## Weight of a Juggler

Posted on: February 27th, 2013 by
16

What is the weight of a juggler?

Clarification: If the juggler of weight W stands on a weighing machine and juggles n balls, each of weight w, what does the weighing machine read? Assume the juggler is fast enough that the weighing machine's reading is the average weight it measures.
-- Via Tom Cover.

#### 16 Responses to Weight of a Juggler

Assuming that he/she is a perfect juggler (i.e. can juggle infinite balls), each of his hands will have one ball at all times. So his weight is W+2w, where W is his own weight and w is the weight of a ball.

No. The system that includes the juggler and all the balls he is juggling has a stationary center of mass. Hence, the "average" weight of the juggler must equal W+nw where W is his own weight, w is the weight of a ball and n is the number of balls.

How can it be more than W + 2w? Are you assuming that he can hold more then one ball in a hand at some point of time?

No. The system's center of gravity must be stationary. So the external force on the system must equal the weight of the system. And the only external force is coming from the weighing machine he is standing on. It means if you dance on the weighing machine, your weight may go up and down, but on an average it has to stay the same. Maybe at the instant the person is holding 2 balls, his weight maybe W+2w, but when he pushes the ball upwards, he exerts a force on it. In that instant, his weight is W+2w+F where F is the force with which he is pushing the ball upwards against it's weight.

Otherwise, at least n-2 balls will be in air at all times, and hence will not contribute to his total weight.

If you look at the instantaneous weight, then at the instant he is holding 2 balls, it is W+2w. At instants when he's holding only 1 ball, it's W+w. At instants when he's propelling a ball upwards, it's W+w+F, and so on. However, mechanical systems like the weighing machine have a slow response time and tend to average out all the forces they experience. If you take the time-average of all forces, they must equal the total weight of the system, so as to prevent the center of gravity from falling. The puzzle asks for average weight. The instantaneous weight depends on a lot more number of variables: the amount of time the juggler takes in propelling a ball up, the action of the juggler (does he apply a constant force throughout or does his force vary while he's propelling it up), the number of balls he's holding, his body movements (if his legs spring, that can have an effect on his measured weight), etc.

Cool, if he is standing on a weighing machine I agree that the reading will vary depending upon the forces he apply on the balls.

But wait, the weighing machine is measuring the weight of the whole system, not his own weight.
Balls not in contact with the juggler should not be considered a part of the system. When they hit his arm the forces they exert should be considered external. Once he catches a ball it becomes part of his 'system'. After he releases it, it again becomes an external object. You are ignoring that.

''System'' here should be understood clearly.
We choose the system based on our convenience. Note that a system of interacting objects does not necessarily mean the objects are connected/touching each other. That perhaps is a misconception that needs to be avoided.
For this puzzle, the best system to consider is the juggler and all the balls that he is juggling. The only external forces on this system are: the weight of the juggler and the balls he juggles, and the upward force exerted by the weighing machine.
Newton's laws can be applied to any system. If we apply it to the system of the juggler and the balls, then the external forces on the system are: the weight of the individual masses, and the upward force exerted by the weighing machine. If for convenience we assume that the center of mass of this system executes a periodic motion, then one can show that the average acceleration of the center of mass is 0. Hence by Newton's 2nd law (Force =mass x acceleration), the average force is 0 (total change in velocity/ total time=0 over a period). Hence the average upward force on this system must equal the average downward force. The only ''external'' upward force on this system of juggler and balls is the force of the weighing machine pushing the juggler up, making sure he's standing and not falling. By Newton's third law, this force has to equal the force that the juggler exerts on the weighing machine and hence the weight measured by the weighing machine. Note that the juggler exerts forces on the balls, but since we cleverly chose our system to include the juggler and the balls, the interaction between the juggler and the balls are ''internal'' forces. Only ''external'' forces on a system need to be considered when applying Newton's laws.

I would also like to point out the difference between ''closed'' and ''open'' systems. The system you described is the juggler and all the balls connected to him. This is an open system since balls may enter or leave the system. The system I defined is the juggler and the set of balls he is juggling. That is a closed system. No external object may enter the system and no object in the system may leave the system. Why is it important to distinguish between open and closed systems? For the simple and plain reason that NEWTON's LAWS DO NOT HOLD FOR OPEN SYSTEMS. A rocket with fuel is an open system. It propels itself forward by ejecting fuel. So it can accelerate forward without the application of an external force. This doesn't violate Newton's law because Newton's law applies only for closed systems. It's clear that for our puzzle, we defined a closed system conveniently so that we can apply Newton's laws of motion.

I am not arguing that a 'system' should always be "a set of connected objects". What I mean to say is that the answer will be W+nw only if we consider the system to be the Juggler and the n balls. The question in that case should ask for the "weight of the system" and not the juggler.

See the clarification posted. I am asking what will a weigjing machine measure. Thats what a layman .means by weight.

I agree! Just that the main question "What is the weight of a juggler?" is incomplete and vague.

Anurag and Gowtham: Here, the assumption is that the juggler starts with all n balls in his hands (and then starts sending one or more balls in the air). So, the weighing machine initially reads W+nw, and then, as Gowtham has been saying, the average weight of the "system" (which is what will be shown by the weighing machine) will always remain to be W+nw, irrespective of how many balls are in contact with the juggler.

But, if the juggler starts empty-handed (let's say there is an outlet on top of the juggler's head which feeds the balls down to the juggler one by one which the juggler immediately puts into juggling motion/rhythm), then what will the weighing machine show?