I was reading through a book that presented the problem:

A disgruntled hockey player throws a hockey stick along the ice. It rotates about its center of mass as it slides along and is eventually brought to rest by the action of friction. Its motion of rotation stops at the same moment that its center of mass comes to rest, not before and not after. Explain why.

I saw a similar question at: Why does an ice hockey stick, when thrown on ice always rotate and translate together before coming to rest? Why not only rotate or only translate? but I didn't feel like I was satisfied with the answer.

The question was in the energy chapter so I tried to solve it with energy. I thought that if the force of friction on the COM was F and it traveled distanced s before stopping, then we can write the equation $\frac{1}{2}mv^2 = Fs$ where the LHS represents the initial kinetic energy. We could also write a similar equation for rotational energy, but I'm not sure what to do after that. Is it even reasonable to attempt to prove it rigorously or just by reasoning?

Another question that came from thinking about this question was how would you measure the torque brought by the friction since it's distributed across the stick, not just at one point?

  • 2
    $\begingroup$ Great question. The alternate scenarios where the stick rotates in place for a bit after its momentum is lost and the stick purely translates along the ice with no rotation still seem plausible to me, if not a bit improbable. I am curious if there is an answer that makes the above scenarios impossible. $\endgroup$ Commented Dec 26, 2020 at 17:22
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    $\begingroup$ I thought spontaneously "cannot be". Was wrong. Learned something today. Thanks. $\endgroup$ Commented Dec 26, 2020 at 19:52
  • $\begingroup$ Note that there are some specific circumstances being glossed over here. This is not always true. For example, a spinning top can be launched while being pushed horizontally, and it will stop travelling horizontally before it stops spinning. I'm fairly certain that for your proposed scenario, the object's footprint (i.e. what touches the ground) has to reasonably approximate the object's size itself (i.e. the width of the spinning mass) $\endgroup$
    – Flater
    Commented Dec 27, 2020 at 0:40
  • 1
    $\begingroup$ Another great example of my above comment is a cue ball. With significant side spin, a ball can remain spinning after it has stopped travelling on the table. It requires a high amount of spin and a low amount of linear velocity (you'd want it to avoid hitting a bank to keep the spin stable), but it is achievable. $\endgroup$
    – Flater
    Commented Dec 27, 2020 at 1:01
  • $\begingroup$ Btw, the problem appears as Question 28 on p. 295 in Physics, 5th ed. by Halliday, Resnick and Krane. $\endgroup$ Commented Jan 22 at 8:49

3 Answers 3


If the stick is a bar, having plain contact with the ground along its length, the friction force opposing the rotation suggests to model it as 2 cantilever beams with uniformly distributed load, fixed in the COM. The friction load is distributed along its length, resulting in max. torque close to COM and zero at the ends.

So for a small area close to the ends, the total torque results only from the load on this area: $$\delta \tau = \delta I\frac{d\omega}{dt}$$ $\delta \tau = \delta Fr$ and the friction force in the element is $\delta F = \mu \delta N = \mu \delta m g$

The moment of inertia $\delta I = \delta m r^2$ and $$\omega = \frac{v}{r}$$

So, $$\mu \delta m g r = \delta m r^2 \frac{1}{r} \frac{dv}{dt} \implies \frac{dv}{dt} = \mu g $$

If we elaluate the force to decrease the average translation velocity in the same region: $$\delta F = \delta m\frac{dv_t}{dt} = \mu \delta N = \mu \delta mg \implies \frac{dv_t}{dt} = \mu g$$

Under the same acceleration, they must decrease together. If it happens for the ends of the bar, all the body will stop spinning and moving linearly at the same time for this model.

But if for example, the central portion have contact but not the ends, it is perfectly possible for the bar keeps rotating, after stopping its translational movement.

  • $\begingroup$ I don't quite understand your argument, partly because I don't understand the formulas with my severely limited physics (you explain tau and I only with formulas; what are they?). $\endgroup$ Commented Dec 26, 2020 at 16:24
  • $\begingroup$ @Peter - Reinstate Monica I edited the answer to define $\tau$ and $I$. The argument is that the same magnitude of acceleration that decreases rotation also decreases translation. $\endgroup$ Commented Dec 26, 2020 at 16:34
  • $\begingroup$ Here's an example of the situation discussed in the last paragraph. $\endgroup$
    – Flater
    Commented Dec 27, 2020 at 0:53
  • $\begingroup$ Don't the end points have larger tangential velocity than the center? Wouldn't this meant that they would stop later as they have the same decelerration? $\endgroup$
    – Gary Song
    Commented Dec 27, 2020 at 23:19
  • $\begingroup$ @Gary Song The way I wrote the equations really can lead to this conclusion, what doesn't make sense. I have to improve the answer. $\endgroup$ Commented Dec 28, 2020 at 16:39

I don't think the statement is correct. In general, a sliding object can stop spinning or translating before the other motion stops. There might only be one.

For example, take a uniform disk. Spin it and set it a rest on the ice. It will spin in place for a while, and come to a stop. Likewise, slide it without spinning. It will slide to a stop.

As the post you linked shows, a hockey stick is special. It has different coefficients of friction at different ends. That can make it spin if you start it just sliding. But it doesn't have to. Slide it with the high friction end in back, kind of like shooting a bow and arrow. It will slide to a stop without spinning.

If you give it a little spin and a lot of velocity, friction can orient it before it slides to a stop.

After reading the other answers (+1 to both), I conclude my intuition was wrong. Rotation and translation do stop at the same time.

  • $\begingroup$ I'm not sure this is correct. Yes, you can in theory spin a resting disk or slide it without spinning. In practice you will probably find though that the spinning disk will (start to) translate slowly, or that a sliding disk will (start to) spin slowly, due to imperfections in the execution or side conditions or imperfect surfaces. And the slow, inadvertent rotation/translation will continue; counter-intuitively, the "faster" movement reduces the friction in the "slower movement". The slower movement is braked slower than if it were the only movement. Just why that is so is the question. $\endgroup$ Commented Dec 26, 2020 at 19:50
  • $\begingroup$ @Peter-ReinstateMonica - Thank you for pointing that out. I see my answer is wrong. $\endgroup$
    – mmesser314
    Commented Dec 26, 2020 at 20:16
  • $\begingroup$ @Peter-ReinstateMonica: While the spinning of an object may cause some translation, that is unrelated to why any other horizontal movement can stop without the spinning itself stopping. Imagine a spinning top, being launched and imparting sideways motion on it. It will stop moving sideways while still spinning. Yes, its spin cause it's tip to slightly translate in a circle, but the sideways movement itself has stopped due to friction. Similarly, a cue ball can spin in place long after it stopped its horizontal movement. $\endgroup$
    – Flater
    Commented Dec 27, 2020 at 0:45
  • $\begingroup$ "I conclude my intuition was wrong. Rotation and translation do stop at the same time." Your intuition was correct in general, they do not have to stop at the same time (counterexample), but I suspect this isn't generally true for hockey sticks because the spinning ends also make contact with the ground/ice, thus "synchronizing" the motions. $\endgroup$
    – Flater
    Commented Dec 27, 2020 at 0:54
  • $\begingroup$ @Flater Yes, the question is not about a top which touches the ground only on a minimal surface area, thus reducing, indeed minimizing the friction torque. The simplest model is probably an infinitesimally thin, flat, disk (because a cylinder with finite height leads to higher normal forces on the front facing side while braking, leading to curved trajectories). $\endgroup$ Commented Dec 27, 2020 at 3:05

Here is a paper which analyses the coupling of sliding and spinning motion with thin disks, and why both motions come to a stop at the same time. The paper establishes a mathematical model and reports results of experiments with a CD on a nylon surface which test the theory.

Edit in response to comments: The mathematical model assumes a flat disk with uniform mass distribution and explicitly ignores effects from its finite height (a non-uniform normal force leading to curved trajectories). The experiments use a CD as an approximation. A top may behave differently, although I can observe that the tip of a top "wanders" along the surface as a result of forces (like precession, a puff of air or an unevenness in the surface) which would by far not suffice to move it if it were not rotating.

The underlying reason for this interaction between spinning and sliding is that dynamic friction is independent of the speed of movement; it only depends on the normal force (here the gravitational force) and material constants. Its direction at every surface point opposes the direction of that point's velocity. With a fast-rotating, sliding object all velocity vectors are almost completely tangential because the rotational part is dominating the vectors. The magnitude of the vectors is very high, but irrelevant: The frictional force does not depend on it. Consequently, the frictional forces are almost entirely tangential as well; they mostly cancel each other out with respect to the forward motion and slow only the rotation.1

In effect, the frictional torque (which slows the rotation) is higher than the linear frictional force (which stops the forward motion) if the rotation is fast compared to the forward motion — and vice versa. This is why the "faster" one of the two movements is braked more, until they align and come to a stop together. The figure below (p. 2 of the paper) shows that interdependence. $\epsilon$ is the quotient of the forward motion and the angular motion, $v/R\omega$. For little linear motion but fast rotation the friction torque dominates (the left side of figure (a)), and for fast linear motion with little rotation the linear friction dominates (right side of the figure):

enter image description here

1 This is somewhat unusual: We often "dissect" velocities or forces into their constituents and consider them individually, independently. In this case though the lateral component influences the longitudinal friction because it changes the direction of the vector, and vice versa: Because the friction in a given direction does not depend on the magnitude of the vector component in that direction. The magnitude of that component is constant, the friction is not. Pretty counter-intuitive.

  • $\begingroup$ Much like cars used to loose all steering capability when you slammed on the brakes (before ABS came along). A tiny asymmetry in the effects of the four brakes was enough to induce a dangerous spin in the car that generally lasted until a) the brakes were released, changing the direction of the car instantaneously to whereever it was currently pointing, b) the car came to a stop, or c) the brakes were released and the car was pointing sufficiently backwards for the tires to regain traction. There's a big reason why we have ABS and EPS today... $\endgroup$ Commented Dec 26, 2020 at 18:33
  • $\begingroup$ So, for a spinning top (with a small finite contact surface), the above is also true. Then does this imply that a spinning top will stop moving at the same time it stops spinning? Or does the shape of the object matter? $\endgroup$
    – Yakk
    Commented Dec 27, 2020 at 0:36
  • $\begingroup$ @Yakk: It depends on what movement you're looking at. A spinning top will stop moving in a given direction (e.g. if you threw it sideways when launching it). In that sense, it does stop moving horizontally long before it stops spinning. However, the spin does cause the tip to move in a circular pattern - and in that sense, it doesn't stop moving horizontally (though in a circular, not linear, pattern) until it stops spinning. $\endgroup$
    – Flater
    Commented Dec 27, 2020 at 0:58
  • $\begingroup$ @Flater nod, but if the hockey stick acted like that, it wouldn't really satisfy the thing that the OP described. So the geometry of the objects does matter, and this answer doesn't talk much about that directly. $\endgroup$
    – Yakk
    Commented Dec 27, 2020 at 1:06
  • $\begingroup$ @Yakk: This is why I mentioned that it depends which horizontal translation you're talking about. If we're focusing on linear translation (i.e. not the circular translation induced from the spin), then the geometry is relevant. Spinning tops and cue balls can keep spinning, hockey sticks can't. But if you include that circular translation into your definition of "horizontal movement" that allegedly stops before the spin stops, then the geometry doesn't matter because the "horizontal movement" will always stop at the same time as the spin. $\endgroup$
    – Flater
    Commented Dec 27, 2020 at 1:11

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