Wet wheel and conservation of momentum

[jartsa]

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]

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.

 

[Dale]

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]

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.

 

[jartsa]

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?

 

[jartsa]

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.

 

[Dale]

Less water on the wheel - less momentum of water on the wheel.

Subtracting 0 momentum doesn’t give you less momentum.

 

[jartsa]

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]

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]

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.

 

[jartsa]

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]

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.

 

[jartsa]

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]

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.

 

[Dale]

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?

 

[jartsa]

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?

 

[sophiecentaur]

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]

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?

 

[Dale]

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?

 

[jartsa]

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.

 

[jartsa]

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]

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.

 

[Dale]

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.

 

[jartsa]

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?

 

[Dale]

So? Did I claim something about momentum of the system?

That it is decreasing.

 

[jartsa]

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.

 

[Dale]

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.

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.

 

[sophiecentaur]

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?

 

[jartsa]

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.

 

[Nugatory]

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.

 

[jartsa]

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.

 

[Nugatory]

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.

 

[jartsa]

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.

 

[jbriggs444]

@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.

 

[russ_watters]

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?

 

[jbriggs444]

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?

I have not seen it nailed down carefully. My take was a wet wheel (or saturated paint roller) losing water (or paint) as it rolls on a dry surface at what is expected to be an unvarying speed while subject to no net thrust.

The puddle that the wheel starts in seems to be window dressing only -- intended to motivate an initial condition of a moving wheel saturated with water.

 

[russ_watters]

The puddle that the wheel starts in seems to be window dressing only -- intended to motivate an initial condition of a moving wheel saturated with water.

Ahh. That's...odd.

 

[Dale]

I claim this bike-paint roller device

I thought that your interest was in the paint/water and explicitly not in the combined system. I have to say that this is pretty annoying. I made a valid point for which you complained that it was not relevant, only to have you now make it explicitly relevant.

Please review and respond to my previous comment which you had previously dismissed as it is now directly relevant.

 

[jbriggs444]

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.

Since you seem to be attempting to eliminate friction with the wall, why not eliminate the wall?

You are a passenger in a weightless environment of some sort. You have a long roll of transparent wrap which you have removed from the box. You look at the roll end on and see that the wrap winds in a long clockwise spiral path from the center toward the outside. You give the roll a hard spin in the clockwise direction. The right hand edge is moving "down" and the left hand edge is moving "up".

You trigger a release of some sort exactly when the leading edge of the plastic is at the right hand side of the roll. The freed wrap proceeds in a straight line downward.

You watch as the cardboard tube inside the plastic roll rises slowly in the air despite the absence of any external force on the tube+wrap system.

Is that a scenario you are after? Did you want to claim any violations of conservation of momentum, angular momentum or energy based on its predicted behavior?

 

[jartsa]

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.

So I guess you are hinting that the Earth does not get the momentum, but instead the bike gets the momentum.

Well that sounds sounds plausible.

(What else should I say? I can't say if it's right or wrong, I don't have such authority here)I have been wondering the following:

Energy has momentum v*E/c². So the battery of a speeding electric car is losing momentum. That momentum has to go to the Earth or to the car. Well maybe it goes to the car. So the device that takes the momentum of the energy and puts it to the car is the wheel, because that's where the energy is going, right?

 

[Nugatory]

So the battery of a speeding electric car is losing momentum. That momentum has to go to the Earth or to the car. Well maybe it goes to the car.

It depends on whether the car is changing speed or not. And there are two ways you can define the speed of the car: Relative to the earth, which is assumed to be immobile; and relative to an arbitrary fixed point in space. The first way is a really excellent approximation and it's easy to calculate with, so it's how we usually work problems involving things moving around on the surface of the Earth - but it is an approximation, and momentum is not exactly conserved by that approximation because it neglects the (very very small) changes in the Earth's momentum.

Either way, in any reasonable electrical vehicle the battery and the rest of the car are solidly connected to one another so move together. Thus, thinking about them as separate systems only complicates things: you have to consider the second-law force pair between the car and the battery, but these forces are internal to the car+battery system so do not change its momentum.

 

[Dale]

Energy has momentum v*E/c2. So the battery of a speeding electric car is losing momentum. That momentum has to go to the Earth or to the car. Well maybe it goes to the car. So the device that takes the momentum of the energy and puts it to the car is the wheel, because that's where the energy is going, right?

Assuming no dissipation and assuming the car’s mass is negligible compared to the earth, then the energy from the battery goes into the KE of the car. The total energy does not go down, and the momentum increases as v increases.

 

[jartsa]

It depends on whether the car is changing speed or not. And there are two ways you can define the speed of the car: Relative to the earth, which is assumed to be immobile; and relative to an arbitrary fixed point in space. The first way is a really excellent approximation and it's easy to calculate with, so it's how we usually work problems involving things moving around on the surface of the Earth - but it is an approximation, and momentum is not exactly conserved by that approximation because it neglects the (very very small) changes in the Earth's momentum.

Either way, in any reasonable electrical vehicle the battery and the rest of the car are solidly connected to one another so move together. Thus, thinking about them as separate systems only complicates things: you have to consider the second-law force pair between the car and the battery, but these forces are internal to the car+battery system so do not change its momentum.

Why is change in Earth's momentum very very small? Is your car moving at speed very close to c on the surface of the earth?

At speeds where Newtonian physics is a good approximation the magnitudes of momentum changes of the car and the Earth are almost the same.

 

[Dale]

Why is change in Earth's momentum very very small?

He meant that the change in the earth’s speed is small.

 

[jartsa]

Assuming no dissipation and assuming the car’s mass is negligible compared to the earth, then the energy from the battery goes into the KE of the car. The total energy does not go down, and the momentum increases as v increases.

So the energy gives the Earth impulse x, and to the car the energy gives an impulse -x + the momentum of the energy itself. This conserves momentum.

 

[jartsa]

Is that a scenario you are after? Did you want to claim any violations of conservation of momentum, angular momentum or energy based on its predicted behavior?

Yes to the first question. No the the rest.

 

[Nugatory]

Why is change in Earth's momentum very very small? Is your car moving at speed very close to c on the surface of the earth?

At speeds where Newtonian physics is a good approximation the magnitudes of momentum changes of the car and the Earth are almost the same.

This has absolutely nothing to do with any relativistic effects, so we can take Newtonian physics to be exact.

The change in the earth’s momentum is exactly equal to and opposite to the change in the car’s momentum. And that change in the earth’s momentum is very small, by many decimal orders of magnitude, compared with what it would it take for us to notice. Thus we normally ignore it and recognize that because we’re not counting it we will seem to lose it somewhere.Only when someone starts a “where did the momentum go?” thread do we point out that this tiny change in the earth’s speed is being ignored.