Does juggling balls reduce the total weight of the juggler and balls?

Suppose you throw the ball upwards at some speed $v$. Then the time it spends in the air is simply:

$$ t_{\text{air}} = 2 \frac{v}{g} $$

where $g$ is the acceleration due to gravity. When you catch the ball you have it in your hand for a time $t_{\text{hand}}$ and during this time you have to apply enough acceleration to it to slow the ball from it's descent velocity of $v$ downwards and throw it back up with a velocity $v$ upwards:

$$ t_{\text{hand}} = 2 \frac{v}{a - g} $$

Note that I've written the acceleration as $a - g$ because you have to apply at least an acceleration of $g$ to stop the ball accelerating downwards. The acceleration $a$ you have to apply is $g$ plus the extra acceleration to accelerate the ball upwards.

You want the time in the hand to be as long as possible so you can use as little acceleration as possible. However $t_{\text{hand}}$ can't be greater than $t_{\text{air}}$ otherwise there would be some time during which you were holding both balls. If you want to make sure you are only ever holding one ball at a time the best you can do is make $t_{\text{hand}}$ = $t_{\text{air}}$. If we substitute the expressions for $t_{\text{hand}}$ and $t_{\text{air}}$ from above and set them equal we get:

$$ 2 \frac{v}{g} = 2 \frac{v}{a - g} $$

which simplifies to:

$$ a = 2g $$

So while you are holding one 3kg ball you are applying an acceleration of $2g$ to it, and therefore the force you're applying to the ball is $2 \times 3 = 6$ kg.

In other words the force on the bridge when you're juggling the two balls (with the minimum possible force) is exactly the same as if you just walked across the bridge holding the two balls, and you're likely to get wet!

I love this class of problem as a fantastic physical example of the mean value theorem. Allow me to describe a specific case that fits the following conditions:

• The man plus the balls has a total weight of $m$
• The entire system (man+balls) starts at rest and ends at rest

From these relatively simple assumptions, I will claim that the average normal force (the force the ground exerts upward) is equal to the weight of the the system. In other words, for a given period of time of length $T$ we have this:

$$ m g = \frac{1}{T} \int_0^T \vec{F}(t) \cdot \vec{n} dt $$

This is a spectacular claim actually. To simplify the notation, consider that $\vec{F}(t) \cdot \vec{n}$ is just equal to the weight a scale would read (this isn't a bad assumption, depending on the scale). Imagine the man is juggling, standing on a scale, and the scale reads a value that depends on time, $w(t)$. The average value the scale reads will be equal to gravity times his mass, including everything he's holding or wearing.

In the story of the man walking across the bridge juggling balls, the total weight is $201 lb$. For every second he weighs $200 lb$, he spends one second weighing $202 lb$ or something similar. The point is that the average value is the same.

Put one ball down. Walk the other across. Go back, get the second ball.

Or, roll the two balls across, then run after them.

Or, the juggler takes off his shoes and walks across barefoot.

This is solved as a "nonlinear thinking" problem, not with "juggling is anti-gravity". The ball-man system must be accelerated downwards with an average of 1 lb of force or the bridge will break. Otherwise you could build a perpetual motion machine from two jugglers on a see-saw who take turns juggling.

(Also, running is like juggling in that the weight is up in the air much of the time--if this could work, you could also just hold the balls and run.)