Can a Fish Move a Ball in Water?

[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.