Moment of Inertia: Thin Spherical Shell

[jake010]

Homework Statement

Derive the formula for the moment of inertia of a thin spherical shell using spherical coordinates and multiple integrals.

Relevant Equations

Moment of Intertia is (2MR^2)/3

Homework Statement: Derive the formula for the moment of inertia of a thin spherical shell using spherical coordinates and multiple integrals.
Homework Equations: Moment of Intertia is (2MR^2)/3

I = (2MR^2)/3

 

[Dr.D]

The word "thin" here is important. That is telling you that all the mass is a the same radius. Do you know how to formulate the differential area on the surface of a sphere in spherical coordinates? If so, then you're practically there. Simply assign a symbol for the mass per unit area, express the necessary radius to a typical point, and integrate over the area, and presto!

 

[jake010]

The word "thin" here is important. That is telling you that all the mass is a the same radius. Do you know how to formulate the differential area on the surface of a sphere in spherical coordinates? If so, then you're practically there. Simply assign a symbol for the mass per unit area, express the necessary radius to a typical point, and integrate over the area, and presto!

The problem is I don't know what I'm doing. The professor introduced double integrals, triple integrals, and moment of inertia today and I need to be able to derive the formula by Friday.

 

[Dr.D]

Do you know what axis is intended for the moment of inertia? They will all be the same for an axis through the center of the sphere, but infinitely many other axes are also possible.

Also, you did not answer my question about whether you know how to express the differential area in spherical coordinates. This is a necessary first step.

 

[jake010]

Do you know what axis is intended for the moment of inertia? They will all be the same for an axis through the center of the sphere, but infinitely many other axes are also possible.

Also, you did not answer my question about whether you know how to express the differential area in spherical coordinates. This is a necessary first step.

In cartesian coordinates, rotation about the z-axis. I understand how spherical coordinates work but I don't really know how to do integrate in spherical coordinates. The professor introduced moment of inertia, integration in spherical and cylindrical coordinates, and double and triple integrals today, so I am very lost. I understand how spherical coordinates work, and I can integrate, find area, volume, etc. in cartesian coordinates.

 

[haruspex]

I understand how spherical coordinates work

So define a surface element in those terms and write an expression for its moment of inertia about the axis. Post however far you get.

 

[Cutter Ketch]

I think calculus class sometimes spends all the time teaching students how to solve integrals and loses track of why you are doing integrals, what the integral means, or how a problem becomes an integral problem.

Here’s the idea: you know the answer for a small piece, and you are trying to write the answer for the whole thing as a sum of all the small pieces.

Do you know what the inertia is for some small piece? Can you describe a thin shell as collection of those small pieces? Can you make the description of the location of each and every piece parametric in the coordinates? Then the inertia of the shell is the inertia of those small pieces summed over the whole shell. Taking the pieces as infinitesimally small, the sum becomes an integral. Yes, you will have to integrate in three dimensions.

 

[Dr.D]

Yes, you will have to integrate in three dimensions.

Actually, there is only the need for a 2D integration. This is where the key word "thin" comes into play. There is no need to integrate radially.

 

[jtbell]

The professor introduced moment of inertia, integration in spherical and cylindrical coordinates, and double and triple integrals today

Have you done any surface integrals in Cartesian coordinates, where the surface element is $dx\,dy$?

 

[Cutter Ketch]

Actually, there is only the need for a 2D integration. This is where the key word "thin" comes into play. There is no need to integrate radially.

One of the integrals is trivial, but does result in multiplying by the thickness. We are used to short cutting the third integral, but the third dimension is required.

 

[Dr.D]

One of the integrals is trivial, but does result in multiplying by the thickness. We are used to short cutting the third integral, but the third dimension is required.

Yes, I agree that most spheres are three dimensional. However, the key word "thin" tells us that the intent was to assume a two dimensional surface for the sphere, thus only requiring two integrations.

 

[Cutter Ketch]

Yes, I agree that most spheres are three dimensional. However, the key word "thin" tells us that the intent was to assume a two dimensional surface for the sphere, thus only requiring two integrations.

If you don’t integrate in the third dimension you get the wrong answer. We know the integrand doesn’t vary significantly in the radial dimension. We know the radial integral is trivial. So we don’t write down the third integral but we DO multiply by the thickness. That thickness is the solution to the third integral. When you multiply by the thickness, guess what, you just integrated the third dimension. Otherwise where did that thickness come from? Just because we are so used to it that we skip steps and jump directly to the answer doesn’t mean all three dimensions are not required. If you tell this poster to only consider two dimensions he will not have the thickness in the answer. The only way he is going to have it in the answer is if he writes and solves the third integral.

 

[jake010]

If you don’t integrate in the third dimension you get the wrong answer. We know the integrand doesn’t vary significantly in the radial dimension. We know the radial integral is trivial. So we don’t write down the third integral but we DO multiply by the thickness. That thickness is the solution to the third integral. When you multiply by the thickness, guess what, you just integrated the third dimension. Otherwise where did that thickness come from? Just because we are so used to it that we skip steps and jump directly to the answer doesn’t mean all three dimensions are not required. If you tell this poster to only consider two dimensions he will not have the thickness in the answer. The only way he is going to have it in the answer is if he writes and solves the third integral.

This is as far as I've gotten:

dI of shell = dI solid sphere/dR
dI of solid sphere =r²dm
=ρr²dv

dv=(rdφ)(dr)(rsinφdθ)

dI=ρr²r²sinφdrdθdφ

I integrate with respect to r from 0 to R, with respect to theta from 0 to 2π, and with respect to φ from 0 to π , then take a derivative with respect to R, and get the wrong answer. What am I doing wrong?

 

[Dr.D]

If you tell this poster to only consider two dimensions he will not have the thickness in the answer. The only way he is going to have it in the answer is if he writes and solves the third integral.

He will, if he knows that "thin" means. But whatever, ... (this is really boring at this point).

 

[Dr.D]

I integrate with respect to r from 0 to R, with respect to theta from 0 to 2π, and with respect to φ from 0 to π and get the wrong answer. What am I doing wrong?

Why would you integrated from 0 to R? Have you thought about what "thin" means? Have you drawn a good picture, showing the whole situation in a perspective drawing? If not, it would help a lot. There is more to this than plug and chug.

 

[jake010]

I know what thin means which is why I'm taking a derivative with respect to R.

 

[Dr.D]

I don't think you are really taking any derivatives at all, but if you think so, ...

I do not seem to be getting through to you, so I'm done here. I'm sure someone else will help you.

 

[jake010]

I don't think you are really taking any derivatives at all, but if you think so, ...

I do not seem to be getting through to you, so I'm done here. I'm sure someone else will help you.

You're not getting through to me because I need detailed explanations of what I need to do. I've made it very clear from the start that I just learned about moment of inertia and triple integrals yesterday and that I'm very lost and need serious assistance. I'm doing my best and if you can't help me please don't bother to reply. In regards to derivatives, this video shows how the moment of inertia for a spherical shell can be found using that of a solid sphere, I understand how what is shown in the video works, so as of now I'm trying to figure out how to find the moment of inertia for a solid sphere.

 

[Cutter Ketch]

If you know the moment of inertia of a solid sphere you can indeed use that to find the moment of inertia of a thin spherical shell. No integrals are required. Unfortunately, your problem expressly asks you to use multiple integrals, so I’m afraid that isn’t going to be acceptable.
I think this is a good thing. In making you add up the pieces for one particular shape they are trying to give you an understanding of how the moment of inertia for any shape comes about.
So, again, integrating is adding up. Can you say what the moment of inertia is for a very small piece of the sphere?

 

[jake010]

If you know the moment of inertia of a solid sphere you can indeed use that to find the moment of inertia of a thin spherical shell. No integrals are required. Unfortunately, your problem expressly asks you to use multiple integrals, so I’m afraid that isn’t going to be acceptable.
I think this is a good thing. In making you add up the pieces for one particular shape they are trying to give you an understanding of how the moment of inertia for any shape comes about.
So, again, integrating is adding up. Can you say what the moment of inertia is for a very small piece of the sphere?

Thanks for the help, I think I'm understanding this better, I got dI=ρr²r²sinφdrdθdφ for a small piece of the solid sphere's moment of inertia but I don't think it's correct because the integral doesn't work out correctly.

 

[jake010]

If you know the moment of inertia of a solid sphere you can indeed use that to find the moment of inertia of a thin spherical shell. No integrals are required. Unfortunately, your problem expressly asks you to use multiple integrals, so I’m afraid that isn’t going to be acceptable.
I think this is a good thing. In making you add up the pieces for one particular shape they are trying to give you an understanding of how the moment of inertia for any shape comes about.
So, again, integrating is adding up. Can you say what the moment of inertia is for a very small piece of the sphere?

Correct me if I'm wrong but is dI actually equal to ρ((rsinφ)^2)(r^2)sinφdrdθdφ because the radius squared in dI=r^2dm is actually (rsinφ )^2 because we're looking for the component of radius parallel to the xy plane?Because the integral works out to the correct answer when dI is that way.

 

[Cutter Ketch]

Correct me if I'm wrong but is dI actually equal to ρ((rsinφ)^2)(r^2)sinφdrdθdφ because the radius squared in dI=r^2dm is actually (rsinφ )^2 because we're looking for the component of radius parallel to the xy plane?Because the integral works out to the correct answer when dI is that way.

Yes, you are getting it. I’m a little worried that “because the integral works out to the correct answer” is usually not an approach conducive to understanding. So if you don’t mind I’d like to coach you a little through the “why” so you can be a little more confident in the future.
Before I do, just so I’m not accused of giving anything away, could you tell me the expression for the rotational inertia of a point mass?

 

[collinsmark]

If you don’t integrate in the third dimension you get the wrong answer.

This isn't true for a problem such as this. (I.e., infinitesimally "thin," hollow shell.)

All that's necessary is to define surface density.

$\rho_s = \frac{M}{A}$

You can think of it as mass per unit area. (If it helps to think of it in familiar units, perhaps think of it as $\left[ \frac{\mathrm{kg}}{\mathrm{m^2}} \right]$.)

For this problem, the density of an infinitesimally thin, spherical shell is $\rho_s = \frac{M}{\mathrm{Surface \ area \ of \ a \ sphere}}$.

This shouldn't be too unfamiliar to you. You've probably done this before for linear density. For example, if you were finding the mass per unit length of a thin rod, you would start with $\lambda = \frac{\mathrm{mass \ of \ the \ rod}}{\mathrm{length \ of \ the \ rod}}$. It's the same idea here with $\rho_s$, except two dimensions are involved instead of just one.

The differential mass, $dm$ is calculated by

$dm = \rho_s (\mathrm{differential \ length})(\mathrm{differential \ width})$

A picture might help here. The following is a picture of a small, differential patch of surface on a sphere of radius $r$, where $\theta$ represents longitude and $\varphi$ represents latitude (well, lattidue with 0 deg and 180 deg defining the poles).

This differential mass can be used in calculating the moment of inertia, keeping care to properly specify the distance of $dm$ to the axis of rotation. The integral can then be carried out integrating over $\theta$ and $\varphi$ . Only two integrals are necessary.

No integration over the radius of the sphere $r$ is necessary. The radius of the sphere treated as a constant. (Don't confuse the radius of the sphere $r$ with the distance from $dm$ to the axis of rotation. They are different entities.)

 

[Cutter Ketch]

This is as far as I've gotten:

dI of shell = dI solid sphere/dR
dI of solid sphere =r²dm
=ρr²dv

dv=(rdφ)(dr)(rsinφdθ)

dI=ρr²r²sinφdrdθdφ

I integrate with respect to r from 0 to R, with respect to theta from 0 to 2π, and with respect to φ from 0 to π , then take a derivative with respect to R, and get the wrong answer. What am I doing wrong?

Oops! My bad! In the previous rapid exchanges I missed the above quoted post. Yes, along with your last post fixing the r^2 you have it completely.
Now let me explain why the r^2 is wrong. You have two different r’s. You have r as in the radial dimension in spherical coordinates. But the r in the expression for rotational inertia mr^2 is not that r. It is the radial distance from the rotational axis.

 

[Cutter Ketch]

This isn't true for a problem such as this. (I.e., infinitesimally "thin," hollow shell.)

All that's necessary is to define surface density.

$\rho_s = \frac{M}{A}$

You can think of it as mass per unit area. (If it helps to think of it in familiar units, perhaps think of it as $\left[ \frac{\mathrm{kg}}{\mathrm{m^2}} \right]$.)

For this problem, the density of an infinitesimally thin, spherical shell is $\rho_s = \frac{M}{\mathrm{Surface \ area \ of \ a \ sphere}}$.

This shouldn't be too unfamiliar to you. You've probably done this before for linear density. For example, if you were finding the mass per unit length of a thin rod, you would start with $\lambda = \frac{\mathrm{mass \ of \ the \ rod}}{\mathrm{length \ of \ the \ rod}}$. It's the same idea here with $\rho_s$, except two dimensions are involved instead of just one.

The differential mass, $dm$ is calculated by

$dm = \rho_s (\mathrm{differential \ length})(\mathrm{differential \ width})$

A picture might help here. The following is a picture of a small, differential patch of surface on a sphere of radius $r$, where $\theta$ represents longitude and $\varphi$ represents latitude (well, lattidue with 0 deg and 180 deg defining the poles).
View attachment 252482

This differential mass can be used in calculating the moment of inertia, keeping care to properly specify the distance of $dm$ to the axis of rotation. The integral can then be carried out integrating over $\theta$ and $\varphi$ . Only two integrals are necessary.

No integration over the radius of the sphere $r$ is necessary. The radius of the sphere treated as a constant. (Don't confuse the radius of the sphere $r$ with the distance from $dm$ to the axis of rotation. They are different entities.)

It is thin but it is not infinitesimally thin. It is a real object. It is thin enough that the integrand does not change much over the thickness making the integral in the third dimension trivial, but the third dimension must be included.

As Jake discovered
“
dI of solid sphere =r²dm
=ρr²dv

dv=(rdφ)(dr)(rsinφdθ)

dI=ρr²r²sinφdrdθdφ
“
Except the first r is the radial distance from the axis and should be r sin(phi)

that is the correct reasoning and the correct integrand with the correct infinitesimal piece of volume. Now answer me this: does the correct moment of inertia for a thin sphere have “dr” in the expression, or does it have thickness? It has thickness. Why? Because that is what dr INTEGRATES to! Yes, you absolutely DO have to integrate in all three dimensions. The fact that the sphere is thin means that the radial integral is trivial, but it is there

\int_r1^r2 dr\ = r2-r1 = thickness

 

[collinsmark]

It is thin but it is not infinitesimally thin. It is a real object. It is thin enough that the integrand does not change much over the thickness making the integral in the third dimension trivial, but the third dimension must be included.

As Jake discovered
“
dI of solid sphere =r²dm
=ρr²dv

dv=(rdφ)(dr)(rsinφdθ)

dI=ρr²r²sinφdrdθdφ
“
Except the first r is the radial distance from the axis and should be r sin(phi)

that is the correct reasoning and the correct integrand with the correct infinitesimal piece of volume. Now answer me this: does the correct moment of inertia for a thin sphere have “dr” in the expression, or does it have thickness? It has thickness. Why? Because that is what dr INTEGRATES to! Yes, you absolutely DO have to integrate in all three dimensions. The fact that the sphere is thin means that the radial integral is trivial, but it is there

\int_r1^r2 dr\ = r2-r1 = thickness

If you are deriving the formula for a spherical shell -- not a thin spherical shell, but a spherical shell that has an inner radius $a$ and an outer radius $b$ -- then yes you would integrate over $r$ from $a$ to $b$. The moment of intertia would not be a function of a $r$ but instead would be a function of $a$ and $b$. In that case a triple integral would be involved.

But the problem statement specifically stated "thin" spherical shell, implying that the thickenss between $a$ and $b$ is negligible. In such a case, the moment of inertia is a function of the radius $r$. Only a double integral is necessary. No $dr$ is involved.

 

[Dr.D]

@Cutter Ketch wants to insist that this problem be treated as a thick shell. He says that this leads to a trivial integration on the radius, but here he is mistaken.

If the thick shell problem is properly formulated, the radial distance from the reference axis to the mass element will vary with the radial position of the mass element. Thus the radial integration will not be the simple integration of dr to get the thickness, but something more complicated. The inclusion of the word "thin" in the problem statement was intended to be significant simplifcation; it is ignored at the expense of considerably more work for no gain.

 

[jake010]

This isn't true for a problem such as this. (I.e., infinitesimally "thin," hollow shell.)

All that's necessary is to define surface density.

$\rho_s = \frac{M}{A}$

You can think of it as mass per unit area. (If it helps to think of it in familiar units, perhaps think of it as $\left[ \frac{\mathrm{kg}}{\mathrm{m^2}} \right]$.)

For this problem, the density of an infinitesimally thin, spherical shell is $\rho_s = \frac{M}{\mathrm{Surface \ area \ of \ a \ sphere}}$.

This shouldn't be too unfamiliar to you. You've probably done this before for linear density. For example, if you were finding the mass per unit length of a thin rod, you would start with $\lambda = \frac{\mathrm{mass \ of \ the \ rod}}{\mathrm{length \ of \ the \ rod}}$. It's the same idea here with $\rho_s$, except two dimensions are involved instead of just one.

The differential mass, $dm$ is calculated by

$dm = \rho_s (\mathrm{differential \ length})(\mathrm{differential \ width})$

A picture might help here. The following is a picture of a small, differential patch of surface on a sphere of radius $r$, where $\theta$ represents longitude and $\varphi$ represents latitude (well, lattidue with 0 deg and 180 deg defining the poles).
View attachment 252482

This differential mass can be used in calculating the moment of inertia, keeping care to properly specify the distance of $dm$ to the axis of rotation. The integral can then be carried out integrating over $\theta$ and $\varphi$ . Only two integrals are necessary.

No integration over the radius of the sphere $r$ is necessary. The radius of the sphere treated as a constant. (Don't confuse the radius of the sphere $r$ with the distance from $dm$ to the axis of rotation. They are different entities.)

Got it. I understand it now. Sorry I haven't been responding, I was studying on my own. Thank you all so so much for the help! I finally get it. I really appreciate you guys taking the time to help me!