 All right, any questions? To try some of the incident center problems? No? They go okay. They go okay once you get used to them. Just make sure that when you're looking for the incident center, you don't use the velocity vectors themselves. You use the perpendicularity of those. It's very easy to get kind of swayed by the problem, visuals of the problem, and accidentally use the wrong center. Sometimes it comes out okay, just because the center happens to be, you know, the problem has to be symmetric in some way. All right, so we're looking at our last little, well, second-last piece of this, I guess. We looked at relative velocity. Now we're going to look at relative acceleration. And as we've seen, the acceleration is a bit more subtle. It's very easy to see velocity when we talk about something moving or even when I have just one of the little pictures, but acceleration is not quite as obvious. So we're going to have to trust the equations on here and not try to outthink what's going on here, Jake. You're going to come up with accelerations, and it's just not going to be obvious to you why that was the case. But that's because it's very difficult, at least for most people, to observe some problem and understand what the acceleration is as easily as we do the velocity. So we have our relative velocity equation that we've been using some in the last little bit, and it shouldn't be any great surprise to use it. If we take the time derivative of that, we get a relative acceleration equation. Talking about here is two points, A and B, on some rigid body. With this body undergoing some kind of general motion, or other, such that the two pieces are following some path like that. I guess I wouldn't do that, because that wouldn't be much of a rigid body if those things are farther apart. So whatever it is they do, A at one moment has some acceleration. That's what this piece is. Simply the translational acceleration of that object at any one point. B might have some other acceleration, and the difference in the two of those give the object both angular velocity and angular acceleration in some direction. Problem dependent, of course. And this one is that acceleration of the point B itself. This piece over here is the one that's going to take a little bit more work. This is the acceleration of point B relative to point A. If you're sitting on point A, looking down at point B on this rigid body, that's the acceleration that you're going to observe. It's going to be apparent to you. Now, the reason that's more difficult is because even though this distance does not change at all. This is a constant distance. That doesn't mean that there can't be acceleration of B towards A along that direction. If we had some rigid body that was actually pinned at point A, then point B is going to go in a circular path from point A, and you know from looked in Physics 1, anything in a circular path is going to have a centripetal acceleration. We called it A sub C. That, however, is going to be one component of the acceleration of B relative to A. All we're going to add to it is the possibility that A itself may already be accelerating. And that is going to take B with it, and so we're going to have to put all those pieces together. So that's part of why it's a more subtle piece. So let's see here. We've got this acceleration of B relative to A. These two generally are a more direct part of the problem. That's just the type of thing we might have looked at in Physics 1, but it's this little part over here that needs more attention paid to it. It's two parts. It's the acceleration of B relative to A in the tangential direction and acceleration of B relative to A in the normal direction. Remember where the tangential direction is defined by whatever direction it happens to be moving. And the normal direction is perpendicular to that towards the center of curvature. Towards the center of whatever curve it happens to be on, even if only momentarily. All right, we've already looked at both of these pieces a little bit before, but let's go over it again. This piece, the tangential piece, has to do with the fact that the object has its own acceleration, rotational acceleration, and the fact that there's some separation between the two. So we can find it by doing that cross product. You can avoid doing the cross product if you already know what the tangential direction is, which in some problems we do, we recognize that a lot of these problems we do are kind of ordinary in the way they're laid out, lots of 1945 degrees in the way. So either one of those will suffice, whichever is easy. It's cool to go through a couple problems and pay attention to that. This one, the normal component one, if you remember, was made up of a couple parts. Has to do with the angular velocity. Remember the centripetal acceleration does have to do with the velocity of the object. If you're in the cross products, that'll be your favorite way to do it. What is in the cross products is that you can already also put those together. For the magnitude, remember that's our little piece that comes out as v squared over r. That will be in the normal direction. So if you already know which is the normal direction, you can say you're doing the cross product by doing that. Or perhaps even easier than that on certain problems is if we recognize that the r, b, a is a vector itself in the opposite direction of the normal component. You can use that advantage, that back to your advantage. So that might be a little bit simpler that way. Any one of those three will work. They're all just variations on a theme, if you will. Whichever one you like, whichever one works best for a particular problem. Is that the last one that you just wrote up? One way is that one, r, b, a is opposite of what? This is because the vector r, b, a, see I need more chalk colors here. The vector, the position vector of b relative to a. If you're at a, where do you have to look and how far do you have to look to c, point b? That is in the opposite direction of the normal component. So that's why e n becomes n a minus if you use the position vector, rather than the position magnitude and the normal unit vector itself. e n points, the unit vector points in the opposite direction of the position vector. All of this based upon a being the reference point, which is just arbitrarily chosen for the picture. Okay, let's put it to work. We'll step through this carefully a couple of times. We already did this problem as a velocity problem, so we're throwing some acceleration and see what happens with it. We did the problem where we had a gear rolling on a surface with another gear attached to it, and then there was a wrap running on that upper gear. Those are end of the term circles. So make yourself a nice big drawing because we're going to have to get a couple things on it. Remember, we did this problem, and we knew the velocity was going to be called that point a. We knew the velocity of point a, and then we figured out the relative velocity of some of the other points, such as the rack itself and some other parts to it. So let's see. I think we call this point b, this point c, which is good, it actually is an instantaneous center. And then we also had over here for a big challenge point b. All right, we're going to throw into it the facts that this whole thing might be accelerating too. So we'll give it some acceleration where this gear itself is accelerating down there. Of course, since this is gearing, it's a no slip condition. All right, the radius of that I think was 150 millimeters. The radius of this one was 100 millimeters, and we had velocity of a was 1.2 meters per second, I believe. Well, we can easily make a full vector out of that by just putting it in the high direction. All right, when we did this, we came up with a couple other things. We also found the angular velocity of the wheel. That, I believe, was 8 radians per second when we did it. And let's see, that would have to be in that direction, or vA to be in that direction, and that makes it then minus k direction. We're going to need some, at least some, if we don't actually do the cross products, we're going to at least need to know what the directions are so we get our minus sign. And then I think we also found that point b, which is the velocity of the rack itself, was, I think, 2 meters per second also in the I direction. That's just all bringing it back up to where we left this problem a little bit ago. All right, let's find a couple things. Find, of course, the angular acceleration of the gear system itself. We know what the acceleration, linear acceleration of the center is. We want to find out what all the other parts are. And then, of course, find the acceleration of the other three points, v, c, and d. All right, I guess the easiest one to find is the angular acceleration itself. Since point c is an instantaneous center and point a has acceleration of a, then we know that that's going to then be the acceleration of a over the distance between those two, which is just the radius of the outer wheel. So that's something you could have done. Oh, I didn't give you AC, sorry. Let's use an acceleration of the, not c, sorry, a, point a. Acceleration point a is 3 meters per second squared in the I direction. So not only is this thing moving along the ground, but it's doing so faster and faster. So we've got all those pieces. You could have done this back as physics wanted, but I've got to go slow because Frank overslept, so we have to bring them in gently. You guys are all warmed up. What's that come on? 20 radians, second squared. In what direction? Because we are going to need it as a vector, whether we do the cross products or not. We want to at least want to get all these things straight. Negative k. So in the negative k direction, I had it all written down as magnitude, so I'll just pin that on there. Okay, so that was the easy part, I think, but we're going to need it. We need alpha for some of these parts, so let's start going through it. Let's do b next. The acceleration of b is going to be the fact that it's rigidly attached to point a. And point a itself is accelerating, plus the fact that the object itself is also accelerating, rotating. That's just our relative acceleration equation. Acceleration of a, the first part, well, we've already got that, so we don't have to worry about that. We'll just bring it in when we need it. But we do need this, and it's made up of those two parts. So we're going to have to do both of those. So one at a time, make sure we got the pieces right. Acceleration of b relative to a in the tangential direction. It's alpha cross rb to a. Let's see, so alpha, we've got it as a vector. And then the vector rb to a, sorry, a to b is, let's see, a to b would be just that little piece there. So probably as easy a thing to do as any is to figure out the resulting direction of this cross product and then just multiply the magnitude rather than doing the full cross product. But either way will work. So let's see, the resulting direction of this cross product, take your right hand, stick it in the direction of alpha, which is into the board. Stick it in the direction of alpha oriented so that you can move your fingers in into the direction of this position vector, the vector a towards b. So into the board, the only way I'm going to be able to get up to the vector in that up direction is if my palm is facing up, that gives my thumb a direction that way. So I know that for this particular component, the tangential direction is simply the i direction. I didn't see many right hands up there doing it with me. Fingers in the direction of alpha cross those into the direction of that vector a towards b. And that only way you can do that is if your palm is up and your thumb is in the i direction. And now we only have to multiply the magnitudes. So it's just going to be alpha r. Alpha we know to be 20 radians per second. The distance from b to a is 100 millimeters and that's in the i direction. The product comes out to be 2.0 meters per second squared i. Comfortable with that? Yeah? No? Why isn't it? If you do the cross product the top line is between 0.10 and the bottom line is 0.0 and negative 20. So why isn't it negative? Should this be negative? You're saying? Well we can do the cross product. Let's check it. It's going to be negative k. Alpha is 0.0 minus 20 because it was 20 in the minus k direction. All of these two lines? Yeah they've got to be in the order of the cross product. Plus as you know even if it's a very simple cross product as this one would be it's still loaded with minus signs. So if you can envision the direction resulting from that cross product by just doing this simple right hand rule then all you have to do is multiply the magnitudes because you've already got the direction and you can save the trouble and the danger I guess of doing the cross product and all its subsequent minus signs and possible troubles and all that. So everybody okay then with the two meters per second squared i? All right let's look at the next part, the normal component. Acceleration of v relative to a in the normal direction. Two cross products there, we can do them and we'll come back to them in a second. It may not be obvious what the normal direction is but this is a pretty obvious way to do it because the position vector of a to b, v relative to a is pretty easy to come up with. So omega squared we have the minus already in there from the direction like minus 8 radians per second squared. That's the, that's the omega squared. Don't really need the minus sign, that was with the direction anyway but it squares out. In fact let's emphasize that we only need the magnitude here, we don't need the direction. Just the direction takes care of all the cross products, that's already in there. So the position vector b relative to a is this little pink one I've got drawn in here. It's where b lies with respect to a, it's got to be in the plus j direction and it's got a magnitude of the 100 millimeters. So it's just .1 meters j. We don't need to do the two cross products, we don't need to figure out ahead of time where the normal direction is because this vector itself is so easy to come up with. Write out the drawn nicely for us and then straight up j direction. So then this becomes minus a 64, 6.4 meters per second squared. Alright, let's put that together. For that inner wheel there's point b. The acceleration of that point is made up of the fact that it's attached to a and a is already accelerating. We know that to be 3 meters per second i. That's the acceleration of a. That's this first part. I'm going to do this draw up the components there, put the vector together, we'll see what the acceleration of point b is. Then we have the acceleration of b relative to a which is also two parts. Another two meters per second squared in the i direction. So that's about two thirds as long. That's the acceleration of b relative to a that's the first part we did right here. Add to it this third part which is minus 6.4 in the j direction. See that's 3 plus 2 that's 5 so we've got to go that distance plus a little more in the minus j direction. That's the vectorial picture of this very first equation to find the acceleration of b. So we know the acceleration of b absolute acceleration of point b at that instant would be something like that. Not I don't think what you would have guessed ahead of time based on intuition. Would you? I mean it's magnitude well it's the square root of 5 squared by 64 6.4 squared so I think I can have it. Yes the whole thing is 812 that's 52 degrees and the magnitude is 8.12. Take your header a little bit. You missed doing the cross products? No? A lot of these especially when we have things laying in 90 degree positions to each other a lot of these cross products is not necessary to do them you can get past them a little bit. Pat okay? Yeah you like that one? No. Let's open and bring it around for the end of the term. No I don't want to. Alright any questions on that before I erase that and we take a look at the acceleration of point c No Jake you don't want them? What else is there to do? Go lie in the sun? I can't get spring fever. Now the thing well here we go we got to do the acceleration now of point c we now know the acceleration of point b from there and the acceleration of point a so we can do point c relative to either one of those points we could do the acceleration of a plus the acceleration of c relative to a or we could do the acceleration of b since we now know it plus the acceleration of c relative to b either one of those would work and they should give the same answer however very likely one of them is easier to use than the other acceleration of a is nicely in the horizontal direction that's made things pretty simple before acceleration of b we have now this vector that we need to use so it's probably true that the first form would be easier to use than the second form but either one would work you have your choice acceleration of c relative to a but that's probably going to be fairly regular as with that because they all have to line up so let's go through the easier way to do it I know that at this point path the acceleration of a we've already got we don't need to do that where that piece comes oh there it is up there blue so the acceleration of c relative to a we need to do these two parts like we just did so tangential part you can do this cross product no trouble or you can predict ahead of time what the tangential direction is going to be by doing the right hand rule on that cross product getting the direction and then just multiplying the parts so you tell me what you think the tangential direction is don't say it out get everybody a chance to come up with it then all we have to do is put in these magnitudes and those are pretty easy we've got those yeah better see those right hand up there curl in some fingers we're going to put them in a fist where's what where do you want the fingers to curl towards work do this fingers start in the direction of alpha factor which is into the board so you're going to start with your fingers pointing into the board right hand cross those in the direction of well not B A but C A where's point C relative to point A here's point C here's point A so that's the vector R C A your fingers start with alpha and then move your hand however you need to so your fingers can next go to R C A in the shortest direction so my fingers are into the board with alpha then palms got to be down to get to R C A that puts my thumb in the minus I direction and you can do the cross bother it's not a big one you can do the cross bother you should get the same thing so that distance between those two points is just the radius of the outer wheel alpha we know to be 20 radians per second squared and we know that this is all in the minus I direction now from doing the right hand rule on the cross product ourselves so that's minus per second squared minus 3 meters per second squared I did you do the cross product really no are you ok with doing it this way alright so point C well we'll draw it together when we've got all the parts it's kind of we're going to do the same kind of drawing we'll get the point B over there alright in the normal direction again I think the easiest way to do it because this vector is going to be so easy to find is the normal way that vector C A I already got drawn there so did you do the same thing to see that no we already did that we've done imagine these cross products and came up with this form we can do it we can double check no no I get it now I think you want to double check oh yeah so omega is going that way so omega is also into the board so we do this cross product first only we have a C instead of an A so omega crossed into RCA is minus I direction so this cross product gives us a minus I direction we cross this with it which means omega into the board cross with the minus I direction puts our thumb straight up to plus J this is minus J this makes it plus J just what we thought we'd get from the cross product why is omega into the board shouldn't it be going like this because if you curl your fingers the way omega is going your thumb's got to be into the board this is not the vector omega because I could have just as easily drawn it down here going that way but the fingers curled in that direction your thumb's going to have to go into the board and so that's the minus K on the omega does anybody want to do the two cross products for us just to double check there'd only be about 7,000 minus signs in there I think all told but maybe 7,001 but you can't miss any of them because you'll get a negative or a positive when you don't want it alright so finish that one because otherwise there'd be no reason to put it down the negative 8 squared is a vector RCA RCA was a big company when I was a kid because even though it was RCA it was a company no RCA RCA is electronics actually I think they I think this they were a cash register company to start with when they first did adding machines mechanical adding machines and cash registers which you also probably never seen none of you have ever been in one of the really old fashioned hardware stores where you give your money to the person they put it in a little cart attach it to a train system overhead and it shoots all the way around the store and up to the cashier that way all the money is off the floor so they can't get robbed then the cashier takes out the receipt the money puts in the change puts receipt back in puts it back on the rail system and comes shooting around right back to the cash register where you were the guy reaches up takes it down gives you your change it's really cool but those don't exist anywhere it used to be one that's connected but went out of business finally there's a type of wonder you go in there and you say I need a 32 widget 407 stainless titanium screw and the guy goes I got that over here he goes to the drawer pulled open knew exactly where everything was man those were the babies now you go to Earl and you get exactly what you need to do and it's massless and stretchiness alright what'd you get for RCA no the vector RCA let me just put that in first 0.15 meters because it's the radius of the outer wheel that's how far the outer part and then what was the vector on that negative j negative j but that comes out to be what two negatives that's positive 9.6 meters per second squared in the plus j direction that's that's what we figured out with our cross products no yet this first cross products negative i then you cross it again you cross the negative i direction with omega so you have omega crossed into negative i is plus j there's two cross products here you can do them if you want you should still get the same thing that that bottom way is just the result of doing that cross plus a couple times so are you comfortable with that then j j okay I see this probably try to do this on my own I know I just make normal mistakes with the right hand rule yeah I'm not very comfortable with doing this it's very difficult to visualize you could try it with your left hand rule of doing the mirror I think there's an example of how to do some practice it does it takes some practice but you also plain and simply you've got to stick your hand out there if I if I was doing this class for a bunch of engineers trying to review for the professional engineering exam or something they all have their hand up there doing this don't try to do it in your head stick your hand up there turn it where you need it however you need to do this right but I guess I don't understand where I need to do it whatever it tells you start with your fingers in this direction and then so that's we can do this one again alpha is into the board so you've got to start that way no you're not using your thumb at all yet you don't use your thumb until you've done this put your thumb away get your thumb out of you go whatever you say that you understand then just to do this example we got laid there vector B relative to A from A to B is in the plus J direction so keeping your thumb put away fingers in the direction of alpha into the board the closest way to get to that direction is to with your palm straight up going in that direction then your thumb has got to be minus I remember you've got to remember this is this is all about 90 degree angles too this is an orthogonal coordinate system so you always um in the direction of your angle the fingers to the and your crossing okay if we're into the board we could get to A to this vector by going the long way around but that's not how we do the right hand roll cross product we do the shortest distance so in there 90 degrees up to there the only way that's going to work is with palm up thumb that way so positive I direction sorry? positive I direction yep and that's for for that vector as written down that's what we had positive I direction okay that was that's this vector right here positive I direction do what you were saying we do the second the normal direction again the right hand thing I don't understand what you're doing it's a crossing point this one okay okay alright we'll do this one again so it's in parentheses you do this cross product first and we'll do this one as written down omega is also in the board minus k so put your thumb away fingers into the board for omega then we have to cross it into well the same vector we just did this rba which is up so into the board the only way my fingers can get there is by turning palm up so that puts this first part in the in the I direction plus I direction then we do this cross product again with that plus I direction so omega is into the board put your thumb away turn your hand so you can move your fingers to the plus I direction that puts your thumb down and so that's why we got the minus in the opposite direction of that position vector so having done that now hopefully believe in this because this is usually an awful lot easier because this position vector is so easy to come up with rather than coming up with it anyway and then doing two cross products so are we alright? yup that went to throw up I guess God's eyes out good shower come back alright so we alright with that everybody comfortable with the two components we have now be careful with them take your time there's not a lot going on there other really than these directions but that if you get those wrong you screwed everything up so here's point C right here at the bottom its acceleration is made up of part of the acceleration of A we know that to be 3 meters per second I no acceleration of point A that's the first part here the velocity of B was 2 but the acceleration of A is 3 then we add on to it a minus 3 meters per second squared I which is well that's what we just put down it just brings us back to where we were so those two negate each other that's a I don't know C A what do I do first? T and then we add on this third part so that's the acceleration and the total acceleration of point C at that instant first two parts cancel each other A A plus A C A T we're saying magnitude opposite direction Colin, okay I don't think you can sit way over there and hide from me Pat should you go through a wall? that's what they thought he did they all thought that I don't know probably when Colin Pat's got a strong constitution he's a pretty tough guy I never do that Bobby, okay? because you guys are going to do point D I've been working really hard here I'm not a tough guy now we got don't you which not yet we still got a whole half hour I got probably eight or nine more problems I can pull out last question where the half hour ago she was with alright so you come up with the last part we're looking for here acceleration of B now think about it a minute because you could do the acceleration of D based on the acceleration of A, B, or C because you have had three of them now one may be easier than the other but they should all give the same answer so I think because Pat is a tough guy he'll base his on the acceleration of B this one definitely on A I don't know which one's easy generally the orthogonal direction is already C wouldn't be bad I don't think A wouldn't be bad B I think would be wouldn't be hard you've got the components all there but why bother so why don't some of you do C and some of you do A will they compare it? I'll do C I'll do C when you do it Colin you should do C Colin C C Colin get it? but if you don't do C you're going to get a C you don't buy that thread because Frank F for Frank that's why he's not even trying how do you do that? oh Bobby's getting a D he's getting a D Jicks getting a J that sucks that's how it gets an A A out they have everything in their right hand up there you're going to do correct with respect to A I'll do C trouble is though the position vector D to C's a little bit more trouble but it's a 45 degree angle so it's not a big deal A is definitely the easier one position vector's easy acceleration vector's easy if you're using this alpha cross r D over A this part? okay in this case we have a D instead of a B alpha's a B negative is no D and D so that's the vector from A to D that's this vector then I should have brought my whole set of chalk so omega into the board put your thumb away until you can find that vector which the only way that's going to work is if you're pulling in that direction that vector that gives you thumb straight up I don't know if it's worth doing C that's going to be confusing for you silly work for me why don't you get your out hit straight down Bobby plus J everybody got the tan here per second J minus and then this position vector D is in the minus J direction so that's two minuses so we have the minus from this and then the minus from that and we can check it because we can do omega cross D A and then omega cross with that plus A that's not very convincing do you think there's only one minus sign here it's like one sign is minus up there we have the point variance per second out of J but it was originally negative and now it's not negative what's originally negative and now it's not negative K this is the magnitude you're not putting in the alpha vector you're putting in the alpha magnitude there's no minus sign it's the magnitude of this vector not the direction the magnitude so that's just that length that's just that length and the J direction comes from doing this figuring out what the tangential direction is by doing that cross product with your right hand the minuses are all part of the vector directions not part of the magnitude so if you put in the minuses again for magnitude you're putting in the direction twice for each time we do these if there's no vector sign we're only putting in the magnitude itself be careful with these it is tempting to overthink these it jacosizes there's point D it's acceleration of made up of A which is 3 meters per second in that direction that was given from the very start we have another I component that's 9.6 so about 3 times that added on 3 in the J direction maybe something like that so the acceleration of point D is something like that with this angle is 13.4 and the magnitude is 13 careful with the acceleration absolute acceleration of these things because I don't know about you I would not have anticipated that on any of these problems B going down like that C going up like that D going up like that so there you go that would make a awesome tattoo it's colors and everything I want to get at a class question let's see hard one or green would have been nice here though too here's the test to deal with the earth as an orbit around the sun so there's the earth Pacific Ocean Atlantic Ocean Africa South America oops go the other way sorry chili alright find the acceleration and if you were launching a rocket you would have to do exactly this type of thing find the acceleration of point A at that instant that's all the equation did no I'll give you anything else you want and we can use just for reference for the vectors at that instant use that correction system next slide so tell me what you need when you get there that's a much better circle find the acceleration of point A find the acceleration of point A as a vector I want to know the distance to the sun distance to the sun from the center of the earth at the radius of the earth's orbit we're going to need the radius of the earth as well sure you're anticipating that so we'll do that radius of orbit is 149.6 times 10 to the 6th kilometers what else do you mean you know earth's radius 742 you know that it goes once around in a year you know that the planet goes once around in a day you can come up with all those pieces what I'm teaching a little bit of geology a little bit of planetary space science a little bit of you're protesting just going to wait this one out yeah I'll speed things up I'll give you some of these intermediate numbers the velocity of the earth in its orbital path is 107.257 kilometers per hour you know that from the circumference of the orbit divided by the fact it's one year so that's yeah that's kilometers per hour then have the the uh okay that's the only one you really need to calculate out that you need it doesn't it doesn't why not because you can easily find that potential velocity in the radius the acceleration of the earth or you can find a point A by subtracting the radius of the earth the radius of the orbit it's like v squared over r you're subtracting the radiuses yeah I'm saying v squared over r where r is the distance between the sun and the planet well however that point is not going in an orbit like that that point is going like this yeah I did not I forgot that the earth was spinning can't believe I'm freaking out if e is the center of the earth let's say the earth is going that way no it's like a spin doesn't matter what else do you need though the you need this part of it we're going to need the position vector that has nothing to do with whichever way it turns omega you'll need of of the line A to E which at that instant is just determined now by that direction so it's everything it got you don't need any particular direction of the direction of the rotation of the earth actually we're just planning to jump over if you were launching a missile you'd need to know the acceleration of the launch point itself at the moment of launch which of course is always at noon that's why the launch places right face in the sun acceleration of e but the speed of the earth is constant for the most part we'll take it as a circular orbit so why is there any acceleration of point e the centripetal acceleration so that would be then the speed of the earth over the radius of the orbit in which direction you use the coordinate system I put there that's in the negative i direction and then the tangential component is r a relative to e alpha e t what's alpha alpha is the angular acceleration of what now point a itself it doesn't have an angular acceleration only a rigid body has acceleration angular acceleration and it's the rigid body on which these two exist remember this equation can never be done on anything but a rigid body so it's the angular acceleration of the earth I'm going to confuse by this term acceleration a relative to e e I'm going to confuse by that good question well put too how did you get that how did you get to r alpha there that's the tangential component I haven't gotten a normal component that's this broken into tangential and normal alright what is alpha it's the angular acceleration of the earth which is zero the earth spins at a regular rate it's not increasing in speed minus omega squared that's the rotational speed of the earth which is once around in one day times the vector in this case a relative to e which is 11 seconds to finish if you want to leave today you have it this is an impossible problem to do you have it in what kilometers per hour squared or meters per second part of it that's the acceleration of the earth you have to also do the tangential component zero like you just said that's the normal component oh go ahead and go we'll talk about this on Monday oh you got it before I'm trying to figure out the direction of omega it's going counterclockwise always then you don't need the direction of omega you just need the magnitude of omega oh but I thought we had to omit this cross product here we're not doing this cross product remember we reduced the cross products down to this much similar term we're going to find out the direction going out of right hand world oh like omit I'm thinking of omega since it's I'm thinking that's counterclockwise so we're going to find out the direction of the cross product oh no let's see omega cross here to that position right there if omega is going the other way it doesn't matter it doesn't matter you get the same you get the same direction either way so when you go plus the i direction no matter which way the earth is turning this way or this way why? well let's first assume that so omega is into the board right? alright so wait am I rotating this clockwise? like that so omega is into the board so omega cross the position vector that's the position vector a relative to e so omega is into the board so omega crossed into a is straight up plus j then we do omega cross plus j omega cross plus j gives us plus i omega is into the board alright crossed with the plus j which is what we got for this just now so omega cross plus j is plus i so that results in a plus i now the other direction that makes omega coming out of the board so omega cross with this vector I have to turn around because now omega is out of the board crossed over to here alright so that makes this part minus j so now we have omega out of the board crossed with minus j so cross to minus j plus i because when we reverse directions we reverse direction on that twice so it's 2 minus 6 which is a plus so we got the same thing alright there you go sounds like you do oh man I didn't stop