Problem: A juggler is juggling a uniform rod one end of which is coated in tar and burning. He is holding the rod by the opposite end and throws it up so that, at the moment of release, it is horizontal, its CM (center of mass) is traveling vertically up at speed $$v_o$$ and it is rotating with angular velocity $$\omega_o$$. To catch it, he wants to arrange that when it returns to his hand it will have made an integer number of complete rotations. What should $$v_o$$ be, if the rod is to have made exactly $$n$$ rotations when it returns to his hand?

Solution: The first thing we can do is figure out what $$n$$ looks like. If $$n$$ denotes the number of rotations, and we know the angular velocity (measured in radians per second), then, assuming $$t$$ is the time during which the rod is in the air, $n=\frac{\omega_o t}{2\pi}$ Next, we need to figure out how long the rod is in the air for a given $$v_o$$. Since the CM is experiencing constant acceleration due to gravity, its velocity when it lands back in the hand of the juggler must be equal to the negative of $$v_o$$. $-v_o = v_o + gt$ $t = -\frac{2v_o}{g}$ Plugging this into the equation for n, $n = - \frac{ \omega_o v_o}{g \pi}$ And now, solving for $$v_o$$ (remember that $$g$$ is negative), $v_o = - \frac{g \pi n}{w_o}$