The hourglass on the scale

Problem

An hourglass is balanced on a scale. When turned over, the sand begins to flow downward. Does the scale read more, less, or the same value? (Hint: think about the motion of the centre of mass.)

Solution

Newton’s second law is applied to the centre of mass of the whole hourglass (glass + sand): $N - M_\text{tot}\,g = M_\text{tot}\,a_{\text{CM},y}$ where $N$ is what the scale reads (the constraint reaction) and $a_{\text{CM},y}$ is the vertical acceleration of the CM.

While the sand is flowing, the falling column of grains is accelerated downward; the dominant effect in the flow regime is that the falling stream decelerates sharply as it strikes the pile below. The overall balance gives a centre of mass whose dynamics require a constraint reaction greater than the static weight: $\ev{\text{the scale reads more}}$ In terms of the CM: the net change in the vertical momentum of the sand (impact at the bottom) exceeds the “deficit” due to the grains in flight, and by the second law for the CM the scale must push harder than the weight alone.

Note: this is a classic and subtle problem; the initial and final transients of the flow can give brief deviations, but in the steady-state falling regime the reading turns out to be slightly higher than the resting weight.

Links

Topics: Centre of mass Concepts: Centre of mass · Newton’s second law Skills: Limiting-case and thought-experiment analysis