13. Rotational Inertia & Energy

Torque with Kinematic Equations

# Torque with Kinematic Equations

Patrick Ford

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Hey, guys. So now that we've seen how to solve some basic torque acceleration questions, we're going to add motion equations into the mix, which will generate a bunch of extra questions problems we can solve. Let's check it out. So you may remember that when we had forced problems, most forced problems were solved using our vehicles in May. You may remember that some of those problems would also involve are three or four equations of motion or cinematics equations. Same thing is gonna happen here with rotation, where some torque problems will require both torque equals Alfa, which is the rotational version of Newton's second law and rotational motion equations. Motion equations Are you am equations Universe? Um, uniformly accelerated motion telematics, equations of motion equation. Whatever you call it, we got 3 to 4 of these guys. Um, so remember, just like how it was with linear motion. The variable that will connect torque equals Alfa to the rotational motion equations, which are these guys here is going to be acceleration. Okay, acceleration. Now, because we're talking about rotation, this means Alfa right. Notice how there's an Alfa here and there's an Alfa here, here and here Okay, so we got Mawr equations and more variables, but it's not harder. Uh, it's just more stuff. So let's check out this example Here, have a solid sphere and I give you the mass and the diameter. Solid sphere is the shape of the sphere. So I know that because of the solid sphere, I'm gonna use the moment of inertia of a solid sphere, which is to over five m r squared. I got the mass. The mass is 200 I have the diameter. Remember, in physics, you're never gonna use the diameter. So as soon as I see diameter, I convert that immediately into radius, which is half of that. So it's 3 m and it spends about an access through its center. This is the regular rotation of a sphere solid sphere, which is around itself. Okay, it says it does this with 180 rpm clockwise. So the rpm is 1 80. What's up with clockwise? Well, clockwise clockwise is going to mean that it is negative. So it's got a rpm of 1 80 now, remember? And as we did in motion problems, rotational motion problems. Whenever you're given rpm vast majority of the time. You're gonna immediately convert that into Omega because most of our equations have omega little W but not rpm in it. Right? So the first thing I'm gonna do here or the next thing I'm gonna do here is convert this into W so w equals or Omega equals two pi F or two pi. Remember, F frequency is our PM over 60. Okay, so this is gonna be two pi negative. 1 80 over 60. This is going to be a three right there. Which means the whole thing will be negative. Six pie radiance per second radiance per second. Cool. So I got that. That's the initial speed. Um, I wanna know how much torque is needed to stop this thing in just 10 seconds. So I'm asking what is a torque to stop it. Right? So that means that this is my initial mega and I wanna have a final omega of zero. And I want to do this in just 10 seconds. So Delta T equals 10. So I hope you'll notice Here you start seeing all these motion variables. And remember the way I solved motion problems is by setting up the curly braces and putting all five motion variables there. So let's do that. Omega initial equals negative. Six pi Omega final equals we wanted to be zero Alfa Delta data in Delta Teeth. Now the tea is 10. Um, and these two guys, we don't have them, okay? And they're also not what we're looking for. But since I saw all these variables, I decided, Hey, let's start sending this up because I know this is coming. But really, what we're looking for is, um, is torque So you might actually have started this question instead of gone here. You might have just written that the sum of all torques equal Scialfa on. That's perfectly fine as well. That's if anything, that's ah Mawr directed way to the answer, right. More targeted, which is fine. There's only one torque here. We're assuming there's one torque. This thing is spinning, and I guess you're applying torque to it, uh, to make it stop so you can assume that there's only one torque. So the sum of all topics will become just the torque that you're looking for. And that is I Alfa. Okay, so if I could have I and I have Alfa. I'm done and that will be my answer. So let's expand I two pi I'm sorry to over five m r squared and Alfa noticed that I have m I have are, But I don't have Alfa. Right? So what you're gonna do is you're going to go over here and try to find Alfa. Okay, so let me plug in these numbers toe were the only thing we're missing is Alfa m is 200. Our is three squared. So as soon as we have Alfa, we can plug it in there. Okay, Now back to the motion equations so the basic ideas get stuck and you go to the other side back to the motion equations. We have three variables, which means we can solve. This is my target and this is my ignored variable. So I'm going to use the Onley equation that does not have a delta theta in it. And the only equation that doesn't have a delta theta in it is the first equation. Okay, so omega final equals Omega initial plus Alfa T and we're looking for Alfa. So Alfa is going to be Omega final minus omega initial divided by t. This, by the way, is the definition of Alfa. It's the change in omega over the changing T. You could have started there as well. That would have worked. Um, this is zero minus negative. Six pie. And the time is 10 seconds. So these cancel end up with six pi over 10. Positive. 65 or 10 which is 1.88 I got a positive, which should make sense even though I'm slowing down. Right. Let's talk about that real quick. My velocity is negative. If I'm slowing down, I'm trying to make my velocity positive. Eso my acceleration should be positive. I'm trying to make my velocity more positive. Another way to think about this that might be even easier is you are You have a negative omega and you're trying to slow down. You have to go in the other direction, the accelerations to go in the other direction. So the acceleration would oppose it because you're trying to slow down. And this is counter clockwise, which would be positive. Okay. Anyway, my acceleration is 1.8 radiance per second squared. Now I can plug this in here, and we are done. So if I multiply all of that, um and then I multiply that by 1.88 I should get, um I get 1354 Newton meter 1354 Newton meter and that's it. That's the final answer. So just to recap real quick, there's basically two parts to this. Um, we were asked for torque so you could have started here. And then you start plugging stuff in and you realize you don't have Alfa, but you have a bunch of motion equations motion variables so you can find Alfa using one of the motion equations. Plug it back in its the classic, um, standard type of physics question where you got stuck with something. We'll look for another variable, plug it in, come back with the value got on dissolved. Alright, so that's it. Hopefully makes sense. Let me know if you have any questions. Let's keep going

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