A popular Phys.S.E question asks how can I measure the weight of my head. One of the answers suggests measuring the moment of inertia.

My suggestion was to construct an apparatus that places the subject on a translatable carrier located on top of a rotating table. Sensors would allow the change of angular velocity to be measured as the carrier is translated relative to the centre of rotation. The subject would begin with their head at the centre of rotation and end with their feet there, many measurements being taken in between.

It would (equivalently) rotate at constant speed and measure the forces necessary to hold the translating carrier in place.

I hoped that this (with x & y translation, if desired) would give enough information to computationally determine a mass density map of the subject.

Q) Are these two approaches basically equivalent?

Q) Would these approaches actually work? A comment on the moment of inertia answer states:

"Isn't the moment of inertia of a rigid body a tensor? More specifically, it's a 3×3 matrix, so all your measurements above can be calculated knowing just nine numbers."

Is this the case, and does this mean that a density image could not be generated?


2 Answers 2


Regarding the second question:

The parallel axis theorem states, that if you have rotation around another axis than a body's center of mass, you only have to add the inertia tensor of a point mass at the location of the body's center of mass. And because of your setup, you will probably not be able to measure more than one component of the person's inertia tensor.

Regarding the first question:

As I understand it, both measurements will yield the same information (one component of the inertia tensor). As mentioned in the linked answer, you will need a model of a human to get the head's weight, which means additional assumptions about anatomy etc.


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: enter image description here

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.

  • $\begingroup$ There is a slight error in your derivation: $\mathcal{L}(g) = \int_0^\infty \omega \xi^2 e^{-s\xi} d\xi = \frac{2\omega}{s^3}$, and not $\mathcal{L}(x-\xi)^2$. Further, if you carry out the inverse Laplacian, you arrive at $\lambda(x) = \frac{1}{2\omega}\frac{d^3}{dx^3}L(x)$. With the parallel axis theorem, this is always zero. So there seems to be something wrong. $\endgroup$
    – M.Herzkamp
    Jul 21, 2014 at 10:55
  • $\begingroup$ Also, to get the full inertia tensor, you need only rotate a body around three orthogonal axes, because each rotation will give you a response vector (angular momentum), which fills three slots in the inertia matrix. $\endgroup$
    – M.Herzkamp
    Jul 21, 2014 at 11:56
  • $\begingroup$ Thanks for the comments! In fact the last equation is correct, but does not follow directly from the one above, I edited it to clarify. $\endgroup$
    – rmhleo
    Jul 21, 2014 at 12:04
  • $\begingroup$ The term $\mathcal{L}(x-\xi)^2$ in your last equation should be $\mathcal{L}(\xi^2)$. $x$ is the convolution parameter and vanishes when making the Laplace transformation. $\endgroup$
    – M.Herzkamp
    Jul 21, 2014 at 12:26
  • $\begingroup$ As for the full inertia tensor, you reduce the number of measurements only if you know the apropriate basis. For example knowing the center of gravity from the start. This matrix is symetric only when you choose the basis correctly, otherwise no constraints can be made between the matrix elements. $\endgroup$
    – rmhleo
    Jul 21, 2014 at 12:27

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.