Two blocks of masses m1 and m2 (m1 > m2) 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.

α = (g/R)(m_{1}−m_{2}) / (m_{1}+m_{2}+M/2)

Two blocks of masses m 1 and m 2 (m 1 > 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.

Hey, guys. Hope you tried this practice problem. It's another classic in physics. Here we have 2 blocks on a pulley and the setup is called Atwood's machine. Looks like this, Right? And the idea is that one block will be heavier than the other, and the system will move in the direction, or go down in the direction of the heavier objects. Here, we're looking for an expression for the angular acceleration of the pulley. We're looking for alpha. Okay. So let's check it out. So we have masses m 1 and m 2. M 1 is the greater one of the 2, which means this is going to move this way. So this acceleration, a 1, goes like this. There's an acceleration a 2 that goes like this. If block 1 goes down, it pulls block 2 up and the disc will spin like this with an alpha. I want to point out that even though I wrote a 1 and a 2, both of these are actually the same so we can just call it a and that's much easier. Okay? Now to solve this question, this is a question that's asking for an acceleration. So we're going to use f equals m a. But because we also have rotation, the disc rotating, we're also going to use torque equals I alpha. Okay. And we're gonna actually write f equals m a twice because we have 2 objects that are moving linear that linear motion that have linear linear accelerations. So I can write f equals m a for each one of these guys. And I'm going to write 1 torque equals I alpha for the pulley. Okay? So let me go ahead and write that sum of all forces for objects 1 equals m 1 a. Sum of all forces for 2 is m two a and sum of all torics equals I alpha. Let me actually space this out a little bit more. Alright. So what are the forces acting on 1? 1 has obviously a tension here. We're gonna call this t one and it has a weight force m one g. Look at the direction of acceleration. It's going down. So that's the direction of positive, which means this guy is positive, this guy is negative. On the second block, I have t 2 and I have m 2g. Look at the direction of the acceleration, it's up, which means this is a direction of positive, which means t 2 is positive and m2g is negative. Okay? Notice how the t's have different signs. The m g's have different signs as well. Now before we start plugging stuff into here, I wanna talk about a very important point, which is these tensions are not the same. K. These tensions are not the same. Let's write this down. T one is not the same as t 2. This is only the case this is only the case. You've solved problems. You might have solved problems where the tensions were the same. That's when police had no mass. If a pulley has mass, if a pulley has mass, it has moment of inertia, it has resistance. So if it has mass, the tensions cannot be the same. So I'm gonna write here since, mass of pulley is not 0. Okay? Before we did, rotational dynamics, we did linear dynamics. You would have solved this problem using f equals MA exclusively, and you would have ignored the mass of the pulley, and the tensions would have been the same. Here, it's not the point. It's not the case. Okay? Now, which tension do you think needs to be stronger? Well, imagine that the pulley is rotating this way, and you can think of the pulley as there's a little bit of a tug of war, happening with the pulley, right? T 1 and t 2 are both pulling on the pulley. Well, if the pulley spins in the direction of t 1, it means that t 1 is greater than t 2. And that happens because m 1 is greater than m 2. Okay. So they can't be the same. If they were the same, the disc wouldn't spin. Okay. You can only say that the tensions are the same if the mass of the pool is ignored but that's several chapters ago. Alright? So let's let's go ahead and plug this in here. The forces for m one are negative t one plus m oneg equals m a. Here, for t dot for object 2, the forces are positive t2+negativem2g equalsma. Now this block, the cylinder here has 2 torques acting on it. This was the direction of acceleration. It has the torque of t one this way, and it has a torque of t two this way. The one that goes in the direction of alpha, this one, t one is going to the positive one because this is the direction of positive, and this one's negative. So up here, I'm going to write that I have a positive torque t 1 minus or negative torque t 2 equals I alpha. Okay. I alpha. Here we have a solid cylinder. Okay. The pull is a solid cylinder So I'm going to replace I with half m r squared and a, and I have alpha here. Remember, we're supposed to rewrite alpha as a over r. Right? Let me remind you real quick. If we have a and alpha, what we're supposed to do is rewrite alpha as a. Okay. What you're supposed to do is rewrite alpha as a. And you do this using a equals r alpha. So alpha equals a over r. In this particular problem, r is going to be radius. That's actually what it is most of the time. R happens to be the radius, so alpha is going to be a over big r. So whatever you see in alpha right here right here, you're going to rewrite it as an a over r. The reason you do this is so that instead of having a, a, and alpha, you have a, a, a, and then you can solve the problem. Now you might be thinking Let me just section this off here. You might be thinking, we're looking for alpha. So why the heck would you get rid of alpha? Like that seems like you're getting farther from the answer. Right? Well, what we're gonna do is we're gonna replace alpha with a, find a, and then go back to alpha. Okay. So the the reason I do this is because if you get a bunch of a's, your r right here will cancel. That's nice. Also, I'm trying to give you a, one sequence of steps to solve problems whether you're being asked for a or alpha, so you're always doing sort of similar things. So even though we're asking for alpha, we're going to find a and then we're going to find alpha. So I propose that you do this no matter what variable you're being asked, a or alpha, always get a first. Cool? Alright. So a over r. Notice, that this r here is going to cancel with this r. K. There's not much we can do with these equations here with 12. That's usually the the the case. And then all of the expansion of variables are gonna happen here. So what I mean is I'm gonna replace t with its torque equation. I'm gonna replace torque of t with the torque equation. Remember, the equation for torque, let me work this out here. Torque of a force is f r sine of theta. In this case, we have a tension, so it's t r sine of theta. Both tensions, if you look up here, are have a little r vector that goes from the middle to the edge. So both little r's are equal to the radius and both of them make 9 degrees of the tension. So here's a tension pulling down on the cylinder and they make 90 degrees. So for both torques, this is going to be big r, this is going to be 90 degrees. So this thing is going to be 1. So the torque of both tensions will just be t r. Okay? So here, you're going to have t one r minus t two r equals half m r a. Notice that there's r's everywhere and now this cancels, which is nice. K. And you end up with t 1 minus t 2 equals half m a. Notice how you cannot simplify this further. Okay? So you have the simplest version of all of these three equations, and now we're going to combine. So you write as many f equals m as you need to, as many torques equals I alpha as you need to. You expand the equations as much as you can, all of them. And when they're fully expanded, fully simplified, we're going to combine all of them. Okay? One thing one way you could do is you could solve for t one here and plug it here, solve for t two here and plug it here. The reason why you would be plugging into this third equation is because it's the equation that has both t 1 and t 2 or, which is much better, you are going to just add up all these three equations. And you can do this for any number of equations you have. Right? Even if you just had 1 f equals m a, 1 torque equals I alpha, you could add up all the equations. So I'm going to write the equations here. T one negative plus m oneg equals ma. Just to be clear, this is monea, this ism2a, so monea. The second equation is positvetwo minus m two g equals m two a. And the third equation right here is t one minus t two equals half big m a. We're going to add up all these 3 equations and look what happens. It's pretty awesome. T 1 positive cancels the t one negative, t two positive cancels t two negative. This should happen a 100% of the time. All your tensions will cancel. Tension here really is a connecting variable. It's just connecting the 2 masses together. It doesn't do anything. You can also think of it as the thing that connects the equation. It connects the motion, so it connects the equations. It just gets you from these 3 different versions, into one version that shouldn't have any t's in it. Okay? So your t's, which should always cancel, which is awesome. So you end up with a very simple final equation. It's going to be m 1g minusm2g 2g equal m 1a+2a+alfm a. Notice I have a a a. So what you would do is you would factor out the a's and solve for it. I'm gonna multiply by 2 actually, you know what? I'm not gonna multiply by 2. I'm gonna handle I'm gonna I'm gonna put up with this one little half instead of having a bunch of twos everywhere. Cool? So a gets factored out. There's 3 things, 3 terms with a. Here are the coefficients in front of a, and those are the things that go in the parenthesis. M 1 +m2+alfbigm equals this stuff here. If you want to get fancy, you can also fact out the gm1 minuteusm2. So a is gm1 actually, I'm going to do it like this, m 1 minus m 2. I'm going to put the g out here. Okay. That's legal. And on the bottom, m one plus m two plus half of m. K. And this is the close to the final answer. We're looking for alpha. We're almost there. But before I find alpha, I want to talk about this real quick. Notice what I have here, a equals a bunch of crap times g. Okay. If you had released, if you release a block without any strings or whatever, it would fall with an acceleration of g. Here, notice that the answer, the form of the answer, in a problem that has rotation is similar to the form of the answer in a problem that has only linear motion. This is linear only, and this is with rotation. They look similar. It doesn't look it actually doesn't seem very similar at first because we have all this crap here. But what this really is is a equals some crap g. So in linear, you have this. In rotational, something shows up in front of the g. The point that I'm trying to make is that you should expect, problems of rotation that have an answer that looks similar in form as it would in, in linear only motion. The other point here is that this crap here, this little box in front of g has to be less than 1. And that's because when you're falling, if you if something's falling and pulling something else, and then it's it's causing it to some rolling and stuff, you're obviously gonna fall a little bit slower. So all of this crap should be less than 1. If you had all these numbers and you multiplied everything, you should get something, a final acceleration that's less than 9.8, and you would know that you're probably correct. If it was more 9.8, you would know that you're definitely wrong. Alright? So that's an interesting point there, that these two look similar in form. Okay? But what we're really looking for, is alpha. Well, alpha equals a over r and then we just have to put an r here. So I'm going to do, I'm going to do g over r. Just going to put the r right there for alpha And then this big old mess here, m 1 minuteus m 2, m 1 plus m 2 plus half of m. K. So what's really hard is finding, not really hard, but the really long part or the longer part is, it's not that bad, is finding a and then finding r, finding alpha, you just divide by r. Cool? So that's it for this one. Hopefully, it made sense. Let me know if you have any questions. Let's keep going.

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