# Hour Glass of Sand and PE + KE

1. Apr 9, 2005

### bjon-07

My proffesor assigned the question to the class.

Will an hourglass filled with sand weigh any different if you tip over (i.e. the sand is on the top falling down vs. the sand is on the bottom not moving)

If the sand is on the top. a few grains of sand will be in free fall (these would be weightless) however when those grains of sand it the bottom of the glass they will have KE which should in theory increase the doward force the hour glass itself is pushing on the scale.

I need to mathmatically prove what will happen in this experiment.

btw I did this in this experiment on a scale in the chem lab of my school. The two weighted the same. (not that the scale I used was not very accurate)

2. Apr 9, 2005

### ramollari

Tricky question. There are two opposing effects: the one you mentioned, i.e. the grains of sand that hit the bottom of the hourglass, and the fact that the grains of sand in the air do not add to the overall weight. Which one has more effect can be measured by experimentation, but I have the belief that it is also determined mathematically.
If you would like to give a try mathematically you should consider the falling sand as a culomn of height h and cross-sectional area A, where mass is flowing at a rate of dm/dt. So, you could determine the force applied on the bottom of the hourglass (dJ/dt). Also you could determine the weight of the culomn that is not adding to the overall weight.

3. Apr 9, 2005

### ehild

It is easy to discuss this problem by the means of the centre of mass. The acceleration of the CM is determined by the resultant of all external forces acting on the system, and the internal forces do not play any role.
The system consist of the hourglass filled with sand. The total mass (hourglass + sand) is M. The external forces are gravity gM and the normal force N from the scale which supports the hourglass.
If the sand is at he bottom and nothing is moving, the CM is in rest, its acceleration is zero, the net force must be zero, N = Mg.

If you tip the hourglass over and some of the sand is falling down the CM also falls. If the mass of the falling sand is m, the acceleration of the CM is

$$a_{CM} = \frac{m}{M}g$$

That means that the net external force on the system is

$$F_{net}=Mg-N=M*a_{CM}=mg \rightarrow N = (M-m)g$$

The scale measures N as weight and it is less now. The falling sand does not contribute to the weight. As for the sand grains hitting the bottom, their force is internal force and does not influence the motion of the CM.

ehild

4. Apr 10, 2005

### bjon-07

I talked with my proffesor about this problem and he said it was wrong. This is not an isolated sytem. He told me to use the equation for a chain falling (we did it for a homework problems) where N=3mgx/l.

5. Apr 11, 2005

### ramollari

How do you know that the CM is accelerating?

You don't explain how you derived that equation. So what is the mass of falling sand (m)? It is like m is a point mass falling at free fall.

6. Apr 11, 2005

### ehild

m is the mass of sand between the top container and the bottom one, moving downward, not supported by anything so being in free fall. I considered a grain of sand a point mass which is in free fall if nothing supports it.

With my very much simplifying assumptions, the sand pours out with constant rate c=Ms/T where Ms is the total mass of sand and T is the time the hourglass has been designed to. If h is the distance between the top and bottom containers (well, I imagine that both containers are flat, which is not true, but as a zero hypothesis, it will do), the time needed that a selected grain reaches the bottom from the top is t = sqrt(2h/g). During this time,
m = ct is the mass of sand that has been poured out and is falling. Assume that h=0.05 m, g=10 m/s^2, T=5 min=300 s. Than t=0.1 s, and m= Ms/3000. The total mass is M=Ms+ Mglass, so the acceleration of the CM would be
a = (g*m+(M-m)*0)/M=g/3000*Ms/(Mglass+Ms), so the relative change of weight is less than 1/3000 for a small 5-minute hourglass.

Well, I might be entirely wrong but I do not see where.

It is an other thing what the scale shows. The plate of the scale is attached to a spring and it is not fixed. When the sand stops pouring, the whole gravitational force from the hourglass, Mg, acts on the plate with the CM of the hourglass still moving. The spring gets compressed and it starts to show higher weight. But the plate does not stop at the equilibrium position, it goes a bit further, so you can read bigger weight as the real one, till the oscillations stop. I just wonder what is the average normal force at the end.

ehild

Last edited: Apr 11, 2005
7. Apr 11, 2005

### ramollari

Your very first assertion seems to be wrong. The mass of sand does not simply fall freely from the upper half to the bottom half! The sides of the glass container prevent it from doing so, but allow only a narrow column of sand.

8. Apr 11, 2005

### ehild

But falls freely in between...

Meanwhile I think now where I did it wrong.

So consider a moment when there is a certain amount of sand which is in free fall, and accelerating. But this same amount should get in rest when it reaches the bottom. The time interval between the first and last grain reaches the bottom is the same as it was needed to the first grain to move from the top to the bottom. The entirely column gets rest with a deceleration of g... So the time average of the acceleration of the sand in between the top and bottom containers should be zero. The CM will not accelerate. The weight do not change except the very short time t=sqrt(2h/g) at the beginning and at the end.

ehild