English Transfer

[Jal]

...Have you ever considered how clean, polished, dirty, chipped, or scratched balls affect the coefficient of friction of balls?

You're right, the coefficient is surely affected and it does make a difference as to how much spin is transferred. But when you express this transfer as a fraction or percentage of the cueball's spin, it's irrelevant. It's irrelevant because no matter what the condition of the balls, there is still a certain spin that will produce maximum throw. And this will result in a 35.7% transfer, albeit a smaller amount of absolute spin for slicker balls. (For slicker balls, the spin which gives you maximum throw is smaller.) In fact, you will get this much of a percentage transfer for any spin up to the one which produces maximum throw. If you increase the spin of the cueball beyond this, the throw and thus the imparted spin will begin to decrease. The percentage transfer will of course drop as well since your increasing the cueball's spin while the OB's acquired spin is diminishing.

The best answer I have seen to this question are answered in Byrne's Complete Book of Pool Shots. In this he explains that the coefficient of friction between two (clean) balls is ~0.02. This means that roughly 2% of the cue balls spin is transferred to the second ball at contact.

The coefficient seems to vary from roughly 0.1 (very slow surface speed), to 0.01 (very fast surface speed). This is based on tests done by Wayland Marlow and the derivation of these values can be found here. Even if these numbers aren't very precise for a particular set of balls, it's probably true that it will vary over a rather large range for them too. But as mentioned in my last paragraph, its actual value doesn't really matter.

For the OP's question (if I understood it correctly) the amount of spin transferred is dependent on the coefficient of friction, the spin (rotational velocity),and Impulse (how long the cue ball is in contact with the object ball and its force in that direction).
http://en.wikipedia.org/wiki/Impulse

... the impulse of a force is the product of the force and the time during which it acts. Although momentum is conserved within a closed system, individual parts of a system can undergo changes in momentum. Impulse has the same units and dimensions as momentum (kg m/s or N·s = Huygens Hy). The impulse of a time-varying force is calculated as the integral of force with respect to time:

Couldn't agree more.

Therefore, if a cue ball is spinning at a certain rotational velocity, it should impart MORE spin on the object ball if it is in contact with it longer, which would be true if it was moving at a slower linear velocity.

Good point but I don't agree with your conclusion. It is true that the slower balls will remain in contact a little longer. But the amount of force between them (compressive and frictional), and the duration of contact, sort of arrange themselves to comply with the conservations laws. If the contact time is increased, the average forces are reduced accordingly, and visa-versa. You still end up with the same impulse which yields the same momentums and energies (ignoring inelasticity and its dependency on speed).

Hope some of this makes sense, and is possibly even right.

Jim