The Juggler Problem

[sjastro]

This problem turned up a few years ago in the selection process for high school students to compete in the Physics Olympiad.

An expert juggler, carrying five juggling pins, has to cross a swing bridge which has a maximum safe load rating of 50 kg. The juggler weighs 47 kg and each of his five pins weighs 2 kg. He believes he can make it across safely in one trip by juggling the pins, so that he is never holding more than one pin. His skill enables him to juggle smoothly without any jerking.
He is
A. incorrect – more information is needed.
B. correct – the total weight will never exceed 49 kg
C. correct – no jerking means no extra weight.
D. correct – the total weight can be made to exceed 49 kg by an arbitrarily small amount.
E. incorrect – the total weight is 47 kg + 5x2 kg = 57 kg.

The challenge here is not only to select to correct answer but to use physics to explain the selection.
Good luck!

 

[Sabertooth]

Intuitively, while his total mass might never exceed 49 kg, his total weight most certainly will.

First, when he catches/decelerates a falling pin and second, when he throws/accelerates that same pin. (Even if he does not jerk in the process.)

 

[miamited]

Hi sjastro,

Well, it's my understanding that a juggler does have at least two of the items being juggled in hand at regular intervals. When he catches one falling he has the next one in his other hand to throw up. Now, there are milliseconds when he wouldn't but if the weight restriction cannot be broken, even for a moment, I'm going with it can't be done. 47kg for the Juggler and 4kg for two pins would break the limit.

However, if we are to then assume that 'safe load limit' can be broken by very small increments before the bridge would actually collapse, then he could. So, D would be the most correct answer in this case. Just as with an automobile tire, the 'maximum psi' limit is usually 44psi. However, anyone can put 50psi in such a tire and the tire will not explode and it will still carry a vehicle. The problem would come if you hit a pothole that would increase the pressure as the tire collapses into itself even beyond the 50psi or you seriously overload the tire.

God bless,
In Christ, ted

 

[tulc]

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)

 

[Snappy1]

More info or clarification is needed

 

[Snappy1]

I don't know physics so I couldn't draw out the math but I might be able to talk this through. Assuming he really is only holding one pin at a time, juggling techniques be darned, his weight won't exceed 49kg. What I don't know about is the force of the 1kg pin added to his 47kg weight. I would think you would have to know the height of the pin hes catching to know if that would push his total weight over 50kg.

 

[Radagast]

This problem turned up a few years ago in the selection process for high school students to compete in the Physics Olympiad.

An expert juggler, carrying five juggling pins, has to cross a swing bridge which has a maximum safe load rating of 50 kg. The juggler weighs 47 kg and each of his five pins weighs 2 kg. He believes he can make it across safely in one trip by juggling the pins, so that he is never holding more than one pin. His skill enables him to juggle smoothly without any jerking.
He is
A. incorrect – more information is needed.
B. correct – the total weight will never exceed 49 kg
C. correct – no jerking means no extra weight.
D. correct – the total weight can be made to exceed 49 kg by an arbitrarily small amount.
E. incorrect – the total weight is 47 kg + 5x2 kg = 57 kg.

The challenge here is not only to select to correct answer but to use physics to explain the selection.
Good luck!

Nice problem. Let's say he throws the pins up with velocity v, and they stay up for T seconds. Then T = 2v/g.

The pins come back down with the same velocity v. Over a time of t seconds he decelerates them and then accelerates them up to launch velocity again, with acceleration a. Then t = 2v/a.

He can only have one pin in hand, so t ≤ T/4. Hence a ≥ 4g.

His maximum weight will be 47 kg + 2 kg + the effect of accelerating the pin, for a total of at least 57 kg.

So I'm calling A as the answer.

 

[Sabertooth]

...+ 2 kg + the effect of accelerating the pin,...

Should the mass of the pin & the effects of deceleration/acceleration be two separate addends?

I would have guessed that the functional weight of the pin would be a product of those two factors.

 

[Radagast]

Should the mass of the pin & the effects of deceleration/acceleration be two separate factors?

I would have guessed that the functional weight of the pin would be a product of those two factors.

No, the upward acceleration is just added to the acceleration due to gravity. Or, if you like, there are two downward forces being added together, F1 = mg due to gravity and F2 = ma due to Newton's third law and the acceleration/deceleration.

So weight = 47 + 2(a + g)/g ≥ 57 kg.

 

[Sabertooth]

...+ 2(a + g)/g...

The underlined portioned looks more like the product that I envisioned (rather than two addends).

 

[Radagast]

The underlined portioned looks more like the product that I envisioned (rather than two addends).

No, that's the sum a + g together with converting the weight (which is actually a force F = mg) back to kg units.

 

[Sabertooth]

No, that's the sum a + g together with converting the weight (which is actually a force F = mg) back to kg units.

Hence a ≥ 4g.

Since a is minimally 4g, option E is the best answer.

So weight = 47 + 2(a + g)/g

= 47 + 2(4g + g)/g
= 47 + 2(5g)/g
= 47 + 10g/g
= 47 + 10
= 57

E. incorrect – the total weight is 47 kg + 5x2 kg = 57 kg.

Juggling five pins has the same effect as carrying five pins.

That answer seems to be true no matter how many pins he is juggling. If he has six pins, that means five are in the air, requiring an acceleration of 5g instead of 4g.

Even if he has two in his hand and three in the air [a > 3g], he will weigh the same (as the OP).

 

[FrumiousBandersnatch]

I think his effective weight will oscillate around the mean of the total weight of juggler and pins.

This is a version of an old puzzle involving a closed truck full of pigeons with a combined weight greater than the maximum load of a bridge (the truck alone is less than the limit). The driver thinks that by getting all the pigeons to fly inside the truck while he drives over, he'll stay inside the weight limit. Is he right?

He can also open the sides of the truck so air can move freely through the chicken wire - will this make a difference?

 

[Radagast]

Since a is minimally 4g, option E is the best answer.

Juggling five pins has the same effect as carrying five pins.

Only if he juggles "optimally." If there is ever a time when both his hands are empty, then the effective total weight will have been over 57 kg.

It is not clear that "no jerking" means "juggling optimally."

 

[Ophiolite]

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, as in this case, across the box canyon.

 

[essentialsaltes]

The challenge here is not only to select to correct answer but to use physics to explain the selection.
Good luck!

I don't really like any of the choices. But barring things like throwing some of them across the whole span, I guess I like E the best.

When you stand on a bridge, gravity is tugging you down, and the bridge is supporting you with what we usually call the normal force. Since you're not accelerating, they are equal and opposite.

If you hold an object in your hand, it's weight is balanced by a normal force provided by your hand. (and the normal force coming from the bridge into your feet has to be that much larger to again keep everything at rest).

If you want to accelerate it up into the air, you increase the normal force from your hand on the object, (and it will increase its force against you hand by the 3rd law) and again that is transferred to your feet and the bridge has to increase its force to support you.

But when you let go, the force between your hand and the object disappears, and the force of the bridge will go back to being just your weight. (I think there are some subtleties here, since your arm itself has some mass and you're flinging it around, but I take the no jerkiness rule to mean that we're neglecting this. Or in catlike fashion she moves her other arm to counteract it (but then there's torsion, egad))

So to some extent the pins in the air don't matter. They're not touching the juggler, so they can't exert any force on her or the bridge.

So the question is... how much additional force does it take to loft the pins in the air? Hard to say. How high is she throwing each one? It doesn't say.

But we can set some boundaries. Those pins in the air don't exert any force on the bridge. But... they aren't just levitating or flying under their own power(*). They're only in the air because of the impetus given them by the juggler. If we imagine her throwing them all the same height as she crosses the bridge at constant speed, the acceleration of the center of mass is zero. We know the total force of gravity is too much for the bridge, and any forces applied by the hand get transmitted to the bridge, so the supporting force has to be equal and opposite for there to be no acceleration. Which is too much for the bridge.

So I would say the average force over time is too much for the bridge, but it probably varies.

And you could imagine a weird situation where she starts by throwing the pins very high, and ends with them at the lowest she can juggle, so the motion of the center of mass of the system actually is going down, so that could reduce the amount of force she has to apply, but not I think enough to save the bridge under any realistic circumstances.

(*)This is why the pigeon transporter is interesting. The pigeons do fly under their own power. if the truck is totally enclosed, though, first of all you're a monster because they'll suffocate. Second, the beating of the wings on the enclosed air will press against the floor of the truck, and so you don't benefit. If air moves freely, however, there is a chance that the breeze can dissipate and less of the flapping force will push down on the truck.

 

[FrumiousBandersnatch]

(*)This is why the pigeon transporter is interesting. The pigeons do fly under their own power. if the truck is totally enclosed, though, first of all you're a monster because they'll suffocate.

For animal welfare reasons, we must assume the truck has a minimal ventillation system

Second, the beating of the wings on the enclosed air will press against the floor of the truck, and so you don't benefit. If air moves freely, however, there is a chance that the breeze can dissipate and less of the flapping force will push down on the truck.

Yes, that's how I see it too. In the enclosed truck, it's a 'black box' weighing (pigeons + container); in the open truck, some of the weight of the birds supported by their flapping is distributed by air movements outside the truck (probably as vortices), so it's a gamble...

 

[Anguspure]

This problem turned up a few years ago in the selection process for high school students to compete in the Physics Olympiad.

An expert juggler, carrying five juggling pins, has to cross a swing bridge which has a maximum safe load rating of 50 kg. The juggler weighs 47 kg and each of his five pins weighs 2 kg. He believes he can make it across safely in one trip by juggling the pins, so that he is never holding more than one pin. His skill enables him to juggle smoothly without any jerking.
He is
A. incorrect – more information is needed.
B. correct – the total weight will never exceed 49 kg
C. correct – no jerking means no extra weight.
D. correct – the total weight can be made to exceed 49 kg by an arbitrarily small amount.
E. incorrect – the total weight is 47 kg + 5x2 kg = 57 kg.

The challenge here is not only to select to correct answer but to use physics to explain the selection.
Good luck!

The juggler and his pins are a closed system. At any one time the juggler is supporting the weight of 5 pins due to his activity, and he maintains this support continuosly and smoothly through a transfer of inertia.

Therefore on average the total weight experience at his feet will be 57 kg, which will fluctuate depending upom how smoothly the juggler is able to maintain his act.

 

[sjastro]

Thanks the comments.
The correct option is (E).
Consider the case of the juggler throwing a single pin without catching it.
If the juggler’s mass is M and the pin mass is m then the total load on F on the bridge is:

F = Mg + ma

Mg is the juggler’s weight, g is the acceleration due to gravity, a is the upward acceleration of the pin in the juggler’s hand and ma is the downward reaction force according to Newton’s third law.
The condition a > g is required to get the pin airborne.

Now consider the simple case of juggling 2 pins on the bridge.
When a pin is in the air the time in flight is t= 2v/g where v is the velocity of the pin when released from the juggler’s hand.
The downward reaction force is somewhat more complicated since when the juggler catches the pin, the downward acceleration has to be overcome before the pin is accelerated upwards and released into the air.
The time in the juggler’s hand T is defined as:

T= 2v/(a-g)

When juggler catches the pin, the pin continues the downward acceleration g with velocity –v before the juggler’s upward acceleration a takes over which eventually leads to the pin being released into the air with velocity v.
Since the criteria for juggling two pins is to have one pin in the air, the condition for the “ideal” acceleration a is governed by the condition T = t.
If T > t the juggler’s acceleration is too low and the juggler will be holding both pins.
If T < t the juggler’s acceleration is too high resulting in a large reaction force that could exceed the load of the bridge.

When T = t, 2v/g = 2v/(a-g)
Solving for a gives a = 2g which is the required acceleration for the ratio t/T = 1.

We can now extend this to the Juggler Problem involving 5 pins with 4 pins in the air during juggling and 1 in the hand.
The ratio is now t/T = 4.
Solving the equation (2v/g)/(2v(a-g)) = 4 gives a = 5g.
The total load on the bridge F becomes:
F = Mg + 5mg

Mg = 47 kg is the weight of the juggler, mg =2 kg is the weight of the pin.
Hence F the total weight is 47 kg + 5x2 kg = 57 kg which is option (E).

What the maths tells us is the juggler’s task to cross the bridge is impossible as the upward acceleration is too high to maintain 4 pins in the air without exceeding the load limit of the bridge.

 

[Radagast]

The correct option is (E).

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.

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.