A Juggler's Weight

[Shnoz (Sophie)]

I was studying for the Physics Olympiad (as you do) when I got a question wrong in the exam paper, and I cannot figure out why.
The question is as follows:



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.



I chose D, because I though that the juggler's weight, plus the weight of the single pin would equal 49kg, but would be slightly exceeded by the force of acceleration.
But the correct answer is E, according to the examination's solution papers. This completely baffled me. How can the pin's weight affect the juggler's weight and the bridge when they contact neither?

[xelasnave]

What do Myth Busters say?
Could we use E=MC^2?
I have no idea, I wonder if the height the pins are throw enters into it...

alex

[sjastro]

It makes sense.

It's Newton's third law in operation. For every action there is an equal an opposite reaction. The maximum reaction force of the bridge is it's load rating.

You need to consider that the juggler is also applying a force to each pin to overcome the effects of gravity in the act of juggling.

The total reaction force of the bridge is therefore the weight of the juggler plus the weight of the pins, plus mass X acceleration of each pin during juggling.

This will exceed 57 kg and the bridge will come tumbling down.

Regards

Steven

[Shnoz (Sophie)]

Ah, I'd overlooked Newton's Third Law! I understand now, thanks for helping.

[OneOfOne (Trevor)]

I don't know if I agree with their answer, I consider it to be the LEAST correct. The jugglers static weight for most of the time will be 47+2kg. At any instant he will need to provide energy to accelerate a single pin into the air, which will temporarily increase his "weight" to something above the static 49kg, this amount being dependent on the height to which he throws the pin, the higher it goes, the more the force, the greater the increase in his weight. Once the pin has been released, his weight will drop to 47kg as all pins will be in the air and so will have no influence on his weight. At some instant, he will need to catch a single pin and his effective weight will increase again, probably by the same amount it increased when he threw the pin into the air. So his maximum weight will average to something around 49kg plus the force to accelerate a single pin....

If you pushed the experiment to the limit, he could throw 4 pins high enough into the air that he would not need to catch them until he gets to the other side and would only need to throw a single pin into the air whilst on the bridge...

[sjastro]

Originally Posted by OneOfOne

I don't know if I agree with their answer, I consider it to be the LEAST correct. The jugglers static weight for most of the time will be 47+2kg. At any instant he will need to provide energy to accelerate a single pin into the air, which will temporarily increase his "weight" to something above the static 49kg, this amount being dependent on the height to which he throws the pin, the higher it goes, the more the force, the greater the increase in his weight. Once the pin has been released, his weight will drop to 47kg as all pins will be in the air and so will have no influence on his weight. At some instant, he will need to catch a single pin and his effective weight will increase again, probably by the same amount it increased when he threw the pin into the air. So his maximum weight will average to something around 49kg plus the force to accelerate a single pin....

If you pushed the experiment to the limit, he could throw 4 pins high enough into the air that he would not need to catch them until he gets to the other side and would only need to throw a single pin into the air whilst on the bridge...

The question is what is the total load on the bridge. The total load is the sum of all the gravitational and inertial forces applied to the bridge. In this context the gravitational force is the weight of the juggler and the pins.

The inertial force needs to be greater than 2 kg-force for each pin. The inertial force acts on the bridge irrespective if the pin is about to leave the juggler's hand or is airborne.

The total load on the bridge exceeds 57kg.

Regards

Steven

[xelasnave]

What force is required to "throw" each pin.
Each pin is 2 kg so to throw one pin say 10 feet up maybe we need more than 2 kgs force and that inertial force may be many fold the weight of the pin...no doubt it can be worked out... and I wonder do we add the force when the pin accelerates from say a height of ten feet to the hands of the juggler... I expect the force of a pin falling from ten feet will be more than its weight of 2 kilos....

I think the problem is more complex than it seems ... well it is to me and I can only observe my reality not the reality as experienced by others...

AND how long is say force from the falling pin time wise... if force needed to be cumulative time separation would be relevant maybe.

alex

I am going out to find a juggler and a bridge

[Shnoz (Sophie)]

I think perhaps what xelasnave said could have a interesting effect upon the answer. Since in this respect, force is going to be transferred into weight upon the bridge, throwing the pins higher would require more force. Therefore, a larger reaction would occur on the bridge. If the juggler threw even a single pin high enough, he could theoretically exceed the bridge's safe load rating.
Perhaps the answer is neither D nor E, but A; because we require to know how much force the juggler is placing upon the pins.

[xelasnave]

Yes if he threw it high enough the pin may smash his hand and the bridge .
alex

[sjastro]

Originally Posted by Shnoz

I think perhaps what xelasnave said could have a interesting effect upon the answer. Since in this respect, force is going to be transferred into weight upon the bridge, throwing the pins higher would require more force. Therefore, a larger reaction would occur on the bridge. If the juggler threw even a single pin high enough, he could theoretically exceed the bridge's safe load rating.
Perhaps the answer is neither D nor E, but A; because we require to know how much force the juggler is placing upon the pins.

Anythings possible. The juggler could be superman and throw the pins out of orbit.

Remember the Juggler is walking across the bridge as he/she is juggling. Assuming that juggling involves throwing the pins vertically, a pin thrown too high will not be retrieved.

The Juggler doesn't have much scope as to how far a pin can be thrown, and while E is not technically correct, it's still probably closest to being the correct answer.

Steven

[OneOfOne (Trevor)]

Yes, I think the solution is rather involved as you would need to know how much force is required to accelerate a single pin to a suitable height, I expect it could exceed the 2kg of the pin itself. The greater the height, the greater the effective increase in the "apparent" weight of the juggler for that instant. Once the pin has left the hand of the juggler, the weight of that pin and the reactionary force generated by the throw is no longer an influence on the bridge. Any pin that is in the air is of no consequence to the equation at all until it is caught again as each one is in free fall.

Myth Busters did a similar experiment with a remote controlled helicopter in an enclosed room and found that the total mass of the room did not change when the helicopter was flying as the force expended in keeping it airborne created a downward pressure onto the floor of the room equal to the weight of the helicopter. However, if the object were in free fall, as the pins are, they do not present a force onto the ground (or bridge in this case) and so are only part of the equation whilst in contact with the juggler.

Do we have a juggler in our midst who has access to a good pair of scales? This would provide the best way to get some real experimental results.

[sjastro]

Originally Posted by OneOfOne

Once the pin has left the hand of the juggler, the weight of that pin and the reactionary force generated by the throw is no longer an influence on the bridge. Any pin that is in the air is of no consequence to the equation at all until it is caught again as each one is in free fall.

That is simply not true. The reaction/inertial force is a function of the juggling action and has nothing to do with the pins. A juggler can generate the same inertial forces with or without pins. It's all in the the acceleration of the arms and upper body. And since juggling is a continuous process the inertial forces are there throughout the entire cycle.

There is also the consideration that walking adds to the inertial forces.

Regards

Steven

[Kal (Andrew)]

This is an old quiz question, a 'Millergram' from Professor Julius Sumner Miller from back in the 60's last century.

Q32: A juggler comes to a foot-bridge of rather flimsy design. He has in hand four balls. The safe load is no more than the juggler himself and one ball. Can he get across the bridge by juggling the balls, always having at most one ball in the hand (and three in the air)?
A: No. A falling ball exerts a force on the hand greater than its own weight.
Rather, a 'thrown' ball exerts greater force than a 'held' one. That is, the additional force equal and opposite to that imparted to a flung ball, in addition to the juggler's mass, would exceed the bridge's tolerance (the bridge can tolerate a juggler and held ball, but not the additional downward force associated with forcing a ball 'up').

The correct answer is E because in order to juggle and have only 1 pin in his hand at a time he will have to impart a minimum accelerating force on that pin equivalent to the gravitational force of 4 pins. Lets assume 1 second in the hand with a constant acceleration for that second, plus 4 seconds in the air for each pin. Therefore to throw the pin and have it come back in 4 seconds you need to accelerate it at 4 x 9.8ms-1. This equals the same force as holding 4 pins in your hand with only the basic 9.8ms-1 acceleration of gravity on them, add the weight of the pin being thrown and you have the 5 pins weight accounted for.

[g__day (Matthew)]

I would have said insufficient information to agree with E) - but its on the right track. You need to know how much time he has each pin in his hand - this will give you the minimum acceleration needed to loft each pin into the air.

If he exceeds the minimum acceleration by a large enough factor he would make it across. For example he throws a pin 500 yards in the air - smoothly - then does the same to three more pins - walks across simply holding the fifth pin - which if he wants he could launch once he's on solid ground - before the first has yet descended. If the bridge is short enough and he is fast and powerful enough - he makes it!

[Terry B]

Originally Posted by g__day

I would have said insufficient information to agree with E) - but its on the right track. You need to know how much time he has each pin in his hand - this will give you the minimum acceleration needed to loft each pin into the air.

If he exceeds the minimum acceleration by a large enough factor he would make it across. For example he throws a pin 500 yards in the air - smoothly - then does the same to three more pins - walks across simply holding the fifth pin - which if he wants he could launch once he's on solid ground - before the first has yet descended. If the bridge is short enough and he is fast and powerful enough - he makes it!

So who can throw anything 500 yards?
It has to be realistic to work.

[Shnoz (Sophie)]

If we assume our juggler is not Superman and must cross the bridge juggling, then if he intends to keep the pins in the air for 4 seconds, then for every pin he accelerates into the air, its inertia causes the pin to experience 5 g's, as Kal said. So each pin weighs as much as five pins when accelerating. So after you apply Newton's third law, the juggler's weight, plus the accelerating pin's weight equals 57kg. This seems like the kind of logic the answer would work on.
Although, I assume the 'weight' of the pins would change depending on how high they were thrown. If the pins only had to stay in the air for 1 second (he must be a VERY skilled juggler), would that make it possible for the juggler to cross the bridge?

[Kal (Andrew)]

If teh pins are in the air for 1 second then he must take exactly 0.25 of a second to throw the pin, with guess what - the EXACT same acceleration as used in my previous example. Acceleration and force is the same, the only thing that changes are length of acceleration while in his hand & how high he throws the pins (and hence how long they are in the air)

[Kal (Andrew)]

And if he throws teh pins 500 yards in the air he is breaking the rule descibed in the original question "His skill enables him to juggle smoothly without any jerking."

[Shnoz (Sophie)]

Ah yes, 1:4 seconds and 0.25:1 seconds are the same ratio. I was never great at maths. So indeed, the acceleration would remain the same, as would the force and reaction, independent of the time taken.
Thanks for helping!

[sjastro]

Hmmm. 4Gs coming out of the hand. I don't think so with a juggling action. A quick calculation shows that if the pin is in the hand for one second (perhaps a reasonable time frame), the pin will exit the hand at over 140 km/hr.

Regards

Steven