Ball Bearing in a Tube

The total mass would be the same, but the weight registered by a balance etc. attached to the tube would be that of the tube without the ball bearing.

Pressure would build beneath the falling ball. If there are horizontal surfaces for this to react against they would display as weight.

Viscosity affects?

I don't think so. Any pressure below the ball would be countered by a vacuum above the ball.

Guys, he did say an evacuated tube, therefore no + or - pressure above or below.

Yeah I had that problem in exams as well.

It will have the same mass, but most balances are actually force balances. When the ball is realeased it will be accelerated by gravity and not in contact with the scale (the scale is not balancing the force of gravity on the ball while it free falls). When it hits bottom there will be a nice impact. Force balances are for relatively static measurements. The last time I dropped a weight onto a scale it did not register anything until it was "in the pan" and supported by the scale.

Yes I have the same idea , the effect of gravity is happen at end ,but before that as the ball in space the scale did not register the weight of the ball.

So, in a word, yes.

This thread has pointed out the fundamental flaw in the way mass is normally measured.

By definition the mass of an object is its ability to resist a change in motion. The problem is that it is difficult to measure this directly. In effect what we end up measuring is the force that is produced by the gravitational attraction between two masses according to the equation

F = G x (M₁ x M₂)/S²

Where G is the universal gravitational constant, M₂ is the mass of the earth, S is the distance between the center of the earth and the centre of gravity of the mass and M₁ is the mass of the object we are measuring.

Straight away we can see that unless all the variables remain constant the resultant measurement of the mass will be inaccurate. The fact that the earth is not a perfect sphere is just one of the factors that will introduce errors in the reading.

A second problem is that the force will only be constant while there are no other external influences and here is where the second problem enters into the calculation. The earth is rotating and depending where you are standing on the earth there is a force that will act against gravity caused by the associated centripetal acceleration.

There are other factors that will affect the force but I will not go into them here.

So JohnDG stated in post #1 and others subsequently reiterated the mass of the system that we are measuring will not change while the ball is free falling. However since we are not actually measuring the mass but rather the force due to gravitational attraction and gravity is accelerating the free falling mass this force will not register during the period of free fall.

For anyone interested, mass can be measured independently of gravity by using an inertial balance. In this, the unknown mass is fixed to a spring and the system set into oscillation. The period of oscillation yields a measurement of the inertial, and hence the mass.

The classic experiment is described at, e.g., http://www.batesville.k12.in.us/Physics/PhyNet/Mechanics/Newton1/Labs/inertial_balance.htm

I believe that astronauts use a device like this to measure the mass of an object. I havn't seen it done recently but remember them doing it on Skylab back in the 70s

what is the use of computing such activities? please tell me so maybe i can force my self to make an experement...

Ask Galileo.

Yes.

This is like the "fly in an airplane" question. I also have a tool called a no bounce hammer. Its a a hollowed out maul head full of lead shot. On impact the shot cases a secondary impact preventing kick-back. Effective as intended, but at no time does the hammer feel any lighter to me. I would venture to say the weight of the tube remains constant.

How heavy is the hammer when you let go? Its weight is zero when you drop it.

When you swing it, you are accelerating its mass, so you are using the "spring" in your arm to sense the mass. YOU are doing the inertial mass measurement.

Hi

Was the ball "weighing" on or down on the tube ... i.e attached somehow to the tube before the release?

So if the "envelope" is just on the tube (wall surface, inside and out) and not including the ball then in the case of the attached ball, on it's release, the "tube translated" weight of itself now will of course weigh less.

If the envelope is inclusive of the ball, then "the arrangement" doe snot weigh less.

It may be I am taking words and riddle too literally?

Thanks,

Ken.

The original post said I have means of releasing the ball to fall inside the tube - also the tube was closed and evacuated - so the original weighing must have included the ball, somehow held at the top of the tube (an electromagnet, perhaps?). If it wasn't attached, how could releasing it make any sense ?

So, if you have a truck load of birds that overload a bridge, just beat on the side of the truck, get the birds flying, then cross the bridge...........practical application, huh ?

The load of birds always fastinated me.

A flock of birds flying over the truck does not add any weight to the truck. It is an open system, the birds are flying in a medium above the truck.

If I have a closed truck (closed system) and the birds are flying inside in a non-accelerated state (they are not in free fall) then we would assume the weight does not change.

However, all "chicken" trucks I have seen have open cages and open sides (we will ignore the fact there is no way they could fly in such tight quarters). If those birds could fly, are they in an open or closed system in balance?

Good morning children, How are you today?.

Today we will play skydivers. Here comes a beautiful airplane full of skydivers. What a nice sight! The airplane with the skydivers weighs 2000 Lbs. And here they jump one by one... each one weighs 150Lbs. and there are 4 of them. OK children, I will give you 10 full minutes to find out how much weighs the airplane now! Gooood work, yes it weighs now 1400Lbs.! Now our nice pilot lands the airplane, and the 4 skydivers climb back into it! How much weighs the airplane now? Goood work. That's enough for today cause I don't want to stress your little brains too much. Tomorrow I promise to teach you the unimportant difference between MASS and WEIGHT.

Wangito.

nice try!!!!! maybe you could answer the real question and not the opinion of others...

The real answer IS given my boy. Am really sorry if you didn't catch the clue. I guess this is why I can't see your scientifically backed answer to the thread?. Now here's your homework:

Fill out the missing and see if you don't express the opinion (what do you mean by "opinion"?) of others.

1. ...........is the pull of the earth on a body.
2. ...........is quantity of matter in a body.

And BTW, ALL the opinions(?) expressed here are opinions of others, Don't you dare blame no one of us here at CR-4 for being original.

Wangito

sorry if you didn't catch the clue. I guess this is why I can't see your scientifically backed answer

didn't catch the clue and scientifically backed answer?? tell you what,, before you think that your answer is perfectly being delivered with scientific explanations you have to find out first what is the difference between a clue and the real answer. or should i say let the definition and meaning differ itself..

think you have the brain?? think again...

anyway for your brain to excercise. try to figure out the tube and not the air plain carrying people. the deference is this,,, the air plain does not have the possible friction that maybe brought about by inbalance force while it takes its way down. imagine the possible velocity and its effect to walls that maybe gives additional weight to its original position.. try to compute for example the weight of ball bearing, the gap between the circumference of its wall, the thickness of ball bearing, finally the length of the tube.

the gentlemen who post the question does not give specific size and weight or even the gap between the tube and ball bearing during its fall.. what if i dropped 120 pounds ball bearing with 16 inches head circumference, 4 inches thickness and a gap of 2 millimeter to its tube and the tube has 1 mm thickness weighing 500 grams.. do you have any idea what is the effect during its fall to the original weights of the tube??

if you could ever figure out then you have the scientific solutions and maybe you could teach in a kinder garten...

you see my friend,, the question is too simple but for you to make a comment you have to think first the other area that maybe contributing to its factor. dont you ever think that youre farv ahead of anybody and nevertheless dont ever make statement that somebody in this room has smaller brain than you are..

and dont call me a boy im a man enough than you are.

The relative size and mass of the ball and tube don't matter - as long as the tube is evacuated and the ball doesn't touch the sides on the way down, Wangito's assertion, and mine, obtain. While the ball is in free-fall it will not contribute to the combined weight of the system.

precisely the point is,, each and everyone of us has a deferent assumptions i only make a reaction cause somebody make a comment of somebody's little brain. back to issue,, for you to assume that the ball bearing wont touch the sides you gave your self a lot of clearance so that you don't have to compute. unfortunately the question it self does not gave such speculations thats why it is for us to comment on the basis that the ball bearing is somehow has relative reaction to the tube otherwise the questions that being posted will be as easy as it is. if your assumption is as correct as what is in the mind of the gentlemen who posted it then there is nothing for us to argue

Dear sir: First let me offer my humble apologies if I hurt your feelings. No offence meant, it is only my way of writing. I do promise not to call you a boy anymore. but please feel free to call ME a boy as I wish I was. At the same token, you can call me "my friend" as you did, title I now carry with great pride. Now down to business:

Let me first introduce you to the term "Conceptual question", is a question that is trying to establish or prove a concept, rather than offer a mathematical or other precise solution. and is often used in natural as well as in social sciences.

Mr. Bob Collins thread is exactly such a question. The concept is simple and therefor no other parameters are given. If we rephrase Mr. Collins question, all he is asking is whether the ball will exert any FORCE on the weighing instrument during it's free fall from top to bottom. and nothing else. It is weight that we are talking about. And the answer is NO. Sir.

For that matter, the two examples given, that of the airplane (and I guess that what you meant when referring to AIR PLAIN, an entirely different matter) and that of the flock of birds doing the Drop-Di-Drop number.

In order to accommodate your theory, you have added all kinds of new parameters and variants that were not included in the original post. Sizes,diameters, ball weight and whether the tube is evacuated or not all of which have nothing to do with the concept and will not affect the final results anyhow.

And last final suggestion, Don't get mad, Get even.

Yours sincerely

Wangito.

I agree with your sentiment.

Sarcasm and ridicule has no place in a disccussion.

One of my professors back in the late 60's used to say "The only stupid question was the one not asked".

Hey, I absolutely agree with you. there's no place for sarcasm. Therefor I will take your example, "The flock of birds" :

• the flock fly over the truck.
• the birds drop the drop over the truck.
• the bird lose weight equal to the drop. immediately after shootout.
• the truck add weight equal to the drop. but only after successful arrival.

wangito

And BTW: No offence meant in any shape size or form.

My professor said that a stupid question almost always results in a stupid answer.

And, where has your sense of humor gone?

Your analysis is same as mine for the flock overhead.

However, I always seem to be trying to locate the flock as the payload hits.

I would have to say the weight will remain the same, since the total weight is measured before the drop, and nothing is added or subtracted during the fall, the weight will remain the same, unless of course if the whole thing (tube and ball) is moved to different leivation.

MidniteFighter

AW ! Cum on..................Difference between weight and mass ?................

"I have means of releasing the ball" is too vague ... needs more definition. If you used a magnet, normal forces pulling the ball upward on the face of the enclosed tube would actually register the tube's weight "lighter" than when the ball is released. During mid-fall, you'll only get the sole weight of the tube.

I would answer the question: does the arrangement weigh less during the time the ball takes to connect with the bottom of the tube?

NO.

A good point!

The Engioneer's Engineer.

Why not use a radio controlled electromagnet?

Or some very fine wires that have sufficent suppleness that they don't materially affect the over all masses / weights involved?

The weight registred by the balance will be less by the weight of the ball bearing during the flight time of the ball bearing