Fiddler on the Proof: Aug 16, 2024

David Ding

Aug 16, 2024

How High Can You Jump?

Explanation:

Let us solve the general problem presented in the extra credit, as the original problem is just a special case of the general problem.

Technically, the problem is 2D, so for center of mass of an athelete's body modeled by an arc, there is the x-coordinate as well as the y-coordinate. Fortunately, by symmetry, the x-coordinate is right over the bar. It remains to find the y-coordinate. Note that when $\varphi =\frac{\pi }{2}$, the arc is a semi-circle.

Without loss of generality, let the semi-circle have a radius of 1. Furthermore, denote the distance between the bottom of the semi-circle (that is, when $\varphi =\frac{\pi }{2}$) and the y-coordinate of the center of mass be $h$. With this construction, we have:

$$\begin{array}{rl}a& =h-\cos \varphi \\ b& =1-h\end{array}$$

Where $0<\varphi \le \frac{\pi }{2}$.

We can find $h$ using the definition of center of mass, as follows.

$$\begin{array}{rl}{\int }_0^{\varphi }(\cos \theta -h)\mathrm{d}\theta & =0\end{array}$$

In other words, the y-coordinate of the center of mass should yield zero when summed over the differences between it and the y-coordinates of all points on the arc, as we are assuming that the athlete's mass is evenly distributed. Solving for $h$ in terms of $\varphi$, we get:

$$\begin{array}{rl}{\int }_0^{\varphi }(\cos \theta -h)\mathrm{d}\theta & =0\\ \\ \sin \theta -h\theta {|}_{\theta =0}^{\theta =\varphi }& =0\\ \\ \sin \varphi -h\varphi & =0\\ \\ h& =\frac{\sin \varphi }{\varphi }\end{array}$$

Thus we have:

$$\begin{array}{rl}a& =h-\cos \varphi \\ \\ & =\frac{\sin \varphi }{\varphi }-\cos \varphi \\ \\ & =\frac{\sin \varphi -\varphi \cos \varphi }{\varphi }\\ \\ b& =1-h\\ \\ & =1-\frac{\sin \varphi }{\varphi }\\ \\ & =\frac{\varphi -\sin \varphi }{\varphi }\end{array}$$

Finally,

$$\begin{array}{rl}\frac{a}{b}& =\left(\frac{\sin \varphi -\varphi \cos \varphi }{\varphi }\right)/\left(\frac{\varphi -\sin \varphi }{\varphi }\right)\\ \\ & =\boxed{{\displaystyle \frac{\sin \varphi -\varphi \cos \varphi }{\varphi -\sin \varphi }}}\end{array}$$

For the original problem, $\varphi =\frac{\pi }{2}$, and so plugging the value yields:

$$\begin{array}{rl}\frac{a}{b}& =\frac{\sin \varphi -\varphi \cos \varphi }{\varphi -\sin \varphi }\\ \\ & =\frac{\sin (\pi /2)-(\pi /2)\cos (\pi /2)}{(\pi /2)-\sin (\pi /2)}\\ \\ & =\frac{1}{(\pi /2)-1}\\ \\ & =\boxed{{\displaystyle \frac{2}{\pi -2}}}\approx 1.75\end{array}$$

For the extra credit, we need to evaluate:

$$\begin{array}{rl}& \underset{\varphi \to 0}{lim}\frac{\sin \varphi -\varphi \cos \varphi }{\varphi -\sin \varphi }\\ \\ =& \underset{\varphi \to 0}{lim}\frac{\cos \varphi -(\cos \varphi -\varphi \sin \varphi )}{1-\cos \varphi }\text{(L'Hôpital's rule)}\\ \\ =& \underset{\varphi \to 0}{lim}\frac{\varphi \sin \varphi }{1-\cos \varphi }\\ \\ =& \underset{\varphi \to 0}{lim}\frac{\sin \varphi +\varphi \cos \varphi }{\sin \varphi }\text{(L'Hôpital's rule again)}\\ \\ =& \underset{\varphi \to 0}{lim}\frac{\cos \varphi +(\cos \varphi +\varphi \sin \varphi )}{\cos \varphi }\text{(L'Hôpital's rule one more time)}\\ \\ =& \frac{2}{1}=\boxed{{\displaystyle 2}}\end{array}$$

Below is a plot of the ratio $\frac{a}{b}$ as a function of $\varphi$ for $0<\varphi \le \frac{\pi }{2}$.