13. Rotational Inertia & Energy

Torque with Kinematic Equations

# Stopping flywheel with friction

Patrick Ford

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Hey, guys. So in this example, we have a flywheel. It's spinning, and we're going to press an object against the wheel to cause the stops. Imagine, got something spinning. And if you squeeze an object against it, if you push an object against it, it would stop. And we wanna know how hard do you have to push against, um, the flywheel so that it stops in a particular amount of time? Okay, so it says here flywheels a rotating disc. This means we're going to use the moment of inertia of a disk, which is half M R Square. Samos. A solid cylinder. Andi. It's used to store energy. Suppose this one has a mass of eight times 10 to the fourth kilograms and has a diameter of m, which means we're going to immediately change the diameter. Is the radius off 2.5? Um, and it's set up vertically. So we got this wheel That's vertical, Um, and it's free to spin around a fixed axis, so there's a fixed access their perpendicular to the wheel. So basically the wheel spins like this around a central axis, um, to slow down the fly. Well, I mentioned this. You push this block here, so there's a force that you apply. Um, I want to point out that once you push here, it's going to have a contact. Therefore, there's going to be a normal force back. And this normal force, we have the same magnitude is your force so action reaction If you push with 10 normal pushes back with 10 um, it says that the coefficients of friction between the block and the wheel right here the coefficients of friction Our 0.6 and 0.8. So I'm gonna write them here. Mu static. Remember, when you have two quick fixes friction, the static coefficient is the greater one. So 10.8 and new kinetic is going to be 0.6. Okay, Onda, we want to know what is how hard do you have to push? So I wanna know what is F um it says here that the wheel is gonna come to a complete stop so Omega Final will be zero from on our PM of 300. So rpm in their show is 300 and it's going to do this in a delta t of 30 seconds. Remember? Rpm almost always gets converted into omega. So I'm gonna do that real quick. I'm gonna say, um, Omega initial equals two pi our PM over 60. That's the equation to convert to. And if you multiply this, you get 10 pie. Okay? 10 pie radiance per second. And notice that Now we have. I'm gonna actually move that over here on dso scratch this outs just because I'm trying to list all my motion variables and I'm trying to make the point to you that as of now, we already have three motion variables, which is good news Means we could solve for the others. The other questions that the other variables that are missing here are Al from and Delta Theta. Okay, so all I'm doing is grabbing the information and sort of fixing it up. Eso how hard do you have to push again? This is a force that's going to cause a torque, right? And so we're gonna use the force. Um, the the some of all torques equals I Alfa Equation. But I want to explain to you what's going on here. So the idea is that the wheel spinning this way, let's say within Omega in issue. And when you push, there is now a normal force here. I had that drawn earlier, and what that means is because there's a normal force and there's friction. There's going to be friction acting against motion. So motion is rotating that way. Friction is going to try to stop the wheel. Which means there's going to be a force of friction this way. Okay, so you got a normal this way on. There is a force of friction this way. What that force of friction does. It causes a torque like this torque of friction, which is opposite to my velocity. So it's trying to slow it down. It's trying to get it to stop. Okay, um, in this particular case just because of the way I drew it, um, this is actually negative. And then this would be positive. As long as they're opposite to each other. You're fine. Okay, so there is a torque. There is an acceleration. This is a force problem with angular acceleration with torque. So we're going to start here and we have to find what this f is. Okay, three Onley. So let's expand this equation real quick. The Onley torque acting on. This is a torque due to friction because you're pushing against this thing. So we're gonna right? Torque of friction, theme, moments of inertia, The moment of inertia. It says here I didn't read this part to you, but it says here you may assume the wheels entire mass is concentrated at its outer rim. This means that this is not a thing is actually not a rotating disc. I apologize. Um, this is not a rotating disc. This is going to be a hollow disc. Okay, so it's going to be hollow disk. So instead of half m R Square, let me just meet this, um, instead of half from our square, it's going to be just m R Square. Okay. Where r is the radius. So I'm gonna put this here m r squared and Alfa. We don't have Alfa, but we could find Alfa if we wanted Thio. Okay, let's take this one step further and expand. This, um, torque is force Little. Our sign of data force in this case is friction. So it's little f r is the distance from the axis of rotation to the point where the force happens. This is the distance that we are talking about. Okay, this is my our vector. Um, because they force happens here. And the axes, obviously, in the middle here, the our vector is as long as the radius of the wheel, because friction happens at the edge there. So we're gonna have f big are and then sign of theta. The angle will be 90 degrees. Notice how they make an angle of 90 degrees Because frictions Shrake down as a result of the pushing against it on. By the way, friction is in situations like this, friction is always gonna make an angle of 90 degrees, and it's always gonna happen at the edge. Right? So sign of 90. So that's nice. Um, and M r square the radius. We have that as well. We're gonna be able to plug it in there and then Alfa and we're looking for F Now, don't get confused. We're not looking for a little f. We're looking for Big F. So what we're gonna do is we're gonna keep expanding this equation. Friction can be expanded. Friction is mu normal. So I could rewrite friction as mu normal. Uh, that's what we're gonna do except there's one change which is normal is the same thing as F. So I'm gonna plug in effort here, and I'm gonna write friction as mu F instead of new Normal. It's the same thing. The reason we do that is because now, finally, our equation actually has are variable, right? Until then, you haven't seen your variable around. Um, so that's that. Let's gonna rewrite this here. I'm gonna cancel This are with this are this is just the one. So it's gonna be m f M u f equals m r Alfa. So f is m r Alfa over mu. And we know these numbers m is or most of them eight times 10 to the fourth radius is 2.5. I gotta go find Alfa. And then we do have, um you which do you think we use kinetic or static? And I I hope you're thinking kinetic because the block is rubbing against the disk. So it's gonna be 0.6, which is kinetic, Okay. And we have to go find Alfa, So let's go do that real quick. So we're gonna go over here and look for Alfa, um, to find out for I can use motion equations. I have three knowns. Uh, this is my target. This is my ignore variable, Which means I can use the first equation to find Alfa Omega Final equals Omega initial plus Alfa T for looking for Alfa. So let's move everything out of the way. You might notice that this is the definition of acceleration of angular acceleration, which is change in omega over Children time. You could have started there as well. The final omega is zero because we're looking to stop. And the initial omega is 10 radiance. Um, 10 pi ratings per second. Now, notice that the way we drew it, it's a negative. So let's plug it in as a negative. Um, and in the time the time we're going forwards seconds. Okay. Now, if you do this, if you do this, you end up with an Alfa of 1.5 radiance per second squared. When you highlight this here. And that's what I'm gonna put right here one point a different color. Um, 1.5 right there. Now, if you multiply once you multiply this whole thing, once you multiply this whole thing, I'm sure I got it right. Yep. You end up with 3.5 times. 10 to the fifth, 3.5 times. 10 to the fifth Newton's. Okay, so that's how much force you have to push against the block with, um and push the block against the wheel so that this thing stops in 30 seconds. All right, so that's it for this one. Thes questions fairly popular. It's a little bit tricky, but again, I think the key thing is to realize you have a force causing a torque causing acceleration that this is the starting place. The trickiest part, I think here is to try to make a connection from this initial equation. And then how we're gonna get f. And sometimes you just have to trust that if you keep expanding the equation like we did here and then we expanded the F here, you have to trust that if you keep expanding, the variable will show up. Um, Alright, so that's it for this one. Let me know if you have any questions and let's keep going

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