A juggler juggles three balls in a continuous cycle. Any one ball is in contact with his hand for one-fifth of the time. Describe the motion of the center of mass of the three balls. What average force does the juggler exert on one ball while he is touching it?

Hint: To describe the motion of center of mass of the three balls, we use the concept of center of mass and the time for which the three balls are touching with the juggler’s hand and the time for which three balls are not in contact with the juggler’s hand.Then, we apply force equations for the two conditions.

Complete step by step answer:
We are given that a juggler juggles three balls in a continuous cycle and one ball is in contact with his hand for one-fifth of the time i.e.
1 ball - $\dfrac{1}{5}$ th of the time $(20\% )$
3 balls - $\dfrac{3}{5}$ th of the time in contact $(60\% )$
And the balls are not in contact for the remaining $\dfrac{2}{5}$ th of the time $(40\% )$. So, we can say that $\dfrac{2}{5}$ th of the time $(40\% )$, the center of mass of three balls is in freefall downwards and $\dfrac{3}{5}$ th of the time $(60\% )$, the center of mass of three balls is in upwards direction. So, Using the force equations, we can write
\[\dfrac{3}{5}a = \dfrac{2}{5}g\]
\[\therefore a = \dfrac{2}{3}g\]
Now, the average force exerted on the one ball while juggler touches it is
$\therefore F = ma = \dfrac{2}{3}mg$, where $m$ - mass of each ball.

Note: Center of mass of a system of particles moves as though it was a particle of mass equal to that of the whole system with all the external forces acting directly on it. Here, the center of mass moves around in a little closed loop, with a parabolic top and likely a circular bottom, making three revolutions for every one revolution that one ball makes.