find-the-indicated-moment-of-inertia-or-radius-of-gyration-a-top-in-the-shape-of-an-inverted-right-3

Question

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

Averell Hause · Accepted Answer

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&#x27;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&#x27;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&#x27;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&#x27;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&#x27;s get a new workbook and say, Let&#x27;s consider one disc of this cone of fitness DX, which we have labeled here in the diagram. And then we&#x27;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&#x27;s erase this grow. We&#x27;re racing this road because, of course, we&#x27;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&#x27;s the end of the solution. Thank you for watching

Anthony Capobianco · Answer

today we&#x27;re going to solve for the moment of inertia of a come were given a uniform solid cone of mass m altitude age with the circular base of Radius R, we begin to calculate its moment of inertia. I by starting with the following definition, I is equal to the integral of our square dia. We can calculate the moment of inertia over this cone by taking an integral over the volume and waiting each infinite testable unit of mass by its distance from the axis squared. We know that the masses m, which means that its density rho is equal to M V, where V equals 1/3 pi r squared age the volume of a cone. Because the corn is uniform, we can rewrite the integral above as I equal row into role of R squared D V integrating over volume instead of mass. To calculate this integral, we must represent the cone as cylindrical shells of infinite testable thickness. D are where the height of each cylinder rule vary with the radius, since a cross section of the cone through the center gives a nice oscillates triangle the height of the triangle of a given distance, from the line of symmetry varies linearly with the distance. This gives us the relation Z of our of each cylindrical shell at Radius R. The height must be zero at the base where a little R equals big are in each at the very top. These constraints give the following relation. Zi fr equals h times one minus little are over big are Then we must find the contribution to the volume of each cylindrical shell. Each shell is an infinite Tess Millay thin sheet of area a equal to pi r z of our. This is simply the surface area of a cylinder, not including the end caps. The thin sheets thicknesses d are so we can write that the volume contribution from this thin slice is Devi equal to two pi r c of r d r The integral becomes I equals two pi row integrating from zero to r r cubed Z of r d r. I&#x27;ve written out the rest of the deprivation for reference here. Following the steps, we can see that the end result is I equal 3/10 m r squared. And that is the moment of inertia for a comb

Carlos Pinilla · Answer

So we have. Ah, you know what Cone one that you like will find its moment of unusual bolding. So we&#x27;re gonna please. We&#x27;ll hide which I want to find its moment of Asia about the suction. So it seems your every plays it up off the con on both. So that, uh, you have something like these Generally diseases there. Nice, eh? On that hide their age, So But we want to make them with the veneration. Would have to doing r squared. Our daddy says so. The taxes on union you cylindrical coordinates. Um, so, uh, well, for But bones are gonna be o r r Who&#x27;s, uh, b r r goes from sear off to a Did you on these for years? Of you, you are. See, you know something like these for our peoples too. Okay, You know that behind is one of the age, so that, uh oh, here divided my well, You can complete this local grease, so those laws should be changed in their age, divided by changing their remember nephew Blufgan, are you going to say you so that the World War Z both from there to there ago? There, So are you see who? I sure these, uh, are h over a to age on the volume elements are we are integrating with respect all the way around this we have through it. Ate that over the moment of inertia is, uh, these internal that enoughto plugging on r squared so that yeah, So you have. So they&#x27;re insulting. Power should be, are cute. And then yeah, So the moment in your jammies Gimme that moment. All right, So, um, go ahead and do it. Seita seems nothing. Their defense, huh? So you just got to buy on dinner for being drill. We&#x27;ll see. We dream. We see. So the travel logs here we see, Just see, one work is leaving. Are I&#x27;m saying you&#x27;re a, you know, seething are no. That&#x27;s the lower body on Love&#x27;s a bridge. One of these is able to age Linus for our age or what? Um, right. So that you exist, Joe. We&#x27;ll see now. Helping to make we sparred are but things are cute. So the remaining into Louisa these So each minute mar Well, e I&#x27;m sorry, Cube. We are on Duh. Not from zero up through. No way. So all these first thing tomorrow. First the jewelry arc you pidge the seven wonders minus R to the full of flour H m c So when we integrate those two, we&#x27;re gonna get our cure result art of their full power. Therefore, his age minus are really each other far too. The fourth Art The fifth power by five sage for a So these volatile in zero on a news. Um hey, to the floor power. They&#x27;re for him. Sage miners. A degree is powered by five. I&#x27;m sage. You&#x27;re a five. So these people do these councils one hour, then you have you before age, then four. Your message Them&#x27;s 1/4 minus one. If so, these have even to five brows. Motive behind that. That was not my thing. Is that so? Five minus four over all times five. So this is a vehicle. Something is able to dad and then get the factor of buys you in there so that the total moment of illusion it&#x27;s quiet should be his number. That is gonna be, But he&#x27;s one. We&#x27;re so 1/4. Power him sage. That was one over tow. Any that is board inside. And then that and you too. Bye. That I&#x27;m still buy. When you gonna simplify that yet by a riff off inch. And this is another thing. So you should be the moment of it. The moment the range of the school.

Surjit Tewari · Answer

caution was 73 like the visa, right? Circular cone. But best 30 s as capital. Are this 30 asses? Yep, a lot. And the total height is that let us God a solid days. Author ideas a small are that mess off This disk is DX and this discus at a distance x from the origin that total I thought the corn is at. So now we can like the density off this or Likoni is roll equal to mass off the phone. We just Yep, the limb upon the volume of the on one by three by oddest. But that&#x27;s And from the similarity off the triangles, we can say odd upon gap in our is equal x upon it. This implies that a small are that this radius is a kwaito x times the base radius upon the total height of the cold. No, the moss off this elemental. This is the m is acquittal area off the days by is what in don&#x27;t the picked off the decks be X in the density off the this which is uniformed. Borda, hold solid cone. And the moment off in a Shia off this diskettes uh, the let that say the mosque is It&#x27;s small then The moment of inertia off the disc about the traditional exists is the artist spread now placing the value of B m. We can die. Uh, in Dubai Adi spot indoor be X Indo density into R squared. This is equal to half into fire into the X in tow density. But I put up our four now substituting the value off our as we have calculated Ah, wow up by be x dot density not X. I do the art by all power for this is a widow by right to grow uh, through the power for upon at stood up our poor that eggs to the power for be X now integrating the both sides We get the movement off Energia off The whole cone is equal to my or by by two into density into our to the power four upon at stewed up our four invigoration from zero to at off explode up our four the x solving this are integrating this We will get bye bye to roll I to the power for upon as student our poor in the one by five in the as student power by now, substituting the value off density Bye bye to indoor the value of density mass upon one by three by oddest. But it indoor I to the power for upon estudio power full in tow as to the power five up on why this will give us three by then. And odd is But this is that acquired moment of inertia.

Donald Albin · Answer

all right. It&#x27;s gonna take us a few pages to get through this one, I think. Um, but let&#x27;s go ahead and do it. Calculate the rotational inertia of a solid uniform. Right? Circular cone of mess M height H and based radius are. So the first thing I want to do is draw a cone. Got a cone here. Okay, there we go. From the base has ah, center. And so through this, I just want to show the access that it&#x27;s rotating around. So we&#x27;re rotating around this axis. Um, we know that I equals the integral of r squared d m. So what I want to do is draw a little volume element. So if we cut right about there the whole way through the cone, then it&#x27;s gonna make a shape that actually looks like a cylinder. I wish I would&#x27;ve drawn that a little bit better. Let me see if I can make that slightly better. Well, that&#x27;s quite a bit better. Okay, so I am going to call, um, and obviously this distance here, from the center to there is going to be our, um if we think about the volume of that cylinder that I&#x27;ve just drawn. Um, I&#x27;m going to call it the D V. And it&#x27;s gonna be, uh, two pi r the circumference of the ah, the base down here. So the whole way around, that&#x27;s two pi r. And then I&#x27;m going to multiply by d r, which is the, um, little tiny with element there the width of that little green line that makes the circle. And then I&#x27;m going Teoh, um, think about how long is this line right here? Well, we can make a proportion understanding that this whole length here is our and our proportion would be that this length whips At least you can see the one I was talking about. I&#x27;m going to call it h sub I for a little incremental part of the, uh of the code. So h sub I is to the total height as this distance here, um, which is r minus are is to this total distance trying to draw their this total distance, which is our absolutely undo that again are so now I can rearrange that and I can write that this little incremental height is age of or are times are minus are so up here. I&#x27;m going to put that in age over. Oh, are times are minus are so basically I have, um if you will length times with times height so the length is the length of the circumference of the circle. The winds is the with that little tiny circle that was created, which is the d r. And the height would be the height of the cylinder. Me, He raced what I just did there. Okay. However, we want d m not Devi. So I also know that the volume of a cone is 1/3 pie r squared h So if we think about the relationship between mass and volume D m would be, uh, em over v devi. So they&#x27;re they&#x27;re proportional to each other. So, um, now I have d m So we need to do the integral of r squared d m. I think I&#x27;m gonna have to go to another page the integral of r squared, which is gonna be D M. Which is going to be the integral of em over v r squared Devi No. Yeah, the integral of em over v r squared D u v. Okay, so let me write that down So the integral of r squared d m. I just replaced d m with m over vdv But now I&#x27;m going to replace Devi by the way em over v r Constance. So I&#x27;m gonna move that outside integral off r squared. So let&#x27;s right the r squared. But now we need to write. Devi Devi is two pi r d r to hi rdr And then we&#x27;ve got more So two pi r d r um, age over our let&#x27;s keep going ar minus far. Okay, so now two pi h over our can all come out because they&#x27;re all Constance So too Hi, h I m two pi each I m over RV into girl Ah of, um got our square there Another ours that would be our to the third power times are minus are d r so that&#x27;s going to give me And I guess I&#x27;ll just keep working down the page to buy h I m over r V. But we have an equation for V, which is 1/3 pi r squared H i r squared H now into girl. By the way, this is from zero to our of our r cubed minus or fourth power parenthesis is around that D r. Okay, so I noticed that the pies cancel out the ages. Cancel out, Um, the three congee Oh, up to the numerator. And so that&#x27;s going to give me six m over our que oh, are are to the fourth power over four because I just integrated. The, um, are to the third cards are the fourth power over four minus are to the fifth power over five from zero to our, um at zero. This is all going to evaluate Teoh zero. So we only have to worry about our. So this gives us six em over. You are to the third power are to the fifth power over four minus are to the fifth power over five. So that&#x27;s going to give us, um six m r squared because our to the fifth power divided by r to the third power is R squared and then we just have 1/4 minus 1/5. So 1/4 minus 1/5. That&#x27;s 5/20 minus 4 20th is gonna give me 1/20 so it&#x27;s going to give us six m r squared over 20 which is the same as three. M r squared over 10 and that is the answer. So I just want to look back at this. The key to this is creating a little devi element of understanding what that Devi element means, Um, putting it in terms of our and then integrating. So I really think that the the greatest difficulty is right here. If you have trouble, go back and listen to my explanation again of, um, how I came up with that devi. But it&#x27;s it was basically considering that little green cylinder, Um and, ah, considering it kind of is rolled out so that it would have a length of width and a height. And so I just took the length and the width and the height of what that would be. All right. So thank you for watching

Problem

Find the indicated moment of inertia or radius of…

Problem 23 Easy Difficulty

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

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