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Acceleration of block on a pulley

Patrick Ford
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Hey, guys. So here we have an example of a torque acceleration problem that has two motions, two types of motion. Let's check it out. So we have a block of mass M attached to a long like rope that's wrapped several times on the pulley. So let me draw the pulley something like this, and you have a rope and a block of Mass. M says the pulley has mass big M and Radius R someone to sort of do this and say M and R and can be modeled as a solid cylinder. Fact that you see a solid cylinder here means we're going to use a moment of inertia equation of half M R Square. And it is free to rotate about a fixed, frictionless axis perpendicular to itself and to its center. Lots of works. I want to talk about some words here. First of all, um, long like rope Long just means you don't have to worry about running out of rope. Um, it's wrapped around this thing, so it's gonna just keep going. Light means that the rope has no mass. Basically, all problems will be like that. So this is just standard language um, wrap several times again, you have to worry about running out of rope. Uh, do free to rotate about a fixed axis. Free to rotate means that the disc can rotate, but it's a fixed access, which means that the access isn't going to move sideways. It stays in place so it spends what is in place. Friction. It's just means that there's no friction due to the axis here, right, there's no rotational friction. Um, perpendicular, perpendicular to itself and through its center through its center is obviously is just through the middle on perpendicular means 90 degrees. It means that the axis of rotation, which is an imaginary line I'm gonna use my finger, um goes is perpendicular to the disc. So it makes 90 degrees of the disc, which means it looks like this. It means that disk spins around this axis, right, So but except at the axis is this way, right? So you have a block hanging here. It's gonna cause it this to do this. That's a eso. All of this stuff is standard language. Hopefully, by now you're getting a hang of what all this crap means with perpendicular access to the center. Um, it just means, like you would expect that you It says that when the block is released from rest, so the block is released from rest the initial go zero. By the way, that also means that the Omega initial of the disk is zero. It begins to fall, causing the pull it to unwind without slipping. So obviously, as he released this, it begins to fall. In other words, as an acceleration down, Um, and because it's connected to the pull it, it's gonna cause it the pulley to spin the disk to spin. So I'm gonna have an Alfa this ways of Well, remember, A and Alpha 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, right? So both of these guys are pluses. Even though this is clockwise, which is usually negative, we're just gonna override that that convention and use our own convention here of saying acceleration is positive. Okay, without slipping unwinds without slipping. Uh, this is standard language, right? Because it unwinds without slipping. We can say that the acceleration off the of the objects equals R, and then the acceleration of the pulley right, which is a special connection between those two that we're gonna have to use here. And by the way, we could also say that the, uh of the rope equals are omega of the disc. These two equations link, uh, the linear variable to its angular equivalent by using little are where little R is the distance to the axis, just like little are in your torque equation. Okay, these two equations are possible. These two equations, these two connections exist because it says that you are rotating without slipping. Now, 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 be way harder, and you wouldn't really be able to solve it without using more advanced physics. So just standard language. I want to get that out of the way. So now how do we solve this? We're looking for both accelerations here. A and Alfa. How do we solve this? Well, first thing we have to do is we have to figure out how many types of motion we have. And then how many types of and then we can figure out which equations and how many equations we're gonna start with. So I have this object has one motion, which is a linear motion. So it has linear acceleration, and this object has one motion as well, which is a rotation. So it has a rotational acceleration. So because I have won a I'm going to be able to write some of all forces equals I may. And this is for the block. And because I have one Alfa I'm going to write. Let's go over here. I'm going to write, um, some of all torques equals I Alfa. Okay, by the way, this is just a process to find a We'll talk about Alfa once at the end. Okay, so we're first looking for a All right. So we have these two equations and what we're gonna do now is expanding to equations as much as possible. So let's look at this block here. What are the forces on the block where there's two forces. I have attention going up, and I have an MG little mg going down. This guy's positive. This guy's negative because of the direction of positive for that object is going down. Okay, so if that makes sense, that's a key part You have to know. So I'm gonna put em g positive plus negative t equals m A. There's nothing else I can do in this equation. This is my target, variable. But I don't have tension. I can't release all this yet. So what I'm gonna do is I'm gonna go to the second equation, expand this torque. Now we're talking about the disk. Obviously, let me write this here. This is for the poli the disk with cylinder. Um, there's only one torque, right. So 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 in 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 gonna cause a torque is t. So this poll it is being pulled down by t. So you have a torque of tea that looks like this, which, by the way, it will also be positive because it moves. It's going the same direction as this. Right? Both of these arrows, They're going this way. Um, this is the direction of Alfa, which is positive. So this is going to be the torque will be positive as well. Okay, so this torque is also positive. So I have the positive torque of tension eyes, the moment of inertia. We have a We're treating this as a solid cinders. So half m r square. Thank Now I have an Alfa 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 Alfa, which is what we have now we're going to replace the Alfa within a and we do this by using V. Sorry. A A equals R Alfa. So Alfa equals a over. Little are in this particular problem. Little are happens to be sorry. I meant to write little art. Um, in this particular problem, little are happens to be big are Okay, so here little are happens to be big art because the distance from the axis of rotation to the point where the rope touches right there is the entire radios because the rope is at the edge of the disc. Okay, so I'm going to rewrite. I'm going to rewrite Alfa as a over our Let me highlight that a over our So notice. Now I have a here and in a here. That's good news. Instead of a an Alfa, have a That's awesome. So let's keep going here, see what we can do. I have to expand this equation here. So the torque of any force torque of any force f his f r sine of data. So here the forces tension are sign of data. Are is this are right here It's the little are from the target equation, which is a, uh, the arrow the factor from the axis of rotation to the point where the force is applied. So this little are happens to be big are here The angle is the angle between these two, right? The angle between R and T, which is 90 which is awesome, because that means this thing becomes a one equals half m r squared a over our okay and really important Notice that this are cancers with this R and this are cancels with this are right. You gotta be careful here. Don't get excited and just start canceling a bunch of crap. Make sure you cancel correctly. All the arts canceled. Good news. I end up with t equals half M A. So I can't simplify this equation anymore. And I can't simplify this equation anymore. But now it can combine the two. I could get this tea and plug it in there. Okay, I can do that. That's what we're gonna do. Now. Notice that in one equation that t is negative and the other equation to tease positive. This tea is positive because this torque was positive on. That's because it's spinning this way. This team is negative because for the mass, the tension is up. So the reason I'm pointing that out so that you don't look at these two ts and freak out why is one positive? The other one's negative. It's fine. Okay, that's actually, how it's supposed to be. So we're gonna put it up there. Um, here. M g minus half m A equals M A. We're looking for a another thing Just to make sure you don't try this. You can't cancel the masses, right? Don't get excited. Decide to cancel the masses. Uh, the EMS referred to different things. Have to be careful. Little m and big m are different things to solve for a we have to combine the ace. I'm gonna move this over here. I have little I may plus half big, um, a little MGI. And now I have this here, Here, I can factor out the A and salt. I'm gonna quickly multiply this whole thing by to to get rid of the fraction there. So to m g equals to m A plus m A. I can factor the A here. So the A has in front of itself to m and one big M. So it's gonna be to m plus Big M equals two mg a equals two mg over two mg over to m plus big. Um, and this is the final answer for part a, um, parte things much simpler Basically, once you find one of the accelerations finding the other acceleration would be much easier for party. We're looking for Alfa and to find Alpha. Just remember, Alpha equals a over. Little are, which in this case, is a over big are because little are happens to be the same as big are a Is this guy right here? So just plug it in. So we're gonna have one over r times two mg to M plus big. Um, okay, this is the final answer for Alfa and that's it. That's it. Finished one. Let me let you have any questions and let's keep going.