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Ch 13: Rotational Inertia & EnergyWorksheetSee all chapters
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Ch 01: Intro to Physics; Units
Ch 02: 1D Motion / Kinematics
Ch 03: Vectors
Ch 04: 2D Kinematics
Ch 05: Projectile Motion
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Ch 07: Friction, Inclines, Systems
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Ch 13: Rotational Inertia & Energy
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Moment of Inertia of Systems
Conservation of Energy in Rolling Motion
Conservation of Energy with Rotation
Moment of Inertia via Integration
Energy of Rolling Motion
Moment of Inertia & Mass Distribution
Intro to Moment of Inertia
More Conservation of Energy Problems
Types of Motion & Energy
Parallel Axis Theorem
Intro to Rotational Kinetic Energy
Torque with Kinematic Equations
Rotational Dynamics with Two Motions
Rotational Dynamics of Rolling Motion

Example #1: Speed of blocks on a pulley (Atwood's Machine)


Hey guys! In this video, I'm going to show you how to solve this very popular question using conservation of energy so this is a conservation with rotation question. Let's check it out. We have two blocks connected by a light string. Here are the two blocks. Light string means that the mass is 0. The string is ran around the pulley as shown. The string is like this, the red line. This set up by the way is called AtwoodÕs machine just in case your professor mentions it. It's a very classic problem Ð pulley with two objects hanging from it. The blocks of mass is 3 and 5, so I'm going to put a 3 here and then I'm going to put a 5 here. The pulley is a solid cylinder. This is telling us the shape of the pulley so we can know the moment of inertia equation to use which is going to be for solid cylinder, _ MR^2. I'm actually going to call this m3 because I have two objects and IÕm going to call this 3 and then R^2. This is the only object that has a radius so I don't have to say R2, I just say R. The mass m3 is 4 and the radius is 8. I you wanted, you could already calculate I, but we're going to do this a little bit later. 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 axis through its center. What does that mean? Let me get a little disc here. The idea is that the pulley is a disc. It's a solid cylinder. There's an axis through the center and perpendicular. Through the center and perpendicular means itÕs making a 90-degree angle like this with the disc. It's free to rotate so when you put the objects of different masses, it's going to tilt towards the heavy one, but the axis is fixed. The disc itself isn't going to move. Imagine that it's like attached to a wall or something and it can only spin but not move. That's what that means. The whole thing is released from rest. The initial velocity is 0. M2 is at a height of 5 meters above the ground initially, so m2 is 5 kilograms but it also has an initial height h2 initial of zero. When you release because it's the heavier one, it's going to go like this and it's going to hit the floor. Its final height is zero. The question is what is the speed of m2 just before it hits the ground? What is v2 final? That's what part A is asking us. What is V2 final? V2 initial is zero because the system starts from rest. The second question, part B, asks what is the pulleyÕs speed. When we talk about the speed of the pulley, we're talking about w. The pulley doesn't have a V. It just rolls around itself. What is w final? It's the only object that rotates so we don't have to say w3 final. We can just say w final. Before we start, we're going to use conservation of energy by the way, but before we start I want to point out that this is a system. Objects are connected and connected objects always move together. They have the same velocity. V1 is at all times the same as V2 of the two blocks. Sometimes you see this called the velocity of the system, V system or simply V because there's no point in differentiating one and two, so we just call them both V. Also whenever a rope is connected and pulling on a pulley, we can write the equation, this is point number one. Pint number two is that we can write the equation V rope equals rw disc, whenever a rope pulls in the disc, where r is the distance to axis. It's the distance between where you pull on the rope, where are you pulling the rope in the axis of rotation. In this case, the rope pulls on the disc from both sides at a distance of the radius. All the way the end means that the distance is the radius. I'm going to be able to write that simply V = Rw. I have these two extra equations to use. By the way, if V = Rw, I can just add that right here. It's really just one big equation. You can think of this as all of the velocities are the same. V = V = Rw. Let's go ahead and write our conservation of energy equation. K initial plus U initial plus work non conservative equals K final + U final. What's going to make this question not harder, but sort of longer and more annoying is the fact that I have three objects. I have to worry about the energy of all three. When I ask is there a kinetic energy in the beginning, you have to think about all three objects. Actually in the 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 going to do is I'm going to write all three: U initial 1, U initial 2, U 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. Kinetic final, everything is moving just before it hit the ground. This guy is going down, this guy is going up and the disc is spinning. I have kinetic final for all three of them. IÕm going to write the potential final as well. Uf1, Uf2, Uf3. Let's analyze this real quick. Does the first guy right here have potential energy at the beginning? The answer is no because it hits on the floor, but it does have potential energy at the end because they flip. The second one has potential energy at the beginning because it's up here but doesn't have it at the end because they flip. Notice how two has it here and then one has it here. What about the disc? 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. 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. The blocks are moving up and down so this is linear motion so they have linear kinetic energy, but the disc is spinning and it only spins around itself. It only has one type of motion. It has rotational kinetic energy. We end up with five terms. We're going to expand all of them. This is going to be mgh and it's for object 2 so IÕm going to put a 2 initial. It's the only energy we have in the beginning. Here we're going to have, this is linear, so it's _ mv^2 + _ mv^2 + this is rotational _ Iw^2. Let's put our coefficients here. This is the first mass and itÕs final. Second mass and itÕs final. This is the moment of inertia of the third mass. I don't really have to put a 3 there because it's the only thing that has a moment of inertia. Then we have the gravitational potential energy which the only one we have is mgh final, this is for the first mass. If you look through this, you might be wondering, can I cancel some stuff? You actually can't cancel anything. You have mÕs everywhere but you don't have one here. More importantly, all the m's are different. You're not going to be able to cancel the masses because they're all different. Let me clean that up. What I want to do here is I want to sort of derive an equation so I'm not going to plug in numbers until the end because IÕm going to show you how some stuff cancels. It's going to look really nasty but you know if you're doing this in a test and your professor doesn't mind, you could start plugging in numbers already. But just check this out for this one time and see some of the things that are going to happen here. One of the things you want to do going forward is you always want to replace I with the I equation which is right here. You always want to replace w with Y. Remember we talked about this. Whenever you have a question that has a V and an w, you always want to replace, rewrite your w in terms of V. That's so that instead of having V and w, you have V and V and that's better because it's fewer variables. The way you do that is you get this equated right here and you say w = V/R. That's what you do. I'm going to do that. There's two things to do here, two things to expand. Nothing else can be expanded so you're just going to leave it alone. I'm going to rewrite this whole thing and then expand this when I get here, m2gh2i. All I'm doing now I rewriting this, it's kind of annoying. Let's make sure we leave some space to finish writing this. That was just rewriting now I have to actually slow down a little bit here. For I, IÕm gonna plug _ m3r^2 and here I'm going to plug for w, IÕm going to plug V/R. V/R and it's big R. Notice what happens. The RÕs cancel but really nothing else is going to cancel. What I like to do at this point is because I don't like fractions, IÕm going to multiply by the smallest number that will make this make all the fractions go away. Basically you have a _ here, you have a 1/2 here. Here you have a _. If you multiply that, that's the lowest denominator so IÕm going to multiply this by four. That way I end up with a 4 here. 4m2gh2i. This becomes a 2m1v1final^2. By the way, one more thing before I continue. All these VÕs are the same. Remember we talked about that. There's really no point in writing v1 final. ItÕs just v final plus 4 multiplies here. This become a 2m2vf^2. Four multiplies against the one quarter, so this becomes a 1. It's going to be m3vfinal^2 plus 4m1gh1final. This is for the first mass right there. Remember what we're looking for. This is extra painful because we're not plugging in numbers just yet. We're looking for a vfinal. Notice that we have vfinals everywhere and no w. That's what you want. You want to have a bunch of vfÕs everywhere and no wÕs. 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. The initial heights of the second block is five and the final heights of the first block is five as well because this guy lost five of heights so this guy can gain five of heights. I'm going to put it here that both of these numbers are five. You know everything. At this point you can plug in numbers and solve. But I'm going to solve this with the letters instead. What you would do is you combine all the V's. There's vf^2 everywhere and I can combine the masses 2m1 + 2m2 + m3. I'm going to throw this guy to the other side. I'm going to clean this up a little bit. I'm going to move this guy here to the other side, so it's going to go -4m2gh2initial minus 4m1gh1final. Remember these numbers are the same so I can just call them h and h. Then you'll notice here that both sides have 4gh. The only thing that's different is the mÕs so IÕm going to write for 4gh. IÕm going to factor that out m2 Ð m1 = vfinal^2. This is just the right side. 2m1, 2m2, m3. I can divide both sides by vfinal. The goal here is to get vfinal by itself. It's going to look like this. Vfinal equals 4gh (m2 Ð m1) / (2m1 + 2m1 + m3) and then I take the square root of that. This is the final answer if you were asked to derive this. I'm doing the harder one which is deriving it and then we could just plug in numbers, get the answer and be done. If you didn't have to derive, what you would do is you would go all the way up to here so that your RÕs cancel and then you just plug in all your variables. Don't be a hero. Don't try to do this the cute way if you don't have to. What you would do here is you would multiply this whole thing and if I did this correctly which I hope I did, we're going to have 4. I'm going to round gravity to 10. Height is a 5. The difference in masses, one mass is of 5, m2 is a 5, m1 is a 3, and then I have 2. The first mass is 3, the second mass is 5, and the third mass is a 4. IÕm going to do this. By the way when I said I hope I did this right, I don't mean the question solution, I mean IÕve multiplied this together here on my paper. I think I actually plugged in the wrong numbers. IÕm going to just sort of do this here live with you guys. This is a 2, 10. 10*10 = 100, itÕs 400 on the top. At the bottom we have 6 + 10 + 4, so this is 20. It is the square root of 20 which is approximately 4.5. The final velocity should be a 4.5 and this is the end of Part A. The good news is, don't freak out, Part B is much faster. For Part B, all we're looking for is w. w is V/R. V is 4.5. The radius in this question is 8. You would just calculate that. I didn't do this on the calculator ahead of here. This is going to be roughly like I'm going to super round this. It's wrong but it's point five. You can do this in the calculator. The number is going to be a little bit different obviously, but it should be very close to 0.5. That's it for this one. Hopefully it made sense. You should try it too 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 RÕs. Make sure you know how to do this. Let me know if you have any questions.

Concept #1: Blocks on a rough table and a pulley


Hey guys! I hope you tried out this practice problem. Let's check it out. Here we have two blocks connected by a light rope which passes through a pulley as shown. Here's a rope right here. It is a light rope which means the mass of the rope is negligible. The pulley is a solid cylinder so the moment of inertia of the pulley will be _ MR^2. Pulleys are always solid cylinders. I give you the mass. I'm going to call this m3. Let's call this guy m1, mm2, letÕs call the pulley m3. m3 = 10 and the radius is 2. The 4-kg block is in a horizontal surface. It shows here and the surface block coefficient of friction is 0.5. There's a friction, here _ = 0.5. This system is released from rest. The initial velocity is zero with the 6-kg block initially 8 meters above the floor. The initial height of h2 initial is 8. I want to know what is the speed that the 6-kg block will have just before hitting the floor. Just before hitting the floor means that this guy will have a final height h2final final 0 and I want to know what is V2 final. V2initial is zero because it starts from rest. That's what we're looking for. 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 as the final velocity of the 6-kg or had I asked for the final velocity of the system. I'm going to write that V1 = V2 = Vsystem or I could just write this as V. Furthermore, that's point one, point number two is the rope is connected to the pulley at its edge. I can write that V rope = R_ pulley or disc. Because it's connected at the edge, R equals big R, little R is the distance between the center of the pulley and where the rope connects. The rope connects all the way at the edge of the pulley so I take distance of radius which means this becomes big R_. V = big R_, which means I can put it right here so this whole thing here is that this big equation here that connects all these things. Obviously this thing is going to move like this. This moves like this and this moves like this. Because I have changing heights, I have friction, so IÕm thinking the work done by friction, velocities are changing. I'm going to use the conservation of energy equation. K initial + U initial + Work non-conservative = K final + U final. There's no kinetic in the beginning because the whole system is at rest. Potential energy. Let's look at potential energy. Instead of writing all three of them, we're going to talk about it. Does this guy have potential in the beginning? It doesn't or even though it's above the floor, the 4-kg is moving sideways, the height doesn't change which means whatever potential energy it has in the beginning, it's going to have at the end. The two of them would cancel in the equation so you can basically ignore the potential energy of this guy. The same thing with a cylinder. The cylinder stays in place, it doesn't change potential energy. The only object that has a potential energy that's worth writing down is the 6. I'm going to say that the only potential energy is going to be m2gh2initial. What about work non-conservative? There's no work done by you, you're just watching. You release this thing from rest, but there is work done by friction because friction is acting right here. So IÕm gonna say that there's the work done by kinetic friction on object one. What about kinetic energy at final? There is kinetic energy at the end. All three objects are moving in some way. The top one is moving horizontal. The 6-kg is moving vertically so they're both moving linear. The disc spins so the disc has rotational energy. I'm going to write _, the first block here has linear so half _ M1V1final^2. The second object right here has linear motion as well, so it's _ MV2final^2. The third one has the disc has moment of inertia so it's _ I_final^2. There is no potential energy at the end. Remember, the block has no potential energy because there is no change of height. The discs, thereÕs no change of height so thereÕs no potential energy. But this guy hits the floor at the end so it has no potential energy. I'm just going to put a 0 here for potential energy. One last thing, one quick adjustment here, these are really the same so I'm just going to write Vfinal and that's actually what we're looking, Vfinal, Vfinal. There's two of them, so we're going to be able to combine them. But before we do that, we have to remember you have to replace I and _ in here, so let's do that. I have m2gh2initial. I also have to expand the work done by friction and I'm going to go off to the side here real quick so we can do that. Work done by friction, kinetic is negative friction kinetic times distance. I want to remind you that friction is _normal. In this particular problem, the 4-kg is on a table like this so mg is down, normal is up. Normal = mg. I'm going to replace _ normal with _ mg. Lastly, I'm going to put this f over here. The work done by friction is negative _mg, that's f right there, d. I'm going to get this whole thing and put it here, negative _mgd. Let's keep going. Hhere I have _ M1Vfinal^2 + _ M2Vfinal^2 + _ the moment of inertia is 1/2M3R^2. Remember, we're supposed to replace _ from this equation here. We're supposed to replace _ with V/R. That's one of the most important parts of these questions. I need you to remember that you're going to replace _ with a V. This becomes V/R. It's Vfinal since it was _final. Notice that this causes the Rs to cancel. What this also does is now we have three V finals and you're going to have to combine them. Here you can't cancel masses. You're not able to cancel the masses because you have different masses everywhere. This one here by the way, it's mass1, I forgot to write it there. If you look around, you'll notice that you have all these numbers. You have the masses. Obviously you have gravity, you have mass, you have gravity, have mass, have mass, have 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 was and the distance that the block, the 4-kg kilogram block, moves. I hope you realized that they are the same. 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. What I'm going to do is I'm going to call this big H and I'm going to call this big H. It's the same exact number. I'm going to have m2gH Ð _ m1gH. On the right side, what I'm going to do is combine all the Vs. Notice that you have this here in front of the Vs. This one is like this. I usually multiply this by a number to get rid of the fractions. I haven't done that yet but I'm still going to do that. Vfinal^2 and then I have 1/2m1 + _ m2 + _ m3. What are we going to do here is multiply everything by 4 to get rid of these fractions. I'm just going to do this here. I'm going to put a 4 in front. These guys will cancel here. Don't put a 4 here. Do not put a 4 there. We're going to already distribute the 4, so this becomes a 2 this becomes a 2, and this becomes a 1. I want to point out here if you want to clean this up, I got gH, I got 4gH and 4gH. That's common so I can factor it out, 4gH. I have m2 - _ m1. Then on the other side I have Vfinal^2, 2m1, 2m2, m3. I'm going to move the masses to the other side, take the square root and we are done. Vfinal will be the square root of 4gH(m2 Ð _m1) / 2m1 + 2m2 + m3. That's it. We're going to plug in the numbers in just a second. This is the hardest way to do this. What I mean by that is it's harder to take it all the way to the end without plugging the numbers. I wanted to show you the harder version of how you might be asked this question which is derive an expression. If your professor doesn't ask you for an expression, if he asks you just to find the number and if he doesn't care that you plug in numbers at any point, you probably want to plug in your numbers somewhere here. Don't worry about taking it all the way to the end unless you have to. I'm showing you how to do that in case you do, but if you don't, donÕt be a hero. Start plugging the numbers here. It's much easier than carrying all these letters around for a long time. If you want to plug all of this, I have this already done. It's 6.47 meters per second is your answer. That's it for this one, pretty common type of problem. Make sure you understand how to do this. Let me know if you have any questions.

Example #2: Speed of a yo-yo


Hey guys! Here's another classic rotation question that we're going to use conservation of energy to solve and it's a yo-yo question. We have a simple 100-gram yo-yo that we're going to release from rest. Simple just means that you're going to 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. That's what simple means. It just means go nuts with simplifications. Mass = 0.1 kg. It starts from rest, V initial = 0. It falls and rolls. Yo-yos do that, so they're falling and rolling all the way down, unwinding the light string around its cylindrical shaft. As it falls, it unwinds a light string. Yoyo has a string around it. Light string means the mass of the string is negligible. Around its cylindrical shaft and that's because a yo-yo has a thing in the middle. The string is here but then the yo-yo usually has sort of an outer casing like that. The idea is that what matters is this inner radius, not the outer radius. The outer radius just covers the outside. Effectively we're going to just worry about this and say that a yoyo looks like this. Let's actually put the little string here. If the string is here and you'll release a yo-yo, it's going to fall and it's going to roll like this. 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 spin with the omega. The radius of the inner, which is what matters, is 2 cm so 0.02 m. It says if the yo-yo can be modeled after a solid disc, in other words treat this thing here as a solid disc which is what we're going to do. In other words, I will be _ MR^2 because it's the I for a solid disc. I want to know what is its linear speed after it drops 50 cm. It drops centimeters, so I can think of this as the height initial. This is 0.5 m. I can think of this as the height initial was 0.5 m and the height final since it falls being zero. Even if the floor is farther below, you can just move the floor up and say I'm going to call this point zero, therefore this point is 0.5. That's how you're supposed to do it. I want to know the final speed. It's released from rest so the initial velocity is zero. I want to know what is the final speed, that's part A. For Part B, I want to 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 going to use conservation of energy to solve this because I have a change in heights, change in velocity. But before I do that, I want to remind you that whenever a rope pulls on a cylinder on a disc or a pulley something like that, V rope = r_ disc, 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 this inner cylindrical shaft, but there's another one here but we're ignoring that. The rope the string is pulling at the edge of this inner cylindrical shaft so weÕre just gonna do this. The distance is the radius because it's at the edge, so you can say that V = R_. That's going to be an equation that we're going to be able to use and we're going to have to use. Let's go up here and we're going to write kinetic initial + potential initial + work non-conservative = kinetic final + potential final. Is there kinetic initial? No, there's no kinetic initial and that's because it's at rest initially. There is a potential energy because you have some heights. There is 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. You don't do any work. There's no friction. There is kinetic at 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 a motion, your potential energy is zero. Basically you have all your gravitational potential goes into kinetic. Let's expand this. mgh initial equalsÉ Here kinetic final, you have to figure out is it linear or is it rotational. In this particular problem, it's both linear and rotational. That same object has linear and rotational energy. What we're going to do is we're going to do _ mv^2 final + _ I_^2 final. The next step is to expand I and _, or rewrite _ I should say. This is going to be mghi = _ mvf^2 plus 1/2I. You're going to see how this cancels really nicely. The answer is going to come out really simple. I is going to be _ MR^2. If you hold off from the temptation to just plug in numbers, this is actually going to become pretty neat, pretty simple, once you see it. remember when you have V and _, what you want to do is rewrite _ in terms of V so that instead of having a V and a _, you have a V and a V. That's fewer variables, that's better. From here, I can write that _ must be V/R. I'm going to plug V/R over here. When you do this, notice that the Rs will cancel. The Rs almost always cancel, pretty much always cancels but youÕve got to be careful about that when you start canceling stuff. Notice also that the masses cancel. Because I have a single object, so all of these Ms are referring to the cylinder, to the cylindrical shaft here, to the yo-yo itself. That mass shows up in all three terms so I can cancel the masses. The M is cancelled and the R is cancelled. Let's clean this up a little bit. IÕm going to multiply both sides by 4 because I want to get rid of this quarter right here. IÕm going to get 4gh initial. This _ becomes a 2V final and then this _ becomes 1 so this whole thing is gone and the only thing that's left out of this whole thing here is the V. It's vfinal^2. This is obviously 3Vfinal^2 and if we're solving for V, you just got to move the three over and take the square root. Vfinal is the square root of 4/3 gh initial. This is pretty cool, borderline cool. Notice that it doesn't depend, IÕm going to 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 only depends on obviously the gravity, it depends on the initial height, the higher you are, the faster the more you drop the faster you will be. This 4/3 here comes from your moments of inertia from your shape. If you have a different shape, this will have a different fraction here. It could have been, I don't know IÕm going to make something up. It could have been 5/2 if you have a different shape. 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. Imagine if you drop a block and after the block falls a 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 2gh. You might remember this. This is for linear motion. This is like a separate thing. Look at how similar these two guys look. This is very similar to this. The only difference is that this has a 2 and this has a 4/3. That's what I mean by the form, the shape. The equations look similar to what you would have expected in linear motion. The difference is that the coefficient is different. Another thing is that the coefficient is less. In rotation, same form the equation the answer will look similar to linear. It has a different coefficient and the coefficient is not just different, but it's a lower coefficient. Coefficient is just a number in front of something else. 4/3 is 1.333 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 only 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. Because it's spending energy into spinning, it actually falls slower. In theory if this yo-yo were to spin at infinity speed, it would actually never fall because it'd be busy spending all of its energy in rotation. 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 2. If you ever do a question like this where you end up with a number that's greater than the coefficient in the linear version if you happen to remember the linear version, then you know that something's wrong. If I gave you for example a question where instead of the yoyo 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. You end up multiplying blah blah, so you end up with the fraction that's different. The 4/3 would be different. But whatever that number is, it would be less than 2. IÕm just trying to give you sort of an intuition in terms of how linear and rotational problems that look very similar also have very similar answers, just slightly different. That's it for this one. Let me know if you have any questions. Let's keep going.

Practice: 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?