# Moment of Inertia Problem

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?

Related Classical Physics News on Phys.org
Doc Al
Mentor
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?

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.

Doc Al
Mentor
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).

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.

Doc Al
Mentor
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?

Doc Al
Mentor
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.

Doc Al
Mentor
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.

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
Mentor
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.

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]

Doc Al
Mentor
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.

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
Mentor
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.

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
Homework Helper
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.

Doc Al
Mentor
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. )

No. See AlephZero's response. (He beat me to it. )
I get it, but, I don't understand why aleph's method doesn't apply to volume integrals.

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?

Revisiting an old thread, does the proof lie in line integrals?

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.