 So far in our general motion, if you remember, this is where we're combining both the rotational aspect of the rigid body and the translational aspect of at least some point on that object. We'll even be a little more specific with that as we get to the kinetics in a little while. We did the kinematics of rigid bodies. We did the absolute motion that works well if at least one point on the body is following a prescribed path, then you can actually prescribe that path and then take the derivative of it to get the velocity and acceleration. Then we did the relative motion, actually we did the relative velocity, which would make you perhaps tremble with fear that we're going to do relative acceleration, and that's indeed what we're going to do today. Then we did instantaneous center, those, well the first one would allow us to do some acceleration. The second one, we didn't even approach the acceleration there, I don't know if anybody was thinking that was missing, but that's what we're going to do today as we finish this up. As has been the case before, once we hit the acceleration, especially with curvilinear paths, things are a little bit more complicated just because there's some extra things going on, and that will indeed be the case today as we get going here. We did have the relative velocity equation that we used in great detail for some of our work a couple of days ago, I think that was actually a week ago I guess it was, and then if we take the time rate of change of that, then we get a relative acceleration equation which is at first blush, not too big a deal, and whichever of these notational forms you prefer is fine, but the general idea is just like it was with velocity, if we know something about the motion, the acceleration specifically of one of the points on the body, we know something about the relationship between the two, we can find out the actual acceleration, true and absolute acceleration of the second point, or other combinations thereof, we know the acceleration of two of the points, we can then use that to figure out the relative acceleration between the two, that will allow us to get the angular acceleration of the body, other viewpoints like that, so there's lots of different ways. So imagine we've got some rigid body here with two points on it of interest, remember it's most important in these rigid bodies that we remember, the distance between those never changed, we could pick any two points, the full definition of rigid body involved in the triangle, but it's sufficient to say any two points possible, no change in distance between any two of them, and imagine because of whatever its general motion is that one point is following some path, and the other point is following some other path, not necessarily are these parallel, and I hope it doesn't really look like that, because remember the, if these paths were parallel, the two paths were perfectly similar to each other, and these are two different points on the body, then we just have translational motion, if the paths are different in some way, and not even parallel, then we do have some kind of rotational component to the motion to describe this, so because of this we might have some angular velocity that this object is undergoing, and there may be some angular acceleration as well, I just have the drawn in that direction, but they could, either one could be in either direction, there's no reason they need to be in the same direction, there's no reason they need to be in that direction, but because of this we've got perhaps a accelerating in some direction that we may or may not know, for ease of argument we usually assume we know the acceleration of a, and we're either trying to find the acceleration of b as a function of these angular components, or we know the angular, or we know the acceleration of b, or we're trying to find the angular components of their velocity, so b might be accelerating in some direction of its own, known or not known depending on what the problem is, but that acceleration itself has two components to it, it has a tangential component, acceleration of b in the tangential direction as the possibility that b is actually accelerating down its own path, it also has a normal component and we'll define that direction as the one directly towards that point a, and in fact then we could be even a little more explicit with this, we can say this is the acceleration of b relative to a in the tangential direction, this is the acceleration of b relative to a in the normal direction, there it is, I told you this is a day of great detail, so hopefully your pictures are nice and big, hopefully your years in color chalk to take notes so that all is making great sense. Alright, so as you can imagine, the one of great interest to us is really this piece here, the other ones are just absolute accelerations, maybe known, maybe we're trying to find them, it depends upon what we have as we go through the problem, but this is then made up of two components each in its own particular direction, so I guess I can just draw it as a vector and just label it as the tangential component, the acceleration vector and the normal component and we've visited those before as we did the rotational motion, I'm just rewriting what I finished with there at the bottom so I'll have a little space to expand it farther, we have these two components of the relative acceleration, remember this is the motion of b as if you are sitting at a and what it looks like b is doing, to a to the observer at a it looks like b is orbiting them because of this rotational component of the motion of the rigid body, if you were sitting at a and doing nothing but looking at b it looks like b is orbiting around you, it may be doing so faster and faster in the tangential direction but then there's also the centripetal component that we have of any kind of orbital motion of any kind, so these two pieces here then we've looked at a little bit before but we can expand on them a little bit now, we've got to make sure we get every part of it in the correct order, so this is the tangential component so we can take care of it as simply the acceleration if you will of the arc length in the tangential direction with our unit vector or our little e's, is that right? That's what we've used in this class of these little e's, some classes use o's, some use of the t itself, if we need to be more explicit with this as we looked at in previous classes with a pure rotational motion, then we can we just need the relative position vector to the final, when we were looking at pure rotation the center of rotation, a had no acceleration so it was just simply the distance from a that we need there but we need to be a little more explicit now because we do have this component b relative to a, so that will give us those two pieces, that will take care of this tangential component, the normal component is a little bit different, if you remember in one form it's r omega squared, if we take out omega and put in r over v, sorry v over r we get the v squared over r, what is familiar with in the normal direction and if we do this as full cross products, it's that double cross product omega cross with the cross product omega cross r v relative to a, we'll be able to do this middle one, it's a little bit simpler because we're working with two dimensional problems, we can figure out very quickly one of the most of our problems just where these things are, if you're in the mood this takes yet a third form which is minus omega squared times a vector, just meaning that r points out from a to b and this is minus because it's a centripetal component where the acceleration is towards the center which as far as the relative motion is concerned the center of the rotation is a, it's a, alright so there's the setup, let's apply it to a problem which was familiar, that rack and gear problem that we did a little bit earlier, so we have this power gear that runs without slipping unless otherwise mentioned all of our problems are subject to the no slip consideration, and then we have this interior, interior gear upon which road this long, and again no slip there either, so imagine either one of those gear that you wish, but there's no slippage going on between the two, make these nice and big because we had, sorry Joe, oh that's pretty big, you can go home and redraw it, what else are you going to do this weekend? If you remember we had a lot of stuff to get on there just dealing with velocity, now we've got more to get on there, now we're going to throw an acceleration, so our original problem was that we knew the velocity of this point a, we called this point b, this point c, and this point d over here, and so we analyzed that problem in terms of what the velocities of all those particular points were, but remember that the velocity of point b actually gave us the velocity of the rack itself at that instant, so what other pieces did we have? This radius is 150 millimeters, this radius was 100, and I think, yeah we came up with, we had a couple more things let's see, the velocity of a was 1.2 meters per second, so as a vector that was, we'll take that as the I direction, we also came up with the angular velocity of the wheel itself, which was minus eight radians per second k, that's something we had to determine when we did that problem, we used the relative motion to do so, and then we also came up with the velocity of b was 2 meters per second, also in the I direction, I think we had the velocity of d, but we're not necessarily going to need it here, okay so that's just the problem we did I believe on Monday, sort of rehashed with some of the results put up there, because now we're going to add to it the possibility that point a has some acceleration, add that on here as one of the givens, let's say it's 3 meters per second squared in the I direction, this drawn just does sort of arbitrary, but it would be horizontal to the lower floor, because that's all point a can do, it can only go in a continued horizontal way, and so some of the things we want to find, the angular velocity of the wheel itself, well we'll go ahead and find the acceleration of point b, c, and d, okay so there's our setup, alright the angular acceleration itself, that's a pretty easy one, because remember point c has no velocity at that instant, and so it's as if everything rotates around that point, so we need the velocity of a relative to c, and then we can just set that up as we had before, where the acceleration of a particular point is r times alpha between them, we don't need it any more than that, we already know the direction of it, there's only one thing it can do, so we can just imagine that at that instant in time point a is in rotational acceleration about point c, at that instant, and so we can figure out then that alpha is the 3.0 divided by the 150 millimeters, which is 20 radians per second then, sorry 20 radians per second squared, and we already know that it's going to be in the clockwise direction which is minus k as a vector, so nothing terrifically common, we don't need the cross product for that one, you just need to remember that at the instant, at any instant shown the contact point has no velocity, point c if it remains with the wheel will go somewhere else in a second, a split second, we're going to find out about that when we found out its acceleration, at that instant time it has no velocity, it may or may not have acceleration, we'll have to find that as we go through this, okay, so we can find the acceleration then point b, we'll relate it to point a which is nice because that's already known, that was given using our relative acceleration equation, so we just need to find out that second part but that second part is made up of these two components here, the tangential normal component, so we just need to do both of those, it's a matter of walking through the pieces, so we'll take the acceleration of a, add to it the tangential component of that acceleration, relative acceleration, and the normal component of that relative acceleration. Doing whichever one of these different ways to do it is more comfortable to you, most of us will find a slight variation of this because it's just the product of the magnitudes to have got, and then we just figure out the direction by simple activation of where the cross product would result, so let's see, let's do these in pieces just for clarity and then we'll bring them back, so the tangential component is the cross product alpha cross r, but since it's two dimensional and all of our vectors are automatically perpendicular then all we need is the product of these, so this will be rb relative to a, which is no trouble, that's the radius of the smaller disk times the magnitude of the angular acceleration which is also no trouble because we just found it, and then in whatever direction that may lay, so let's see what the deal is here. Alpha is in the minus k direction, remember we figured out based upon this acceleration that alpha itself is like that, so that's the minus k direction, so the alpha vector straight into the wall, cross that with the vector that goes from a to b, so we have straight in crossed with the vector that goes from a to b, so that puts our thumb in the i direction, and so we're done with the cross product, we know that this will be in the i direction, that is the tangential direction for point b relative to point a, everybody comfortable with that, if you do the cross product itself with these two vectors put in that should be exactly what you get, and we got those bits, the magnitude of b relative to a is 100 millimeters, and alpha we just found is 20 radians per second squared, not the minus, the minus we already used when we established the direction of the alpha vector, the angular acceleration vector, and then used the cross product on it, that gave us this positive i, so we don't want to put the minus 20 in here, this is magnitude only, the minus went with the direction, and then that's i, so this is b, did I have it meters per second, yeah, so this would be then two meters per second square in the i direction, so we've got this one because it was given, we've got this one because we just did it, and only you had to mentally do the cross product, didn't have to actually do it, if you're relieved with that, stand up and shout, no stand up, that's all right, so we can now do the other part, let's see if we can do it this way, same radial distance between the two points, and if we know the angular velocity of the wheel, which we do, so we can do this one in the same kind of way, okay, the relative acceleration between the two points in the normal direction is the same RVA that we already used, omega squared, and we can, it's probably easier to do this one than it is to do two cross products, but you can check it, it's very important which order you do these cross products in, because you're just the wrong direction if you don't, so minus omega squared, that's no big deal, the vector RB to A is that one, B relative to A is in that direction, and we're opposite that direction, which would make sense if B is moving around A, it's got to have a centripetal component, which must be towards A, but this vector is away from A, so then minus sign makes sense, we can do, so we're opposite that direction which is plus J, so this is now minus J, everybody follow the logic, you're frowned, did you follow the logic? We know the direction of this vector, it just goes from A to B, which is in the plus Y direction, so we're in the minus J direction with the vector, and all we need to do is then put in the magnitudes, which we know, so that's 100 millimeters, omega squared, the magnitude is 8, so that is squared, oh and then minus J direction, running out of board here, that's minus 6.4 J, meters per second squared J, okay so let's put those all together, we have to keep track of some of the points as we're going here, erase some when we don't need them, remember we're looking for the acceleration of B, acceleration of point B, that's the fact that A itself is accelerating and they're tied together, so that affects each other, that's 3 meters per second I, that was just given, all of these will be meters per second squared, so I'll put the units in at the end, plus the tangential component of the acceleration of B about A, which was 2 meters per second squared I, so I'll just add that into the I vector I already had from A, this is the acceleration of A, this is the acceleration of B relative to A tangential, and then I have the normal component that we just figured out, which is minus 64 J, 6.4 J, and all of that is meters per second squared, so let's see what we got here, draw a nice big picture, there's B, there's the rack that runs on point B, and point B has 5 meters per second squared component in the I direction, so that's A B in the X direction, I guess, our I direction, and minus 6.4 in the J direction, so it's about that Y, plus a little bit more, so that's acceleration of point B in the Y direction, and then of course it's actually the two of those put together, so that then is the acceleration of point B at that instant due to the fact that the wheel is rotating and it's accelerating as it rotates, I don't know about you, I couldn't have looked at this problem, looked at point B and anticipated that, maybe on some of the other points we could, but just not that one, you freaking out, no, you're okay? Yeah, not only is point B accelerating forward, but it's also accelerating down, and you can tell that because a split second later, that point is going to start coming down the rim of that inner gear, because that is marking the point B, that point B is on the gear itself, not on the rack, and so that kind of anticipates for us that it would be somewhat accelerating that downward direction, I don't think it's obvious that it's greater in the Y direction, minus Y direction, but that has to do with what the angular velocity and the accelerations themselves were, and these are just made up numbers. Everybody comfortable with doing this step as a cross product, rather than actually working through the cross product, because actually laying up the cross product I think is more work, we can do this because it's a two dimensional problem, and these vectors that we're concerned with are automatically perpendicular to each other, and a perpendicular cross product is the easiest of all of them because the magnitude is just the product of the magnitudes of the other vectors going into it, and then we can just figure out by observation, doing the right hand rule what the direction is. Alright, I need board space, can I erase that, we can go on point C, that's point B, we'll do point C now, but we'll do it very much the same way, so we can step through it again as another exercise in doing this relative acceleration. Alright, so that's acceleration on point B, acceleration on point C, we'll do it the same way. It's tied to point A, and point A is accelerating. We could do this with point B instead of A, now that we know, well we used to know the acceleration of point B, you have it written down, I know, but the acceleration of point A is a lot simpler since it's linear, and then the acceleration of C relative to A, and that being as it was before, both a tangential component and a normal component, both of which we figure out in the same way we just did for point B, so we'll step through that together and see if we all get the same thing. Acceleration point A we know, that's the easy part, so we'll do the acceleration of point A in the tangential direction, that's R, not B relative to A, but C relative to A, we're doing a different point now, this was written down for point B, so now we're doing point C, so this is R, C relative to A, alpha in the tangential direction. So again, let's see, alpha's into the board, the direction of C relative to A is in the opposite direction than what we had before. Written down here, this is A relative to C, we want opposite that, so we have alpha into the board, R, C relative to A, straight down, that means it's in the minus I direction, we'll put in the minus sign, put in the I sign, and then we can put in the pieces, which gives us minus 3 meters per second squared in the I direction. A lot easier to do it that way than to do the cross products, taking advantage of the fact that these are all perpendicular vectors. If they weren't, we'd do the cross product. Alright, so now we have the acceleration of A and the acceleration of C relative to A in the tangential direction, now we'll do it in the normal direction, this is probably the easier one to use, minus omega squared, we've got omega, it's still the same angular velocity it had before, and the vector R, C relative to A, remember this is written for B on this side, we're now doing point C, so the vector C relative to A which goes from the center down, we're in the opposite direction of that, so that was minus J, this makes it a plus J, we can just write a plus, and we've got those, omega squared times RCA, which is the 150 millimeters, we do that up, you get 9.6 meters per second J, positive, we've got all three parts, you put it back together and tell me what the acceleration of point C is, we've got all three parts, we've now got the acceleration of A which was given, it started the problem anyway, now we've got the two components of the relative acceleration, you can put them in, put the whole vector together, then sketch it like we did for point C, minus our point B, and see if it makes any sense, so here's the bottom of the wheel, riding on the ground, no slip, there's point C, let's see what it looks like, should do it, is it amazing, I know it's beautiful, this is all beautiful, but is it amazing too, beautiful and amazing, and elegant, my goodness, we're all becoming poetic, poetic, well, I don't know, did you expect that, did you, well I didn't, I was, even having taught this color, I didn't come back this, oh yeah, now I remember the size of Phil's drawing, he's not messing around here, did you expect that for the acceleration of point C, yeah, guys, I don't suppose, notice, let's see, we have a point A that has a plus 3 I acceleration, the tangential component has a minus 3 I, so the I component cancels, we're left with just 9.6 meters per second squared J, which is simply centripetal component, and everybody swears that they saw that coming, but then you think about that, remember that point's not moving at the second, it's in pure rotation relative to point A, uniform to the motion at that instance about point A, alright, do point D, do point B and C now do point D, yes sir, it's D right there, do point D and then sketch it out, see what it looks like, see if it makes some sense, guess this also makes some sense, an instant later, remember the wheel has rolled a little bit farther and that point C will have just gone up at that second, it's also gone forward a little bit by then, but I can't draw an actual instant in motion, I can draw a macroscopic delta T, so you lay out, now the acceleration of point D, you can pick any of the other points you want, both for obvious reasons, since everything's been so obvious to you guys going along here so far today, that point A would be the easier one to pick, but you can use either B or C, why would A be easier, no it's not that, it's remember all these cross products, A is just a much simpler vector than any of the other vectors, remember you need the position vectors for these and the position vector of D relative to A is much simpler than B to D or C to D, not much simpler, but simpler, so why give yourself headache when you're not getting me extra credit from me for giving yourself headaches, that's my job, so you don't have to use point A if you don't want to, we know the acceleration vector of the other ones, you can use point B or C, we know both of those now, just remember if you don't use point A here you can't use point A at those other two places, these letters have, these subscript letters have to always match, so figure out the acceleration of point D, nobody will find it remarkable because you're all clairvoyant evidently, did you know I was going to say that, no, haha you're not that clairvoyant then, your head's a little bit I guess, alright so figure out what the acceleration is, we already have an acceleration of point A, it's a simple list of the acceleration vectors we've got, well C would be pretty simple but then the position vectors are a little more difficult, so figure out those two components, if we get the same thing we'll just keep moving if we don't, is that it? I believe so, we'll see, is that what your crystal ball told you? Yeah, we just figured something out, I'm going to say what it is so I don't get it right We'll see, did you agree with her? Joe, you okay? You run out of steam at the end of the week here? Alright, the tangential component, you're going to do that way just instead of B we have a D here now, the normal component is easy enough done that way For the normal component, D relative to A, the vector that places D relative to A will be that one, that's R, C relative to A, yeah we want the opposite of that, which makes sense Remember, that's the centripetal component and if we're doing it relative to point A it should point to point A and if you do this double cross product you should get the same thing I don't want to do that Because you don't have two right hands, I understand So you have to figure out the direction of this vector first, let's see, omega is into the board Cross with D relative to A, so that's plus J and then omega again is into the board Cross with plus J is in that direction D headed right towards A Yeah, but that is an easier way to do it for these kind of problems and these isn't the kind of problems we're doing Phil, did you get it? You mean you're going to start now? Did you guys agree? Yeah After some indecision? Oh discussion, you were wrong? No, I don't want to do this method Yes, Milady, you are the correct and you should have changed yours I wouldn't have done it when we were doing it You okay? You getting it, Tom? On your head Chris, you got it? No? Phil? Boycotting? So, let's make that phase Alright, let's see Acceleration of point A, we already know that's three I Then I think the normal component has a direction and it has a component in the I direction Well, let's write it down and see what you got The tangential component, I got three J So this is 3.0 meters per second squared J And the normal component, 9.6 I So we can add that to the I component we've already got there So this is 3 plus 9.6 in the I direction That's the first and the third parts And then the second part is the 3 J and that's all meters per second squared So we get 12.6 second squared And let's see, so point D Right here, it's got a strong component horizontally Even more than the acceleration of the wheel itself So that component is 12.6 meters per second And a little one fourth of that component in the J direction Three J So that's the acceleration of point D It's accelerating in the horizontal direction with the whole wheel It's got a centripetal component about point A anyway And then it's headed on its way up the wheel as the wheel rotates So that's the vertical component that we see there Because a split second later when the wheels moved a little bit farther Point B, point D will have moved up that wheel a little bit So that's the vertical component that's headed in that direction Anybody need to see the steps it took to get any one of these? These two pieces here? Do the tangential one Do the tangential one with 3 J? Yeah, just going to the Okay So that's A D relative to A The acceleration of D relative to A in the tangential direction That's the magnitude of these parts Both of which we have D, remember not B here, this was set up for B D is 150 millimeters away from point A That's the R part And then alpha is just the angular speed of the wheel What was that? Negative 20, yeah But we only need the magnitude of 20 Radians per second squared The direction by doing alpha cross R D relative to A Alpha if you remember is into the board Because it's angular acceleration is clockwise So that's A, sorry alpha is into the board Crossed with the vector that goes from A to D Is plus J What's the point of your fingers to the right or to the left? I put my fingers in the direction of alpha first The only way I can get to the direction for this vector Is if my thumb is pointing out My palm has to face B That puts my thumb up Positive J direction And that's just what we had here then So a bit easier way to do the cross product Because all of our vectors are normal If we had, if you'd done this relative to point C You would have had a different vector Then those wouldn't have been quite as easy as this is Joe How's the opposite of the board again? Because it's like going a little on the board No, that's the way we drop But if we put our fingers in that direction Thumb has to be into the board I can't curl my fingers in that direction To get my thumb to point anywhere else Not on my right here I thought I did cross A with something to get a direction But A you're saying A The force of A is going into the board But A itself is spinning in a different direction No, there's A and there's alpha Which one are you talking about? Alpha Alpha Alpha The wheel that's wrote is accelerating In a clockwise direction And the only way I can curl my fingers In a clockwise direction is if my thumb is into the board So the vector So alpha has a vector Is what was it? 20 radians per second Square is in the minus K direction And that's the right hand rule There's two ways we can draw a rotation We can do it with these curvy little arrows But that's not a vector Because vectors do not curve Vectors are straight That's just a photographic representation Of the acceleration If we want the vector We have to use our right hand rule And that puts the angular acceleration vector Into the board Which is nice Makes it automatically perpendicular To all the other vectors we've got No problem Okay Questions on that Before we clean up And do a new problem You'll like this It's a beam problem Not a beam But we don't have to go shopping for the beam Because we already did that in the previous class We've got 3,000 of them On a loading dock We're just going to use one of them Here's the 7 I'll leave that Because that's a fair reference Alright So here's one of those beams We just bought At two points Not evenly spaced This is 3 feet That's 4 feet This is 3 feet At two points evenly spaced It's suspended from wires That run up to winches So there's our beam Nice and level Don't want to drop it Look at this And maybe not as a beam But as one of those window washer platforms And so you want these things nice and level So you don't lose A window washing fluid You can lose window washers But they're replaceable This day and age is putting it down in the paper You've got 60 people applied So we want to keep it up over the fluid So here's the deal So this is now being lowered At a steady rate But then the brakes are put on the winches To slow the rate at which it's lowering However, the winches don't have the same breaking power So one breaks better than the other So I'll label this one B and C We're going to take a look At two end points Two A and D So the acceleration As it's dropping But we put on the brakes At the winches That causes an upward acceleration Because those points are slowing down And B is a little bit greater Than the acceleration of D So I'll drop that one So there's the acceleration of point B There's the acceleration of not point B Point C And the magnitude per second squared And this is three feet per second squared So we were lowering at a constant velocity Needed to stop lowering it Put on the brakes That's the same as an upward acceleration Of those two points But due to an imbalance in the winches The acceleration wasn't quite the same So that causes an angular acceleration Of the beam in that direction It's going to tilt such that A is higher than D An instant later So we want to find out some of those things We need to find that angular acceleration But then also find the acceleration of points A and D We know the acceleration of points B and C Want to find out the acceleration of points By looking at the acceleration Of what it would cause that to do Because of the greater acceleration of point B Than point C A second later The beam is going to be like that Which means it's rotating in that direction It wasn't before It was lowering at steady rate So it had no angular acceleration Now point B is being slowed down At a greater rate than its point C So it's going to tilt then So you're going to go in this direction So it must have an angular acceleration In that direction What do you mean two different ones? It's a rigid body It only has one angular acceleration A split second later That angular acceleration may be different Because it not only has to do with the magnitude of these But how far apart those are And as it starts to tilt Those aren't the same distance apart As they are at this instant It will come to a stop And the window washers are gone And we're on the news Alright, so see what you can do with it You need to find alpha in one way Once we have alpha Then we can use it for any of the parts Because alpha shows up here Might need an omega squared A couple different ways you can do it It should come up with the same answer either way But you need to relate it to the acceleration Of one or the other of those Because that's known Different ways you can do this First part find an alpha And then once you find an alpha We can do the other points Because we'll need those for those particular points You got it already? Let's see In the unit you want This is getting serious This is doing good thinking You're cracking knuckles So let's see Maybe I'll get you started Let's write down the acceleration of point B Relative point B Just setting things up Relative motion equations With all the different formulas We know the acceleration of B If we can find the acceleration Of D relative to B in the tangential direction That could give us alpha Because we know the distance between them The acceleration of D relative to B In the normal direction We could find We know the distance between them We need the angular velocity of the beam Which we don't have So is this going to work? It is Remember when we did free fall We dropped an object as soon as we let go of it It instantly has acceleration But it has no velocity yet Because it hasn't had any time That acceleration is there But it hasn't had any time to work That's the same thing here Rotationally It has this rotational acceleration But it hasn't had any time to work yet It's still level It was level just before So this is actually zero Because omega is zero at that instant So since that depends upon omega And omega is zero So we can then set that up We could have done it for other points Because that would have happened with any of the points I just picked B and D Because we need to accelerate some point D Sometime anyway So in no sense doing this between point D and C You can find alpha But then you still have to back out You can find D anyway What are you thinking of that? Claire Hoyn's voice told you That was the case No, this is how you're going to find alpha Because alpha is in here You know this And you know point D Oh, wait I've got different letters in here Let me reuse you Point B Hang on, hang on, hang on Which is easy Change the letters here Or change the letters there Over there Okay So this is Because I wrote A, B, D and E I don't know where point C went on this My notes here So this is C Yeah, we're doing point C relative to B So that makes more sense It doesn't materially Materially change anything It just makes more sense I guess now Because we know the acceleration of C We know those two accelerations D and C That makes more sense Now we can find alpha We know these two We can find this as a function of alpha And that's the only thing that we don't have So remember these are vectors So this is 3 feet per second squared J B is 5 feet per second squared J The beam was moving downward Now they put on the brakes So that's equivalent to an acceleration upward And then the tangential relative acceleration We can do this way You can do the cross product with your right hand And see where it acts That will give us a chance to find alpha So this is alpha Times the distance between the two points Which is B and C Which is 4 feet And then the direction Alpha cross C to B C relative to B So that's back, back So alpha's into the board Put your fingers in that direction Into the board You've got to face your palm towards point B And that forces your phone down It's minus J So we have a plus J Everything's in the J direction So it all adds up together really well And you can find that alpha In fact, we can cancel the J vector So 3 equals 5 Minus 4 alpha is our equation Because everything was in the same direction And the unit vectors would cancel As well as all the units Would you bet minus J? From doing this Alpha, alpha With that kind of rotation Angular acceleration is into the board So alpha's into the board We do alpha cross Not B to A But C to B So that's alpha across C to B That's minus J And so alpha then equals 0.5 Actually the units don't all cancel Because we only have feet there And feet per second squared there So the feet cancel We get 1 over second squared there 1 over second squared there And that will become radians per second squared So the units do work out You just have to be careful with them as you do Okay, now we know alpha Now you can find the acceleration of A and D And you got 6 minutes to do it You can do it David's already done Because David knew I was going to ask that He did it yesterday That's the way to do school David didn't Just work a couple of weeks ahead And then stop coming You know what I'm going to put on the final exam Just write it out and give it to me You're all done It's easy being clairvoyant Alright, so now we can find the acceleration Of the other points Let's see, we'll do one of them I'll give it to you as a get out of class question Actually, when you do one of them You can actually Yeah, I'll let you do one of them And then I'll show you an even easier way to do it But David already knows Well, it's graphical In nature So you don't draw pictures You think You either think in pictures in your head Or you think in words But you don't draw them down Right down pictures very often So acceleration of B Could be or acceleration of C It's just one we know And then the usual two components The usual two components we have to pick up there We already know the acceleration of B It's plus 5J It's in the same order AB AB So the vectors won't be the A Which vector? This vector? Yeah, we've switched the order of the letters here So we have to switch the order of the letters everywhere else To get the right vector What about this component? Zero So Anthony was smart He did the easy one first And I can just take the rest of the day off And let's see acceleration of B we've got We just need to do that Last little bit here And so it's R, AB What makes sense? Alpha See alpha's into the board The position of A Relative to B That's that vector Alpha to B So that's up So that's Plus J B is also plus J So this is going to be even greater Plus J So what's R alpha? For that one 1.5 Yeah, so 6.5 Yeah, so this is 1.5 Feet per second Square J So the acceleration of A Is B Which is 5 Bore Is 6.5 Feet per second Square J Now, two minutes left Well, without doing this Could you find the acceleration of point D? Without doing the cross product Is that can you do it By looking at the picture now that you've got three vectors? In fact we could have done it with two of the vectors We've got these two points We know they're accelerations That one's 5 That one's 3 This one's We now know it would be 6.5 It's not going to work Unless you do a scale drawing Which this isn't quite Phil, you see it? You're nodding, Chris You see it? If we draw a straight line Through these points That will give us the acceleration Of that other point The sketch won't work Well, you could draw it to scale and do similar triangles And get it But you get a little tiny bit left over here I believe it's 0.5 Feet per second Square And you could prove that Just by finding the acceleration of that point Oh, it's 1.5 Sorry, 1 That takes us into the weekend It's time to relax Oh, that's One thing at the end of the video Oh, that's the ANR Yeah, that's the one Any questions? Take some practice Be methodical about this Because if you do it too quickly You're going to lose a minus sign You're going to get A direction and an I direction And we're going to get a square Be patient with these The negative, remember Was with the direction On the K vector We only need When we do this We only need the magnitude We've already got the direction As opposite to that So you don't want to do Well, here's the square So it doesn't matter The tangential direction is already defined The minus 20 Was part of that This is pure magnitude These numbers