A clock is under free fall for $60$ seconds and its second hand makes exactly one revolution during that period of time. It begins at rest with its second hand facing upwards. Given that the second hand has a length of $r>0$ and that the downward gravitational acceleration is some constant $g>0$, what is the total distance traveled by the tip of the second hand?
I came up with this little problem myself and got stuck at the end with a possibly impossible integral to evaluate. Here's a quick graphical interpretation of the problem: https://www.desmos.com/calculator/ck8gxjughf
So, I started with the parametric equations representing the path of the tip of the second hand:
$$ \begin{align} x = r\sin \left( \frac{\pi t}{30} \right), \quad y = r\cos \left( \frac{\pi t}{30} \right) - \frac12 g t^2 &&\ &&\ \text{for } 0 \leq t \leq 60 \end{align} $$
Then the total distance traveled can immediately be written as $$ \begin{align} \int_0^{60} \sqrt{ \dot x ^2 + \dot y ^2} dt & = \int_0^{60} \sqrt{ \left( \frac{\pi r}{30} \cos \left(\frac{\pi t}{30} \right) \right)^2 + \left( - \frac{\pi r}{30} \sin \left(\frac{\pi t}{30} \right) - gt \right)^2} {\rm d}t \\ & = \int_0^{60} \sqrt{ g^2t^2 + \frac{\pi r g}{15} t \sin \left( \frac{\pi t}{30} \right) + \left( \frac{\pi r}{30} \right)^2} {\rm d}t \end{align} $$
Which you can quickly approximate, giving a semi-satisfying answer to the problem, but I doubt has a closed-form solution in terms of $r$ and $g$.
Since having the revolution take $60$ seconds seems to make things messy, and that we're more interested in evaluating these integrals, consider instead $$ \begin{align} x=r \sin(t), \quad y = r \cos(t) - \frac12 gt^2 &&\ &&\ \text{ for } 0 \leq t \leq 2\pi \end{align} $$
Are there any closed-form solutions of this integral for specific values of $r$ and $g$? $$ \int_0^{2\pi} \sqrt{g^2t^2 + 2rgt\sin(t) + r^2} {\rm d}t $$
So, the tip of the seconds hand will generate an "accelerated" cycloid.
Being accelerated, it is no longer periodic, and the length will depend on the starting position.
The presence of the term $t \cdot \sin t$ under the square root seems to make very unprobable that the integral might be evaluated in closed form.