How high does a slide need to be if the person at the bottom grabs a pole and rotates the pole 72 degrees?

Question

My attempt:

$mgh = \frac{1}{2}I \omega^2$

$\omega^2 = 2 \alpha\Delta \theta$, so

$mgh = \frac{1}{2}I 2 \alpha \Delta \theta$

$mgh = I \alpha \Delta \theta = \tau \Delta\theta$

$\tau = rF\sin \theta = Lma$ ( $L$ for the length of the string, $m$, the mass of the person, and $a$ for the person's linear acceleration)

$mgh = Lma\Delta\theta$

$h = \frac{Lma \Delta\theta}{mg} = \frac{La \Delta\theta}{g}$

I don't know what to do now since I'm not sure I can solve for $a$.

Answer 1

This is one of those three part dynamics questions.
For the first part you need to use energy conservation to work out the horizontal speed of the person just before hitting the pole.
The second part is the application of the conservation of angular momentum about the pole's pivot point when the person grabs hole of the pole.
Note that the collision between the person and the pole will be inelastic - kinetic energy will not be conserved.
In the third phase again apply energy conservation but note that the person is now part of the ascending pole.