Moment of Inertia via Integration: Study with Video Lessons, Practice Problems & Examples

The moment of inertia quantifies an object's resistance to rotational motion and can be derived using integration. For a disk, the moment of inertia is calculated as 12mr^2. In contrast, for a ring, all mass is at a constant radius, leading to I = mr^2. Understanding these principles is essential for analyzing rotational dynamics and applying concepts like angular momentum and torque effectively.

Finding Moment Of Inertia By Integrating

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.

Moment of Inertia of A Non-Uniform Disk

Hey, guys. Let's do this problem. Alright. We want to find the moment of inertia about an axis through the center perpendicular to a disk, but we want it to find it for a non-uniform disk. Okay. The mass distribution is given by this. Okay? So the moment of inertia, as we know, is just going to be r2dm. The problem is we can't just pull the r2 out and integrate m because we know that different masses are going to be located at different radii. So we have to consider r as changing with m. So it's not a constant. It can't be pulled out. Okay? Let's look at this disk. 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 disk. Okay? And the trick here is to say that sigma, which is the mass per unit area is going to be a small infinitesimal amount of mass contained in that ring divided by the area of that ring. So this little infinitesimal amount of mass that we want right here is just going to be σdA. Okay? Where dA is just the radius of this little ring that we're saying, contains an amount of mass dm. Okay. Now the trick here is that you want the infinitesimal area to just be the circumference of this ring times the thickness and the thickness of this ring is just going to be dr. It's going to be of infinitesimal thickness because we want to consider the ring to be at r. Okay? So if it's at one position, it can 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 dA is going to be the circumference, which is 2πr, 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 αr2. So this is αr22πrdr where now alpha is the constant. So I can write this as 2παr3dr. And now I can rewrite my moment of inertia integral. This is r22παr3dr, where the 2πα is a constant. I can simply pull that out of the integral. And this is just going to be r5dr from 0 to capital R, the radius of the disc, and that's just 2πα16R6. 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 alpha right here. So we are not done. Okay? However, we can find 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 of dm is equal to the integral of σdA, and the integral of dm is just the total mass. Okay? And this becomes well what's sigma? It's αr2. And what's dA? It's the same dA that we saw. So 2πrdr, right, where alpha and 2π are constants. So this is 2παr3dr and we are integrating from 0 to the rim. So this is 2πα14R4. Let me minimize myself here. That 2 cancels with that 4, we get a 2 in the denominator. And so this is 12παR4. That's what mass equals. So what we need to do is we need to solve for alpha entirely in terms of the mass and the radius, which we can do with this equation right here. So all that this means is that alpha I am going to multiply the 2 up and divide everything else over 2mπR4. So now what I need to do is I need to take this result and I need to plug alpha into that. The first thing I am going to do though, really quickly, cancel the 2, make that a 1 third. So this is going to be 13π times alpha, right? Where alpha is 2mπR4 times R6. Okay? So, what are we going to lose? We're going to lose a π, and we're going to lose a factor of 4. So the moment of inertia is 23mR2. Hold on. 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 m. This is going to be little m. This is going to be little m, this is going to be 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 alpha r2, 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? But this is the gist of how to solve a problem where the object is not uniformly distributed. Alright? Thanks so much for watching guys. Good luck with everything you've got going on this semester.

Here’s what students ask on this topic:

How do you find the moment of inertia of a disk using integration?

To find the moment of inertia of a disk using integration, you start with the integral formula for moment of inertia: $I={\int }_{r^{2}dm}$. For a disk, the mass is uniformly distributed, so you express $dm$ in terms of the surface mass density $\sigma$ and the area element $dA$. The surface mass density $\sigma$ is $m/{\pi }^2$. The area element $dA$ for a thin ring is $2\pi rdr$. Substituting these into the integral, you get $I=2\pi \sigma {\int }_{r^{3}dr}$. Solving this integral from $0$ to $R$, you get $I=1/2m{R}^2$.

What is the moment of inertia of a ring and how is it derived?

The moment of inertia of a ring is derived by considering that all the mass is concentrated at a constant radius $R$. The formula for moment of inertia is $I={\int }_{r^{2}dm}$. For a ring, $r$ is constant and equal to $R$, so you can pull $R^2$ out of the integral: $I=R^2{\int }_d$. The integral of $dm$ over the entire mass $m$ is just $m$, so the moment of inertia is $I=m{R}^2$.

Why is integration necessary to find the moment of inertia for most objects?

Integration is necessary to find the moment of inertia for most objects because the mass distribution is not uniform or concentrated at a single radius. The moment of inertia is given by $I={\int }_{r^{2}dm}$, where $r$ varies with each infinitesimal mass element $dm$. For objects like disks or spheres, the distance $r$ changes continuously, requiring integration to sum the contributions of all infinitesimal mass elements. This process accounts for the varying distances and provides an accurate measure of the object's resistance to rotational motion.

What is the difference between the moment of inertia of a disk and a ring?

The moment of inertia of a disk and a ring differ primarily due to their mass distributions. For a ring, all the mass is concentrated at a constant radius $R$, leading to a moment of inertia $I=m{R}^2$. In contrast, a disk has its mass uniformly distributed across its area. The moment of inertia for a disk is derived using integration and is given by $I=1/2m{R}^2$. The key difference is that the disk's mass distribution requires integration to account for the varying distances of mass elements from the axis of rotation.

How does the surface mass density affect the calculation of the moment of inertia for a disk?

The surface mass density $\sigma$ affects the calculation of the moment of inertia for a disk by providing a way to express the infinitesimal mass element $dm$ in terms of the area element $dA$. For a uniformly distributed disk, $\sigma$ is given by $m/{\pi }^2$. The area element for a thin ring at radius $r$ is $2\pi rdr$. Substituting these into the integral for moment of inertia, $I=2\pi \sigma {\int }_{r^{3}dr}$, allows you to solve for the moment of inertia, resulting in $1/2m{R}^2$.