Hey guys. In this video, we're going to talk about how to actually find the moment of inertia of various objects by using integration. Alright? Let's get to it. Now, the moment of inertia of any object that are typical moments of inertia are going to be given by a formula. Okay? So let's say that you have a disc rotating about an axis through its center perpendicular to the surface, the moment of inertia is just going to be 12mr2 times the mass of the disk times the radius of the disk squared. That's an example. You could be given the moment of inertia for a rod about its center, about its edge, for a ring, for a solid sphere, hollow sphere, etcetera. Just a bunch of different scenarios. Okay? However, what if you don't have the scenario, the equation for a particular scenario, or the problem intends for you to find the moment of inertia from scratch. Like, how does this equation even come about? Okay. In order to do that, we need to use integration. Okay. Now the way to figure out the moment of inertia for any solid object is to consider it point mass by point mass. For an infinitesimal mass, right, just one point let's say that I'm considering a disk again. There is one little point on that disk that has a mass dm. Some infinitesimal mass. And it is some distance little r away from the rotational axis. Okay? Its moment of inertia is going to be, once again, infinitesimal moment of inertia. Because it's an infinitesimal mass, it just contributes a very very tiny, almost zero amount equal to just r2dm. So about this disc rotating, the moment of inertia due solely to that little point mass is just dI=r2dm. Now, if we want to add up all of those different little masses, each of which is at a different radius, and that's the key here, each at a different radius. We aren't going to assume that they all exist at the same radius. Only in a few cases is that true. Okay? And then simply to find the moment of inertia, all we have to do is add up all of those contributions of those little dm at their r2, okay, across the entire surface. So basically just across the entire surface of this disc, which is just the process of integrating. Okay? So the moment of inertia about some axis for some object is going to be the integral of r2dm. Okay. Now the thing to notice about this, the most important thing, is that r is typically going to change with m. So we can't simply pull r^2 out and integrate dm and say it's m. The only scenario that we can do that is if all the mass is at one radius. Okay? So let's consider this problem right here. Let me minimize myself. What's the moment of inertia of a ring? Okay. A ring is exactly this scenario. Right? For a ring, all of the mass, all dm, at r equals the radius of the ring. Right? So absolutely when I'm trying to calculate the integral, this r2dm, r^2 is going to be a constant because all of these little masses exist at the radius. 1 we could say exists here, 1 we could say exists here, exists here, etcetera. They all exist at the radius. So absolutely, we can pull r^2 out and then it's just the integral of dm. Okay? And we are told that it's a mass little m and a radius little r. So technically, this is only true when little r equals capital R. And then the integral of dm is just m. Right? So this is m r^2. Okay? And that is the moment of inertia. If you guys look at a table of moments of inertia given to you in your book, or if you just look up a list of moments of inertia, that you will find is exactly the moment of inertia for a ring. And that's just because all of the mass is concentrated at one radius. So you absolutely can pull the radius out in this problem. Okay? So like I said, this usually isn't possible. Typically, it's not as simple as just pulling r^2 out and saying that the integral of dm is m. This is only in the special case where all of the mass is at the rim of a ring. Okay? So let's see another example where this isn't necessarily true. Okay? Let me minimize myself again. What's the moment of inertia of a disc? Okay? The mass is uniformly distributed. And that's going to be important. Okay? So the moment of inertia, what we're going to start with is this right here. But now we cannot pull r^2 out because dm's located at different r's. So the distance r absolutely does change with the distance. Okay? So let's look at a disk. And how are we going to tackle this integral? Well, we need to rewrite this is always the goal rewrite in terms of r. Okay? We want to rewrite dm in terms of r. Okay? Well, the mass is spread uniformly across this entire disk. Okay. So we're going to have some mass per unit area which I'm going to call sigma. Sigma is typically the surface density for any quantity. In this case, it's the surface mass density and it's just going to be the mass m, right, over the area of this disk pi r^2. Right? Radius r. Okay? So let me actually change the way we're looking at this disk. Okay? If we want to integrate for the entire disk, what we need to do is we need to consider a single radius r and we need to find how much mass is contained at that radius. The problem is that it's not just one point that's at that radius, right? This is a circle so it's a whole ring that is at that radius. Okay? Now remember that sigma is a mass times an area. So sigma, we can say, is equal to some tiny mass divided by some tiny area that it's spread across. And we can say that because it's uniform. Okay? Or we can say that this tiny little mass is just sigma times this tiny little area. This tiny little area is just the area of this little thin ring at radius r. We want to figure out how much mass is contained in this little ring. Okay? So, first, we have to figure out what the area is. And the trick that you guys are going to use is for a very, very thin ring, the area is going to be the circumference times the thickness. Okay? And this works for very very small areas like this. Okay. The circumference is clearly just 2 pi r, right? We're talking about a very very thin ring at radius r. So the circumference is 2 pi r. But what is the thickness? We are going to consider this ring at r, right? Which means that it has to have an infinitesimal thickness, dr. Okay? So it's the circumference times the thickness. So this is zm our area. Okay? So this integral becomes the integral of r^2. Dm becomes sigma dA. Okay, so sigma first of all is a constant because this is a uniformly distributed sphere. Okay, so the density doesn't change around the disk, so I can pull that sigma out. Now what did we say was dA? Well, it's 2 pi r dr. Once again, 2 pi is just a constant that can come out. So 2 pi sigma integral of r^3 dr. Now notice our entire integral is in terms of r, which is perfect because that's exactly what we want, And we're going from a radius of 0 from the center of the disc out to the rim of the disc. Okay, this is just 1 fourth r to the 4. Okay? And now what we need to do is we need to plug in sigma. Remember what is. Okay? So this is 2 pi times m over pi r^2 times r^4 over 4. Okay, so we lose this 2, that becomes a 2. We lose this r^2, that becomes r^2. And we lose this pi. So what's the moment of inertia? One half m r^2. Okay. And if you looked in an index, a table of moments of inertia, you would find that that is exactly the moment of inertia for a uniformly distributed disc if we are considering an axis through the center of the disc. Plus I had said towards the start of this video that this was the moment of inertia for a disc. Alright, guys. That wraps up this video. Thanks so much for watching. Good luck with everything you've got going on this semester.
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Moment of Inertia via Integration
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