Angular momentum w/ changing moment of inertia A man of mass m1 is standing on a disk with radius R and mass M at a distance r

The man starts walking around the disk with constant angular speed w1 and as a result the disk begins to rotate in the opposite direction with constant angular speed w2 (angular momentum is conserved). At the same time, the point mass m2 begins to move slowly towards the edge of the disk. The mass then falls off the edge of the disk.
The question is this: if the man were to stop walking after the point mass falls off of the disk, what direction will the disk rotate? Will it stop rotating, continue rotating in the same direction, or reverse directions?
I'm basically going between 2 thought processes with this problem, and I dont know which is right:
1) If the point mass os treated as part of the disk, then when it falls, the moment of inertia of the disk will be reduced. However, angular momentum is conserved, so the angular speed of the disk will increase as a result. This would then imply that if the man were to stop, then the disk would also stop.
2) As the disk slides away from the axis of rotation, the angular momentum of the system increases in the direction of rotation of the disk. After the disk falls off, the angular momentum stays at the elevated level, so that when the man stops, the disk continues to spin in the direction it was originally spinning.
 A: You have three relevant objects here: the man, the disk, and the mass. Each have its own angular momentum in terms of its angular speed:
$$L_1=I_1\omega_1$$
$$L_d=I_d\omega_2$$
$$L_2=m_2r_2^2\omega_2$$
where the moment of inertia of the point mass is $I_2=mr_2^2$ and its angular speed is the same as that of the disk, as you specified in the comments. Also specified is that the man will adjust walking such that $\omega_1$ is constant. Therefore, the only two time dependent values here are $r_2$ and $\omega_2$
Since the total angular momentum of the system is conserved, it must be that
$$\frac{\text d}{\text d t}L_{total}=0$$
$$I_d\dot\omega_2+m_2r_2^2\dot\omega_2+2m_2r_2\dot r_2\omega_2=0$$
or rearranging for $\dot\omega_2$:
$$\dot\omega_2=-\frac{2m_2r_2\dot r_2\omega_2}{I_d+m_2r_2^2}$$
Now, without knowing how $r_2$ changes over time we cannot determine what this expression becomes in general. However, you can use what you know about the signs of the relevant variables to determine if $\omega_2$ ends up increasing or decreasing as the ball moves outward. I will leave this to you.
Then from this you can think about just the angular momentum of the man and the disk. Consider what would happen if $\omega_1$ was suddenly reduced to $0$, and what that would then mean for the disk using the fact that angular momentum must be conserved between just the man and the disk during the stopping process.

If the point mass is treated as part of the disk, then when it falls...angular momentum is conserved, so the angular speed of the disk will increase as a result. 

This is not the case. The ball is going to leave the disk is such a way that it does not impart/take away any angular momentum. Therefore, when the ball leaves the man and the disk won't experience any sudden change in angular momentum. You are probably mixing this up with some intro physics problems where a kid leaps off of a merry-go-round, in which case you have to consider the angular momentum transfer between the kid and the merry-go-round. The ball is not leaping off the disk here, it is just move off of it.
