# Juggling a ball

I have the following exercise:

Explain the progress of the force you need to juggle a $$5\ \rm{kg}$$ ball as a function of time.

Now since the question is kind of imprecise to me, I have problems solving it.

My approaches:

If I hold the ball in my hand, I need a force of $$mg$$ right? Because of gravity. Now if I want to throm it in the air I need to apply a force bigger than $$mg$$. Are these thoughts correct?

And now how to write this as a function of time?

Since this is a homework-type question, I'm not going to write the entire answer:

In juggling, the position of balls with respect to time looks like this.

Let's denote the length of your forearm $$d$$, the breadth of your shoulders $$b$$, let's assume the highest point the balls achieve is at your chin, and let's denote the height difference between your chin and your hand at the moment of release $$h_r$$, at the moment of catch $$h_c$$, the height difference between your chin and your elbow $$h$$ and the angle between your forearm and a horizontal line $$\theta$$ (we'll assume $$\theta$$ is small).

$$h_r = h + d\sin\theta\\ h_c = h - d\sin\theta$$

Let the time the ball is flying upwards $$t_1$$, the time downwards $$t_2$$ and the net time $$t$$:

$$t_1 = \sqrt\frac{2(h+d\sin\theta)}{g}\\ t_2 = \sqrt\frac{2(h-d\sin\theta)}{g}\\ t = \sqrt\frac2g(\sqrt{h+d\sin\theta} + \sqrt{h - d\sin\theta})$$

Let's denote the horizontal velocity during the travel through the air as $$v_x$$, vertical velocity when leaving one hand $$v_{y1}$$ and when being caught by the other hand $$v_{y2}$$:

$$v_x = \frac{b}t = \frac{b\sqrt\frac{g}2}{(\sqrt{h+d\sin\theta} + \sqrt{h - d\sin\theta})}\\ v_{y1}=gt_1 = \sqrt{2g(h+d\sin\theta)}\\ v_{y2}=gt_2 = \sqrt{2g(h-d\sin\theta)}$$

The speed $$v_c$$ when being caught is (we approximate $$\sin^2\theta\approx0$$)

$$v_c = \sqrt{v_x^2+v_{y2}^2}=\sqrt{\frac{b^2g+16gh(h-\sin\theta)}{8h}}$$

and the speed $$v_r$$ when released (again we approximate $$\sin^2\theta\approx0$$)

$$v_r = \sqrt{v_x^2+v_{y1}^2}=\sqrt{\frac{b^2g+16gh(h+\sin\theta)}{8h}}$$

Let's denote $$T$$ the period of the motion and assume that during the time the ball is in your hand, its speed increases linearly from $$v_c$$ to $$v_r$$:

$$\pi d = v_c\frac T 2 + \frac12\frac{v_r-v_c}{\frac T 2}\left(\frac T 2\right)^2$$

From that we have

$$T = \frac{4\pi d}{v_c + v_r}$$

Assume that during the time the ball is in your hand, its speed increases linearly from $$v_c$$ to $$v_r$$, the dependence of the speed on time is therefore:

$$v(t) = v_c + \frac{v_r - v_c}{\frac T 2}t = v_c + \frac{v_r^2 - v_c^2}{2\pi d}t$$

For your next steps, look up centripetal force.