Center of Mass, Moments and Slicing

Okay, I begin. Ah, could I have it? Could I have your video, please? We begin with the usual question: Any homework? Ah, sorry, any question from homework or from the previous session? We're about to finish talking about the application to the integral to volumes. Any question at all? Give me a quick answer.If it's a no, and still say no, so that we pass up. No. Oh, sorry, I didn't. Um, my speakers was connected to the wrong thing, so I couldn't hear you. Oh, sorry, I was asking the usual question for starter meaning any question from homework, any question from other previous sessions, about the material in general.No, I don't think so. Not that long, huh? I'm finding my pen. Alright, now we're about to finish doing the volume. The whole point is cutting it into pieces. Now I want to show you something interesting. This is a later when you learn physics. There's a trick to find the center mass. So today it's about a clever application.Of something, it's not the AP requirement, but it's so useful and it's a good exercise. So I do want to cover it. It's not going to take very long. This is a quick way to find the center mass. What do we mean by the center mass? Yeah, generically, I'm going to define it just overall. By the way, what happens to Eddie? Did he actually say something or not? Not being able to come. One second, I'm quickly checking.What's happening on WeChat? Yeah, he's coming. Good. Hey, good to see you, Eddie. We're just getting started. I was about to send you your mom a reminder on WeChat to see what you're coming in, and good to have you. Well.Today it's all about a simple application to center of mass, and it's useful in physics. It's also a straightforward integral application. The definition of the center of mass, intuitively, we all know what it is, but if you're given a region, let's suppose at the very beginning this is a two-dimensional first, and the way to find the center of mass is to set any but anywhere as the origin of the universe. For example, I can pick this is the origin.And for every single particular little region on the disk here, and we can find the so-called displacement vector. You can call that basically the r, or regarding the point p now, and or you can just think of it as a complex number, or you can think of this actually the x y coordinates separately. Then the definition of this center mass rcm now as a vector, and again you can consider.Is it as a complex number or a pair of coordinates, which is the x m and the y m, would simply equal to? We're actually weighing every single displacement r p multiplied by the dm now, but the so-called mass. If we're only doing geometry, in the future you could load the area with a different mass density, then you really have a physical center mass, but you were only looking at the geometric center mass now. So suppose the so-called. Mass really just means the area, so you could think of this actually as the area. You do such a overall integral. Fundamentally, what you're doing is that you're weighing the RP by the area. But when you do the integral, you have to eventually divide it back up because after the integration in units, you don't get meter, you get meter cubed because the area would bring in the area. So what would you do? You divide by the total area.This is the meaning of center mass. Well, to have an intuitive understanding of that would be why don't I multiply both sides over, and the RCM is going to serve as actually a representative, as if all the total area is divided a single point. Now, I would say we do have some geometric intuition about for this region where the center mass would be. Usually, if you are looking at a neat geometric region, the center mass would be.Exactly at the geometry center. Is RCM like the length of the R vector? Yes. No, it's not the length. It's a whole vector. It's a vector itself, not just a magnitude. It has two components, or you can think of it as a complex number having the real and the imaginary parts. So if that's the case, you can think of its representative as if the total area multiplied by it weighing.That RCM now really equal to the entire integral DA. Well, I want you to be sort of sufficiently confused at what's going on now, but we want to look at several intuitive cases. On, for example, if I just give you a route which is placed this way, this is a simple route. Okay, where is the center mass intuitively?It's a place where underneath you put an fulcrum, so that the whole rod would be balanced. The middle. Yeah, exactly at the middle. How do we bury it out using that integral? I actually want to do it to demonstrate it is true using that integral here. Then what would we do? Well, there are several ways to do it.For example, without loss of generality, why didn't we choose to set up the coordinate? I'm going to actually choose to set up the origin of the universe at the center and choose my axes along the direction of the road. It's my choice. And then we are to prove that RCM would equal zero, and not mention it's one-dimensional problem now. Before the mass or or the road, it's concentrated on the x-axis. So this is really.Reducing to the x m would really equal to the integral of the x, which is really for arbitrary point p, and this p would really go all the way from negative half of the l to positive half of the l, right? If you consider integrating over all the spread of the point p now, and then times the area, but the total area is just a lump, so you can actually the dx, and then you're integrating from negative.Positive 就完底。Okay. Doesn't make sense. I'm so confused on like the RCM formula. All the integral. The RCM is a weighted average, so you're adding basically. Oh, think this way: before we do the integral now, why don't we think about the bunch of these great points.Okay, if you actually, I'm giving you here is, ah, these are mass points. Okay, this is a mass of the M one, and I to the location in our object definition of the universe, and this is R one vector, and this is the location of the second mass point. They don't have area; they just it's a mass point here. And here you got actually M two R two, and then you have actually another M three R three.And finally, there's going to be M four R four. If you look at the four mass points spread out as a system, and you want to sort of find the representative center, and then what you do, you do weighted average. You can actually look at the so-called RCM, just equal to the M one over. I'm going to actually define the total sum of the mass here as the sigma of all the M's going from one to four.This is the total mass, and you're gonna actually take the you put the percentage of this much on the R one, and add onto the M two over the total mass here, R two, add onto the M three total mass, R three, and add M four total mass of the R four.Does it help at all to really show from the discrete point of view what's the intuitive understanding of the social center mass? For example, if I start really really simply, I'll just give you two points here: in space, arbitrarily in space, I don't want to pretend that they are arranged on a pretty nice direction. So if I put five kg of the mass here and fifteen.g of the mass here. Can you intuitively tell me, or hopefully rigorously tell me, where is the center of mass for a two object case?```Yes, I don't think I really remember this part. It's not about remembering; it's about learning it. I have just shown you. And start building the intuition. And if there's something confusing, you ask. I would answer. Addy, what's your thought? Do you know the concept of the center mass?So, like, it's the weighted average of all the mass like in the object. Very good, and apply to these two points where is the center mass.12.0Oh wait, no, no, um, what what are the location of the point? I have drawn it on the board. You just illustrate on the board. How do we find well where that center of mass is? Now the system here, which are these two points. These are the only.Only two points we're dealing with. Do you understand this equation I'm giving above? It is weighted average. Why do we start simple? What if I really just aMake this a sentence.In a system, if you just add R one plus R two and then divide by two, now don't you just get the midpoint in between? Yeah. Alright, well, what if I actually make that fifteen kg? So intuitively, does the center mass shift to the point A or to the point B? B. Yes.How much closer? Where would be the center mass now? We won't give equal weights to the two anymore, but instead, according to the formula I'm giving you. What weights are we giving to the two points? ```We use five kg for the M one and fifteen kg for M two. Of course. Then what conclusion do you have? The five.Over twenty R one plus, um fifteen over twenty R two. Yeah, that's right. You do the weighted average. You give it more proportion to R two, but graphically where did that land? Exactly, it's commercially a quarter.Meaning that ratio would be one to a three. Remember, it's the weighted average. If you're closer to B, then you take the bigger weight to account for B. Now, that's my point. Does I come back to you? And I wrote it as a vector, which is fine. It could actually mean the two coordinates. Remember, if you're given the two coordinates of the A and coordinates of the B to find the where is that cross sector. Meaning, I would really just.The quarter of the point now. That's exactly what you apply to both of the coordinates, isn't it? Yeah. Okay. Now do we understand how that generalizes from a two point system to many point system? You still do this weighted average, and eventually it applies to a continuous distribution of point, so the summation becomes an integral now.Instead of summing up infinitely many points, now we write it here and as an integral. That's what the top is. Replacing the area, whatever the mass here, we religiously invested in a certain point as the infinitesimal area at that location, and eventually mapping this area or the mass into the area. The total this denominator.Do we see the comparison between this integral definition now and this summation definition?Yeah, you're still confused with. Uh. Uh. Is the area total the MS? Yes. That's equivalent. But I'm still confused on the numerator. The numerator what's the meaning of that integral? A integral is really a summation of infinitely many pieces. The symbol of da here would be actually really just equivalent to whatever the little mi that we had earlier. A integral is a sum. This is really.It's summation of infinite many such little ais, and you're doing the RP weighted by the AI, which is equivalent to the MI. Although each one because it's an infinitesimal piece, so it's a differential. And when you're adding infinite many pieces, remember, isn't that how you define integral? Yeah. So the numerator is really a limiting case of the discrete case.Okay. Are you like? Are you totally on the same page? I think I get.Why, like the sum turns into the integral, but I don't really understand like the numerator of the integral. Hey, well, correspondingly, this R I is the R I here, and this M I is actually the D A here. That grand sum borrowing from the numerator of each of this sum here, like you said, that is actually.Equal to the integral, and they all share the same denominator, which is the total mass. So called total mass, which is really the total area. Well, it's because all the points are like little parts of the areas, right? Yes, that's right. I think I get it now. Beautiful, I did you. Well, let's apply to a single rod now. You know, it's your choice to set up your coordinates.Neatly, so that you don't deal with a nasty two D case. So that's how we chose the coordinate. And then, do we understand now in this one D case, the so called D area? It's really length. So, do we understand this integral we have written down here, which is a direct substitution of the definition of the center mass?嗯，Can I think about it for a little bit? Of course.``` Let me post it when you think. So that I may be able to help you think.So the x is a x coordinate of the point. Yes, that's actually the RP. In this case, is reduced to a single coordinate. If it is two D, oh sorry, you go ahead. So this is one D. Yes, so the.Area would just be replaced by the length of the rod. Correct. Okay. It is actually the meaning of this is really the average of the function value. You think of the r as a function value defined by space now, and you're looking at what is the average over a area. So you do the integral and then you divide by the total area.Like usually, what you do by average value of function, remember, and back in our definition of the average function, y equal to the x now. Remember, we visualize it as a frozen river when it defrizes and the whole thing actually evens out, give you a certain height now. It's the same weighted average. And if you find the average now, don't you do the integral of the f of x dx undivided by the entire length of the integral from a to b.B minus a now. This is the same as as if we're actually thinking this other x takes a special case of the displacement. I'm replacing it by the location of this particular x now, which you can think of as the length in our case, and that gives you actually this. It just gives you the average location where the rod is.Yeah, and can we go ahead and finish the calculation, see whether it confirms your intuition. Okay. One second. Ah, I need to bring my charger. Sorry. I'll be back in thirty-five minutes.中文嗯。Thank you. Alright, I got zero. Yes, we're anticipating. That's only because we chose a coordinate very smartly. We're just confirming. That in fact, the center is nicely located at the center. That is the center mass. Had you chosen somewhere else as a coordinate here, then you wouldn't get zero. You're just going to get the vector pointing to the very center. Are we good? Um, I'm going to the bathroom. You're getting what? May I go to the bathroom? Of course. Okay, let's see. Alright, let's actually do a non-trivial one and.I hope this is also becoming trivial later, but at the moment it's an untrivial question now. And if I'm gonna actually give you this, ah, you know we're gonna start with a right triangle first. It's a very famous conclusion in geometry, but I do not assume that you know it. And you're getting this right triangle. It's a triangular board where the evenly distributed the mass now. I want to find the water center mass of the entire board. Well, in this.It is, it is decently easy to set up the coordinate system, simply because I give you right triangle. So that actually means maybe we can choose this point as the center of the coordinate system, and we're going to calculate what is the center mass cm and what is the y cm separately, because the the vector a definition I gave you, which is r cm, equals to the r vector p da and over the.Here area here applies to the two coordinates as a vector. It always carries the both parts, or you could combine them into a complex number. It's up to you. Let's do one at a time. If you want to find what is x m now, you're only concerned with the horizontal coordinate of the area. Then why don't we just organize them into slices? Because they all share the x m now. So we're gonna actually look at one slice of the area at a time. So this is my.CA```嗯，Can you repeat what you said about using the DA? We're saying that since I'm interested in looking at the weighted average of the x now, so I organize the area according to the same x.So I cut it into slice instead of a infinitesimal little patches. There, there's no need to to really do the integral point by point. Why? Because this whole slice here they share the same x, right? That's what we call the x p. Yeah. So you need to realize that your da now can be written. Now this is not one d anymore. Earlier it's one d, so a that was just becoming. In length, that's how you wrote down the dx. Now, now this is genuinely a part of the area, piece of the area, so you have to calculate what that little piece of area is. What is that da? Since we are interested in integrating the x, now it looks like x will be the integrating variable, so we aim at writing the da in terms of x variable. What we do know that this is a sliver of the base now, just a dx.But the point is, what is a height, right? Yeah, at this moment, you might realize something. It's easier had we set this point as the origin. Might be counterintuitive to you, but I'm going to make that advice now. We're going to actually set the origin backward, like this. You see, when you write your equation, would look like would it look nicer.So this is my x, and that's my y coordinate. So the x-coordinate y is measured from that side, which is totally fine.Sorry.We try finding the height by using Pythagorean theorem. Why do we need the Pythagorean theorem? Do we actually know the hypotenuse better than we know the height? This is similar triangle. Remember when you did relative rates? Now you know the big dimension. Right? Okay, you know the big dimension of the triangle. Now you know the base, and you'll find correspondingly, what is that height.We are finding what's the weighted average or the center mass of this right triangle board, and we set up the corner system in this favorable manner, and then we're doing the two corners one at a time.We set up proportion using the base.Yes. So, um, the x p over a, then make it equal to the. Let's go there to the y p. Okay, the y p over b. Yeah. Very good. Then you cross multiply. You could simplify. Yeah. So.Continue.Yes, the area. Which actually equal to the y p, and then times the x, right? And when you bring it in, it just equal to go ahead. And now you you might as well just work out the integral for us. So according to definition, xm would equal to integral over the total area, which we know is half of the ab. How do I write the integral?The B X P over A D X. Correct, and the boundary from. From zero to a.Yes, exactly. Well done. Go ahead and find it out. What are we getting? You forgot something, actually. Ivy, this is only the DA, isn't it? E o.The. Wait. If you just do the integral, you're going to get one because the numerator simply would give you one half of the AB—that just the total area.Do we need to put the XP?You're weighing the DA by X year. Are we all on the same page? Do every day? Does every day agree with this integral? Please. Yeah. Very good. Continue now. ``````Uh, I got. Um. Two A or three. Do all of you agree? I'm still working it out. I think that's correct. You can go ahead and work out why, or you can be.Smart about it, then immediately know what must be the way. In fact, the way we're setting the coordinate now will not be most favorable for working on the y coordinate. Instead, you want to do something else. Nevertheless, I want you to notice the symmetry. But anyway, just find out what is the y m. If you're done with finding the x m, however, I highly encourage encourage you to look back and see what works. Okay, and do it smartly.Get your hands busy, Eddie. Please. I actually have a quite a lot to say after this now. Okay. Thank you.````````` 嗯，或者你 like 写的。X P in terms of Y. A, you. What do you mean by X P in terms of the Y? Because you find the Y CM, you're really integrating over the Y coordinate. Ah, I mean by definition, they should actually equal to the Y P times the da divided by the total area, wouldn't that? Yeah, and the integral would be depending.How you set up your coordinate, but no matter what, probably it will be from a to b. Then we're not gonna write anything in terms of x anymore because we're really interested in what's the weighted average of the y coordinate. And I would say, probably you want to isolate a different kind of da to investigate to begin with, right? Aren't we interested in this kind of da now? Because they all share the same y coordinate. Yeah.Good. Indeed, that's what we're gonna do. This is actually the particular DA. But if you want to see a see through a way that you don't have to redo the integral, that would be nice. And there is certainly such a way.嗯，如果我 can't see through it，Can I just.Redo the integral. Yeah, sure, but you probably want okay. Feel free to you might want to reset the coordinate slightly because after all, x and y are symmetrical. You know, why don't you just flip it? If you're gonna make it stand up like this now, and this is a b and that's an x and that's the y, right? Y direction. Then, aren't you do the same integral again? That's what I meant by really.You don't have to redo the whole job. You can just choose to align the y equal the y direction in the x direction. Makes sense. Yeah, even on the AP at this moment, it is permissible for you to simply say similarly. You can just say similarly, or you can just say by the same logic. YCM just equal to a two b over three. There is no need to actually have to.Write up using the same coordinate system. However, just in case, okay, if I really want to stick to that coordinate system, sometimes AP tests give you such a problem. They set everything up for you. They want you to really just go ahead and write what is that the expression of the DA as a function of y. If you're really forced to, given that picture above, now then let's write it. For the sake of just writing it, and then what is the numerator?呃，呃，why.Times, the y a over b, d y. Hmm. So, good. You're identifying this part is simply by definition the y, and then you're writing the d a, and that's the so called the base, and the so called the height. That's this part now. And repeat again. What did you give me? The y a over b. That's.It's not the same. Because remember, actually, I V, at when you plug in the very small y, in fact, you get a very, very big so-called height. As y increases here, whatever is the span of the slice, it's decreasing. Eventually, when you plug in the b now, and you're getting zero. Correct. Yeah. So you have to, if you really want to get it right, you have to change that.However, I sort of just illustrated what's a supplementary homework. Can we finish it? Write it out and actually do the integral to confirm for yourself. And also, this is the center mass, which is at the centroid. As if you're connecting the median, they intersect right here, which is two-thirds. Check out that fact. Really, just search up what is the center mass of a triangle and how that applies to general sides of the ABC, meaning it doesn't have to be.Right triangle. Now our next session. I'm gonna pick up from here and do interpretations of the results. Let's homework clear. Yeah, please get it right. Yes. Alright, take care. Bye bye.

Center of Mass, Moments and Slicing

Center of Mass, Moments and Slicing

Overview

Focus Areas

Lecture Video

Generated transcript