Can a Fish Move a Ball in Water?

[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]

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.]

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.]

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]

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 http://discovermagazine.com/2003/mar/featscienceof 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

 

[Dale]

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.]

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]

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]

...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.]

...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]

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]

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]

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?

 

[phinds]

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]

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/

 

[.Scott]

 

[phinds]

View attachment 247458

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

 

[Anachronist]

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?"

 

[georgir]

Thread has gone somewhat tl;dr but the fish can make part of the water rotate in one direction and the rest of the whole ball rotate in another direction, yes. So it can make the bowl roll.

 

[metastable]

One last idea... The fish is simplified to a powered (both extension and contraction) 2-part telescoping rod of constant volume and same density as water. The fish starts in a contracted state against the “side” of the glass ball 90 degrees from the ball’s contact point with the ground. The fish has constant mass but adjustable frontal area, drag coefficient, and the water’s fluid density is adjustable as well (in different scenarios). A list of the bodies involved are: the glass, the water, the front part of the rod, the back part of the rod and the ground. At 0 time, the back part of the rod, the glass and the water comprise mass #1. The front part of the fish is mass #2. During powered extension of the rod utilizing a certain amount of energy/work against the glass, the front part of the rod acquires kinetic energy relative to the ground. The change in kinetic energy over time of the front part of the fish relative to the ground depends on the front part and rear part of the fish’s mass, acquired velocity relative to ground, frontal area, drag coefficient and the fluid density. If the frontal area and drag coefficient are arbitrarily small, an arbitrarily small amount of energy over time (power) is transferred to the water. Since the front part of the fish acquires kinetic energy relative to ground, and the rear part is extending directly against the glass, as mentioned before rear part of rod, glass and water are M1 and front part of rod is M2. Since M2 acquires momentum in the rod extension M1 acquires momentum in the opposite direction as well via:

$M_1V_1=M_2V_2$

Since the ball ($M_1$) has friction with a point on the ground, the ball can’t slide it must roll instead. If the ball were resting in a tiny dimple at the top of a ramp, perhaps this small initial inpulse might be enough to intiate a roll down the ramp where at the bottom a rocky outcrop beside the ocean offers the fish a potential route of escape from the ball...

 

[jbriggs444]

One last idea... The fish is simplified to a powered (both extension and contraction) 2-part telescoping rod of constant volume and same density as water.

If the fish changes shape and/or position, the water is constrained to do so as well. The center of mass of the two together does not change. Accordingly, the total momentum of the two together does not change. All of the invocations of kinetic energy and of momentum and all of the equations you may write down do not change this.

Let us formalize a proof.

The ball is full of water with a fish in it. No air bubbles. No voids. The water and fish are of uniform and equal density. The ball is spherically symmetric and also of uniform density. Assume that the ball starts at rest on the floor with the contents also at rest.

If the ball does not leak, it is clear that the center of mass of the fish plus water is always located at the center of the ball. The center of mass of the ball alone is also always located at the center of the ball.

Let $v_b$ be the velocity of the ball and $v_c$ be the velocity of the center of mass of the contents. The fact that they are always co-located allows us to write down an equation:
$$v_b=v_c$$
Suppose that there is no net external force on the ball plus contents. Then momentum of the ball+contents system is conserved.
$$m_b v_b+m_cv_c = 0$$
But since $v_b$ and $v_c$ are equal, we can substitute in one for the other yielding
$$m_b v_b + m_c v_b = 0$$
$$(m_b + m_c) v_b = 0$$
Unless both ball and contents are massless, it follows that $v_b = 0$ and then that $v_c = 0$.

Conclusion: The only way the ball moves is if you can contrive to get an external force to move it. This is do-able as has been described earlier.

 

[metastable]

Isn’t the force M1 transfers to the water defined by $F=(1/2)C_d*rho*A_f*V^2$ meaning you can adjust it by reducing the frontal area and/or drag coefficient?

 

[Vanadium 50]

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?"

No, it's not.

 

[jbriggs444]

Isn’t the force M1 transfers to the water defined by $F=(1/2)C_d*rho*A_f*V^2$ meaning you can adjust it by reducing the frontal area and/or drag coefficient?

It's all a connected system. Every force you can think of between fish, ball and water is internal to the system. You cannot get a net input of momentum using internal forces. That's Newton's third law in action.

 

[.Scott]

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

Ahhh. I interpreted "filled" as "filled normally" vs. "filled completely".

That makes it much more difficult. The fish can swim in circles - leaving the water to flow in the opposite direction. The friction between the water and the bowl would push the ball forward. But when the fish stopped, the ball would stop rolling as well - and it would brake to a stop.

Angular momentum must be conserved - but not the angular position. So a displacement in the position of the ball is possible.

 

[metastable]

It's all a connected system. Every force you can think of between fish, ball and water is internal to the system. You cannot get a net input of momentum using internal forces. That's Newton's third law in action.

In the system as previously described, if the telescoping rod pushes with enough work/energy, with low enough frontal area and drag coefficient, there are scenarios in which it can "coast" across the entire glass ball and bump into the other wall after pushing off from the opposite side.

If the ball doesn't roll, and the fish gets enough momentum relative to the ground pushing off the wall to coast horizontally to the other side of the ball, and the fish didn't push off the ground, then what mass acquired momentum in the opposite direction to satisfy $M_1V_1=M_2V_2$?

(I specify that during the coasting phase, all of the fish's kinetic energy relative to ground is not transferred to the water or ball until it bumps into the other side)

 

[eltodesukane]

If the fish has the same mean density as water, the hamster method will not work. But theoretically it could induce a flow that could spin the ball via friction on the inner walls, resulting in rolling.

Just like a kid on a swing can go from rest to max swing just by moving his legs.

 

[metastable]

Since the front part of the fish acquires kinetic energy relative to ground, and the rear part is extending directly against the glass, as mentioned before rear part of rod, glass and water are M1 and front part of rod is M2. Since M2 acquires momentum in the rod extension M1 acquires momentum in the opposite direction as well via:

$M_1V_1=M_2V_2$

Since the ball ($M_1$) has friction with a point on the ground, the ball can’t slide it must roll instead. If the ball were resting in a tiny dimple at the top of a ramp, perhaps this small initial inpulse might be enough to intiate a roll down the ramp where at the bottom a rocky outcrop beside the ocean offers the fish a potential route of escape from the ball...

 

[jbriggs444]

If the ball doesn't roll, and the fish gets enough momentum relative to the ground pushing off the wall to coast horizontally to the other side of the ball, and the fish didn't push off the ground, then what mass acquired momentum in the opposite direction to satisfy $M_1V_1=M_2V_2$?

The water, of course. You pushed off against the wall. The wall rigidly encloses the water.

(I specify that during the coasting phase, all of the fish's kinetic energy relative to ground is not transferred to the water or ball until it bumps into the other side)

The fish's momentum is transferred immediately to ball and water. It is equal and opposite to the sum of the other two at all times. Kinetic energy is not a conserved quantity. It does not have to be transferred anywhere.

 

[metastable]

The fish's momentum is transferred immediately to ball and water. It is equal and opposite to the sum of the other two at all times.

Im having a great difficulty distinguishing any difference in the following 2 scenarios (except that the fish will run into the wall after a certain amount of time in the scenario on the left)...

(In both scenarios the fish uses their tail fin to push directly off the glass)

 

[jbriggs444]

Im having a great difficulty distinguishing any difference in the following 2 scenarios (except that the fish will run into the wall after a certain amount of time in the scenario on the left)...

The scenario on the left is incorrect because the fish cannot make the ball move by pushing on it. All it can succeed in doing is making the water within the ball flow backward just enough to match its forward momentum.

 

[metastable]

 

[cmb]

Surely it is easier to ask whether a fish can be made to move in a ball of water by spinning the ball?

I reckon if you roll a ball along the ground full of water , the water and the fish would start spinning. Ergo, the fish just has to swim to oppose that motion, which I am sure they can do, and if so then that would have the counter effect and decelerate the ball, i.e. can move it.

I think the transfer function of momentum between the fish's swimming strokes and ball would be quite limited though (through viscous effects between water and ball), they'd probably get exhausted quickly to get anywhere and if it was filled up completely to avoid the question of a half-filled ball, then they'd suffocate without a supply of re-oxygenated water.

So I'd say, yes ... but don't expect your fish to survive a roll around the garden with you.

 

[jbriggs444]

[nothing but a drawing]

It is not clear what point you are trying to make.

 

[metastable]

It is not clear what point you are trying to make.

I believe this one may be more clear.

I believe you are saying the depiction on the left is not possible if the fish pushes off the glass (all fish water balls are floating in water in the following diagram). So what about the two depictions on the right? On top, the fish pushes off the glass, the fish accelerates, the glass doesn't move at all during the coasting phase of the fish. The energy transferred to the water by the fish per second can be stated (where rho is the fluid density of the water) $W=(1/2)*C_d*rho*A_f*V^3$, and the fish retains some of the kinetic energy it acquired in the push when it hits the opposite wall. On the bottom right, The energy transferred to the water by the fish per second can also be stated (where rho is the fluid density of the water) $W=(1/2)*C_d*rho*A_f*V^3$, but there is a hole for the fish to escape and so the glass ball acquires momentum via $M_1V_1=M_2V_2$.

Is the scenario on the bottom right valid?

 

[jbriggs444]

I believe you are saying the depiction on the left is not possible if the fish pushes off the glass.

Correct.

On top, the fish pushes off the glass, the fish accelerates, the glass doesn't move at all during the coasting phase of the fish.

That one is fine. The ball never moves. Not during the push off. Not during the coasting phase. Not at impact. Never.

Is the scenario on the bottom right valid?

There is a hole through which the fish escapes. You have accounted for that momentum flux.

Through that hole water enters. You have not accounted for that momentum flux.

In the external environment there may be pressure changes associated with the circulation that accompanies the motion of fish and displaced water. You have not accounted for that momentum flux.

 

[metastable]

That one is fine. The ball never moves.

In the depiction on the top right the ball is floating freely except for viscous forces and it isn't connected to the ground. The fish acquires momentum pushing off the glass (not from directly pushing off the water molecules as in typical fish swimming). The fish arrives at the opposite side retaining most of the kinetic energy it had immediately after separation of contact with the glass. Assuming the fish is almost as long as the ball is wide, only a tiny amount of water is displaced on the fish's journey to the opposite side. Since the fish's kinetic energy isn't converted entirely to motion in the water before it hits the other side, then the kinetic energy of the reaction force from the push must have gone somewhere else. The only other place the energy can go is the glass, which can then transfer to the bulk water since there are no voids and water is considered incompressible. Since the glass isn't rigidly attached to the ground it can move. What step in this logic is flawed?

 

[jbriggs444]

What step in this logic is flawed?

All of it. Kinetic energy is not a conserved quantity. You should be working with conserved quantities if you want to reason about where a quantity has to go. Momentum is a conserved quantity.

You also need to be quantitative. "only a tiny amount of water" is not quantitative. If you were working with momentum, it would be obvious that the momentum of the water and of the fish are equal and opposite.

 

[metastable]

So if the glass has 50% of the mass, the fish has 49.9% of the mass, the water has 0.1%, the fish is long enough that it only moves 0.1% of the ball width to get to the other side (& the water mass is insignificant). The mass of the fish acquires momentum relative to the glass during the push. Suppose the fish has a hole through it so the displacement of the water as it travels through the fish is insignificant compared to the displacement of the fish. So if the fish acquires momentum, and the water mass and displacement during the push is insignificant, what else acquires momentum to satisfy $M_1V_1=M_2V_2$ if not the glass (until the fish reaches the other side)?

 

[jbriggs444]

the water mass is insignificant.

Don't just say it. Show it. Hint: you cannot.

 

[metastable]

I struggle to see how it's any different from the following scenario:

2 flatbed train cars, one with a open topped fish aquarium secured to the deck, the other has a crane arm, with a robotic water density fish on a water density pole that can reach over and position the tail of the fish against a side of the aquarium on the other carriage, then the crane arm is locked in position on the other carriage. A special cover and seal on the top of the aquarium ensures the robotic arm can move through the water without significantly displacing the water surface. The robotic fish delivers a push on the side of the aquarium on the other car. Both cars move briefly in opposite directions until the fish crashes into the opposite side of the aquarium. During the time both cars are in motion, why is it not equivalent to the fish pushing directly off the wall inside the ball causing the ball to roll a brief and limited distance in the opposite direction as the motion of the fish?

 

[jbriggs444]

Snip Rube Goldberg scenario.

You have a robotic arm applying a force to a fish in a tank and you do not see that this is an external force?!

 

[metastable]

The batteries and motor are within the fish and equivalent to the fish's muscles.