An hourglass H weighs h. When it's placed on a scale with all the sand rested in the lower portion, the scale reads weight x where x = h.
Now, if you turn the hourglass upside down to let the sand start to flow down, what does the scale read?
I imagine initially, when the sand starts to fall but before the first batch of grains touch the bottom of the hourglass, these grains of sand effectively are in a state of free fall, so their weight would not register onto the scale. The weight at this point has to be less than h. However, what about the steady state when there is always some sand falling and some sand hitting the bottom of the hourglass? In the steady state, although we are having some sands in the free fall state and thus decrease the weight of H, there are also sands that are hitting (decelerating) the bottom of the hourglass. This deceleration should translate increase the reading on the scale more than the actual weight of those impacting sands. To illustrate the last point, imagine a ball weighing 500g rested on a scale. If you drop this ball from a mile high onto the same scale, on impact, the scale would read higher than 500g. in the same way, in our hourglass question, will the decreasing effect of weight due to free-fall cancel out exactly the increasing effect of weight due to sand impacting? does it depend on the diameter of the opening? does it depend on the height of the free-fall? Does it depend on the air pressure inside the hourglass?
Best Answer
Analyzing the acceleration of the center of mass of the system might be the easiest way to go since we could avoid worrying about internal interactions.
Let's use Newton's second law: $\sum F=N-Mg=Ma_\text{cm}$, where $M$ is the total mass of the hourglass enclosure and sand, $N$ is what you read on the scale (normal force), and $a_\text{cm}$ is the center of mass acceleration. I have written the forces such that upward is positive
The center of mass of the enclosure+sand moves downward during process, but what matters is the acceleration. If the acceleration is upward, $N>Mg$. If it is downward, $N<Mg$. Zero acceleration means $N=Mg$. Thus, if we figure out the direction of the acceleration, we know how the scale reading compares to the gravitational force $Mg$.
The sand that is still in the top and already in the bottom, as well as the enclosure, undergoes no acceleration. Thus, the direction of $a_\text{cm}$ is the same as the direction of $a_\text{falling sand}$ . Let's just focus on a bit of sand as it begins to fall (initial) and then comes to rest at the bottom (final). $v_\text{i, falling}=v_\text{f, falling}=0$, so $a_\text{avg, falling}=0$. Thus, the (average) acceleration of the entire system is zero. The scale reads the weight of the system.
The paragraph above assumed the steady state condition that the OP sought. During this process, the center of mass apparently moves downward at constant velocity. But during the initial "flip" of the hour glass, as well as the final bit where the last grains are dropping, the acceleration must be non-zero to "start" and "stop" this center of mass movement.