B Can the fish move the ball?

metastable

So assuming a neutrally buoyant fish can’t make the ball roll... then a scuba diver in a 55 gallon drum, submerged in a particular liquid in which they are neutrally buoyant... shouldn’t be able to put an arm against each side of the drum and make the drum tip over or traverse a single millimeter, simply by pushing their body mass violently from side to side?

rcgldr

Homework Helper
I'm not good at this conservation of angular momentum, but won't the fish's angular momentum in one direction be exactly canceled by the water's momentum in the opposite direction? This isn't like other situations, where a rider could "grab" the machine and manipulate it.
I updated my answer with a simpler explanation unrelated to angular momentum.

The fish swims in place at some fixed (relative to ball) point in front of and above the contact point of the ball. I'm assuming there's enough friction or something like vanes between the water and interior surface of the ball so that the spinning water also spins the ball, so there is resistance to the water spinning, at least while the ball is accelerating.

Since the fish is in front of the contact patch, then the center of mass of the ball, fish, and water is also in front of the contact patch, and gravity's downward pull at the center of mass and the floor's upward push at the contact point coexist with a forwards torque on the ball.

This situation is similar to a ball with an internal electric motor inside, except for the interface between the water and the interior surface of the ball. Both require the center of mass to be kept in front of the contact point in order to accelerate. Once at speed, then there only needs to be enough torque to overcome rolling resistance.

A.T.

Science Advisor
So assuming a neutrally buoyant fish can’t make the ball roll...
I was taking about a fish with the same density as water. A neutrally buoyant fish with non-uniform density can move the CoM relative to the ball and thus can produce some roll motion that way.

A.T.

Science Advisor
Since the fish is in front of the contact patch, then the center of mass of the ball, fish, and water is also in front of the contact patch...
This assumes that the fish is more dense than the water, which makes it trivial (hamster method). The method I described in #5 would work with a fish of the same density as the water, so the CoM is always exactly above the contact point. But it wouldn't work continuously against resistance, because the fish would have to swim faster and faster. This is analogous to an internal reaction wheel, that spins one-way, while the ball rolls the other way.

pinball1970

Gold Member
It is a difficult situation. Our expectation is that if a rigid object bumps into something that a high impulsive force results. But the water complicates things. In this case we have zero net momentum both before and after the collision. Does a high impulsive force actually result? I am having a hard time wrapping an intuition around the situation.

We have an upward flow of water prior to the collision. So a good question would be what happens to that flow. One answer is that it should stop. But if it stops, there has to be a force making it stop. That force is negative pressure. Negative pressure that should be centered on the impact point and that should negate the impulse from the collision. I think we're going to have to get some cavitation going before we can impart much net force. And even then, it would only be temporary.

[Negative pressure is not unreasonable. If the water is under atmospheric pressure we can have up to 15 psi of negative gauge pressure before we hit zero absolute pressure. Even negative absolute pressure is physically reasonable. Water has surface tension. In the absence of nucleation sites, it will resist forming voids. A quick trip to Google yields this article which is less than authoritative, but quite readable. This hit is more authoritative]
A football filled with water (just the top pole cut off) and an angled coat hanger with a make shift fish on the end. Anything that can go on the end of the cost hanger but still get through the north pole. Once in rotate slowly in the horizontal plane.
Possible issues are the symmetry of the ball is reduced (a little) so centre of gravity a little skewed. Also some extra turbulence from the wire but a thin wire should not impact too much.
I think it's better
This is not a closed system, because the container is in contact with the floor. As long as the fish can swim away from the center of the ball and induce a circular flow of water, then the angular momentum of the water and ball have to be offset by the angular momentum of the earth (the only point of interaction is at the contact point between ball and floor).

If the experiment was tried in space, free from any external forces, then the center of mass of the now closed system could not be moved.

I don't see how the fish in a ball differs that much from any other "self-propelled" vehicle, such as a unicycle, bicycle, motorcycle, car, ..., other than it's inefficient.

update - A better explanation is that the fish can swim forwards in the water to keep the center of mass of ball, fish, and water in front of the contact point. Then the pull of gravity and the upwards force from the floor combine to exert a forwards torque on the ball.
I think this is what I was thinking about in terms of creating an Eddy, the centre of gravity would shift as the fish swam round. The ball is resting on a relatively small point on the south pole so very little friction to overcome. If the the fish swam round fast enough the forces on the edges could move enough to cause the ball to oscillate. Even if it is small

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Dale

Mentor
A better explanation is that the fish can swim forwards in the water to keep the center of mass of ball, fish, and water in front of the contact point.
If the fish is the same density as the water and if the ball is full to the top then the center of mass cannot go in front of the contact point. This was exactly my confusion since that is the mechanism for the hamster, but that mechanism doesn’t work for the fish. But even without moving the center of mass we can still produce an external torque by friction.

A.T.

Science Advisor
If the fish is the same density as the water and if the ball is full to the top then the center of mass cannot go in front of the contact point.
This is true if the fish has uniform density. If the fish has the same mean density as water but non-uniformly distributed the total CoM can be moved relative to the ball.

It seems there are two general mechanisms:

1) If the CoM can move relative to the ball: Translate it in front of the contact (hamster method, works continuously against resistance)

2) If the CoM cannot move relative to the ball: Spin something inside so the ball spins the other way (flywheel method, continuous work against resistance limited by the max flywheel speed)

metastable

Suppose the fish weighs 5kg and the water and ball weigh 5kg, and fish floats to the “side” of the ball (90 degrees from the ball’s contact point with the ground) where it uses its tail fin to push off the edge of the glass with an energetic impulse of 250 joules. When it reaches the other side it pushes with 500 joules. At the other side again 1000 joules. The energy of the impulse doubles with each push off the opposite sides.

To calculate the motion of the fish and the ball after the first interaction can I use:

m1v1 = m2v2

where m1 is the mass of the fish and m2 is the mass of the ball and water?

Dale

Mentor
1) If the CoM can move relative to the ball: Translate it in front of the contact (hamster method, works continuously against resistance)
My understanding is that the whole point of the exercise with the fish is to remove mechanism 1, since that is obvious from experience with hamsters.

metastable

...the fish crawls with its fins along the edge of the glass, applying tangential force directly to the glass via the friction with its fins. the “normal force” holding the fish against the glass in this case would the fish’s forward velocity combined with the curvature of the glass...

jbriggs444

Science Advisor
Homework Helper
...the fish crawls with its fins along the edge of the glass, applying tangential force directly to the glass via the friction with its fins. the “normal force” holding the fish against the glass in this case would the fish’s forward velocity combined with the curvature of the glass...
This is a workable mechanism. Simplify and turn the fish into an octopus that attaches its tentacles to the wall and pulls itself along. The result is that we have the fish (and water due to viscous drag) rotating one way and a resulting torque tending to rotate the ball the other.

A.T.

Science Advisor
...the fish crawls with its fins along the edge of the glass, applying tangential force directly to the glass via the friction with its fins. the “normal force” holding the fish against the glass in this case would the fish’s forward velocity combined with the curvature of the glass...
It could also fix itself to the glass by suction, and use the fin to push water backwards. Both are variants of the flywheel method, with the water being the flywheel.

jbriggs444

Science Advisor
Homework Helper
Suppose the fish weighs 5kg and the water and ball weigh 5kg, and fish floats to the “side” of the ball (90 degrees from the ball’s contact point with the ground) where it uses its tail fin to push off the edge of the glass with an energetic impulse of 250 joules. When it reaches the other side it pushes with 500 joules. At the other side again 1000 joules. The energy of the impulse doubles with each push off the opposite sides.

To calculate the motion of the fish and the ball after the first interaction can I use:

m1v1 = m2v2

where m1 is the mass of the fish and m2 is the mass of the ball and water?
If the water plus fish fill the ball then, when the fish pushes off, water fills the void he leaves behind. The total momentum of fish plus water remains constant and zero. Accordingly the momentum of the ball is zero.

Doubling zero many times still leaves zero.

metastable

If the water plus fish fill the ball then, when the fish pushes off, water fills the void he leaves behind. The total momentum of fish plus water remains constant and zero. Accordingly the momentum of the ball is zero.
So if the ball is submerged in an aquarium, with a fish inside the ball and a fish outside the ball...

The fish outside the ball pushes its tailfin off the ball deriving an impulse in one direction, and the ball moves the opposite direction.

Scenario 2 is when the fish inside the ball pushes its tail off the glass with the same amount of force. If I understand correctly you are saying in Scenario 2, the ball won’t translate by even a millimeter at any point in time.

jbriggs444

Science Advisor
Homework Helper
If I understand correctly you are saying in Scenario 2, the ball won’t translate by even a millimeter at any point in time.
Correct.

metastable

So the frontal area of the fish, its drag coefficient and fluid density of the water its submerged in has no bearing? I thought if we were to reduce the frontal area and drag coefficient of the fish—

then fish (as opposed to the water) retains more kinetic energy after the fish travels X distance, after it pushes off from the side of the ball with its tail fin.

DaveC426913

Gold Member
Since the fish is in front of the contact patch, then the center of mass of the ball, fish, and water is also in front of the contact patch....
Woah.
We assumed the fish has the same density as water.
So CoM is centre of ball.

[ edit: OK, I'm late to the table ]

Immelmann

The ball, water, and fish are a closed system. What external force do you think there is that would make the ball move?
do not know. why do you think asked the question in the first place?

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phinds

Science Advisor
Gold Member
do not know. why do you think asked the question in the first place?
I was trying to get you to think.

Immelmann

not here to think buddy, just get an answer

Immelmann

thanx guys. did not think this was such an involved question. lol. you all went over my head with your techo-language! had to go google some of those words! lol

phinds

Science Advisor
Gold Member
not here to think buddy, just get an answer
Ah. Then you have come to the wrong forum. We're not really a Q&A type forum. We take it to be more helpful to teach people how to think on their own rather than spoon feeding them answers. If you really want to just ask a question instead of learning how to figure it out for yourself, there are Q&A forums that will be more effective for you.

Here is a perfect example of what you can expect on this forum:
https://www.physicsforums.com/threads/question-about-pulleys.975458/

Homework Helper

phinds

Science Advisor
Gold Member
And what does that have to do with this thread? This thread is about a full sphere, not a half-full one

Anachronist

Gold Member
This is basically the same problem as "If a bunch of birds are sitting in a closed container and then start flying around, does the weight of the container change?"

Want to reply to this thread?

"Can the fish move the ball?"

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