, because this thing depends on r squared. This thing depends on 1 over r squared, right? So these things do cancel. If this thing was much larger, then how would these change?

I would go up, right? And omega would go down. So a larger marble would spin less rapidly but wouldn't absorb any more kinetic energy. Is there a limit on the size of the marble for which our calculation is correct? Well, 36 can't be greater than 10 meters. Yeah, because if the radius is greater than 36 meters, it's just going to roll over that pit anyway.

OK, now it says, solve part A again for a block that slides without friction instead of a rolling marble. So how do we do that? We just remove this part, right? And we can solve for it again.

And so here, right, so here the height is 22.6 meters. In the case of a block with no rolling, this is 16.6 meters.

OK, so this is including kinetic energy that gets absorbed into the rotation. If it's just a block and it's not rolling at all, then you need less height. The way you solve that is you just remove this term from the kinetic energy. You see how the same velocity meets a smaller height.

Yeah, it says the block slides without friction. But you're right, because to roll without slipping needs friction, right? OK, there's more example problems in the notes. Take a look. And I'll see you guys in 10 days.

But next week, Simona and Mircea will be here.