Struggling with this problem....

[Fab Foodie]

Dears

This was on R4 the other morning and I have not heard the answer:

If you have an hour-glass/egg timer and you weigh it, then turn it over to allow the sand to run, does it weigh any less because some of the sand particles are in mid-air as they fall?
I think it will weigh very slightly less. @Hill Wimp thinks it will weigh the same.

Please respond with your thoughts UNLESS you heard the answer, then please keep it to yourself for a while OR hide it as a 'Spoiler'.

Cheers
FF

 

[Deleted member 26715]

Same, as even if they are in free fall they are still within the boundaries of the hourglass

Edit:- I no longer listen to Suicide Radio so mine is a guess

 

[midlife]

Mass is the same but weight will be slightly less as some grains of sand are in freefall. It's the Mass versus weight thing IMO.

 

[Moon bunny]

Meanwhile, a related problem, if everyone on a crowded ship jumped upwards, would the ship stay at the same level, rise slightly, dive slightly, or dive then rise?

 

[DCLane]

Meanwhile, a related problem, if everyone on a crowded ship jumped upwards, would the ship stay at the same level, rise slightly, dive slightly, or dive then rise?

Dive then rise as the passengers pushed down before jumping to get lift.

As for @Fab Foodie 's question - what @midlife said.

And do you lot not have any work to do? If so I've a pile of marking you have a go at instead

 

[Moon bunny]

Finished it all for the week. Can I go out to play now?

 

[Deleted member 26715]

Finished it all for the week. Can I go out to play now?

Have you shown your working out?

 

[roubaixtuesday]

My gut reaction is that it makes no difference as all momentum transfer is internal to the egg timer.

A thought experiment to explore:

Imagine a very light, very tall cylinder, say Nelson's column size but near zero mass and containing a perfect vacuum. The cylinder has a cannonball at its base, and is instantaneously upended, and placed on a weigh scale. What would the scales observe?

I think:

1. Initially, just the weight of the cylinder. No force exists between cannonball and cylinder, so it's existence cannot be detected by the weigh scale.
2. A very high spike as the cannonball hits the base, transferring momentum. Much more than the weight of the cannonball itself.
3. A constant weight equal to the cylinder plus cannonball.

So I change my gut reaction and propose that the weight does, indeed, reduce by the weight of the sand in flight.

 

[DCLane]

Finished it all for the week. Can I go out to play now?

Quick marker, eh?

I used to work with a colleague who'd simply check the reference list and then drop the assignment down the stairwell.

Floated down slowly: 30 % and a fail
Landed quietly: 40 %
Landed quietly with some references: 50 %
Landed with a thud: 60 %
Landed with a thud and more than 10 references: 70 %

He didn't take to me asking him to mark them properly

On the other hand I did some marking alongside an academic called Nigel Bradley, now sadly deceased. He took 2 minutes per exam paper, then went to the pub. Me? I spent 20 minutes marking the same papers as part of the standardisation process we had to do. When he'd returned we looked at the sample of 10 we'd both marked and there was no more than 2 marks difference per paper, no grade boundary differences

 

[classic33]

I'd say it increases slightly, the impact force would be greater than the weight of the sand now in freefall.

 

[newfhouse]

Quick marker, eh?

I used to work with a colleague who'd simply check the reference list and then drop the assignment down the stairwell.

Floated down slowly: 30 % and a fail
Landed quietly: 40 %
Landed quietly with some references: 50 %
Landed with a thud: 60 %
Landed with a thud and more than 10 references: 70 %

He didn't take to me asking him to mark them properly

Did the building weigh less while the assignments were falling? What if the pages became unbound? What does entropy weigh?

 

[Rocky]

@Fab Foodie and @Hill Wimp

Physics has your answer:

https://www.europhysicsnews.org/articles/epn/pdf/2010/03/epn2010413p25.pdf

A special feature of an hourglass is that the mass flow m [kg/s] of sand through the narrow orifice is fairly independent of the sand level. This is caused by the granular nature of the sand. The flow m is present in the free-falling stream of sand but also in the slowly moving dense mass in the upper compartment. If the free-fall time of the sand is τ, the weight of the free-falling mass is g.(m. τ). When hitting the surface in the lower compartment, it exerts a momentum per second, i.e., a downward force (gτ) on the surface. This force exactly compensates the missing weight of the sand that is in free fall. This was the beautiful argument to conclude that the weight should stay the same.

However, in 1985 Shen and Scott showed that the centre of mass (c.o.m.) of the sand is not moving down at constant speed, but is decelerating during the steady operation of an hourglass. That causes a subtle gain of its weight on a balance. With an ingeniously constructed hourglass they even managed to confirm the presence of excess weight.

So........having said that, what has one said? .......it appears there is an answer but I'm not clever enough to understand it. I think we can confidently say that it might weigh the same, it might be heavier or it might be lighter.

 

[Salar]

This reminds me of the budgie flying around in an aircraft question.

A good few years ago I jokingly convinced a work colleague that the more information he downloaded onto his hard disk, the heavier his computer would be.

I was surprised to find the answer when I googled it.

 

[MartinQ]

I suppose like most things, a simplification is in order. So assume the hour glass has no weight, height h is cylindrical in shape and the sand is rplaced by a cannon ball of mass M.

Turning it upside down, the hour glass (height h) would have no weight when the ball is in flight. Simple suvat gives a speed of impact at
sqrt(2gh)
So the impulse (force*time) due to the change in momentum is
M*sqrt(2gh)
When the ball is at rest at the bottom, the weight is
Mg.

So in a very simplified scenario, the weight of the hour glass will change as there is no reaction force between some of the sand and the container, so its effectively taken out of the system. How it will change with the multiple impulses is probably covered above.

 

[EltonFrog]

This all too much for me, you lot'll break the internet if you're not careful.