# Moment of Inertia Problem

by greswd
Tags: inertia, moment
 P: 147 Here's a derviation from HyperPhysics: He says: $$dV=πy^{2}dz$$ However, if we're finding the surface area of the sphere: $$dA=2πRdz≠2πydz$$ If we cannot use $$dA=2πydz$$, how come $$dV=πy^{2}dz$$ is still applicable?
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P: 41,457
 Quote by greswd He says: $$dV=πy^{2}dz$$
That's the volume of those circular disks.

What does finding the surface area of the sphere have to do with this?
P: 147
 Quote by Doc Al What does finding the surface area of the sphere have to do with this?

If I want to find the MoI of a hollow sphere.

Based on the same assumptions $$dA=2πydz$$, it should be like that, but it's not .

I'm trying to calculate the MoIs of some hollow objects, not sure what kind of area formula I should use.

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P: 41,457
Moment of Inertia Problem

 Quote by greswd If I want to find the MoI of a hollow sphere. Based on the same assumptions $$dA=2πydz$$, it should be like that, but it's not .
Not sure what you mean by "the same assumptions". To find the surface area of that ring, you need the arc length, not the vertical distance (dz).
P: 147
 Quote by Doc Al Not sure what you mean by "the same assumptions". To find the surface area of that ring, you need the arc length, not the vertical distance (dz).
Yup, that's the perplexing part of the formula $$A=2πRz$$

As z tends to zero, that particular segment of a sphere should tend toward a segment of a cone. And for a cone, when z tends to zero

$$A→2πrz$$

The difference between the conical area and the area of a cylinder wrapped around it should decrease until they are equal when z tends to zero.
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P: 41,457
 Quote by greswd Yup, that's the perplexing part of the formula $$A=2πRz$$ As z tends to zero, that particular segment of a sphere should tend toward a segment of a cone. And for a cone, when z tends to zero $$A→2πrz$$
Again, it's the length of that side that must be used. And that length is not simply z. (It's at an angle.)

 The difference between the conical area and the area of a cylinder wrapped around it should decrease until they are equal when z tends to zero.
Their ratio remains finite. They aren't the same.
 P: 362 Is this for the moment of inertia of a sphere?
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P: 41,457
 Quote by HomogenousCow Is this for the moment of inertia of a sphere?
A spherical shell.
 P: 362 I'm pretty sure it's for a full sphere, that dI there represents the moment of inertia of a single disk at height z, and then he integrates from the bottom to the top.
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P: 41,457
 Quote by HomogenousCow I'm pretty sure it's for a full sphere, that dI there represents the moment of inertia of a single disk at height z, and then he integrates from the bottom to the top.
Yes, that would be correct if he were finding the moment of inertia of a solid sphere, but he's not.
P: 147
 Quote by Doc Al Again, it's the length of that side that must be used. And that length is not simply z. (It's at an angle.)
That sounds correct.

But why does this not apply to volume integrals?
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P: 41,457
 Quote by greswd But why does this not apply to volume integrals?
Any slant with the side will be a second order correction to the radius and will have no effect on the volume element.

The volume is the area times thickness:
dV = area * dz = πr2dz

If we vary the radius by dr, we get:
dV = π(r + dr)2dz = π(r2 + 2rdr + dr2)dz
dV = πr2dz + 2πrdrdz + πdr2dz

The higher order differentials can be ignored.
P: 147
 Quote by Doc Al Any slant with the side will be a second order correction to the radius and will have no effect on the volume element. The volume is the area times thickness: dV = area * dz = πr2dz If we vary the radius by dr, we get: dV = π(r + dr)2dz = π(r2 + 2rdr + dr2)dz dV = πr2dz + 2πrdrdz + πdr2dz The higher order differentials can be ignored.
So ignoring the higher order differentials, we still get an exact answer?

What if the external area is perimeter times thickness:

dA = perimeter * dz =2πrdz

Varying the radius by dr, we get:

dA = 2π(r+dr)dz =(2πr+2πdr)dz = 2πrdz + 2πdrdz
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P: 41,457
 Quote by greswd So ignoring the higher order differentials, we still get an exact answer?
Yes, that's how it's done. In the limit as they go to zero, (dx)2 can be ignored compared to dx.
 What if the external area is perimeter times thickness:
But the surface area is not perimeter times thickness; it's perimeter times the length of the side.
P: 147
 Quote by Doc Al Yes, that's how it's done. In the limit as they go to zero, (dx)2 can be ignored compared to dx. But the surface area is not perimeter times thickness; it's perimeter times the length of the side.
ah, that makes sense.

but then volume can also be considered area times some funny ratio and not just the vertical height dz
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P: 41,457
 Quote by greswd but then volume can also be considered area times some funny ratio and not just the vertical height dz
You can express the length of the side in terms of the height. But that will also depend on the angle.
P: 147
 Quote by Doc Al You can express the length of the side in terms of the height. But that will also depend on the angle.
yeah it will. so you've already explained why we can use dz for volume integrals, by ignoring the higher order differentials.

but for external area, you said we must account for the slant, I used (r + dr), doesn't that do the job?

I'm afriad my mathematical understanding is really sketchy.
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