I find your proposal the best because:
- makes the least assumptions about the person (body model independent)
- simplistic and yet can be as accurate as desired (only limited by the precision of your measurement devices)
- doesn't rely on a person keeping a position or relaxing or stuff like that
Q)Would these approaches actually work?
R)Yes, here is how:
We only care about the mass distribution of a person along one axis $\lambda_{(x)}$. So the person is laid down (along the radius of our rotating table) and the axis at some distance $\xi$ from the head (or feet doesn't matter). Below a schematic representation:
If we knew $\lambda_{(x)}$ we could calculate the angular momentum $L_{(\xi)}$ for this case as:
$$L_{(\xi)}=\int_0^l \omega (x-\xi)^2 \lambda_{(x)} dx$$
where $l$ is the length of the person (the height). Let us put $g_{(x-\xi)} = \omega (x-\xi)^2$ which yields
$$L_{(\xi)}=\int_0^l g_{(x-\xi)} \lambda_{(x)} dx = g \ast \lambda$$
the convolution product these two functions, so we will have
$$\mathcal{L} L_{(\xi)} = (\mathcal{L} g) (\mathcal{L} \lambda)$$
$$\lambda = \mathcal{L}^{-1} \left( \frac{\mathcal{L} L}{\mathcal{L} g} \right)$$
Where $\mathcal{L}$ and $\mathcal{L}^{-1}$ are the direct and inverse Laplace transformation operations, which are more convenient for this case.
So now we have a formal expression for $\lambda_{(x)}$, and integrating it through any segment of the body, will give us the mass contained in it.
Now, how do we find $L_{(\xi)}$ and $g_{(x-\xi)}$? They are the measurables in your experimental proposition:
If your turning table keeps constant $\omega$, then measuring for different $\{\xi_i\}$ the torques appearing when displacing the person from one position to the other you can easily use them to obtain your set of angular momentum steps integrating them through the time spent in displacement (I think this integral could be measured directly) and thus have $\{\xi_i,L_{(\xi_i)}\}$, plus $g_{(x-\xi)}$ would just be a parabola calculable from its expression.
Equally, just measuring the torques appearing on the person for rotation in the different positions (stopping the machine for each displacement) you get the momentums desired measuring the torques needed to hold the person in place.
But the one I think is more feasible is coupling your turning table to a spring, and measuring the different rotation frequencies, which will directly give you the momentums of inertia $I_i = k \Omega_i^2$*(see below) with respect to the different positions of the person $\{\xi_i\}$. In this case you don't really need to strap the person, nor measure torques which is a bit more challenging, just the table and clocks. And the expression above would be:
$$\lambda = \mathcal{L}^{-1} \left( \frac{\mathcal{L} I}{\mathcal{L} (x-\xi)^2} \right)$$
Some final notes:
a) The momentum of inertia w.r.t. a rotation axis is just a scalar value characterizing this specific rotation, while the momentum of inertia tensor allows you to calculate the momentum of inertia of the rigid body for an arbitrary rotation axis. This tensor for a person would be hard to measure since you need rotate her along several axis (9) and not all rotations are suitable for keeping the person fixed, adding complexity to the experiment which is not needed.
b) The parallel axis theorem is very useful, but you need to know the momentum of inertia of the body in question w.r.t. the axis passing by its center of mass, which for a person you would need to devise a different experiment to find it, then measure $L_{cm}$ for the center of mass, and easily calculate the different $\{\xi_i,L_i\}$ for the deconvolution of $\lambda_{(x)}$, because you still have to do this to separate head mass from body mass. But your method is self sufficient without finding $L_{cm}$, and it can be argued that the same amount of measurements is needed.
*Here $k$ is the elastic constant of the spring and $\Omega_i$ is the oscilation frequency of the person about the axis located in $\xi_i$, and its relation to $I_i$ given above follows from solving the Harmonic Oscilator for this case.
