5 December 2013

The weight of an hourglass

This seems to be a classical problem [1]: what is the weight of an hourglass? Careful consideration shows that the weight is larger while running than at rest!
 Figure 1: Running hourglass.
The overall approach is simple enough: determine the center of mass position $$z_{\text{CM}}(t)$$ (for the sand only; the container is immobile and plays no role in this problem). Defining the acceleration $$a_{\text{CM}}=\ddot{z}_{\text{CM}}$$ and the total sand mass $$M$$, the force on the scale $$G$$ and the rest weight $$G_0 = M g$$ are related by: $$G=G_0-M\, a_{\text{CM}}$$. We just need the sign of $$a_{\text{CM}}$$.

Here is where quick reasoning gave me the wrong result:

All the mass is essentially at rest, except for the flowing sand, which accelerates downwards. The overall acceleration then has the same sign (positive) and $$G < G_0$$.

This is wrong, because the contribution of the flowing sand is negligible, while the sand in the upper part is not at rest (the upper surface moves downwards with a velocity $$v$$.) As Shen & Scott put it [1], the effect of the flow during time $$\text{d}t$$ is to transfer a mass $$\text{d}m$$ from the top (at a velocity $$v$$) to the bottom (at rest). The additional force due to this momentum variation is then:
\label{eq:short}
\Delta G = G-G_0 = \frac{v \, \text{d}m}{\text{d}t} = K v
with $$K = \frac{\text{d}m}{\text{d}t}$$ the rate of mass flow (taken as constant). The result \eqref{eq:short} is valid up to a numerical prefactor defined by the geometry of the setup.

 Figure 2: Simplified setup (adapted from [2]).
Let us now do the full analysis for the center of mass motion. Instead of the traditional-shaped hourglass in Figure 1 we will consider the simpler setup shown in Figure 2 [2]. $$z_{\text{u}}$$ and $$z_{\text{l}}$$ are the heights of the centers of mass for the sand in the upper and lower compartments, respectively. $$m_{\text{u}}$$ and $$m_{\text{l}}$$ are the corresponding masses.

The compartments are separated (in $$z=0$$) not by one orifice, but by a fine sieve, such that the sand flow is homogeneous over the cross-section. The center of mass is at a height:

\label{eq:zcm}z_{\text{CM}} = (z_{\text{u}}m_{\text{u}}+z_{\text{l}}m_{\text{l}})\, /M \, .The masses vary linearly in time: $m_{\text{u}} = M- Kt \quad \text{and} \quad m_{\text{l}} = Kt$ with $$K$$ defined above.

For the heights we need to introduce the density of the sand $$\rho$$, the cross-section area $$A$$, the position of the bottom $$z_{\text{b}}$$ and the height of sand in the two compartments,  $$h_{\text{u}}$$ and $$h_{\text{l}}$$ (see Figure 2). We can now write:
$z_{\text{u}} = -\frac{h_{\text{u}}}{2} = - \frac{m_{\text{u}}}{2 \rho A} \quad \text{and} \quad z_{\text{l}} = z_{\text{b}} - \frac{h_{\text{l}}}{2} = z_{\text{b}} - \frac{m_{\text{l}}}{2 \rho A}$Substituting in \eqref{eq:zcm} and taking the second derivative (only terms quadratic in time contribute to it) yields:

\label{eq:final}
\ddot{z}_{\text{CM}} = \frac{2}{M} \left [ - \frac{K^2}{2 \rho A} - \frac{K^2}{2 \rho A}\right ] \Rightarrow G-G_0 = -M \, \ddot{z}_{\text{CM}} = 2 K \frac{K}{\rho A} = 2 K v
where the last step follows from $$K = \frac{\text{d}m}{\text{d}t} = \frac{\text{d}}{\text{d}t} (\rho A h) = \rho A v$$. We have retrieved the result \eqref{eq:short}, up to a factor of 2.

This geometrical prefactor is related to the velocity ratio between the upper and lower interfaces and thus to the ratio of their areas, see [1] for the complete derivation.

[1] K. Y. Shen & B. L. Scott, The hourglass problem. Am. J. Phys. 53, 787 (1985).
[2] F. Tuinstra & B. F. Tuinstra, The weight of an hourglass. Europhys. News 41, 25 (2010).