# Moment of Inertia of a hollow sphere with Mass 'M' and radius 'R'.

#### Kaushik

Homework Statement
Derive the formula for moment of inertia of a hollow sphere.
Homework Equations
Required answer $\frac{2MR^2}{3}$
Homework Statement: Derive the formula for moment of inertia of a hollow sphere.
Homework Equations: Required answer $\frac{2MR^2}{3}$

Consider a Hollow sphere.
At an angle $Θ$ with the vertical, consider a circular ring whose moment of inertia is given by $MR^2$.
The most basic question I have is, should we consider the volume of the small ring or the area?
After this let us have further discussion.

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#### phinds

Gold Member
Homework Statement: Derive the formula for moment of inertia of a hollow sphere.
Homework Equations: Required answer $\frac{2MR^2}{3}$

Consider a Hollow sphere.
At an angle $Θ$ with the vertical, consider a circular ring whose moment of inertia is given by $MR^2$.
The most basic question I have is, should we consider the volume of the small ring or the area?
After this let us have further discussion.
Perhaps I'm missing something but I'm at a loss to see how that approach is any help at all, trying to reduce the entire thing to just a ring. I think you need to explain your reasoning before anyone can offer any further help. This IS the homework section, after all, and you are expected to do some work on your own.

#### RPinPA

Homework Helper
At an angle $\Theta$ with the vertical, consider a circular ring whose moment of inertia is given by $MR^2$.
That's not correct. If $M$ and $R$ stand for the mass and radius of the whole sphere, then you can't use $M$ and $R$ to also mean the mass and radius of this ring.

The most basic question I have is, should we consider the volume of the small ring or the area?
You should consider its mass. All of that mass is at the same distance $r$ from the axis, so you can express the moment of the inertia as $m_\text{ring} r^2$ as the hint above said (with the wrong symbols).

So the two questions you need to answer are: (1) How can you express the mass of the ring in terms of $M$ and $\Theta$? and (2) How can you express the radius of the ring in terms of $\Theta$?

The small ring's mass is some fraction of the whole sphere's mass. What would seem to you a reasonable way to derive an expression for that fraction?

#### mpresic3

You should consider its mass. All of that mass is at the same distance rrr from the axis, so you can express the moment of the inertia as mringr2mringr2m_\text{ring} r^2 as the hint above said (with the wrong symbols).
This is either misleading or wrong. The mass in the hollow sphere is not all at a distance r from the rotation axis. The mass near the pole of the hollow sphere is close to the polar axis (assumed to be the axis of rotation), and the mass near the equator would be far from the polar axis (assumed to be the rotation axis.). Without loss of generality, because of the spherical symmetry of the hollow sphere, you can assume the polar axis is the rotation axis.

Kaushik's suggestions are still good, but you may want to consider suggestion2 to obtain the radius first. THen use his suggestion 1.

Many times in questions like this, I find it convenient to define a (areal) density (here the total mass of the shell divided by the area of the spherical shell). Do any integrations in terms of the density. Last reintroduce the mass from the (inverted) equation you have defining the density,

#### Delta2

Homework Helper
Gold Member
I find it more straightforward to apply directly the generic definition of moment of Inertia
$$I=\int r^2 dm$$ where r is the distance from the axis of rotation ( and not the distance from the center of the sphere), and dm is the mass element.

In the case of a hollow sphere it is $dm=\rho dA$ where $dA$ the surface element of the sphere which in spherical coordinates for a sphere of radius R is $dA=R^2\sin\theta d\theta d\phi$ and $\rho=\frac{M}{4\pi R^2}$ the surface mass density of the hollow sphere.

Now one critical question you should answer is " What is the distance $r(R,\theta)$ from the axis of rotation of the surface element dA, in terms of R(the radius of the sphere) and $\theta$"

and then all you got to do is calculate the surface integral (double integral on $\theta,\phi$
$I=\int r(R,\theta)^2 dm=\oint r(R,\theta)^2 \rho dA=\int_0^{2\pi}\int_0^{\pi} r(R,\theta)^2\rho R^2 \sin\theta d\theta d\phi$.

Again i recommend extra caution in order not to confuse the distance from the axis of rotation r, with the distance from the center of the sphere R.

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#### phinds

Gold Member
GUYS --- I think we need to stop with the hints and let Kaushik do some of the work. You've just about spoon fed him the answer.

#### RPinPA

Homework Helper
This is either misleading or wrong. The mass in the hollow sphere is not all at a distance r from the rotation axis.
Nor did I say it was. The question I was responding to was about the moment of inertia of a small ring at angle $\Theta$. You might have noticed why I introduced the symbol $r$ to distinguish it from $R$.

#### Orodruin

Staff Emeritus
Homework Helper
Gold Member
2018 Award
This is either misleading or wrong. The mass in the hollow sphere is not all at a distance r from the rotation axis.
This is wrong. In the post you quoted he explicitly introduced the mass $m$ and radius $r$ of the ring precisely to avoid confusion with the mass $M$ and radius $R$ of the sphere and clearly stated why he did so.

Apart from that I agree with #6 - let the OP work with the problem.

#### mpresic3

I misunderstood your reference to "its mass" as the mass of the hollow sphere and not the mass of the ring.
I see you also use mring in the later equation to avoid this misunderstanding. I was too hasty in reading your post. Sorry, for the confusion.

#### Delta2

Homework Helper
Gold Member
You should consider its mass. All of that mass is at the same distance $r$ from the axis, so you can express the moment of the inertia as $m_\text{ring} r^2$ as the hint above said (with the wrong symbols).
I doubt about the part texted with bold letters. The axis of rotation of the sphere will (in the generic case) make an angle with the axis of the ring so its moment of inertia wont just be $m_\text{ring}r^2$. Or the distance r (from the axis of rotation) of all parts $dm_\text{ring}$ of the ring wont be the same.

#### Orodruin

Staff Emeritus
Homework Helper
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2018 Award
I doubt about the part texted with bold letters. The axis of rotation of the sphere will (in the generic case) make an angle with the axis of the ring so its moment of inertia wont just be $m_\text{ring}r^2$. Or the distance r (from the axis of rotation) of all parts $dm_\text{ring}$ of the ring wont be the same.
The OP has explicitly chosen a part of the sphere that has a fixed angle to the rotational axis. This is quite clearly a ring of points that are equidistant to that axis.

#### Delta2

Homework Helper
Gold Member
The OP has explicitly chosen a part of the sphere that has a fixed angle to the rotational axis. This is quite clearly a ring of points that are equidistant to that axis.
Ok suppose we have a sphere centered at the origin and take the rotation axis to be the y-axis. I choose a ring of the sphere that lies on the xy plane. Are you dead sure that all points of the ring are equidistant from the y-axis?

#### jbriggs444

Homework Helper
Ok suppose we have a sphere centered at the origin and take the rotation axis to be the y-axis. I choose a ring of the sphere that lies on the xy plane. Are you dead sure that all points of the ring are equidistant from the y-axis?
While one could choose to use rings that do not exploit the symmetry of the situation, the OP has chosen to use rings that do.

#### Delta2

Homework Helper
Gold Member
While one could choose to use rings that do not exploit the symmetry of the situation, the OP has chosen to use rings that do.
yes well might be the case, it was just not so clear to me from his post. A figure would definitely clear it up. But sometimes we got to live without figures.

#### RPinPA

Homework Helper
I doubt about the part texted with bold letters. The axis of rotation of the sphere will (in the generic case) make an angle with the axis of the ring so its moment of inertia wont just be $m_\text{ring}r^2$. Or the distance r (from the axis of rotation) of all parts $dm_\text{ring}$ of the ring wont be the same.
I'm sorry, but I disagree. The rings are not at an arbitrary angle. The rings are defined by you, the person solving the problem. If you have such an awkward situation, you've made an incorrect choice.

But OK, I guess maybe it's not always obvious that you should take advantage of symmetry, though it was pretty clear to me that the OP was doing so.

So first note that you are free to define any coordinate system that is convenient to you for integration. Now, let's explicitly define our coordinate system so that we define "vertical" as the axis of rotation and "horizontal" as normal to that. And let's chop the sphere into horizontal slices, parametrized by a central angle $\theta$. It is clear that everything on one thin slice is the same distance $r$ from the axis, and that $r$ depends on $\theta$, and that we will integrate over $d\theta$

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#### Delta2

Homework Helper
Gold Member
I'm sorry, but I disagree. The rings are not at an arbitrary angle. The rings are defined by you, the person solving the problem. If you have such an awkward situation, you've made an incorrect choice.

But OK, I guess maybe it's not always obvious that you should take advantage of symmetry, though it was pretty clear to me that the OP was doing so.

So first note that you are free to define any coordinate system that is convenient to you for integration. Now, let's explicitly define our coordinate system so that we define "vertical" as the axis of rotation and "horizontal" as normal to that. And let's chop the sphere into horizontal slices, parametrized by a central angle $\theta$. It is clear that everything on one thin slice is the same distance $r$ from the axis, and that $r$ depends on $\theta$, and that we will integrate over $d\theta$
yes , well I don't know what I was thinking, it is obvious as you said that we should take advantage of the symmetry. Maybe if a figure was provided I would 've understood better.

#### RPinPA

Homework Helper
yes , well I don't know what I was thinking, it is obvious as you said that we should take advantage of the symmetry. Maybe if a figure was provided I would 've understood better.
No problem. I was surprised that there was so much negative reaction to what I thought were fairly elementary observations, so I was apparently not as clear as I thought I was.

I think "take advantage of symmetry" and "choose a convenient coordinate system" are tricks that I've forgotten have to be learned. I think as a student I did plenty of problems the hard way that could have been vastly simplified by saying something like "let the x-axis be the direction the particle is moving".

"Moment of Inertia of a hollow sphere with Mass 'M' and radius 'R'."

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