Moment of Inertia of a Solid Sphere: Revisiting the Derivation from HyperPhysics

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SUMMARY

This discussion focuses on the derivation of the moment of inertia (MoI) of solid and hollow spheres, referencing HyperPhysics. The volume element is defined as dV=πy²dz, while the surface area is incorrectly assumed as dA=2πydz. The participants clarify that for surface area calculations, the arc length must be considered, not just the vertical distance. The correct moment of inertia for a solid sphere is derived using integration, yielding the formula 2/5 MR², while the moment of inertia for a hollow sphere is 2/3 MR².

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  • Familiarity with the concepts of moment of inertia and its physical significance.
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greswd
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Here's a derviation from HyperPhysics:

sph2.gif


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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greswd said:
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?
 
Doc Al said:
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.
 
greswd said:
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).
 
Doc Al said:
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.
 
greswd said:
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.
 
Is this for the moment of inertia of a sphere?
 
HomogenousCow said:
Is this for the moment of inertia of a sphere?
A spherical shell.
 
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.
 
  • #10
HomogenousCow said:
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.
 
  • #11
Doc Al said:
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?
 
  • #12
greswd said:
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 + [STRIKE]2πrdrdz[/STRIKE] + [STRIKE]πdr2dz[/STRIKE]

The higher order differentials can be ignored.
 
  • #13
Doc Al said:
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 + [STRIKE]2πrdrdz[/STRIKE] + [STRIKE]πdr2dz[/STRIKE]

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 + [STRIKE]2πdrdz
[/STRIKE] :confused:
 
  • #14
greswd said:
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.
 
  • #15
Doc Al said:
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
 
  • #16
greswd said:
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.
 
  • #17
Doc Al said:
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.
 
  • #18
greswd said:
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.
 
  • #19
greswd said:
but for external area, you said we must account for the slant, I used (r + dr), doesn't that do the job?
No. See AlephZero's response. (He beat me to it. :smile:)
 
  • #20
Doc Al said:
No. See AlephZero's response. (He beat me to it. :smile:)

I get it, but, I don't understand why aleph's method doesn't apply to volume integrals.
 
  • #21
I think I got it now.

Let's say I'm trying to find the circumference and area of a circle by using rectangles to approximate.


As the no. of rectangles approaches infinity, their combined height will always be equal to the diameter, and will never approximate the circumference.

However, the combined area of those rectangles will approximate the area of the circle.


This is why dz works for volume and not area integrals.

It's a geometric proof, but how do we prove it mathematically?
 
  • #22
Revisiting an old thread, does the proof lie in line integrals?
 
  • #23
greswd said:
Revisiting an old thread, does the proof lie in line integrals?

Yes. Integration rocks ! As always. Also, as I can see by your thread, you are inquisitive of finding moment of inertia of a solid sphere, right ? I did not follow the hyperphysics derivation, but I derive it using moment of inertia of hollow sphere, which is 2MR2/3..

I am sorry. You ought to find moment of inertia of hollow sphere ?
There are two ways:

1. Integration (difficult)
2. Coordinate geometry (easy!).. which I think you have been told here.
 

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