A flowing hourglass changes its weight in the course of time because of the of its center of mass. While this insight is not new, it is frequently said that the effect is tiny and hardly Here, we present a simple experiment that allows the monitoring of weight as a function of time, and that shows that there are different stages of the weight variation. The experimental result is in quantitative agreement with theory.

The question whether or not the weight, shown by a scale, of a running hourglass differs from the weight of the hourglass at rest is of great pedagogical value since experience shows that this question leads to controversial and instructive discussions. In our opinion, it teaches that a systematic of a physical situation is superior to a phenomenological in a certain sense.

Phenomenologically, when the sand starts pouring part of it is freely falling and is thus not supported by the scale, which leads to a reduced weight shown by the scale. But when the grains arrive at the bottom, they come to rest in the lower chamber by transferring momentum and, consequently, force to the balance. A simple calculation1 shows that these two effects cancel out; thus, it may be argued that except for short intervals of time at the very beginning and at the very end of the the scale shows the same weight as if the sand were at rest. It has been shown that this argument is wrong (or insufficient);2–5 nevertheless, it is still taught in high schools and universities. Interestingly, this flawed solution can be found in well recognized and journals (e.g., Refs. 6–10, to mention just a few out of many), and it even made it to the National Physics Olympiad of Singapore.11

A related problem is the change of weight of an Atwood machine when it is in This classical experiment by Poggendorff12 (repeated many times in the literature, e.g., Refs. 13 and 14) can be explained easily by evaluating the of the center of mass. Thus, the idea that leads us immediately to the correct solution is simple: look to the of the center of mass of the hourglass! This more systematic approach reveals that the solution sketched above fails at least for the following arguments:

• The in the upper and lower containers, as well as in the freely falling stream of sand, is not at rest, and sand of time-dependent mass moves with a certain velocity determined by the outflow rate and the geometry of the container. Thus, in the course of time, moving (in the upper container) comes to rest at a different location (in the lower container), which implies an

• At a given time t, during the small interval of time dt, sand moves effectively from a certain height hu in the upper container to a height hl in the lower container (see Fig. 1), where $h u − h l$ is a decreasing function of time. This corresponds to an whose time dependence is determined by the geometrical shape of the hourglass.

Fig. 1.

(color online) (a) In an hourglass, the mass of moving in the upper container and in the stream sand changes over the course of time. This comes to rest in the lower container, exerting a force during deceleration. (b) In the regime of steady during time dt the mass dm is effectively transferred from $h u ( t )$ to $h l ( t )$, where $Δ h ≡ h u − h l$ is a function of time and depends on the geometry of the hourglass. This transfer involves of mass and, thus, a time-dependent force.

Fig. 1.

(color online) (a) In an hourglass, the mass of moving in the upper container and in the stream sand changes over the course of time. This comes to rest in the lower container, exerting a force during deceleration. (b) In the regime of steady during time dt the mass dm is effectively transferred from $h u ( t )$ to $h l ( t )$, where $Δ h ≡ h u − h l$ is a function of time and depends on the geometry of the hourglass. This transfer involves of mass and, thus, a time-dependent force.

Close modal

From the of the of the center of mass,2–5 it follows that the weight of the hourglass is a function of time that depends on the specific details including the geometric shape of the hourglass, the outflow rate, and the angle of repose of the sand. In the literature, the effect was described as tiny, perhaps smaller than the change of gravity when turning the hourglass over and thereby changing the distance of the center of mass of the sand from the center of Earth,3 an effect that is on the order of some μg and cannot be with usual physics lab equipment. Even in an experiment with a very large device of height $120 cm$, emptying in only $10 s$, a maximal change in weight of only $49 mg$ was 2 and the time dependence of this effect could not be determined. Using a multiple-orifice setup5 where the upper cylindrical container was separated from the lower cylindrical container by a sieve, a quantitative was presented. For the given geometry, however, a constant increase of the weight is expected from theory whereas the shows a clear trend that may be understood either from the large uncertainty (error bars) of the or from the non-stationary rate. Indeed, unlike fluids granular matter shows at a rate independent of the filling height. Whether this result, shown for a single orifice using different approaches,15–17 holds true also for through a sieve remains unclear. But even for the simplified experiment5 the full time dependence of the change of weight could not be due to perturbations of the opening mechanism.

In this paper, we present experimental results that show that the difference of weight of a flowing hourglass, in comparison with the same hourglass in which the sand is at rest, is clearly even with simple equipment. When the weight of the hourglass as a function of time, we observe different stages of behavior that can be understood qualitatively in comparison with a high speed video recording. While the simple phenomenological argument mentioned above fails to explain the change of weight, by means of a systematic of the physical process (namely, the of the center of mass) we derive a theoretical description without any free parameters that is in quantitative agreement with the experiment.

In Secs. II and III, we will present the experimental results and the corresponding theory, respectively. Section IV discusses the stages of the apparent in the time dependence of the change of weight of the hourglass. Unlike earlier references, this paper will show that even with simple equipment the weight of the hourglass can be as a function of time, and that this time dependence can be explained theoretically.

The experiment is shown in Fig. 2. The upper container consists of a cylinder and a cone characterized by $R 1 = 2.75 cm , R 2 = 1.0 cm$, and $z 2 = 3.5 cm$, initially filled up to $z 1 = 8.2 cm$ with $M = 800 g$ bronze powder (our “sand”) of density $ρ = 5 , 230 kg / m 3$, a grain size of approximately $150 μ m$, and angle of repose $α = 28.3 °$. This filling was chosen for its high density and in order to avoid electrical charging of the that might disrupt homogeneous The lower compartment consists of a cylinder characterized by $R 3 = 2.75 cm$ and $z 3 = − 21.5 cm$. Both upper and lower compartments are manufactured from polycarbonate Unlike common hourglasses, the upper and lower vessels were not sealed, in order to avoid a difference of air pressure in the upper and lower compartments; in closed, non-evacuated hourglasses the of sand through the orifice carries a small amount of air with it leading to increased pressure in the lower chamber.18 It has been shown that the counter-flowing air leads to termed “ticking of hourglasses,” which has been extensively studied in the literature.19–22 The open construction of our experiment avoids pressure differences caused by flowing air and thus provides a steady of sand at rate $F = 720 g / s$.

Fig. 2.

(color online) Experimental setup (left). The sketch (right) introduces the geometric parameters, defines the sub-volumes I–VI, and shows α, the angle of repose. The sketch shows a cut through the symmetry axis.

Fig. 2.

(color online) Experimental setup (left). The sketch (right) introduces the geometric parameters, defines the sub-volumes I–VI, and shows α, the angle of repose. The sketch shows a cut through the symmetry axis.

Close modal

The hourglass was placed on a scale using a load cell with $2 kg$ capacity (LCB70, ME-Meßsysteme). The electrical signal from the strain gauge of the load cell was amplified and digitized using an ADS1281 analog-to-digital converter. The weight of the hourglass was sampled at a rate of 250 Hz.

Initially, the orifice of the upper container was closed by a lid held by an electromagnet. At time t = 0, the orifice was opened and the pouring of the sand led to an of the center of mass, and thus to a change in the weight by the scale. Figure 3 shows the weight of the hourglass during the run, with the weight of the apparatus subtracted (this figure is enhanced online with a movie of the experiment). A running average over six consecutive force measurements was applied. It is worth mentioning that the experiment is highly reproducible; the deviations between independent repetitions of the experiment are tiny. These deviations and all the fine structure of the data are no more than the line width of the curve in Fig. 3.

Fig. 3.

Weight of the granulate inside the hourglass as a function of time (the weight of the apparatus has been subtracted). At time t = 0, the orifice opens. We observe a characteristic evolution that can be understood in detail from the theory presented in Sec. III. The dashed line shows the increase in weight due to the of the center of mass (enhanced online) [URL: http://dx.doi.org/10.1119/1.4973527.1].

Fig. 3.

Weight of the granulate inside the hourglass as a function of time (the weight of the apparatus has been subtracted). At time t = 0, the orifice opens. We observe a characteristic evolution that can be understood in detail from the theory presented in Sec. III. The dashed line shows the increase in weight due to the of the center of mass (enhanced online) [URL: http://dx.doi.org/10.1119/1.4973527.1].

Close modal

The details of the experimental result will be discussed below. In Sec. III, we compute the weight of the hourglass theoretically with some model assumptions. Note that the theory presented does not contain any free parameters. Comparing experiment and theory in Sec. IV, we obtain quantitative agreement. Noticeable deviations between the experiment and theory can be seen only in the very beginning and very end of the experiment. These deviations are due to model assumptions, are well understood, and will be discussed in detail in Sec. IV.

The geometry of our model hourglass is sketched in Fig. 2. As for a typical hourglass, the upper container consists of a cone and a cylinder and the lower container has the same shape as the upper container. For the model presented here, we assume that the total volume of the sand only fills the cylindrical portion of the lower container, so that we do not need to include the cone of the lower part in our calculation.

Our strategy for the computation of the vertical position of the center of mass (com), Z(t), is to subdivide the volume into geometrical primitives such as cylinders, cones, and annuli, whose geometrical extensions are known for each time t of operation. The mass and center of mass of these primitives are either known or can be obtained by combining other primitives of positive and negative mass. The coordinate of the center of mass of the hourglass as a function of time is then obtained from the combination of the masses and centers of mass of the primitives.

Following this strategy, as illustrated in Fig. 2, we subdivide the volume into sub-volumes I–VI of time-dependent mass $m i ( t )$ and center of mass $Z i ( t )$, with $i = I , … , VI$, so that the overall com of the sand is
$Z ( t ) = ∑ i = I VI Z i ( t ) m i ( t ) ∑ i = I V I m i ( t ) = 1 M ∑ i = I VI Z i ( t ) m i ( t ) ,$
(1)
where M is the total mass of the sand. Knowing Z(t) as an analytic function, we can compute the time-dependent force acting on the balance as
$F w = M ( g + d 2 Z ( t ) d t 2 ) .$
(2)
The masses and centers of mass of the sub-volumes are functions of time of the form
$m i ( t ) = { m i 0 = const . for t ≤ t i 0 m i ′ ( t ) for t i 0 ≤ t ≤ t i e m i e = const . for t ≥ t i e ,$
(3)
$Z i ( t ) = { Z i 0 = const . for t ≤ t i 0 Z i ′ ( t ) for t i 0 ≤ t ≤ t i e Z i e = const . for t ≥ t i e ,$
(4)
where $i = I … VI$ and $t i 0$ and $t i e$ are, respectively, the times for each sub-volume at which the starts and stops. For the initial conditions, we assume that the upper container is filled up to the height z1 and the upper surface of the is flat. At time t = 0, the starts flowing through the orifice at rate F. In contrast to a fluid, the granular-material rate F is independent of filling height.23–33 (The independence of pressure on filling height is known as Janssen's law;34 the independence of rate as Beverloo's law.15) The functional form of the rate (Beverloo's law) follows from a dimensional and is given by35,36
$F = d m d t = C F g ρ A 5 / 4 ,$
(5)
with CF a dimensionless constant, g the gravitational field strength, ρ the density, and A the cross-sectional area of the orifice. In this paper, we assume that the rate F is independent of time as long as sand This is a good approximation except, perhaps, for the very first instant where the of the granulate is limited by its inertia, and the very last moment when the filling height of the upper container is on the order of the diameter of the orifice. For the quantitative comparison in Sec. IV, we the value of F under this assumption.

In Secs. III B–III F, we compute mi and Zi and the corresponding times $t i 0$ and $t i e$ for all sub-volumes, and subsequently, in Sec. IV, the function Z(t) and the corresponding weight of the hourglass. For these calculations, we will introduce the time-dependent filling level z(t), defined as the highest vertical position occupied by sand, and some temporary variables, indicated by an asterisk (e.g., $x ∗$). Such variables are only relevant in the context of the present section.

Fig. 4.

Scenarios (a) and (b) for the outflow of sub-volume I.

Fig. 4.

Scenarios (a) and (b) for the outflow of sub-volume I.

Close modal
Fig. 5.

Top: com as functions of time for scenarios (a) (dotted line) and (b) (solid line); bottom: mass of sub-volume I as a function of time.

Fig. 5.

Top: com as functions of time for scenarios (a) (dotted line) and (b) (solid line); bottom: mass of sub-volume I as a function of time.

Close modal

For the quantitative plots, we take the parameter values: $g = 9.81 m / s 2 , ρ = 5 , 230 kg / m 3 , R 1 = 2.75 cm , R 2 = 1.0 cm , z 1 = 8.2 cm , z 2 = 3.5 cm$, $α = 28.3 °$, and $F = 0.69 kg / s$, in agreement with the experiment described in Sec. II.

At the initial time, t = 0, sub-volume I (see Fig. 2) has the shape of a cone. This gives us
$m I 0 = ρ π 3 R 1 3 tan α ; t I 0 = 0 ; t I e = m I 0 F , m I e = 0 ; m I ′ ( t ) = m I 0 − F t ; Z I 0 = z 1 − 1 4 R 1 tan α .$
(6)
There are two possible scenarios for the emptying of sub-volume I: In case (a), shown in Fig. 4(a), a crater of constant slope α forms in the middle of the sub-volume. The crater grows until its radius approaches R1 at $t = t I e$. In case (b), shown in Fig. 4(b), a crater of radius R1 and increasing depth forms. At $t = t I e$, its slope approaches the angle of repose α. For both cases, we describe the body of mass $m 1 ′ ( t )$ sketched in Fig. 4 as a superposition of the initial cone whose mass and center of mass given in Eq. (6) and the cone of (negative) mass $m ∗ ( t ) = − F t$ and com
$Z ∗ ( t ) = { z 1 − 1 4 ( 3 F tan 2 α ρ π ) 1 / 3 t 1 / 3 case ( a ) z 1 − 3 4 F ρ π R 1 2 t case ( b ) .$
(7)
The com of sub-volume I for $t I 0 ≤ t ≤ t I e$ is then
$Z I ′ ( t ) = 1 m I ′ ( t ) [ Z I 0 m I 0 + Z ∗ ( t ) m ∗ ( t ) ] .$
(8)
Figure 5 shows ZI for both cases. It may be surprising that the limits $t → t I e$ are different for the two cases. When we express for case (a) $Z I ′ ( t )$ as a function of the instantaneous radius $r ∗$ of the smaller cone we obtain
$Z I e = lim r ∗ → R 1 [ z 1 − tan α 4 R 1 4 − ( r ∗ ) 4 R 1 3 − ( r ∗ ) 3 ] = z 1 − R 1 tan α 3 .$
(9)
For case (b), we express $Z I ( t )$ in terms of the position z of the tip of the smaller cone, and here we obtain
$Z I e = lim z → z 1 − R 1 tan α [ z 1 − 1 4 ( z 1 − R 1 tan α − z ) ] = z 1 − R 1 tan α 2 .$
(10)
The different limits (see Fig. 5) are an interesting mathematical peculiarity that has its origin ultimately in the continuum description of granular matter, which is problematic in the limit of mass approaching zero. This issue is not a problem for our calculation, however, because the mass vanishes $m I ( t ) → 0$ as $t → t I e$, and therefore the product $m I ( t ) Z I ( t )$ does not contribute to Eq. (1) for $t → t I e$.
Initially, sub-volume II (see Fig. 2) consists of a cylinder of height $z 1 − z 2$ and radius R1, closed at the top by a cone of negative mass and an identical cone, but of positive mass, at the floor. For this sub-volume we have
$m II 0 = ρ π R 1 2 ( z 1 − z 2 ) ; t II 0 = t I e ; t II e = t II 0 + m II 0 F m II ′ ( t ) = M II 0 − ( t − t II 0 ) F ; m II e = 0 .$
(11)
For $t II 0 ≤ t ≤ t II e$, the filling level reads
$z ( t ) = z 1 − F ρ π R 1 2 ( t − t II 0 ) ,$
(12)
and we write for the com of the cylinder
$Z cyl ∗ ( t ) = z ( t ) + z 2 2 .$
(13)
For the upper (cu) and lower (cl) cones, we obtain
$m cu ∗ = − ρ π 3 R 1 3 tan α ; Z cu ∗ ( t ) = z ( t ) − R 1 tan α 4 m cl ∗ = − m cu ∗ ; Z cl ∗ ( t ) = z 2 − R 1 tan α 4 .$
(14)
Consequently, we have that the com is at
$Z II ′ = 1 m II ′ [ m II ′ Z cyl ∗ + m cu ∗ Z cu ∗ + m cl ∗ Z cl ∗ ] = z ( t ) + z 2 2 − R 1 tan α 3 ,$
(15)
and in particular
$Z II 0 = z 1 + z 2 2 − R 1 tan α 3 ; Z II e = z 2 − R 1 tan α 3 .$
(16)
Figure 6 shows the functions $Z II ( t )$ and $m II ( t )$.
Fig. 6.

com (top) and mass (bottom) of sub-volume II as functions of time.

Fig. 6.

com (top) and mass (bottom) of sub-volume II as functions of time.

Close modal

Sub-volume III is located below sub-volume II; see Fig. 2 and the expanded details in Fig. 7, where the nomenclature used in this section is introduced. In Fig. 7, we define the conical segments $S 1 , … , S 4$, of negative and positive masses, whose superposition equals sub-volume III. Note that only S3 and S4 depend on time.

Fig. 7.

(a) Sketch of sub-volume III to introduce the nomenclature. (b)–(e) Segments of volumes of positive ⊕ and negative ⊖ masses used in the calculation. Their superposition results in volume III.

Fig. 7.

(a) Sketch of sub-volume III to introduce the nomenclature. (b)–(e) Segments of volumes of positive ⊕ and negative ⊖ masses used in the calculation. Their superposition results in volume III.

Close modal

#### 1. Segment S1

Sub-volume S1 (see Fig. 7) is a time-independent cone of positive mass with height and radius
$z a ∗ = R 2 T ; R ( z a ∗ ) = z a ∗ tan α ,$
(17)
respectively, where we define for later reference
$T ≡ 1 tan α − R 1 − R 2 z 2 .$
(18)
Consequently, the mass and com are
$m 1 ∗ = ρ π 3 tan 2 α ( R 2 T ) 3 ; Z 1 ∗ = 3 4 R 2 T .$
(19)

#### 2. Segment S2

Segment S2 is a time-independent cone of negative mass with the same radius as S1 and height $z a ∗ + z 2 R 2 / ( R 1 − R 2 )$, so that with Eq. (17) we have
$m 2 ∗ = − ρ π 3 tan 2 α ( R 2 T + z 2 R 2 R 1 − R 2 ) 3 Z 2 ∗ = R 2 T − 1 4 ( R 2 T + z 2 R 2 R 1 − R 2 ) .$
(20)

#### 3. Segment S3

Segment S3, of negative mass, is a cone with radius R(z) and height $R ( z ) tan α$; therefore, for $t III 0 = t II e ≤ t ≤ t III e$
$m 3 ∗ ( z ) = − π ρ 3 [ R 2 + R 1 − R 2 z 2 z ( t ) ] 3 tan α Z 3 ∗ ( z ) = z ( t ) − tan α 4 [ R 2 + R 1 − R 2 z 2 z ( t ) ] .$
(21)
The filling height z(t) and $t III e$ are computed later from the total mass of sub-volume III. For $t → t III 0$ when $z → z 2$, we obtain
$m 3 ∗ ( z 2 ) = − π ρ 3 R 1 3 tan α ; Z 3 ∗ ( z 2 ) = z 2 − 1 4 tan α R 1 .$
(22)

#### 4. Segment S4

Segment S4 is a cone of mass and com given by
$m 4 ∗ ( z ) = π ρ 3 [ R 2 + R 1 − R 2 z 2 z ( t ) ] 2 [ z ( t ) + z 2 R 2 R 1 − R 2 ] , Z 4 ∗ ( z ) = 3 4 z ( t ) − 1 4 z 2 R 2 R 1 − R 2 .$
(23)
The first bracketed expression in Eq. (23) specifies the radius $R ( z ( t ) )$ of the cone, and the second bracketed expression, its height. The initial mass and com, for $z → z 2$, are
$m 4 ∗ ( z 2 ) = π ρ 3 R 1 2 z 2 [ 1 + R 2 R 1 − R 2 ] Z 4 ∗ ( z 2 ) = z 2 4 [ 3 − R 2 R 1 − R 2 ] .$
(24)

#### 5. Mass and COM of sub-volume III

Using the results of Secs. III D 1–III D 4, we compute the mass and com of sub-volume III by superposition of the segments $S 1 , … , S 4$. From the initial mass in sub-volume III, we have that
$m III 0 = m 1 ∗ + m 2 ∗ + m 3 ∗ ( z 2 ) + m 4 ∗ ( z 2 ) = F ( t III e − t III 0 ) ,$
(25)
and because $m III e$ obviously vanishes, we obtain the time $t III e$. The total mass as a function of time then follows from:
$m III ( t ) = m III 0 − F ( t − t III 0 ) .$
(26)
For the computation of the com via the coms of the segments,
$Z III ′ ( t ) = 1 m III ′ ( t ) [ Z 1 ∗ m 1 ∗ + Z 2 ∗ m 2 ∗ + Z 3 ∗ ( z ) m 3 ∗ ( z ) + Z 4 ∗ ( z ) m 4 ∗ ( z ) ] ,$
(27)
we need the filling height z(t), since $m 3 ∗ , m 4 ∗$, $Z 3 ∗$, and $Z 4 ∗$ depend on time through z(t). The filling height can be found from
$m III ( t ) = m 1 ∗ + m 2 ∗ + m 3 ∗ ( z ( t ) ) + m 4 ∗ ( z ( t ) ) ,$
(28)
with the segment masses given in Eqs. (19)–(21) and (23), and with the time-dependent mass given in Eq. (26). The left-hand side is a third-order polynomial of the filling height. By solving the third-order equation, we obtain z(t) and subsequently $Z III ′$ as an explicit function of time via Eq. (27). Figure 8 shows the resulting functions $Z III ( t )$ and $m III ( t )$.
Fig. 8.

com and mass of sub-volume III as functions of time.

Fig. 8.

com and mass of sub-volume III as functions of time.

Close modal

#### 1. Time-independent rate

Sub-volume IV is a tiny annulus below sub-volume III, and which is hardly visible in Fig. 2. The magnification shown in Fig. 9 defines the variable $z a ∗$, which is the same as in Sec. III D. One can argue whether it is necessary to compute the contribution of this sub-volume since its mass is small and the computation requires some effort. We believe it is necessary for the sake of consistency. Recall that we need the second derivative of the com to obtain the weight of the hourglass [see Eq. (2)]; thus, small rapidly changing masses may lead to significant contributions to the weight. For $t ≪ t IV 0$, the area at z = 0 is covered with sand such that Beverloo's law, Eq. (5), is a good approximation. For $t ≈ t IV 0$, it is certainly not justified to assume constant flux F; however, in the absence of solutions in the literature for this situation, we here also assume constant flux F. Moreover, sub-volume IV is certainly tiny as compared to all other sub-volumes so that the details of the for this volume are of minor importance for the total weight of the hourglass, provided that the large jumps in the second derivatives, discussed above, are avoided.

Fig. 9.

Sketch of sub-volume IV at time $t ≤ t IV 0$ (left), and at a later time $t IV 0 < t < t IV e$ (right).

Fig. 9.

Sketch of sub-volume IV at time $t ≤ t IV 0$ (left), and at a later time $t IV 0 < t < t IV e$ (right).

Close modal

#### 2. Mass of sub-volume IV

Sub-volume IV combines a frustum (height z(t), radii R(z) and R2) of positive mass and a frustum (height z(t), radii R(z) and $R ∗$) of negative mass, as shown in Fig. 9. Its mass is, therefore,
$m IV ′ ( t ) ρ π = z 3 [ R 2 ( z ) + R ( z ) R 2 + R 2 2 ] − z 3 [ R 2 ( z ) + R ( z ) R ∗ + ( R ∗ ) 2 ]$
(29)
with
$R ( z ) = R 2 + z ( R 1 − R 2 ) / z 2 , R ∗ = R 2 − z T ,$
(30)
and with T defined in Eq. (18). The result for the mass is then
$m IV ( t ) ρ π = T R 2 z 2 + 1 3 T ( R 1 − R 2 z 2 − T ) z 3 .$
(31)
At time $t IV 0 = t III e$, we have $z = z a ∗$ and Eq. (31) simplifies to
$m IV 0 = ρ π R 2 3 ( 1 T − 1 3 tan α T 2 ) .$
(32)
We then find the mass of sub-volume IV to be
$m IV ′ ( t ) = m IV 0 − F ( t − t IV 0 ) ; t IV e = t IV 0 + m IV 0 F$
(33)
and $m IV e = 0$.

#### 3. COM of sub-volume IV

For the computation of the com, we need the filling height z(t) in explicit terms. This can be obtained by solving a third-order equation obtained from equating $m IV ′ ( t )$ in Eqs. (31) and (33). With z(t) from this procedure, the com of sub-volume IV (frustum) is given by
$Z IV ′ ( t ) = z ( t ) 4 R 2 ( z ) + 2 R ( z ) R 2 + 3 R 2 2 R 2 ( z ) + R ( z ) R 2 + R 2 2$
(34)
and $Z I V e = lim z → 0 Z IV ′ = 0$. Figure 10 shows the functions $Z IV ( t )$ and $m IV ( t )$.
Fig. 10.

com and mass of sub-volume IV as functions of time.

Fig. 10.

com and mass of sub-volume IV as functions of time.

Close modal

The calculations for sub-volumes V and VI (sand stream and lower container) are closely related, therefore we present them together in this section. Sub-volume V is the falling stream bound by the orifice at z = 0 and the filling level $z l ( t )$ of the lower container. Sub-volume VI consists of the sand resting in the lower container, which is a cylinder of radius R3 with its bottom at z3, as shown in Fig. 2. Based on experimental observations, we assume that the upper surface of the sand is flat, due to the large rate in our experiment. The same calculation can be performed for any chosen angle of repose at the top of the lower container. The close relation of the sub-volumes V and VI originates in the fact that their common boundary changes with time. Indeed, the surface at vertical position $z l ( t )$ moves upward in time, therefore, despite constant rate F the rate of sand settling at the floor is different from F.

#### 1. Density, mass, and com of the sand stream in general

Particles enter volume V at a certain initial velocity $v ∗$, which follows from continuity between volumes IV and V. Since the flowing sand stream is not confined by lateral walls, here we cannot rely on Beverloo's law15 of Eq. (5), but rather assume that the is proportional to its cross-section and hence that its velocity is given by
$v ∗ = − F ρ π R 2 2 .$
(35)
At time $t = 0 = t V 0$, the lid opens and the grains start falling at velocity $v ∗$. Using
$F = d m d t = d m d z d z d t = d m d z v ( z ) ,$
(36)
we can compute the linear density of the sand stream $d m / d z$. The velocity of the grains at vertical position z is
$v ( z ) = − ( v ∗ ) 2 + 2 g z ,$
(37)
and thus
$d m d z = − F ( v ∗ ) 2 + 2 g z 0 ≥ z ≥ z l ,$
(38)
with the filling height $z l ( t )$ given below.
In terms of the as yet unspecified upper and lower boundaries of the stream, zL and zU, respectively, we obtain general equations for the mass of the stream,
$m jet ( z L , z U ) = ∫ z L z U d m d z d z = F g [ ( v ∗ ) 2 + 2 g z U − ( v ∗ ) 2 + 2 g z L ] ,$
(39)
and its com,
$Z jet ( z L , z U ) = 1 m jet ( z L , z U ) ∫ z L z U z d m d z d z = − F 3 g 2 m jet ( z L , z U ) [ ( z L g − ( v ∗ ) 2 ) ( v ∗ ) 2 + 2 z L g − ( z U g − ( v ∗ ) 2 ) ( v ∗ ) 2 + 2 z U g ] .$
(40)

#### 2. Vertical position of the interface zl (t) between sub-volumes V and VI

Sub-volume VI becomes active when the first grain arrives at the floor at position z3 at time
$t VI 0 = 1 g ( ( v ∗ ) 2 + 2 g z 3 − v ∗ ) .$
(41)
For $0 = t V 0 ≤ t ≤ t IV e$, the mass that has flowed through the orifice up to any particular time is either located in the stream or as a sediment in the lower container, so that
$F t = π R 3 2 z l + m stream ( z l , 0 ) .$
(42)
With Eq. (39) for $m stream$ we obtain a quadratic equation for $z l ( t )$.

#### 3. Mass and com of sub-volume V (sand stream)

Using the general results from Secs. III F 1 and III F 2, we can compute the evolution of mass and center of mass position of sub-volume V. The scenario is the following: At time t = 0, the lid opens and the sand stream gains mass rapidly. The first grain hits the floor at z3 a short time later. From this moment on, the mass of the stream decreases slowly since the lower filling level zl increases. When the last grain enters the stream at the orifice, the mass of the stream decreases rapidly and vanishes when the last grain arrives at the top of the sand in the lower container. In quantitative terms:

• For $0 = t V 0 ≤ t ≤ t VI 0$,
$m V ′ ( t ) = m stream ( − g 2 t 2 + v ∗ t , 0 ) Z V ′ ( t ) = Z stream ( − g 2 t 2 + v ∗ t , 0 ) .$
(43)
• For $t VI 0 ≤ t ≤ t IV e$,
$m V ′ ( t ) = m stream ( z l , 0 ) Z V ′ ( t ) = Z stream ( z l , 0 ) ,$
(44)

with $z l ( t )$ given by the solution of Eq. (42).

• For $t IV e ≤ t ≤ t V e$,

at time $t V e$ the last grain comes to rest at vertical position $z l e ≡ z l ( t V e )$. The terminal filling height $z l e$ is obtained from equating the initial mass of sand in the upper container of the hourglass with the final state, where the sand is contained in the lower container,
$m I 0 + m II 0 + m III 0 = m V e$
(45)
yielding
$z l e = z 3 + R 1 2 R 3 2 ( z 1 − z 2 ) + z 2 3 R 1 2 + R 1 R 2 + R 2 2 R 3 2 ,$
(46)
and the corresponding time
$t V e = t VI e = t IV e + 1 g ( ( v ∗ ) 2 + 2 g z l e − v ∗ ) ,$
(47)

with $v ∗$ given in Eq. (35).

For completeness, we add the initial and final values: $m V 0 = m V e = 0$; the com's $Z V 0$ and $Z V e$ are both undefined but finite; that is, they do not contribute to the total com of the system. Figure 11 shows the functions $Z V ( t )$ and $m V ( t )$.

Fig. 11.

com and mass of sub-volume V as functions of time.

Fig. 11.

com and mass of sub-volume V as functions of time.

Close modal

#### 4. Mass and com of sub-volume VI (lower container)

Sub-volumes V and VI have a common interface at $z l ( t )$, and therefore $t VI e = t V e$. The mass and com of volume VI can be found directly from the evolution of the lower filling height $z l ( t )$, the solution of Eq. (42). In terms of $z l ( t )$, we then have
$m VI ′ = ρ π R 3 2 [ z l ( t ) − z 3 ] , Z VI ′ = 1 2 [ z l ( t ) − z 3 ] .$
(48)
The initial and final values are
$m VI 0 = 0 m VI e = m I 0 + m II 0 + m III 0 Z VI 0 ≡ lim t → + t V I 0 Z VI ′ ( t ) = z 3 Z VI e = 1 2 ( z l e − z 3 ) .$
(49)
Figure 12 shows the functions $Z VI ( t )$ and $m VI ( t )$.
Fig. 12.

com and mass of sub-volume VI as functions of time.

Fig. 12.

com and mass of sub-volume VI as functions of time.

Close modal

The total center of mass Z(t) is obtained via Eq. (1), where the masses and coms of the sub-volumes are given in the form of Eqs. (3)–(4) with the details for each sub-volume computed in Secs. III B–III F. The solution Z(t) obtained here is rigorous and entirely analytical, albeit given in piecewise form, which makes mathematical operations with this function rather cumbersome. Computer algebra systems such as Maple can be used to conveniently compute the second derivative of this function to obtain the weight Fw of the hourglass via Eq. (2), with the total mass given by $M = m I 0 + m II 0 + m III 0$.

Figure 13 shows the final result of the the center of mass Z(t) and the force resulting from its second derivative according to Eq. (2). While Z(t) does not seem impressive, the force shows distinct characteristics that are similar to the experimental data. In particular, we recognize three distinct stages: (i) the initial interval $0 < t ≲ 0.2 s$, where the stream and the cone develop; (ii) the stage of approximately constant $0.2 s ≲ t ≲ 1 s$; and (iii) the final stage, $t ≳ 1 s$, when the upper container and the stream run empty and the comes to rest. In all three stages, we find qualitative and even partially quantitative agreement between theory (solid line) and experiment (dotted line) without employing any adjustable parameters in the model. In particular, in stage (ii) where the solution discussed in the introduction fails, we see good quantitative agreement of theory and experiment.

Fig. 13.

(color online) com of the sand flowing in the hourglass as a function of time, and weight of the hourglass due to the according to Eq. (2). The horizontal dashed line shows the weight of the hourglass when the is at rest. For comparison, the dotted curve shows the experimental result from the experimental setup pictured in Fig. 2. The horizontal bars indicate the intervals of time when the respective sub-volumes are active, $( t i 0 , t i e )$ with $i = { I , … , VI }$. The colors of the bars (online) correspond to the colors in Fig. 2.

Fig. 13.

(color online) com of the sand flowing in the hourglass as a function of time, and weight of the hourglass due to the according to Eq. (2). The horizontal dashed line shows the weight of the hourglass when the is at rest. For comparison, the dotted curve shows the experimental result from the experimental setup pictured in Fig. 2. The horizontal bars indicate the intervals of time when the respective sub-volumes are active, $( t i 0 , t i e )$ with $i = { I , … , VI }$. The colors of the bars (online) correspond to the colors in Fig. 2.

Close modal

Despite the overall good agreement of theory and experiment, there are some sizeable local deviations that will be discussed in detail. First, we notice a sharp negative peak at t = 0 in the theoretical curve, which appears less sharp in the experimental data. The reason for this peak is the assumption that at t = 0 the sand starts pouring instantaneously; that is, a finite mass is set into downward at finite velocity instantaneously, which implies a δ-function shaped negative Looking at the experimental data, we notice that there is a similar behavior, however, the negative peak is of finite value and duration due to the fact that the sand is not set into instantaneously, but needs a finite time to When the total amount of sand is reduced, the mass at t = 0 decreases and the peak gradually disappears. Figure 14 shows the experimentally weight in the initial interval $t < 0.22 s$ for different amounts of

Fig. 14.

Left: Experimentally weight in the initial time interval $t < 0.22 s$ for different amounts of (magnification of Fig. 13). The figure illustrates the appearance of the initial negative peak due to the of the total amount of after the lid is opened at t = 0. Right: Magnification of the plateau in Fig. 13. The deviation of the experimental and theoretical result is about 10%.

Fig. 14.

Left: Experimentally weight in the initial time interval $t < 0.22 s$ for different amounts of (magnification of Fig. 13). The figure illustrates the appearance of the initial negative peak due to the of the total amount of after the lid is opened at t = 0. Right: Magnification of the plateau in Fig. 13. The deviation of the experimental and theoretical result is about 10%.

Close modal

Second, at $t ≈ 0.25 s$ in Fig. 13 we observe a damped oscillation in the experimentally weight that does not appear in the theoretical result. The reason for this deviation is the splash of the in a short period after the instant when the stream arrives at the bottom at $t = t VI 0$. In the theory, we assume that the comes to rest instantaneously. The movie of the experiment (see video enhancement to Fig. 3 online) shows that this assumption is well justified for the entire experiment, except for a short interval after $t VI 0$. (A different small oscillation can be seen, e.g., in the interval $0.05 s < t < 0.1 s$ in Fig. 14, which is due to a resonant frequency of the balance.)

A third significant deviation between experiment and theory can be seen at $t ≈ t VI e$ when the experiment terminates. While theory predicts an instantaneous drop of the weight, the experiment shows a smooth decay. The reason for this deviation is the assumption of the Beverloo-law constant rate F in Eq. (5). According to Janssen's law,34 the pressure, and thus the rate, are independent of the filling height except when the filling height is too small and approaches the size of the orifice or even drops below it. But this is the case for $t ≈ t VI e$, so that the assumption of constant is not justified close to the very end of the experiment. For the same reason (the assumption of constant rate), in the interval $( t VI 0 , t VI e )$ the small positive peak at $t = t VI 0$ appears in the theory (see the discussion in Sec. III E 1).

The motivation of the division into sub-volumes I to VI is purely to provide a convenient calculation of $M d 2 Z ( t ) / d t 2$. This division is by no means unique and so does not lead to any useful physical interpretation of the isolated contributions to FW. The horizontal bars in Fig. 13 indicate the time intervals, $( t i 0 , t i e )$ with $i = { I , … , VI }$, when the respective sub-volumes are active; that is, when $m i ≠ const .$ Comparison with the curve for the com and the force $F W ( t )$ shows that an interpretation of the isolated contributions is, at the very least, not straightforward. This is especially true since these contributions cannot exist independently of one another.

From Fig. 2, we see that the sub-volumes are of rather different size. Therefore, one might argue that some of their contributions are negligible and can be disregarded for the computation of FW. However, recall that the change in weight $F W − M g$ follows from the second derivative of Z(t) in Eq. (2). Neglecting any of the sub-volumes would cause non-smoothness of Z(t), at least since summation in Eq. (1) would exclude part of the total. Even for small imperfections of smoothness, the second derivative of Z(t) would reveal wild jumps. We can see such a behavior in Fig. 13 at $t ≳ 0$ and $t ≈ 1.15$ s due to imperfections of our model assumptions, as discussed above. Although both assumptions do not violate conservation of mass but correspond only to tiny inaccuracies in the rate, they leave a clear fingerprint in the force $F W ( t )$. Therefore, we believe that it is necessary to consider the contributions of all sub-volumes before computing the second derivative, even if it requires some technical effort.

The weight of a running hourglass deviates from the weight of the hourglass at rest due to the of the sand. While this insight is not new (but still not generally acknowledged), in several references it was claimed that the effect is tiny and hardly using standard laboratory equipment. In the present paper, we perform the experiment using very simple equipment and demonstrate quantitative agreement with theory.

While in many cases phenomenological approaches to physics problems are successful, in the case of the weight of an hourglass such an approach is misleading and gives erroneous results that can be found in and lecture notes. More systematic approaches, such as the one presented here (namely, to consider the of the center of mass), are certainly more laborious and less elegant. However, they are certainly less prone to fail, which may be considered as the educational message of this paper.

We thank the German Science Foundation (DFG) for funding through the Cluster of Excellence Engineering of Advanced Materials, ZISC and FPS.

1.

At each instant of time, the total weight is a sum of contributions of the material in the upper container, the material in the lower container, the force exerted when the falling sand grains come to rest, thus changing their momentum. Assume at time t = 0 the sand starts pouring at constant flow rate $f≡dm/dt$. At this time, all sand is in the upper container. The first grain will come to rest at time t1 in the lower container where it arrives at velocity v = gt1. For $0≤t≤t1$ the measured mass is $M(t)=M0−ft$. For $t≥t1$ we obtain $M(t)=M0−ft+f(t−t1)+1gdpdt$, where dp / dt is the rate of momentum change of the pile of sand at the bottom. This momentum change is $d(mv)/dt=m dv/dt+vrel dm/dt$, where $vrel$ is the relative speed between the falling grains and the pile at the bottom. Since $dv/dt=0$ and $vrel=gt1$, we end up with $M(t)=M0−ft1+ft1=M0=const.$ As a result of this (flawed) solution, we obtain that apart from the very beginning $0 and the very end of the flow (where an equivalent calculation can be done), the balance shows M0, and hence is independent of whether the sand in the hourglass flows or rests. The problem with this calculation is that the accelerated motion of the center of mass is disregarded, and this is what the present paper is about.

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