Angular momentum of the center of gravity of yo-yo Sorry for my broken English.
I am a physics undergrad, and I know only a basic stuff about the subject.
Yesterday I was taught about the angular momentum of yo-yo and
my prof said that when the yo-yo hits the end of the string, the angular momentum of the center of gravity of yo-yo is $Mvr_a$ ($M$  is the mass of yo-yo, $v$ is the velocity of the center of gravity downwards, and $r_a$ is the radius of the axle).
But what really happens to the yo-yo is that part of the mechanical energy is lost due to the shock when it hit the end (I can understand this), and the angular momentum of the center of gravity of the axle becomes zero, because the direction the center of gravity moves is on the straight line with the string. I cannot understand the latter part because in my opinion the direction of the move of the center of gravity should be vertical to the string as it rotates around the point string is attached. Maybe I misunderstood what he said.
Can anybody teach me this?
 A: Well, I'd say the thing is wrong for a general yo-yo. Where it works is the following: the yo-yo is just a hollow cylinder of mass $m$, which at the same time is your axes of radius $r$. In this case the moment of inertia is $I = m r^2$. If the yo-yo is rolling with speed $v$ over this axes the angular velocity is $\omega = v / r$. The angular momentum is 
$$L = I \omega = m r^2 { v \over r} = m v r $$
i.e. your result. This then is true everywhere and not only at the end. 
Note, it would be the angular momentum with respect to the centre of gravity not the angular momentum of the centre of gravity (that would be zero). 
A: This discussion is for a falling/returning yo-yo with no extra energy input from the holder (e.g., no thrust downward from the top or pull upward from the bottom).
The motion of the yo-yo can be segregated into two parts: the motion of the Center of Mass (CM), and the rotation about the center of mass.  The acceleration of the CM is due to the vector sum of the external forces: gravity and the tension on the string.  The rotation about the CM is due to the torque from the tension in the string.
While the yo-yo is falling downwards the acceleration downwards is $a = {(mg - T)\over m}$ where $a$ is acceleration, $m$ is mass, $g$ is the acceleration of gravity, and $T$ is the tension in the string.  The motion of the yo-yo about its CM is due to the external torques about the CM; this is true even if the CM is accelerating as proven in more advanced physics mechanics tests.  The torque about the CM is due to the tension on the string; gravity provides no torque about the CM since gravity act at the CM.  Assuming the yo-yo rolls down the string without slipping, the angular acceleration of the yo-yo $\alpha = {a \over r}$ where $r$ is the distance from the CM to where the string pulls on the yo-yo.  $\alpha = {{Tr} \over I}$ where $Tr$ is the torque due to tension and $I$ is the moment of inertia of the yo-yo about the CM.
Once the yo-yo reaches the end of the string (the bottom of its motion) the force of tension increases and the CM is stationary.  Assuming the string is loosely wound the yo-yo continues to rotate but it slips with respect to the string and friction slows the rotation down and eventually rotation ceases.

(If you pull up sufficiently hard on the yo-yo when it is at the bottom the increase in tension is an increase in friction and the yo-yo stops slipping and will climb back up the string.)
There are numerous good descriptions of the motion of a yo-yo on the internet, One good one is: How do yo-yos work? Why do they come back up after being thrown down?   at https://www.physlink.com/Education/askExperts/ae18.cfm
