Hourglass Redux

Let $m_1$ be the mass of sand in the top and $m_2$ the mass of sand in the bottom of the hourglass. Then the height of the sand in each part is $y_i = m_i/\rho A$ . We calculate the moment $my_{cm}$ for each section: $\begin{array}{rccc} \text{part} & \text{mass} & y_{cm} & my_{cm} \\ \hline \text{top} & m_1 & L + m_1/2\rho A & m_1(L + m_1/2\rho A) \\ \text{bottom} & m_2 & m_2/2\rho A & m_2^2/2\rho A \\ \hline \end{array}$ Thus for the total amount of sand (ignoring the sand that is flowing) the center of mass is given by $My_{cm} = m_1L + \frac{m_1^2 + m_2^2}{2\rho A}.$ The time derivative describes the speed of the center of mass. Note that $dm_1/dt = -q$ and $dm_2/dt = q$ : $Mv_{cm} = -qL + \frac{q(m_2 - m_1)}{\rho A}.$ Taking the derivative again shows that the (downward) acceleration of the center of mass satisfied $Ma_{cm} = \frac{2q^2}{\rho A},$ and this is the additional amount shown by the scale.

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What about the sand in the middle? Note, first of all, that $dm_2/dt > q$ because the amount of sand in the middle part of the container decreases. To analyze this in more detail, model the falling sand as free fall motion from rest. In reality, the sand has an initial speed of $q/\rho A$ , but this we assume to be small enough to ignore relative to the final speed of the sand.

The sand falls over a distance $h - y$ ; the time needed to do so is $\Delta t = \sqrt{2(h-y)/g}$ . This means that the amount of falling sand is $m_3 = q\Delta t = q\sqrt{\frac{2(h-y)}g}.$ The average speed of the falling sand is half of its final speed, $\bar v_3 = \tfrac 12 g \Delta t = \sqrt{\frac{g(h-y)}2}.$ Now consider again the three parts of the sand and their masses and average speeds: $\begin{array}{rccc} \text{part} & \text{mass} & \bar v & m\bar v \\ \hline \text{top} & m_1 & q/\rho A & m_1q/\rho A \\ \text{bottom} & m_2 & 0 & 0 \\ \text{falling} & q\sqrt{2(h-y)/g} & \sqrt{g(h-y)/2} & q(h-y) \\ \hline \end{array}$ Now the total mass of sand is constant: $m_1 + m_2 + q\sqrt{2(h-y)/g} = M$ . Take the time derivative and use $m_2 = \rho A y$ : $\frac{dm_1}{dt} + \frac{d}{dt}\rho A y + \frac{d}{dt}\left(q\sqrt{\frac{2(h-y)}g}\right) = 0;$ $-q + \rho A \frac{dy}{dt} - \frac{q}{\sqrt{2(h-y)g}}\frac{dy}{dt} = 0;$ $\frac{dy}{dt} = \frac{q}{\rho A - q/\sqrt{2(h-y)g}}.$

Now the speed of the sand's center of mass is $Mv_{cm} = \sum m\bar v = \frac{m_1q}{\rho A} + q(h-y),$ and the time derivative is $\begin{aligned} Ma_{cm} & = \frac{q^2}{\rho A} - q\frac{dy}{dt} \\ & = \frac{q^2}{\rho A}\left(1 + \frac{1}{1 - q/\rho A \sqrt{2(h-y)g}}\right) & = \frac{q^2}{\rho A}\left(1 + \frac{1}{1 - q^2/\rho A m_3 g}\right). \end{aligned}$ Compare the expression in brackets to the constant 2 in the original solution. This constant is justified if $q^2 << \rho A m_3 g$ , i.e. if the rate at which the sand falls is slow compared to the size of the container and the amount of falling sand.

The difficult step required to find the difference between the "resting" mass of the hourglass and the steady state "flowing" mass is identifying the additional force that would be present during the flow.

This force is due to the motion of the center of mass of the sand. The center of mass is shifting from the top of the hourglass to the bottom. If this motion of the sand's center of mass is accelerating , then it is causing a force equal to the product of the acceleration and the total mass of the sand.

$F_{flow} = m_{sand} \frac{d^2 y_{com}}{d t^2}$

First we need to find an expression for the hourglass's center of mass. Let's start with the moment shown in the problem statement:

We can use the equation for the center of mass of two particles:

$y_{com} = \frac{y_{top} m_{top} + y_{bottom} m_{bottom}}{m_{sand}}$

and we can treat the top and bottom chunks of sand vertically centered at their halfway point:

$y_{com} = \frac{\left( h+ \frac{(L-y)}{2} \right) m_{top} + (\frac{y}{2}) m_{bottom}}{m_{sand}}$

The mass of the top, bottom, and total portions of sand can be calculated by multiplying the cross-sectional area by the density and height of each section. At the point in the image, the top section has a filled height of $(L-y)$ , while the bottom is filled to $y$ . The total volume of sand is a full section, which has height $L$ .

$y_{com} = \frac{\left( h+ \frac{(L-y)}{2} \right) (H-y) A \rho + (\frac{y}{2}) y A \rho}{L A \rho}$

This can be simplified and results in a quadratic, linear, and constant term with respect to the height $y$ :

$y_{com} = \left( \frac{L}{2} + h \right) - \left( 1 + \frac{h}{L} \right) y + \frac{y^2}{L}$

Equipped with this expression for the center of mass, we need to express the filled height $y$ as a function of time $t$ . We know that the mass flow rate is $q$ , which tells us that the total mass at time $t$ is $qt$ . Using the density, we can find the volume as a function of time $v(t)$ , which is the filled height times the cross sectional area.

$\begin{aligned} v(t) &= \frac{m}{\rho} \\ v(t) &= \frac{qt}{\rho}\\ y(t)A &= \frac{qt}{\rho}\\ y(t) &= \frac{qt}{\rho A} \end{aligned}$

This expression can be substituted into our center of mass expression to find the center of mass of the falling sand:

$y_{com}(t) = \left( \frac{L}{2} + h \right) - \left( 1 + \frac{h}{L} \right) \frac{qt}{\rho A} + \frac{q^2 t^2 }{\rho^2 A^2 L}$

To find the acceleration of the center of mass, we take the double derivate with respect to time. We only need to concern ourselves with the quadratic term:

$\begin{aligned} a_{com} &= \frac{d^2 }{d t^2} \frac{q^2 t^2 }{\rho^2 A^2 L}\\ &=\frac{2 q^2}{\rho^2 A^2 L} \end{aligned}$

Finally, we can substitute this acceleration into $F = ma$ to find the corresponding force.

$\begin{aligned} F_{flow} &= m_{sand} \frac{2 q^2}{\rho^2 A^2 L}\\ &= L A \rho \frac{2 q^2}{\rho^2 A^2 L}\\ &= \frac{2 q^2}{\rho A} \end{aligned}$

Thus the total force on the scale is the force of gravity, plus this force of the moving center of mass due to flowing sand:

$F = Mg + \frac{2 q^2}{\rho A}$

Arsenio Meneses - May 25/2018 ........ Lets consider that the top and bottom part have area A and height L and are connected directly. So, the center of Mass of the system will move a distance L/2 + L/2 = L The time to follow will be M/q . So the acceleration of the C.M. will be given by L=(1/2)at^2 => a=2Lq^2/(M^2) The force acting on the C.M. while falling is Ma ===> F= 2Lq^2/M ===>> F= 2Lq^2/(δ.xVol) ==> F= 2Lq^2/δ.AL ==>. F= 2q^2/(δ.x A) δ - density The reading of the scale will be Mg + that F , or Mg+ 2q^2/(δ.xA), as the second option

(h-y)=(1/2)g( t^2)----1

Mass flow rate is q,so average velocity

v=(q)/(density *A)

t=(h-y)/( (q)/(density*A ) Substituting t in 1

We get,

(h-y)g=2 (q^2)/((density A)^2)-----2 We know extra force=

A(h-y)(density*g)

Multiply both sides in 2 with (A*density)

So extra force=2(q^2)/(density*A)