Obtaining the moment of inertia of a sphere and spherical shell

[hmparticle9]

Homework Statement

Finding the moment of inertia of the sphere/spherical shell.

Relevant Equations

x^2 + y^2 + z^2 = R^2

I have made an attempt, which I give below. There are two issues. The first is that I am missing a factor of 2 in the MOI of the sphere. The second issue is that the integral I am getting for the spherical shell is very complex. I have tried symbolab and it chucks out some horrible things. This is making me doubt myself. The axis is to be taken through the centre of the sphere.

I know I could google solutions, but I just would like to know where I am going wrong.

 

[erobz]

Lets focus on the solid sphere.

Lets work off of this diagram. What do you think the differential slice you are trying to take looks like here where we are looking at a sphere directly into the page?

 

[erobz]

To give a bit more of why I'm asking you to do this is because its natural (from my own struggles with it) we falsely convince ourselves of results about this slicing mechanism that we first practice with 2D shapes (in the First and Second Moments of Area). In 2D when you slice a shape parallel to the axis the entire element resides at the same distance from that axis.

Because of this, in 3D we are then instinctually led to believe that the entire element you have conceived by slicing shape parallel to $z$ axis is at a distance of $y = \sqrt{R^2 - z^2 }$. But when viewed from the top down, looking into the $z$ axis, you will see this:

The distance distribution of mass along that sliced element is not actually the same over the entire element like it is with 2D shapes. (Instead imagine the differential cuboids $z dy dx$ that sum together to construct your slice in the first view).

This is why with mass moments of inertia in intro physics books you see the label of a "thin rod" or a "thin plate" etc... They are distinct from the mass moment of inertia of their general bodies because of a approximate derivation.

With the sphere you aren't going to get away with simplifications that neglects the difference in distance across an element so easily. You can use Triple Integration $\iiint r^2 \rho dV$ or you can have some tactics with defining differential mass moment of inertias elements $dI$ as they do in this example https://www.miniphysics.com/uy1-calculation-of-moment-of-inertia-of-solid-sphere.html

I hope that helps.

 

[hmparticle9]

I realise why my answer makes no sense now.

For instance my integrand "sums" over $z^2$. In the moment of inertia calculation

$$I = \sum_i m_i r_i^2$$

$r_i$ is the location of the mass $m_i$. In my "version" all my mass is located along the the $z$-axis.

What I am getting is it that
$$m_i = \frac{M}{V} \pi (R^2 - z^2) $$
and
$$r_i = z$$

As well as the issues you mentioned. Am I correct in my understanding?

 

[erobz]

I realise why my answer makes no sense now.

For instance my integrand "sums" over $z^2$. In the moment of inertia calculation

$$I = \sum_i m_i r_i^2$$

$r_i$ is the location of the mass $m_i$. In my "version" all my mass is located along the the $z$-axis.

What I am getting is it that
$$m_i = \frac{M}{V} \pi (R^2 - z^2) $$
and
$$r_i = z$$

As well as the issues you mentioned. Am I correct in my understanding?

I think you have the error figured out.

Just to be sure, looking at a more "straight" example. What you were doing on accident is saying that the entire mass of the differential element "slice" is at the distance of the purple arrow in this diagram.

So in this example you want to instead apply:

$$ I_z = \rho \iiint ( x^2 + y^2) dx~ dy~dz $$

to an element like this

You effectively must "build up" your correct element "slice" that you are trying to take by integrating first over $x$, and $z$ over some limits for the shape, leaving you with a single integral in $y$ and $dy$ to evaluate.

 

[erobz]

For the sake of completeness, the "slice" element at some distance $y$ for this box is found by this process:

$$ I_z = \rho \int \int_{-h/2}^{h/2} \int_{-b/2}^{b/2} ( x^2 + y^2 ) dx~dz~dy $$

$$ = 4 \rho \int \int_{0}^{h/2} \int_{0}^{b/2} ( x^2 + y^2 ) dx~dz~dy $$

$$ = 4 \rho \int \int_{0}^{h/2} \left[ \frac{1}{3}x^3 + y^2 x \right]_{0}^{b/2} dz~dy $$

$$ = 4 \rho \int \int_{0}^{h/2} \left( \frac{1}{24} b^3 + y^2 \frac{b}{2} \right) dz~dy $$

$$ = 4 \rho \int \left[ \frac{1}{24} b^3 z + y^2 \frac{b}{2} z \right]_{0}^{h/2}~dy $$

$$ = 4 \rho \int \left( \frac{1}{12 \cdot 2 } b^3 \frac{h}{2} + y^2 \frac{b}{2} \frac{h}{2} \right)~dy $$

Now, bring in the factor of 4 from changing the limits of integration earlier and you should recognize the integrand as the Second Moment of Area of a rectangle about its axis through its center ( $z$ direction), and the next term is the application of the Parallel Axis Theorem about the $z$ axis.

$$ =\rho \int \left( \frac{1}{12 } b^3 h + y^2 bh \right)~dy $$

You'll have to extend this to the sphere geometry if you are trying to slice it as you originally have.

 

[FactChecker]

You acknowledge that you made a mistake and recognize what it is. IMO, the problem assumes that polar coordinates will be used. You should be aware that using the wrong coordinate system or wrong approach can lead to errors, no matter how finely the approximation is divided up. (see this)