Friday, August 1, 2008

Answer: Hourglass
New Problem: Square Wheel

It's hard to say exactly what the plot would look like, if you try to account for things like the varying pressure on the sand, the varying height of the pile in the bottom of the hourglass, etc. We can say generally what the weight-vs-time graph looks like by finding an important property. Start with

F = dp/dt
Fdt = dp

Integrating, we find that the total force-seconds is equal to the change in momentum. But the hourglass has zero momentum when the sand is all up top, and again zero when the sand is all in the bottom. So the integral of force over time for the whole experiment must be zero.

The force on the hourglass is from gravity and from the scale. Because the force of gravity isn't changing, that means the force read on the scale must, on time-average, be equal to the weight of the hourglass. Any "bumps" in the scale's readings must be canceled out by "valleys".

That tells us about the entire outline of the plot. As for the details, it depends on the acceleration of the center of mass of the hourglass (F=ma). Just as the sand starts falling, the center of mass is certainly accelerating down. The force of gravity is stronger than the force exerted by the scale, and so the reading goes down for a time. When the sand hits the bottom, it accelerates up. If the rate of sand flowing were constant, and all the sand fell the same distance, and the sand in the top weren't sinking, then the total acceleration of the hourglass would be zero. There are complications so that while the sand is falling, the center of mass might have some nonzero acceleration, but it would be small.

As the last bits of sand fall, the acceleration of the center of mass must be upwards - it's going towards a state where it's no longer falling. So the basic idea of the outline should be something like this:

Square Wheel
A square wheel rolls on a bumpy road smoothly, and without slipping. What is the shape of the road?

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