Addition of moments of inertia

[etotheipi]

The construction of any object from two pieces is covered by the parallel axis theorem even if done sequentially in time. And the derivation of the parallel axis theorem must contain the case where the axes of the two objects coincide. I prefer one theorem not three.

In the case where the axes coincide, the parallel axis theorem reduces to $I_1 = I_2$; i.e. the MoI of the same distribution about the same axis is, of course, the same. However I don't see how this helps.

As @robphy notes, the parallel axis theorem doesn't cover the addition of the MoI's of different components about some axis to obtain the MoI of the whole configuration about that axis. That is instead an example of the principle of superposition.

I wrote up the two derivations in #24 and they are quite distinct operations!

 

[hutchphd]

My guess is that the parallel axis theorem contains all of the information necessary to define the moments of inertia for any arbitrary object, rendering this discussion moot.
If not, then your point is taken...

 

[robphy]

In practice my use of the theorem always involves construction of a new object from lesser simpler pieces. If you enjoy a hundred special rules then you are wired different from me.

The construction of any object from two pieces is covered by the parallel axis theorem even if done sequentially in time. And the derivation of the parallel axis theorem must contain the case where the axes of the two objects coincide. I prefer one theorem not three.

In this problem given by the OP (adding of shells with a common center), the parallel-axis theorem is a vacuous step since the translation from the center of mass of the object is the zero vector.

Even if it were not the zero vector, the parallel axis theorem tells you about the moment of inertia of an object of mass M_1 about a parallel axis. Once armed with that information,
you still have to use superposition
to add that moment of inertia of the first object
to the moment of inertia of a second object with mass M_2 (possibly also obtained with help of the parallel axis theorem) .

So,

<br /> \begin{align*}<br /> I&#039;_{\underbrace{M_1+M_2}_{superposition}} &amp;=<br /> \underbrace{<br /> \underbrace{\left(M_1\Delta d_1^2+I_{M_1}\right)}_{\mbox{ parallel axis thm} }{{\Large +}}<br /> \underbrace{\left(M_2\Delta d_2^2+I_{M_2}\right)}_{\mbox{ parallel axis thm} }<br /> }_{\mbox{superposition}}<br /> \end{align*}<br />In the OP, d_1=0 and d_2=0 since concentric shells are being used.
So, the parallel axis theorem alone doesn't help the OP.

Update:
as @Ibix suggests below and @etotheipi suggested above,
the parallel axis theorem relies on superposition.

 

[Ibix]

The derivation of the parallel axis theorem that I'm familiar with is that for a body of density $\rho(x,y,z)$ rotating (without loss of generality) about an axis parallel to the z direction and passing through $x=x_0, y=0$ has moment of inertia$$\begin{eqnarray*}
I&=&\int_V\rho(x,y,z)\left((x-x_0)^2+y^2\right)dV\\
&=&\int_V\rho(x,y,z)(x^2+y^2)dV\\
&&-2x_0\int_V\rho(x,y,z)xdV\\
&&+x_0^2\int_V\rho(x,y,z)dV
\end{eqnarray*}$$
The first term is the moment of inertia about the origin, the second term is zero if the center of mass lies at the origin, and the third term is just $x_0^2M$, where $M$ is the mass of the object.

It seems to me that this derivation relies on an assumption that you can add moments of inertia for point particles rotating about a common axis - i.e. that you can write the integral at all. If you can do that then both the parallel axis theorem and the answer to the OP's question follow.

Edit: beaten to it by both @etotheipi and @robphy, I see

 

[etotheipi]

It seems to me that this derivation relies on an assumption that you can add moments of inertia for point particles rotating about a common axis - i.e. that you can write the integral at all.

And I suppose this really just comes from how the MoI is defined in the first place. E.g. for a rigid body rotating in the $x$-$y$ plane about the $\hat{z}$ axis, each particle has $\vec{L}_i = \vec{r}_i \times m_i (\vec{\omega} \times \vec{r}_i) = m_i {r_i}^2 \omega \hat{z}$, so it follows $\vec{L} = (\sum m_i r_i^2) \omega \hat{z}$ from which we define $I_z = \sum m_i r_i^2$, i.e. as a summation over mass elements. And same goes for the integral as the limit of this sum for mass elements $dm$.

 

[robphy]

Indeed, without superposition, one can't add vectors [and tensors].
(The moment of inertia is really a tensor.)