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

In summary: but then volume can also be considered area times some funny ratio and not just the vertical height...
  • #1
greswd
764
20
Here's a derviation from HyperPhysics:

sph2.gif


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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  • #2
greswd said:
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?
 
  • #3
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 [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.
 
  • #4
greswd said:
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).
 
  • #5
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 [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.
 
  • #6
greswd said:
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.
 
  • #7
Is this for the moment of inertia of a sphere?
 
  • #8
HomogenousCow said:
Is this for the moment of inertia of a sphere?
A spherical shell.
 
  • #9
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.
 

1. What is Moment of Inertia?

Moment of Inertia is a measure of an object's resistance to rotational motion. It is a physical property that depends on the object's mass distribution and the axis of rotation.

2. How is Moment of Inertia calculated?

Moment of Inertia (I) can be calculated by summing the products of the mass of each particle in the object and the square of its distance from the axis of rotation (r^2). This can be represented by the equation I = Σmr^2, where m is the mass and r is the distance from the axis of rotation.

3. What are the units of Moment of Inertia?

The units of Moment of Inertia are kg*m^2 or kg*cm^2. These units reflect the mass and distance components in the calculation of Moment of Inertia.

4. How does Moment of Inertia affect an object's rotation?

The higher the Moment of Inertia of an object, the more resistance it has to rotational motion. This means that it will require more torque to cause the object to rotate, and it will also rotate at a slower speed compared to an object with a lower Moment of Inertia.

5. How is Moment of Inertia used in real-world applications?

Moment of Inertia is an important concept in engineering and physics, and is used in various applications such as designing machinery, analyzing the stability of structures, and calculating the motion of rigid bodies. It is also used in sports, such as in the design of equipment like golf clubs and tennis rackets.

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