Questions on the forces on a Yo-Yo

[aaaa202]

Played with a metalring with a string rolled around it today - i.e. i played with an oversized yoyo. I had some trouble understanding the physics behind it all though.
We assume that yoyo would roll giving us enough information to solve the equations of motions which gave us that the force of the string would be half the force of gravity during the phase, where the yoyo unrolled. That was all good, I have trouble however understanding how the string "knows" how to only exert an upwards force of ½mg!
Let me elaborate:
Let's assume that the string is not stretched to start off with. The ring will then be acted on by the force of gravity only and accelerate down. But then after having fallen the necessary height for the string to get tight the ring will exert a downwards force on the string making the string exert an upwards force on the ring. But how does this become ½mg. I can see if from the equations but it seems like a mystery to me how it suddenly knows only to exert a force of ½mg when the reaction that causes this is actually a downwards force mg.
I hope that made sense - probably didn't and if so please tell me where to elaborate, because this really pains me.

 

[rcgldr]

Assume a massless string that is frictionless, then the ring would free fall downwards and the tension in the string would be zero. Now assume the massless string is glued to the ring and the ring doesn't move (assume the ring isn't swinging), the tension in the string is m g, the weight of the ring.

For the original case, the string has enough friction that the ring is force to roll downwards on the string as it "falls". Angular inertia in the rolling ring reduces it's downwards rate of acceleration to 1/2 g. The tension in the string is then = m g - 1/2 m g = 1/2 m g.

Replace the ring with a solid uniform disk and the downward rate of acceleration is 2/3 g, and the tension in the string would be m g - 2/3 m g = 1/3 m g.

 

[Danger]

I don't know any of the science stuff that you're discussing, but I do know that the primary principle involved in the workings of a yo-yo is conservation of angular momentum. If the string and the yo-yo were both frictionless, it would go down and stay down. Once yo-yo/string friction engages, it returns. (Witness the "sleeper" manoeuvre wherein you can let the yo-yo spin freely at the bottom of the trick, then give it a wee jerk to re-engage the string to the axle and have it climb back up.)