# How long does a coin take to slide off a turntable?

So I was helping my daughter prepare for a AP physics test. I studied Physics at university many many years ago and always assume I'll be able to figure out any reasonable question about mechanics.

I created a problem for her to solve, and overshot what she could do - however I subsequently realized I think I've overshot what I can do as well.

The problem is as follows (the initial part I see on this forum comes up frequently):

I have a turntable with a radius of 2 meters. I have a coin or object on the turntable 1 meter from the center. Static coefficient of friction 0.8, dynamic/kinetic coefficient of friction 0.5. The turntable increases angular velocity very gradually.

a) What is the angular velocity of the turntable at which the coin starts to slide.

b) Assuming the turntable maintains this angular velocity, how long does it take for the coin to slip off the edge of the turntable.

Now part a) is easy and has been covered elsewhere. Part b) is a lot more difficult (for me). I'm looking for a pointer in the right direction, at least what I should be searching for to get a better understanding.

I thought initially I could find the function a(r), to give me the acceleration at any radius a, considering the kinetic friction and centripetal(/centrifugal) force. Then integrate that over the 1 meter it's going to slide to find the time.

This in itself then started getting much more difficult, since I need to integrate over time to get the velocity, but over distance to total up the time. So this is pointer request 1. Assuming a variant of this problem in which it is a collar on a rotating rod, and hence is constrained to move along the rotating radial line, how would I go about doing this?

Then the second issue I started to realize is that it wouldn't even move along the (rotating) radial line - since it is now sliding it will curve backward from the perspective of the observer on the turntable. How would one describe this behavior? What would be the search terms to find more info on this? I can do it numerically but can't figure out how to do this analytically.

• Consider changing the dynamic coefficient of friction to zero. The resulting problem would be comparable to the difficulty of AP Physics problems. – DrSheldon Sep 29 '18 at 1:40
• @DrSheldon, this than becomes way too easy and simply solved by kinematics of the inertial frame. I do agree we should keep the OP away from becoming a test-problem writer... – npojo Sep 29 '18 at 8:39
• @npojo: Could you please clarify the context of "easy"? If you meant too easy for us here on Physics.SE, I would agree with you. If you meant for a student preparing for an AP test, I think my modification is at the right level. The stumbling block for most AP problems is usually figuring out what kind of problem it is; the rest of the solution is typically straightforward. – DrSheldon Sep 29 '18 at 11:25

This does seem like a highly nontrivial problem. Looking from the perspective of the non-inertial, rotating frame of the turntable, we see what we traditionally call fictitious forces, so in addition to friction, you assign a centrifugal force acting on the coin. The threshold beyond which the coin starts to slide is that in which the centrifugal force equates the maximum static friction force, and that's probably how you solved part (a) of your problem.

When the coin starts to move, however, you need to add another fictitious force, called Coriolis force, which goes like $$\mathbf{F}_C = -2m\mathbf\,{\mathbf{\Omega}}\times\mathbf{v}$$ Where $$m$$ is the mass of the coin, $$\mathbf{\Omega}$$ is the angular velocity of the turntable, and $$\mathbf{v}$$ is the velocity of the coin in the non-inertial frame. This is the term that accounts for the fact that the trajectory won't stay radial from the point of view of the turntable.

But there is still another term that we haven't accounted for, which ir the kinetic friction. Because of the nature of friction, this force is always changing direction, in order to be always pointing against the direction of motion. Thus, if the kinetic friction coefficient is $$\mu$$, the kinetic friction force is something like

$$\mathbf{F}_{fr} = - \mu mg\frac{\mathbf{v}}{v}$$

where $$v = |\mathbf{v}|$$. Thus, in the non-inertial frame, the full equation of motion (accounting for centrifugal and Coriolis forces, and friction) is $$\frac{d\mathbf{v}}{dt} = -\mathbf{\Omega}\times(\mathbf{\Omega}\times\mathbf{r})-2\mathbf\,{\mathbf{\Omega}}\times\mathbf{v}- \mu g\frac{\mathbf{v}}{v}$$

This is not trivial at all to solve; I think you can only do this numerically. So maybe you did overshoot what you could do ;)

• I am not certain this is actually the complete physical picture. You initially equate centrifugal force to static friction force. The coin moves outwards, creating an orthogonal Coriolis force which initially is very small (being dependent on the radial velocity component only). Along the orthogonal axis, force is too small to overcome static friction force. So the single dynamic coefficient $\mu$ is not sufficient to solve the problem. – npojo Sep 29 '18 at 8:35
• @npojo You mean I would have to consider dynamic and static friction simultaneously? I hardly think that's the case, and for me the idea is simple: if the particle is moving, it's dynamic friction, and that's it. The force points against the velocity vector, and its magnitude is proportional do the dynamic coefficient. – Bruno De Souza Leão Sep 29 '18 at 12:12

If anyone is curious for the answer, here goes. From part a) we get that $$\Omega = 2.8$$ rad/s. We can simplify the equation in the other answer (which is missing minus signs) by breaking it into vector components:

\begin{aligned} \frac{d\mathbf{v}}{dt} &= -\mathbf{\Omega}\times(\mathbf{\Omega}\times\mathbf{r})-2\mathbf\,{\mathbf{\Omega}}\times\mathbf{v}- \mu g\frac{\mathbf{v}}{v}\\ \Leftrightarrow \left<\ddot{x},\ddot{y}\right> &= \Omega^2\left - 2\Omega\left<-\dot{y},\dot{x}\right> - \mu g\frac{\left<\dot{x},\dot{y}\right>}{\sqrt{\dot{x}^2+\dot{y}^2}} \end{aligned}

Then we use new functions $$q$$ = $$\dot{x}$$ and $$p = \dot{y}$$ to make 4 equations, those 2 and:

\begin{aligned} (3)\quad \dot{q} &= \Omega^2 x + 2\Omega p - \mu g\frac{q}{\sqrt{q^2 + p^2}}\\ (4)\quad \dot{p} &= \Omega^2 y - 2\Omega q - \mu g\frac{p}{\sqrt{q^2 + p^2}} \end{aligned}

so that we can solve numerically with MATLAB, which gives 0.91 seconds.

• Thanks for pointing out the misplaced minus signs, I already corrected them. Besides, nice to see the actual numerical solution, +1 :) – Bruno De Souza Leão Sep 29 '18 at 12:17

Perhaps is also interesting to see also the path of the coin

Figure 1: $$\Omega(t)$$=const.

Figure 2: $$\Omega(t)=f(t)$$.