 Welcome back to our lecture series Math 1220 Calculus II for students at Southern Utah University as usual today I am your professor Dr. Andrew Misseldine This is the first lecture for lecture 23 for which we are going to focus on the concept of a centroid now to give us understanding what that means Think of the following basic idea We're all familiar with the concept of balance, right? Human beings we walk around and we fall over all the time and we try not to fall over We use this idea of balance an even distribution of weight Enabling someone or something to remain upright and sturdy at all times right and as children we often even played with balancing toys and attempt to make them balance a very Interesting or difficult positions, right? We often have one of those little birds hanging on our finger at one point We like these idea of a balancing toy And so while trying to balance the toy or pencil or whatever There appears to be a single point which if supported the toy will balance perfectly on its own And as children we were searching for this center of balance or in science It's sometimes called the center of mass or center of gravity In mathematics, we're gonna we're also interested in the center of mass And this is what we mean by when we say centroid like in this context This center of gravity is our centroid that we're looking for now suppose that we have a Uniform plank that is to say a plank of uniform density Maybe like we're playing teeter-totter or seesaw, you know back in the day when Children were allowed to play dangerously nowadays. These are thought of as a child catapult, right? If you were trying to balance this seesaw, where might the sinner be? Well, since everything is completely uniform the idea is to put The fulcrum the balancing point right here Under the midpoint the place that is halfway Halfway across halfway in the middle good if it's completely uniform It's gonna be the midpoint of this corresponding line segment now suppose that this same plank What we're using for a seesaw now has two children playing together on the board, right? So let's draw this again Now we have two children in I'm just gonna draw them as blobs, right? you get one blob over here and If the two children are equal in their mass then again the board will be balanced and you can put the fulcrum in the middle But what happens when you have the sad little child who goes to the playground all by Herself she doesn't have anyone to play with well dad's gonna play with her woohoo, right? But you put dad on the other side the problem is well dad is much more massive than That his daughter right here and so therefore this shifts the center of gravity, right? So that if we place the center of gravity in the middle when the daughter goes up, right? she doesn't have enough mass to To bring herself back down because the dad over here is just too massive in which case she kind of gets stuck in the air Well dad realizing it's not really fun for her his little girl to be stuck in the air He kind of jumps himself in the air thus propelling her downward But then of course the problem is when she's down and dad's up She doesn't have enough torque to keep her down and therefore the child catapult occurs and we lose We lose little Susie right here that that's a problem, right? and so the issue here is that for the typical seesaw if the If the mass has been moved if we have a greater mass on the right the Sincer of gravity also has to kind of shift over so these things are balanced And there were some seesaws teeter-totters, whichever you call them Which you could actually move the fulcrum towards the heavier child towards the parent that way it would be balanced And therefore children could recreate in a more desirable fashion, right? And so we have some personal experience perhaps with this You know if you blew up if you grew up in a in an environment where playgrounds were considered acceptable Many of us live a very sheltered life nowadays But we can generalize this idea that is if we consider the mass of objects in the in the center of gravity We'll then gravitate towards the more massive object consider. We have indistinct objects in objects, which They're going to be what we're going to orient them along the x-axis of some kind so we have indistinct objects distributed over Well, not necessarily the x. Well, yeah, we'll keep it with the x-axis in objects And so things like this we have our x-axis so we got One object here one object here one object here another object here And so we have these elements x one x two x three and x four at the moment I don't necessarily claim that these are uniformly distributed across the x-axis We just have these these objects here now each of these objects has a mass associated to it So we get mass one mass two mass three and mass four So you can see by the way i drew drew these objects little mass two. It's such a durable little thing It's so cute and small and then mass four over here so big so massive These objects have different masses associated to them And so in order to find the center of gravity What we have to do is we have to calculate a so-called weighted average The center like if we want to place the fulcrum, where is it going to go here here here? It's hard to tell right where does this fulcrum go? We don't know We're looking for this so-called x bar here. We're using x bar. This is a notation We borrow from statistics where the bar over variable indicates we're finding the center or the mean or average of that thing And so to find x bar here what we're going to do is we're going to take two sums So first we're going to take a sum From i equals one to n where we're going to take together all of the values mi times xi xi is the location of the objects Mi is the weight of that object, right? So we're going to take this sum on the top this weighted sum And then on the bottom we're going to divide it We're going to divide by the total number of weights that are in play We add together all of the mi's and so this gives us a so-called weighted average And so for a typical student, we're probably used to this type of idea because Weighted averages is how grades are calculated in a course of some kind And which case you take you take like your average like oh, I had 90% on homework. I had 30% on quizzes. Yeah, I know big drop there, but I had 87% on exams And so you take the weights like well, maybe homework was 10% quizzes were 15% Exams were whatever was left Uh, and so we have this weighted average going on there And so the weighted average can be used to find the center of mass We could do this for x bar. We could also do the same thing for a y bar, right? Like if we had things distributed across the y axis, we could find the weighted average y bar And so this can be useful to help us find the center of gravity So imagine as you can see here Let's find the center of mass for the system of three objects you can see on the screen These objects have a mass of three four and eight and they're located at negative one one two negative one And three two respectively So the green object is the lightest of all of them. It's only worth three units of mass It's located at negative one one The red object is slightly more massive, but not by much It's looking to be located at two negative one And then the yellow object is the most massive it's it's worth eight units of mass And so one can ask where is the center of mass which you can see it already indicated in our drawing right here But and intuitively that does seem to be like the middle But how does one calculate such a thing? Well using the notation we did before let's figure out what the total mass of the system is So we're going to take the sum where i equals one to three just three objects here We're going to add up the m is and so this is just three Excuse me three plus four plus eight We see that the total mass in this system here is going to be well four and eight is 12 plus three is 15 So there's 15 units in play right now to find x bar We're going to compute on top the so-called moment mi Times xi and then we divide this by the total mass Which is the sum of the mi's right here And so we know there's going to be a 15 on the bottom So on the top we're going to take three times its location negative one Plus four times the location of the red point two and then add that eight times the location of the third one three We're just looking at the x coordinates right now So we get negative three plus eight plus 24 over 15 And so we can see that we can simplify this of course Um eight takeaway three is a five plus 24 is 29 So you get 29 over 15 This is our x bar. This would be the y or the x coordinate of the center of mass 29 15 is just a little bit shy of 30 15's which will be two And so you're going to see that the center of mass Is slightly to the left Uh of x equals two That gets you x bar y bar is computed similarly. We first start off with this moment mi yi Over the total mass In which case again, this is all over 15 This time though, it's giving me a similar sum. It's just that we're going to use y coordinates instead of x coordinates this time We get three times one for the green one We're going to get four times negative one for the red one and we're going to get eight times two for the uh for the yellow one In which case we got three minus four plus 16 Right all over 15 Uh three minus four is negative one plus 16 is 15 So we get 15 over 15, which is equal to one and so the y coordinate of the center of mass here would be one so in particular our Centroid for this system of three objects would be 29 over 15 comma one And so we calculate this weighted average of the x coordinates We calculate the weighted average of the y coordinates and this gives us the center of mass Many of us in the past have learned a formula for the midpoint the midpoint of two points x bar y bar. This is going to just be the x one x two over two and then y one plus y two over two So in this system where you calculate the midpoint This is just a special case of this weighted average we're doing right now in this situation Each of the points has the same weight of one So kind of like when we had the seesaw before when you had two Equally mass children on the two sides They each have a mass of one thus the total mass is two You add them together divide by two that gives you the x bar the y bar And therefore the center of mass will be found in the middle the geometric midpoint But in like in space if you have like three Uh three stars working on each other in terms of gravity Their center of mass their center of gravity would not necessarily be the middle Geometrically speaking it'll be pushed closer to the more massive object. And this is how we can find Uh, this is how we can find the center of gravity in sort of this simple uh discrete setting