 All right, everything we've been doing so far, we're going to go back and go over again a little bit in a slightly different way. So far, we've been looking at particle dynamics, and in fact, if you remember, this is very much the same sequence we went through in Physics 1. We did the kinematics first. Remember what that was, the kinematics? Fill, kinematics. Remember that? Well, how do we start? How do we start class? What do we look at? Both here and in Physics 1. Well, yes, specifically linear, I guess we started, in fact, we started with 1D, but what we were looking at then back in kinematics is nothing more than position, velocity, and acceleration, and the interchange between these two, or these three, and of course the fact that all of this happens over some time period. We started with linear, then went to curvilinear, did a little bit with 3D just to touch on it some. But that's all we looked at the first couple weeks in here, and spent even more than that in Physics 1 doing just that. Then we went to the kinetics, which is where we bring in the business of how do we make sure we know what the accelerations are and how those other things are affected. And we had three different ways that we approached kinetics problems. Our first was Newton's law, which worked very well for general type problems, especially if forces were constant, because an acceleration's constant is a pretty straightforward problem. But it also worked well for any problem that directly involved acceleration and or forces. Then we looked at the work energy method for solving kinetics problems that worked real well for position dependent problems, especially those that had changes in height in the gravitational field, which was a change in the gravitational potential energy. But also, problems where the position of the object in the problem affected how much of a spring might be stretched in those problems that worked real well for the energy method. Then we just had a little bit with the impulse momentum method and how that affected our ability to solve certain types of problems worked very well for time dependent problems. Now we're going to kind of quickly go over almost all of kinetics in a different way in that now we're going to do rigid body dynamics. So some of this will be as a lot of what we've done has been a bit of a review from Physics 1. However, if you remember in Physics 1, when we looked at rigid bodies, we only looked at them as objects that could rotate and that was all. What we're going to do in this class, we're going to quickly review pure rotation. But then we're going to put the two together and get a more general motion solution to these types of problems where we have objects that will move to different places. But we're going to take account of the fact that the object is a real size object and its particular orientation can very much affect the problem and its solution there. But then we're going to go over these things again with strict rigid body dynamics put together in general motion where we look at objects that can translate and rotate which the simplest example is of your car tires. They rotate as they move, but they also move from one place to another. So we'll do exactly that in a little bit. So to open with let's define what we mean by a rigid body. Because if we don't have a good definition of that, it's not going to be clear just what we're talking about. Especially when we get to certain problems in a little bit where we're going to have not one, but two, three, maybe even four rigid bodies in the problem. And we have to understand why they're rigid bodies, but the union of them is not particularly a rigid body. So imagine we've got some object here. A rigid body is that such that if we place on it three arbitrary non-colonial points they of course will make up a triangle. Whenever you have a triangle you have lengths of certain sides, little letters for that I guess. You'll have angles, alpha because it's near A, we'll call this one beta and I guess we'll call that gamma alpha beta gamma. I don't know what the Greek letter for C is if there was one. I think gamma was G, but there was no C. It doesn't matter as long as we've got some labeling system. So a rigid body is that such that these things are always invariant. Length of the sides, I guess I'd have to put a symbol for line segment. The line segments A, B, and C are constant as well as are the included angles, alpha, beta, and gamma are also constant. The values measured at any time during the problem for those three things will always be the same. In other words, what we're learning in strength materials about strength does not apply. Oh yeah, this is not an elastic solid like we're looking at in strength materials where that whole class depends upon the fact that real objects deform. So this is very, very much different than that. But don't forget in strength materials we don't have acceleration of any of those things. We're always under static equilibrium. In this class we're rarely under any situation of static equilibrium. We'll have some situations where the acceleration might be zero but only in a particular direction or in a particular way but there's always at least something accelerating in this class nothing accelerates other than our ability to absorb no material. Nothing accelerates in strength materials which is why it follows directly on the statics. So what isn't necessarily constant, those things are always constant. What may or may not be constant and depends upon just what the problem is, the location of each of those points might not be constant. If we had a, one of those points was the point about which an object rotates then, for example if it was point A, then position vector point A would be constant but as the piece rotates none of the others are constant. That's where we're going to get the object movement and then of course follow from that is that the velocities of each of those pieces may or may not be constant. And then of course the accelerations not only may or may not be constant may or may not be zero. Any one of those might be constant but any one of them might not be either in this class. However, what is inviolate is the shape of the object itself but what can change with the second part is that the object's orientation at any time can change. Remember those points A, B and C are arbitrarily placed and that's sufficient to define them in the entire object. So that's our strict definition of rigid body. We're not going to really bother looking at it in any detail because most of the things we're going to look at are just so obvious and rigid bodies that we'll be fine with it. But we will be looking at different objects linked together as we look at linkages and different machines and how they move, the dynamics of those type of things and we'll be concerned with then the interaction of several rigid bodies. So of course a rigid body can move in translation. What that means strictly to us is that given any two points on the object we don't need to define three that's sufficient for the shape for defining translation we only need two and there could be any two objects any two points on that body it's in translation if the kinematics of those two points are equal to each other. So if A moves from one place to another then that's the same movement that B undergoes for any two points on the object which is sufficient to define translation from the entire object. And if that's always true then of course it follows that the velocity of A and B are always the same acceleration for A and B are the same and we don't need to go over that because that's no different than the particle motion we just looked at. If all points move exactly the same then let's not look at them all, let's look at only one and that's exactly what we did back here as particle kinematics so we don't need to go over that again in any way. What is going to be a little bit different as we get a little bit farther in this is there are some motions that don't appear to be translational type motions. Don't forget that the path from one point to the other from A earlier to A later could be a curvilinear path as well. One good example of that is a type of example if we have a sign that can swing on two cables or two linkages of some kind even though the points don't undergo straight line paths as they go from one place to another they'll go kind of a circular path each point undergoes exactly the same path so the entire sign is in translation even though the cartoon picture doesn't quite look like it that is a true translational motion problem in fact a curvilinear one and we don't need to look at all the different points we would have been sufficient to just look at one particular point and know exactly what all of the sign is doing so a lot of students think that that's a rotational problem rather than a translational problem but it's not so then that brings us directly to the rotational motion that we're going to look at and we'll look at pure rotation today just to warm things up a little bit and then we'll put the two together translation and rotational motion together as we look at a more general type approach to these problems so rotational motion most easily described with a simple circular object rotating about a single axis of some kind our reference points then are some line on that object will move through some angle so our position is governed by the angle that makes for some reference line just like our position vectors make some position with respect to some reference point and then of course we're concerned with the changes in those angles as this object undergoes some kind of rotation these are vectors because there's a different direction things can move in rotation and in others we have to take that into account we will use the right hand rule for that vector notation which is common throughout all of science and engineering as far as I know there's rumors that in Russia they use the left hand rule but with the interaction of the sciences between the Eastern and the Western I'm sure that we all agree by now but remember that the deal with that is put your fingers, curl your fingers in the direction of rotation that automatically orients your thumb in a particular direction in this case it would be up and then that will be the direction of that vector that represents rotation so we'll be able to represent it in several different at least two different ways where either the direction of a regular vector or with a little circular arrow type notation either way is sufficient but we will be getting to several cross products and cross products are much easier to figure out with a straight line vector than they are to figure out with a little curly representation of the very same type of motion once we establish a position of an object and it's changing position then of course we're interested in its velocity the angular velocity is defined as well exactly what it was in physics one and the first part of this equation which is this semester which is the change in position with respect to time time rate of change position we look at it in both average terms and instantaneous terms and this is no different than what we did with our translational motion as we looked at these things and we can also use that dot notation so this should all look very very familiar there's no difference between the concepts and the notation with this rotational motion that we used in translational motion the first part of the term other than the fact we're looking at a different type of motion so we use a different symbolism for that but the concepts the general notation the general mathematical notation is exactly the same and then of course we want to pay attention to how the velocity changes and then that brings us to acceleration which again is laid out in exactly the same way it was in translation as an object changes its velocity we can look at the average change and we use an alpha to represent that make sure that your alphas and your a's that we use in translational acceleration don't look too much the same put a little artistic flair on these symbols so that you make sure you know and you make sure I know what it is you've written so the instantaneous acceleration was the average acceleration as measured in this and the time and that's exactly as before being on the first derivative of the velocity most of what we did in physics one if not all of it and a lot of what we did earlier was constant acceleration a lot of that will be the same here but we'll look at some non-constant acceleration problems just because you guys are that much brighter and then remember that the double dot notation applies in cases where the acceleration is constant as before then the average acceleration and the instantaneous acceleration are equal to each other because they're never changing that's kind of our definition of average all right, if you remember too as we go through everything else all our kinematic symbols between translational motion and rotational motion are perfectly analogous to each other and you can get from all the translational equations to the rotational equations by simply swapping out the variables one for one position for position velocity and acceleration and you get exactly the same things all right, so we'll do a couple we'll do a little bit here today with pure rotation and then on Monday we'll step into rotation and translation together or what we just call general motion so let's look at the type of problem we're going to see in this class imagine some kind of link arm that rotates around a particular point it may or may not be joined to some other object and we need to see how that moves this is exactly the type of things you see in robotics but we're for now just going to look at a single arm you can imagine that that part itself is indeed a rigid body but the object as a whole depending upon what the other arm does how it moves what happens at the other end of the object the two together is not a rigid body because we could certainly imagine a triangle made out of the ends of each of those and that triangle is going to change shape radically as we go through the problems so imagine we have an arm here whose angular velocity is changing uniformly such that it goes from 10 radians per second to 25 in 10 seconds and if you remember from physics one all of the angles when used in calculations must be in radians rather than in degrees we might talk about them in terms of degrees I don't know about you I'm a lot more comfortable talking about angles as measured in degrees and radians when I'm just thinking off the cuff but for the calculational purposes we need to keep it in radians so given that change in velocity this arm is one foot long want to find out something about this point here we'll call it point P want to find the velocity of point P actually that's the speed and also find the acceleration of point P we can do that in vector terms actually we can do velocity of point P as well in vector terms I'll see what you think we should do with that so if you can remember because those are actually translational quantities that come from the rotation of the object then also find the change in angle all of the distance traveled by the point P as well so want to find those three quantities this is mostly just review from what we've done before think about this velocity though remember it's directly a function of the angular velocity of the piece and the distance from the center of rotation so if you remember it looks something like that and that's going to be different at different times because the velocity is changing remember what is this in terms of a vector quantity how do we give direction to that I think you said it no? unit vectors we need some coordinate system that's going to work for this and I don't know if there's a better one than the normal tangential coordinate system we established some time ago the unit vector and the tangential direction depending upon which direction the object is actually moving and the normal component perpendicular to that and always directed towards the center of rotation and that's all that point is doing it's in curvilinear motion about some center point defined by a rigid body pure rotation about the same point so we can call that then in the normal direction tangential direction it turns out and this is not something we used in physics one but I expect you to be able to handle this here this can also be determined by the product of omega cross r that locates the p the point p from the center and let's double check that let's see for this picture our fingers in the direction of rotation that puts my thumb into the board so omega is directly into the board the position vector I orient my right hand such that I go from omega into the board towards r going out to point p that orientates my hand in this way and that's exactly the velocity I would expect the point p for that rotation the direction of velocity for that rotation so this is not something we could have done without a good clear idea of what that omega vector actually represents this cannot be crossed with rp itself we need the right hand representation of that kind of motion itself with changing omega from 10 to 25 seconds we can't really work that number out we could pick it at a particular time we're not going to bother with that however when we do the acceleration we'll also need a particular time so we might as well pick one and so let's say at 5 seconds what is then the velocity of that point and then also use that same time to find the acceleration of those points too hopefully it's obvious that at 5 seconds we're going to be midway in between here because I did say the change in speed was uniform so halfway between there is what just 17 and a half so our velocity then would be rp is a distance of 1 foot the speed at 5 seconds is 17.5 radians per second and the direction is in the tangential direction which we don't necessarily know the specifics of the point because we don't know the angle on the arm itself but that's sufficient to say the tangential direction since the velocity was changing we needed to pick a particular time so what the heck, it equals 5 seconds what about the acceleration of point p that arm speeding up which means point p is going to speed up as well we need to figure out by how much and again let's just pick it and solve it at 5 seconds so what's the acceleration of point p, how are we going to find that and then let's just go through the mechanics of it with the numbers well since the arm is spinning faster and faster with this changing angular velocity then it's linear velocity is going to be changing and since the angular velocity is increasing the linear velocity is going to increase too that'll be in the tangential direction so we'll have the tangential component as nothing more than the time rate of change of that tangential velocity well that v dot t is just the acceleration so we can figure out how the speed of the point itself is changing is that sufficient or do we need something else to go with it we're going to have a centripetal component for this as well that's exactly what we looked at in physics one we only looked at uniform circular motion but no I guess we did a little bit of accelerated circular motion but not much so we need to take into account that was the very same thing we did when we looked at the normal tangential particle motion anyway we're looking essentially at the motion of a single particle at point p so those things we can all find out there are slightly different ways to take care of that though too that are a little more sophisticated than we've had before the acceleration of point t sorry the tangential acceleration of point p all of these are point p is the distance point p is from the center times the angular acceleration we did that in physics one as well and then this will be in the tangential direction this is also a cross product in the same way that the velocity we're going to have that a little bit greater level of sophistication that we didn't have in physics one as we look at all these different points so let's see what that means it's spinning faster in the clockwise direction it was going 10 radians per second I said in that direction it was going 25 so the direction the speed is increasing is in the same direction as omega is that puts the acceleration vector also directly into the board rps was as it is before so the direction of this cross product is the same as the direction of this cross product because omega and alpha both have the same direction into the board in this case and that's what we'd expect if alpha is in the board cross it with r that gives us an acceleration in the tangential direction in the same direction as the velocity at that instant so that also makes sense and that number we can figure out because we've got enough information to figure out the acceleration, angular acceleration is there and we know that the distance is one foot from the center the normal component we can figure out in a couple of different ways we can do the v squared over rho I don't remember if our book uses rho or r my recollection is they use rho but r is sufficient same thing but this is also oops sorry I need a vector sign over that one vector sign over this one as well so this is in the normal direction another way to find that out is if we make this swap r omega equals v we can put that in here we get r squared omega in a normal direction this is also however a cross product and it's a little bit more involved one it's actually two cross products it's omega cross omega cross r where don't forget when writing cross products the order of the two vectors is crucial if you swap the order of these then you've got the negative of that so you have to do it in the right order so we need those parentheses in there what's the term in that case it's not commutative or it's an associative whichever we need the order in which we do these cross products so this one's harder to imagine because it's two cross products however this omega cross r the one we have in that second parentheses we've already got and that's the tangential velocity so I'll keep it a little more in general rather than say point p it's tangential to rock velocity at any time and so that one we can do let's see remember omega's into the board the tangential velocity is slightly down as I've drawn it so when I cross those omega into the board crossed into t gives me my thumb directed right down towards the center which is exactly what we know of the normal component of the acceleration should indeed do so by whatever means you wish to do that you're free to I think the easiest thing to do is well we've already got v the velocity of the point calculated we know what the radius of the rotation is the distance that point p is so I think it's just as easy to use that form right there with all the pieces and not do the cross products we'll do the cross products at times it's just probably not the most efficient way to do it here since we don't have any specific angle on that arm at any one time we just know how it's changing so take a couple seconds figure out what this velocity is sorry that acceleration is and leave it in normal and tangential components and we'll be fine and then we'll come back when you've got that to figure out the distance that point s actually travels now is that supposed to be r omega squared or is it r squared omega r r omega squared right thanks for staying away calm yeah r omega squared yeah the units wouldn't have worked out and that's probably what you were looking at notice that the same components that are here are also the scalar parts of the cross products because these by definition by definition these vectors are all perpendicular anyway so we've already got v at 5 seconds remember do this all at 5 seconds because it is changing changing value the main part you need is alpha got it already ok that we already have so that one's easy and then you need to figure out what the acceleration of the point p is but that's the most easily done from r the rotational acceleration and we have enough information right here to figure out what the rotational acceleration is remember I told you that it was changing speed uniformly which is the same thing same constant acceleration got the parts filled already not hard 5 seconds 10 seconds I had to pick some number since I'm the boss I had to pick which number so I picked 5 seconds that number that's just enough to tell you what the acceleration is the angular acceleration and then we had to pick some point because all of these values depend upon what's happening at a particular instant since it's constantly changing so it's just an arbitrary arbitrary choice you should be used to the fact now that adults are always arbitrary in the decisions they make certainly your parents are minor sorry about my own dad alright David let's see v squared over rho is the easiest way to do this so you should be getting a magnitude for the normal acceleration of point p as 17.5 radians per second quantity squared over the radius or the distance from the center of rotation which is the one foot and notice we get units then of what seconds what seconds no we don't yeah we get it yeah no wait we don't like those yeah no no this isn't radians this is feet that's where the trouble was units will always tell us what we're screwing up now we get feet per second squared because that was 17.5 this is just the one foot I messed that part up so now we're okay we caught it that's 306 feet per second squared 306 feet per second squared for that part and then the tangential magnitude and then we're just going to multiply by the unit vectors so this is all we need is just the magnitude's normal part I need just r alpha and alpha is the change in velocity by the time it took to make that change 5 minus 10 radians per second over the 10 seconds it took to do that so that gives us again feet per sorry feet per second squared and that's 1.5 feet per second squared and so we've got now the entire acceleration let's put it back together 1.5 in the tangential direction plus 306 in the normal direction so that's far and away the greater part of it per second squared so now we know the acceleration of point P what's going to be very important to us in a little while is we're going to need the acceleration of point P so we can figure out what the acceleration or end of the other arm is that's exactly the type of thing that's done in robotics you need to know the acceleration of all the points assuming that all of them are rigid bodies linked together so the last little bit I asked you to find was the change in or the distance traveled by point P in 5 seconds we know the time we know the acceleration we know the initial velocity sounds to me like a constant acceleration of waging 1.5 A T squared plus B I T then we've got all those numbers there comes out to be 175 feet the change in angle that went with that it's not what I asked for now and then done rp times delta theta the change in the angle is 175 but you could have done that in physics 1 which brings us to the next part of how all these things interrelate to each other and a lot of it comes out to be just what we've had before any one of these or alpha d omega dT then it's also true that's the differential form it's also true that the integral form applies as well if we don't have constant acceleration if we have acceleration that changes with time then we're going to have exactly that piece there to it and the same thing the velocity part using our definition of velocity and differential form we're going to make it into integral form and it's those times when either of these or both are constants that they just come out of the integral and that's most of what we did in physics 1 but not necessarily a luxury we'll enjoy it here because you're a more sophisticated student if you remember we did have one little piece that I warned you was a part of this that was often forgotten so I'll give you the same warning now if we solve this one for dT we get dT equal to d omega over alpha if we solve the second one in the same way we get dT is d theta over omega since those are both equal to dT if we combine those two then we get alpha d theta equals omega d omega if you remember I gave you A dS equals B dV when we were talking about translation and that's one of the pieces useful for solving some of these problems that students often forget especially when acceleration is a function of time and then you can integrate this in effect it integrates to the kinetic energy but don't forget those again as a piece of the solutions that we need to as we go through these real briefly just so you remember for constant acceleration problems which several of ours will be especially when we get to the point of general motion and we're going to be combining rotational acceleration and letting one or both of them be constant to allow us to get our feet wet and those type of problems that we have looked at before so it's still very useful to know what the constant acceleration equations are and they're exactly the same form as were the constant accelerations in translational motion for example the bulk change in position angular position with time but that's also equal to the straight arithmetic average of the two numbers this is also one of the constant accelerations equations that students tend to forget just because it's so simple to calculate the average students sometimes forget a very nice thing to do we also have as one of our equations alpha equals d dt r omega 2 minus omega 1 over delta dt remember the deal with these constant acceleration equations is that each of them has four components in them if you have those same four components that's the equation you use to solve this one has omega 1 omega 2 time and change in position this one has acceleration the two velocities and time and the only reason I'm giving them to you in this order is the order I gave them to you in physics 1 below those many years ago the change in position is delta s equals v i delta t plus 1 half at squared only this time we're using our angular kinematics symbols rather than our translation and then the last one which almost no student forgets is v2 equals v1 plus 2a delta s only we cast it in angular terms rather than in terms of translation exactly the same in translational form units all work out just right keep an eye on them we will have the additional parts though that I've already given you of the translational velocity of a point is the cross product of a vector is defining its motion and the tangential acceleration of some point is also a cross product and the normal is yet a third but it comes in two flavors just in case you're not in the mode to do three cross products because this one you've got to do this first cross product results in a vector then you've got to do the second cross product with your result and you can't mess up a single minus sign you can't mess up a single multiplication you've got to get it all right alright so let's look at a non-constant angular acceleration problem imagine a small motor here used to run a much larger blower of some kind by means of a belt going around the two actually this is a constant acceleration problem we'll do this and then we'll do a non-constant acceleration problem alright so the motor has an angular acceleration of two radians per second squared the radius of this let's call this one a and this one b the radius of the motor 0.15 meters and the radius of the blower pulley is 0.4 meters so this is a constant acceleration problem one still but it's getting a little more complicated now we have two rigid bodies connected plus it's going to bring up one that briefly addressed in physics one but we didn't need to put a lot of weight into it other than the fact that we considered it true so I want to figure out the acceleration and velocity of 0.p after two revolutions of the motor pulley a so instead of defining that by time a point in time we can find it by a particular position so this is a constant acceleration one it's not going to take anything more than the type of stuff we had in physics one and then this time it can be a little more specific with the vector since we're taking that point at the bottom there the acceleration of 0.p a normal component and a tangential component and we know something about how to figure out both the loads and then the velocity of 0.p is going to depend upon whatever its velocity is at that time but we're going to need that for the normal component anyway for the v squared over r that's probably the easiest way to do it with the vectors if you want across products give that a simple coordinate system okay so with a couple minutes allotted to you see if you can't chug through through that rather quickly notice that there's a a ratio between the angular velocity of a and that of b and the angular acceleration of a and that of b so the velocity and angular performance you need in the equations are those of the blower pulley b rather than the motor pulley a at that phase speaking German do you speak German yeah so it can't be that I don't either so it would sound terrible so what we're given here with this is 2 revolution that 2 revolution is you're sort of given delta theta not specifically because remember revolutions is not a measure of angle we need radians invert that 2 revolutions 2 pi radians per revolution and then you have it gradients remember all of our angular measures as used in the equations our radians these are sometimes referred to as as engineering they're very common to be speaking in terms of revolutions per minute etc that's got to be a picture of that like our little ten and I don't know if it gets all the way up to the board one important assumption we need to take here and it's going to be very important throughout the next the rest of the class is we're assuming through all of this and it's not always true and you can confirm that with things you do with your car we're assuming that the belt does not slip that there's a no slip condition between the belt and the fluids if it does slip things are all together different so we have to assume for this that the belt does not slip that's not necessarily the case when you're starting up a motor which we argued because we have an acceleration on the motor and that's the squealing you'll sometimes hear in your car if your fan belts and other clay belts are slipping especially when you turn your power steering you turn your wheels bomb has that happened to you recently? the wheels are pretty far to the side it really puts a lot of strain on the power steering motor and you'll hear the belt slipping don't try to pick up a girl in your car with squealing belts that's what I always say is that not working? certainly we can multiply 2 times 2 pi together this is delta theta so what this really is is delta s since you have the distance in there what's that one? once you've got delta theta then you can find the distance that point traveled by doing r delta theta you can come up with the velocity of point p based upon the velocity of any one of the points on the motor pulley because they're intimately connected by a non-slipping pulley that's the purpose of using a chain instead of a belt insurers non-slipping because of the teeth on the gears and the rollers on the chain so it's many of the same type of things we had to figure out before the velocity of this point p which you can figure out the velocity of any point on the motor pulley since they're going to do the same with the belt non-slipping but then you'll also need the acceleration of point p and that you've got to get using the ratio of the diameters for any the acceleration of the motor pulley I put in the work order to get that fixed but I just couldn't get over this morning I really found it this morning because we get reflection we have the sort of shiny window sills over there and they cause enough reflection on the board that it's a distraction for the camera to believe me we're not amateurs we do have a professional in the room does that make sense Phil, alright you could do it with these cross products I don't know that it makes much sense to do that it will for some complicated problems where we've got lots of different positions on these things but we need a couple pieces as usual we need the velocity of point p of course in the tangential direction obviously it's going to be in that direction and that will be the same as the velocity of any of the other points because they're all connected in the built movement so we can figure that out r omega a remember though that's after two revolutions so we're going to have to find that part and that also because rb omega b but r omega a and omega b are related by the ratio of their radii pretty quick fix so the angular velocity of a since it's constant velocity we can find that out after two revolutions from one of our constant acceleration equations we're given the acceleration and the change in position and assuming it's starting up from rest you didn't ask I said when starting up it won't slip but that was the discussion about belt slippage I was making sure that you listened to me at all times rather than just occasionally stopping by and listening once you find the angular speed pulley a after the two revolutions then you can find the angular speed of b and the velocity of point p and then once you've got the velocity of point p then you can start figuring out the acceleration components the tangential acceleration of point p remember the angular acceleration of b times the radius of b and the angular acceleration of b is again related by the ratio of the radii and those are pretty straightforward to come up with and then the normal component is the velocity of point p squared over the radius of b now lots of different things gears, pulleys, link arms and all kinds of things in mesh so there's going to be lots of radius in these problems, lots of radii be careful pay attention to the fact that you've got to have the right ones in the right place that should give you all the pieces and then you can even figure out the magnitude of the acceleration of point b we're getting there at the end so I guess we've got to make this a get out of class question I'll update the solution it could have gotten out early Chris you got some stuff we'll check it and see if your weekend is starting if you have your doubts alright acceleration of point p is that the total acceleration I'm not quite what I got what'd you get for the components where it goes you know I didn't get the point 6 at point 3 at 2.81 not 2.26 check that you're using the right radii for those pieces what's the difference in p for the velocity of point p for the velocity of point p 1.06 1.06 meters per second w a2 2.7 could be w a I have 2.66 let's make sure we're looking at the right number this one that's after 2 revolutions okay 7 I had 7.09 how do you square them no that's un-squared I had 7.09 make sure we're looking at the same things we'll make it a yeah 7.09 yeah but in engineering we need real numbers we don't work with pies in math class it is not in engineering class it is frying your brain let's see other pieces to look at delta s I didn't figure out directly in the s form well we're getting right near the end let's put these together and then you don't have to stay the whole weekend with me so this should give an omega a 7.09 radians per second that's the speed of a after 2 revolutions at that acceleration at constant acceleration so then you can use that then to find the angular velocity of b at the same time and since a is little it's going to run a lot faster than is b by the ratio of their radii and I had 266 on that one did you get that too Tom ok and then that one we can use it might be the easiest way then for finding the velocity sorry the acceleration of point b because then that would be rb omega b square and we have those two numbers that should be 2.81 mu per second square anyway we have that number there let me make sure that's the normal yeah normal velocity and then for the other one the acceleration um the acceleration of uh the acceleration of point p is the same as the acceleration of any point on the belt so we could have looked at it on the upper piece then 2 we can get that with ra alpha a equal rb alpha b we know alpha a we know the two radii we can find alpha b in fact we just need rb alpha b we don't need to take that apart we can just use it like that so that should have been 0.3 let me show them right now the right one that's the tangential 0.3 which is exactly what this piece is rl and then you can put those two magnitudes together plus 2.81 and you should get 2.83 actually that's meter square seconds to the fourth those are squared and then we square root them notice that once again the centripetal component is the dominating part of the acceleration does that end up what it's okay whether that's the normal acceleration right here but just below it you can either calculate it using b squared over r or you can calculate it using r omega squared whichever you have easier we haven't have omega so it was just easier to do that since the acceleration of any point on that belt is the same as anywhere else since the belt's not slipping then you can set these as even that's also sort of a subtly that students tend to just forget the confusion of looking at new problems right here right there that's the acceleration of point p in the tangential direction of the acceleration of the motor yeah tom you get those and I'm at 2.85 okay yeah that's probably just right alright the 1.065 that's the velocity 1.065 you're trying to get omega from that but that's not what the velocity is oh yes I just didn't have to write it down this velocity okay it's just that angular velocity is the radius there