We solve analytically the differential equations for a skier on a hemispherical hill and for a particle on a looptheloop track when the hill or track is endowed with a coefficient of kinetic friction μ. For each problem, we determine the exact “phase diagram” in the twodimensional parameter plane.
I. INTRODUCTION
Two classic homework exercises in an elementary mechanics course are the skier on a hemispherical hill (Fig. 1) and the particle on a looptheloop track (Fig. 2).^{1} Both problems illustrate nicely the use of conservation of energy (to find the speed as a function of height) followed by $ F = m a$ (to find the normal force).
It is interesting to consider what happens when the hill or track is endowed with a coefficient of kinetic friction μ. Somewhat surprisingly, the exact differential equations turn out to be analytically solvable.^{2–11} Our purpose here is to provide a unified treatment of the two problems, using only elementary methods that are easily accessible to undergraduates (e.g., linear firstorder differential equations). Though most of our results have been obtained previously—as we shall document in detail—they are somewhat scattered in the literature. It may thus be of some modest value to have a complete elementary derivation collected in one place.
The skier and looptheloop problems give rise to very similar differential equations, which differ only by some sign changes. However, these sign changes lead to significant differences in the qualitative interpretation of the solutions. Since the skier problem turns out to be somewhat simpler, we treat it first and give a complete solution; in particular, we determine the exact phase diagram in the twodimensional parameter plane. For the looptheloop, we solve the differential equations only up to the first time (if any) that the particle halts or completes one cycle of the loop, so we obtain only a partial phase diagram. The full phase diagram will (as we explain later) contain an infinite sequence of bifurcations, and we leave its computation to a reader who wishes to take up where we have left off.
II. SKIER ON A HEMISPHERICAL HILL
When $ \mu > 0$, by contrast, the normal force is no longer a decreasing function of θ, nor is it guaranteed to reach zero within the interval $ 0 \u2264 \theta \u2264 \pi / 2$. Indeed, $ d N / d \theta  \theta = 0 = 2 \mu ( 1 \u2212 \lambda ) m g > 0$, so the normal force is initially increasing.
In this way, we have obtained a phase diagram that divides the $ ( \mu , \lambda )$ plane into three possible qualitative behaviors:

For $ 0 \u2264 \lambda < \lambda \u22c6 ( \mu )$, the skier halts after a finite time at some angle $ \theta halt ( \mu , \lambda )$: This angle is an increasing function of λ that runs from 0 to $ arctan\u2009 \mu $ as λ runs from 0 to $ \lambda \u22c6 ( \mu )$.

For $ \lambda = \lambda \u22c6 ( \mu )$, the skier comes to rest asymptotically as $ t \u2192 + \u221e$ at the angle $ \theta = arctan\u2009 \mu $.^{20}

For $ \lambda \u22c6 ( \mu ) < \lambda < 1$, the skier flies off the hill at some angle $ \theta fly ( \mu , \lambda )$: This angle is a decreasing function of λ that tends to 0 as $ \lambda \u2192 1$.
The curve $ \lambda \u22c6 ( \mu )$ thus forms the boundary between the “halt” phase and the “flyoff” phase (see again Fig. 4).^{21} In particular, the skier always either halts or flies off; she never reaches angle $ \pi / 2$.
Some typical curves of $ \Lambda ( \theta )$ for all three scenarios are shown in Fig. 3. Note, in particular, that $ \Lambda ( \theta ) = \Lambda \u2032 ( \theta ) = 0$ when $ \lambda = \lambda \u22c6 ( \mu )$ and $ \theta = \theta \u22c6 ( \mu )$; and note the fundamental qualitative difference between the curves for $ \lambda < \lambda \u22c6 ( \mu )$, which reach the Λ = 0 axis, and those for $ \lambda > \lambda \u22c6 ( \mu )$, which do not.^{22}
Some typical curves of $ \theta halt ( \mu , \lambda )$ as a function of λ are shown in Fig. 5, and some typical curves of $ \theta fly ( \mu , \lambda )$ as a function of λ are shown in Fig. 6. Please note the discontinuous change in behavior as the phase boundary $ \lambda \u22c6 ( \mu )$ is crossed: $ \theta fly ( \mu , \lambda \u22c6 ( \mu ) )$ (the dotted curve in Fig. 6) is much larger than $ \theta halt ( \mu , \lambda \u22c6 ( \mu ) )$ (the dashed curve in Fig. 5).^{23} This is a very simple example of sensitive dependence to initial conditions, giving rise to a discontinuous phase transition—a phenomenon pointed out already by James Clerk Maxwell in 1876.^{24}
Since the proofs of all the previous claims involve some slightly intricate calculus, we relegate them to Appendix A in the supplementary material.^{25}
Let us remark, finally, that by the same methods one can study the more general problem in which the coefficient of kinetic friction is an arbitrary function $ \mu ( \theta )$ of the position along the hill: The Eq. (5) is still a firstorder inhomogeneous linear differential equation for the unknown function $ N ( \theta )$—albeit now one with nonconstant coefficients—so can still be solved by the method of integrating factors (though the result may not be analytically expressible in terms of elementary functions). We leave it to interested readers to pursue this generalization.
Some recent related articles are Refs. 10, 11, and 26, which study a particle sliding down an arbitrary curve in the presence of kinetic friction; Ref. 27, which uses the Lagrangian formalism with Lagrange multipliers to analyze a particle sliding without friction down an arbitrary concave curve; and Ref. 28, which studies a ball rolling (initially without slipping, later with sliding and kinetic friction) on an arbitrary curve in the presence of gravity, including an experimental realization.
III. PARTICLE ON LOOPTHELOOP TRACK
The looptheloop problem is more complicated than the skier, for three reasons: The particle can cycle around the track; it can reverse direction; and it can halt due to static friction. Each time the particle reverses direction, we need to apply Eq. (22) with a new value for $ sgn ( \theta \u0307 )$; this repeated switching between different equations seems quite complicated, and probably needs to be handled by numerical solution.^{29} To simplify matters, we will here follow the block only until it first reaches $ \theta \u0307 = 0$ or falls off the track; we, therefore, have $ \theta \u0307 \u2265 0$.
The solution (25) must, therefore, be supplemented by the two inequalities $ N ( \theta ) \u2265 0$ and $ N ( \theta ) \u2265 m g \u2009 cos \u2009 \theta $. (Please note that, unlike in the skier problem, both of these inequalities point in the same direction; this radically changes the nature of the qualitative analysis.) The block comes instantaneously to rest when $ N ( \theta ) = m g \u2009 cos \u2009 \theta $ or falls off the track when $ N ( \theta ) = 0$, whichever happens first; if neither happens for $ \theta < 2 \pi $, then the block completes one full cycle of the looptheloop. Now, the inequality $ N ( \theta ) \u2265 m g \u2009 cos \u2009 \theta $ is the more stringent one in the lower half of the looptheloop (that is, $ \u2212 \pi / 2 \u2264 \theta \u2264 \pi / 2$ modulo $ 2 \pi $), while the inequality $ N ( \theta ) \u2265 0$ is the more stringent one in the upper half of the looptheloop (that is, $ \pi / 2 \u2264 \theta \u2264 3 \pi / 2$ modulo $ 2 \pi $). Therefore, the block can come instantaneously to rest only in the lower half of the looptheloop, and it can fall off the track only in the upper half of the looptheloop.
In the presence of friction ( $ \mu > 0$), the analysis proceeds as follows:
 The first step is to determine the conditions under which the particle halts in the first quadrant ( $ 0 \u2264 \theta \u2264 \pi / 2$). The particle halts at angle θ when $ \Lambda ( \theta ) = 0$, i.e., in case the initial velocity satisfies$ \lambda = \lambda halt ( \theta , \mu ) \u2009 = def \u2009 2 1 + 4 \mu 2 \u2009 [ ( 1 \u2212 2 \mu 2 ) + e 2 \mu \theta [ 3 \mu \u2009 sin \u2009 \theta \u2212 ( 1 \u2212 2 \mu 2 ) \u2009 cos \u2009 \theta ] ] .$Since$ \u2202 \lambda halt ( \theta , \mu ) \u2202 \theta = 2 e 2 \mu \theta ( \mu \u2009 cos \u2009 \theta + sin \u2009 \theta ) ,$$ \lambda halt ( \theta , \mu )$ is an increasing function of θ in the interval $ 0 \u2264 \theta \u2264 \pi / 2$ (as is intuitively clear: To reach a larger angle, more initial velocity is needed). In particular, the particle reaches $ \theta = \pi / 2$ with $ \theta \u0307 > 0$ if and only if$ \lambda > \lambda halt ( \pi / 2 , \mu ) \u2009 = def \u2009 2 \u2212 4 \mu 2 + 6 \mu e \pi \mu 1 + 4 \mu 2 .$
 If the particle reaches angle $ \pi / 2$ without halting, the next step is to determine the conditions under which the particle flies off in the second or third quadrant ( $ \pi / 2 \u2264 \theta \u2264 3 \pi / 2$). The particle flies off at angle θ when $ N ( \theta ) = 0$, i.e., in case the initial velocity satisfies$ \lambda = \lambda fly ( \theta , \mu ) \u2009 = def \u2009 2 \u2212 4 \mu 2 + 3 e 2 \mu \theta ( 2 \mu \u2009 sin \u2009 \theta \u2212 cos \u2009 \theta ) 1 + 4 \mu 2 .$Note that $ \lambda fly ( \pi / 2 , \mu ) = \lambda halt ( \pi / 2 , \mu )$. Since$ \u2202 \lambda fly ( \theta , \mu ) \u2202 \theta = 3 e 2 \mu \theta \u2009 sin \u2009 \theta ,$we see that $ \lambda halt ( \theta , \mu )$ is an increasing function of θ in the interval from $ \pi / 2$ to π and then a decreasing function in the interval from π to $ 3 \pi / 2$. The first of these facts is again intuitively clear: To survive to a larger angle without flying off, more initial velocity is needed. The second fact implies that if the particle reaches angle π without flying off—that is, if$ \lambda \u2265 \lambda fly ( \pi , \mu ) \u2009 = def \u2009 2 \u2212 4 \mu 2 + 3 e 2 \pi \mu 1 + 4 \mu 2,$
then it also reaches angle $ 3 \pi / 2$ without flying off. This is intuitively clear when there is no friction, but not so obvious in the presence of friction. This implies—analogously to what happens in the skier problem—a discontinuous change of behavior as λ passes through $ \lambda fly ( \pi , \mu )$. See Fig. 7 for plots of $ \lambda halt ( \theta , \mu )$ and $ \lambda fly ( \theta , \mu )$ versus θ for some selected values of μ.
 If the particle reaches angle π (and hence also angle $ 3 \pi / 2$) without halting or flying off, the next step is to determine what happens in the fourth quadrant ( $ 3 \pi / 2 < \theta < 2 \pi $). The particle halts at angle θ in case λ equals the quantity $ \lambda halt ( \theta , \mu )$ defined in Eq. (31). From Eq. (32), we see that $ \u2202 \lambda halt ( \theta , \mu ) / \u2202 \theta $ is negative at $ \theta = 3 \pi / 2$ and positive at $ \theta = 2 \pi $, with a unique zero at $ \theta = 2 \pi \u2212 arctan \mu $. So $ \lambda halt ( \theta , \mu )$ is decreasing in the interval $ 3 \pi / 2 \u2264 \theta \u2264 2 \pi \u2212 arctan \mu $ and increasing in the interval $ 2 \pi \u2212 arctan \mu \u2264 \theta \u2264 2 \pi $. Its maximum value in the interval $ [ 3 \pi / 2 , 2 \pi ]$, therefore, lies either at $ \theta = 3 \pi / 2$ or at $ \theta = 2 \pi $. Since we are in the situation $ \lambda \u2265 \lambda fly ( \pi , \mu ) > \lambda fly ( 3 \pi / 2 , \mu ) = \lambda halt ( 3 \pi / 2 , \mu )$, the only relevant question is whether λ is larger than $ \lambda halt ( 2 \pi , \mu )$ or not. If it is, then the particle reaches angle $ 2 \pi $ without halting. If it is not, then the particle halts at some angle in the interval $ ( 2 \pi \u2212 arctan \mu , 2 \pi ]$, namely, the unique angle where $ \lambda = \lambda halt ( \theta , \mu )$. The first of these cases always occurs when $ \lambda fly ( \pi , \mu ) > \lambda halt ( 2 \pi , \mu )$, i.e., when $ 0 \u2264 \mu < \mu crit \u2248 0.713 \u2009 089$. (See Appendix B in the supplementary material^{25} for the proof that there is a unique such value $ \mu crit$.) When $ \mu \u2265 \mu crit$, then there is a “halt in fourth quadrant” phase at $ \lambda fly ( \pi , \mu ) \u2264 \lambda \u2264 \lambda halt ( 2 \pi , \mu )$ and a “survive to angle $ 2 \pi $” phase at $ \lambda \u2265 \lambda halt ( 2 \pi , \mu )$. We record the formula$ \lambda halt ( 2 \pi , \mu ) \u2009 = def \u2009 ( 4 \mu 2 \u2212 2 ) ( e 4 \pi \mu \u2212 1 ) 1 + 4 \mu 2 .$
 If the particle survives to angle $ 2 \pi $, then it has there a forward velocity corresponding to a value,$ \lambda new \u2009 = def \u2009 \Lambda ( 2 \pi ) = \lambda e \u2212 4 \pi \mu + ( 2 \u2212 4 \mu 2 ) ( 1 \u2212 e \u2212 4 \pi \mu ) 1 + 4 \mu 2,$$ = e \u2212 4 \pi \mu [ \lambda \u2212 \lambda halt ( 2 \pi , \mu ) ],$$ \u2265 0 .$
Since $ \lambda halt ( 2 \pi , \mu ) > 0$ in the survive to angle $ 2 \pi $ phase, we have $ \lambda new < e \u2212 4 \pi \mu \lambda $: Thus, the kinetic energy is reduced by at least a factor $ e \u2212 4 \pi \mu $ at each revolution. The subsequent motion can then be found by repeating the foregoing analysis with λ replaced by $ \lambda new$.
The resulting phase diagram is shown in Fig. 8. Since $ \lambda halt ( 2 \pi , \mu )$ grows extremely rapidly with μ, we have used $ \lambda $ instead of λ on the vertical axis, to compress the plot. This phase diagram agrees with the one found by Kłobus (Ref. 8, Fig. 2); the value of $ \lambda halt ( 2 \pi , 1 )$ also agrees with his. All three phase boundaries are increasing functions of μ: See Appendices B1–B3 in the supplementary material.^{25}
Of course, this phase diagram only follows the particle up to the first time that it reaches $ \theta \u0307 = 0$ or $ \theta = 2 \pi $. A more complete analysis would show that the phase “survives to angle $ 2 \pi $” is itself divided into subphases “halts in the first quadrant” ( $ 2 \pi < \theta < 5 \pi / 2$), “flies off the second quadrant” ( $ 5 \pi / 2 < \theta < 3 \pi $), “halts in the fourth quadrant” ( $ 7 \pi / 2 < \theta < 4 \pi $), and “survives to angle $ 4 \pi $”; and this latter phase is further divided into subphases; and so on infinitely. We leave it to interested readers to work out the details of this infinite sequence of bifurcations.
ACKNOWLEDGMENTS
The authors are extremely grateful to three referees for their detailed and helpful comments on several versions of this paper.