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Physics question...propulsion in a swimming pool

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Old 07-02-2012, 06:33 PM   #1
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Physics question...propulsion in a swimming pool

Question about jet propulsion...


So over the weekend, we were floating around drunk in my swimming pool arguing over stupid stuff. Well, we starting talking about physics and it got ugly - even had my donut raft flipped over in rage!

Let me paint the picture for you...

You are sitting on a raft in the center of the pool - completely out of the water. You are holding a foam water blaster:

You suck water in from your right side and shoot the water back into the pool on the same right side.
Will it move your raft in the opposite direction (to the left)?

Would it change the outcome if you:
A) fired with the tip of the blaster IN or OUT of the water?
B) fired with the tip of the blaster OUT of the water but hit the pool side with the stream of water?
C) sucked water in from the left side and shot it out on the right side?
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Old 07-02-2012, 06:37 PM   #3
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not a whole lot to do there eh?
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Old 07-02-2012, 06:39 PM   #4
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Old 07-02-2012, 06:43 PM   #5
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You're talking such minor amounts of energy that you wouldn't notice a difference no matter what you did. If you could do it in rapid succession, then you'd actually have some measurable propulsion.
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Old 07-02-2012, 06:46 PM   #6
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Quote:
Originally Posted by Pugga View Post
You're talking such minor amounts of energy that you wouldn't notice a difference no matter what you did. If you could do it in rapid succession, then you'd actually have some measurable propulsion.
ive discovered pugga is generally non bias and smart when it comes to smart things. therefor, i agree with that ^^

but from a normal stupid person stance. . . i'd think the most effective way out of your options is the one where you suck it up on one side and blow it out the other underwater. . . . . that sounds wrong. . .anyway, most of that energy isn't going to transfer past your wrists, so it's pointless really. oh yeah, pugga already said that
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Old 07-02-2012, 06:47 PM   #7
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LOL you sure you're not still drunk?

Shooting out of either side will propel you in the opposite direction. If you keep your arm still enough, you'll get more propulsion out of hitting the side of the pool, because equal and opposite reaction of water hitting concrete/brick is more concentrated and direct than the water in the gun just shooting and mixing with the rest of the water in the pool.

BTW, doesn't matter what side you take the water from. That thing can only hold so many ounces of water and takes it in at such a slow rate that it won't make a difference.
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Old 07-02-2012, 06:50 PM   #8
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same basic principle as the waterjet used for propulsion in personal watercraft... newton laws of motion, more specifically bernoule's equation, ...thrust is generated in a direction opposite the direction of the exhaust stream..

..so, if the velocity and flow rate of the stream is the same for the two scenarios (discharged under water), and (discharged above water), ...the thrust should be the same... one thing to note, ...the pressure on the exhaust side, would technically be higher underwater than above (hydrostatic pressure vs. atmospheric pressure above the surface), this would affect the velocity/flow ..so you might argue a slight advantage if the discharge is above water...
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Old 07-02-2012, 06:54 PM   #9
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I agree with the statement of minor amounts of energy, however to answer the question. You would stay in the same spot (sucking water in = 1 unit of negative energy, blowing it out = 1 unit of positive energy)
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Old 07-02-2012, 07:15 PM   #11
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Quote:
Originally Posted by spares View Post
I agree with the statement of minor amounts of energy, however to answer the question. You would stay in the same spot (sucking water in = 1 unit of negative energy, blowing it out = 1 unit of positive energy)
Not necessarily true, spares. If you sucked water in at a fixed rate and then expelled it with much greater force, you would travel in a direction opposite of that greater force. This crudely describes the principle behind several propulsion systems (jet engines, boat propellers, etc). Of course, his foam water blaster could not generate enough force to significantly propel anything as large as a person (at least not without a whole lot of very fast pumping).
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Old 07-02-2012, 07:19 PM   #12
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For some reason I overlooked the whole adding energy from the person... I will blame it on Canada Day and the equal and opposite reaction that comes with the celebrations
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Old 07-02-2012, 07:21 PM   #13
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air has density, therefor the moving water in the air, will cause you to move slightly.
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Old 07-02-2012, 07:27 PM   #14
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You don't tell us how fast you squirt it out.

It's a physics problem - so let's look at the relevant physics laws.

A body at rest will tend to remain at rest, unless acted upon by a Force.

Force = Mass * Acceleration

In any closed system, momentum is conserved.

Momentum = mass * velocity.

Total Momentum at time 0: zero. (velocity of raft, water, all zero)

At time 1 - the water is "slurped."
Total Momentum at time 1: still zero. (momentum is conserved!)

0 = mass(raft) * velocity(raft towards edge of pool) - mass(slurped water) * velocity(slurped water)
OR
velocity1(raft) = (mass(slurped water)/mass(raft)) * velocity(slurped water)
HOWEVER - since the "slurped" water is stationary at the time it is slurped (the speed of the plunger is irrelevant!) V1 (raft towards the edge of the pool) remains zero.

At time 2 - the water is "squirted."

Total Momentum at time 2: still zero. (momentum is conserved.)

0 = mass(raft) * velocity(raft) - mass(squirted water) * velocity(squirted water)
OR:
velocity(raft) = (mass(squirted water)/mass(raft)) * velocity(squirted water)

However, at this time, the raft velocity is now greater than zero, away from the edge of the pool.

What happens to the water after it leaves the squirter is irrelevant. (It doesn't need the edge of the pool to "push against;" that's a layman's fallacy.)

If the squirter is off-center from the raft when squirted, the raft will have be less linear velocity and more angular velocity. But that's a problem for another day.
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Old 07-02-2012, 07:31 PM   #15
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One more thing to think about - when you load a gun, there's no recoil from the act of picking up and loading. Duh right?

But why is that?

It's because the initial velocity of the bullet when you pick it up off the table is zero.
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Old 07-02-2012, 07:47 PM   #16
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Quote:
Originally Posted by bjmoose View Post
One more thing to think about - when you load a gun, there's no recoil from the act of picking up and loading. Duh right?

But why is that?

It's because the initial velocity of the bullet when you pick it up off the table is zero.
when you pick up the gun you exert a force on it, ...and it exerts the same force on you (newton's 3rd law)..

..the "recoil" felt from firing the gun is basically the same thing, ...but the impulsive force resulting from accelerating the projectile and the burning powder exiting the barrel is much higher, albeit shorter in duration...
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Old 07-02-2012, 07:49 PM   #17
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Quote:
Originally Posted by wileyC View Post
same basic principle as the waterjet used for propulsion in personal watercraft... newton laws of motion, more specifically bernoule's equation, ...thrust is generated in a direction opposite the direction of the exhaust stream..

..so, if the velocity and flow rate of the stream is the same for the two scenarios (discharged under water), and (discharged above water), ...the thrust should be the same... one thing to note, ...the pressure on the exhaust side, would technically be higher underwater than above (hydrostatic pressure vs. atmospheric pressure above the surface), this would affect the velocity/flow ..so you might argue a slight advantage if the discharge is above water...
I must disagree with you there. It would be the opposite. If the squirter were discharged above the water, there would be LESS advantage, due to resistance. Picture this, if you push your arms in front of you, and there is only air, the resistance is effectively zero. Do this under water, and you move in the opposite direction (this is how we can swim) If there were no resistance, we couldn't move, because we have nothing to press against.

Quote:
Originally Posted by MXscott14 View Post
air has density, therefor the moving water in the air, will cause you to move slightly.
The density of air is minimal compared to the density of water. This again means that the resistance is greater when the squirter would be discharged under water. The reason that there is increased movement of the raft this way is because the raft is floating on TOP of the water. The resistance to movement would only be that of the "drag" of the raft moving across the water.
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Old 07-02-2012, 07:53 PM   #18
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Quote:
Originally Posted by wileyC View Post
when you pick up the gun you exert a force on it, ...and it exerts the same force on you (newton's 3rd law)..
I think we're saying the same thing here. Since it's initial momentum is zero, and your initial momentum is zero - the momentum of both after you pick it up remains zero.

Quote:
..the "recoil" felt from firing the gun is basically the same thing, ...but the impulsive force resulting from accelerating the projectile and the burning powder exiting the barrel is much higher, albeit shorter in duration...
Right. But to solve the problem, we need not concern ourselves with the "Force" involved at all. We can calculate the final velocity of either the bullet or the shooter experiencing the recoil (assuming he's wearing roller skates) by knowing the ratio of their masses and the final velocity of one or the other.
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Old 07-02-2012, 08:02 PM   #19
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Quote:
Originally Posted by BamaToy1997 View Post
I must disagree with you there. It would be the opposite. If the squirter were discharged above the water, there would be LESS advantage, due to resistance. Picture this, if you push your arms in front of you, and there is only air, the resistance is effectively zero. Do this under water, and you move in the opposite direction (this is how we can swim) If there were no resistance, we couldn't move, because we have nothing to press against.



The density of air is minimal compared to the density of water. This again means that the resistance is greater when the squirter would be discharged under water. The reason that there is increased movement of the raft this way is because the raft is floating on TOP of the water. The resistance to movement would only be that of the "drag" of the raft moving across the water.
from a fluid dynamics perspective, ...i.e. control volume w/ mass flows in/out, ..the velocity and flowrate, all other things the same, is all that matters... ever see that new jet suit, that sucks up water from the surface and pumps it up to a jetpack on the guys back thrusting him upward? could extenuating things alter the velocity and flowrate - sure, ...if for instance the difference in the pressure on the exhaust side (which is why boat engines can produce more power by dumping their exhaust above water rather than below, i.e. "through-hull exhaust)...
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Old 07-02-2012, 08:09 PM   #20
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Quote:
Originally Posted by bjmoose View Post
I think we're saying the same thing here. Since it's initial momentum is zero, and your initial momentum is zero - the momentum of both after you pick it up remains zero.



Right. But to solve the problem, we need not concern ourselves with the "Force" involved at all. We can calculate the final velocity of either the bullet or the shooter experiencing the recoil (assuming he's wearing roller skates) by knowing the ratio of their masses and the final velocity of one or the other.
yes, i see what you are saying, ...newton's laws serve as an abstract for these behaviors (except for parasitic losses, heat exchange, etc..), ...but you'll find that fluid dynamics, and, more broadly, thermodynamics give more insight into what's going on and why... actually the laws of thermodynamics, especially the 1st law, expressed in long form for control volumes, quite perfectly envelops all that goes on in the physical world...
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