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13. Rotational Inertia & Energy

Moment of Inertia via Integration


Finding Moment Of Inertia By Integrating

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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.

Moment of Inertia of A Non-Uniform Disk

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Hey, guys, let's do this problem, All right? We want to find the moment of inertia about an access through the center perpendicular to a disk, but we want it to find it for a non uniformed disk. The mass distribution is given by this. Okay, so the moment of inertia, as we know, is just going to be r squared D M. The problem is, we can't just pull the r squared out and integrate M because we know that different masses, they're gonna be located at different radi I So we have to consider are as changing with him. So it's not a constant it can be pulled out. Okay, let's look at this disc. What we essentially want to do is we want to choose from the central axis some radius and figure out how much mass is contained within this radius. Okay, Now, for those of you guys who watched the concept video, you saw me do exactly this for a uniformly distributed disc. Okay? And the trick here is to say that sigma, which is the charge per unit. Sorry, not the charge. The mass per unit area is going to be a small, infinite testable amount of mass contained in that ring divided by the area of that ring. So this little infinite, testable amount of mass that we want right here is just gonna be Sigma times D a. Okay, where d a is just the radius off this little ring that that were saying, um contains an amount of mass D m. Okay, now, the trick here is that you want to The infinitesimal area is just going to be the circumference of this ring. Times the thickness and the thickness of this ring is just going to be D r. It's gonna be of infinitesimal thickness because we want to consider the ring to be at our okay, So if it's at one position, it could only have an infinitesimal thickness. Otherwise it wouldn't be at one position. It would have an inner radius and an outer radius. Okay, so d a is going to be the circumference, which is two pi r times that infinitesimal thickness. The main difference in this problem from the problem that you saw in the concept video is that sigma is not a constant in this problem, right? Sigma is Alfa R squared. So this is Alfa R Squared times two pi r d r Where now Alfa is the constant. So I can write this as two pi alfa R cubed d r And now I can rewrite my moment of inertia. Integral! This is R squared times two pi alfa r cubed D R. Where the two pi Alfa is a constant I can simply pull that out of the integral. And this is just gonna be our to the fifth d r from zero to capital r the radius of the disk. And that's just two pi alfa times 1/ are to the six. Okay, now the trouble here is that the problem explicitly says give your answer entirely in terms of the mass and the radius and we have Alfa right here. So we are not done. Okay. However, we confined the total mass of the cylinder. Sorry, the disk By integrating sigma. Okay, remember this equation right here? Well, this equation can be rewritten as the integral off D M is equal to the integral of Sigma D A. And the integral of D m is just the total mass. Okay? And this becomes well, what sigma? It's Alfa r squared And what's the d A. It's the same d a that we saw. So two pi r d r right where Alfa and two Pi or Constance. So this is two pi Alfa integral of our cubed d r. And we're integrating from zero to the rim. So this is two pi alfa times 1/ are to the fourth my minimize myself here that to cancels with that four, we get a two in the denominator. And so this is one half pi Alfa are to the fore. That's what mass equals. So what we need to do is we need to solve for Alfa entirely in terms off the mass and the radius, which we can do with this equation. Right here, let me keep this. Yeah, OK, so all that this means is that Alfa, I'm gonna multiply the two up and divide everything else over to em over pi R to the fourth. So now what I need to do is I need to take this result, and I need to plug it in Plug Alfa into that. The first thing I'm gonna do, though, really quickly canceled the to make that a one third. So this is going to be one third pi times Alfa, right where Alfa is to em over pi R to the fourth times are to the sixth. Okay, so what are we gonna lose? We're gonna lose a pie, and we're going to lose a factor of four. So the moment of inertia is two thirds m or square. Let me just check one thing really quickly. Sorry, guys. I gave the mass as little M. So let me just change the notation so that it all matches up. This is going to be little limb. This is going to be little in. This is going to be little m. This is going to be a little m little m just so that the notation is consistent. Okay, So this is how you would tackle a problem where the object is not uniformly distributed, but the mass distribution has a particular function. In this case, we were told that the mass distribution had the function of Alfa R squared where this was the mass per unit area. If you're doing an entire volume instead of being given a surface area, you could be given a volume density. Okay, um, but this is the gist of how to solve a problem where the object is not uniformly distributed. All right, thanks so much for watching guys. Good luck with everything you've got going on this semester.