Hey, everyone. So let's check out this practice problem here. Now, this might already be a pretty long video, but I think this is a great problem because it combines a lot of ideas about rotational motion and forces and moments of inertia. So let's check this out. We'll work it out together.
We have a 1,000-kilogram disc. That's this blue thing over here, and it's got a 5-meter outer radius. Alright? So when I say that this mass here is 1,000 and the radius to basically the center is going to be 5 meters. So if the axis is sort of like this, then what we got is that this distance here is \( r_{\text{outs}} \) is equal to 5. Alright. It's mounted on an inner axle that also has mass. So this is sort of like a rod thing and this thing also has mass. I'm going to say that \( m_{80} \) and it also has a radius. Now, this diagram is kind of small, so what I am going to do here is I like to draw sort of like a top view. So let me just draw this out really quick for the top view and it's basically just going to look like 2 sort of concentric circles. Right? So it's going to be sort of like you have an inner axle like this and then you got sort of like the outer disc that looks like that.
There we go. So we've got that this distance here is going to be \( r_{\text{outs}} \) and then this distance here is going to be \( r_{\text{in}} \). So this is equal to 1, this is equal to 5 and that's basically what you got. Right? Alright. So there's a motor that's on the axle and it's going to speed up or slow down the system, but the motor stops working and the system is initially spinning at 70. Some of the words the \( \omega_{\text{initial}} \) is equal to 70. Now what happens is you want to slow this whole system from spinning so you're going to apply a friction force to the surface of the axle.
Now that's important because the surface of the axle is actually going to be over here. It's not going to be over here. That's the disc. Okay? So you have a friction force that's going to be acting over here to sort of slow this thing down and bring the whole thing to a stop. So we want to calculate how many revolutions the system will take to stop. So what does that mean? It really means that we're looking for \( n \) which is the number of revolutions and we want the \( \omega_{\text{final}} \) to be 0. Right? We want no angular speed. So if you want, \( \omega \) or sorry, if you want \( n \) we're going to have to relate that to one of our rotational motion of variables. Number \( n \) doesn't really appear in our equations but \( \Delta \theta \) does. \( \Delta \theta \) is equal to \( 2\pi \) radians times \( n \) which is the number of revolutions. So we want to calculate \( n \), we're really going to have to figure out what is \( \Delta \theta \). So I'm going to put that over here. Hopefully, you guys realize we have 3 out of the 5 motion variables. So I'm just going to go ahead and list out the other 2. That's going to be \( \alpha \) and the time, which we actually don't know. Okay?
So in order to figure out \( \Delta \theta \) to figure out \( n \), I need actually one other variable. I only have 2 out of 5, but I need 3 out of 5 to solve my problem. So how do I do this? Well, there's one other piece to this problem which was that we know that the friction force is equal to 200 Newtons and it's going to act right over here, the friction force. Now, whichever way this thing is spinning, right, it could be spinning this way or this way. It doesn't matter because the friction force is going to oppose that motion whichever way it's going. Right? Could be up or could be down depending on how the thing is spinning. So this friction force, which is some distance away from the axis of rotation. Right? If you're looking at it from the top the axis goes straight through the center, then what happens is this friction force produces a torque and this torque is going to generate an angular acceleration Alpha. Remember, sum of all torques equals \( I \alpha \).
Alright? So that's really what's going on in this problem. To look at to figure out this final variable, we're going to have to look at the force related to a torque and that's going to produce an angular acceleration. Alright. So that's really what we're doing. So let's go ahead and get started with that part which is that the sum of all torques is going to equal \( I \alpha \). There's only one torque in this problem and it's going to be the torque from the friction force. So that's going to equal \( I \alpha \). So remember this is just going to be friction times the distance times the sine of the angle equals, and this moment of inertia here. Right? So we have the friction force, it's \( f \). What about the distance that it is from the axis of rotation? Well, it's easiest to see from this diagram here that this distance, this \( r_{\text{inner}} \) and this distance, the distance away from the axis of rotation are the same.
So really the distance away is actually just the \( r_{\text{inner}} \). It's the inner radius of the axle itself, which we know is 1 meter. What about the angle? Well, like I said here, whichever way this thing is spinning, it could be spinning this way. Right? So it could be spinning this way or this way. The friction force is always going to oppose a direction of motion and the angle between the, axis of rotation and friction is 90 degrees. Alright. So this is 90 and that's okay because the sine of 90 turns turns into 1. What about this moment of inertia here? Now, you gotta be careful because there's actually 2 objects in this problem. There's this sort of solid cylinder that's mounted or that's sort of around a disk. Right? So you have a disk that's on that's mounted onto a solid cylinder.
Alright? So what I'm gonna do here is I'm gonna say that this moment of inertia because there's 2 \( I \)s is gonna be the \( I \) of the axle plus the \( I \) of the disc and this is gonna be times \( \alpha \). Okay. So here's what we've got here. So we've got I'm gonna start plugging in some numbers. We've got 200 times 1 and then times 1 is gonna equal and then what is the equation that we use for the eye of the axle? You might think Right? You might think it's gonna be one of these because this sort of looks like a rod, but remember the rod these equations pertain to something that's spinning sort of perpendicular to the center. Right? It's gonna be spinning like this. But that's not what this axle is doing. This axle is sort of spinning around itself right this way.
So it's actually not gonna be either one of these equations. Alright. So again, it's kind of like dare to trick you. It's actually gonna be a solid cylinder. Right? You can kinda just imagine this is a long cylinder like this and it's spinning this way. So this is actually gonna be \( \frac{1}{2} \) and remember the mass of the axle is gonna be \( m \). So I wanna use the little \( m \times r_{\text{inner}}^2 \) plus and now this guy is gonna be this guy is gonna be for the disc. This is gonna be sort of like a thick-walled cylinder. Right? This is gonna look sort of like a doughnut shape. And the doughnut shape, remember, has 2 radii \( r_{\text{inner}} \) and \( r_{\text{outer}} \) and this is the equation we're gonna use for that one.
So this is gonna be \( \frac{1}{2} \), except now I'm not gonna use little \( m \), I'm gonna use big \( M \). And then this is gonna be \( r_{\text{inner}}^2 + r_{\text{outer}}^2 \). Alright. So kind of tricky there, but there's two moments of inertia and they are kind of like related to each other. Alright? So that's your moment of inertia times \( \alpha \). Okay. So we've got the 200 equals and when you work this out, what you're gonna get here is you're gonna have \( \frac{1}{2} \) of 80, \( \frac{1}{2} \) of 80 and then times one squared plus, this is going to be \( \frac{1}{2} \) of 1,000 times, this is gonna be 1 squared, that's our inner, plus 5 squared, that's the our outer and then times \( \alpha \). Now if you work this out, you're gonna get is the 200 is equal to 13,000, 13,040 and this is gonna be kilogram meters squared. This is gonna be Newtons and this is gonna be times your \( \alpha \). Alright? So remember, we're looking for this \( \alpha \) over here because that's the last of the That's the third variable that we need. So if you work this out, we're gonna get is 200, divided by 13,040 is gonna equal your \( \alpha \) which is gonna be 0.015. This is gonna be radians per second squared.
The last thing that we need though is we're gonna stick a negative sign in here. Right? And this is just really has to do with the signs, but basically what happens is the thing is gonna be slowing down. So we wanna include this negative sign because that \( \alpha \) is gonna slow things down. Alright. So this is the last of our variables and this is gonna be 0.015. Okay. So now that we've got 3 out of our 5 variables here, now we can go ahead and figure out a kinematic equation that is gonna give us \( \Delta \theta \). Alright? So hopefully, you guys realize that if you have the omegas and \( \alpha \), the equation we're gonna use here is that \( \omega_{\text{final}}^2 \) is \( \omega_{\text{initial}}^2 + 2 \alpha_{2} \times \Delta \theta \).
Alright? So we've got here is that 0 squared equals 70 squared plus 2 times negative 0.015 times, \( \Delta \theta \). Alright. So we've got is negative 70 squared divided by negative 0.015. The negatives will cancel and that's good and you're gonna get Oh, I'm sorry. I forgot a 2 here. So there's a 2 here. This is negative two times 0.015 is equal to \( \Delta \theta \). And when you work this out, what you're gonna get here is a \( \Delta \theta \) of a thousand 163,333 and this is gonna be in radians. Now we're not quite done yet. This is your \( \Delta \theta \). Now we just have to plug this back into this equation in order to figure this out. So, I'm just gonna go ahead and this is gonna be your final answer over here. Your \( \Delta \theta \) divided by your \( 2\pi \) is gonna equal your \( n \). So in other words, a 163,003,33 divided by \( 2\pi \) and the number of revolutions is gonna be about 26,000. So there's 26,000 revolutions necessary to sort of get this thing to stop. Alright? So this is your final answer over here. Hopefully, you guys got it. You know, hopefully, that made sense to everybody. It's a really great problem. It sort of pulls together a lot of, you know, stuff about moments of inertia and forces and motion. That's it for this one, folks.