Hey, guys. So in this video, we're going to start talking about torque acceleration problems, also known as rotational dynamics problems, where we have 2 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. So it's going to have a combination of both of them. Let's check it out. So remember, that when we have problems where a torque causes an angular acceleration, where torque causes an angular acceleration α, we use the rotational version of Newton's second law. So instead of F=ma, we're going to write torque, sum of all torques equals Iα. Okay? So then when we have torque α questions. That's how we solve it. But in some problems, we're going to have more than just α. We're going to have rotational and linear motion. So 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. So the block falls in linear motion and the pulley accelerates around its central axis in rotational motion. We have both. And in these questions, we're not going to use just torque equals Iα, but instead, we're going to use both torque equals Iα and F=ma. Okay? So we're going to write sum of all forces equals ma for each acceleration we have, and we're going to write sum of all torques equals Iα for each α that we have. So for each a, we write F=ma and for each α we write torque equals Iα. Now what do I mean by each a and each α? 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. And then here I have another object with another type of motion. So there's 2 motions in total. Okay? But if you had more types of motions and I'll get to that in the example below, you would use more than just 2 equations. Alright? We'll get to that. When you do this, when you combine F=ma with torque equals α, you end up with an a and an α. That's two variables. One of the techniques we're going to use to solve these questions is instead of having a and α, we're going to replace α with a. And by doing this, instead of having a and α, I'm going to have a and a. Imagine in this, second equation here. Imagine if that α somehow became an a, then you would have a there and a here. And that's good because instead of having 2 variables, you now have one variable. This is a key part in solving this question is going from, α to a. Okay. And the way we do this is by remembering that a and α are connected. They're connected by this equation. a=rα, where r is the distance between, it's the distance between the force and the axis of rotation. So it's our r vector from the torque equation if you remember. So where r is distance, I'm going to call this distance to the axis. Okay. Distance to the axis from the force. Okay. But this is actually not the equation we're going to use because what we're looking for is we're trying to replace α. So what we're going to do is we're going to say α=a/r. Wherever we see an α, we're going to replace it with a/r and that's going to simplify. Okay. So we're actually going to use in these questions a combination of 3 equations. F=ma, torque equals Iα, and we're going to use this one here to link the 2, the 2 first equations. Alright. The last point I want to make here is that the signs for a and α, as well as the signs for v and ω must be consistent must be consistent. What do I mean by that? So I'm going to give you, one example that allows me to talk about these 4 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. Okay. So imagine that if you are going this way. Right? And then you're going this way. Okay. Your I'll draw it over here. You're spinning like this and then you go up the hill spinning like this. So that's your v and this is your ω. Okay? 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. And a and this means your α is actually like this. Okay. Because if your velocity if you're going up like this, means that your ω is like this. Then an acceleration that's down, means that your α is like this. Alright. And all of these signs have to be consistent. So in most of these p
Rotational Dynamics with Two Motions - Online Tutor, Practice Problems & Exam Prep
Rotational Dynamics with Two Motions
Video transcript
Acceleration of block on a pulley
Video transcript
Hey guys. So here we have an example of a torque acceleration problem that has 2 motions, 2 types of motion. Let's check it. So we have a block of mass m attached to a long, light rope that's wrapped several times around the pulley. So 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. So I'm going to sort of do this and say m and 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 half mr2. And 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 don't have to worry about running out of rope. "Free to rotate about a fixed axis" 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. Right? There's no rotational friction. "Perpendicular," perpendicular to itself and through its center. "Through its center" is obviously, it's 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, goes is perpendicular to the disc, so it makes 90 degrees of the disc, which means it looks like this. It means the disc spins around this axis, right? So, but except that the axis is this way, right? So you have a block hanging here. It's gonna cause this to do this. Right? That's it. So 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 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 0. By the way, that also means that the ω initial of the disc is 0. It begins to fall causing the pulley to unwind without slipping. So obviously, as you release this, it begins to fall. In other words, there's an acceleration down, and because it's connected to the pulley, it's gonna cause the pulley to spin, the disc to spin. So I'm going to have an α this way as well. Remember, a and α 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 going to override that convention and use our own convention here of saying acceleration is positive.
Without slipping, unwinds without slipping. This is standard language, right? Because it unwinds without slipping, we can say that the acceleration, of the object equals r and then the acceleration of the pulley, right, which is a special connection between those 2 that we're going to have to use here. And by the way, we can also say that v of the rope equals rω of the disk. These two equations link the linear variable to its angular equivalence by using little r, where little r is the distance to the axis, just like little r in your torque equation. Okay. These two equations are possible. 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 would be way harder, and you wouldn't really be able to solve it without using more advanced physics. Cool. So just standard language, I wanna get that out of the way. So now, how do we solve this? We're looking for both accelerations here, a and α. 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 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 1 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 1 α, I'm going to write let's go over here. I'm going to write, sum of all torques equals Iα. Okay. By the way, this is just a process to find a. We'll talk about α once at the end. Okay? So we're first looking for a. Alright. So we have these 2 equations and what we're gonna do now is expand these 2 equations as much as possible. So let's look at this block here. What are the forces on the block? Well, there's 2 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. Cool? So if that makes sense, that's a key part here to know. So I'm 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. So what I'm going to do is I'm going to go to the second equation, expand this, torque. Now we're talking about the disc, obviously. Let me write this here. This is for the pulley, the disc, the cylinder. 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 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. So this pulley is being pulled down by T. So you have a torque of T that looks like this, which by the way, will also be positive because it moves, it's going the same direction as this, right? Both of these arrows are going this way. This is the direction of α 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. I is the moment of inertia. We have a we're treating this as a solid cylinder, so halfmr2. Okay. Now I have anα here. Here. Remember, and this is key. This is really, really important. Most important part of this question is when we have an a and an α, which is what we have now, we're going to replace the α with an a. And we do this by using v sorry, a. a equals rα. So α equals a over little r. In this particular problem, little r happens to be, sorry, I meant to write little r. In this particular problem, little r happens to be big R. Okay. So here, 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 disk. Okay? So I'm going to rewrite, I'm going to rewrite α as a over r. Let me highlight that, a over r. So notice now I have an a here and an a here. That's good news. Instead of a and α, I have a and a. That's awesome. So let's keep going here and see what we can do. I have to expand this equation here. So the torque of any force, torque of any force f is fr sin of θ. So here the force is tension, r sin of θ. 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. So this little r happens to be big R here. The angle is the angle between these 2. Right? The angle between r and t, which is 90, which is awesome because that means this thing becomes a 1, equals halfmr2 a over r. Okay? And really important, notice that this r cancels with this r, and this r cancels this r. Right? Gotta be careful here. Don't get excited and just start canceling a bunch of crap. Make sure you cancel correctly. All the r's cancel. 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 I can combine the 2. I can get this t and plug it in there. Okay? I can do that. That's what we're going to do. Now 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. So the reason I'm pointing that out is so that you don't look at these 2 t's 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 going to put it up there, put it here, mg minus half m a equals ma. 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 and decide to cancel the masses. The m's 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's. So I'm going to move this over here. I have little ma plus half big m a little mg. And now I have this here. Here, I can factor out the a and solve. I'm going to quickly multiply this whole thing by 2 to go to the fraction there. So 2mg equals 2ma plus M a. I can factor the a here. So the a has in front of itself 2m and 1 big M. So it's going to be 2m+big M equals 2mg. a equals 2mg over 2mg over 2m+big M. And 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 α. And to find α, just remember, α equals a over little r, which in this case is a over big R because little r happens to be the same as big R. a is this guy right here. So just plug it in. So we're going to have 1 over R times 2 mg2m+big M. This is the final answer for α, and that's it. That's it for this one. Let me know if you have any questions, and let's keep going.
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(m1−m2) / (m1+m2)
α = (g/R)(m1−m2) / (m1+m2+M)
α = (g/R)(m1+m2) / (m1+m2+M)
α = (g/R)(m1−m2) / (m1+m2+M/2)
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.
