13. Rotational Inertia & Energy

Moment of Inertia via Integration

# Moment of Inertia of A Non-Uniform Disk

Patrick Ford

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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.

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