okay so remember last time we were talking about rotation right and we defined our new coordinates theta right and we try to map everything we could

from our understanding of linear motion to rotational motion right and we did things like we define the new velocity right and we define a new acceleration and then we wrote all the kinematic equations which are very

similar to the translational kinematic equations you just replace one variable for another all right so that was very helpful today we're going to talk about force okay rotational force okay rotational force we essentially

called torque okay so why do we need a concept of rotational force why do we need to define rotational force okay so we have rotational acceleration

and therefore okay or we observe that there is acceleration in the world and we wonder how do we describe it right another comment that's right yeah we

see rotation in the world and we're gonna have to do some calculations like if you apply this amount of force to this object how quickly does it rotate all right so we need to be able to relate this to this acceleration to force and so we need a concept of rotational force okay now rotational

force is a little bit trickier than linear force right in the same way that remember we also defined this is our moment of inertia inertia and how is

moment of inertia different than mass what's the critical difference here yeah it depends on where you calculate it right you have an object and if it's not and it works you have an object like a sphere it has one moment of inertia around this axis a different moment of inertia around this axis right

it'll rotate differently if you push it around one axis or another axis the same thing will be true for force all right imagine you have some nut you're trying to turn right and you have your wrench right if you apply a

certain force here it's not going to turn the nut as much as if you apply the same force at a different radius right that's why we have these wrenches that's the whole idea right to be able to apply a force with a longer lever arm so the key here is the distance between the point of rotation

and where the force is applied of the lever arm okay so we need to build into our concept of rotational force the fact that it's more effective if you have a longer lever arm okay so let's think about that a little bit

more generally I mean say you have some weird object an egg or something I don't know and you're going to rotate your your yolk right you're going to rotate it around the yolk right take your egg you put a pin in it or something now what happens if you apply a force like this is it

going to rotate why isn't it going to rotate I'm applying a force it's the forces applied at a point which is different which is pretty far away from the axis of rotation or am I not going to get any rotation here no component

of the force is along the direction of rotation well I haven't defined the direction of rotation right just define an axis and put a force on the force is parallel to the lever arm well let's think about what the lever arm is right so this is exactly the idea one way you could think about it

is to say well here's the lever arm all right it's like this I draw it from the point of rotation to the place I'm putting the force and he's exactly right the force is parallel to the lever arm here the force is perpendicular lever arm that's why I was effective if I had drawn a force

like this I wouldn't be turning the screw would I right so this case if so now we've learned if the force is parallel to the lever arm you don't get any action another way to think about it is to extend the line of action of the force say well I'm gonna draw a dotted line through where the

force goes and then I'm gonna take a lever arm to be the distance from the axis of rotation to this line of action which in this case would be zero

so it's just two different ways to think about the same concept I'm not trying to teach you two concepts just one concept from two different perspectives one is calculate the lever arm from draw the lever arm from the point of origin to the point where the force is applied right and there you have a nonzero lever arm but you have an angle you have no angle

between these two so it disappears and we'll talk about the mathematical formulation in a moment or extend the line of action as close as you can to the point of rotation and look for the lever arm and there you see it zero so if instead you apply your force here for example right then you can either

draw a line of action of the force and say aha I have a nonzero lever arm and a nonzero force and as your intuition suggests this will give you a rotation of your head right or similarly if you apply the force like this right

up here also you could extend the line of action and then calculate the lever arm okay now in this case call the lever arm L if these two things are

perpendicular as you extend the line of action so that the force and the lever arm are perpendicular then we can write the torque as the magnitude of the torque as the magnitude of the force times the lever arm okay in this case

when the force and the lever arm are perpendicular to each other okay now to talk about the more general mathematical form for torque that you can use in for any angle of lever arm and force for example if you wanted to

draw the lever arm like this and then calculate the torque then you need to just think for a moment about the cross product and I'm 95% sure that everybody here knows how to do a cross product but just in case let's just review it briefly so that we're all talking the same language some of

you might know this as the inner product but in general so we're stepping down to a mathematical universe outside of physics if you have a vector a and a vector B okay and you want to calculate their cross product then the cross product is perpendicular to both okay this is C is a

cross B sometimes people call this the outer product to contrast it with the dot product sometimes called the inner product interesting things about the cross product is that you get a vector number dot product a dot B you get a

number it's a scalar just a value this kind of direction this has a magnitude right and a direction and direction is perpendicular to both so a and B two vectors are enough to define a plane and the cross product is perpendicular to that plane okay and the magnitude of C is the

magnitude of a magnitude B times sine of the angle between you remember for the for the dot product we have the cosine here here it's the sine so what

does that mean it means that you only get nonzero values you only get nonzero values if these two things are yeah that's right sorry so if these

two things are parallel to each other you get zero right so for example a a they're parallel to each other a cross a magnitude is equal to zero because the angle between them is zero so you get the maximum value when

there's a 90 degree angle between your two vectors okay when they're view for everybody okay so let's come back and think about and use cross

product to define a more general concept of torque okay so let's take a concrete example you have some lever literally a lever with an arm on it okay and you apply a force up like this okay and so this is some angle five

so one thing you can do is say and this is a vector of length R so one thing you can do is say well the lever arm here is R sine Phi okay so

torque the magnitude of the torque is the magnitude of F times R sine 5 right I've extended the line of action of my from the point here to here to get

the minimum lever arm and I found that it's R sine Phi because it's just this component of the triangle and the magnitude of the torque is just FR sine Phi okay another thing I can do is break up this force into two components I can break this one up into two components I could say along the

lever and perpendicular to the lever right this one is F sine Phi and this

is F cosine Phi now which of these two contributions is going to give me rotation this one right you pull on the lever really hard this way push on that way it's not going to turn it so this is the only one that gives me

rotation so I can say that torque so now these two things are perpendicular the force is perpendicular to the lever on them so I can use the complete lever arm here I can say the magnitude of torque is equal to F sine Phi times R same answer right nice to live in the universe where

math works or I can say well I don't want to bother about taking all those shortcuts I'm just going to have my vector and have my lever arm which is also a vector and I'll say that the torque is F cross R okay this is the

most generic formulation of the torque regardless of the angle between the force and the lever arm now you can see how this easily reduces for example to this one if the two are perpendicular you just get the magnitude met this times the magnitude of this times the sine of their angle

but they're perpendicular the center of the angle is just one you can see how you don't like to think about vectors and cross products you can revert to the situation where the two angles are perpendicular either by breaking the force into the useful component and the component that's a radial and then using this 90-degree angle or by drawing the line of action

of the force down here and using this 90-degree angle okay why does that

matter why does that matter is a cross B the same as B cross a no

they're opposite each other right this is a cross B and this is B cross that's

right in this case the torque vectors in it is in or out of the page depending on the sign yeah so R cross F it's out of the board so use your right hand to determine the direction of the result of the cross product so if it's a cross B you do curve from A to B and the result is

that way here's R cross F so it's on page okay so let's use our clickers just get a little bit of practice with that okay so these four forces all

have the same magnitude f1 f2 f3 f4 there's the origin and it's not a point of rotation which of these forces produces the greatest torque about this

point so why is it a and that matters because right so let's compare them one

by one f3 and f4 have the same bar right but f3 is perpendicular to the lever arm or f4 is less than that so clearly f3 is more torque than f4 what about f2 f2 is right along the lever arms is gonna be no torque there

f1 is clearly greater than f2 f3 was greater than f4 f1 has to be more torque than f3 because it's the same angle and a longer lever arm so it's f1 all right what about this one which of these four forces produces a torque

about though that's directed out of the plane so towards you use your right hand all right so some disagreement here why don't you discuss it with your

neighbor from the origin to the point right back so in this case ours we're here

okay everybody vote again please this point you've either been persuaded or you're un persuadable okay so consensus says D some holdouts say E

who thinks it's more than one of these who knows how to press D so well let's

think about it two of these point in the same direction right f1 and f3 and at the same lever arm direction of the same lever arm so their torque is going to be pointed in the same direction so f1 and f3 point in the same direction f2 is zero length torque and f4 is going to point the

opposite direction from f1 and f3 right so either the answer is f4 by itself because it's the only one with unique direction or the answer is E more than one of these right so it has to be either D or B all right you guys already got that so then it's just a question of the direction so R goes from O to the point of action right and then f4 is down so my right hand

says it's towards you guys so I agree it's f4 all right let's do some

examples have some calculations it used to get some practice and they'll do some actual problems and then we'll talk about bicycles and stuff like that

okay so somebody gives you this problem and say you got a wrench right

around something the wrench is 0.8 meters long this angle is 19 degrees and you put a force down here by putting your foot on it of 900

the question is how much torque is there torque is cross F right so we're

just interested in the magnitude of the torque we can say this is R negative F times sine of theta okay so I know what this is it's 0.8 meters I know

what this is it's not a newton's what about this one what's theta so I use 19 degrees no what angle do I need I need the angle between R which is this one and this guy here right this is my R vector an angle between this one

it's going to be so this is 90 and this one's going to be 19 right so this angle is what I want is a hundred and nine degrees so this one is 109 so I

get 680 Newton meters the units of torque are force and distance of course okay how else could we have done this one all right so decompose this

force into this component and this component right and then just take the

component that was actually being helpful all right so when this guy's stepping on the wrench he's doing a lot of putting a lot of force which is not actually very helpful to rotation it's the angle between R and F right this one's R this one's R and this one's F oh I'm sorry I see what

you mean yes it's this angle thank you so this one's R right F is like

this okay I extend this here this is 90 degrees this one is 19 I need this

angle here which is 90 plus 19 this is 109 degrees the angle between R and

what happens if I use this angle right so this is 71 degrees right what if I

done 71 degrees why is that I get the same answer because the answer is the same well the reason is that sign is symmetric right and so it doesn't

really matter and also because this is what you would do to get that component okay anyway you guys understand so that's how you calculate torque now let's relate torque to to motion right and to acceleration

because that was our goal in the beginning remember is to understand the connection between rotational acceleration and rotational force what is that relationship and well if you have some object rotating around axis with our object maybe today you guys burn all your creativity in three-day

weekend elephant you say elephant every single time a zebra okay so we have a zebra girl on a stick you're disgusting who would even say such a

thing talk to your psychiatrist up your medication okay I forgot this video tips you can actually just go back and look at can't really deny it can I all right so here's our zebra

okay so and we have some so let's think about the point of point of

rotation of some point on the zebra okay part of the zebra we're going to calculate the point of rotation about the nose okay so if here's our origin right we have an R vector okay and then we have some force all right

somebody applies a force to this zebra okay so this thing is rotating like this around this axis and there's a force applied to the zebra right some force like this now this force is going to have components that are tangential to the rotation right like this and forces that are radial you

know that providing that move towards the center all right so we can decompose into the radial and and tangential components and then this is the z component so you have the z component you have the radial component

and the tangential component okay and since we're interested in not in the motion of the zebra up and down the spit or its motion relative to the axis we're interested in how fast it's rotating it's this piece here that we're interested in right this tangential force on the zebras nose

okay and we know something about the relationship between force and acceleration right so we can write down what we know and we know something about acceleration linear acceleration right this is the linear acceleration

of the nose and rotational acceleration is related by the radius right so now we have a relationship between the force that's been applied and the angular acceleration of the zebras nose all right and so let's say

so let's combine this and our alpha right and here I just plug that in now what happens if I multiply both sides here by R times R is M R squared alpha

right now what is F times R if if the force is defined to be tangential and therefore perpendicular to the radius and this is just the torque right and what is M R squared it's the moment of inertia right okay so F equals

MA for linear motion just gets translated to tau equals I alpha for rotational motion exactly the same concepts force is mass times acceleration right so last time we talked about identifying inertia moment

of inertia as rotational inertia right because we identified in the kinetic energy formula as he said kinetic energy of rotation is one-half I omega squared which reminded us of one-half MV squared for linear motion so my

argument was let's just group everything between the one-half the omega squared and call that rotational inertia or a moment of inertia and here is validation that this is a sensible thing to do because it also gives us a sensible definition for torque in terms of acceleration all right so the

same quantity that lets us calculate kinetic energy and analogy to translation of motion allows us to relate force rotational force and rotational acceleration in exactly the same way as translational motion okay and that's very handy if you want to calculate things like angular

acceleration given a force so if you have for example a pulley that's attached to a table and somebody says I'm applying a force of 9 newtons to this thing and this is 0.06 meters and the mass of the pulley is 50

kilograms and somebody says okay if I apply this force 9 newtons and tangent to the edge of the pulley what is my angular acceleration so how do you

calculate alpha there well the torque alpha appears in this equation right so and we know what I is for this kind of thing we can just look it up in

memorize but it's mr squared over 2 okay we know are we know em we know 2 so we need alpha we just need to calculate the torque somehow how do we calculate the torque well we're given the force and the radius we know

they're applied perpendicular so if you can just say the force times the radius is I alpha right now we know how to relate force to angular acceleration so alpha is FR over I is FR over mr squared over 2 one of these

are as cancels and we know how to calculate that and if somebody says

well what's the tangential acceleration right we could just remember okay and we can do a similar one similar thing with this kind of

problem which we see a lot you got a pulley I got a rope you got a mass this guy has radius R and mass M this guy's mass little M okay and this

is some height above the water okay the question is what is the acceleration of mass M okay now if this was not attached to the string

what would be the acceleration of mass M right no big deal just due to gravity why is it going to be different here is there another force the torque

provided fully it's a single system I think the rope connects these two so if you're going to accelerate the mass M you also have to spin the pulley think

about in terms of energy you turn these but the mass is the only thing with potential energy right you turn that potential energy into kinetic energy it doesn't just anymore go into kinetic energy this guy also goes in a kinetic energy rotation of this guy right so his velocity is not going to be as great as it would be otherwise because some of the energy is spent

over here okay so let's calculate it what's the acceleration well at first let's just think about this one here okay so free by diagram for the pulley as as you guys said torque here right now I could also say well there's

gravity down but then there's just gonna be a normal force here fixing it notice that no those neither of those two are going to be relevant anyway to its rotation it's not moving anywhere so I don't really have to think about those okay and for the mass my free body diagram just has

tension T and G right so for this one I say well the torque is the radius times the tension the tension is what's providing the rotation it gives me a tangential right this is equal to I alpha right and I can just look up

I and say is M R squared over 2 so this is M R squared alpha over 2 right

so I have RT is equal to M R squared alpha over 2 so I can solve for the tension is an alpha over 2 so now I have a relationship between the tension on the string and the angular acceleration well how am I going to find out what the tension on the string is well I have this guy here right another

equation the sum of the forces here I have mg minus T what is this gonna equal are those two equal to zero no because this thing is in motion right some of the forces is equal to the mass times the acceleration right now I

can plug in this guy here right I know what the tension is nice m alpha over 2 equals MA now what I do what do I do I have a single equation because I've

already used this one up I know little m I know big M I don't know what a is that's what I'm trying to figure out right it asked us what is the acceleration that's M I also don't know what alpha is but two unknowns in one

equation this one here RT oh yeah it's really hard to edge with algebra on the

board here oh I see yes but you're saying T and alpha okay let's take a

step back I is MR square root of a 2 this one's correct all right and then I just equate that to RT this one's correct then I just divide by R so yeah

should have one power of R there yes looks good okay so but we're back to the same problem what do we do with this alpha and this a the rope has a single

value of its acceleration right doesn't stretch doesn't pull doesn't squish so the tangential acceleration here is gonna be equal to the acceleration of the mass right which means that we can say a is equal to our alpha right so

here's how we connect the linear and rotational motion right we say a and alpha are not two free parameters are connected to each other by the radius right right and that's where that art that are goes so then I can plug that

in over here and I get mg minus m over 2 and then our alpha over there

just becomes a is equal to MA and now I have an equation with single unknown one plus m over 2m so let's think about whether this has the right

right what happens when a times m over 2 cross at the ends okay so that looks

right okay so does it make sense tell me why

becomes zero yeah the bottom is gonna be really large and the acceleration is going to be zero and that if the lean mass little n goes to zero then there's no acceleration so if little mass goes to zero this the bottom of the whole of large acceleration goes to zero okay the big mass is large then

it's gonna be a big bottom all right so what happens when the pulley gets infinitely massive this thing gets large so the acceleration goes to zero all right what happened to the pulleys mass lists what's that the

objects of free fall okay okay so there's an example of that's a classical kind of problem we've been doing a lot this is turning rotational and potential energy into rotational into kinetic energy in this case with two components okay let's do some thought experiments so here we have a

glider mass m so it's on a frictionless horizontal track connected to this object and two by massive swing but over this pole okay the glider accelerates to the right the object accelerates downwards and the string rotates the pulley what's the relationship between this tension

and this tension and Mg which is m2g which is the force that the the hanging object would fall with if it were in free fall okay so this is a tricky one

all right so talk to your neighbor vote again all right somebody was persuaded of me nobody thought he was true before but somebody persuaded their neighbor nicely done okay so first let's break this into pieces first let's think

about m2g versus t2 we're going to make an argument from the relationship between those two so that the system is accelerating so m2g has to be greater than t2 so let me ask another way in under what circumstance with could

m2g be equal to t2 right because then the net force on that mass would be zero and there would be no acceleration of that mass right so t2

has to be less than m2g in order for m2 to be accelerating how do we know is accelerating it says somewhere I can't find but the object accelerates downwards okay so t2 definitely has to be less than m2g okay so that leaves

us with BC or E right okay so vote again BC or E the C's have it but not by much so tell me about the relationship between t2 and t1 is it equal

to are they equal to each other who thinks they're equal to each other okay who wants to make arguments but equal to each other is it the same

string I mean you're thinking it's the same string yes it's connected but this is not a this is this is a spinning pulley right this is not a rope sliding over the pulley it's not just a guide for the rope right imagine the

problem was slightly different and here's the pulley and we have two strings one attached here and one attached here with the physics change at all no right it's really even they should think of it as two different strings okay so who wants to argue that they're not the same

some of the force from t2 is going to be used to rotate the pulley what does that mean before she's used does it get disappear after five seconds or

something I see where you're going but formulated precisely think about it this way what happens the limit when the wheel is infinite I like this things made out of a neutron star or something what happens

won't turn right which means there's gonna be no force on on t1 right the only situation in which they're equal is when this thing doesn't rotate right and the rope just slips over it understanding rotation lets us solve

tricky problems by separating complicated motion into two pieces all right remember the asteroid problem we had where we show that the kinetic energy of two asteroids could be written as the kinetic energy of the center of

motion plus the kinetic energy relative to the center of motion you can do the same thing here if you have for example a you know cheerleaders baton right she throws it up it spins and moves in a right it's got this

trajectory plus it's spinning all right you can think of this as you know the baton just moves this is the center of mass motion right it doesn't matter about its rotation the rotation is not going to affect the motion of its

center of mass why is that remember we did this problem where you launch a cannon off the ship doesn't matter if it's a piano doesn't matter if it's spinning right you apply force to the center of mass that determines the motion of the center of mass right there's only one external force which is

grabbing that dictates its motion okay plus the rotation right rotation of this thing relative to the center of mass right so this is the center of mass right this thing rotates relative the center of mass so this is

complex motion if you wanted to track for example one part of this thing say where does this guy go well this trajectory would be pretty complicated all right you know it's like well if I had drawn it to spin more than once

it'd be more complicated yeah but you get the idea so you can write the kinetic energy here as one half m v squared where this is the center of mass plus one half I omega squared so motion of the center of mass plus

motion relative to the center of mass so with that in mind let's think about what happens when something rolls on a flat surface okay so let's take a bicycle wheel for example okay a

bicycle wheel right here's the center of mass of the bicycle wheel right and it moves with some velocity the velocity of the bike and because we're not considering rotation here just considering motion the center of mass

every point on the wheel here has the same velocity right it's all just velocity of the center of mass right now that's its motion this linear

motion of the center of mass and then hopefully your bicycle wheel is also rotating all right unless you slammed on the brakes now here we have some

angular velocity right and every point on the wheel should have the same angular velocity right sorry the wheel has a single angular velocity every point along the rim has the same tangential velocity right every same

linear velocity okay so let's add these guys up right now in order for this bicycle wheel to roll smoothly what you actually need is for this velocity here to balance this philosophy what do I mean by roll

smoothly I mean roll without slipping okay if you're slipping like on our initial problems a box sliding along a surface the two surfaces are moving with respect to each other here I'm talking about moving about rolling without any slipping how do you roll without slipping there's no sliding at

all which means the point of contact has no have no relative velocities and the motion comes from the rolling of the object along the surface rather than it's sliding along the surface that means that this velocity and this velocity have to cancel so you have no velocity here at the point of contact

okay how does that happen well it happens if these two guys are equal right this one right so you just need to choose the angular velocity to be

related to the center of mass math velocity the center of mass by the radius in which case this guy there's no velocity this one here has two VCM right this one has a velocity like this and this one has a velocity like

this where this one is two components VCM VCM and this one is two components okay first time I thought about this is took me a little while to absorb it but a bicycle wheels basically always falling forwards it's not sliding

at all it's infinitesimally falling on to the next little bit of wheel right or you can think about it as rotating around this point but then the point of rotation moves so in order to roll without slipping you need friction

this cannot happen if you don't have friction you have no friction imagine you're standing on a bicycle on a perfectly frictionless surface you spin the wheels you're going nowhere right or you don't have to spin the wheel somebody gives you a push you slide without without spinning the wheels in order for this to happen you have to have friction you can't have this happen in a frictionless universe all right cuz it'd be very difficult to

maintain the rotation at exactly the right speed you could in theory that's supposed to do it you're pedaling it just the right velocity to move the wheels okay so if somebody says to you in a problem rolling without slipping that's what this means it means that you can confidently relate

the motion of the center of mass of the object to the tangential motion to the angular velocity of the object okay rolling without slipping so for example you have a yo-yo okay use your finger if yo-yo here it starts here

with zero velocity BCM equals zero Omega equals zero and ends up down

here after a moment it's dropped I H and it has some velocity of the

rotational velocity Omega the question is what is the speed of the center of mass of the yo-yo when it's dropped a certain distance now if this was just a problem of dropping an object and not spinning if easy you turn

potential energy into kinetic energy you can take the same approach here but remember we have two kinds of kinetic energy now kinetic energy of motion of the center of mass plus the rotation of the yo-yo about the center of mass right so we can say MGH is 1 half and V center of mass and squared plus 1 half

by Omega squared right we would only have this problem initially if it wasn't rotating but now we have this as well and so we were given H and the mass we're asked to find V we can look up what I is but then we

this is where knowing that it doesn't slip on the road is critical because it lets us relate the velocity of the center of mass to omega right now we've done something like this before we related tangential acceleration to to

rotational acceleration right because we have an object on the surface and that's fine here we're relating the motion of the center of mass to the angular velocity this is slightly different right and this only happens when the object is rolling without slipping okay so this is 1 half MB

squared here we put in what I is which is mr squared over 2 and we replace this with velocity of the center of mass over R so we remove this

from the equation because we've connected this velocity in this rotation by knowing that it rolls without slipping down the road and then it's just a little bit of algebra the R's cancel this is R squared this is R

squared this becomes MV squared over yes for right so this is 3 fourths and V

squared see it would have been 1 half MV squared but we add a contribution from its rotation which turns out to be in this case the dominant contribution right the kinetic energy of rotation is twice the kinetic energy of motion of the center of mass I went from I'm sorry I got the other way around this

is a half plus fourth so the motion of the center of mass is twice the motion of the kinetic energy of rotation okay and from here you can easily solve for what V squared is let's do a couple more problems with

rotation it's a very similar problem here we have a lightweight string wrapped around a small hoop okay if the free end of the hoop of the string is held

in place and the hoop is released from rest the string unwinds and the hoop descends how does the tension in the string compare to the weight of the hoop so what is the weight of the hoop just MG right just the force that would be applied okay everybody click so what do you guys think okay but what

about the okay so the center of mass of the hoop is accelerated yeah so what's

the net force on the center mass of the hoop right but the center of mass

of the hoop the hoop is moving downwards the center of mass is moving down so the sum of the forces is down so there's nonzero acceleration down right what are the forces on the hoop

And, all right, so what do you think is the relationship between t and w? Aaron?

But it's attached to this string, does that matter? Why not? You're taking that into account from the center, what do you mean?

It doesn't matter where it's attached to me, what if I attached it to the center? Well, it doesn't matter. Well, you would get a different rotation, right, you get different motion, you're talking about different magnitudes, but the key to this problem is that they're telling you it's moving down.

So they're telling you the net force on the center of mass because they tell you it starts from rest and it moves down, right? So you don't have to worry about any of the other complicated stuff with rotation. You just need to know that there are two forces on the center of mass. Do you have a question or comment? This one is a lot less tricky than it looks.

It's a trick trick question. It looks like a trick question, okay? Alright, how about these guys? Here you have a bunch of things rolling down. Okay, somebody's having a race. Okay, everything has the same mass, but of course different moments of inertia.

And I can't remember them all so I just wrote them down for you. Okay, there's a sphere, a loop, a cylinder of radius r, a hollow sphere, so the fourth one is a hollow sphere. And then there's a cylinder with the same mass but a smaller radius. So you cylinder with radius over two. The question is, in what order do they arrive at the bottom?

Who wins the race and who loses? Okay, how do we figure this one out? Yes?

I ended up finding the reciprocal. We ended up getting 3, 4, 5 times the reciprocal. Reciprocal or what? Okay, so you're saying a larger high means a smaller velocity at the center of mass. So explain that to me physically.

Why does that make sense? Why does a larger moment of inertia mean a smaller velocity at the center of mass? Anybody else? In the very back. It's harder to turn it, right? You have a fixed amount of potential energy because you start at a certain height. The potential energy is going to get moved into rotational energy and translational energy.

The larger the moment of inertia, the larger the fraction is going to go through rotational energy. The smaller the moment of inertia, the less goes through rotational energy. What if you have something up there with zero rotational inertia?

Just like an infinitely thin cylinder. All of the energy would go into its motion, right? That's the fastest limit. So what you want is things with The thing that arrives first has the smallest moment of inertia. The thing that arrives last has the largest moment of inertia. So just looking at it physically, which one is going to have the smallest moment of inertia?

We all have the same mass, so it's a question of the distance of that mass from the axis of rotation, right? So smallest moment of inertia comes from smallest r. So the thin cylinder is going to win because it has more of the mass concentrated close to the axis of rotation, right?

After that is the sphere, right? It's got a lot of stuff very close to the center. After that is the fat cylinder. The sphere and the fat cylinder is not easy to compare intuitively. In fact, they're pretty close. One is one half, one is two fifths. So it's not a big difference. After that is the hollow sphere, which has all the mass at a certain radius, right?

And then the loop. And the loop is, of course, the largest because it has all the mass at a radius r, right? In the hollow sphere, some of the mass is at the axis of rotation, right? Some of it is actually touching the axis for the loop. None of it is. It's the maximal moment of inertia.

Or if you want to think about it mathematically, right, the kinetic energy has these two contributions and you can write the moment of inertia in each of those objects as some coefficient times m r squared and then plug that in.

Then, as we were learning, the velocity is proportional to one over this coefficient, right? So in this case, a larger c means a larger moment of inertia, so a smaller final velocity. Alright, and then if you look at the coefficients for each of these objects, then they arrive in this order.

Questions? Alright. So that's all we have for today. There's a few more example problems.