Mass moment of inertia of a spherocylinder

[UPDATE 22/08/2014: Corrected a misprint in the formula for $I_x$. See comment below.]
Using the moments of inertia calculated in the previous post for the hemisphere (and taking for granted those of the cylinder), we can now determine those of a spherocylinder.

The height of the cylinder is $h$, while its radius (and that of the spherical caps) is $R$.

$I_z$ is easy to determine by summing the corresponding moments of the cylinder and of the two hemispheres: $I_z= \rho \, m_c \frac{R^2}{2} + 2 \rho \, m_h \frac{2}{5} R^2$. Developing $m_c$ and $m_h$ and introducing the aspect ratio $\gamma = 1+ \frac{h}{2R}$ yields:
$$I_z = \pi \rho R^5 \left [ (\gamma -1) + \frac{8}{15} \right ]$$

Mass moment of inertia of a hemisphere

I recently had to calculate the moments of inertia of various solid bodies (e. g. the spherocylinder). They can be obtained starting from the definition; this is the kind of calculation described in papers as "tedious but straightforward". I tried to simplify the process as much as possible, using the symmetry properties of the objects and the parallel axis theorem. The latter is of course useful for "composite" objects (combinations of simple units) but, less intuitively, also for sections of such units, as we'll see below for the (solid) hemisphere.

Saint-Venant's principle and its relatives - part 2

In part 1 of this discussion we had concluded that the Laplacian was a "zero-sum game", i.e. that a static modulation along one space dimension was exactly matched by a decay in the perpendicular dimension: $q_x^2 + q_z^2 = 0$.

What happens for a time-dependent field distribution? For simplicity, let us assume a purely harmonic dependence: $f(x,z,t) = F(x,z) \exp(i \omega t)$, with translation invariance along $y$, where $f$ stands for a component of the electric or magnetic field (in the scalar wave approximation).

The field now obeys:

$$\begin{equation} \label{eq:dalemb} \Box \, f = \frac{1}{c^2} \partial _t^2 f(x,z,t) - \underbrace {( \partial _x^2 + \partial _z^2)}_{\Delta} f(x,z,t)= 0 \end{equation}$$

where $c$ is the speed of light and $\Delta$ is the Laplacian operator discussed in part 1. The wave operator $\Box$ is often called d'Alembertian.

The Carvallo paradox - part II

I mentioned the Carvallo paradox in a previous post. Here, I will give a simpler version and some comments. Consider that, instead of using a spectrometer, we only use the dispersive element (e.g. a prism). After refraction, the signal components (the various colors) are projected onto a perfectly absorbing screen:

From Wikimedia; under CC-SA 1.0 License.

Even though the incoming signal $s(t)$ is of finite length (and thus energy), each component (of infinite length) $f_i(t)$, where $s(t)=\sum_i f_i(t)$, has finite and constant power, meaning that over a period of time larger than the duration of the original signal the screen will receive a higher amount of energy (which can in fact be made arbitrarily large). The Carvallo paradox can then be restated as follows:

1) A finite-length signal is a sum of infinite-length components,
2) which can be separated and manipulated individually.

There is obviously a problem with 1) or 2) (or both!), since accepting them would break both causality and energy conservation. Note that invoking quantum mechanics does not solve the paradox: we can envision a similar setup using for instance sound waves.

Whether 2) is valid or not, the "separation" step is non-trivial, as one can see from the power spectrum of the signal: $P(\omega) = |\tilde{S}(\omega)|^2 = \left | \sum_i \tilde{F}(\omega)\right |^2$, where the uppercase and tilde combination denote the Fourier transform. On the other hand, the power spectrum after separation (e.g. that absorbed by the screen) is: $P'(\omega) = \sum_i  \left | \tilde{F}(\omega)\right |^2$. What I called separation thus amounts to decorrelating the various signal components.

Houlgate

Saint-Aubin church

Cabourg

The Grand Hotel.