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Speed of blocks on a pulley (Atwood's Machine)

Patrick Ford
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Hey, guys. So in this video, I'm gonna show you how to solve this very popular question using conservation of energy. So this is a conservation with energy with rotation question. Let's check it out. So we have two blocks connected by a light strain here. The two blocks light string means that the mass is zero and the string is ran around the pulley as shown. So the string is like this the red line, Um, this set up, by the way it's called. It's called Atwood's Machine. Just in case you're Professor mentions it, It's a very classic problem pulling with two objects hanging from it the blocks of masses three and five. So I'm gonna put a three here on gonna put a five here. The pulley is a solid cylinder. This is telling us the shape of the pulley so I can know the moment of inertia equation to use, which is gonna be for solid cylinder half M r squared. I'm actually gonna call this M and three because I have two objects and 12 and I'm gonna call this three, um, and then r squared. This is the only object that has a radius, so I don't have to say are three just say are the mass? M three is four and the radius is eight now. If you wanted, you could already calculate I What? We're gonna do this a little bit later? Um, some interesting stuff happens. If you leave everything in terms of letters, as I will show you, it is free to rotate about a fixed perpendicular access through its center. So what does that mean? So let's get a little desk here. Three ideas that the pulley is a desk. It's a solid cylinder. There's an access through the center in perpendicular, so through the center in perpendicular means making a 90 degree angle like this with the disc, Um, it's free to rotate. So when you put the masses of different when you put the objects of different masses, it's gonna tilt towards the heavy one. Um, the heavier one. And but the axis is fixed, so the disc itself isn't going to move. Imagine that it's like attached to a wall or something, and it doesn't It can only spend but not move, so that's what that means. The whole thing is released from rest initial velocity zero and M two is at the height of 5 m above the ground. Initially, So I'm to is 5 kg, but it also has an initial heights H two initial of zero when you release because it's the heaviest one, it's the heavier one. It's gonna go like this, right, and it's going to hit the floor. So it's final height. Is zero all right? The question is, what is the speed of them to just before it hits the ground? So what is V two final? That's what part A is asking us. What is V two final? Okay, the two initial is zero because the system starts from rest. The second question, Part B asks what is the police speed Now, when we talk about the speed of the poorly, we're talking about Omega. The poorly doesn't have a V, it just rolls around itself. So what is Omega final again? It's the only object that rotates, so we don't have to say Omega three final. We could just say omega final. Okay, before we start, we're gonna use conservation of energy, by the way. But before we start, I wanna point out that this is a system objects they're connected and connected objects always move together to have the same velocity. So V one is at all times the same as V two of the two blocks. Um, sometimes you you see this called the velocity of the system the system right? Or simply because there's no point in differentiating one and two, we just call them both feet. Also, whenever a rope is connected and pulling on a pulley, we can write the equation. So this is point number one. Um, point number two is that we can write the equation the rope equals are omega disc. Whenever rope pulls in disk where R is the distance two axes, It's the distance between where you pull on the rope where you pull on the rope and the axis of rotation. In this case, the rope pulls on the disc from both sides at a distance off the radius, right, So all the way at the enemies that the distance is the radius. Okay, so I'm going to be able to write that simply the equals big are omega. Okay, so I have these two extra equations to use. And by the way, if v equals R Omega. I could just add that right here. So it's really just one big equation. Okay, You can think of this as all the velocity is the same. Vi equals vi, which equals R o makeup. Cool. All right, so let's go ahead and write our conservation of energy equation K initial plus you initial plus work non conservative equals K final. Plus you final. What's gonna make this question? Not harder, but you're sort of longer. More annoying is the fact that I have three objects I have to worry about the energy of all three. So when I ask, is there Connecticut? In the beginning, you have to think about all three objects and actually beginning the system is not moving. So there is no kinetic energy potential energy. There's three of them. And for now, what I'm gonna do is I'm gonna write all three. You initial one. You initial too. You initial three and we'll talk about that in just a second. There's no work non conservative because there's no work done by you. You're just watching. There's no work done by friction. Okay. And kinetic final. Um, everything's moving. Just before I hit the ground, right? This guy's going down. This guy's going up in the disks spinning. So I have kinetic final for all three of them. And I'm going to write the potential final as well. You have one. You have to. You have three. Now let's analyze this real quick. Does the first guy right here have potential? In the beginning? The answer is no, because it's it's on the floor. But it does have potential energy at the end because they flip right. The second one has potential energy the beginning because it's up here but doesn't have it at the end because they flip. Okay, so notice how to hazard here, and then one has it here. What about the disk? The height of the disc doesn't change in the beginning. It's up here at the end. It's up here so we can cancel the potential energies like this. Okay, All three of them have kinetic energies, but it's not enough to know that it has kinetic energy. You have to know what type of kinetic energy it has. Well, the blocks air moving up and down. So this is linear motion. So they have linear, kinetic energy but the disk is spinning and it on Lee spins around itself. It only has one type of motion. And so he has rotational kinetic energy. Cool. Now we end up with five terms. We're gonna expand all of them. So this is going to be MGH, and it's for object to. So I'm gonna put a two initial. It's the only energy we have in the beginning. Here we're gonna have This is linear. So it's happened. The square plus half MV squared. Plus, this is rotational half I Omega Square. Let's put our coefficients here. This is the first mass in its final 1st 2nd mass in its final. Um, this is the moment of inertia of the third mass. I don't really have to put a three there because it's the only thing that has a moment of inertia. And then we have the gravitational potential energy, which the only one we have is M g H final. This is for the first mass. Cool. Now, if you look through this, you might be wondering, Can I cancel some stuff? You actually can't cancel anything You have ends everywhere, but you don't have one here and more importantly, all the EMS are different, right? So you're not gonna be able to cancel the masses because they're all different when you clean that up. Um, and what I want to do here is I wanna sort of deriving equation. So I'm not gonna plugging numbers into the end because I'm gonna show house, um, stuff cancels. Okay, so it's gonna look really nasty, but, you know, if you're doing this in a test and your professor doesn't mind, you could start plugging numbers already. Alright, but just check this out for this one time and see some of the things that are gonna happen here. So one of the things you want to do going forward is you always want to replace I with the equation, which is right here. And you always want to replace Omega with V. Remember, we talked about this. Whenever you have a question that has a V and on Omega, you're always going to you always want to replace. Rewrite your omega in terms of V, and that's so that instead of having V and Omega, you have V and V. And that's better because it's fewer variables the way you do that is. You get this equation right here and you say Omega equals the over our. So that's what you do. Okay, So I'm gonna do that. There's two things to do here. Two things to expand. Nothing else can be expanded. So you're just gonna leave it alone? So I'm gonna rewrite this whole thing and then expand this when I get here. Okay? M two g h two I All I'm doing now is rewriting this. It's kind of annoying. All right, stop right there. Let's make sure I'm gonna leave some space. Finished writing this. Okay, that was just rewriting. Now I have to actually slow down a little bit here, for I am gonna plug half m three are square. And here I'm gonna plug for Omega. I'm gonna plug the over arm so the over are and it's big our notice. What happens? The ours canceled. Um, but really nothing else is going to cancel. All right. What I like to do at this point is because I don't like fractions. I'm going to multiply by the smallest number that will make this make all the fractions. Go away. Basically, have a half here. You have a half here. He didn't have a 1/4. If you multiply that, that's the lowest denominator. Someone to multiply this by four. Okay. And that way, I end up with a four here for M two g h two I This becomes a to to m one V one final square. Um, by the way, 11 more thing before I continue all these, these are the same. Remember, we talked about that. So there's really no point in writing. V one. Final is just the final. Okay, plus four multiplies here. This becomes a two to em to the Final Four multiplies against the one quarter. So this becomes a one. So it's gonna be m three v final squared. Plus four m one G H. One final. This is for the first mass right there. Who? All right, So remember what we're looking for. Okay, this is extra painful because we're not playing the numbers just yet. We're looking for the final notice that we have the finals everywhere in No Omega's. That's what she wants. Okay, You wanna have a bunch of VFW's everywhere and no Omega's notice. Also, if you do sort of an inventory of everything you have. You know, the masses. You obviously know G. And you know the heights, right? The initial heights of the second block is five in the final heights of the first block. This FIBA's well, right, because this guy lost five of heights, so this guy could gain five of heights. So I'm gonna put it here that both of these numbers are five, you know, everything. So this point, you can plug in numbers and solve, but I'm going to solve this with the letters instead. And what you would do is you combine all the V s. So there's VFC squared everywhere, and I can combine the masses to m one plus two m two plus M three. Okay, Now I'm going to, um I'm going to throw, um, this guy to the other side. I'm gonna clean this up a little bit. Boom. I'm going to move this guy here to the other side, so it's gonna go negative. Four m two g h two initial minus four m one G h. One final. Okay, um, remember, these numbers are the same, so I could just call them H and H and Then you notice here that I have both sides have four g h four g h. The only thing that's different is the ends. So I'm gonna write four g h when a factor that out m two minus m one equals the final square. This is just the right side two and one to m two m three. We're almost there now. I can divide both sides by the final. The goal here is to get the final by itself. So it's gonna look like this v final equals four G h m two minus m one over to m one plus two m two plus m three. Then I take the square root of that. This is the final answer if you were asked to derive this. Okay, so I'm doing the harder one, which is deriving it. And then we could just plug in numbers, get the answer and be done. Now, if you didn't have to derive, what you would do is you would go all the way up to here so that your arse cancel. And then you just plug in all your variables, right? Don't be a hero. Don't try to do this a cute way if you don't have to. All right. So, um, now, what you would do here is you would multiply this whole thing. And if I did this to get correctly, which I hope I did, we're gonna have, um, four. I'm gonna round gravity to 10. Height is a five. The difference in masses. One masses of five empties of five M three empty, um, one is a three. Okay. And then I have to. The first mass is three. The second mass is five. And the first Matt and the third mass is a four. I'm gonna do this. Okay? By the way, when I said I hope I did described I don't mean the question solution. I mean, I've multiply this together here in my paper. Um, and I think I actually plugged in the wrong number. So I'm gonna just sort of do this here, live with you guys. So you got this is a two 10 um, 10 times 10 hundreds, 400 on the top. And at the bottom, we have six. 10 and this is four. So this is 20. So it is the square of 20 which is approximately 4.5. Okay, so the final velocity should be Ah, 4.5. And this is the end of part A. Uh the good news is, uh, don't freak out. Part B is much faster for Part B. All we're looking for is Omega. Right? So Omega is V over R V s, 4.5. The radius in this question is eight. So you would just calculate that I don't actually have, um I didn't do this on the calculator ahead of here. Um, things is gonna be roughly like point. Hmm. I'm gonna super around this. It's wrong, but it's 0.5, right? So you could do this in the calculator. Uh, the number is gonna be a little bit different, obviously, but it should be very close to five. So that said for this one, hopefully make sense. You should try to at the very least, make sure you know how to get to this equation here and then plug in your numbers as soon as you've replaced I and got rid of the ours. Can't make sure how to do this. Let me Do you have any questions?