Rotating space station I've come across a question of which I can't fully understand the solution:

A space station is located in a gravity-free region of space. It consists of a large diameter, hollow thin-walled cylinder which is rotating freely about its axis. The cylinder is of radius $r$ and mass $M$.
Radial spokes, of negligible mass, connect the cylinder to the centre of rotation. If astronaut (mass $m$) now climbs halfway up a spoke and lets go, how far along the cylinder circumference from the base of the spoke will the astronaut hit the cylinder?
With the astronaut at the centre, the cylinder spins with angular velocity $\omega_0^2 = g/r$.


Attempt at a solution:
With the man at the centre, the moment of inertia of the system is
$ I=Mr^2  $ spinning at $\omega_0^2$.
With the man at $r/2$, the moment of inertia of the system will be:
$$I' = Mr^2 + m(r/2)^2 = (M +   m/4)r^2 $$
By conservation of angular momentum, the cylinder will now be spinning at angular velocity $$\omega' = \frac{{Mr^2}}{{(M+m/4)r^2}}\omega_0 $$ The tangential velocity of the astronaut at the point of release is $v_\mathrm{man}=\omega'(r/2)$.
However, when the man lets go, the MoI of the system returns to $I$, spinning at $\omega_0$.
The man sweeps out an angle $\alpha = \pi/3$ along the circumference, and travels a distance $S_\mathrm{man} = \sqrt{r^2-(r/2)^2} = (\sqrt3/2)r$, taking a time $S_\mathrm{man}/v_\mathrm{man} = \frac{{\sqrt3}}{{\omega}}(1+\frac{{m}}{{4M}})$.
In this time, the base of the spoke travels a distance $S_\mathrm{spoke} = v_\mathrm{spoke}r\times t = \omega rt = \sqrt3r(1+\frac{{m}}{{4M}})$. and so the difference in distance is $$S_\mathrm{spoke} - S_\mathrm{man} = \left(\sqrt3\left(1+\frac{{m}}{{4M}}\right) - \frac{{\pi}}{{3}}\right)r.$$
However, the answer given is $(\sqrt3 - \frac{{\pi}}{{3}})r$.
 A: 
However, when the man lets go, the MoI of the system returns to $I$, spinning at $w_0$. 

Not true.  
Just before letting go the angular momentum of the system (man and cylinder) is the sum of the angular momentum of the man and the angular momentum of the cylinder.
That division of angular momentum will be the same the moment after the man lets go so the angular momentum and hence the angular velocity of the cylinder will not change.
A: At the point of release the angular velocity of the space station doesn't change. Intuitively you can see this by the fact that during the release you are just letting go, so there is little you can do to change the angular velocity. To make it a little more rigorous remember that the angular momentum of freely moving objects is always conserved. After the astronaut releases the spoke his angular momentum stays constant, meaning the angular momentum of the space station must also stay constant to preserve total angular momentum. The space station's moment of inertia doesn't change so it's angular velocity also doesn't change.
A: What I notice at a glance:
$w' = \frac{{Mr^2}}{{(M+m/4)r^2}}w_0^2$ should be $w' = \frac{{Mr^2}}{{(M+m/4)r^2}}w_0$.
"when the man lets go, the MoI of the system returns to I
, spinning at $w_0$ "
This should be "spinning at $w'$".
Then double check every step.
A: No angular momentums are needed; pure kinematics will do:
At the moment right after the astronaut lets go of the spoke, his velocity equals that of that point in the spoke. He now moves rectilinearly until impact while the cylinder keeps spinning at constant speed about its geometric center which also moves rectilinearly.
(Before the astronaut let go, he and the cylinder were together rotating rigidly about their common center of mass, which is a complex situation -- but fortunately not one we need to handle here).
Let's work in a coordinate system where the cylinder's center is stationary at the origin. Choose units of length such that the cylinder's radius is $1$ and units of time such that its angular velocity is $1$.
The length of the astronaut's path during the fall is then $\frac12\sqrt3$, and the event where he lands is $\pi/3$ radians ahead of the spoke's initial position.
The astronaut's speed is $\frac12$, so it takes him $\sqrt3$ time units to fall. In that time the spoke has moved $\sqrt3$ radians.
The distance between the foot of the spoke and the astronaut at impact is therefore $$ (\sqrt3 - \pi/3) \text{ radians} $$
