The Juggler Problem

[essentialsaltes]

In this case:
dp/dt = d/dt(mv + m(-v))=0
Hence F=0.

What. the. heck. are you doing? This is not even wrong.

The force on an object at some particular time equals the time derivative of the momentum at that time. You can't just add the same equation at two different times and have that equal 'the' force on the object. At best you would have F1 + F2 = something or other.

Just before the hand releases the pin (at time 1 where the variables are labeled appropriately):

F = d/dt(mv1) = m d/dtv1 = ma1

How big is that? Depends how rapidly the hand is accelerating the pin at that moment.

Just after the hand catches the pin (at time 2):

F = d/dt(mv2) = m d/dtv2 = ma2

How big is that? Depends how rapidly the hand is accelerating the pin at that moment. (the fact that the velocity is up in one case and down in the other is irrelevant. The force depends on the acceleration (or vice versa if you like). Yes v1=-v2. But the equation that gives us the force depends on the change in v, not v itself. This is elementary.

What this tells you it doesn’t matter how high or low the juggler throws the pins, whether he uses biomechanics, the dynamical system or a combination of both, he cannot affect the load on the bridge through juggling.

No, that is not the case. Impulse doesn't get used very often, but here maybe it is useful. Impulse is the integral over time of applied force. And it equals the change in momentum.

If you catch a falling pin moving at -v and throw it back up with velocity v, then you've given it an impulse of 2mv. This is situation 1.

If you throw it up faster than v, then the impulse is greater than 2mv. For simplicity, if you use constant force and your catch/throw motion takes the same amount of time in both cases, to impart a greater impulse requires using a greater force than in situation 1.

Just think of it as recoil. Instead of throwing a pin, you're firing a rifle up, and there's a recoil impulse directed down into you and ultimately into the bridge. If you use a gun with a larger muzzle velocity, there will be a greater recoil. If you throw the pin higher, there will be more 'recoil'. If you throw it lower, there will be less. You can control how much force you use (and thus change the amount of force between the bridge and your feet, so that it is not necessarily identical to the total weight of the system)

It’s no coincidence either this ties in with the fact the total work of a juggling cycle is zero.

If you throw the pins higher so that the COM of the system moves higher, you have 'lifted' it against the force of gravity. You have performed work.

Where did I explicitly associate negative work with downwards acceleration?
This is what I stated.

“Now that it has been established a pin is accelerated downwards as per the video, positive and negative work is performed during the cycle.”

It sounded like you needed to establish this fact in order to make your conclusion. So we'll just go with my first take, with which you seem to agree. If the pin is accelerated downward momentarily or not is irrelevant to the problem.

What I do know however is the total work must be zero as gravity is a conservative force

Gravity is not the only force in the problem.

and the COM follows a closed loop.

It doesn't have to. We can throw pins higher or lower and change the COM with every cycle.

In your view on the other hand work can never be negative as the resultant force is always in the same direction as motion.

What? I explained my view quite clearly:

"(In my view, if a continuous upward force is applied, the force and displacement are in opposite directions during the first half of the hand/pin interaction as the pin is being slowed down from its fall. Negative work is done, and predictably, the pin slows down to rest. During the second half of the interaction, positive work is done during the upward acceleration leading up to the release.)"

A pin that is falling is moving down. An upward force is in the opposite direction to this motion. Yes, this produces negative work. [Note that the resultant force is obviously also upward, since the pin is accelerating upward. Gravity is losing the fight to pull the pin to the ground because the juggler's force is larger than the force of gravity.]

As previously explained while the juggler is free to do anything he wants he adds no momentum to the juggling process hence F=0.

Oh boy. I think what you were trying to say before was something about delta-p (i.e. the impulse). Deltas are differences. If the pin falling at (-v) is then thrown up with velocity v, the change in the momentum is not zero. delta-p = mv2-mv1 = m(v - (-v)) = 2mv.

Momentum is a directed quantity. If you catch a thing moving down, and throw it up, you have added momentum to it.

In a conservative gravitational field the sum of the kinetic and potential energies is constant anywhere along the pin’s trajectory in flight.

In flight, yes. But that is not the complete cycle. There is another force. One supplied by the juggler. (which is not zero, nor does it average to zero)

This cannot be true if the total work is non zero.

Why not?

For example at the zero level of gravitational potential energy, the level at which the pins are released and caught, the magnitude of the velocities of the pins for throwing and subsequent catching cannot be equal.

My lack of god, it's almost as though the person decided to throw the pin higher (or lower) and threw it with a greater (or smaller) speed than it was moving when he caught it. Congratulations, you have solved the problem of how a nonzero amount of work could be done by the juggler! You catch a pin falling down with one speed, and throw it up with a different one! You've given it more (or less) kinetic energy! You've done work on it! You exerted a net force on it.

You have accurately described exactly how one would go about doing that.

 

[essentialsaltes]

Again focus on the interesting part of the problem. Gravity is boring. It can't do any work. Sure the pin will fall down at the same speed as it was sent up, but we didn't expect that to be the interesting part of the cycle. The interesting part is the juggler's actions.

He does not have to throw it up at the same speed that it was caught.

 

[sjastro]

What. the. heck. are you doing? This is not even wrong.

The force on an object at some particular time equals the time derivative of the momentum at that time. You can't just add the same equation at two different times and have that equal 'the' force on the object. At best you would have F1 + F2 = something or other.

Just before the hand releases the pin (at time 1 where the variables are labeled appropriately):

F = d/dt(mv1) = m d/dtv1 = ma1

How big is that? Depends how rapidly the hand is accelerating the pin at that moment.

Just after the hand catches the pin (at time 2):

F = d/dt(mv2) = m d/dtv2 = ma2

How big is that? Depends how rapidly the hand is accelerating the pin at that moment. (the fact that the velocity is up in one case and down in the other is irrelevant. The force depends on the acceleration (or vice versa if you like). Yes v1=-v2. But the equation that gives us the force depends on the change in v, not v itself. This is elementary.



No, that is not the case. Impulse doesn't get used very often, but here maybe it is useful. Impulse is the integral over time of applied force. And it equals the change in momentum.

If you catch a falling pin moving at -v and throw it back up with velocity v, then you've given it an impulse of 2mv. This is situation 1.

If you throw it up faster than v, then the impulse is greater than 2mv. For simplicity, if you use constant force and your catch/throw motion takes the same amount of time in both cases, to impart a greater impulse requires using a greater force than in situation 1.

Just think of it as recoil. Instead of throwing a pin, you're firing a rifle up, and there's a recoil impulse directed down into you and ultimately into the bridge. If you use a gun with a larger muzzle velocity, there will be a greater recoil. If you throw the pin higher, there will be more 'recoil'. If you throw it lower, there will be less. You can control how much force you use (and thus change the amount of force between the bridge and your feet, so that it is not necessarily identical to the total weight of the system)

You have totally misunderstood the mathematics.
The maths describes the conservation of momentum!!
The equation F = dp/dt= d/dt(mv+ m(-v))=0 should have made this perfectly clear which states if momentum is conserved the resultant force is zero.
Juggling for any given cycle is a physical process describing an initial momentum state where the pin is released from the hand with momentum mv, to a final state where the pin is caught with momentum m (-v).

As a result your post is largely irrelevant and incorrect as it is been based around your misunderstanding of the maths.
A classic case of barking up the wrong tree.

My lack of god, it's almost as though the person decided to throw the pin higher (or lower) and threw it with a greater (or smaller) speed than it was moving when he caught it. Congratulations, you have solved the problem of how a nonzero amount of work could be done by the juggler! You catch a pin falling down with one speed, and throw it up with a different one! You've given it more (or less) kinetic energy! You've done work on it! You exerted a net force on it.

You have accurately described exactly how one would go about doing that.

I’ll ignore the insulting tone and comment why your reasoning is so emphatically wrong.
The motion of the pin while in the juggler’s hand, is not like say pendulum motion, where the resultant force is a smooth function and work can be calculated for any given distance interval.
Instead after the pin is caught it is momentarily brought to rest; followed by being accelerated upwards and released.

You can’t use the work-energy equation to calculate the work performed over your distance interval for the simple reason the function is not smooth at the point where the pin is brought to rest.

Even if this was possible the other problem is that you are ignoring the trajectory of the pin while airborne.
You need to loop integrate over the entire pin trajectory in which case the work is found to be zero.

Using the work-energy relationship for a pin that is thrown and then caught (ie when the pin is only airborne) might suggest a similar problem where the trajectory of the pin in the juggler’s hand is ignored, but as I have stressed this is taken care of by the pin’s initial velocity which is determined by the juggler’s actions.
The pin’s initial velocity “links” both the airborne and hand held pin trajectories and the work-energy equation holds for the entire cycle.

 

[essentialsaltes]

You have totally misunderstood the mathematics.
The maths describes the conservation of momentum!!
The equation F = dp/dt= d/dt(mv+ m(-v))=0 should have made this perfectly clear which states if momentum is conserved the resultant force is zero.
Juggling for any given cycle is a physical process describing an initial momentum state where the pin is released from the hand with momentum mv, to a final state where the pin is caught with momentum m (-v).

Conservation of momentum would state that the change in the momentum is zero.

i.e. ∆p = 0

But in this case, given your initial and final points:

∆p = mv(final) - mv(initial) = m(-v) - m(v) = -2mv

This is not zero.

Momentum is a vector. If momentum is conserved, then the momentum is the same in the initial state as in the final state. They aren't if the velocity is v in once case and (-v) in the other.

You can’t use the work-energy equation to calculate the work performed over your distance interval for the simple reason the function is not smooth at the point where the pin is brought to rest.

Well, you can't evaluate the integral without knowing the particulars of F(t). There is nothing impossible with the simplifying assumptions I made. That F is constant. Regardless, even if we can't calculate the integral, the work-energy equation is still true. If I throw it faster than I caught it, which is obviously not impossible, it has more kinetic energy afterwards, because I have done non-zero work on it.

Even if this was possible the other problem is that you are ignoring the trajectory of the pin while airborne.

No, you and I both agree that the pin's total energy is the same when it is thrown as when it again falls down to the same height.

You need to loop integrate over the entire pin trajectory

Exactly. While it's in the air, no work is done. But when it's in the hands of the juggler, he can choose to throw it faster or slower, which will change its energy, because more or less work will be done on it. For the whole cycle the total change in work is 0 + however much work was done by the juggler.

in which case the work is found to be zero.

Nope.

Using the work-energy relationship for a pin that is thrown and then caught (ie when the pin is only airborne) might suggest a similar problem where the trajectory of the pin in the juggler’s hand is ignored

It certainly does.

but as I have stressed this is taken care of by the pin’s initial velocity which is determined by the juggler’s actions.

Yes, but the juggler can clearly choose how hard to throw it. Think about where the choice occurs.

He throws the pin up at v.

It flies through the air.

If he catches it at the same height, it is moving at -v at that point. Boring and no work is done.

He cannot now go back in time and choose to have thrown the pin at a different speed. But he is entirely free to pump some work into the pin and throw it up faster than v.

 

[sjastro]

Conservation of momentum would state that the change in the momentum is zero.

i.e. ∆p = 0

But in this case, given your initial and final points:

∆p = mv(final) - mv(initial) = m(-v) - m(v) = -2mv

This is not zero.

Momentum is a vector. If momentum is conserved, then the momentum is the same in the initial state as in the final state. They aren't if the velocity is v in once case and (-v) in the other.

The juggler throws a pin upwards with initial momentum mv which slows downs and is momentarily brought to rest with zero momentum.
The impulse Ft = Δp1 for this interval is (mv - 0) = mv.
The pin then falls and to be caught in the jugglers other hand.
The impulse Ft = Δp2 for this interval is (0 – m(-v)) = mv.

So when the juggler catches the pin the momentum has increased to mv which is the same when it left his other hand.
Hence momentum is conserved.

Alternatively we both believe that energy is conserved and that the kinetic energy KE=0.5mv² is the same at the point when the juggler throws a pin and catches it.

KE = 0.5mv²
=0.5pv

Since KE ≥ 0, 0.5pv is an absolute value whose sign doesn’t change irrespective of whether v is positive or negative.
In this case if p is not conserved then neither can KE be conserved.

Well, you can't evaluate the integral without knowing the particulars of F(t). There is nothing impossible with the simplifying assumptions I made. That F is constant. Regardless, even if we can't calculate the integral, the work-energy equation is still true. If I throw it faster than I caught it, which is obviously not impossible, it has more kinetic energy afterwards, because I have done non-zero work on it.

The particulars of F(t) may be unknown but we do know the derivative F’(t) is undefined at the point where the pin is momentarily at rest.
This is where the downwards acceleration “stops” and the upwards acceleration “starts”.
Applying the work energy equation is wrong as you are assuming the same force F applies over the distance interval where the pin is caught and released.
There is no dependence between the two forces with regards to work.
This is clearly illustrated in the very first cycle.
The juggler has performed work in throwing the pin before the pin has even been caught.

Consider your crouching juggler example with the speed of the pin leaving the juggler’s hand.
Can you establish a force function between the two processes which allows the calculation of the work done?
Clearly the answer is no as there are two distinctly different forces at play.

No, you and I both agree that the pin's total energy is the same when it is thrown as when it again falls down to the same height.



Exactly. While it's in the air, no work is done. But when it's in the hands of the juggler, he can choose to throw it faster or slower, which will change its energy, because more or less work will be done on it. For the whole cycle the total change in work is 0 + however much work was done by the juggler.



Nope.



It certainly does.



Yes, but the juggler can clearly choose how hard to throw it. Think about where the choice occurs.

He throws the pin up at v.

It flies through the air.

If he catches it at the same height, it is moving at -v at that point. Boring and no work is done.

He cannot now go back in time and choose to have thrown the pin at a different speed. But he is entirely free to pump some work into the pin and throw it up faster than v.

This thread is now tending towards repetition and obfuscation with line by line responses.

The juggler problem where the pin remains in his hand for part of the cycle can be tested indirectly.
All you need is a set of bathroom scales preferably the analogue type that can rapidly show changes in load.
You don’t have to juggle weights; your hands are the pins.
Simulate the juggling action.
What you will find is the readout oscillates around your weight.
The faster you juggle the higher the frequency of oscillation but it is still centred on your weight.

If that’s not convincing try keeping one hand still and the other hand moving up and down to simulate the condition of a pin moving in one hand at various speeds relative to the other hand.
Once again irrespective of how fast the hand movement, the frequency of oscillation changes but it is still centred on your readout weight.
The behaviour of the scales is contrary to what you are suggesting and is simply explained as the tension and load reaction forces generated by movement trying to cancel each other out.

 

[essentialsaltes]

The juggler throws a pin upwards with initial momentum mv which slows downs and is momentarily brought to rest with zero momentum.
The impulse Ft = Δp1 for this interval is (mv - 0) = mv.
The pin then falls and to be caught in the jugglers other hand.
The impulse Ft = Δp2 for this interval is (0 – m(-v)) = mv.

Therefore the total change of momentum is the first change plus the second change, which is 2mv, which is nonzero.

So when the juggler catches the pin the momentum has increased to mv which is the same when it left his other hand.

No, it isn't. Because at one of those two points it is moving with velocity v, and at the other with (-v). Momentum is a vector quantity. moving up at v is not the same as moving down with -v.

Hence momentum is conserved.

No, it clearly isn't. Why would we expect it to? For the part of the process you're so focused on, gravity is the only force acting on it, and is always pointing down.

Since KE ≥ 0, 0.5pv is an absolute value whose sign doesn’t change irrespective of whether v is positive or negative.
In this case if p is not conserved then neither can KE be conserved.

Sure it can. Because p is a vector and KE is a scalar. It makes it quite easy. You drive 100kph east and then turn so you drive 100 kph south. Momentum has changed, but KE has not.

A planet in a perfectly circular orbit has its momentum changing continuously, but its KE stays the same.

The faster you juggle the higher the frequency of oscillation but it is still centred on your weight.

The important point is that the amplitude of the oscillation will increase. if you throw it faster, the momentum recoil is larger, and you will gain more 'weight' than if you throw it slower.

 

[sjastro]

Therefore the total change of momentum is the first change plus the second change, which is 2mv, which is nonzero.



No, it isn't. Because at one of those two points it is moving with velocity v, and at the other with (-v). Momentum is a vector quantity. moving up at v is not the same as moving down with -v.



No, it clearly isn't. Why would we expect it to? For the part of the process you're so focused on, gravity is the only force acting on it, and is always pointing down.

So you are now trying to read your own conclusions into my post.
It’s your comment on gravity which is telling. Not only is gravity a conservative force, it can be either an internal or external force.
When the pin is thrown in the air there is the gravitational force between the pin and the Earth.
The pin and Earth form a closed system. Under these conditions gravity is an internal force.
The absence of gravity as an external force is the reason why momentum is conserved; which is supported by my previous post if you ever try to read it objectively.

Sure it can. Because p is a vector and KE is a scalar. It makes it quite easy. You drive 100kph east and then turn so you drive 100 kph south. Momentum has changed, but KE has not.

A planet in a perfectly circular orbit has its momentum changing continuously, but its KE stays the same.

The argument “p is a vector and KE is a scalar” is a logical fallacy in supporting your case as it implies KE and p cannot be conserved in the same process. Elastic collisions are examples where KE and p both conserved.
Since you believe p is not conserved in the juggling process it’s up to you to show that either gravity is not acting as an internal force or overturn Newtonian physics to demonstrate internal forces do result in the non conservation of momentum. Your maths is not the saving grace because it assumes gravity is an external force.

The important point is that the amplitude of the oscillation will increase. if you throw it faster, the momentum recoil is larger, and you will gain more 'weight' than if you throw it slower.

If you actually performed the test you would find the amplitude remains fairly constant with velocity.
As I stated previously the load and tension reaction forces try to cancel each other out, leaving the weight of the system.
The amplitude is due to the statistical variations in the load and tension reaction forces.

 

[essentialsaltes]

The argument “p is a vector and KE is a scalar” is a logical fallacy in supporting your case as it implies KE and p cannot be conserved in the same process.

It implies no such thing.

Since you believe p is not conserved in the juggling process

You have focused on #1: the pin's momentum, and #2: the part of the process when the pin flies under only the influence of gravity.

At the start it flies up. At the end it flies down. Momentum is not the same at beginning and end.

(If you want to suddenly include the earth, that's fine, but this is in contradiction to your previous analysis of the situation, where you focus solely on the pin's momentum. "The juggler throws a pin upwards with initial momentum mv")

 

[sjastro]

It implies no such thing.



You have focused on #1: the pin's momentum, and #2: the part of the process when the pin flies under only the influence of gravity.

At the start it flies up. At the end it flies down. Momentum is not the same at beginning and end.

(If you want to suddenly include the earth, that's fine, but this is in contradiction to your previous analysis of the situation, where you focus solely on the pin's momentum. "The juggler throws a pin upwards with initial momentum mv")

What a desperate attempt at obfuscation.

The reason why I was “focused” on the pin’s momentum is that the points where the pin is thrown and caught, the momentum mv is entirely taken up by the pin. When the pin is in its upward motion the momentum lost by the pin is gained by the Earth; for the downwards motion the reverse occurs.
This alone provides a very simple explanation as to why momentum is conserved and the behaviour of gravity as the impulse FΔt = 0 ⇒ F = 0, i.e. gravity is this case is an internal not an external force.

I note you are cherry picking my post as to avoid responding to my request while engaging in the argument by repetition fallacy.

Let me remind you of the request:

Since you believe p is not conserved in the juggling process it’s up to you to show that either gravity is not acting as an internal force or overturn Newtonian physics to demonstrate internal forces do result in the non conservation of momentum. Your maths is not the saving grace because it assumes gravity is an external force.

Admittedly my request is difficult to respond to so let’s look at a simple problem which is along the same lines as the juggler momentum issue.

Consider dropping an elastic ball of mass m onto a hard surface.
As you should know momentum is conserved in elastic collisions.
Let’s use your method to calculate the momentum change.
The ball has a momentum p1=mv, when it hits surface the collision is elastic hence it rebounds with a momentum p2=m(-v).
But (p1-p2) = 2mv ≠ 0 hence momentum is not conserved.
Try explaining how your method can cause such a glaring discrepancy.

 

[sjastro]

Since Essentialsaltes seems to have disappeared without addressing the keynote of his argument;
Δp = m(-v) - mv= -2mv implying momentum is not conserved; let me address the flaw in this logic.

m(-v) – mv = -2mv
⇔ -mv – mv = -2mv
⇔ mv + mv = 2mv

The vector diagram for this last equation is of the form:

Both vectors mv and mv which are represented as arrows of equal length representing magnitude are collinear and are in the same direction.

This contradicts the initial and final momentum for the pin and elastic ball which are in opposite directions.


The vector diagrams for the pin and elastic ball are of the form:

Since the vectors are of equal magnitude and in opposite directions they cancel.

Δp = m(-v) + mv =0

Momentum is therefore conserved.

Essentialsaltes argument can also be refuted using simple physics.
In the case of the elastic ball when it hits the surface the ball exerts a force on the surface.
By Newton’s third law the surface also exerts a force on the ball of opposite and equal magnitude.
The forces cancel each other out and the resultant external force is zero which means there is no momentum change during the collision.

The same principles apply to a pin in flight.
There is an internal gravitational force between the pin and Earth.
By Newton’s third law the magnitude of the gravitational force on the pin and Earth are equal and in opposite directions.
These cancel out and gravity as an external force is zero and the momentum change between the pin and the Earth for any time interval in the pin’s flight is zero.

 

[essentialsaltes]

Since you believe p is not conserved in the juggling process it’s up to you to show that either gravity is not acting as an internal force or overturn Newtonian physics to demonstrate internal forces do result in the non conservation of momentum. Your maths is not the saving grace because it assumes gravity is an external force.

One can either describe the situation including the earth or not. Although the description differs, obviously the underlying reality will be the same.

If you exclude the earth, then momentum is not conserved, since gravity is treated as an external force. (This is why your posts #43 and #45 are hilariously wrong, as it is only in #47 that you suddenly have a desire to include the earth.)
If you include the earth, then momentum is conserved. Hurray. We now are in agreement that your posts 43 and 45 were wrong, since you concluded that momentum was conserved by only calculating the momentum of the pin and claiming it was 'the same' at beginning and the end, although they were in different directions.

But the problem is not about momentum, but about the force between the feet and the earth. #1, whether it is large enough to break the bridge. And #2 (with regard to how accurate choice (E) is) whether that force is exactly equal to the total weight (I will even grant that this need only be true in some time averaged fashion). My contention is that the juggler can choose to change the height to which he throws the pins, and that this will change the force.

If we eschew acrobatics and anatomic complexities, then the force between the hand and the pin is the same in magnitude as between the feet and the bridge.

Throwing it higher can be accomplished by using a greater force. The force of the throw is in the control of the juggler. Changing the force between the hand in the pin changes the force between the feet and the bridge. So a choice to throw the pins higher or lower will affect the force between the feet and the bridge.

Consider dropping an elastic ball of mass m onto a hard surface.
As you should know momentum is conserved in elastic collisions.
Let’s use your method to calculate the momentum change.
The ball has a momentum p1=mv, when it hits surface the collision is elastic hence it rebounds with a momentum p2=m(-v).
But (p1-p2) = 2mv ≠ 0 hence momentum is not conserved.
Try explaining how your method can cause such a glaring discrepancy.

Ironically, this is exactly the error you perpetrated in posts #43 and #45. If we focus only on the ball (or pin), then momentum is not conserved. But you expressly stated that it was in those posts.

"So when the juggler catches the pin the momentum has increased to mv which is the same when it left his other hand.
Hence momentum is conserved."

As we now both agree, mv is not the same as -mv.

The 'solution' to your 'glaring discrepancy' of course is that for a collision you need at least two things to collide. As in the juggling situation, the other thing is the earth. The ball has its momentum changed by 2mv in the collision (or throwing process), since the problem states the collision is elastic. (and thus the earth responds by changing by -2mv)

But in the case of the juggler, we are not told that he is equivalent to an elastic collision. As I alluded earlier, he is not a perfect 'spring' that is forced to throw the ball at the same speed he caught it. If he chooses to catch it at one velocity and throw it at another, the change in momentum will be m(v1 + v2). He can't control v1 since that is fixed by his previous throw(*), but he has a choice of v2. So he can adjust the momentum transferred to the pin. And that means the recoil momentum of the earth is not fixed. A simple way to adjust the velocity is by using different amounts of force. Again, this means the force applied to the earth is not fixed.

(*) actually he can by catching the object higher or lower than the height at which is was released, but we will make yet another simplifying assumption.

 

[essentialsaltes]

Since Essentialsaltes seems to have disappeared without addressing the keynote of his argument;

Sorry, I was at an international science conference in Brussels.

Δp = m(-v) - mv= -2mv implying momentum is not conserved; let me address the flaw in this logic.

m(-v) – mv = -2mv
⇔ -mv – mv = -2mv
⇔ mv + mv = 2mv

The vector diagram for this last equation is of the form:

Both vectors mv and mv which are represented as arrows of equal length representing magnitude are collinear and are in the same direction.

This contradicts the initial and final momentum for the pin and elastic ball which are in opposite directions.

I keep hoping this is some elaborate joke.

This is not a contradiction. Focus on the prize. We are discussing the change in the momentum, yes? Difference. Subtraction. You have $5 and are given $10 more --> the change is +$10, so to find the change we subtract the initial amount ($5) from the final amount ($15). Change = final - initial.

We have three states of interest.

A) leaving the hand with velocity v
B) 'pausing' at the top of the trajectory with velocity 0
C) just before the catch with velocity -v

The mass of the pin remains the same at m during the entire process.

From A to B.

Change in momentum = m(0) - m(v) = 0 - mv = -mv

From B to C.

Change in momentum = m(-v) - m(0) = -mv - 0 = -mv

The total (unlike change, total means addition) change in momentum from A to C is:

Δp = -mv + -mv = -2mv

The vector diagrams for the pin and elastic ball are of the form:

Since the vectors are of equal magnitude and in opposite directions they cancel.

Δp = m(-v) + mv =0

Wow, you put two vectors tip to tail and added them. This is not how we calculate change.

Momentum is therefore conserved.

Nope. Now you've gone back to your error in #43 and #45, and contradicted your argument in #47. (In order to get the total momentum change to be zero, you have to add the earth. As you have correctly explained before, but now apparently forgotten.)

The forces cancel each other out and the resultant external force is zero which means there is no momentum change during the collision.

Oh dear, I guess I charitably misread what you said in #47. Sorry. I apologize for saying you contradicted yourself between 43/45 and 47. You have been consistently wrong.

These cancel out and gravity as an external force is zero and the momentum change between the pin and the Earth for any time interval in the pin’s flight is zero.

Between? A strange word to use. The momentum change of the total system of the pin and the earth is zero. But again in all your calculations above, you only calculated the momentum change of the pin. You did not include the earth. This is why you have to calculate it incorrectly to get it to be zero.

 

[sjastro]

Let me summarize your two posts as follows.

It is a sad state of affairs that you have to resort to outright lying in order to establish supposed contradictions in my posts.
Given this is the obfuscation process at work to deflect from your own errors the motivation is not surprising.

Your nonsense that momentum is subtracted rather than added in order to define whether it is conserved or not, is refuted by the principle of the conservation of momentum equation.

R = ΣF = d/dt(m1v1+ m2v2 + m3v3 + …… + mnvn) = 0 if momentum is conserved.

Let me remind you the specific form of this equation was explained in this post which you completely misunderstood.
What do you notice about momentum terms in the brackets, they are summed not subtracted.
The reason is blindingly obvious, the derivative of each momentum term is an individual force and the resultant force R is the sum of the individual forces.

Then there is your take on the elastic ball issue.
I find it mildly amusing that while you acknowledge the physics results in momentum being conserved, you refuse to admit your calculations are wrong as the momentum change is 2mv (or -2mv) and not zero.
Instead you engage in false dichotomies as if these automatically correct the errors in your calculations.
If you used the correct method of adding the momentum instead of subtraction, as per the conservation of momentum equation, the contradiction disappears.
The same reasons apply to the juggling problem.

The ridiculous outcome of your method if the magnitude of the initial and final momentum is the same but in opposite directions, momentum can never be conserved.

 

[essentialsaltes]

Your nonsense that momentum is subtracted rather than added in order to define whether it is conserved or not, is refuted by the principle of the conservation of momentum equation.

R = ΣF = d/dt(m1v1+ m2v2 + m3v3 + …… + mnvn) = 0 if momentum is conserved.

What do you notice about momentum terms in the brackets, they are summed not subtracted.

What I notice about the summed terms is that they involve different masses m1, m2, etc. You sum up the momenta of all of the masses in the system. However, each one only appears once.

Therefore to add the momentum of the pin at one time to the momentum of the pin at a different time is just garbage.

What I also notice is that it is not just a big sum of momenta, but right in front there is a big fat differential operator. The time derivative operator d/dt is, of course, a limit version of Δ/Δt. And these deltas are as I've described with final value minus initial value.

I find it mildly amusing that while you acknowledge the physics results in momentum being conserved, you refuse to admit your calculations are wrong as the momentum change is 2mv (or -2mv) and not zero.

It is zero only if you also include the momentum of the earth, which you have steadfastly refused to consider in the calculations, even when you mention its importance. This is where the addition comes in in the formula your provide. m1 is the pin. m2 is the earth. Since you want to talk about particular start and end states, the differential formula is not so applicable, and the version with impulse and Δp is simpler.

The momentum change of the pin is 2mv and that of the earth is -2mv. This sum is zero. But the change in the momentum of the pin alone is not zero, as you claim.

The ridiculous outcome of your method if the magnitude of the initial and final momentum is the same but in opposite directions, momentum can never be conserved.

But surely you recognize that the pin is indeed going up when it leaves the hand, and going down just before it is caught? Yes or no?

 

[sjastro]

What I notice about the summed terms is that they involve different masses m1, m2, etc. You sum up the momenta of all of the masses in the system. However, each one only appears once.

Therefore to add the momentum of the pin at one time to the momentum of the pin at a different time is just garbage.

What I also notice is that it is not just a big sum of momenta, but right in front there is a big fat differential operator. The time derivative operator d/dt is, of course, a limit version of Δ/Δt. And these deltas are as I've described with final value minus initial value.



It is zero only if you also include the momentum of the earth, which you have steadfastly refused to consider in the calculations, even when you mention its importance. This is where the addition comes in in the formula your provide. m1 is the pin. m2 is the earth. Since you want to talk about particular start and end states, the differential formula is not so applicable, and the version with impulse and Δp is simpler.

The momentum change of the pin is 2mv and that of the earth is -2mv. This sum is zero. But the change in the momentum of the pin alone is not zero, as you claim.



But surely you recognize that the pin is indeed going up when it leaves the hand, and going down just before it is caught? Yes or no?

It’s clear you don’t understand the conservation of momentum equation can be simplified.

Let’s use the dropping of the elastic ball onto the rigid surface example where we both agree momentum is conserved.

R = ΣF = d/dt(m1v1+ m2v2 + m3v3 + …… + mnvn) = 0; if momentum is conserved.

In this case F are the forces acting on the ball and the Earth and are defined as Fi = d/dt(mivi).
In the case for the elastic ball, m1 = m2 = m, v1 = v, v2 = -v.
The Earth’s momentum can be ignored as the velocity imparted to the Earth by the colliding ball is so ridiculously small that v(Earth) = 0 is a valid approximation.
We can also ignore the momentum exchange effects between the pin and Earth while the ball is in flight (like the juggling pins) as for any given momentum for the pin (mu) and Earth (Mu) there is a corresponding m(-u) and M(-u).

By pairing and adding, these momentum terms vanish.
The equation is therefore simplified to:

R = ΣF = d/dt(mv + m(-v)) = 0

In this case mv is the initial momentum state of the ball from when the ball is dropped to the point of collision and m(-v) the final state of the ball after collision.

Now for the important point, the (mv + m(-v)) = 0 term in the equation is vector addition not subtraction.
You are emphatically wrong in using vector subtraction.
Vector subtraction is (mv – m(-v)) = 2mv which contradicts the conservation of momentum.
Furthermore R ≠ 0 as d/dt(2mv) ≠ 0 is also a contradiction as collisions from an isolated system where R is zero.

 

[essentialsaltes]

The Earth’s momentum can be ignored as the velocity imparted to the Earth by the colliding ball is so ridiculously small that v(Earth) = 0 is a valid approximation.

I'm sorry, you're just ludicrously wrong. The reason v(earth) is small is because m(earth) is very big. When you multiply them together, you'll get something on the order of mv. Because momentum is conserved.

Again, surely you recognize that the pin is indeed going up when it leaves the hand, and going down just before it is caught? Yes or no?

 

[sjastro]

I'm sorry, you're just ludicrously wrong. The reason v(earth) is small is because m(earth) is very big. When you multiply them together, you'll get something on the order of mv. Because momentum is conserved.

And I am sorry to point out your attempts at ridicule have backfired (again) as what you are describing is a 100% inelastic collision.

Here is the proof:
Momentum before collision:- mv + M.0 where m and M are the mass of the ball and Earth respectively.
Momentum after collision:- (M+m)V (no rebound of ball occurs due to 100% inelastic collision).

V = mv/( M+m) ≈ mv/M as M>>m (momentum is conserved in inelastic collisions even though KE isn’t).
Therefore MV ≈ Mmv/M = mv.

Given the example is for an elastic collision where the ball loses no KE energy, the Earth’s momentum should in fact be zero.
You couldn’t be more comprehensively wrong.

Again, surely you recognize that the pin is indeed going up when it leaves the hand, and going down just before it is caught? Yes or no?

The best you can do is to repeat this lame question.
The answer is yes.
Now that the question has been answered, why don’t you address the issue which you have been avoiding like the plague.
If your method of vector subtraction is correct then explain why it shows the momentum of a bouncing elastic ball is not conserved when even you agree the conservation of momentum does occur.

 

[essentialsaltes]

And I am sorry to point out your attempts at ridicule have backfired (again) as what you are describing is a 100% inelastic collision.

No, it isn't.

The best you can do is to repeat this lame question.
The answer is yes.

Excellent. Then since the momentum points up at one point and down at the other, there is no way the momentum can be the same. Because up is not the same as down when we're talking about vector quantities like momentum.

Therefore your statement "the momentum change between the pin and the Earth for any time interval in the pin’s flight is zero" is wrong, particularly if (as you have) you are going to assume the velocity of the earth is zero (and therefore the earth has no momentum worth calculating). If the only momentum of interest in the problem is that of the pin, then yes it would be clear that momentum is not conserved, because up is not down.

If your method of vector subtraction is correct then explain why it shows the momentum of a bouncing elastic ball is not conserved when even you agree the conservation of momentum does occur.

I'll get to it in a bit. The dinner gong has sounded.

 

[sjastro]

No, it isn't.

Really?
A one line line response lacking any sort of detail is clear enough you have nothing to offer in refuting the maths.

Excellent. Then since the momentum points up at one point and down at the other, there is no way the momentum can be the same. Because up is not the same as down when we're talking about vector quantities like momentum.

Therefore your statement "the momentum change between the pin and the Earth for any time interval in the pin’s flight is zero" is wrong, particularly if (as you have) you are going to assume the velocity of the earth is zero (and therefore the earth has no momentum worth calculating). If the only momentum of interest in the problem is that of the pin, then yes it would be clear that momentum is not conserved, because up is not down.

Amazing.
It's as if last 20-30 posts discussing your misconceptions never happened.

I'll get to it in a bit. The dinner gong has sounded.

I hope it's of better quality than "No it isn't".

 

[essentialsaltes]

I'll get to it in a bit. The dinner gong has sounded.

The elastic ball, explained.

We hold a perfectly elastic ball about 1.25 meters above the surface of the earth. It is at rest. As is the earth.

We gently release the ball. At the moment of release it is still motionless, as is the earth.

The initial momentum of the ball is zero.

It begins to fall. Its momentum is no longer zero.

The change in its momentum is nonzero.

Its momentum is not conserved.

Nothing could be so obvious.

But, as we know, momentum is conserved in an isolated system. How are we to explain this? Obviously, the ball is not isolated. It is being acted on by an external force. There is gravity. The earth.

As sjastro has correctly described but incorrectly calculated, if we treat the earth as part of the system, then there are only internal forces, and momentum is conserved. But we must actually treat the earth as part of the system. If there is an internal force of the earth on the ball, then by Newton's Third Law, there is an internal force of the ball on the earth. And they are equal and opposite.

Consider the ball (of mass m) after it has fallen for a time t

The force acting on it is the force of gravity = -mg (I shall take up to be positive)

Its acceleration obviously is -g

Its velocity is -gt

Its momentum is -mgt

Now the earth (of mass M) falls up to meet it.

The force on it is equal and opposite to the force on the ball, so that is +mg

To calculate the acceleration of the earth a(E)

F = Ma(E) = mg

Therefore a(E) = (m/M)g

and v(E) = (m/M)gt

and the momentum of the earth is Mv(E) = M * (m/M)gt = mgt

Miraculously, the momentum of the earth is equal and opposite to the momentum of the falling ball at every moment during the fall. Thus the total momentum of the system is zero. Exactly the same as when we released the ball, when both the ball and the earth were at rest.

After about 0.5 s (let's call that moment t=T) the ball hits the earth and collides elastically. For both momentum and KE to be conserved (as in an elastic collision), it is obvious that both the ball and the earth flip the sign of their respective velocities/momenta.

Before the collision:
v(ball) = -gT
v(E) = (m/M)gT

Afterwards:
v(ball) = +gT
v(E) = -(m/M)gT

Now, this situation is (as sjastro again correctly points out, but misunderstands) very like the juggler situation. The difference between the initial and final situations is that the velocity of the ball/pin has changed direction (and therefore so has the momentum).

sjastro focuses only on the pin, or ball, and somehow suggests that the momentum has not changed, when in fact it has changed by twice its amount.

initial momentum of the ball is -mgT
final momentum of the ball is +mgT

Change in momentum is 2mgT

This is no cause for concern, because we need to account for the earth. Its momentum flips from mgT to -mgT, changing by -2mgT. So the total change of the system is zero. Momentum is conserved and there is joy throughout the universe.

Now imagine if we had chosen to release the ball from a height of 0.6 meters or 1 meter. It would take longer to fall to the earth, and would reach a greater velocity. And the same applies to the earth. To conserve momentum, it too would reach a higher speed. When they collide, there would be a greater change in the momentum of each (though the total as we see is always zero).

That change in momentum is the impulse described up-thread, which is related to the force between the two objects. In the one case, the force of contact between the ball and the ground. In the other, the force between the juggler's hand and the pin (which is then transmitted to the bridge via the juggler's feet). Choosing to throw the pin higher than the previous throw can be achieved by increasing the amount of force used, which will then be transmitted to the bridge. Since the juggler has control over the amount of force applied, the amount of force transmitted to the bridge is not fixed, as implied by choice (E).

Again, it's simple to think about this in terms of recoil. Throwing a pin up is like firing a bullet up. If you fire a bullet up with a greater velocity, you're giving it more momentum, and there will be a greater recoil. If you throw the pin up with greater velocity, there will be a greater recoil force directed into the bridge.