13. Rotational Inertia & Energy

Moment of Inertia via Integration

# Finding Moment Of Inertia By Integrating

Patrick Ford

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Hey, guys, In this video, we're gonna talk about how toe actually find the moment of inertia off various objects by using integration. All right, 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 access through its center perpendicular to the surface. The moment of inertia is just gonna be one half times the mass of the disc times the radius of the disk square. That's an example you could find. You could be given the moment of inertia for a rod about its center, about its edge for a ring for, um Ah, 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 toe figure out the motive inertia for any solid object is to consider it point Mass by point Mass for an infinitesimal mass, right? Just one point. Let's sit down. Considering a disc again. There is one little point on that disk that has a mass d M, some infinite testable mass. And it is some distance. Little are away from the rotational access. Okay, its moment of inertia is going to be once again infinitesimal moment of inertia because it's an infinite visible mass. It just contributes a very, very, very tiny, almost zero amount equal to just r squared d m So about this disc rotating Sorry about this axis. The moment of inertia due solely to that little point mass is just D I equals R squared D M. Now, if we wanna 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 existed the same radius Onley 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 dems at there are squares 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 access for some object is going to be the integral off r squared D m. Okay, now the thing to what? Notice about this. The most important thing is that our is typically going That's terrible underscored, typically going to change with them So we can't simply pull r squared out and integrate d m and say it's m the Onley scenario that we can do that is if the all the masses 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 d m at r equals the radius of the ring. Right? So, absolutely. When I'm trying to calculate the integral this r squared d M r squared is going to be a constant because all of these little masses exist at the radius. One we could say exists here. One we could say exists here exists here, etcetera. They all exists at the Radius. So absolutely we can pull our squared out and then it's just answerable of GM, okay? And we're told that it's a mass little M and a radius Little are so technically, this is Onley true when little r equals capital are and then the integral of d m is just m Right, So this is m R squared, 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 masses, um, 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 a simple is just pulling r squared out and saying that the integral of d. M. Is m. This is only in the special case where all of the mass is at the, um, 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 masses uniformly distributed, and that's gonna 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 squared out because dems located Uh huh. At different ours. So the distance are absolutely does change with the excuse me with the distance. Okay, so let's look at the disk. And how are we going to tackle this integral? Well, we need toe rewrite. This is always the goal rewrite in Terms of are okay. We want to rewrite D M in terms of our Okay, well, the masses spread uniformly across this entire disc. Okay, so we're going tohave 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 density, and it's just going to be the mass m right over the area of this disc. Pi r squared, right radius R. Okay, so let me actually change the way we're looking at this disc. Okay? If we want to integrate for the entire, um Disc, 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 uniforms, 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 to figure out what the area is, And the trick that you guys were 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 a very, very small areas like this. Okay, The circumference is clearly just two pi R, right? We're talking about a very, very thin ring at Radius R. So the circumference is two pi r. But what is the thickness? We're going to consider this ring at our right, which means that it has toe have an infinitesimal thickness D r. Okay, so it's a circumference. Times the thickness. So this is our area. Okay, so this integral becomes the integral of r squared d m becomes sigma D A. 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 D A. Well, it's two pi r d are once again to pie is just a constant that can come out. So two pi sigma integral of our cubes d are now notice. Our entire integral is in terms of our which is perfect, because that's exactly what we want. And we're going from a radius of zero from the center of the disc, out to the rim of the disc. Okay, this is just 1/4 are to the fore. Okay, And now what? We need to do is we need to plug in Sigma, remember what sigma is. Okay, so this is two pi times m over pi r Squared times are to the fourth over four. Okay, so we lose this to that becomes a to we lose this r squared that becomes r squared and we lose this pie. So what's the moment of inertia? One half m r squared. Okay. And if you looked in an index Sorry a table off moments of inertia. You would find that that is exactly the moment of inertia for a uniformly distributed disk if we're considering an access 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 disk. 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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