# What's Wrong with My Line of Reasoning? Calculating a Hollow Sphere's Moment Inertia

1. Feb 29, 2008

### Eus

Hi Ho!

I know that many books show the way to derive the moment inertia of a solid and a hollow sphere in many ways, each according to the lines of reasoning of their authors.

I also have my own line of reasoning that I have successfully applied in finding the moment inertia of a solid sphere, but not the one of the hollow sphere.

I have asked two of my lecturers to help me find the mistake in my line of reasoning. But, both just showed me their own lines of reasoning. The first one directly pointed out to me that I should had used a spherical coordinate system when I had barely said, "Sir, I have a problem in deriving the moment inertia of a hollow sphere..." The other one said that I should had consulted a mathematician instead after he explained his way of deriving the moment inertia. Therefore, would you please help me find the mistake in my line of reasoning that I write down below?

Moment inertia of a point object is defined as: $I=M\ R^2$.

Because I know that $dx$ means a very very small quantity, firstly I tried to calculate the moment inertia of a solid sphere $\left ( I=\frac{2}{5}\ M\ R^2 \right )$ as follows:

1. I used a 3-D Cartesian coordinate and I put a solid sphere having a radius $R$, a total mass $M$, and a uniform mass distribution, centered at the origin and rotated about the $z$-axis.

2. Because the solid sphere was symmetric with respect to the $x-y$ plane, I only considered the upper half of it, which was located in the $+z$ direction.

3. I thinly sliced the upper half part of the sphere parallel with the $x-y$ plane. This resulted in many circular disks with varying diameter depending on their distances from the $x-y$ plane. One such disk was depicted in Figure 1 as a blue rectangle. This particular disk, which was as thin as $dz$, was located at distance $z$ units from the $x-y$ plane, and therefore it has a diameter of $\sqrt{R^2-z^2}$.

4. Because the thin disks were not point objects, I could not apply the formula yet. Therefore, I took one such thin disk and started to thinly slice it into many rings with varying diameter. One such ring was depicted in Figure 2.

5. Again, because the thin rings, which were as thin as $dr$, were not point objects, I could not apply the formula yet. Therefore, I took one such thin ring and started to thinly slice it into many small boxes of size $dz$ x $dr$ x $ds$ as depicted in Figure 3.

6. Because the boxes were very very small, they can be regarded as point objects. Therefore, I could apply the formula now.

7. First, the formula called for the mass of the point object. The mass of the solid sphere could be related to its volume as $M=\rho\ V$. Because the solid sphere had a uniform mass distribution, the boxes had very very small masses $dm$ that could be related to its very very small volumes $dv$, where $dv=(dz)(dr)(ds)$, as $dm=\rho\ dv$.

8. Looking at Figure 3, $di_{box}=r^2\ dm$. To get the total moment inertia for the thin ring, the moment inertia of each box along the ring was summed up as follows:

$$di_{ring}=\int^{2\pi}_{0}di_{box}$$

$$di_{ring}=\int^{2\pi}_{0}r^2\ dm$$

9. Because the integration over the whole ring had to be in terms of $ds$, the $dm$ term was decomposed as follows:

$$di_{ring}=\int^{2\pi}_{0}r^2\ (\rho\ dv)$$

$$di_{ring}=\rho\ \int^{2\pi}_{0}r^2\ dv$$

$$di_{ring}=\rho\ \int^{2\pi}_{0}r^2\ (dz\ dr\ ds)$$

$$di_{ring}=\rho\ r^2\ dz\ dr\ \int^{2\pi}_{0}ds$$

10. Therefore, $di_{ring}=\rho\ r^2\ dz\ dr\ (2\pi\ r)$, which was simply $di_{ring}=2\pi\ \rho\ r^3\ dz\ dr$.

11. Next, the moment inertia of each thin ring was integrated over the whole thin disk in terms of $dr$ to get the moment inertia of each thin disk as follows:

$$di_{disk}=\int^{\sqrt{R^2-z^2}}_{0}di_{ring}$$

$$di_{disk}=\int^{\sqrt{R^2-z^2}}_{0}2\pi\ \rho\ r^3\ dz\ dr$$

$$di_{disk}=2\pi\ \rho\ dz\ \int^{\sqrt{R^2-z^2}}_{0}r^3\ dr$$

$$di_{disk}=2\pi\ \rho\ dz\ \left (\frac{1}{4}\ \left \vert r^4 \right \vert ^{\sqrt{R^2-z^2}}_{0}\right )$$

$$di_{disk}=\frac{1}{2}\ \pi\ \rho\ dz\ \left(R^2-z^2\right)^2$$

12. Finally, the moment inertia of the upper half of the sphere could be found by integrating the moment inertia of each thin disk over the whole length $R$ along the $+z$-axis as follows:

$$I_{\frac{1}{2}solid\ sphere}=\int^{R}_{0}di_{disk}$$

$$I_{\frac{1}{2}solid\ sphere}=\int^{R}_{0}\frac{1}{2}\ \pi\ \rho\ \left(R^2-z^2\right)^2\ dz$$

$$I_{\frac{1}{2}solid\ sphere}=\frac{1}{2}\pi\ \rho\ \int^{R}_{0}\left(R^4-2R^2\ z^2+z^4\right)\ dz$$

$$I_{\frac{1}{2}solid\ sphere}=\frac{1}{2}\ \pi\ \rho\ \left \vert R^4\ z-\frac{2}{3}\ R^2\ z^3+\frac{1}{5}\ z^5\ \right \vert ^{R}_{0}$$

$$I_{\frac{1}{2}solid\ sphere}=\frac{1}{2}\ \pi\ \rho\ \left(R^5-\frac{2}{3}\ R^5+\frac{1}{5}\ R^5\right)$$

$$I_{\frac{1}{2}solid\ sphere}=\frac{1}{2}\ \pi\ \rho\ \left(\frac{8}{15}\ R^5\right)$$

$$I_{\frac{1}{2}solid\ sphere}=\frac{4}{15}\ \pi\ \rho\ R^5$$

13. To find the moment inertia of the whole solid sphere, the moment inertia of the upper half of the solid sphere was multiplied by two as follows:

$$I_{solid\ sphere}=2I_{\frac{1}{2}solid\ sphere}$$

$$I_{solid\ sphere}=2\left(\frac{4}{15}\ \pi\ \rho\ R^5\right)$$

$$I_{solid\ sphere}=\frac{8}{15}\ \pi\ \rho\ R^5$$

14. As the last step, the moment inertia of the whole solid sphere was given in terms of its mass $M$. The relation of $M=\rho\ V$ where $V=\frac{4}{3}\pi\ R^3$ was substituted as follows:

$$I_{solid\ sphere}=\frac{8}{15}\ \pi\ \rho\ R^5$$

$$I_{solid\ sphere}=\frac{8}{15}\ \pi\ \left(\frac{3}{4}\ \frac{1}{\pi}\ \frac{M}{R^3}\right)\ R^5$$

$$I_{solid\ sphere}=\frac{2}{5}\ M\ R^2$$

Yup, I got the right result!

Now since I was able to calculate the moment inertia of a solid sphere with the idea of $dx$ as a very very small quantity, finally I tried to calculate the moment inertia of a hollow sphere (a thin spherical shell), which was actually only a very very thin plate as the surface of a hollow sphere, $\left (\frac{2}{3}\ M\ R^2\right )$ as follows:

1. I used a 3-D Cartesian coordinate and I put a hollow sphere having a radius $R$, a total mass $M$, and a uniform mass distribution, centered at the origin and rotated about the $z$-axis.

2. Because the solid sphere was symmetric with respect to the $x-y$ plane, I only considered the upper half of it, which was located in the $+z$ direction.

3. I thinly sliced the upper half part of the sphere parallel with the $x-y$ plane. This resulted in many circular rings with varying diameter depending on their distances from the $x-y$ plane. One such ring was depicted in Figure 4 as a red rectangle. This particular ring, which was as thin as $dz$, was located at distance $z$ units from the $x-y$ plane, and therefore it has a diameter of $\sqrt{R^2-z^2}$.

4. Because the thin rings were not point objects, I could not apply the formula yet. Therefore, I took one such thin ring and started to thinly slice it into many small plates of size $dz$ x $ds$ as depicted in Figure 5.

5. Because the plates were very very small, they can be regarded as point objects. Therefore, I could apply the formula now.

6. First, the formula called for the mass of the point object. The mass of the hollow sphere could be related to its surface area as $M=\rho\ A$. Because the hollow sphere had a uniform mass distribution, the plates had very very small masses $dm$ that could be related to its very very small areas $da$, where $da=(dz)(ds)$, as $dm=\rho\ da$.

7. Looking at Figure 5, $di_{plate}=(\sqrt{R^2-z^2})^2\ dm$. To get the total moment inertia for the thin ring, the moment inertia of each plate along the ring was summed up as follows:

$$di_{ring}=\int^{2\pi}_{0}di_{plate}$$

$$di_{ring}=\int^{2\pi}_{0}(R^2-z^2)\ dm$$

8. Because the integration over the whole ring had to be in terms of $ds$, the $dm$ term was decomposed as follows:

$$di_{ring}=\int^{2\pi}_{0}(R^2-z^2)(\rho\ da)$$

$$di_{ring}=\rho\ \int^{2\pi}_{0}(R^2-z^2)\ da$$

$$di_{ring}=\rho\ \int^{2\pi}_{0}(R^2-z^2)(dz\ ds)$$

$$di_{ring}=\rho\ (R^2-z^2)\ dz\ \int^{2\pi}_{0}\ ds$$

9. Therefore, $di_{ring}=\rho\ (R^2-z^2)\ dz\ (2\pi\ \sqrt{R^2-z^2})$, which was simply $di_{ring}=2\pi\ \rho\ (R^2-z^2)^{\frac{3}{2}}\ dz$.

10. Finally, the moment inertia of the upper half of the hollow sphere could be found by integrating the moment inertia of each thin ring over the whole length $R$ along the $+z$-axis as follows:

$$I_{\frac{1}{2}\ hollow\ sphere}=\int^{R}_{0}di_{ring}$$

$$I_{\frac{1}{2}\ hollow\ sphere}=\int^{R}_{0}2\pi\ \rho\ (R^2-z^2)^{\frac{3}{2}}\ dz$$

$$I_{\frac{1}{2}\ hollow\ sphere}=2\pi\ \rho\ \int^{R}_{0}(R^2-z^2)^{\frac{3}{2}}\ dz$$

It can be calculated that:

$$\int (a^2-x^2)^{\frac{3}{2}}\ dx=\frac{1}{8}((5a^2-2x^2)(x\ \sqrt{a^2-x^2})+3a^4\ \arcsin(\frac{x}{a}))+c$$

Therefore, the calculation of $I_{\frac{1}{2}\ hollow\ sphere}$ can be continued as follows:

$$I_{\frac{1}{2}\ hollow\ sphere}=2\pi\ \rho\ \frac{1}{8} \left \vert (5R^2-2z^2)(z\ \sqrt{R^2-z^2})+3R^4\ \arcsin(\frac{z}{R}) \right \vert^{R}{0}$$

$$I_{\frac{1}{2}\ hollow\ sphere}=2\pi\ \rho\ \frac{1}{8}\ (3R^4\ \arcsin(\frac{R}{R}))$$

$$I_{\frac{1}{2}\ hollow\ sphere}=2\pi\ \rho\ \frac{1}{8}\ (3R^4\ \frac{\pi}{2})$$

$$I_{\frac{1}{2}\ hollow\ sphere}=\frac{3}{8}\ \pi^2\ \rho\ R^4$$

11. To find the moment inertia of the whole hollow sphere, the moment inertia of the upper half of the hollow sphere was multiplied by two as follows:

$$I_{hollow\ sphere}=2I_{\frac{1}{2}\ hollow\ sphere}$$

$$I_{hollow\ sphere}=2(\frac{3}{8}\ \pi^2\ \rho\ R^4)$$

$$I_{hollow\ sphere}=\frac{3}{4}\ \pi^2\ \rho\ R^4$$

12. As the last step, the moment inertia of the whole hollow sphere was given in terms of its mass $M$. The relation of $M=\rho\ A$ where $A=4\pi\ R^2$ was substituted as follows:

$$I_{hollow\ sphere}=\frac{3}{4}\ \pi^2\ \rho\ R^4$$

$$I_{hollow\ sphere}=\frac{3}{4}\ \pi^2\ \left (\frac{M}{4\pi\ R^2}\right )\ R^4$$

$$I_{hollow\ sphere}=\frac{3}{16}\ \pi\ M\ R^2$$

I got the wrong result this time. But, why? Isn't that I had successfully applied my line of reasoning in finding the moment inertia of a solid sphere? What's wrong with my line of reasoning? Did I miss some important points in finding the moment inertia of a hollow sphere?

Thank you very much.

Best regards,
Eus

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2. Feb 29, 2008

### HallsofIvy

Took me a while to read through that!

Your "line of reasoning" appears good to me but I think I see a technical error.

In your paragraph 7, you have
The ring does not have unit radius but radius $\sqrt{R^2- z^2}$, circumference $2\pi\sqrt{R^2- z^2}$ so I believe this should be
$$di_{ring}=\int^{2\pi \sqrt{R^2- z^2}}_{0}(R^2-z^2)\ dm$$

3. Mar 1, 2008

### Eus

Hi Ho!

Thank you very much for reading my line of reasoning.
Although it seems long and complicated, the idea is actually very simple, isn't it?

I don't get the reason why the fact that the ring in the second case does not have unit radius affects the calculation in any way. Isn't that the ring in the first case also does not have unit radius but radius $r$?

I have tried your suggestion as follows:
$$di_{ring}=\int^{2\pi\ \sqrt{R^2-z^2}}_{0}di_{plate}$$

$$di_{ring}=\int^{2\pi\ \sqrt{R^2-z^2}}_{0}(R^2-z^2)\ dm$$

8. Because the integration over the whole ring had to be in terms of $ds$, the $dm$ term was decomposed as follows:

$$di_{ring}=\int^{2\pi\ \sqrt{R^2-z^2}}_{0}(R^2-z^2)(\rho\ da)$$

$$di_{ring}=\rho\ \int^{2\pi\ \sqrt{R^2-z^2}}_{0}(R^2-z^2)\ da$$

$$di_{ring}=\rho\ \int^{2\pi\ \sqrt{R^2-z^2}}_{0}(R^2-z^2)(dz\ ds)$$

$$di_{ring}=\rho\ (R^2-z^2)\ dz\ \int^{2\pi\ \sqrt{R^2-z^2}}_{0}\ ds$$

$$di_{ring}=\rho\ (R^2-z^2)\ dz\ \left \vert s \right \vert^{2\pi\ \sqrt{R^2-z^2}}_{0}$$

$$di_{ring}=\rho\ (R^2-z^2)\ dz\ (2\pi\ \sqrt{R^2-z^2})$$

The result does not change anything for the next step (i.e., Step 9).
Therefore, the error is not in the part that you pointed out, is it?

Thank you.

Best regards,
Eus

4. Mar 1, 2008

### D H

Staff Emeritus
Eus,

The underlying problem is that you modeled the surface of a sphere as a stack of infinitely short cylinders. This approach doesn't even yield the correct surface area for a sphere. The surface area of the sphere by your approach is

$$A = \int_{-R}^R 2\pi\sqrt{R^2-z^2} dz$$

Using the substitution $z=R\sin\theta$,

$$A = \pi \int_{-\pi/2}^{\pi/2} 2 \cos^2\theta d\theta = \pi R^2\left(\frac1 2 \sin 2\theta + \theta\right)\biggr|_{-\pi/2}^{\pi/2} = pi^2 R^2$$

The surface area of a sphere is of course $4\pi R^2$. So what's wrong here? The problem is with the cylindrical model. A model that yields the correct answer is a stack of infinitely short conic frustums. The incremental surface area is, with some abuse of terminology,

$$dA = \pi (r+(r+dr))\sqrt{dz^2 + dr^2} = 2\pi r \sqrt{1+\frac {z^2}{r^2}}\,dz = 2\pi r \frac R r dz = 2 \pi R dz$$

Integrating from z = -R to R yields $A=4\pi R^2$, the correct answer.

Using the conic frustum model as opposed to the cylindrical model does yield the correct answer for the moment of inertia for a hollow sphere. I'll leave that as an exercise for you to complete.

Your ultimate question is "Why does the solid cylinder model yield the correct answer for the moment of inertia but not for the hollow sphere?" The answer is that the difference between a cylinder and a conic frustum is a little wedge of material in terms of volume and a slant in terms of area. That little wedge of material is inconsequential compared to the volume of the cylinder. The slant, on the other hand, is very significant when it comes to area.

Last edited: Mar 1, 2008
5. Mar 2, 2008

### Eus

Ah, what a good way to check whether a model is correct or not!
I was riveted in finding the moment inertia of a solid and hollow sphere so that I forgot to validate the underlying model.
Thank you for pointing this out.

Infinitely short conical frustums never came to my mind before because I was riveted to the model that uses many infinitely short cylinders to find the volume of any solid. Besides that, I did not see the reason why many infinitely short cylinders would not do the job.

I still don't get how you can transform the LHS into the RHS of the equation above.
Would you please kindly elaborate more on how you transform it?

Thank you very much for pointing out the mistake in my line of reasoning: the underlying model.

Now I have a final question:

I had known that as far as I could partition any object into many elements I could approximate (and truly got the exact) area, length, surface area, volume, moment inertia, and so on.

Now I know that I must be really careful in my choice of the shape of the infinitely many elements that make up the object.

What I don't know is how should I make the choice. For example, before this I thought that there is no difference between using an infinitely short cylinder and an infinitely short conical frustum because both appeared to be a 2-D circle. But, it turned out that the slant had a very profound effect in finding the surface area of a sphere, whereas the wedge didn't have any impact whatsoever in finding the volume of a sphere.

Since the application of calculus is very broad, how should I determine the choice?
For example, what kind of rule-of-thumb that I should have had to be able to tell my self that I needed to use an element that had a slant (i.e., a conical frustum) to find the surface area of a sphere?
Another example is that why I could not have used a conical frustum to find the volume of a sphere? The slant should have approximated the boundary of the sphere better, shouldn't it?

Therefore, how should I determine the choice so that I can give the reason for my decision?

Thank you very much.

Best regards,
Eus

6. Mar 3, 2008

### D H

Staff Emeritus
I did say "with some abuse of terminology". Since I'm already guilty of abuse of terminology,

$$\sqrt{dz^2 + dr^2} = \sqrt{1+\left(\frac{dr}{dz}\right)^2}dz$$

Since $r^2+z^2 = R^2$, $dr/dz = -z/r$, and the rest follows. Note that this is just the equation for arc length.

7. Mar 4, 2008

### Eus

I don't see any abuse of terminology in the way you derive $dA$ because for me it is pretty good. I would be glad if you can point out which terminology you abused.

Ah, I got it now! Please let me show you that I have understood your derivation below:

$$dA=\pi (r+(r+dr))\sqrt{dz^2+dr^2}$$

$$dA=\pi (2r+dr)\sqrt{dz^2+dr^2}$$

$$dA=(2\pi r+\pi\ dr)\sqrt{dz^2+dr^2}$$

Because $\pi\ dr$ is infinitely small, it can be dropped, can't it? So,

$$dA=2\pi r\sqrt{dz^2+dr^2}$$

$$dA=2\pi r\sqrt{dz^2\left (1+\frac{dr^2}{dz^2}\right )}$$

$$dA=2\pi r\ dz\sqrt{1+\frac{dr^2}{dz^2}}$$

$$dA=2\pi r\ dz\sqrt{1+\frac{dr}{dz}^2}$$

Since $r^2+z^2=R^2$,

$$r^2=R^2-z^2$$

$$r=\sqrt{R^2-z^2}$$

$$\frac{dr}{dz}=\frac{d\left (\sqrt{R^2-z^2}\right )}{dz}$$

$$\frac{dr}{dz}=\frac{d\left (\sqrt{R^2-z^2}\right )}{d\left ( R^2-z^2 \right )}\ \frac{d\left ( R^2-z^2 \right )}{dz}$$

$$\frac{dr}{dz}=\frac{1}{2\sqrt{R^2-z^2}}\ (-2z)$$

$$\frac{dr}{dz}=\frac{-z}{\sqrt{R^2-z^2}}$$

$$\frac{dr}{dz}=\frac{-z}{r}$$

Therefore,

$$dA=2\pi r\ dz\sqrt{1+\left ( - \frac{z}{r}\right )^2}$$

$$dA=2\pi r\ dz\sqrt{1+\frac{z^2}{r^2}}$$

$$dA=2\pi r\ dz\sqrt{\frac{r^2+z^2}{r^2}}$$

$$dA=2\pi r\ dz\frac{\sqrt{r^2+z^2}}{r}$$

$$dA=2\pi dz\sqrt{r^2+z^2}$$

$$dA=2\pi dz\ R$$

So, $dA=2\pi R\ dz$. Nice!

Thank you very much.

Best regards,
Eus

8. Jul 20, 2008

### PAL_PHYSICS

Re: What's Wrong with My Line of Reasoning? Calculating a Hollow Sphere's Moment Iner

Dear Friend,
I DID THE SAME WAY WHAT YOU HAVE DONE. BUT I DIDN'T USE ANY CALCULUS. BUT I USED THE SOLID SPHERE MOMENT OF INERTIA EQUATION TO PROVE THE MOMENT OF INERTIA OF A HOLLOW SPHERE. HEREWITH I WILL UPLOAD THE WORK PAPER. PLEASE GO THROUGH IT AND LET ME KNOW..... THANK YOU

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9. Jul 20, 2008

### Redbelly98

Staff Emeritus