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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?

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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.

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