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Immelmann
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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
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.Immelmann said:not here to think buddy, just get an answer
And what does that have to do with this thread? This thread is about a full sphere, not a half-full one.Scott said:
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.metastable said: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.
No, it's not.Anachronist said: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?"
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.metastable said: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?
Ahhh. I interpreted "filled" as "filled normally" vs. "filled completely".phinds said:And what does that have to do with this thread? This thread is about a full sphere, not a half-full one
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.jbriggs444 said: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.
A.T. said: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.
metastable said: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...
The water, of course. You pushed off against the wall. The wall rigidly encloses the water.metastable said: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 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.(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)
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)...jbriggs444 said: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.
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 said: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)...
It is not clear what point you are trying to make.metastable said:[nothing but a drawing]
jbriggs444 said:It is not clear what point you are trying to make.
Correct.metastable said:I believe you are saying the depiction on the left is not possible if the fish pushes off the glass.
That one is fine. The ball never moves. Not during the push off. Not during the coasting phase. Not at impact. Never.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.
There is a hole through which the fish escapes. You have accounted for that momentum flux.Is the scenario on the bottom right valid?
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 said:That one is fine. The ball never moves.
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.metastable said:What step in this logic is flawed?
Don't just say it. Show it. Hint: you cannot.metastable said:the water mass is insignificant.
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 said:Snip Rube Goldberg scenario.
If the arm extends outside the tank then you have an external force. At this point, your responses are going well past tiresome and silly into the arena of obnoxious.metastable said:The batteries and motor are within the fish and equivalent to the fish's muscles.
Nobody says it is. See post #5 and #103. They show how to do this with a closed rigid container without changing the center of mass relative to the container.metastable said:Changing the center of mass within the tank isn't necessary...
metastable said:...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 said: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.
It's already been acknowledged that the fin action on the right circled in red (applying tangential force directly to the glass via the friction with its fins) causes the ball to roll.A.T. said:Nobody says it is. See post #5 and #103. They show how to do this with a closed rigid container without changing the center of mass relative to the container.
But your robotic arm violates the closed rigid container condition, so it's trivial.