B Wet wheel and conservation of momentum

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A cyclist coasts along a road, he drives across a small puddle of water, after which the wheels leave wet lines on the road.

Now we concentrate our attention to the linear momentum of the water on a wheel. It decreases. Momentum is conserved, so what got the momentum that the water had?
 

Nugatory

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A cyclist coasts along a road, he drives across a small puddle of water, after which the wheels leave wet lines on the road.

Now we concentrate our attention to the linear momentum of the water on a wheel. It decreases. Momentum is conserved, so what got the momentum that the water had?
The earth. Same general situation as your recent question about the magic potato planting device.
 
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Now we concentrate our attention to the linear momentum of the water on a wheel. It decreases.
Why do you say that? The part of the tire touching the ground has zero momentum so the water on it also has zero momentum and continues to have zero momentum. It doesn’t change at all that I can see.
 

sophiecentaur

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The back of the wheel will have an initial vertical acceleration as it leaves the ground and it will follow a cycloid curve. So will the water. Vertical / all momentum is conserved as ever and the Earth is part of the system.
 
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The earth. Same general situation as your recent question about the magic potato planting device.

I think that would not conserve energy.

When cyclist applies brakes, bike's momentum becomes earth's momentum and bike's kinetic energy becomes heat energy in brakes. It's an inelastic collision.

Maybe you can tell me where the kinetic energy of the water goes, when the momentum of the water supposedly goes to earth?
 
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Why do you say that? The part of the tire touching the ground has zero momentum so the water on it also has zero momentum and continues to have zero momentum. It doesn’t change at all that I can see.
Less water on the wheel - less momentum of water on the wheel.
 
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The back of the wheel will have an initial vertical acceleration as it leaves the ground and it will follow a cycloid curve. So will the water. Vertical / all momentum is conserved as ever and the Earth is part of the system.
Yeah, but I would be interested to hear the answer.
 

Nugatory

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Maybe you can tell me where the kinetic energy of the water goes, when the momentum of the water supposedly goes to earth?
If the momentum of the earth changes, the kinetic energy of the earth also changes. Any loss of kinetic energy of the entire ball/water/bicycle system in an inelastic interaction will show up as heat (and conservation of momentum is how we calculate the post-interaction speeds to see how much kinetic energy went into heat).

We need to back up and consider two more basic cases.
1) I throw a ball (mass ##m_B##, velocity ##v_0##) at a wall. The collision is elastic, so the ball rebounds with no loss of kinetic energy, and the ball’s momentum changes as the ball reverses direction.
2) I throw the same ball at the same wall. This time the collision is completely inelastic so that the ball sticks to the wall.

In both cases, momentum is conserved and this requires that the earth change speed slightly. It’s worth calculating the post-collision speed of the earth in both cases (but when you do, use the symbol ##m_E## for the earth’s mass instead the numerical value - easier to appreciate the underlying physics that way).
 

sophiecentaur

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Yeah, but I would be interested to hear the answer.
I have read answers in several places in the thread. What terms do you actually want the answer in? Water goes one way, Earth is pushed in the other - via the bike wheel.
If you can accept the conservation of momentum then it all follows. But Energy is not conserved and you have to bear that in mind.
 
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Subtracting 0 momentum doesn’t give you less momentum.
Yeah, but I still think that less water means less momentum of water.

If all water has left, is there any momentum of water left? No.

If the water left some momentum behind when it left, that momentum is not "momentum of water" anymore.
 

sophiecentaur

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Yeah, but I still think that less water means less momentum of water.

If all water has left, is there any momentum of water left? No.

If the water left some momentum behind when it left, that momentum is not "momentum of water" anymore.
That's sort of true but how relevant is it? Of course each drop of water will have a different amount (and direction) of momentum; some will barely lift off the ground and some could shoot right over your head.
 
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In both cases, momentum is conserved and this requires that the earth change speed slightly. It’s worth calculating the post-collision speed of the earth in both cases (but when you do, use the symbol mEm_E for the earth’s mass instead the numerical value - easier to appreciate the underlying physics that way).
subscript w means water, subscript e means earth, r is water/earth mass ratio, p is momentum, v is velocity, E is kinetic energy


##\frac{E_w}{E_e} = \frac{p_w * v_w}{p_e * v_e} ##

Now we substitute earth's momentum by water's momentum and earth's velocity by r times water's velocity.

## \frac{p_w * v_w}{ p_w * r * v_w} = \frac{1}{r} ##

The kinetic energy of earth is a tiny fraction of the kinetic energy that the water had.
 

Nugatory

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The kinetic energy of earth is a tiny fraction of the kinetic energy that the water had.
Are you now considering the completely inelastic case in which the water ends up at rest relative to the earth? This is the same situation as a thrown ball hitting a wall and sticking, and as your potato-planting device from your earlier thread. Because the collision is inelastic, kinetic energy is not conserved and the "missing" kinetic energy ends up as waste heat one way or another.

And you are right that the earth gains very little kinetic energy in these situations; that's because the change in the earth's speed is very small and the kinetic energy goes as the square of that so is even smaller. Another way of saying this is that we can approximate the completely inelastic collision in which the ball ends up stuck to the wall by saying that the change in the earth's speed and momentum is zero and all of the initial kinetic energy ends up as heat.
 
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Yeah, but I still think that less water means less momentum of water.
How can you subtract 0 and get a nonzero change? Math still works for water on bicycle wheels.

If all water has left, is there any momentum of water left? No.
Sure, but it cannot have left in the way you are claiming that it left. So how did it leave?
 
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Sure, but it cannot have left in the way you are claiming that it left. So how did it leave?
A small patch of wet tyre touches the dry road, some water moves from the patch to the road. That was the idea - or claim.

If that is not possible, then what is @Nugatory talking about?
 
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sophiecentaur

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the problem needs to be specified better else we have to assume the footprint and the water starts off stationary in contact with the ground. It’s a rolling wheel situation. No momentum when in contact.
What is the precise question?
 

Nugatory

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A small patch of wet tyre touches the dry road, some water moves from the patch to the road. That was the idea.

If that is not possible, then what is @Nugatory talking about?
This is why I’ve been trying to get you to properly understand the less complicated cases first, so that you can see all the implications of momentum conservation within the entire system. But if you do want to grind through this more complicated problem.... what happens to the wheel if there is water adding weight to the leading edge but not the trailing edge (because it leaves the wheel for the ground)? What assumptions have you been making that are invalidated by this effect?
 
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A small patch of wet tyre touches the dry road, some water moves from the patch to the road. That was the idea - or claim.
The loss of momentum doesn’t happen there. Can you see where it happens?
 
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The loss of momentum doesn’t happen there. Can you see where it happens?
Between the highest position, where the patch has the maximum momentum, and the lowest position, where the patch has the minimum momentum.
 
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This is why I’ve been trying to get you to properly understand the less complicated cases first, so that you can see all the implications of momentum conservation within the entire system. But if you do want to grind through this more complicated problem.... what happens to the wheel if there is water adding weight to the leading edge but not the trailing edge (because it leaves the wheel for the ground)? What assumptions have you been making that are invalidated by this effect?

Well that causes a torque on the wheel, which causes the bike to gain some speed and momentum, and the earth to gain opposite momentum. Didn't think about that ... but doesn't invalidate any assumptions.
 

Nugatory

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Well that causes a torque on the wheel, which causes the bike to gain some speed and momentum, and the earth to gain opposite momentum. Didn't think about that ... but doesn't invalidate any assumptions.
It invalidates the assumption that the bike is coasting, at least as the word is usually understood: "not exchanging momentum with the earth" or equivalent. That's the assumption that I had in mind.

But are you now at the point where you have the answer to the question you started the thread with, namely "what got the momentum that the water had?" One way or another the total momentum is conserved, total energy is conserved although some kinetic energy may disapear as heat, and if you construct a complicated enough problem it may take a while to spot the mechanism by which momentum is transferred between earth and bicycle.
 
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Between the highest position, where the patch has the maximum momentum, and the lowest position, where the patch has the minimum momentum.
The forces there are internal to the water+bicycle system, so they cannot affect the momentum of the system.
 
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The forces there are internal to the water+bicycle system, so they cannot affect the momentum of the system.
So? Did I claim something about momentum of the system?
 

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