PILOT SPIN
Spin Zone => Spin Zone => Topic started by: Rush on November 21, 2024, 05:37:15 AM
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Left but I sense that’s wrong. Not an engineer.
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Left. The ping pong ball is full of air, which makes it want to float. And since it's attached to the bottom, it will want to lift the scale when it floats. The left steel ball is being held from above and while it displaces water, it has no force acting upon the scale itself. So that's my reasoning.
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I think we need to make the following assumptions which are presumably implied in the problem:
*The amount of water on each side is equal and it is balanced absent any balls.
*The “problem” is what happens when you introduce the balls.
*After ball introduction you wait until the system settles and becomes static again. No water has splashed out.
*The ping pong ball is not made of steel. In other words, it is much less dense (lower mass) than the steel ball. But the mass is not zero.
So what has changed? You have added the weight of the ping pong ball (however light) plus its support to the right side, while adding no weight to the left since the steel ball is supported outside the system.
Therefore the scale will tip to the right.
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So what has changed? You have added the weight of the ping pong ball (however light) plus its support to the right side, while adding no weight to the left since the steel ball is supported outside the system.
Therefore the scale will tip to the right.
This would be true if the ping pong ball was full of water. No weight from the steel ball since it's suspended outside the system. Weight from the ping pong ball shell. That also assumes that the thing attaching the pp (I'm tired of typing ping pong) ball to the cup is weightless and rigid.
Now, is the water-filled pp ball shell lighter or heavier than water? if lighter, then tip to the left. If heavier then tip to the right. I'm not sure, but generally plastic is heavier than water, so it goes left.
But going simply, the iron ball and the ping pong ball both displace the same amount of water. But with the water level the same between the two cups, the right cup is lighter, so it would go left. (As JB said)
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it will balance since the water weight is the same on both sides. the balls only displace the water
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42….
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it will balance since the water weight is the same on both sides. the balls only displace the water
This is true if the weight of the ping pong ball and it’s support is negligible.
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Left. The ping pong ball is full of air, which makes it want to float. And since it's attached to the bottom, it will want to lift the scale when it floats. The left steel ball is being held from above and while it displaces water, it has no force acting upon the scale itself. So that's my reasoning.
It wants to float and so it’s pulling on the bottom but the water still pushing down (atmospheric pressure on its surface) is an equal and opposite force. It goes nowhere, assuming the rotational force from the weight on the opposite side of the fulcrum remains the same which you are correct in saying it will. Also assuming the pp ball and support have no mass.
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42….
I give up. What’s the significance of 42?
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I give up. What’s the significance of 42?
Hitch Hikers Guide to the Galaxy.
All questions were asked and all data was input and it turned out that 42 is the answer to everything.
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Hitch Hikers Guide to the Galaxy.
All questions were asked and all data was input and it turned out that 42 is the answer to everything.
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I never saw the movie. I just read the book(s) I think there were 4 or 5 of them, and that was somewhere around 40 years ago.
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Hitch Hikers Guide to the Galaxy.
All questions were asked and all data was input and it turned out that 42 is the answer to everything.
Ah! I never even read the books. No idea why I didn’t get into it, everybody else did.
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I give up. What’s the significance of 42?
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Simple experiment to perform if you have a kitchen scale handy on which you can place a glass of water. Place glass of water on scale. Note weight or use Tare to zero the amount. Then place finger in water. (This mimics the left side with the steel ball suspended from the ground.) Does the scale weight display change?
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Simple experiment to perform if you have a kitchen scale handy on which you can place a glass of water. Place glass of water on scale. Note weight or use Tare to zero the amount. Then place finger in water. (This mimics the left side with the steel ball suspended from the ground.) Does the scale weight display change?
And then to mimic the right side, cut my finger off and place it in the glass?
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And then to mimic the right side, cut my finger off and place it in the glass?
Only if your finger has much lower density than water.
;D
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Simple experiment to perform if you have a kitchen scale handy on which you can place a glass of water. Place glass of water on scale. Note weight or use Tare to zero the amount. Then place finger in water. (This mimics the left side with the steel ball suspended from the ground.) Does the scale weight display change?
That would work if you could remove water to keep its level the same before and after the fingering.
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But, will the airplane takeoff?
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it will balance since the water weight is the same on both sides. the balls only displace the water
Ding ding ding!
My source for this exercise (husband, who else?) says the official answer is there will be no change. You assume the ping pong ball and its support have no weight (typical conditions in engineering and physics classes), you’re only considering the water displacement which is equal, and the weight of the steel ball which is irrelevant since it’s supported outside the system. The fact that the steel ball is trying to sink and the ping pong ball trying to rise is also irrelevant because they’re both being held in stasis.
I’m sure there are professors somewhere that will argue with this.
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OK, I see that. They both displace the same amount of water. What's in the balls makes no difference. Hm... pretty cool thought experiment.
Also, the speed of the treadmill has no effect on the outcome.
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What if a fly lands on the treadmill?
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What if a fly lands on the treadmill?
Great question! If going in the opposite direction as the treadmill, does the fly have a shorter landing roll? Say the fly is going 5 knots left and the treadmill is going 5 knots right, does the fly land shorter than on a stationary treadmill?
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Ding ding ding!
My source for this exercise (husband, who else?) says the official answer is there will be no change. You assume the ping pong ball and its support have no weight (typical conditions in engineering and physics classes), you’re only considering the water displacement which is equal, and the weight of the steel ball which is irrelevant since it’s supported outside the system. The fact that the steel ball is trying to sink and the ping pong ball trying to rise is also irrelevant because they’re both being held in stasis.
I’m sure there are professors somewhere that will argue with this.
that's a relief, I don't have to return my diploma!
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Ding ding ding!
My source for this exercise (husband, who else?) says the official answer is there will be no change. You assume the ping pong ball and its support have no weight (typical conditions in engineering and physics classes), you’re only considering the water displacement which is equal, and the weight of the steel ball which is irrelevant since it’s supported outside the system. The fact that the steel ball is trying to sink and the ping pong ball trying to rise is also irrelevant because they’re both being held in stasis.
I’m sure there are professors somewhere that will argue with this.
The steel ball is not outside the system. It is supported both by its tether and its buoyancy.
Buoyancy is experienced by both balls, trying to push them upward. But the ping pong ball is attached to the bottom of its container, so the upward force is balanced by a downward force on its tether. All the forces between the ball and container are internal. But the steel ball's upward buoyancy is transmitted to the water and downward on the container. If you measured the force on the steel ball's tether you'd get a smaller weight than if the ball were hanging in the air. If you doubt this, imagine the ball was a water balloon instead of steel. If you hang it in the air you get, say, 1 lb tension on its tether. If you lower it into water you get 0 lb tension. But now the container has an extra 1 lb of water in it that isn't supported by the tether.
I can do the experiment if anyone doubts and video the results. I could 3-D printing two identical shapes, one hollow and one solid (cones are easiest to print that can be hollow and watertight; spheres are a real pain.) Then set things up.
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The steel ball is not outside the system. It is supported both by its tether and its buoyancy.
Buoyancy is experienced by both balls, trying to push them upward. But the ping pong ball is attached to the bottom of its container, so the upward force is balanced by a downward force on its tether. All the forces between the ball and container are internal. But the steel ball's upward buoyancy is transmitted to the water and downward on the container. If you measured the force on the steel ball's tether you'd get a smaller weight than if the ball were hanging in the air. If you doubt this, imagine the ball was a water balloon instead of steel. If you hang it in the air you get, say, 1 lb tension on its tether. If you lower it into water you get 0 lb tension. But now the container has an extra 1 lb of water in it that isn't supported by the tether.
I can do the experiment if anyone doubts and video the results. I could 3-D printing two identical shapes, one hollow and one solid (cones are easiest to print that can be hollow and watertight; spheres are a real pain.) Then set things up.
Uh oh. It’s possible I’m wrong. More importantly it’s possible my husband is wrong!
I did a little experiment. I put a cup of water on a scale, it weighed 247 grams. Then I lowered the closest thing to a solid sphere I could find (a glass teardrop shaped keychain bauble) into the water and continued holding it suspended midway. The weight stabilized at 256 grams.
Oh dear.
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Uh oh. It’s possible I’m wrong. More importantly it’s possible my husband is wrong!
I actually agreed with your analysis back in post #3. Then I did the experiment that I described in post 15 and that you have now done:
I did a little experiment. I put a cup of water on a scale, it weighed 247 grams. Then I lowered the closest thing to a solid sphere I could find (a glass teardrop shaped keychain bauble) into the water and continued holding it suspended midway. The weight stabilized at 256 grams.
Oh dear.
The shape of the embedded object doesn't matter - just the volume of water it displaces.
The only mistake I see your husband may have made was assuming the steel ball was outside the system for purposes of pressures and weight.
I'll see if I can do an actual experiment that replicates the problem diagram as closely as possible.
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I actually agreed with your analysis back in post #3. Then I did the experiment that I described in post 15 and that you have now done:
The shape of the embedded object doesn't matter - just the volume of water it displaces.
The only mistake I see your husband may have made was assuming the steel ball was outside the system for purposes of pressures and weight.
I'll see if I can do an actual experiment that replicates the problem diagram as closely as possible.
The mistake we both made was thinking the steel ball was not “in” the system, when its contact with the water itself puts it in the system. Doh! 😣 I guess I have to relinquish my engineering degree.
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that's a relief, I don't have to return my diploma!
Looks like you might have to after all. At least you’re not alone.