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. |

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 \(G < G_0\).

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

\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.

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.

*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]). |

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.

^{[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).↩}

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