(III) Determine the moment of inertia of a uniform solid cone ...

Question

(III) Determine the moment of inertia of a uniform solid cone whose base has radius R₀, height L and mass M. The axis of rotation (𝒵) is the symmetry axis perpendicular to the base, Fig. 10–66. [Hint: Think of the cone as a stack of infinitesimally thin disks of mass dm, radius R, and thickness dz.]
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Accepted Answer

Welcome back, everyone. In this problem, we want to find a moment of inertia of a uniform solid cone about its central axis which is the axis of symmetry perpendicular to the base. The cone has a base radius RB height H mass M. And we're assuming that the cone is made of aluminum. For our answer choices is is that it's a third of Mr B squared B half of Mr B squared, C 3/10 of Mr B squared and D 13 10th of Mr B squared. Now to find a moment of inertia I of a solid cone about its central axis. We can use the formula for the moment of inertia of a disk about an axis perpendicular to its surface passing through its center. And we can use that because we are considering the cone as as composed of a stack of infinite decimal thin disks each having a different radius that decreases linearly from the base to the tip. OK. So the moment of inertia for those discs or for one of those disks is going to be equal to a half of Mr squared. OK. Where M is the mass of the disk and I is its radius to figure out the moment of inertia. For the coin, we can integrate these contributions along the height of the coin. But there are a few things that we have to make note of. First, we need to express our radius in terms of the radius of the base. And we know that R is going to be equal to the radius of the base multiplied by one minus Z divided by H. Next, we know that the infinite decimal mass DM of a disk at height Z with thickness DZ is going to be D MD V DM with respect to V being equal to DM. OK. Divided by pi R squared DZ. Now, when you think about it, this makes sense because DV represents a small change in volume for our disk or circular disk, its volume is going to be pi R squared. H. Thus, that's why we represent it as PI R squared DZ. We also know that mass divided by volume gives us our density. So our small mass divided by a small volume is still going to represent the same density role. OK. So we can rearrange this then to say that our small mass DM is going to be equal to row PI R squared DZ. In this case, we can use our capital R here or radius. OK. Now, what else do we know? What do we know about our identity law? Well, we also know that row, OK is equal to the mass divided by the volume and four or cone. Its volume is going to be a third of pi R the radius of the base RB squared H which we can rewrite as three M divided by pi RP squared H. OK. So with that in mind, for our small mass, if we can put our formula for row into our expression for DM, and we can also put our formula for R into our expression in terms of the radius of the base. Then that can help us, OK, we can simplify our problem even further. So thus, what this is telling us then is that DM. OK. DM is going to be equal to row, that's three M divided by pi RB squared, H multiplied by pi multiplied by R squared. So that's gonna be RB multiplied by one minus Z hr squared. That's gonna leave us with RB squared by one minus Z divided by H all squared. Then we're going to write back DZ. Now when we look at our expression Pi Kans pi RB squared, KS RB squared. So that leaves us with our small mass DM being equal to three M divided by H multiplied by one minus Z divided by H all squared DZ. OK. Now remember we know that our moment of inertia equals half Mr squared. So to find the moment of inertia for us, we can find the moment of inertia for a small piece and then integrate it. That was the plan that we said we had before to integrate it along the height of the cone. So for the moment of inertia, then and our four small moment of inertia or small change in the moment of inertia is going to be equal to a half of a small change in the mass multiplied by R squared. We have an expression for a small change in mass. So we can put it into our equation. So that's not gonna be a half multiplied by all of this. You know what, I'm just gonna copy it here. OK? And carry it down because it's a lot to write. So it's gonna be a half multiplied by all of that, that represents DM OK, multiplied by R squared. We already have our expression for R squared, RB squared multiplied by one minus Z divided by hr squared. Now, at this point, if we're going to figure out, I let's go ahead and integrate. So we're going to integrate on the left hand side with respect to I and on the right hand side with respect to Z here, we can simplify a bit, we can uh simplify some terms here, group or constants together. So on the left hand side, the integral of D I is I, but we have a bit more work to do on the right. So now we're gonna be finding the integral OK of three Mrb squared divided by two H OK, multiplied by one minus Z divided by H raised to the fourth DZ. Now we're integrating with respect to Z. So we can take our variables, our constants outside of the integral. So that's gonna be three M three Mrb squared all divided by two H multiplied by the integral of one minus Z divided by H raised to the fourth DZ. And now all we need to do is to integrate our function here. OK. So we're gonna integrate with respect to Z. Now, how can we do that? Well, first, OK. First, let me put this in blue here. Let's let you be equal to our function one minus Z divided by H OK. Then that means DZ is going to be equal to negative one divided by H and thus, DZ, OK, DZ is going to be equal to negative H DU. So what that means then we can substitute DZ with respect to you or in terms of D sorry. And we can substitute U into our function, OK. To get our moment of inertia, I to be equal to three Mrb squared divided by two H multiplied by the integral of U to the fourth multiplied by negative HDU. OK. Now, if in, in integrating what we have here, OK, when we do that, we're going to get our moment of inertia I to be equal to this constant three Mrb squared, I probably should have copied that multiplied by negative H U to the fifth divided by five. Now, if we take this entire expression, if we, if we put back our value for you into our expression and simplify, OK? Put this here. If we put back this into our expression and simplify, then we notice that we're going to get I, we're going to end up with I being equal to three Mrb squared. OK? And I'll leave you to simplify this on your own divided by two H multiplied by each to the fifth, sorry H divided by five. My apologies. OK. Now, at this point, notice that cancers and we are left then after integrating with our moment of inertia to be equal to three MRP squared divided by two multiplied by five or 10. In other words, 3/10 of Mr B squared. Thus, that is going to be the moment of inertia for our uniform solid cone about its central axis. When we look back at our answer choices that tells us then that C is the correct answer. Thanks a lot for watching everyone. I hope this video helped.

Key Concepts

Moment of Inertia

Infinitesimal Mass Element (dm)

Integration in Physics

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