Free Fall Scenario: 100kg Piston Kinetic Energy After 20m Drop

[beckerman]

Merely pointing out that the following is a self-contradicting statement.



They won't accelerate at the same velocity because the drag gives each a different interaction. Also, things don't accelerate at a velocity; they accelerate at a rate of velocity change.

Ok poor wording on my part. But all freefalling objects accelerate at a rate of 9.8mpss. I still do not understand your comparison with the bubbles. I was stating that if you have two objects of identical size, and one is heavier than the other, and they were both dropped, the heavy one would reach the bottom of the drop quicker because its weight allows it to overcome drag forces more effectively. So if we can say our piston weighs 1kg at the beginning of the descent but its weight increases as the water is set to motion, then perhaps we would have a 50kg piston after a couple meters of drop. As the water moves it has less and less of an effect on the pistons bouyancy.And as the piston becomes less buoyant and more heavy its acceleration would be hindered less and less by drag.

 

[beckerman]

I was thinking along the same lines when I was moving toward a "hollow donut" idea.

Perhaps this should be addressed to the OP. A device operating on an "unbalanced wheel principle" cannot be used as a power source, because you expend as much energy lifting the weight back to the top.

Even though OP's idea is somewhat far from an unbalanced wheel, the principle still applies. He's still simply "storing" the energy as potential energy. Drop the weight to convert it into useful energy.

But it's still nothing more than a storage device, one that has to be "recharged".

I suppose this is similar to an unbalanced wheel. Also your hollow doughnut applies, if it was oval shaped with long straight sides, so one side could be used as our cylinder. Call it what you will. Does anyone agree with me that the piston will fall "fast" and not "slow?" Can anyone prove either way?

 

[NoTime]

Ok poor wording on my part. But all freefalling objects accelerate at a rate of 9.8mpss. I still do not understand your comparison with the bubbles.

The keyword here is freefall.
The only time you get freefall is in a vacuum.
A feather obviously does not fall as fast as a BB.
And a hot air balloon with people in it, having a mass much greater than a feather, doesn't fall at all.

I was stating that if you have two objects of identical size, and one is heavier than the other, and they were both dropped, the heavy one would reach the bottom of the drop quicker because its weight allows it to overcome drag forces more effectively.

This is simply not true.
Take a roll of aluminum foil, drop it.
Now unroll the foil and drop it.
Same mass, but in the unrolled state, it's going to take a long time to drop.
It's all about density here.

So if we can say our piston weighs 1kg at the beginning of the descent but its weight increases as the water is set to motion, then perhaps we would have a 50kg piston after a couple meters of drop. As the water moves it has less and less of an effect on the pistons bouyancy.And as the piston becomes less buoyant and more heavy its acceleration would be hindered less and less by drag.

Consider my frozen water example.
The water/piston is one system. Not two.
Bouyancy has no meaning here.

 

[NoTime]

I suppose this is similar to an unbalanced wheel. Also your hollow doughnut applies, if it was oval shaped with long straight sides, so one side could be used as our cylinder. Call it what you will. Does anyone agree with me that the piston will fall "fast" and not "slow?" Can anyone prove either way?

No, I don't agree.
Look up plumbing. Read up on how pipe diameter affects water flow.

 

[beckerman]

I said two objects of identical size with different mass. That means the density of the two objects are different. Your foil scenario is the same as my scenario. There both right. Read and think before you make corrections. Also are you picturing a 100kg object in a piece 3/4" pipe? Pipe diameter was never specified because I was hoping you and others would visualize a best case scenario and not what you might be limited to at Home Depot.

 

[NoTime]

I said two objects of identical size with different mass. That means the density of the two objects are different. Your foil scenario is the same as my scenario. There both right. Read and think before you make corrections. Also are you picturing a 100kg object in a piece 3/4" pipe? Pipe diameter was never specified because I was hoping you and others would visualize a best case scenario and not what you might be limited to at Home Depot.

Yep, I mispoke myself.
Substitute the word aerodynamic for density.

Pipe is pipe. Doesn't matter where you buy it.
And Home Depot sells pipe up to 4" in diameter.

You don't seem to be paying attention.
Your question is no longer interesting.

 

[beckerman]

Consider a 20cm diameter cylinder. 1 linear meter = 3.14 cubic m. So for the piston to descend 1 meter, 3.14 m3 of water must exit. Likewise for the piston to descend 1m in 1 sec the water would have to flow at 3.14 cubic meters per second. In a 20 cm wide pipe I really don't think the back pressure generated to push that volume would be that great. The volume or pipe diameter of the plumbing used to rout the water back to the top of the cylinder can be any size. It does not have to match the cylinder size. Therefore I feel the resistance in this part of the system could be kept quite small. If I was convinced 100% I knew what I was doing I would not be here. I appreciate everyones help, and hope you realize if I am arguementative I am just trying to be certain.

 

[Danger]

You're entitled to your opinion, of course, but you're quite mistaken about NoTime. Everyone gets a bit frustrated at finding new ways to say the same thing over and over again. That naturally leads to an occassional slip-up in wording.
I'm going to have to bail out of this thread as well. It's not because I'm no longer interested; rather, I've exhausted my knowledge in the subject. It looks like a job for Clausius.

 

[NoTime]

Consider a 20cm diameter cylinder. 1 linear meter = 3.14 cubic m. So for the piston to descend 1 meter, 3.14 m3 of water must exit. Likewise for the piston to descend 1m in 1 sec the water would have to flow at 3.14 cubic meters per second. In a 20 cm wide pipe I really don't think the back pressure generated to push that volume would be that great. The volume or pipe diameter of the plumbing used to rout the water back to the top of the cylinder can be any size. It does not have to match the cylinder size. Therefore I feel the resistance in this part of the system could be kept quite small. If I was convinced 100% I knew what I was doing I would not be here. I appreciate everyones help, and hope you realize if I am arguementative I am just trying to be certain.

Thanks Danger.

Ok, If you are interested in learning.

1) calculate the correct volume of 20cm diameter by 1m length of water. You are off by a lot.
Show all of the math you use.

2) give the mass (weight) of 3.14 m^3 of water. Your current answer.
If you live in the US give the answer in lbs.
Show all of the math you use.

3) give the length of your 20cm diameter piston.
Show all of the math you use.

 

[beckerman]

Fair enough,
r=10
r sqrd=100
3.14*100=314 square centemeters
314/100 = 3.14 square meters
3.14 sq meters * 1 meter long =3.14 cubic meters
weight of 314 ccs water 3.14 kg (I live in the US incedentally)
Length was given, 20m
If you are interested I posted this question on yahoo answers. It is titled Free falling piston scenario. There are some different views posted you might like to read.

 

[Danger]

r=10
r sqrd=100
3.14*100=314 square centemeters
314/100 = 3.14 square meters
3.14 sq meters * 1 meter long =3.14 cubic meters
weight of 314 ccs water 3.14 kg (I live in the US incedentally)
Length was given, 20m
If you are interested I posted this question on yahoo answers. It is titled Free falling piston scenario. There are some different views posted you might like to read.

Wanna recheck that?

 

[beckerman]

The 314 square centemeters = 3.14 sq meters part? Would I not divide by 100 since there are 100 centemeters in a meter? Help me out here.

 

[Danger]

There are 100 centimetres to a metre, but you forgot to square it. A square metre equals 10,000 square centimetres.

 

[beckerman]

Of course thank you.

 

[beckerman]

Another way to look at this scenario would be to compare the cylinder and plumbing to a balanced see-saw. If you had two tubes standing vertical with their bottoms and tops attatched, the water in one would balance the water in the other. Moving a cubic meter of water down in one would mean moving one up in the other, and vise versa. There would be no force required because opposite the cubic meter of water in the left tube, there is a cubic meter in the right tube that is directly opposite it. Moving any mass in either tube would be like pushing one side of a perfectly balanced see-saw. If you had a perfectly balanced see saw and threw 1kg on 1 side it would push that one side down very quickly.

 

[DaveC426913]

If you had a perfectly balanced see saw and threw 1kg on 1 side it would push that one side down very quickly.

No, it wouldn't. You're forgetting about inertia. How fast do you think you could set 2,000kg of water in motion? Can you see a seesaw with 1,000kg sitting on each end bouncing up and down as if empty?

 

[beckerman]

No, of course not. What I see happening is consistent with what I have been saying all along. The piston would accelerate slowly at first (to break the inertia of the stationary water) but once it set the water into motion the piston would accelerate at a normal 9.8mpss (less losses due to friction between the water and the plumbing)

 

[NoTime]

Fair enough,
r=10
r sqrd=100
3.14*100=314 square centemeters
314/100 = 3.14 square meters
3.14 sq meters * 1 meter long =3.14 cubic meters
weight of 314 ccs water 3.14 kg (I live in the US incedentally)
Length was given, 20m
If you are interested I posted this question on yahoo answers. It is titled Free falling piston scenario. There are some different views posted you might like to read.

1) Danger helped you on this one, but I havn't seen your corrected answer anywhere.

2) 314 cc of water does not weigh 3.14Kg.
Neither does 3.14 cubic meters.

cc usually stands for cubic centimeter, also written as cm^3.
How did you get from 3.14 cubic meters (m^3) to 314 cubic centimeters (cm^3).
Pay attention to what Danger told you here.

How long is a meter in feet.
hint: It's more than a yard.
Just how big is a chunk of water 3.14meter x 3.14 meter x 3.14 meter (or 3.14 cubic meters) in cubic feet?

3) You gave "The piston weighs 100kg but only weighs 1kg under water." in the original post.
The 20m was for the pipe.

I asked how long is the piston.
It's not 20 meters.
With a 20cm diameter it has a specific length.
You might shout Eruka! when you figure it out.

 

[beckerman]

Dear No Time,
When an object descends in a cylinder filled with water, which is directly linked at top and bottom by an identical cyliner which is also filled with water, describe the chain of events that must take place for 1chunk of water measuring 1 cubic foot to move 1 linear hand!

 

[beckerman]

Erukaaaaaaaaa!

 

[NoTime]

Dear No Time,
When an object descends in a cylinder filled with water, which is directly linked at top and bottom by an identical cyliner which is also filled with water, describe the chain of events that must take place for 1chunk of water measuring 1 cubic foot to move 1 linear hand!

Sorry, this isn't an answer to anything I asked you.

 

[beckerman]

Lets remember who asked the original question

 

[beckerman]

Lets also observe who seems to have had very little to say in opposition to my most recent posts.

 

[rcgldr]

You keep mentioning the piston is accelerating, but seem to imply that once the water starts moving, the water isn't accelerating. However, in order for the piston to acceleration, the water also has to accelerate, and the acceleration is limited by the mass of both the piston and the water.

In addition to momentum of the water resisting acceleration, as the water speeds up, viscosity of the water resists the flow, slowing the rate of acceleration, until the system reaches a terminal velocity.

 

[beckerman]

I have acknowledged that the piston and the water have to accelerate in unison, and also that there would be a terminal velocity that would be slower that the terminal velocity would be in air. I have also acknowledged that the 1kg of weight circulating the water through the system would do it slowly at first. What I am objecting is the theory that the object would descend exactly as it would in a pool. I made a comparison earlier to a balanced see saw. Dave commented that my reasoning was not accurate because I was ignoring inertia. He was right to an extent but I do not feel he discredited my reasoning. His comment was that you would have to use a balanced see saw with weight at either end equal to the weight of the water in each cylinder as a model. If you had a balance beam with 1000kg at either end and added 1kg to one side that one side would go down, and if there were nothing to stop its travel it would continue to accelerate.

 

[rcgldr]

What I am objecting is the theory that the object would descend exactly as it would in a pool.

You're correct here, it's not the same. In a pool, fluidic drag is resisting the movement of the weight. In the weight in a closed filled tube case, the momentum and viscosity of the water are resisting the movement of the weight. In a vertical, open tube, both the water and the weight are near free fall with only aerodynamic drag against the water and weight resisting the movement.

 

[DaveC426913]

What I am objecting is the theory that the object would descend exactly as it would in a pool.

As the one initially propsed the comparison to sinking in a pool, let me say that I did not mean to suggest that it would happen just as in a pool. I was merely saying that it would be no faster than that.

I concede though that it's a lousy comparison. In a pool, the weight is always fighting stationary water. In your tube however, the water picks up momentum and the weight is essentially carried along.

Why don't you sketch and upload a rough diagram? I'd like to ensure we're talking about the same things. Example, I keep going back to a circular donut, rather than an elongated 'paperclip' shape, but both introduce other problems.

 

[beckerman]

Ok, I can do that if you direct me how. The "paper clip" shape you described is the shape I was invisioning. I am a little confused about your comment. You stated that in stationary water the piston would "always" be fighting stationary water. Conversely in the tube scenario it does not fight the water during the whole length of its travel. Saying the descent would be just as slow as the descent in a pool seems contradictary to these observations.