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Intro to Moment of Inertia | 30 mins | 0 completed | Learn |

Moment of Inertia via Integration | 19 mins | 0 completed | Learn |

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Moment of Inertia of Systems | 23 mins | 0 completed | Learn |

Conservation of Energy in Rolling Motion | 45 mins | 0 completed | Learn |

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Parallel Axis Theorem | 14 mins | 0 completed | Learn |

Conservation of Energy with Rotation | 36 mins | 0 completed | Learn |

Torque with Kinematic Equations | 59 mins | 0 completed | Learn |

Rotational Dynamics with Two Motions | 51 mins | 0 completed | Learn |

Rotational Dynamics of Rolling Motion | 27 mins | 0 completed | Learn |

Concept #1: Rotational Dynamics with Two Motions

**Transcript**

Hey guys! In this video we're going to start talking about torque acceleration problems also known as rotational dynamics problems where we have two types of motions. What that means is that on top of having a rotation, we're also going to have objects moving in a linear direction. It's going to have a combination of both of them. Let's check it out. Remember that when we have problems where a torque causes an angular acceleration _, we use the rotational of version of Newton's second law. Instead of F = ma, we're going to write torque, sum of all torques, equals I _. When we have torque-_ questions, that's how we solve it. But in some problems we're going to have more than just _. We're going to have rotational and linear motion. For example here in this picture, if you have a block that's connected to a pulley, if you release the block it's going to accelerate this way. But because it's connected to the pulley, it's also going to cause the pulley to accelerate this way. The block falls in linear motion and the pulley accelerates around its central axis in rotational motion. We have both. In these questions, we're not going to use just Torque = I_, but instead we're going to use both Torque = I _ and F = ma. We're going to write sum of all forces equals ma for each acceleration we have. We're going to write sum of all torques equals I _ for each _ that we have. For each a, we write F = ma, and for each _, we write Torque = I _. What do I mean by each a and each _? You have to count how many motions exist in the problem. Here I have one object with one acceleration, one type of motion, linear motion. Then here I have another object with another type of motion. There's two motions in total. But if you had more types of motions and I'll get to that in the example below, you would use more than just two equations. We'll get to that. When you do this, when you combine F = ma with Torque = I _, you end up with an a and an _. That's two variables. One of the techniques we're going to use to solve these questions is instead of having a and _, we're going to replace _ with a. By doing this, instead of having a and _, I'm going to have a and a. Imagine in this second equation here, imagine if that _ somehow became an a, then you would have a there and a here. That's good because instead of having two variables, you now have one variable. This is a key part in solving this question and itÕs going from _ to a. The way we do this is by remembering that a and _ are connected. They're connected by this equation, a = r _ where r is the distance between the force and the axis of rotation. It's our r vector from the torque equation if you remember. Where r is distance, IÕm going to call this distance to the axis from the force. But this is actually not the equation we're going to use because what we're looking for is we're trying to replace _. What we're going to do is we're going to say _ equals a/r. Wherever we see an _, we're going to replace it with an a over R and that's going to simplify. We're actually going to using these questions a combination of three equations, F = ma, Torque = I _ and we're going to use this one here to link the two first equations. The last point I want to make here is that the signs for a and _, as well as the signs for V and w must be consistent. What do I mean by that? I'm going to give you one example that allows me to talk about these four variables and it's this one. You have a disc that's rolling. Actually, let me draw it. You have a disc that rolls up a hill. Imagine that if you are going this way and then you're going this way, you were spinning like this and then you go up the hill spinning like this. That's your V and this is your w. But if you're going up a hill, gravity is pulling you down so your acceleration is downhill. It's going to be like this. This means your _ is actually like this because if you're going up like this, it means that your w is like this then an acceleration that's down means that your _ is like this. All of these signs have to be consistent. In most of these problems because these are dynamics problems, theyÕre problems about acceleration, we're going to say that the direction of positive will follow acceleration. Let me add that here. Direction of positive will follow acceleration, so this should be positive which means V is negative. If V is negative, w has to be negative as well because V and w must have signs that are consistent with each other and _ must be positive. So a and _ are positive together. In this particular situation V is opposite to a, so it's negative, and w goes with V. So w goes with V and the sign of a goes with _. It's very important those signs are consistent. Let's do an example here. Before we solve an actual full length problem, what I want to do is show you some of the things that I've talked about here in a variety of different examples. Let's check it out. Here for each one of these, we want to solve for acceleration. First thing I want to know is which equations would you start with? How many equations and what kinds of equations? Then we're going to sketch a diagram. Remember what I told you that you write this point right here. We write F = ma for each a and we write Torque = I_ for each _. What that means is that you look at this system here. Let's say that this mass is bigger than this, so that the system tilts this way. This means you have an acceleration here and this thing is going to accelerate up and this thing is going to accelerate this way. These are linear accelerations. The pulleys are also going to accelerate in this direction. Notice that all the accelerations are consistent with each other. IÕm going to call this positive which means this is positive, which means this is positive, positive, positive. The direction of positive is this. You end up some weird stuff like this guy is positive down and this guy is positive up. That's okay. This is the same direction because it's a system and they're moving together. Because I have three a's and two _Õs, I can write five equations. I have three equations that I can write three F = ma, one for each one of these blocks and two Torque = I _, one for each of the disks. One last point I want to make here is that all of these aÕs are the same, a1 = a2 = a3 and all the _Õs are the same. We're going to call this a and all the _Õs are the same; _1 = _2 so I can just refer to _. Lastly, not only are all the aÕs the same, all the _Õs are the same, they're also connected: a = r_. All the accelerations are connected somehow. What I want to do quickly is for part B is sketch a diagram showing forces acting on these things. Let's start with this guy over here. Let's call this 1, 2, 3 and 1, 2. The guy at the bottom here, you have m1 is being pulled down by m1g and then there's a tension here, t1. If you are this pulley right here, IÕm going to just draw it over here. You have a tension, the same tension here pulling it down, t1, and then you have this tension here pulling you the other way, t2. Those are the only two forces that matter. If you are the block over here, let's call this guy m2, then you have t2 pulling you to the left and t3 pulling you to the right. If you're the disc, you're being pulled to the left by t3 and then you're being pulled down by t4. Then if you're the block, you're being pulled up by t4 and then being pulled down by m3g. This is m3. Obviously there is normal and m2g here. But these guys don't really matter because they just cancel each other. They don't really affect motion. Here's all this stuff. These are all the different forces. If you want to, we're going to quickly put some signs here. Remember, the direction of acceleration is this. When I write F = ma for this guy, this is positive and this is negative. This is positive and this is negative because it follows this direction. This is positive and this is negative. That being said, when you look at the torques, look at these guys here, these discs. The discs will have torques acting on them. There is a torque pulling this disc this way. If you got a disc, t1 is pulling the disc like this down. This is the torque due to t1 and this is the torque due to t2. We have to be consistent with signs. Spinning this way is negative in this problem because everything is going that way, so this is positive and negative. We're almost done. Got to squeeze these stuff in here. I got the torque of t4. You barely see that there and then I have the torque of t3. The t4 one is positive and t3 is negative. Hopefully you don't actually get a question like this but you might get a much simpler version. I'm just trying to show you. For particular case like this, what matters is the direction of the signs and all that stuff. Next one is much simpler. It's a simple yoyo, so the idea is that you have a cylinder here that you let loose and it spins. It accelerates down this way. Imagine if you're holding, if the string is here and the yo-yo is here then it starts falling, it's going to roll like this. It's going to roll like this that's because the string is on that side. If you had a yo-yo here, then you let it go. It's going to accelerate like this and it's going to accelerate like this. In this case, there are two different situations here, this is positive and this is positive. In this case this is positive and this is positive. Notice that in both cases, a down is positive but here the acceleration is positive this way. Here the acceleration is positive this way. It depends on the situation you have. But the key thing is that these two guys match each other. They gotta match. The accelerations have to both be positive together. How many equations do we have here? How many types of motion? I have one object with two motions. I have the same object has an a and an _. That means that IÕm going to write F = ma once for that object and IÕm going to write sum of all torques equals I _ once. Those are the number of equations. Let's look at all the forces. You only have one object. You have a force of tension that's pulling you up. Acceleration is down, _ is this way. This is positive, this is positive. Notice the tension was going up so tension is negative and tension causes a torque in this direction. Let me draw this a little bit differently. Let's do tension over here, it's negative and it causes a torque of tension this way. Does a torque of tension spin in the same direction as the _? It does. _ spins like this and the torque of tension spins that way as well so this is positive. Getting the sign here is very important. If you screw up the sign, you get the wrong answer. Something interesting that happens here is your tension is negative but the torque of tension is positive. That's okay. It's supposed to be like that. Don't freak out. Just because tension is negative doesn't mean that the torque of tension is positive. They're going to be different. LetÕs set up for this one and let's do this last one here. I have a cylinder rolling downhill. It's a similar situation as the simple yo-yo and that I have one object with two motions. I have one object that has both an a and an _, that's because it's rolling and moving at the same time. Therefore I can write sum of all forces equals ma once and I can write sum of all torques equals I _ once. Before we can start solving these questions you have to know which equations, how many equations you have to write. We have this object and I'll tell you, we'll talk about this more later, but IÕll tell you that the forces you have are mgx. Obviously, you have an mgy as well but mgy just cancelled with normal. These two guys cancel so they don't really do anything. There's a friction that goes this way. The friction is responsible for the only torque you have. There is a torque of friction. In this case you are going this way, V, and you have an acceleration this way. You're falling this way and you're getting faster this way. Your w is this way and your _ is this way as well. Down and spinning this way is positive. This guy is positive, this guy is negative. This torque here is in the same direction as _. All of these guys are positive, so this torque is positive as well. Notice that just like what we had here, tensional is negative but the torque of tension was positive. Here, friction is negative but the torque due to friction is positive. We'll check out these two examples later. But for now, I just want to introduce this. Again, just a big introduction. We're going to do more of these. Hopefully this makes sense. Let me know if you have any questions and let's keep going.

Example #1: Acceleration of block on a pulley

**Transcript**

Hey guys! Here we have an example of a torque acceleration problem that has two motions, two types of motion. Let's check it out. We have a block of mass m attached to a long light rope that's wrapped several times on the pulley. Let me draw the pulley, something like this and you have a rope and a block of mass m. It says the pulley has mass big M and radius R, and can be modeled as a solid cylinder. The fact that you see a solid cylinder here means we're going to use a moment of inertia equation of _ M^2. It is free to rotate about a fixed frictionless axis perpendicular to itself and through its center. Lots of words. I want to talk about some words here. First of all, long, light rope. Long just means you don't have to worry about running out of rope. It's wrapped around this thing so it's going to just keep going. Light means that the rope has no mass. Basically all problems will be like that so this is just standard language. Wrapped several times, again you donÕt have to worry about running out of rope. Free to rotate about a fixed axis; free to rotate means that the disc can rotate but it's a fixed axis which means that the axis isn't going to move sideways. It stays in place, so it spins but stays in place. Frictionless just means that there's no friction due to the axis here. There's no rotational friction. Perpendicular to itself and through its center; through its center is obviously just through the middle and perpendicular means 90 degrees. It means that the axis of rotation which is an imaginary line, I'm going to use my finger, is perpendicular to the disc so it makes degrees with the disc which means it looks like this. It means the disc spins around this axis but except that the axis is this way. You have a block hanging here. It's going to cause the disc to do this. ThatÕs it. All this stuff is standard language. Hopefully by now you're getting a hang of what all this crap means, with perpendicular axis to its center, it just spins like you would expect it to. It says that when a block is released from rest, so the block is released from rest, V initial equals zero. By the way, that also means that the w initial of the disc is zero. It begins to fall causing the pulley to unwind without slipping. Obviously, as you release this, it begins to fall. In other words, it has an acceleration down. Because it's connected to the pulley, it's going to cause the pulley to spin, the disc to spin. I'm going to have an _ this way as well. Remember a and _ have to match. Remember also that we're going to choose the direction of positive to be the direction of acceleration, which means this has to be a plus and this has to be a plus. Both of these guys are pluses even though this is clockwise which is usually negative, we're just going to override that that convention and use our own convention here of saying acceleration is positive. Without slipping, unwinds without slipping. This is standard language. Because it unwinds without slipping, we can say that the acceleration of the object equals r and then the acceleration of the pulley, which is a special connection between those two that we're going to have to use here. By the way, we can also say that V of the rope equals rw of the disc. These two equations link the linear variable to its angular equivalent by using little r, where little r is the distance to the axis just like little r in your torque equation. These two equations are possible. These two equations, these two connections exist because it says that you are rotating without slipping. That being said, you're always going to be rotating without slipping so you don't really have to worry about that. You don't have to look at that and say well what am I supposed to do with this? Nothing. This is standard language. That's always going to be there. In fact if it wasn't there, you wouldn't be able to use these equations. This question will be way harder and you wouldn't really be able to solve it without using more advanced physics. This is just standard language. I want to get that out of the way. How do we solve this? We're looking for both accelerations here, a and _. How do we solve this? First thing we have to do is we have to figure out how many types of motion we have and then we can figure out which equations and how many equations weÕre going to start with. I have this object has one motion which is a linear motion, so it has linear acceleration. This object has one motion as well which is a rotation, so it has a rotational acceleration. Because I have one a, I'm going to be able to write sum of all forces equals ma, and this is for the block and because I have one _, I'm going to write sum of all torques equals I _. By the way, this is just the process to find a. WeÕll talk about _ once at the end. We're first looking for a. We have these two equations and what we're going to do now is expand these two equations as much as possible. Let's look at this block here. What are the forces on the block? There's two forces. I have a tension going up and I have an mg, little mg, going down. This guy is positive this guy is negative because of the direction of positive for that object is going down. I hope that makes sense. It's a key part here you have to know. IÕm going to put mg positive plus negative T equals ma. There's nothing else I can do in this equation. This is my target variable but I don't have tension. I can't really solve this yet. What I'm going to do is I'm going to go to the second equation, expand this. Torque. We're talking about the disc obviously. Let me write this here. This is for the pulley, the disc, cylinder. There's only one torque. There's a force here. There's an mg that pulls this thing down but it doesn't cause a torque because if a force acts on the center of mass of an object, it doesn't cause a force. There's some force holding this thing up so it doesn't fall but that force also doesn't cause torque. The only force that's going to cause a torque is T. This pulley is being pulled down by T. You have a torque of T that looks like this which by the way will also be positive because it's going in the same direction as this. Both of these arrows are going this way. This is the direction of _, which is positive, so the torque will be positive as well. This torque is also positive. I have the positive torque of tension. I is the moment of inertia. We're treating this as a solid cylinder so _ MR^2. I have an _ here. Remember and this is key. This is really, really important. Most important part of this question is when we have an a in an _, which is what we have now, we're going to replace the _ with an a. a = r_, so _ = a/r. In this particular problem, little r happens to be big R because the distance from the axis of rotation to the point where the rope touches right there is the entire radius because the rope is at the edge of the disc. I'm going to rewrite _ as a/r. Let me highlight that, a/r. Notice now I have an a here and an a here. That's good news. Instead of a and _, I have a and a. That's awesome. Let's keep going here, see what we can do. I have to expand this equation here. The torque of any force F is Frsin_. Here the force is tension r sin_. r is this r right here. It's the little r from the torque equation which is the arrow, the vector, from the axis of rotation to the point where the force is applied. This little r happens to be big R here. The angle is the angle between these two, the angle between R and T, which is 90 which is awesome because that means this thing becomes a 1, equals (_ MR^2)(a/r). Really important, notice that this r cancels with this r and this r cancels with this r. Gotta be careful here. Don't get excited and just start canceling a bunch of crap. Make sure you cancel correctly. All the rÕs cancel. Good news. I end up with T = _ Ma. I can't simplify this equation anymore and I can't simplify this equation anymore but now I can combine the two. I can get this T and plug it in there. I can do that. That's what we're going to do. Notice that in one equation the T is negative and the other equation, the T is positive. This T is positive because this torque was positive and that's because it's spinning this way. This T is negative because for the mass, the tension is up. The reason IÕm pointing that out is so that you don't look at these two T's and freak out. Why is one positive, the other one is negative? It's fine. That's actually how it's supposed to be. We're going to put it up there. mg Ð _ Ma = ma. We're looking for a. Another thing, just to make sure you don't try this, you can't cancel the masses. Don't get excited and decide to cancel the masses. The Ms refer to different things. You have to be careful. Little m and big M are different things. To solve for a, we have to combine the a so I'm going to move this over here. I have little ma + _ Ma = mg. I have this here. I can factor out the a and solve. I'm going to quickly multiply this whole thing by 2 to get rid of the fraction there. 2mg = 2ma + Ma. I can factor the a here, so the a has in front of itself 2m and one Big M, so it's going to be (2m + M) = 2mg. a = 2mg / (2m + M). This is the final answer for Part A. Part B is much simpler. Basically, once you find one of the accelerations, finding the other acceleration will be much easier. For Part B, we're looking for _. To find _, just remember _ equals a/r, which in this case is a/R because little r happens to be the same as big R. a is this guy right here, so just plug it in. WeÕre going to have 1/R(2mg / (2m + M)). This is the final answer for _ and that's it. That's it for this one. Let me know if you have any questions and let's keep going.

Practice: Two blocks of masses m_{1} and m_{2} are both attached to a long, light rope that is wrapped several times around a pulley, as shown below. The pulley has mass M and radius R, can be modeled as a solid cylinder, and is free to rotate about a fixed, frictionless axis perpendicular to itself and through its center. When the block is released from rest, it begins to fall, causing the pulley to unwind without slipping. Derive an expression for the angular acceleration of the pulley.

Practice: When you release a simple 100-g yo-yo from rest, it falls and rolls, unwinding the light string around its cylindrical shaft, which is 2 cm in radius. If the yo-yo can be modeled after a solid disc, calculate its linear acceleration.

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