Wet wheel and conservation of momentum

In summary, the conversation discusses the concept of conservation of momentum and its relation to kinetic energy in a cyclist riding over a puddle of water. It is concluded that in an inelastic collision, the kinetic energy of the water is converted into heat and the momentum of the water is transferred to the Earth, resulting in a slight change in the Earth's speed. This is similar to the case of a thrown ball hitting a wall and sticking, as well as the potato-planting device mentioned earlier.
  • #1
jartsa
1,576
137
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?
 
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  • #2
jartsa said:
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.
 
  • #3
jartsa said:
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.
 
  • #4
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.
 
  • #5
Nugatory said:
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?
 
  • #6
Dale said:
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.
 
  • #7
jartsa said:
Less water on the wheel - less momentum of water on the wheel.
Subtracting 0 momentum doesn’t give you less momentum.
 
  • #8
sophiecentaur said:
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.
 
  • #9
jartsa said:
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).
 
  • #10
jartsa said:
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.
 
  • #11
Dale said:
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.
 
  • #12
jartsa said:
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.
 
  • #13
Nugatory said:
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.
 
  • #14
jartsa said:
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.
 
  • #15
jartsa said:
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.

jartsa said:
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?
 
  • #16
Dale said:
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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  • #17
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?
 
  • #18
jartsa said:
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?
 
  • #19
jartsa said:
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?
 
  • #20
Dale said:
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.
 
  • #21
Nugatory said:
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.
 
  • #22
jartsa said:
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.
 
  • #23
jartsa said:
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.
 
  • #24
Dale said:
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?
 
  • #25
jartsa said:
So? Did I claim something about momentum of the system?
That it is decreasing.
 
  • #26
Dale said:
That it is decreasing.

That I have not claimed. I have said that we concentrate on the water, and the momentum of water decreases.

Then I asked what gets the momentum that the water had.

Then I have been told that the Earth gets the momentum that the water had.
 
  • #27
jartsa said:
That I have not claimed. I have said that we concentrate on the water, and the momentum of water decreases.
Ah, my mistake. I thought the other question was the interesting one.

jartsa said:
Then I have been told that the Earth gets the momentum that the water had.
In the ideal case, during the time that the water interacts with the Earth its momentum does not change. It is only during the time that the water interacts with the wheel that its momentum changes. The thing that gets the water’s momentum should be clear.
 
  • #28
jartsa said:
Then I have been told that the Earth gets the momentum that the water had.
Yes but how does it get that momentum? It has to be because the wheel is slowing the water as the point on the wheel aims at the ground and any force on the water is coming from the Earth via the wheel. i.e. the water is not in contact with the Earth until it's placed there (at nearly zero velocity) when its portion of the periphery of the wheel gets to the bottom. There will be a step change in radial (centripetal) force on the wheel due to the rotating mass being reduced as the water loses contact with the wheel.
Imagine that we replace the water by a series of lead masses, attached around the periphery of the wheel. The masses become detached when they get to the road but they have zero momentum at that point because their velocity is instantaneously zero.
This is getting very turgid- and to what end, I wonder?
 
  • #29
sophiecentaur said:
Yes but how does it get that momentum? It has to be because the wheel is slowing the water as the point on the wheel aims at the ground and any force on the water is coming from the Earth via the wheel.
Just a small clarification: I do not believe that water's momentum goes to the earth.A cyclist riding a massless bike coasts slowly past a long table, picks a mug of water from the table, then puts the mug back nicely next to another mug.

The pick up event: mug's momentum increases, cyclist's momentum decreases the same amount. Earth's momentum does not change.

The put back event: mug's momentum decreases, cyclist's momentum increases the same amount. Earth's momentum does not change.
 
  • #30
jartsa said:
I do not believe that water's momentum goes to the earth.
If there's no friction between mug and table and you aren't requiring that the bicycle moves at a constant speed, then yes. That's a different setup with a different mechanism for transferring momentum between the various components of the system so can lead to different momentum transfers.
 
  • #31
Nugatory said:
If there's no friction between mug and table and you aren't requiring that the bicycle moves at a constant speed, then yes. That's a different setup with a different mechanism for transferring momentum between the various components of the system so can lead to different momentum transfers.
Well I guess I need to make some design changes:

A cyclist has a paint roller attached to the end of the handlebar, he drives past a wall onto which the roller paints a line.

Now I claim this bike-paint roller device is a kind of a rocket, the propellant is the paint, the momentum of the propellant goes to the rest of the rocket and nowhere else.
 
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  • #32
jartsa said:
Well I guess I need to make some design changes:

A cyclist has a paint roller attached to the end of the handlebar, he drives past a wall onto which the roller paints a line.

Now I claim this bike-paint roller device is a kind of rocket, the propellant is the paint, the momentum of the propellant goes to the rest of the rocket and nowhere else.
If I’m understanding you properly: the roller is loaded up with paint and at rest relative to the bicycle. It starts turning when it contacts the wall, and as it rolls it unloads its paint onto the wall to leave a line. Can we assume for simplicity that the roller itself is massless (so we don’t need to calculate its moment of inertia or mess with its rotational kinetic energy and it can spin up instantaneously) and that it rolls without slipping (so we don’t have to mess with friction) and that the paint doesn’t spatter?

With these assumptions, the interaction between the bicycle and the wall reduces the mass of the bicycle and increases the mass of the earth; the change in each is of course equal to the mass of the paint transferred from one to the other. The paint starts out moving at the initial speed of the bicycle and is slowed to the speed of the wall; therefore the wall exerts a force on the paint to decelerate it, and by Newton’s second law the paint exerts an equal and opposite force on the wall.

Working in a frame in which the wall is initially at rest: the speed of the bicycle does not change. The wall starts at rest and is very slightly accelerated. The paint starts out with some momentum because it’s moving at the same speed as the bicycle; this momentum is transferred to the wall with a line of paint on it.
 
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  • #33
Nugatory said:
If I’m understanding you properly: the roller is loaded up with paint and at rest relative to the bicycle. It starts turning when it contacts the wall, and as it rolls it unloads its paint onto the wall to leave a line. Can we assume for simplicity that the roller itself is massless (so we don’t need to calculate its moment of inertia or mess with its rotational kinetic energy and it can spin up instantaneously) and that it rolls without slipping (so we don’t have to mess with friction) and that the paint doesn’t spatter?

Yes those sound like good assumptions.We could also put motors, brakes, sensors, and computers on the roller, surely that technology can eliminate forces on the wall. ... No let's put the sensors on the wall, and a wireless communication system between the sensors and the paint roller.
The paint tool must be a roller, not a brush. Spray can works too, if correctly aimed. Brush does not exert a force on the paint, so paint has to exert a force on the wall.
 
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  • #34
@Dale seems to have hinted pretty strongly at the situation in #27. The elephant in the room is the mistaken assumption that the water on the wheel or the paint on the roller is uniformly distributed around the circumference of the device.

If the wheel is rolling and losing water/paint, this assumption is clearly counter-factual. There is less paint/water on the wheel behind the contact point than on the wheel in front of the contact point.

You can try to repair this problem by losing water very slowly or applying a very thin coat of paint. But that does not help. You reduce the imbalance and the rate at which momentum is transferred due to the imbalance. But you increase the length of time for which the imbalance is present.

As has been suggested, if you find yourself in a situation where conservation laws are being violated, you've made a mistake. You need to find that mistake.
 
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  • #35
Isn't this scenario exactly the same as picking up a bucket of water, moving it 2m to the right and putting it back down?
 
<h2>1. What is the Wet Wheel experiment and how does it demonstrate conservation of momentum?</h2><p>The Wet Wheel experiment is a simple physics demonstration where a spinning wheel is partially submerged in water. As the wheel spins, the water is pushed outwards due to the centrifugal force. This results in a change in the wheel's momentum, but the total momentum of the system (wheel + water) remains constant, demonstrating the principle of conservation of momentum.</p><h2>2. How does the Wet Wheel experiment relate to real-life situations?</h2><p>The Wet Wheel experiment is a simplified version of real-life situations where conservation of momentum is at play. For example, the motion of a spinning top, the recoil of a gun, or the movement of a rocket in space all follow the same principle of conservation of momentum.</p><h2>3. What factors affect the outcome of the Wet Wheel experiment?</h2><p>The outcome of the Wet Wheel experiment can be affected by several factors, including the speed and direction of the wheel's spin, the amount of water in the container, and the shape and size of the wheel. These factors can change the amount of force and momentum exerted on the water, resulting in different outcomes.</p><h2>4. How does the Wet Wheel experiment demonstrate the law of action and reaction?</h2><p>The Wet Wheel experiment also demonstrates the law of action and reaction, also known as Newton's Third Law of Motion. As the wheel pushes the water outwards, the water exerts an equal and opposite force on the wheel, causing it to spin in the opposite direction. This shows that for every action, there is an equal and opposite reaction.</p><h2>5. Can the Wet Wheel experiment be used to calculate the amount of momentum in a system?</h2><p>While the Wet Wheel experiment can demonstrate conservation of momentum, it is not a precise method for calculating the exact amount of momentum in a system. Other factors, such as friction and air resistance, can also affect the outcome of the experiment. To accurately calculate momentum, more precise methods and measurements are required.</p>

1. What is the Wet Wheel experiment and how does it demonstrate conservation of momentum?

The Wet Wheel experiment is a simple physics demonstration where a spinning wheel is partially submerged in water. As the wheel spins, the water is pushed outwards due to the centrifugal force. This results in a change in the wheel's momentum, but the total momentum of the system (wheel + water) remains constant, demonstrating the principle of conservation of momentum.

2. How does the Wet Wheel experiment relate to real-life situations?

The Wet Wheel experiment is a simplified version of real-life situations where conservation of momentum is at play. For example, the motion of a spinning top, the recoil of a gun, or the movement of a rocket in space all follow the same principle of conservation of momentum.

3. What factors affect the outcome of the Wet Wheel experiment?

The outcome of the Wet Wheel experiment can be affected by several factors, including the speed and direction of the wheel's spin, the amount of water in the container, and the shape and size of the wheel. These factors can change the amount of force and momentum exerted on the water, resulting in different outcomes.

4. How does the Wet Wheel experiment demonstrate the law of action and reaction?

The Wet Wheel experiment also demonstrates the law of action and reaction, also known as Newton's Third Law of Motion. As the wheel pushes the water outwards, the water exerts an equal and opposite force on the wheel, causing it to spin in the opposite direction. This shows that for every action, there is an equal and opposite reaction.

5. Can the Wet Wheel experiment be used to calculate the amount of momentum in a system?

While the Wet Wheel experiment can demonstrate conservation of momentum, it is not a precise method for calculating the exact amount of momentum in a system. Other factors, such as friction and air resistance, can also affect the outcome of the experiment. To accurately calculate momentum, more precise methods and measurements are required.

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