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: $$\begin{equation} \label{eq:short} \Delta G = G-G_0 = \frac{v \, \text{d}m}{\text{d}t} = K v \end{equation}$$ 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:

$$\begin{equation} \label{eq:zcm}z_{\text{CM}} = (z_{\text{u}}m_{\text{u}}+z_{\text{l}}m_{\text{l}})\, /M \, .\end{equation}$$ 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:
$$\begin{equation} \label{eq:final}  \ddot{z}_{\text{CM}} = \frac{2}{M} \left [ - \frac{K^2}{2 \rho A} - \frac{K^2}{2 \rho A}\right ] \Rightarrow G-G_0 = -M \, \ddot{z}_{\text{CM}} = 2 K \frac{K}{\rho A} = 2 K v  \end{equation}$$ 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.

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^{[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).↩}