The Juggler Problem

[Goonie]

Clearly the only way to resolve this is practical experimentation.


Any volunteers?.......

 

[Radagast]

Clearly the only way to resolve this is practical experimentation.

Solvitur ambulando, in fact.

 

[sjastro]

Well, no.

Your analysis is actually identical to mine, except that a > g in your analysis is a+g > g in mine (i.e. my a is your a−g). Either way, the maximal weight is ≥ 57 kg.

Claiming the maximal weight is ≥ 57 kg makes no sense at all.
When the juggler accelerates the pin upwards he/she is performing work against the gravitational field.
Since there is no upper limit in your case, the work W=Fx can be infinitely large since the acceleration a (and F) have no maximum value.
This violates the conservation of energy due to the work-energy principle.

However, you are assuming that the juggler is juggling in an optimal way, with exactly one pin in hand at all times. There is nothing in the problem formulation, as you presented it, which says so. Therefore there are no grounds for saying that the weight is precisely 57 kg.

This is not correct either.
The 57kg value is an expectation value not a precise value.

This can be proven as follows:
From my previous post the load on the bridge was found to be F= (M+5m)g =57kg.
The expectation value <F> can be calculated by using the time frames for when a pin is in the air, in the juggler's hand and the total cycle time.
If 2t1 is the time for the pin in the juggler's hand, 2t2 is the time for a pin in the air then 2(t1+t2) is the total cycle time where the factor 2 takes into account the symmetry of the upward and downward times.
Furthermore if the upward speed of the pin in the juggler’s hand is at1, it becomes the initial velocity for the pin when it becomes airborne and t1 and t2 are related by the equation t2 = at1/g.

The expectation value <F> becomes.
<F> = {2t1[(M+5m)g + ma] + 2t2(M+4m)g}/{2(t1+t2)}

The [(M+5m)g + ma] term is the load on the bridge when the juggler accelerates a pin in his hand for the time period 2t1.
The (M+4m)g term is the load on the bridge when the pin becomes airborne for the time period 2t2.
The 2(t1+t2) term is the total cycle time.

Eliminating t2 by using the substitution t2 = at1/g
<F> = {[2t1(M+5m)g + ma] + 2at1/g(M+4m)g}/{2(t1+at1/g)}

Rearranging terms:
<F> = [t1(M+5m)g+t1(M+5m)a]/[t1(g+a)/g]
= t1[(M+5m)(g+a)]/[t1(g+a)/g]
= (M+5m)g

Hence <F> = (M+5m)g= 57kg is an expectation value.

What the maths and physics tells you is that regardless of what the juggler might do a 57 kg load is the most likely outcome.

 

[Radagast]

Claiming the maximal weight is ≥ 57 kg makes no sense at all.

It's the truth -- see my derivation. Equality only holds if the juggler always has exactly one pin in hand. If there is ever a time when all 5 pins are in the air, then the launch acceleration is higher than optimal, and the maximal weight is strictly greater than 57 kg.

When the juggler accelerates the pin upwards he/she is performing work against the gravitational field.
Since there is no upper limit in your case, the work W=Fx can be infinitely large since the acceleration a (and F) have no maximum value.
This violates the conservation of energy due to the work-energy principle.

As the launch times approach zero, the acceleration a and the maximal weight do indeed both approach infinity. But that violates nothing (it only indicates that you hit limits imposed by human physiology).

The 57kg value is an expectation value not a precise value.

"Expectation value"? What do you mean by that? There's no probabilities here. Do you mean average weight? If so, then yes, the average weight is exactly 57 kg.

But it's the maximal weight that's important here, because that's what breaks the bridge.

If you impose the "optimal juggling" constraint (the juggler always has exactly one pin in hand) then the weight is exactly 57 kg all the time.

If you don't, then the weight is 47 kg when all pins are in the air, and > 57 kg during the periods when pins are being launched.

I don't think you understand the problem, quite frankly.

 

[essentialsaltes]

Nothing in the set up prevents the juggler from throwing it 'harder than she has to'.

Also, not that anyone has said this, she is not a 'spring' bouncing pins back up into the air by storing potential energy in her arm and then releasing it. The energy she uses to loft them into the air came from her Wheaties. She can control the F and x to impart an arbitrary amount work up to some practical limit.

 

[AirPo]

A, more information is needed. Like, how wide is the space he has to cross? Is it narrow enough he could simply toss the pins over and walk across?
tulc(likes to think outside the box)

Or walk the pins across one at a time.

 

[Not_By_Chance]

Now I remember why I hated physics at school.

 

[Waterwerx]

Unless I've messed up the equation...
First step would be to determine the maximum downward force allowed, which is determined by 50kg x 9.8m/s^2 = 490kgm/s^2. The force of the juggler is known(460.6kgm/s^2).
What must be determined is how much additional force in conjunction to the juggler will be exerted during the process of accelerating(downward and upward) the pins. Obviously a pin is going to be exerting more than 2kg x 9.8m/s^2 = 19.6kgm/s^2 when it contacts the juggler's hand.
Its been a while since I last dealt with physics problems, but I'm fairly certain the additional force of a single pin falling from a height of just 1 meter would be enough to make the total force > 490kgm/s^2. So the pin sizes & height the pins reach during juggling would have to be exceedingly/abnormally small(not to mention the juggler needing to be ridiculously fast), which the presented problem does not indicate.

 

[sjastro]

It's the truth -- see my derivation. Equality only holds if the juggler always has exactly one pin in hand. If there is ever a time when all 5 pins are in the air, then the launch acceleration is higher than optimal, and the maximal weight is strictly greater than 57 kg.



As the launch times approach zero, the acceleration a and the maximal weight do indeed both approach infinity. But that violates nothing (it only indicates that you hit limits imposed by human physiology).

For the acceleration to approach infinity the change in velocity has to be instantaneous, unless of course you think the velocity itself approaches infinity.
It seems to me you have confused yourself into thinking the juggling process is defined by a smooth mathematical function.
The process is defined by four separate time intervals, two intervals for the pins moving up and down in flight, and two for the pins in the juggler’s hand moving up and down.
None of the time intervals are infinitesimally small and the velocity is finite in these intervals hence the acceleration does not approach infinity.

The problem also explicitly states that the juggling action is smooth without jerking which excludes the possibility of high accelerations let alone infinite acceleration.

It is up to you demonstrate mathematically where the acceleration approaches infinity in the juggling cycle instead of making unsupported comments.

"Expectation value"? What do you mean by that? There's no probabilities here. Do you mean average weight? If so, then yes, the average weight is exactly 57 kg.

Since you don’t understand what the expectation value of a function is this might help.
The probability distributions are based on how long the pins are in the juggler’s hand or in the air divided by the total cycle time.
The equation for the expectation value <F> in my previous post clearly illustrated this.

You are also contradicting yourself.
If you agree the average weight is 57kg then your inequality cannot be correct as the 57kg is the minimum value.

But it's the maximal weight that's important here, because that's what breaks the bridge.

If you impose the "optimal juggling" constraint (the juggler always has exactly one pin in hand) then the weight is exactly 57 kg all the time.

If you don't, then the weight is 47 kg when all pins are in the air, and > 57 kg during the periods when pins are being launched.

By repeating this “optimal juggling” argument after I went into detail to show the expectation value <F>=57kg indicates you have no comprehension of the mathematics.
Let me make it as simple as possible for you with no maths and very little physics.
A 47kg juggler’s centre of mass is located in a certain position.
When he picks up the pins he now weighs 57 kg and his centre of mass shifts.
When he juggles the pins his centre of mass on average does not shift as the as the upward acceleration of the hands in shifting the centre of mass is countered by the downward acceleration of the hands and vice versa.

Since the average centre of mass doesn’t change his total weight on average remains at 57kg.

I don't think you understand the problem, quite frankly.

Coming from you that is a stunning retort.
Perhaps you are unaware this problem has been around for years well before I was aware of it.
The fact is Option (E) is the correct answer unless you want to argue the examiners who posed the problem didn’t understand it either.

 

[essentialsaltes]

When he picks up the pins he now weighs 57 kg and his centre of mass shifts.

The system has a mass of 57 kg, and the center of mass of the system is at some location.

When he juggles the pins his centre of mass on average does not shift as the as the upward acceleration of the hands in shifting the centre of mass is countered by the downward acceleration of the hands and vice versa.

But if he should choose [as nothing in the problem as written seem to preclude this] to throw the pins higher or lower, the center of mass of the system can move up or down. The center of mass could (on average) accelerate upward or downward, which would affect the normal force between the bridge and the feet (because yes, until the bridge breaks, the juggler is not accelerating).

(Or to look at it another way. We know that internal forces cannot accelerate the center of mass. But nothing prevents the juggler from throwing pins higher and higher. Which means there must be an external force. Since the center of mass is going up, it can't be gravity. Ergo, the normal force from the bridge must be increasing above and beyond the weight of the pins + juggler.)

Or take a simple limiting case. The juggler holds an anvil, and the total weight of juggler plus anvil is 100 N shy of breaking the bridge. Holding it fixed, no problem. If he pushes the anvil up with 50 N more force, the anvil will start to accelerate up. But the anvil pushes down on the person with an extra 50 N by the Third law. To keep the person from falling, the bridge will have to press up 50 N harder as well. So far so good. But if the juggler chooses to push up on the anvil with 250 N, the bridge will break.

Perhaps you are unaware this problem has been around for years well before I was aware of it.
The fact is Option (E) is the correct answer unless you want to argue the examiners who posed the problem didn’t understand it either.

I think the question is not posed very well. So yes, I will take the examiners to task.

 

[sjastro]

The system has a mass of 57 kg, and the center of mass of the system is at some location.



But if he should choose [as nothing in the problem as written seem to preclude this] to throw the pins higher or lower, the center of mass of the system can move up or down. The center of mass could (on average) accelerate upward or downward, which would affect the normal force between the bridge and the feet (because yes, until the bridge breaks, the juggler is not accelerating).

(Or to look at it another way. We know that internal forces cannot accelerate the center of mass. But nothing prevents the juggler from throwing pins higher and higher. Which means there must be an external force. Since the center of mass is going up, it can't be gravity. Ergo, the normal force from the bridge must be increasing above and beyond the weight of the pins + juggler.)

Or take a simple limiting case. The juggler holds an anvil, and the total weight of juggler plus anvil is 100 N shy of breaking the bridge. Holding it fixed, no problem. If he pushes the anvil up with 50 N more force, the anvil will start to accelerate up. But the anvil pushes down on the person with an extra 50 N by the Third law. To keep the person from falling, the bridge will have to press up 50 N harder as well. So far so good. But if the juggler chooses to push up on the anvil with 250 N, the bridge will break.



I think the question is not posed very well. So yes, I will take the examiners to task.

One can take a snapshot of any given cycle such as throwing a pin upwards which imparts a load reaction force on the bridge.
By the same token a snapshot of another cycle can show a pin being accelerated downwards in the hand which results in a tension reaction force of equal and opposite magnitude.
The point being that juggling over a large number of cycles these opposing configurations have a tendency of evening out with the load and tension forces cancelling each other leaving as you put it the weight of the system.

Option(E) was never meant to be a snapshot view for any given cycle but an expectation value which equals the weight of the system.
I have no problems with the examiners.

 

[essentialsaltes]

One can take a snapshot of any given cycle such as throwing a pin upwards which imparts a load reaction force on the bridge.
By the same token a snapshot of another cycle can show a pin being accelerated downwards in the hand which results in a tension reaction force of equal and opposite magnitude.

The juggler presumably is always exerting an upward acceleration on the pin. Both when the pin is 'decelerating' and when it's being thrown up.

Regardless, if the juggler decides to start throwing higher or lower, the 'tension' and the load do not have to be equal in magnitude.

The point being that juggling over a large number of cycles these opposing configurations have a tendency of evening out

These subtleties and wiggle words about tendencies and expectation values are not in the question.

 

[essentialsaltes]

Regardless, if the juggler decides to start throwing higher or lower

Or hey, he could gradually crouch down at a mild acceleration, lowering the center of gravity of the system.

 

[sjastro]

The juggler presumably is always exerting an upward acceleration on the pin. Both when the pin is 'decelerating' and when it's being thrown up.

You haven't looked at the context of the entire cycle.
When a juggler catches a pin, the following movement is a downwards acceleration.
The cycle is not simply the upwards acceleration on the pin.

Regardless, if the juggler decides to start throwing higher or lower, the 'tension' and the load do not have to be equal in magnitude.

Who said they should be exact?
The important point however the disparity between the two is nowhere near as great as what you think for the simple reason that ignoring atmospheric resistance, gravity is a conservative force.
A juggler’s centre of mass moves in a “closed path” during a cycle.
Since the gravitational field is conservative the work performed over a closed path is zero so it’s not at all surprising one would expect the tension and load reaction forces to be near equal.
It’s the “variations” in the juggling process plus atmospheric resistance on the pins that result in deviation which is dealt with by considering the expectation value.

These subtleties and wiggle words about tendencies and expectation values are not in the question.

I’m sorry you have such a negative opinion of expectation values as a wiggle words despite the mathematical framework described in this post.
Quantum physicists would be similarly unimpressed as the expectation value of quantum mechanical operators plays an important role in the theory.

 

[essentialsaltes]

You haven't looked at the context of the entire cycle.
When a juggler catches a pin, the following movement is a downwards acceleration.

"show a pin being accelerated downwards"

The arm/hand may accelerate downwards when the pin hits the hand, but the acceleration of the pin is going to be upward when it meets the hand.

The cycle is not simply the upwards acceleration on the pin.

When the pin is free in the air, there is only gravity, and the pin is accelerating downward. When the pin is touching the hand, the acceleration of the pin will be upward (unless this is a very strange juggling movement).

Who said they should be exact?

You wrote "a tension reaction force of equal and opposite magnitude"

A juggler’s centre of mass moves in a “closed path” during a cycle.

Once again, it is the COM of the system that is relevant. If the juggler throws the balls higher or lower, he can move the center of mass.

Furthermore, the juggler's own COM does not necessarily move in a closed path, if he crouches down during the course of his walk in order to keep the bridge from breaking. The 'closed path' situation is an assumption not stated in the problem.

Quantum physicists would be similarly unimpressed as the expectation value of quantum mechanical operators plays an important role in the theory.

Certainly, but this is not a problem in QM, but a mechanics problem posed to high school students.

 

[sjastro]

"show a pin being accelerated downwards"

The arm/hand may accelerate downwards when the pin hits the hand, but the acceleration of the pin is going to be upward when it meets the hand.



When the pin is free in the air, there is only gravity, and the pin is accelerating downward. When the pin is touching the hand, the acceleration of the pin will be upward (unless this is a very strange juggling movement).

Take a look at this juggling video.

The video clearly indicates when the pin is caught it is accelerated downwards in the hand including the wrist snap.
In case you are entertaining the idea this is the result of the hand being pushed back by the momentum of the pin, the final frames shows the hands remaining stationary in catching the pins to complete the routine.

You should also note the final throw in the juggler’s right hand where the downward acceleration of the pin is most apparent, corresponds not surprisingly to the highest toss as more potential energy is converted into kinetic energy which leads to a higher initial velocity when the pin is released.

You wrote "a tension reaction force of equal and opposite magnitude"

You are taking me out of context.
This is what I wrote.

“By the same token a snapshot of another cycle can show a pin being accelerated downwards in the hand which results in a tension reaction force of equal and opposite magnitude.”

I was addressing your issue about having the tension and load equal in the same cycle.
The quote pertains to comparing the load reaction force of one cycle to the tension reaction force in another in the context of defining the expectation value.

Once again, it is the COM of the system that is relevant. If the juggler throws the balls higher or lower, he can move the center of mass.

Furthermore, the juggler's own COM does not necessarily move in a closed path, if he crouches down during the course of his walk in order to keep the bridge from breaking. The 'closed path' situation is an assumption not stated in the problem.



Certainly, but this is not a problem in QM, but a mechanics problem posed to high school students.

The juggler’s COM moves because the juggler is performing work.
Now that it has been established a pin is accelerated downwards as per the video, positive and negative work is performed during the cycle.
Assuming a perfect world where the juggler’s technique is flawless in the absence of atmospheric resistance, total work performed in a cycle is zero.
If the total work is zero then the juggler’s COM does follow a closed path which is also consistent with gravity being a conservative force.
In an imperfect world the total work for any given cycle may not be exactly zero but it is still the most likely outcome.

It also knocks your idea on the head that crouching would make any difference.
All you are doing in this case is shifting the origin of the COM; the COM will still follow a closed path around this origin.
Even if crouching was gradually performed over the bridge how many steps would it would take before this was no longer possible?

Two valid assumptions can be made about the problem is that the bridge is long and juggling is normal without performing any party tricks.

 

[Waterwerx]

Some of you are taking a simple physics problem and making it more complex than what it actually is...

Will the additional upward/downward acceleration of ONE pin cause the juggler to exceed the maximum tolerance of the bridge? The answer was yes.

Questions like this on exams/tests are pretty much designed to weed out people that would, for example, fill a tank of air up with 10 PSI of air to fill a car tire to 32 PSI that already has 22 PSI air in it.

 

[essentialsaltes]

Take a look at this juggling video.

As the video clearly indicates when the pin is caught it is accelerated downwards in the hand including the wrist snap.

Certainly the head of the pin dips down after being caught, but I doubt that there is much, if any, downward acceleration of the pin. To a first approximation the COM of the pin makes a parabola under the influence of gravity with the acceleration pointing down, and an inverted parabola under the influence of the hand with the acceleration pointing up.

But even if it exists, this downward acceleration is not of much relevance to the solution of the problem. If the pin is falling toward the earth with velocity -v, and the hand accelerates it more downwards slightly to make it travel even faster toward the ground -(v+some amount), all that means is that the hand will have to apply even more upward acceleration to get the velocity back up to +v (under the assumption the juggler wants to throw the pin to same height).

Fatally to your later argument about positive and negative work, if the hand is accelerating the pin downward, and the pin is moving downward, the force and displacement are in the same direction, so positive work is being done.

(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.)

You should also note the final throw in the juggler’s right hand where the downward acceleration of the pin is most apparent, corresponds not surprisingly to the highest toss as more potential energy is converted into kinetic energy which leads to a higher initial velocity when the pin is released.

No no no. Although the arm musculature has some characteristics of a spring, it is just not the case that the arm is storing potential energy to throw the pin back up.

This again is why I discuss the juggler's choice of how high to throw the pins. The muscles and the forces applied are under the control of the juggler. They are not regulated by 'storage of potential energy'.

[To throw it higher, one thing you might do is to exert your force over a longer time/distance. One way to do this is to start the throw from lower down. This, I submit, is why the pin dips lower on that throw. The hand is allowing the pin to fall further (rather than actually forcing it down)]

The juggler’s COM moves because the juggler is performing work.
Now that it has been established a pin is accelerated downwards as per the video, positive and negative work is performed during the cycle.

The negative work cannot come from a downward acceleration as you describe.

Assuming a perfect world where the juggler’s technique is flawless in the absence of atmospheric resistance, total work performed in a cycle is zero.

No, if the juggler chooses to throw the pins higher and higher each time he touches it, he performs positive work in each cycle. And vice versa if he throws them lower each time.

It also knocks your idea on the head that crouching would make any difference.
All you are doing in this case is shifting the origin of the COM; the COM will still follow a closed path around this origin.

No, if he skillfully continuously moved his center of mass, he could see to it that the center of mass was accelerating. This would imply there is a net force. If he crouched lower, his acceleration would be downward. This means that the normal force supplied by the bridge is less than the weight of the system, since the acceleration is downwards.

I agree it's somewhat absurd, and I didn't think of it initially, but the idea of throwing the pins to different heights did occur to me, and seems much less absurd. And would have the same effect of causing a net acceleration that would alter the forces involved.

Two valid assumptions can be made about the problem is that the bridge is long and juggling is normal without performing any party tricks.

They're valid because you say so? I mean, I grant you that 99% of high school students facing that question are going to ignore air resistance. A common unstated physics assumption. But some of these others are not so immediately obvious. Nor are they stated in the problem. As I said before, the best answer is E (assuming you're not going to wimp out and say A) but I think the problem was not very carefully written.

 

[Strathos]

Clearly the only way to resolve this is practical experimentation.


Any volunteers?.......

You could just juggle while standing on a scale...

 

[sjastro]

Certainly the head of the pin dips down after being caught, but I doubt that there is much, if any, downward acceleration of the pin. To a first approximation the COM of the pin makes a parabola under the influence of gravity with the acceleration pointing down, and an inverted parabola under the influence of the hand with the acceleration pointing up.

But even if it exists, this downward acceleration is not of much relevance to the solution of the problem. If the pin is falling toward the earth with velocity -v, and the hand accelerates it more downwards slightly to make it travel even faster toward the ground -(v+some amount), all that means is that the hand will have to apply even more upward acceleration to get the velocity back up to +v (under the assumption the juggler wants to throw the pin to same height).

Ok I accept the fact that a juggler’s release velocity of the pin is also determined by factors involving biomechanics which affect the release velocity v out of the juggler’s hand.
You are still ignoring the gravitational field is conservative.
A consequence of this if the juggler throws the pin up at any velocity v, it will come back at a velocity –v at the same vertical level.

Now for a bit of maths:
F=ma
= mdv/dt
= dp/pt where p=mv is the momentum.

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

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.

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

Fatally to your later argument about positive and negative work, if the hand is accelerating the pin downward, and the pin is moving downward, the force and displacement are in the same direction, so positive work is being done.

(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.)

The negative work cannot come from a downward acceleration as you describe.

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.”

The trajectory of the pin in the juggler’s hand is an undefined curve as are the general vector for the tangential force and the parametric equations describing the curve.
Calculating the work performed anywhere along a length interval of the curve is unknown.
What I do know however is the total work must be zero as gravity is a conservative force and the COM follows a closed loop.

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

No, if the juggler chooses to throw the pins higher and higher each time he touches it, he performs positive work in each cycle. And vice versa if he throws them lower each time.

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

The total work W = ∫ F.dx = 0 since F=0.

Furthermore the gravitational field cannot be conservative if the total work is non zero for the cycle.
In a conservative gravitational field the sum of the kinetic and potential energies is constant anywhere along the pin’s trajectory in flight.
This cannot be true if the total work is non zero.
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.

No, if he skillfully continuously moved his center of mass, he could see to it that the center of mass was accelerating. This would imply there is a net force. If he crouched lower, his acceleration would be downward. This means that the normal force supplied by the bridge is less than the weight of the system, since the acceleration is downwards.

I agree it's somewhat absurd, and I didn't think of it initially, but the idea of throwing the pins to different heights did occur to me, and seems much less absurd. And would have the same effect of causing a net acceleration that would alter the forces involved.

Since you have employed a biomechanical argument let me return the compliment.
Can you explain how a juggler progressively crouching while walking over the bridge while simultaneously trying to juggle does not push off the bridge with his feet thereby increasing the load?
Then there is the other inconvenient point that the juggler can only travel a limited number of steps before he can no longer crouch any further.
What happens if the bridge is longer than the range at which the juggler can crouch?

They're valid because you say so? I mean, I grant you that 99% of high school students facing that question are going to ignore air resistance. A common unstated physics assumption. But some of these others are not so immediately obvious. Nor are they stated in the problem. As I said before, the best answer is E (assuming you're not going to wimp out and say A) but I think the problem was not very carefully written.

I’m sticking to the script of the problem.
You on the other hand are trying to rewrite it with crouching jugglers and extremely short bridges.
Ironically your attempts in trying to beat the Juggler Problem are only confirming the validity of Option (E).