 And it is a spot where it causes some problem. So we've looked at the kinetics of pure translation of rigid bodies where now, even though the object is not rotating, it still has the effect of some moments that can cause changes in some of the forces. We looked at that in the like the airliner problem we looked at where because the thrust of the engines was not directed directly through the center of gravity, it caused a change in the reactions on the landing gear of the aircraft even though there was no rotation of that aircraft during the taxi. Then we also looked at just pure rotation where objects were not subject to any kind of translation at all, just rotation. Now we're going to put the two together and do general plane motion where we have the possibility of both translation and rotation going on. Of course the easiest example is a tire rolling on the ground. Either with no slippage or maybe with some partial slippage, it could still happen. But the point is that the wheel, the center will translate to some other positions and the angular position of the wheel will have changed as it rolls. The compound nature of both of these usually too are looking at acceleration. So there'd be acceleration of the center of gravity but also some kind of angular acceleration. And so we'll look at a couple of things with that today. The basic idea being if we've got some object subject to several forces and all kinds of things are possible into several moments of some kind in some direction and the compound effect of those is going to be some angular velocity perhaps some angular acceleration and some translation of the center of gravity wherever that might happen to me. So if we then, oops, can't have the thing change from yellow to pink the compound effect of all of those is such that the forces will cause some bulk acceleration of the body. In effect if we put it in true kinetic terms we might have m times a. So essentially the net force there might be some net acceleration as well. So we have the three equations of motion. Subscripts. Do the same thing in the y direction. Those components are the pieces of this resultant vector in whatever coordinate directions we need some of the moments also to be then of course resulting in some acceleration. Now to help Alex with the one question he had if we do this about some other point besides the center of gravity maybe about some point p that's no trouble if we know the moment of inertia with respect to p which we may or may not it depends on what the object is we're talking about what other information we might have about it but if we apply the parallel axis theorem to that anyway it then sort of backs it up to the angular acceleration with respect to the center of gravity plus this turn m a g d s the acceleration I had up there and that d is reduced from the line of action of the net force that m a g to that point p noticing too that we need to take into account the directions of these things that this would have an angular this would have this part here would have an angular direction at least as I pictured it in a clockwise direction we have to take that into account with any other moments that are in the problem so there's not anything terribly new there other than what we've got what we've had already however if you're short of equations then you can always add into it any kinematics that are available in the problem such as the no slip condition that's our most common one of the kinematics equations that we can add in there but any of those other ones the relative velocity relative acceleration any of those type of equations are fair game if necessary to solve the problem necessary and useful to solve the problem so let's do a couple problems there's nothing really new up there just we're getting a chance now to put it together so that we might have both angular acceleration and linear acceleration in a problem now okay so we'll look at a type of problem then I'll give you a special in class exercise to see who's been paying attention if anybody so imagine a spool with cord wrapped securely around the outside about a separate cord wrapped around the inside five meters inside only two meters, mass of eight kilograms radius and gyration of 0.35 meters so the spool is made up of a couple of different masses all on the same axis but in general we don't have that type of thing in the book so the radius and gyration is an indication of just how much of the mass is in the interior spool part and how much is in the flanges on the outside okay we want to find alpha we can do it in two ways to help illustrate that very problem that Alex was bringing up and it doesn't matter if we have a general motion problem or any of the other type problems the dealing with it is the same okay free body diagram diagram which is one that essentially shows the net result of what we have going here obviously the two cords exert a force however in that one attached to the ceiling we don't know how big that is that has to be one of our unknowns we do know how big F is though of course it's also got a weight those the net effect of all of those together is going to be some acceleration of the center of gravity so that would be essentially the net result of the force and then we've got to figure if it's doing that because that's pulling that way there'll be angular acceleration in that direction and that's what we're supposed to find how many unknowns? unknown what was the other one? let's assume if we're given mass let's not take weight as unknown because it's just so straightforward if we're given radius of gyration let's not take moment of inertia as unknown because again it's just a very straightforward calculation so we don't know this tension and we don't know the angular acceleration doesn't mean we need to find all three but to find any one of them at least we're going to need three equations what are the three equations? center of gravity is there and that will equal well be specific IGL there's two equations what's the taking the moment of inertia is one of the unknowns so take that as given that's just a straight calculation don't take that as an unknown take that as an equation so what's the third equation? well that's not going to help there are no forces in the x direction so all we're going to have is zero equal to zero so the kinematics equation is available so we need a kinematics equation for useful purposes called that A so what is the equation that we use there? it's got to be an equation that has at least one of these in it I don't want to introduce any more unknowns but that's all the ones we have anyway T will be in here T will also be in here and G is in there and G is in the other one so hopefully something that relates one of the two pairs possible that is as simple as possible because I think that's what Pat was heading towards you have to assume that this cable around the outside is not slipping so it's a no-slip condition and what'd you say, Jay? before you jacosize your answer? it's a no-slip we're assuming that the cable is not slipping and since this part of the cable is not moving itself then what's true about point A? it's what? it's not accelerating well it's an instantaneous center and so that's R1 we can then use that very same type of relation we used before AG is going to equal R1 alpha as a no-slip condition which I think is what you said, Jay, isn't it? so there's your third equation so set up these two equations we don't necessarily need to go through the actual solution of these it's not an algebra class what part is that? or is that totally different? what part did Bob say? that AG first two equations, make sure we got them right Frank, you have to get them over there I'll make them get them over there two components to this one of them involving omega we're assuming starting from rest so that there is no omega at that instant spare your equations won't you get them? forces in the y-direction fairly straightforward I hope plus T they're in the same direction no question about that I wouldn't think W's down that's all the forces and so then we'll have the acceleration of the center mass so there's F equals M in the y-direction if you want you can then put in the no-slip condition what do you have there for WG? I'll remember it's good practice in these to indicate your positive direction so that you can lay these out properly we expect acceleration up to go with a clockwise so that might as well pick that as the plus direction save ourselves a minus sign so what do you have for some of the moments about G then? R2 yeah it would be and W goes the way of course goes right through G IG which is K squared M M times AG I just made this substitution in fact that also didn't do it on that and you can eliminate T between them if you want and I don't need you to solve it that's an exercise in algebra but you should get that was to be 19.8 the weight itself is 78 so not much of the tension not much of the weight is left on that on the cable anymore so that's where the algebra are eliminating T between these two equations leaving you with one equation in alpha and solve for it and fill in the rest from there any questions on that part? I think so because for some it's pretty obvious for this problem that the acceleration and the angular acceleration are in the same direction but that problem coming up where it's not as obvious and it's going to be quite helpful in fact it'll keep you from losing the minus sign if you have that kind of drawing some of the problems are so simple you can just put them on the same drawing you don't need to do separate ones just some kind of different indication of what the acceleration is alright now in direct response to Alex's question let's do the some of the moments about some other point in fact we'll just pick point A the solution in some of the moments about some other point we also have the moment of inertia about that point which we don't however we do have it about the center the moments about point A again we'll take that direction positive we know alpha is going to be in that direction so we'll give us a clean answer we have what moments being caused about point A what moments caused about F F is a distance of both radii away R2 and it is in that direction alright what else W of course in the opposite direction what about is that R1 that's alright and notice we don't need to worry about the tension which is the advantage of doing it this way if you wish it just eliminates one of the unknowns IG we've got K squared M alpha we're looking for but let's see do we add MAG or do we subtract MAG sorry MAGT do we add it or subtract it that's our positive direction about point A AG is going to do with a positive direction it's also going to be in that positive direction with respect to point A and in fact there's the D we need is the minimum distance between the line of action of MAG and that point A this is MAG times D which is also we have the effect of the way in there but it's done this is the sum of the moments this is the result of that sum of the moments okay so I should have like a plus W on the other side nope nope just that and then notice that now you have a single equation with a single unknown alpha the one that was asked for and when you solve that you get the same alpha you did before that helped a little bit Alex illustrating that do you see the advantage to doing this it's not huge but it does eliminate one of the unknowns if the tension had been asked for there might not be any advantage in doing this but since the tension wasn't asked for then there definitely could be an advantage to eliminating it from the problem we're going to do an example with something like a link arm or something nope not yet you've got to a little in the stumble before you can walk in GR that could be negative as well as you said yes if I had chosen the other direction it would have put minuses on those remember this is just an arbitrary choice I chose it that way because I knew alpha would be in that direction and that would give me a positive alpha it's nice if you know that direction so if we happen to have a problem where if AG is that way but I've picked that way is my positive direction that will give a minus MAGD because that's my positive direction the action of AG is against that it's as simple as that we've got another one coming up where it is negative because it's a little bit trickier problem alright any questions before I clear up and give you a mental challenge like you've never faced before you got for it Bob I bet you wish you brought a chocolate egg now there's some energy for a bribe chocolate eggs go a long way to extra credit points that's for sure alright what I've got is a simple simple little spool thing here with some string wound around the and I'm going to do two things and you have to predict what's going to happen both times so first thing I'm going to do is let the string wrap around such that it comes out that way and you need to predict what the spool is going to do and to help with that even though it may be obvious anyway it's going to be a free body diagram yep that's what that means, Jake what else does that mean do you think that T stands for turn around go the other way does that make more sense start from rest I'm going to pull on the string T as shown free body diagram and then predict with it the kinetic diagram these circles just aren't filled in it's going to look like you guys there's nobody right at the center and no value to this you have to predict the magnitude of any of this just predict it's action free body diagram then the kinetic diagram the kinetic diagram is the diagram that has MAG on it which is the sum of the forces and then has IG alpha on it which is the sum of the moments shows what you believe the sum the result of those actions is going to be but he's right so far with the kinetic diagrams oops I mean free body diagrams we got one correct free body diagram get this stuff out of there it is not the object that's why can I not go to the ground do you make these in a smaller frame I'm going to go to the next diagram I'm going to go to the next diagram the correct free body diagram obviously the tension is in there however hard I choose to pull on the cord any others Jake you got something else what else the weight I hope nobody disputes that it's only cardboard Bobby is there any other forces a normal force because it's sitting on the table there Frank in which direction my attempt is to pull it that way friction is going to act in the reverse direction to that to the left so that's the free body diagram what do you think is the kinetic result of that first is t greater than f or f greater than t t's got to be greater than that there's no way friction is greater than the force causing it the sum of the forces will be that way what about the sum of the torques in fact don't we have some of the torques here almost a couple it's not a couple since those aren't the same magnitude but it's like a couple which will have some net result that way the two of those together is most certainly going to cause it to turn counterclockwise you did so anybody have anything significantly different than either of those especially you folks that didn't have the friction and the normal force in alright so let's run the test let's see if I pull this way that's exactly what it does what we think this is so exciting you just never know what's going to happen like being at the science museum look at that that was a beautiful demonstration and did exactly what we thought it was going to do alright fake deal I could ask some dope on the street to help me down alright so let's do the same thing and you predict what's going to happen however I'm going to put the cord on the bottom now so on a free body diagram and a prediction via kinetic diagram of what's going to happen I'm going to pull it so it doesn't slip if I pull it with and it slips and the guy knows what's going to happen I said you were ready then you say you're not finished you got a feeling of what it's going to do Bobby you ready you're ready to lay down your life for this Alan did you finally commit to I'm going to pull it so it doesn't slip so it doesn't slip yeah I said that right I said that to those listening it's kind of soft so it doesn't slip if I pull hard enough the thing will just go like that and we don't have a very interesting problem and we like interesting problems well I need a bigger block in the chair and that's an extra pleasure so when you take this again next year we'll be ready Alex you said alright let's see what the free body diagram looks like of course there's going to be some tension there what else still weighs something I hope there's no dispute in that at the bottom there any friction this time yeah why not otherwise if I pulled on it it would just simply slide along turning it all there won't be some friction which direction yeah there's only two choices right and it's left which direction you said right right you said left left with a question mark imagine it's on not only ice greasy slimy ice the slickest thing known to man absolutely zero coefficient of friction static or kinetic if that was the case and I pulled on it what would it do it would go to the right since I don't have a greasy slimy snotty surface then friction must oppose that so friction must be to the left you got it right and you don't understand it imagine what it would do if there was no friction if I pulled on it it would go that way it wouldn't turn but it would just go that way it might turn a little bit just because the cable's off center but in general it would just slide that way what? no? that means the friction would add to that force it would take off that way even faster friction doesn't do that does it? let's see maybe this will help let's imagine this if you're not buying the no friction argument let's imagine the infinite friction argument let's imagine this is a geared wheel with a rack to the ground there's no possibility of slipping there so if I pull on it that way about an instant center there isn't that a met torque on it that way and it's going to roll who says that's going to that pretty small torque compared to that so if we sum the torques about the center of gravity looking for a sense of these that's a very small torque about G whereas that if we have that kind of friction and that's all the gears could do the wheel gears would push against the the rack gears the rack gears are going to push back that's a much, much bigger torque this one goes that way this one is much smaller and goes that way how is this diagram really any different from the diagram we just had other than the fact that the forces moved down a little bit well you don't mind it so let's see it cable coming out, strings coming out from underneath I pull it lightly so it doesn't slip no it's a magic trick it's a magic trick if you try to impose something on it that's not going to be there if you impose a friction in the opposite direction which shouldn't be there there's a force to the right there must be friction to the left and it takes, it's amazing how little tension it takes to make it move don't do it with just the weight of it off the edge so now what you need to do because you guys are schooled in this and most of you got it wrong now when you're old enough this is a good bar bet right here you don't pick some guy all muscly and you've gotten more money than brains say hey buddy, I'll bet you for a beer which way this will go when I fall really got it right because he doesn't know how to draw diagrams, he pulled himself most people look at that and think a few will say it'll do what it did, most will say it'll roll the other way start with this way, bet him this way and then he gets a free beer and then you say oh gee we better go double enough now and then put it that way and pull it no, what everybody's thinking is that this is the predominant effect and if the cable's underneath it's got to turn counterclockwise but the friction has a much greater torque the other way remember it's the sum of the torques that causes the angular acceleration so if that radius was smaller would it go left like what would we do to make that go left why do we think it's going to go left because obviously for something obviously we've seen something that's done that it's because you can see this force I guess it's because you can see this force and so you think this is harder this is always harder students have always had a little bit more trouble placing the friction when they do other forces there's no question about this once you, if you thought friction was to the left then there's no other possibility but it going that way because we still have this clockwise motion between the two forces and we have the net force that way, oh we didn't do the kinetic diagram there's a plan of the area the sum of the forces has got to be that way then the sum of the forces that way even if you thought the friction was going the other way that just makes two forces going that way and if I said it wasn't going to slip you should have even been more convinced that it would roll for the right like it didn't anyway you can try that if you want just see how easy it is it's amazing how simple how light a force it takes to move that way and you can do it with just a spool of thread too you don't have to have something cool like that is that was my spool we have my kite kite line another problem let's step up here now this one also you may need to argue with your intuition a little bit imagine my truck notice my truck is not pink even though my car is and they look kind of the same there's my truck upon it I have maybe this spool that I was transporting all right give the truck an acceleration let's say 5 feet per second squared the spool is not tied down unsecured so a fine angular acceleration of the spool we'll give you some of the details on it it's weight is 200 pounds it's radius of gyration it's 2 feet it's actual radius is 3 feet so it's a little more mass towards the rim than it does towards the axle so the radius of gyration is more towards the rim than the axle all right now oh there is another part we need the coefficients of friction between the spool and the truck bed and the level must be some grease on that truck bed I'll have to have it clean and that weight is weight of the spool a free body diagram of the spool free body diagram of the of the spool of roll of paper or burrito obviously it's got a normal force to accelerate right through the truck bed friction in what direction where you can sit out there do you remember the question at the base of the spool as the truck accelerates or I guess the possibility is zero there is no friction couldn't possibly be because in the extremes it sits right there and the truck accelerates away then it's dragging across the bed of the truck there's got to be friction also as it accelerates it accelerates with the truck in that case they're going to have to be friction because otherwise the truck wouldn't be dragging it along so somehow there's friction either way can't be no friction what do we think then to the right or the left right Jake doesn't know Jake wants a new force vector for our classes like that sorry Jake not going to happen Colin you voting yet Bobby you said right Pat you changed to right why do you think it's right now roll to the left now what you have to think is not what the spool is going to do because that's not maybe as obvious as we just saw what you have to think about is what's causing that it's the truck trying to go to the right it's going to have to drag the surface in contact with in the same direction as it so that's the friction force because that's the way the truck bed is going to move it's going to drag that with it now what happens from there isn't as clear if there's no slip then it'll pull the the spool along in that direction it'll start to turn in this direction which means it'll roll a little farther back on the truck which kind of makes sense that's why they tie these loads down at least part of it I assume there's other reasons if it does slip where the truck pulls away so quickly there's actually some slipping there's still going to be this little bit of friction it'll still tend to roll backwards there's our kinetic part for the rotation I think is everybody comfortable that it's going to do that how about though for MAG where's the center of gravity going to go what you have to imagine is you're standing way out here on the side of the road and you're looking and you're seeing this kind of thing happen see the truck pulled away you start to see the spool roll off the truck however if you're looking just at the spool itself not looking at the spool and the truck bed if you're looking just at the spool itself will the spool a little bit later be here in the same place or here no matter where the truck is we have these possibilities here's where it starts a second later the truck has accelerated to there let's draw it a little longer so we can get it on there there's three possibilities that the spool will have gone forward itself a little the spool will have stayed where it was anyway or the spool will actually go backwards some nope it's all there on the board remember looking for where do I put the arrow head on the MAG vector that vector is the sum of the forces there's the only force in that direction so it must be in this direction so this is a case where the spool is going to move forward but it's going to roll backwards because of the motion of the truck bed itself relative to the truck if you were sitting on the truck if you were sitting right there you would see the spool start to roll off the back of the truck but as we look at it out here we're going to see that the spool go forward even though it's rotational motion will be backwards because the object it's on is also moving forward so you have to trust the diagrams there's only one force going in the horizontal direction that's got to be the resultant acceleration of the center gravity so then the equations of motion as a starter two things we just mentioned by the way of course the center of gravity right in the middle there we do have forces in the y direction however there will be no acceleration in all that direction so all that tells us is n is w in this case so we can put those together to say u, w is that fair? because we don't know if it's going to slip at this point or not if the truck goes fast enough it will actually skid backwards relative to the truck bed a little bit if this truck doesn't go too fast then this will be a static friction situation so which is it because now we've got the coefficient of friction in there and we have two choices what we can do assume one or the other then we can solve for this the forces involved and see if our assumption was right that's especially easy with the static one because remember the static coefficient the static friction is quite variable it goes from a minimum of 0 up to a maximum oh sorry Jake don't stop me ever as a fellow alright let's back to tape up a little bit did you think I was talking to you let's assume no slip we've worked with that most often anyway it gives us an extra kinetic equation therefore the maximum possible friction is if we're right at the limit and we got all that stuff maximum possible static friction is 30 pounds if we solve this problem and let's see we've got what three unknowns F alpha and AG but we've got that kinematics kinematics equation for alpha and AG if we assume no slip so we can solve for F check it if it's less than the 30 pounds then this assumption is good if it's greater than the 30 pounds then the assumption that there's no slip won't apply and we'll have to read you the problem then alright then we also need to sum the moments we can do it about G we can do it about the contact point either one's going to work it's your choice about G it's a little bit about the contact point do tend to prefer that how many D stuff tends to be a little bit tricky at times how would you sum it about the other points the contact point you've got all three forces go through there I'm saying I saw a little one right there in fact I hadn't thought of that before well let's finish this and then then we'll talk about it alright we sum the forces about G the only one we have is friction pick a positive direction and we don't have to it only does one thing what F times R go right through there radius of gyration another small problem though here bed itself has its own acceleration the acceleration of the center of gravity is going to be the acceleration G relative to T that's the business we've got that's the no slope condition we can apply there and in this case we know that these are going to be opposite so we're going to need a minus R alpha there because this forward acceleration match the backward with everything we have here we can solve for this force of friction we've got alpha is unknown we've got that to link alpha and AG because AT is known we can solve for the force if that force of friction is less than this 30 pounds then our assumption was good because we have up to 30 pounds of static friction available so over that it breaks loose and starts to slip we've got enough here to solve for we can take AG and eliminate we're now assuming static friction that eliminates AG wait now I'm just going to start we don't want that that's at the static limit we want we want to leave it like that put that into there 0.15 the static coefficient of friction you don't want to assume we're at the static limit because we don't know that we are we just want to know if we're under the static limit okay but then there's no assumption well this is the assumption no slip here but there's no way to check it we can put that in there that single equation for alpha now when you get alpha you can put it in here and see what f is and see if it's greater than the 30 we can assume no slip we can't assume that we're at the static limit alright so you can double check this everything here is known except alpha you can solve for it 1.15 and then Friday we'll double check whether or not our no slip condition is okay what is what? this? no on the left side same equation double I put AG into here for f and then I put f into here so this is f times r f times r the right side of the equation is the same unchanged because that eliminated f from the equation to give me alpha I'm just unsure where this equation came from because normally I think a friction is mu times the normal force at the limit yes we're not at the static limit that's what this is here's the static limit we're assuming that we're not slipping but we're not assuming we're at the static limit remember the coefficient maybe the friction goes up to that point anywhere below that is static friction is a no slip condition if we exceed that which we now know it would be 30 pounds then things start to slip so you can't assume we're right there you can assume we're somewhere in here which is what we assume with this this is the no slip condition otherwise that wouldn't have been true how do you know you're not at that I don't know I'm not that's what I'm going to check if I solve for f and I get 30 pounds I'm right at the static limit and I know if I do one tiny little bit more of acceleration they'll be slippage I said I could assume it's anywhere in here I can't specifically assume one place or the other