- The weight is constant
- The weight decreases and then increases
- The weight increases
- The weight increases and then decreases
Solution :
Friday, 5 August 2011
Physics GRE - #4
Consider an hourglass on a scale pictured below at times t=0, 0.001, and 1 hour. What happens to the scale's measure of weight of the hourglass plus the sand combination as the sand falls?
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Wouldn't the answer also depend on how much sand is in free-fall at once? If the impulse of the sand is not enough to overcome the loss of the weight from the sand in free-fall behind it, then I agree it will decrease, then increase to W presumably.
This also doesn't consider the loss of weight when you pick up the system to flip it, plus the increase in weight when you put it back down.
Seems like a better way to write this problem would be to have a solid mass falling in a cylindrical container.
I was assuming that the last speck of sand hits the bottom of the hourglass at exactly one hour. This would mean that the scale reads more than its supposed to, since to increase and then decrease to W, you need some split seconds of adjustment time (however you are right in that it'll decrease, increase, and then decrease to W).
As for whether it depends on if how much sand is in free-fall, at 0.001 seconds, no sand has reached the bottom of the hourglass, so we don't need to worry about that if we're just evaluating the 3 specific times they gave us.
And yes, it would be much better to rewrite this problem as a solid mass falling thorough a trap door that opens at some time (maybe I'll do this in the future once you guys have forgotten all about this problem :) ).
What kind of interaction do you mean?
Also, think about this:
If you were standing on a scale, and you bent down a little, how does the scale reading change?
Air is somewhere in between, but at drag forces would counter-act some of the missing weight from sand in free fall. I just don't know how much. I need a scale and an hourglass. :P
Hmm. Ok. I'll check it out.
Geez! You're right! The comment box just disappeared! Thanks for the heads up!
I've switched it back to the old system for now.
Comments for the new posts should be working.
Thinking on your feet: You had it at decreasing, only one answer left.
This question seems kind of misleading. I agree with your analysis until the very end. At 1 hour, the last speck of sand will increase the weight as it delivers an impulse, but once that speck of sand is at rest, the weight returns to W. So it will decrease, then increase, then decrease to W. I guess it depends on whether you think that last grain of sand is at rest at t=1.
Wouldn't the answer also depend on how much sand is in free-fall at once? If the impulse of the sand is not enough to overcome the loss of the weight from the sand in free-fall behind it, then I agree it will decrease, then increase to W presumably.
This also doesn't consider the loss of weight when you pick up the system to flip it, plus the increase in weight when you put it back down.
Seems like a better way to write this problem would be to have a solid mass falling in a cylindrical container.
Wouldn't the sand interacting with the air compensate for the missing weight of the sand in free fall? Wouldn't it remain ~constant throughout operation?
@Dave:
I was assuming that the last speck of sand hits the bottom of the hourglass at exactly one hour. This would mean that the scale reads more than its supposed to, since to increase and then decrease to W, you need some split seconds of adjustment time (however you are right in that it'll decrease, increase, and then decrease to W).
As for whether it depends on if how much sand is in free-fall, at 0.001 seconds, no sand has reached the bottom of the hourglass, so we don't need to worry about that if we're just evaluating the 3 specific times they gave us.
And yes, it would be much better to rewrite this problem as a solid mass falling thorough a trap door that opens at some time (maybe I'll do this in the future once you guys have forgotten all about this problem :) ).
@orsonscott:
What kind of interaction do you mean?
Also, think about this:
If you were standing on a scale, and you bent down a little, how does the scale reading change?
Well, if the medium were honey the weight would remain constant the entire time. If the medium were a vacuum, the weight would drop and then rise back to the total weight.
Air is somewhere in between, but at drag forces would counter-act some of the missing weight from sand in free fall. I just don't know how much. I need a scale and an hourglass. :P
Nevermind. I was making the mistake of ignoring the fact that because the sand is accelerating in free fall is essentially unsupported.
Any time sand is falling, the center of mass of the hourglass is moving. The CoM starts and ends at rest, so it must have both accelerated and decelerated within the hour. Thus, the net force on the hourglass has been both downwards and upwards within the hour.
The commenting on the newer posts is broken.
Hmm. Ok. I'll check it out.
Geez! You're right! The comment box just disappeared! Thanks for the heads up!
(8692910415860994666):
I've switched it back to the old system for now.
Comments for the new posts should be working.
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