Rotational Dynamics with Two Motions - Video Tutorials & Practice Problems

On a tight schedule?

Get a 10 bullets summary of the topic

1

concept

Rotational Dynamics with Two Motions

Video duration:

16m

Play a video:

Hey, guys. So in this video, we're going to start talking about Tory acceleration problems, also known as rotational dynamics problems where we have two types of emotions. What that means is that on top of having a rotation, we're also gonna have objects moving inland linear direction, so it's gonna 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 Alfa we use the rotational version of Newton's second law. So instead of Africa's, they may. We're gonna write torque. Some of all torques equals I Alfa Okay, so that we have torque Alfa questions. That's how we solve it. But in some problems, we're gonna have more than just Alfa. We're going to have rotational and linear motion. So, for example, here in this picture, um, if you 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 that central access in rotational motion, we have both. And in these questions, we're not going to use just, um, Tourky, Cozy Alfa. But instead, we're gonna use both Torque equals that Alfa and efficacy may. Okay, so we're going to write some of our forces equals m A for each acceleration we have, and we're going to write some of all torques equals Alfa for each Alfa that we have. So for each A, we write articles I may. And for each Alfa we write, torque equals Alfa. Now, what do I mean by each day and each Alfa, we have to count how many emotions exists 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 two motions in total. Okay, But if you had Mawr types of motions and I'll get to that in the example below, you would use more than just two equations. All right, we'll get to that. Um, when you do this, when you combine f equals a made with torque equals Alfa, you end up with an A and an Alfa. That's true variables. One of the techniques we're gonna use to solve these questions is instead of having A and Alfa, we're going to replace Alfa with a And by doing this instead of having A and Alfa, I'm going to have a and A imagine in this second equation here. Imagine if that Alfa somehow became an A then you would have a there in a here. And that's good, because instead of having two variables, you now have one variable. This is a key part in solving this question is going from, um, Alfa Teoh A. Okay, And the way we do this is by remembering that a and Alfa are connected. They're connected by this equation a equals R Alfa, where r is the distance between, um it's the distance between the force and the axis of rotation. So it's our our vector from the torque equation, if you remember. So where are is distance? I'm gonna call this distance to access. Okay, distance to the access 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 Alfa. So what? We're gonna do is we're gonna say Alfa equals a over our Wherever we see an Alfa, we're gonna replace it with a over our and that's going to simplify. Okay, so we're actually going to using these questions a combination of three equations because it may torque equals Alfa. And we're gonna use this one here to link the two. The two first equations. Alright, The last point I wanna make here is that the signs for a an Alfa as well as the signs for V in omega 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 four variables, and it's this one. You have a disk that's rolling. Um, actually, let me draw it. You have a disk that rolls up a hill. Okay, so imagine that if you are going this way, right, and then you're going this way. Okay, you're drive over here. You are spinning like this and then you go up the hill spending like this. So that's your V. And this is your omega. 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 hey, and this means your Alfa is actually like this. Okay, Because if your velocity if you're going up like this, means that you're Maiga's like this, then I'm acceleration. That's down means that you're alphas like this, Alright? And all of these signs have to be consistent. So in most of these problems, because these are dynamics problems, they're problems about acceleration. We're going to say that the direction of positive will follow acceleration. Okay, so let me add that here, um direction off. I meant to make a little star direction of positive will follow acceleration. So this would be positive, right? Which means V is negative. And if these negative omega has to be negative as well, because via Omega must have signs that are consistent with each other and Alfa must be positive. OK, so an Alfa positive together in this particular situation, V is opposite way. So it's negative and Omega goes with V. So omega goes with V and the sign of a goes with Alfa. Okay, It's very important. Those signs of consistent All right, let's do an example here before we solve an actual full length problem. What I wanna 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. So here, for each one of these, we want to solve for acceleration. First thing I wanna know is which equations would you start with? How many equations? What kinds of equations. Alright, so and then we're going to sketch a diagram. So remember what I told you. You write. Remember what I told you? That you write this point right here. We write f equals M A for each A and we right, 20 goes Alfa for each for each alfa. So what that means is that you look at this on the system here. Let's say that this mass is bigger than this so that the system tilts this way. Right? This means you have an acceleration here. Um and this thing is gonna accelerate up. And this thing is going to accelerate this way. These air linear accelerations, the pulling they're also gonna accelerate in this direction. Notice that all the accelerations are consistent with each other. Okay, I'm gonna call this positive, Which means this is positive. Which means this is positive. Positive, Positive. So the direction of positive is this Okay, you end up some weird stuff like, this guy is positive down and this guy's positive up. That's okay. This is the same direction because it's a system and they're moving together. So because I have three days and two alphas, I can write five equations. Alright, So I have three equations that I can write three Africans. They may one for each one of these blocks and to torque equals Alfa, um, one for each of the disks. Okay, One last point I wanna make here is that all of these aides are the same. A one equals a two equals a three and all of the officers the same. So it's gonna call this a and all the officer, the same Alfa one equals Alfa too. So I could just refer to Alfa. And lastly, not only are all the is the same all the office the same they're also connected A equals r alfa. So all the accelerations are connected somehow. Okay, Now what I wanna do quickly is for part B is sketched a diagram showing forces acting on these things. So here I have. Um let's start with this guy over here. Let's call this one, 23 one to Okay. So the guy at the bottom here you have m one is being pulled down by M one g. And then there's a tension here. T one. Okay, if you are this pulley right here, I'm gonna just draw it over here. Um, you have attention. The same tension here, pulling him down t one. And then you have this tension here pulling the other way. T two. Okay, Those are the only two forces that matter if you are the block over here, Um, let's call this guy. Mm, too. Then you have t to pulling you to the left and t three, pulling you to the right. If you're the disk, you're being pulled to the left by t three. And then you're being pulled down by, um, tea four. And then if you're the block, you're being pulled up by t four and then being pulled down by mm. Three G. This is entry. Obviously, there is normal and m two g here. But these guys don't really matter because they just cancel each other. They don't really affect motion. Okay, so here's all the stuff now, these are all the different forces. If you want Thio, we're gonna quickly put some signs here. So remember the direction of acceleration is this Okay? So when I write, f equals I made for this guy, this is positive, and this is negative. Okay, this is positive. And this is negative because because it follows this direction, this is positive, and this is negative. Okay, Now, that being said, when you look at the torques look at these guys here. These discs, the disks who have torques acting on them. So there is a torque pulling this disc this way. T one right here. If you got a disk, T one is pulling the disk like this down. Right s. So this is the torque due to t one. And this is the torque duty to 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, and we're almost done. Got to squeeze this stuff in here. Um, I got the torque of t four barely see that there. And then I have the torque off T three. The t 41 is positive. NT three is negative. Okay, 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 all the you know, for a particular case like this, what matters? Use the direction of the signs and all that stuff. OK, eso next one is much simpler. It's a simple yo yo. So the idea is that you have a cylinder here that you let loose and it spins so it accelerates down this way. Now imagine if you're holding the If the string is here and the yo yo is here and then it starts falling. It's gonna roll like this. Okay, so it's going to roll like this now. That's because the strings 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, they're two different situations here. In this case, this is positive and this is positive. In this case, this is positive. This is positive notice that in both cases, a down is positive. But here the acceleration is positive. This way. Here, the accelerations. Positive. This way. All right, so it depends on the situation you have. But the key thing is that these two guys match each other. Okay, you got a match. So the acceleration have to both be positive together. All right. So how many equations do we have here? Well, how many types of motion I have? One object with two motions. One object with two motions. I have the same object has an A in an Alfa. So that means that I'm gonna write f equals in May once for that object. And I'm gonna write some of all torques equal Scialfa once. Okay, Those are the number of equations. Now, let's look at all the forces. You only have one object. You have a force of tension that's pulling you up. Um, acceleration is down. Alfa is this way. Okay, so this is positive. This is positive. Noticed. Attention is going up, so tension is negative and tension causes a torque. Tension causes a torque in this direction. Um, let me draw this a little bit differently. Um, how do I put this trying to move this away from here? Let's do this. Let's do tension over here. It's negative, and it causes a torque of tension this way. Okay, Now does the talk of tension spinning the same direction is the Alfa. It does Alfa spin like this, right? And the torque of tension spends that way as well. So this is positive. Okay, so getting the sign here is very important if you scrub to sign to get the wrong answer, Okay, Something interesting that happens here is your attention is negative, but the torque of tension is positive. That's okay, right? So 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. So that's that's that's set up for this one. And let's do this last one. Here I have a cylinder rolling downhill. Um, it's a similar situation as the simple yo yo and that I have one object with two motions. Okay, so I have one object that has both in a in an Alfa. That's because it's rolling and moving at the same time, right? therefore, Aiken. Right? Um, some of our forces, because it made once and I can write some of the torques equals I Alfa once. So before you can start solving these questions, you have to know which equations how many equations you have to draw. You have to, right? Okay, now, here we have this object here, and I'll tell you, we'll talk about this more later. But I'll tell you that the forces you have are m g X. Obviously you have an M g y as well, but MG just canceled with normal. These two guys cancels, they don't they don't really do anything. And there's a friction that goes this way. And the friction is responsible for the Onley torque you have. So there is a torque off friction. Now, in this case, you are going this way. The ends. Actually, you're like this. You have a V this way, and you have, um, on acceleration this way. Okay. You're following this way. You're getting faster this way. So your omega is this way, and your Alfa is this way as well. Okay, so down and spinning this way is positive. So this guy is positive. This guy's negative. Um, this torque here, this torque here is in the same direction as Alfa. Right? All of these guys were positive. So this torque is positive as well. Notice that. Just like what we had here. Tension was negative, but the torque of tension was positive here. Frictions Negative, but the torque due to friction is positive. Okay, well, check out these two examples later, but for now, I just wanna introduce this cool. So again, just a big introduction. We're gonna doom or of these? Hopefully, this makes sense. Let me know if you have any questions and let's keep going.

2

example

Acceleration of block on a pulley

Video duration:

11m

Play a video:

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.

3

Problem

Problem

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.

A

α = g(m_{1}−m_{2}) / (m_{1}+m_{2})

B

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

C

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

D

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

4

Problem

Problem

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.

A

3.3 m/s^{2}

B

4.9 m/s^{2}

C

6.5 m/s^{2}

D

Not enough information to calculate

Do you want more practice?

We have more practice problems on Rotational Dynamics with Two Motions