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?
Blocks on a rough table and a pulley
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Hey, guys, I hope you tried out this practice problem. Let's check it out. So here we have two blocks connected by a light rope. Um, which passes through a pulley, as shown. So here's the rope. Right here it is a light rope, which means the mass of the rope is negligible. Okay, Um, the pull It is a solid cylinder. Eso The moment of inertia of the pulley will be half M R square police. They're always solid cylinders. I give you the mass mass. Uh, I'm gonna call this m three. Okay, let's call this guy. I am one m to its called a poli m three m three equals 10 and the radius is to the 4 kg block is a horizontal surface. It shows here, and the surface block coefficient of friction is 40.5. So there's a friction here. Mu equals 0.5. Um, the system is released from rest. The initial velocity is zero. And the 6 kg block with the 6 kg block, initially 8 m above the floor. So the initial heights of 22 h two initial is eight. Okay. And I want to know what is the speed that the 6 kg block will have just before hitting the floor. So just before hitting the floor means that this guy will have a final heights, a final heights H two, Final of Zero. And I want to know what is V two final. The two initial is zero because it starts from rests. Alright, so that's what we're looking for. Um, remember that this is a system. Everything moves together. Therefore all the velocities are the same. Had I asked for the final velocity of the 4 kg, it would have been the same exact thing. Is the final velocity the 6 kg or had I asked for the final velocity of the system? So I'm gonna write that V one equals V two, which equals V system, Or I could just write this as a V. Furthermore, that's 0.1 point Number two is the rope is connected to the pulley at its edge, so I can write that the rope equals our omega pulley or disc. Okay, now, because it's connected at the edge, r equals big. Our little R is the distance between the center of the polling and where the rope connects the road connects all the way at the edge of the polling. So it's a distance of radius, which means this becomes big. Are Omega okay? V equals big art omega, which means I can put it right here. So this whole thing here is that this big equation here that connects all these things. Now, obviously, this thing is gonna move like this. This moves like this. And this moves like this because I have changing heights. I have friction. So I'm thinking the work done by friction velocity. They're changing. I'm gonna use the conservation of energy equation. So okay, initial. Plus you initial plus work non conservative equals K Final. Plus you final. There's no kinetic energy. The beginning because thes objects are the whole system is at rest potential energy. Let's look at potential energy. Now, instead of writing all three of them, we're gonna talk about it. So does this guy have a potential energy in the beginning? It doesn't, um or even though it's above the floor, the 4 kg is moving sideways. The heights doesn't change, which means whatever potential energy has in the beginning, it's goingto have at the end, the two of them would cancel the equation. So you can basically ignore the potential energy of this guy. The same thing with the cylinder. The cylinder stays in place. It doesn't change potential energy. So the Onley object that has a potential energy that's worth, um, writing down is the six. Okay, so I'm gonna say that the only potential energy is going to be M two g h two initial. What about work? Non conservative? Well, there's There's no work done by you. You're just watching you release the thing from rest. But there is work done by friction because friction is acting right here. Okay, Someone say that there's the work done by kinetic friction on objects. One Alright, what about kinetic Final? At the kinetic energy file, there is kinetic energy at the end. All three blocks are all three objects. Air moving in some way. The top one is moving horizontal. The 6 kg is moving vertically, so they're both moving linear. Um, this expense So the disk has rotational energy, so I'm gonna rights. Half the first block here has linear so half m one view, One final squared. The second object right here has linear motion as Well, so it's half M V two final squared, and the third one has The disk has moment of inertia. So it's half I Omega Final square. There is no potential energy at the end. Remember, the block has no potential energy because it doesn't change height. The disk doesn't change heights. There's no potential energy. But this guy hits the floor at the end, so it has no potential energy. So I'm gonna put a zero here for potential energy. One last thing, one quick adjustment here. These V s are really the same. So I'm just gonna write the final, okay? And that's actually what we're looking for. The final, the final. There's two of them, um, eso we're gonna be able to combine them. But before we do that, we have to remember you have to replace I and W in here. So let's do that. I have m two g h two initial. I also have to expand the work done by friction, and I'm gonna go off to the side here real quick. Eso weaken. Do that work done by friction. Kinetic is negative. Friction kinetic times distance. I want to remind you that friction is mu normal. And this particular problem, the 4 kg is on a table like this. So m g is down. Normal is up normal equals m g. So I'm going to replace Mu normal with Mu m. G. And lastly, I'm gonna put this f over here. So the work done by friction the work done by friction is negative. Um, mu mg, that's f right there. D. Okay, so I'm gonna get this whole thing and put it here. Negative, Mu M g d. All right, let's keep going. So here I have half m one the final squared, half em to the final squared. Plus half the moment of inertia is another half m three r squared. And remember, we're replacing. We're supposed to replace Omega right from this equation here, we're supposed to replace Omega with V over R. That's one of the most important parts of these questions. I need you to remember that you are going to replace Omega with a V. Okay, so this becomes V over. R. It's the final since it was Omega final Notice that this causes the arts to cancel. What this also does is now you have three V finals and you're gonna have to combine them. Okay. Here, you can't cancel masses. You're not able to cancel the masses because you have different masses everywhere. Okay, this one here, by the way, it's mass one. I forgot to write it there. All right, so if you look around, you notice that you have all these numbers. You have the masses. Obviously you have gravity, have mass of gravity, have mass, have mass of mass. We're looking for the velocities. I'm giving the coefficient of friction. So the two guys we haven't talked about here is the initial height that you drop or the total height you drop or what, the initial height, waas and the distance that the block, the 4 kg block moves. And I hope you realized that they are the same, right. This block is moving down and pulling this guy, so they're both moving the same distance. So this distance here D is the same as this height. So what I'm gonna do is I'm gonna call this Big H, and I'm gonna call this Big H, and it's the same exact number. Okay, so I'm gonna have m two g big H minus mu M one g big H. And on the right side of what? I'm gonna lose combined all the V's. So notice that you have, um, this here in front of the V's thesis ones like this. Now, I usually multiply this by a number to get rid of the fractions. I haven't done that yet, but I'm still gonna do that. Um, the final squared, then I have half m one plus half m two plus one quarter m three. Okay, So what we're gonna do here is multiply everything by four to get rid of these fractions. Um, and here, I'm gonna end up with I'm just gonna do this here. I'm gonna put a four in front of four in front. Um, these guys will cancel here. Don't put a four here. Right. Do not put a four. There were going toe already distributed. Four. So this becomes a to this. Becomes a two on this becomes of one. I wanna point out here. If you want to clean this up, I got g h. I got four g h and four g h. That's common, so I can factor it out. Four g h I have m two minus mu M one and then on the other side. I have the final square two and one to, um to one. M three. Okay, I'm gonna move the masses to the other side, Take the square root, and we are done. The final will be the square root of four G h. M two minus me. Um, one divided by two M. One plus two m two plus m three. Andi, that's it. Okay, we're gonna plug in the numbers in just a second. Now, this is the hardest way to do this. Um, in terms off, it's harder. What I mean by that is that it's harder to take it all the way to the end without plugging numbers. I wanted to show you the harder version of how it might be asked this question, which is derived an expression right now, if you're professor doesn't ask you for an expression, if he asks you just to find a number and if he doesn't care that you plug in numbers at any point, um, you probably wanna plug in your numbers somewhere here, Okay? So don't worry about taking it all the way to the end unless you have to. All right. So I'm showing you how to do that in case you do. But if you don't, don't be a hero. Start plugging the numbers here. It's much easier than carrying all these letters around for a long time. Quote. Now, if you were to plug all of this, I have this already done. Um, it's 6.47 meters per second. Is your answer cool. That's it for this one. Um, pretty common type of problem. So make sure you understand how to do this. Let me know if you have any questions.
Speed of a yo-yo
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Hey, guys. So here's another classic rotation question that we're going to use conservation of energy to solve. And it's a yo yo question. Right? So we are. We have a simple 100 g, Yo yo, that we're going to release from rest now. Simple just means that you're gonna be able to make some assumptions to simplify the yo yo yo. Yos are actually more complicated than how we're going to solve them. But here we're going to simplify it. That's what simple means. It just means, you know, go nuts with simplifications. Alright, so mass equals 0.1 kg. It starts from rest Venus and go zero. It falls and rolls. Yo, Yo's do that right? So they're falling and rolling on the way down. Um, unwinding the light string around. It's cylindrical chef. So as it falls, it unwinds. A light string. Yo yo has a string around it. Light string means the mass of the string is negligible around it's cylindrical shaft. And that's because the yo yo has, um, I also has a thing in the middle. The cable. The string is here, but then the oil usually has sort of an outer casing like that, right? Three idea is that what matters is this inner radius, not the outer radius. The radius just covers the outside. So effectively, we're gonna just worry about this and say that I owe you. Looks like this. Okay, let's actually put the little string here. Now, if the string is here and you release a yo yo, it's going to fall and it's going to roll like this. So if the string is on this side, it's going to go like this. This is the velocity of the center of mass, and it's also going to spend with the Omega. Okay, So the radius of the inner, which is what matters, is two centimeters. So 0.2 m. It's as if the union can be modeled after after a solid disc. In other words, treat this thing here as a solid disc, right, which is what we're gonna dio. Um, in other words, I will be half m r squared because that's the eye for a solid disk. I want to know what is its linear speed after drops 50 centimeters. So the drops 50 centimeters. Um so I can think of this as the heights Initial, right. This is 0.5 m. I can think of This is the height initial was 0.5 m and the heights final since it falls, um, being zero. Okay, Even if the floor is farther below, you can just move the floor up and say, I'm gonna call this 00.0. Therefore, this point is 0.0.5. Okay, that's how you're supposed to do it. All right? I want to know the final speed. It's released from rest. So the initial velocity zero, I want to know what is the final speed. That's part game for part B. I wanna know what's the final angular speed. So what is the linear speed and what is the final angular speed? So what is Omega Final is what Part B is asking. We're gonna use conservation of energy to solve this because I have a changing heights change in velocity. But before I do that, I want to remind you that whenever a rope pose on a cylinder on a disk, um, or in a police, something like that V rope equals are omega disk, where R is the distance between the center and where the rope pulls in this case, the rope is pulling all the way at the edge of the of this inner cylindrical shaft room. There's another one here, but we're ignoring that, Um, the the rope the string is pulling at the edge of this cylindrical inner political shots or is gonna do this The radius, the distance is the radius because it's at the edge. So you can say that the equals R Omega. Okay, so that's gonna be an equation that we're going to be able to use. We're going to have to use. All right, So let's go up here and we're gonna write Kinetic Initial plus potential initial plus work, Non conservative equals Kinetic Final, plus potential final. Okay. Is there a Connecticut issue? No. There's no kinetic in issue, Andi. That's because it's at rest. Initially, there is a potential energy because you have some heights. Um, there's no work non conservative. You're not doing anything. Even though you're holding the string, you're not actually giving energy to the system, right? So your work, you don't do any work, there's no friction. There is Connecticut, the end, and there is no potential at the end because we're moving the ground up, so to speak. And that's the lowest point. And at the lowest point of emotion, your potential energy is zero. So, basically, you have all your gravitational energy goes into potential audio gravitational potential goes into kinetic. So let's expand this MGH initial equals. Now here, kinetic final, you have to figure out, Is it linear or is it rotational? And in this particular problem, it's both linear and rotational. That same object has linear and rotational energy. So what we're gonna do is we're gonna do half MV squared Final plus half I Omega Square final. Now, the next step is to expand I in Omega, right or rewrite Omega. I should say so. This is gonna be m g h. I equals half MV Final squared plus half, and you're going to see how this cancer is really nice. And the answer is gonna come out really simple. And I is gonna be half m r squared. And if you hold off on the temptation to just plug in numbers, this is actually gonna become pretty neat. Pretty simple once you see it. So if remember when you have V N w when you have V and w What you wanna do is rewrite W in terms of V so that instead of having a d n a w have a V and V, that's fewer variables. That's better. So from here, I can write the W Omega must be V over R. So I'm gonna plug V over R over here. And when you do this, notice that the ours will cancel. Right? The artist almost always canceled. Um, pretty much always canceled. But you gotta be careful about that. Don't get trigger happy and start canceling stuff. Notice also that the masses cancel. Right, Because I have a single objects. So all of these ems air referring to the cylinder, uh, to the cylindrical shaft here to the yoyo itself. Um, and that mass shows up in all three terms so I can cancel the masses. So the EMS canceled in the arse, canceled. Alright, so let's let's clean this up a little bit. I'm gonna multiply both sides by four because I want to get rid of this quarter right here. Okay, so I'm gonna get four g h initial. This half becomes a to V final, and then this quarter becomes one So this whole thing is gone, and the only thing that's left out of this whole thing here is the V. So it's V final squared. This is obviously three V final squared. And if we're solving for V, he's gotta move the three over and take the square root. Okay, so the final is the square root of 4/3 G h initial, and this is pretty cool. Borderline cool. Um, in that notice that it doesn't depend. I'm gonna write this here because it's really important. Actually, it doesn't depend on M or R. So the final velocity of a yo yo has nothing to do with its radius has nothing to do with its mass. It on Lee depends on obviously the gravity. It depends on the initial height. The high you are, the faster the more you drop, the faster you will be. And this 4/3 here comes from your moments of inertia from your shape. So if you have a different shape, this will have a different fraction here. Okay, So it could have been I don't I'm gonna make something up. It could have been 5/2. If you have a different shape. Okay, but so that's one point. A second point that I want to make is that in these rotation problems, the form of the answer is very similar to their linear counterparts. What the hell did I just say? Imagine if you drop a block and after the block falls the height of H, we can calculate using motion or energy. But you may remember this that the velocity at the bottom is the square root of two GH. You might remember this, right? This is for linear motion. Okay, this is like a separate thing. Um, now, look at how similar these two guys look. This is very similar to this. The only difference is that this has a two and this is a 4/3. So that's what I mean by the form, the shape the equations look similar, um, to what you would have expected in linear motion. The difference is that the coefficient is difference. Another thing is that the coefficient is less so in rotation, same form the equation. The answer will look similar. Two linear. It has a different coefficient, and the coefficient is not just different, but it's a lower coefficient. Quite efficient is just a number in front of something else. Okay, 4/3 is 1.33333 which is lower than two. What this means is that if you drop a block and a yo yo, the block will fall faster than the yo yo. And that's because the yo yo has two types of energies. The block on Lee has linear the yo yo is converting its potential energy into both linear kinetic energy, which makes it fall fast and rotational kinetic energy, which makes it spin. So because it's spending energy into spinning, it actually falls slower. In theory, if this yo yo were to spin at this like an infinity speed, it would actually never fall because we'll be busy spending all of its energy in rotation. So hopefully that makes sense. The reason why I want to point this out is because you can use this as a way to verify if you're correct. For example, this fraction could never be greater than two. And if you ever do a question like this, um, where you end up with a number that's greater than the coefficient and linear version. If you happen to the middle layer version, then you know that something's wrong. If I gave you, for example, a question where instead of the yo yo being a solid cylinder but instead we had a hollow cylinder, the solution will be exactly the same. The only difference is that this I over here, this number in front of the I would be different. And that difference would sort of work its way through the problem. That's this half right here. This half is this right here, right? And you end up multiplying, blah, blah, blah. So you end up with a fraction that's different. Four or three would be different, but whatever that number is, it would be less than two. Okay, so just trying to give you sort of an intuition, intuition in terms of how linear and rotational problems that look very similar also have very similar answers. Just slightly different Cool. That's it for this one. I mean, if you have any questions, let's get going
A small 10-kg object is connected to the right end of a thin rod of length 4 m and mass 5 kg. The rod is free to rotate about a fixed perpendicular axis on its left end, as shown below. The rod is initially held at rest, horizontally. When the rod is released, it falls, rotating about its axis, similar to a pendulum. What is the speed at the rod’s center of mass when the rod is vertical? BONUS:What is object’s speed when the rod is vertical?