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Find the indicated moment of inertia or radius of gyration.A top in the shape of an inverted right circular cone has a base radius $r$ (at the top), height $h$, and mass $m$. Find the moment of inertia of the cone with respect to its axis (the height) in terms of its mass and radius. See Fig. 26.57

Calculus 2 / BC

Chapter 26

Applications of Integration

Section 5

Moments of Inertia

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So we need to find the the moment of inertia of this cone so we can just draw the cone. Here we have the radius from the center of the cone to the edge. Well, actually, label this this capital r, and then we have this access going through. Put it here. And then we have this fitness associated with this cone and this thick. This changes as you go up the cone. So we can say that this is the X we're gonna see This angle here is going to be considered Alfa. And then we have the height of this entire cone. This is going to be considered a tch height. And this Vertex here, we're going to label this very tex, eh? As you go down the cone as you go down the cone, the radius changes, so we're gonna have to model it with radius are lower case R. And at this point, we can tackle the question. So we have the mass of this cone is going to be equal to the, uh that's the cone Times. The volume of the cone and the volume of the cone is going to be pi r squared h over three. So we can say that the mass ah times equals equals the density times the volume. This is the definition for mass and this would be equal to Ty row R squared H over three. Given that the volume of a cone is again pi r squared h over three. So in this case, we would have to solve for row. So Roe is going to be three em over pi r squared age. So in this case, we can say OK, let's get a new workbook and say, Let's consider one disc of this cone of fitness DX, which we have labeled here in the diagram. And then we'LL say at a distance away from Vertex, eh? And at this point, we can say Okay, our is going to be equal to X ten Alfa. So this is where it gets a bit complicated in the sense that we need to find the volume of this piece. So the volume of this peace would be, of course, pi R squared times the thickness so dx and then we can salt. We can plug the end for our and say that this is gonna be pie X squared ten squared of Alfa d X. At this point, we can say that M is going to be equal. The mass of this cone is going to be equal to pi A sorry ro times pi X squared tangent at ten and squared of Alfa Times again dx I The moment of inertia of this cone is going to be the mass times the radius squared, divided by two. And at this point, we can say Okay, if this is if this is true, we can plug in for our mass and we have pie rather row pi X squared attention squared of Alfa the Ex all over two times x ten of Alfa squared That would be our radius term. And at this point, we can say that I was in an equal who row pie of x to the fourth, ten to the fourth Alfa D X all over too. And at this point we can melt, we can integrate so we can say that I was going to be equal two zero toe h of of Rho pi ten to the fourth Alfa all over, too Times X to the fourth DX. Now this is on ly because we have DX here. So in order to find the true, um, moment of inertia, if we were to integrate this from zero to the full height of the cone, we can essentially find the moment of inertia with respect to X, where X is again that thickness. So we simply have to go for through the full length integrate folks, integrate the full length of the cone with respect to X, and we would find the moment of inertia. So this is going to be row pie ten to the fourth Alfa invited by two of age to the fifth over five, and we can say I equals ropey over, wrote wrote Pi Ro. Divided by two times are to the fourth over eight to the fourth, and this has given that ten of Alfa equals R over a tch. This is from the cone. This is a given from the cone. So if we know that this is a given from the diagram of the cone, we can substitute this and say that this is going to be our to the fourth over H to the fourth and then times eight to the fifth over five and this is going to be equal to pi. Grow our to the fourth H all over ten. At this point, we know that road is going to be equal to three em over pi r squared h So at this point, we know that I is going to be equal to Pi Ro are to the fourth H over ten times and rather let's erase this grow. We're racing this road because, of course, we've been a substitute. Four row and row here is three em over. Pi r squared h this h this h cancels out this pie. This pie cancel out and we are left with three em are squared all over ten. So this would be the moment of inertia of this cone here, and that's the end of the solution. Thank you for watching

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