# FAQ overflow

#### QUESTION

Question shortly: How far would a hockey puck slide in two different cases:

1. The puck is sliding (translation) on ice and spinning on its flat surface.
2. The puck is sliding on ice without spinning.

Other conditions are the same in both cases.

Simplifications: no air resistance, the coefficient of friction is const while sliding(spinning)
Let's formalize the problem:

Given a disc with diameter D = 8 cm, mass m=170 g, the coefficient of friction on ice let be $\mu=0.02$ Initial values:

1. $v_0=10 \frac{m}{s}$ and $\omega_0=100 \frac{rad}{s}$
2. $v_0=10 \frac{m}{s}$ and $\omega_0=0 \frac{rad}{s}$

Obviously the part 2) is trivial. I am working currently on part 1). This seems to be very difficult. Instead of a solid disc i took a thin ring at first. But even in that case the solution seems to be an integro-differential equation. If you have some ideas let us know.

The answer is they will stop simultaneously, and this does not depend on the initial translational and angular velocities. The origin is the intrinsic friction-mediated coupling between rotation and sliding. If rotation is too fast for a given speed $v$, the translational friction will be very small. And vice versa, if rotation is too slow, the rotational deceleration is small. Thus, there exists a "magic" value of $\epsilon=v/R\omega$ towards which every initial condition is attracted. In the paper cited above this magic value was found to be approximately 0.653.
Finally, if you change the shape of the pack (as you in fact did by taking a ring instead of a disc), the magic value of $\epsilon$ also changes. See another paper that discusses this: Phys. Rev. Lett. 95, 264303 (2005). See also this comment to this old paper on puck motion on the ice.