Experimental measurements of orientation and rotation of dense 3D packings of spheres

Abstract

Many recent advances in the study of granular media have stemmed from the improved capability to image and track individual grains in two and three dimensions. While two-dimensional systems readily yield both translational and rotational motion, a challenge in three-dimensional experiments is the tracking of rotational motion of isotropic particles. We propose an extension of the refractive index matched scanning technique as a method of measuring individual particle rotation. Initial measurements indicate that shear-driven rotational motion may stem from gear-like motion within the shear zone.

Appendix: Moments of inertia

Recall the moment of inertia for a solid sphere of mass $M$ and radius $R$,

$$\begin{aligned} I_{s} = \frac{2}{5} M R^2 \end{aligned}$$

For the purpose of these calculations, it will be more convenient to express this in terms of density $\rho$,

$$\begin{aligned} I_{s} = \frac{8}{15} \rho \pi R^5 \end{aligned}$$

(2)

First, we start with the axis along the hole, about which rotation is visually indeterminant under the current setup. For convenience, the moment of inertia of the material which is removed will be subtracted from the solid sphere inertia.

$$\begin{aligned} I_1 = I_s - I_{hole,1} \end{aligned}$$

The moment of inertia of a differential mass element of the holed-out material is given by that of a thin solid disk,

$$\begin{aligned} \mathrm{d}I_{hole,1}&= \frac{1}{2} r^2 \mathrm{d}m\\ \mathrm{d}I_{hole,1}&= \frac{1}{2} \rho \pi r(z)^4 \mathrm{d}z, \end{aligned}$$

where

$$\begin{aligned} r(z) = \left\{ \begin{array}{ll} h, &{} \quad |z| \le \sqrt{R^2-h^2} \\ \sqrt{R^2-z^2}, &{} \quad |z| > \sqrt{R^2-h^2} \end{array}\right. \end{aligned}$$

(3)

$h$ is the hole radius in the interior of the sphere and $z$ is the coordinate along the drill axis. Then, after performing the integrals,

$$\begin{aligned} I_{1} = I_s \sqrt{1 - x^2} \left( 1 + \frac{1}{2} x^2 - \frac{3}{2} x^4\right) \end{aligned}$$

(4)

In the intermediate steps, $I_s$ is substituted in using Eq. 2 and $x$ is defined to be the hole size ratio, $x = \frac{h}{R} < 1$. Next, we move to the moment of inertia for rotations about the axes perpendicular to the hole.

$$\begin{aligned} I_{2,3} = I_s - I_{hole,2,3} \end{aligned}$$

We again use a thin disk as the mass element for the hole. The differential moment of inertia is then defined by applying the perpendicular and parallel axis theorems,

$$\begin{aligned} \mathrm{d}I_{hole,2,3}&= \frac{1}{2} \mathrm{d}I_{hole,1} + z^2 \mathrm{d}m\\ \mathrm{d}I_{hole,2,3}&= \frac{1}{2} \mathrm{d}I_{hole,1} + \rho \pi r(z)^2 z^2 \mathrm{d}z \end{aligned}$$

Now, we evaluate the integrals using the same limits for $z$ and definition of $r(z)$ from Eq. 3. Again, $I_s$ is substituted for the solid sphere inertia and $x$ is substituted for the hole size ratio.

$$\begin{aligned} I_{2,3} = I_s \sqrt{1 - x^2} \left( 1 - \frac{3}{4} x^2 - \frac{1}{4} x^4\right) \end{aligned}$$

(5)

For our grains in particular, $\rho = 1.18$ g/cm$^3$. The grains have a diameter of 0.6 cm. For a solid sphere, Eq. 2 gives

$$\begin{aligned} I_s = 4.80 \cdot 10^{-3}\hbox { g} \cdot {\hbox {cm}^2} \end{aligned}$$

The hole diameter is 0.15 cm, which gives a size ratio of $x = 0.25$. From Eqs. 4 and 5,

$$\begin{aligned} I_1&= 4.77 \cdot 10^{-3}\hbox { g} \cdot {\hbox {cm}^2}\\ I_{2,3}&= 4.43 \cdot 10^{-3}\hbox { g} \cdot {\hbox {cm}^2} \end{aligned}$$