A problem inspired by the ice hockey game

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.

Answer 1

Your problem is deeper than it might seem at the first glance. Apart from numbers, it involves a remarkable phenomenon, which was studied in detail only a few years ago in the paper Phys. Rev. Lett. 90, 248302 (2003).

Suppose the puck is an infinitely thin disc, and it is sent across the ice sliding and rotating. What will stop first, translational or rotational motion?

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.

This value fully characterizes the asymptotic motion of the puck and allows you to calculate the friction force distribution, deceleration and finally the path (which is a straight line in this approximation). In fact, a part of that was already found in the paper cited above.

However, this concerns only an infinitely thin disc, for which the pressure distribution is equal everywhere under the puck. With a thick disc you have more pressure under the front part of the puck, which additionally complicates the problem (still, see some discussion in the paper).

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.

Answer 2

This is a tricky question, one that I encountered in a similar situation a few years ago when working on a similar simulation (for a curling game, not hockey, but fundamentally similar).

A few things jump out at you immediately about the problem; the other poster is correct that the translational and rotational motion indeed must stop simultaneously. My approach to this was a bit more intuitive than his described answer though; the reason is simply that due to the motion of one or the other, the transition from kinetic to static friction is never made independently of the angular or translational velocity, but made simultaneously. This is true of both ideal and more complex simulations. (Note, however, that that does not mean that the angular velocity cannot become zero, particularly in situations where the coefficient of friction is not constant at all points on the traveling surface; however, it the angular velocity will not remain precisely zero while the translational velocity is non-zero.)

As to the distance the "puck" will travel when rotating or not... consider the situation where the translational velocity is zero, and the rotational velocity is non-zero. Assuming constant kinetic (and static) friction at all contact points, the puck will not move; therefore, when translational energy is added to the system via a force vector, there is no disturbing force to perturb the original energy.

Interestingly, this seems not to be precisely correct in the indication that the angular force will not affect the translational force, as the case of curling shows quite clearly; the "rock" (equivalent to the puck) will turn (or curl, hence the name of the game) in the direction of the rotation of the leading edge of the rock. I don't have precise equations for this, but intuition indicates that the complete path traveled by a puck (rock) imparted with an angular velocity is the same distance as a rock (puck) without an angular velocity, but that the puck (rock) with the angular velocity travels a curved path (the total length of which is equal to the distance the unrotating rockpuck would travel in a straight line).

As to the reasons why a curling rock curls... I'm uncertain, but you've reminded me to ask a question about that. :-)