 You've got to accuse me that I was off at him. You know, he said he's been off me, and I ended up being a rax, more veteran than I believed in, finally. I was like, I have no demonstrations of that kind at all. You were launching your sessions? No, no, no. See, you guys, when you go elsewhere in your lives, everything you do in physics doesn't occupy every second of your thoughts. Sometimes you just do stuff. You're not thinking, oh, this should be a collision and all that, so much fun calculating. Just do something once in a while. Let's make it a little more. All right, if you remember, on Monday we had just gotten to the point. We were working on this business that for every linear piece we had, how do you like today's new job? Is that all right in the back of the room? Want to get some sunglasses? Do you have a green? What? Do you have a green? Nope. It's not on me. I do in my collection. I'm working through the colors that for every linear variable we had, we come up with an exact complement of it for the rotational motion, so much so that all we had to do is swap the variable, use the very same equations, and keep going. The ideas weren't terribly different, a little bit different, but not terribly different. Every single thing we did had a rotational equivalent from position to velocity, acceleration, and of course time, and I'm not going to bother with there's no such thing as linear or rotational time, but then the constant acceleration equations took exactly the same form. Everything was spot on. Was there anything else we had in there? Something else we'd come across so far. No, we didn't actually get to kinetic energy, we were just on the verge of that right now. All the units had a parallel zone, but I guess that was it, I guess that's all we've managed so far, but then we were working on kinetic energy. So we had this idea that if something rotates, as it rotates, every single little piece of it rotates, every single little piece, and we took a look, we picked a little piece of mass right there, and depending on how far away it is from the center and whatever the angular speed was of this thing as it rotates, we can figure out the velocity of that little piece of mass, M1. However big or however small it might be, we can say that. We've looked at a couple other pieces, we have to look at them along the same line, but they don't need to be, they could be anywhere. We could take a piece over here, and we could figure out based upon how far, oops, R1, based upon how far it was away, R2 from the center, we could then figure out its velocity at that instant. And so on, we could do so many little pieces of mass of this thing that we end up describing the entire thing. And we can say however small we want those little pieces to be, we could let them be a centimeter on the side if we want, or we could say, ah, that's too big even, let's go down to an atom, let's call each little piece of mass an atom, and so do all that math, atoms that make up this entire thing, every single little piece of them. For extra credit, I'd like someone to do that, atom by atom by atom. Write it up and turn it in. We could do that for all these little pieces, anywhere they exist. We could figure out how fast each piece of mass is going. Every single piece there has some kinetic energy, and we could add up all of those pieces to get the entire kinetic energy of the, of the, whatever the object is. We'll add up the first piece all the way up to the n-th piece. However big we want n-th, maybe a hundred would satisfy us, a hundred pieces. Divide the thing that's rotating up into a hundred pieces, figure out the kinetic energy of the n-th piece, or maybe we want n to be ten thousand, or ten million, or what's a bigger number than ten million? Ten million. That's too big. If it was infinity, we'd be here all week. Eleven million. Eleven million is bigger than ten million. So, however small we want these pieces to be, the smaller they are, the bigger n is, but the point is, characterized, accounted for, and it's the tendon kinetic energy with every single little piece. Even the fact that the pieces in close aren't moving very fast. They still have some kinetic energy, not a lot maybe, but they still have some. The pieces farther out are moving faster. They have more kinetic energy, but we total it all up for every single little piece. I won't put these indices back in, but they're all there. And we then know the entire kinetic energy of that object, even though the object itself isn't going anywhere, all the little pieces that make it up are going somewhere. Every little piece has kinetic energy, even though the object at a hole is not going anywhere with its own velocity. The center of mass of the object isn't going anywhere because it's still basically right there. So we were just working on that business a little bit. The summation thing, any common factors come out, so of course that's the one-half comes out, and then we've got this v the velocity of each piece in there, but we'd also realize what the velocity of each piece depends upon how far that piece is away from the center and how fast the object is rotating. So we replaced v with r omega and we had then r i omega squared. That's all and this is sort of a quick summary of the last 10 or 15 minutes of the class on Monday. And omega of course is the same for every piece on there. They're all part of the same object. That object is rotating at some angular speed omega. It's got to be the same for all the pieces. Otherwise some parts of it would rotate at some speed and some parts would rotate at another speed so the omega can come out and we're left with the sum of m i r i squared. Oh yeah, omega squared down. Well it isn't. Now it is. Now, we're just getting to the point where we can take a look at that thing there. That sum of m i r i squared. Let's do what you did. The kind of thing. Let's do the kind of thing you did in calculus which is why I want you to have taken it before you take this class. So you're familiar with this kind of thing that we're going to do. Let's let m i go smaller and smaller until we can get the absolute smallest possible we can. This is the same thing we did with delta t when we were working on instantaneous acceleration. We just let delta t go so small we let it go to zero and use that as an accurate picture of everything that was going on because the size of delta t was so small it didn't matter. We'll do the same thing with m i. We'll let m i for the most part go to to zero. Maybe, maybe well what happened with that when we did that with the time is what we were really talking about was not delta t but then it became d t as we got the differential. So this then becomes this piece here then becomes r i well we don't even need the i in a second because of what's going on r r squared dm where the mass is so small it's now a differential element remember that from calculus that's our little time that's essentially zero and remember when you did this calculus you didn't sum up all those pieces you actually integrated them all. So that sum then becomes the integration of over the entire object of each little tiny piece and how far it is away from the center the square how far it is away from the center and we can do that for any object we've got whether it's a nice circle like this we could do that if we were letting something like a big cylinder rotate maybe about a point through its axis and we could do that integral for all of those pieces as well we could do it for any kind of shape actually if we wanted to I know how you love integration so I'm sure you would like to do all those integrals every time we have an object rotating in a problem you integrate this so that we could figure out the kinetic energy of it as it rotates Phil you love to integrate I bet you'd probably do it at breakfast for the rest of us who'd rather not integrate let's let somebody else do these integrations of discs and cylinders and spheres and axles and all those regular type of things we see that rotate let somebody else do those integrations and then just put them in a table for us and we'll just look them up while Phil's off integrating at breakfast we'll just look up this integration because why should we do it when somebody else is already done and so the table of these integrations is in your book and in fact all these integrations have a nice name this is known as capital I the moment of inertia term moment is more a mathematical term and inertia remember has to do with the mass of an object so for very regular ordinary everyday solids this is in your book and our book has it on table 12-3 page 382 take it out and take a look better take a look alright look something like that there's some nice regular everyday solids page 383 sorry page 382 page 382 there's some nice everyday solids that look very much I guess I'll put it up here on the projector for everybody to see since people like Patrick evidently bring his book I'm not done with it yet didn't go anywhere it's right there it's not gone anywhere all I just see I put a square thing under something means I'm working here I'll be back soon in a minute you don't ever want to ask me a question while I'm waiting for the projector to warm up because I have time to answer you ask me questions when I don't have time I can care less oops alright for those of you that don't have your book Patrick there's the deal here over here on the right that column is just simply the result of all these integrations look some of them actually come out as simple as MR squared some others like MR squared but to get some constant in front just the result of the integration Phil will give us a little report after tomorrow's breakfast on that what you have to be very careful with is notice that some of the objects or for particular objects like these first two it's the very same thing it's a thin hoop in both cases but the way it's rotating is very very different in the first thing it's rotating kind of like a wheel is in the second one it's rotating kind of like a wheelism you know it's rotating like everybody's done that thing where you've taken a coin and flipped it on the table top so it starts spinning twirling on its edge that's kind of the picture that's here it's rotating about an axis that goes up across a diameter notice the moment of inertia for those two things is very very different that's because some of the mass is at a different radius in one than it is from the other it's that distribution of the mass with relationship to how the objects rotating where it's rotating around that determines its moment of inertia where that mass is where it rotates changes everything a disc or a cylinder it doesn't even matter how long it is if the rotation axis goes down the center of the cylinder if it's wobbling about a perpendicular rotation to its own axis then the length of it matters that kind of makes sense I guess if you've ever grabbed anything simple like a broomstick and twirled it you can tell that it takes some effort to do that with it and in fact that's even the type of picture we have down here if you grab the boom stick in the center and twirl it's not too big a deal if you grab it at one end it's going to take a lot more effort for you to do so because it's moment of inertia as you'll notice from the calculation is four times bigger when you grab the very same object from one end than if you do it through the center because there's more mass a lot farther away at a bigger R than there is from this from the center then the last two sphere about a center and then a thin spherical shell that'd be like a very thin like an egg shell but they're not they're kind of like eggs but notice the moment of inertia for those two situations is quite different even if the mass and the radius were the same because it's a very different distribution of mass with respect to that center of rotation so we get these different values for the moment of inertia and now if I put this all back together sorry to make you wait for it but here it is I get the kinetic energy of an object rotating even though the object itself is really not going anywhere it's still right where it is which we normally would have taken to be a v of zero when we're looking at particle motion now just the fact that it's rotating gives it some kinetic energy rotational kinetic energy notice how similar this looks to a linear kinetic energy there's linear k there's the rotational k look how similar they look you swap out v put in omega just like we did for all our other equations take out v put in omega but now we can do the same type of thing with mass when we had a discussion on the linear kinetic energy of mass we now have a rotational mass and we can now figure out the kinetic energy using our table now to make the swaps that we made before so something doesn't actually have to be going anywhere as long as it's doing something it'll have some kinetic energy this is the very idea behind a flywheel if you've ever heard of such a thing it's the idea if you can get this big heavy mass spinning up to some speed you can use that to store energy for you until you need it there have been car designs that have tried to use that where part of the braking of the car to bring it to a stop is done by using the loss of kinetic energy of the car as it comes to a stop to increase the kinetic energy of the flywheel so that when you get up to the stop sign the car is not going anywhere the car sitting there kinetic energy but there's a whole bunch of rotational kinetic energy all stored up in the flywheel then when the light turns green they engage the wheels of the car to that spinning flywheel and use that spinning flywheel now to bring the car back up to speed as the spinning flywheel slows down very efficient use of kinetic energy you're transferring the linear kinetic energy to rotational and back again over and over and over through stops and starts the trouble is those flywheels are very very heavy so not only do they have to be accelerated even if they're spinning down as they speed the car up that mass of the flywheel still needs to be accelerated so you've got this very heavy car the engine and the wheel and your mother-in-law the seat there and a brother-in-law in the back that's a lot of mass to move so it's not the most practical way to use kinetic energy that you lose coming to a stop to speed up the car more practical way to do it is to use that loss and kinetic energy the generator charge the battery then use the energy saved in the battery to make the car go when the light is turned green and that's what typical hybrid cars like the Prius and the Honda Civic and the others all do now for their to use this loss and linear kinetic energy coming to a stop to then increase the kinetic energy of the car but we could do it with flywheels it's just not all terribly practical so let's do a little a couple little things with this kinetic energy I mean with this moment of inertia in general it's that integral but in specific we don't want to do that that's already done here for a whole bunch of regular solids also if there are a couple objects that are rotating we can figure out the moment of inertia of each one of them individually and then just add them up and that'll give us the total moment of inertia of an object for example imagine we have something sort of like a space station or some kind of satellite might be maybe we can model it very simply as four rather heavy pieces maybe each one is a different experiment or module on a space station or one's a telescope optical telescope one's a radio telescope trash can and the other is an mp3 player all the things you need on the space station and imagine they're hooked together by very light structural members just to hold them in the place they need to be, remember there's no gravity in that space so you don't have to actually hold anything up you just have to keep it from floating away so let's imagine each one of these modules has a mass and is separated by a distance of 2R so we have these very heavy masses connected by very very light structural members and imagine we decide that we better see what it's like if we expect it to rotate on that axis sometimes are given a little bit of a spin to help stabilize them so we want to figure out the moment of inertia of this entire space station assembly this very simple model of one we need to figure out how much kinetic energy do we need to give it to spin at a certain rate where we can get that energy those kind of things are our calculations we might make alright so first thing we do at our table is there in there the moment of inertia of a 4 module space station there it is oh you have the special edition because they're special no there's not however we we can say we've got these masses at a distance from the center let's just use the original form of this we have and just sum up for each of the pieces the mass of each piece and how far away it is because we don't have these things are spheres but the spheres in the table is for a rotation about an axis that goes right through their center that's not the case we have here these are rotating about an axis off of their center but what we'll do is say well these are small enough we'll just take them to the point masses at some distance away from the axis and we'll leave it at that that will give us a good first estimate of the moment of inertia of this space station model let's see this one has a moment of inertia of m times its distance from the axis rotation which will take to be r so the first one has a moment of inertia of mr squared it's a pretty small mass it's small enough that it's far away we don't really care about its size compared to everything else if we needed to if this was a big sphere we could figure out how much the fact that it's all rotating at a different distance but this is a small enough mass will take it all to be at some distance r squared so it will be pretty good what's the moment of inertia of this little piece so that we can sum them all together it's also mr squared in this space in this space so it's just 4 mr squared simple as that you'll have a couple homework problems where you've got some relatively small object of some mass rotating at some distance away from a center of axis and just doing the sum of mr squared will be fine because that particular thing doesn't have to exist in the table so you have to do something else now maybe you in a fit of brilliance comes by and says let's take that same space station same scientific modules attached by some very slender structural members very very light they're so light notice we didn't even have to take them into the calculation they're that light oh sorry John didn't know what they were doing for 10 minutes about to do anybody else but you have the the brilliant idea of doing an axis and rotation over there on the edges let's do that see if maybe that's not going to be a better way to do it I goes down with this model we'll save on some kinetic energy getting it to spin at some certain speed if I goes up it's going to cost us more kinetic energy and maybe it's an idea we don't want to follow the moment of inertia for this model what's the moment of inertia for this first piece it's m but it's r is 0 and same for this piece so these two pieces don't count into it we're assuming that the radius of that piece is much less than the distance r of these structural members so it just doesn't really contribute so what's the moment of inertia contribution of this one right there this one out here at a distance 2r it's m times its distance squared but its distance squared is 2r so it's 4mr squared what about this one the same so there's 2 4mr squared to account for that means its moment of inertia for that structure is 8mr squared so it's going to take more energy to get this rotating to a certain angular speed in fact it's going to take twice as much as it would get this one rotating up to the same speed and that energy in outer space is going to have to come from somewhere probably from a tanker truck a tanker shuttle it's going to lean out pumping it up getting a little propeller going no, propellers don't work in outer space billion is that right oh yes so do you have a thought about this is nice and even just by the center of that well you can just calculate each piece if somebody else comes along and says let's do this space station is wherever the axis of rotation is let's say it's there each piece is a certain distance away from that axis of rotation this one happens to lay right on and so you just add all those up it's the mr squared for each one of those little pieces that's when the size of this is small compared to that distance there you want it as a little point mass and get a good first cut now that's a quick estimation but for the actual space station they need to know the moment of inertia about every little piece in pretty and they need to be pretty accurate about it they don't want to lose control of the international space station someday because they don't quite know what the moment of inertia is they have to know it in three-dimensional rotation because at any time they might need to change its position in any possible direction the pat brother-in-law comes out gets on one end that end tilts down then their pool table doesn't work right because the pool table is tilted then they're all angry they slam their beers down half of them are all up there all the way all the people man what? you don't think NASA could move the pool table to work in space you bet they could that's NASA man did you have any in the balance? yeah I was wondering if it was very equal in space actually at any time the size of those beers was more significant could you sort of supplement no no there's another technique we have to use where if we have an object and we know it's moment of inertia about an axis through its center but we want to figure out the moment of inertia about an axis parallel to that we use what's called a parallel axis theorem but we don't need that here well I don't know if it's even given in the book I know none of the homework problems if the book doesn't have it if we do a class called dynamics that I teach in the spring sometime we will figure out a more accurate presentation of this if this size isn't negligible compared to the distance away from the axis which we want now alright let's see oh yeah we got a little time we'll do a quick problem with it where we can take this all into account imagine we have into the wall a big pulley not a big pulley a big pulley we have a big pulley attached to it a mass so you know that when you let go of that mass it'll start to accelerate as it falls is this free fall right there gravity pulls it down is that free fall neglect air resistance it's not going to be moving very fast depending on what the mass is it might take a while to fall to the bottom but is that a free fall problem what Joe? Joe says yes who thinks Joe is full of Louis on this point not in general is everybody free fall problem yes Len says yes Mike says yes we were talking about free fall problems yesterday I redefined it for you what did I say the only force in the problem is gravity what about the rope isn't there going to be some tension in that rope if you cut that rope are you saying nothing would change the rope's wound around there it's not sliding around it it's just wound around it so there's no sliding going on there's static friction to keep it from unwinding I guess it can't be a free fall problem there's got to be some tension in that rope unless this thing has no mass then the rope would have no tension but other than that there's no tension so it's not a free fall problem so let us figure it out let's see we'll give this cylinder what we'll say this is a cylinder pulleys pulleys are essentially very very flat cylinders maybe sometimes there's a little groove in them but we won't worry about that just take it to be a cylinder 8 kilograms radius 12 centimeters through its center like that we'll take this to be 117.7 newtons and it's going to fall a distance of 4.2 meters what I'd like us to find is just the split second the split split split second before the mass hits the floor what's the angular velocity of the pulleys we've got an object that's dropping which means it's losing gravitational potential energy where's that gravitational potential energy going? into the kinetic energy of what? Alan said the mass you said the pulleys both because before now both are moving and so both will have kinetic energy the mass will actually have translational kinetic energy 1.5 mv squared no sweat 1.5 will look up in the book m well we can get m from the weight v squared well that has to do with how fast it's falling that has to do with how fast that's spinning it's going to actually work quite well for us there's no spring in the system so I won't even put that in what work is being done in this problem why not what is there to do any work? well gravity is there to do some work but we take care of gravity over here don't account for it twice only account for it once there is no work going on in this problem there's no change in kinetic energy yeah the mass had none now it's got some and the pulleys had none now it's got some so there's definitely a change in kinetic energy is there any change in potential energy of what? the mass only the pulleys is still where it was when it started it didn't fall down any but the mass most certainly did we'll assume that the rope itself doesn't have any significant mass I got it over an ace they sell massless, scratchless ropes over there, they're perfect for it delta K of what again, what has some kinetic energy change? both things do so let's figure out the change in kinetic energy for the pulleys and the change in kinetic energy for the mass and just add them together that's what's kind of cool about this equation, it's just an accounting problem whatever's in the problem, count them all up and add them all together as long as you don't miss anything it works out great change in kinetic energy of the pulleys what's that going to look like that's of course K e2 minus K e1 for the pulleys and it's what? kinetic energy of the pulleys one half i omega squared remember that i is the MR squared integration already done for us for this pulleys which will take to be a very flat cylinder so delta Kp then is what? it's delta Kp wouldn't it be one half i would factor out of both terms because there's going to be Kp2 minus Kp1 half i is in the set both of them and it's common to both it doesn't change it comes out times V 2 squared sorry, omega 2 squared minus omega 1 squared does that look right? for delta the change in kinetic energy of the pulleys fill that doesn't look right what looks better, what looks wrong let's write it out more completely so that's Kp2 minus Kp1 right? the change in kinetic energy of the pulleys which is Kp2 is one half i omega 2 squared right? one half i omega 1 squared one half and i are constant common to both terms factors out leaving behind omega 2 squared minus omega 1 squared is that the same as omega 2 minus omega 1 quantity squared we talked about it yesterday didn't we? in this case it is because one of them is 0 but otherwise no, if i were you in the way i looked at the wheel of the red pen i wouldn't make that change yet omega 1 is 0 because it starts from rest the whole system starts from rest so delta Kp is one half i omega 2 squared omega 2 we are looking for what's the kinetic energy change of the mass itself it starts from rest some amount of time later it just is getting to the bottom with some velocity v2 so what's its change in kinetic energy same kind of thing Km2 minus Km1 what's the initial kinetic energy of the mass 0 it's not moving yet it starts from rest so it's only the kinetic energy it ends up with 1 half mv squared 1 half mv v2 squared because in this case both of the initial values drop down doesn't always happen but in this case it did so that's the delta K term what's the delta UG term the mass the pulley doesn't go anywhere it's mounted to the wall look at that that's high tensile strength concrete with volts put through it that's not going anywhere so it's mg delta H what's mg that's mg right there it fits as newtons that must be its weight so there's mg right there mg 9.8 meters per second squared that will give us a would that be the right units no I dealt it I was putting in mg again I'm so excited I'm so used to doing mg I didn't know I already had it so what do I want in there that's different from what they said wow how does it feel to say that Alan all right minus cause it's dropping that you can figure out directly especially now that I wrote it down correctly where is it who's got it it's minus 4 minus minus 494 newton meters so now our work energy equation is 0 equals one half comes out of both terms i omega squared plus m v square o2 minus 494 is that right is that the work energy equation now okay couple things we still need we need i what is i how do we find it so it's either the third picture or the fourth picture which is it which one of those pictures is spinning like this one spinning it's the third picture in fact that looks like a pulley if you turn it 90 degrees that's exactly the picture we got there in our pulley it looks just like it and so its moment of inertia is one half m r squared one half kilograms and r we got everything else in meters so we're going to need that in meters so what's that come out to be what's that come out to be 0.06 0.06 do you agree that those digits 0.0 somebody either confirm it or deny it how hard is one half 8 times 0.12 squared what class should be 0.06 finally confirm after 20 or 30 minutes 0.06 what are the units that's the units for moment of inertia remember we're not talking about just mass moving it's also important where that mass is and that looks awful like m r squared that we got moment of inertia from this place so there's the moment of inertia of the object what's this m it's the mass of this falling weight when you have it there 12 kilograms alright so now our work energy equation one half i i is 0 6 per second squared times omega 2 squared which we're looking for we're looking for that plus m v2 squared m is 12 kilograms v2 squared we don't know how fast the pull is falling either the mass is minus 494 all of this Newton meters that's just plugging in the values we now have i and how many unknowns in there two unknowns we're looking for omega 2 we don't have v2 we don't know how fast the object hits the deck either we need a second equation two unknowns we need two equations there's one equation we need two equations what's our second equation we don't know the acceleration we would if it was a free fall problem but it's not we don't know the acceleration conservation of momentum well that's not going to work how much momentum does the system start with 0, just sit there ends up with some momentum that's because it picked up energy which increased the momentum too what's our third equation I mean sorry our second equation we've got one equation two unknowns it doesn't have to be but that would sure help us we had something that related those two velocities is there anything that relates the speed of this pulley to the speed of the mass wouldn't it be true that the speed of the mass is directly related to the angular speed of the pulley by the radius of the pulley what's the equation that relates those two V equals R omega that's only true if the rope isn't slipping then that velocity relationship won't hold will it because the rope would be slipping off at some other speed we wouldn't know about so there's our second equation V2 equals R omega2 we're using capital R for the radius of the pulley there's our second equation now we can put them together we're looking for let's take out V2 put in R omega then we can just solve for omega2 which is just an algebra problem now so I won't make it do it 65.5 gradients per second with that Joey, you're not comfortable your thumb hurts maybe I will but this was attached to a site of an elevator or something we'd have to take that into account or if the rope wasn't a rope but a very heavy chain we'd have to take that into account because we've got to get it turning around it's all mass that needs to be accelerated that's what's happening here where it's mass that needs to be accelerated as well as that mass needs to be accelerated and we have to take that into account too alright any questions with this or take it down and do the next thing Mike so about the velocity and the omega move at the same time like there's not a change there's going to be you can always write the velocity to omega if if there's no slipping going on which we're assuming there's not assuming that that rope is nice and snug on there okay next thing I need to look at let's see in fact it's related to this problem in a way and in fact we're going to this will be next week's lab our last lab for the tournament very much like that alright remember it was sort of at this point when we were looking at particle motion when we made our shift from kinematics to kinetics what did we start talking about when we started talking about kinetics we had and still have this business of every linear equivalent has a rotational equivalent we even found that out to be for mass not this idea of rotational mass not just how much mass there is but where it is we even had kinetic energy we get just by swapping the variables we had about but when we were done talking about these things started talking about kinetics that's when mass finally came into the picture but what else came into the picture sorry John what'd you say yeah force we started talking about force we started talking about how do we get an acceleration that we need or how do we figure out what the acceleration is and we brought in this idea that f equals ma so far everything we've done in a linear fashion had some rotational equivalent you might expect that's the same thing here let's see if we can figure out what it is we already know m would be i and a would be alpha but we don't know what to do with force yet so let's see if we can figure out what the deal is with force and rotational motion alright let's see we've got some object here pinned so that it can rotate freely at the center and we're going to apply a force to it and see what happens we're going to apply a force alright like that this thing has some mass and it has some radius what's going to happen what then will be the acceleration of that object nothing remember it's pinned at the center it can rotate it can spin but it can't go anywhere it can't translate in a linear fashion so it's obvious that whatever we had a force here causing accelerations we can't just put a force there and say that causes acceleration and we're all set so it's got to be some other some other idea to it so what if I took the same object, same mass same radius even the same force except I moved the force a little bit what if I take the force and instead apply it over there will that make that thing start to spin will that give that thing an alpha without even knowing anything about this kind of business you just show up here, you know, just fell off the turn up truck you can figure that that alpha is going to be in that direction because that's what that force is going to do to spinning things alright let's see a step farther this one might not be as obvious same object same mass same force only this time I apply it a little bit differently to some spot maybe right there is that going to cause that thing to accelerate at the same rate it's going to have some kind of angular acceleration in that direction, it's still going to do that but suspicion is it wouldn't be as big that's the general feeling Alan you feel like that that's true a couple of the others feel like that's true and that's indeed the fact if we're able to do this experiment in some way we find exactly this so it appears that the distance the force is from the point of rotation has an effect if it's zero we get no acceleration if it's big we get a lot of acceleration if it's somewhere in between we get somewhere in between acceleration and that's exactly the deal we call this kind of name a force applied at some distance from a center rotation we call this torque it's the French word for I want to write that some symbol for it every other time except for mass when we used I but every other time we came up with a Greek letter so we're going to have a Greek letter here it's called tau it sort of looks like a one-legged pi give or take a little bit remember when you write reports we don't write out TAU because we don't need to remember the word processors torque is equal to the force being applied times the perpendicular distance that force is away from the center which for this picture was r for this picture was zero for that picture maybe that's r over 2 something like that so I'll just call it d that's the perpendicular distance the force is from the center of rotation this is known as the moment arm and so we now have our rotational equivalent of f equals m a there could be more than one torque because we could have one force at one distance and another force at some other distance so we have to take into account all of the torques there might be some torques trying to spin it counterclockwise like friction might be doing and we've got to add all those up to see what the effect is going to be so we're going to have to sum all the torques that will cause an object with some moment of inertia to accelerate and there's our rotational equivalent of f equals m a it actually is a vector because this could happen in three dimensional space a force could be off in any direction in any direction itself and the object could be oriented in any way so it really is a three dimensional thing we'll take it to be true dimensional the objects we talk about can only spin in the plane of the board or in the plane of your paper the force is only going to lie in that plane as well the radius is only going to lie in that plane so we'll take it just to be a simple two dimensional thing in that the acceleration is either clockwise or counterclockwise that's all it can happen so we've got this idea of force at some distance so let's put it to work real quick in a quick example imagine we're lifting some girder or something maybe you're going to put in a new deck so I've got this you're coming down to help me for some extra credit so we've got this 10 meter girder I like a big deck 30 foot deck so we're lifting that by means of some crane here just pulling up on a wire attached to one end let's put this thing at 38 degrees awesome something tension in the line is 20,000 newtons my question is how much torque is being applied to that girder because if it's being lifted like that it's going to tend to rotate around that end so we have an object that's in some kind of rotational tight motion so let's figure out the torque now before we get going too far let's think about what we got oh I shouldn't let me put one of those back real quick we had this kind of thing it's a perpendicular distance happens to be the radius in this same example so I'll put them both down because that would be one of the same here that was our original picture for torque we have something a little bit different here because this force is not perpendicular to the line that connects it to where the point of rotation which would be right back here somewhere you tell me what are the units on force newtons what are the units on distance meters newtons meters however this is not joules because remember joules was from work where the force and the distance were in the same direction as each other here the force and the distance are perpendicular to each other that's very different so this is not a joule here as we use it only a fool will call that a joule that's the only physics wrap there is it's just that long to it it's only so we have a slightly different situation here because this force is not perpendicular to this distance back to the point of rotation it's not a big deal for us because we can take that force break it into perpendicular components so there's one part of the force that goes that way that is perpendicular to the line that goes back to the point of rotation and there's one part of the force that's right there let's call this one the force parallel the force perpendicular because that's what we need we need force perpendicular to the line going back to the point of rotation how much torque is this part contributed how much torque is this piece contributing it's perpendicular to the line going back to the point of rotation we know the length of that line find that component the torque contributed by it would be F5 L is the length of the ladder itself you tell me how much torque is contributed by this part of the force because this one force remember is no different than these two forces doing the same thing at the same time how much torque does that force contribute why none that force points right through the point of rotation so it's distance perpendicular distance from the point of rotation is zero any force that goes right through the center doesn't do any torque that was the very first picture we had up there anyway so no torque from that part only a torque from the perpendicular part and you can figure out what that component is because we know from the trigonometry what the deal is here turns out that angle there is also 38 degrees you can you can pull that together if you think about the geometry a little bit perpendicular component is that big and we've got those numbers and the distance back to the center rotation is the 10 meters I guess we're calling it oh call it L here D we had over there whatever the L just makes more sense for the length of a girder D doesn't make any sense for the length then dang yeah that's something worth writing down to do how much torque is being applied to can we also confirm that 158 158 kilonewton meters time to go but what we can do real quick on Monday is we'll take this problem again we can dog give you some mass of this we'll figure out how much angular acceleration of the girder that will cause we're pulling up on it it's going to rotate about that point so we'll figure that out on Monday we'll need a mass of that girder and then it's angular I'm sorry it's moment of inertia