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Moment of Inertia Problem

by greswd
Tags: inertia, moment
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greswd
#1
Nov24-12, 11:04 AM
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Here's a derviation from HyperPhysics:



He says:

[tex]dV=πy^{2}dz[/tex]


However, if we're finding the surface area of the sphere:

[tex]dA=2πRdz≠2πydz[/tex]


If we cannot use [tex]dA=2πydz[/tex], how come [tex]dV=πy^{2}dz[/tex] is still applicable?
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Doc Al
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Nov24-12, 11:31 AM
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Quote Quote by greswd View Post
He says:

[tex]dV=πy^{2}dz[/tex]
That's the volume of those circular disks.

What does finding the surface area of the sphere have to do with this?
greswd
#3
Nov24-12, 12:11 PM
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Quote Quote by Doc Al View Post
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 [tex]dA=2πydz[/tex], 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.

Doc Al
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Nov24-12, 02:49 PM
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Moment of Inertia Problem

Quote Quote by greswd View Post
If I want to find the MoI of a hollow sphere.

Based on the same assumptions [tex]dA=2πydz[/tex], 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).
greswd
#5
Nov30-12, 09:05 AM
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Quote Quote by Doc Al View Post
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 [tex]A=2πRz[/tex]



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

[tex]A→2πrz[/tex]

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.
Doc Al
#6
Nov30-12, 09:19 AM
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Quote Quote by greswd View Post
Yup, that's the perplexing part of the formula [tex]A=2πRz[/tex]



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

[tex]A→2πrz[/tex]
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.
HomogenousCow
#7
Nov30-12, 09:47 AM
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Is this for the moment of inertia of a sphere?
Doc Al
#8
Nov30-12, 09:49 AM
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Quote Quote by HomogenousCow View Post
Is this for the moment of inertia of a sphere?
A spherical shell.
HomogenousCow
#9
Nov30-12, 09:56 AM
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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.
Doc Al
#10
Nov30-12, 10:01 AM
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Quote Quote by HomogenousCow View Post
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.
greswd
#11
Dec2-12, 02:23 AM
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Quote Quote by Doc Al View Post
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?
Doc Al
#12
Dec2-12, 07:11 AM
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Quote Quote by greswd View Post
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.
greswd
#13
Dec2-12, 07:47 AM
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Quote Quote by Doc Al View Post
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
Doc Al
#14
Dec2-12, 07:56 AM
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Quote Quote by greswd View Post
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.
greswd
#15
Dec2-12, 09:37 AM
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Quote Quote by Doc Al View Post
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
Doc Al
#16
Dec2-12, 09:55 AM
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Quote Quote by greswd View Post
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.
greswd
#17
Dec2-12, 10:29 AM
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Quote Quote by Doc Al View Post
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.
AlephZero
#18
Dec2-12, 11:34 AM
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Quote Quote by greswd View Post
but for external area, you said we must account for the slant, I used (r + dr), doesn't that do the job?
No. Make a drawing of a cut through the center of the sphere. The cut surface is a circle radius r. You have a little right-angled triangle, with the horizontal side length dr, the vertical side length dz, and the slant (the actual surface) is length ##\sqrt{dr^2 + dz^2}## by Pythagoras's theorem.

Of course you can use trig and get the length in terms of an angle, if you want.


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