Today Puzzle #756
Puzzle No. 756鈥� Monday 8 June
My perfect soft-boiled egg takes exactly three minutes and I use the 3-minute sand glass egg-timer that my granny gave me. This morning as I was watching the sand fall from the top to the bottom I wondered about that bit of sand that is free-fall in mid-air. Is it weightless? So does the hour glass actually weigh less during these three minutes than when it is just sitting doing nothing?
Today’s #PuzzleForToday has been set by Hugh Hunt, Reader in Engineering Dynamics and Vibration at Trinity College, Cambridge
Click here for the answer
No, it weighs the same.
This is because the impulse delivered by the sand as it hits the bottom is exactly equal to the weight of the sand in mid-air. Consider the time t it takes for sand to fall a distance s under gravity g. Use the standard "SUVAT" formulae to get s=½gt2 so t=√(2s/g)If the sand is falling at a rate of mdot (kg per second) then the mass of sand in the air is mdot*t and multiplu by g to get the weight of sand in the air which is mdot*√(2gs)
The impulse delivered by the sand at the bottom comes from Newton's 2nd law which says F=ma , or put another way, "Force is rate of change of momentum".If the sand is moving at speed v when it hits the bottom then the rate of change of momentum is mdot*vand we get v from SUVAT again, v2 = 2gs so v=√(2gs) and the force is therefore mdot*√(2gs) - exactly the same as the weight of the sand airborne.
At the moment the sand starts to fall there will be airborne sand with no impulse - so the hourglass will be lighter. Then at the end there will be no airborne sand but continued impulse so the hourglass will be heavier. But these are small effects, and anyway on average the weight will be unaffected.
There are other small effects owing to the change of shape of the sand pile with time - you can ponder these at your leisure!
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