Slipping on an Arbitrary Surface with Friction

Felipe Gonz´alez-Cataldo∗ and Gonzalo Guti´errez† Departamento de F´ısica, Facultad de Ciencias, Universidad de Chile, Casilla 653, Santiago, Chile.

Julio M. Yáñez‡ Departamento de F´ısica, Facultad de Ciencias, Universidad Católica del Norte (Dated: August 22, 2018) The motion of a block slipping on a surface is a well studied problem for flat and circular surfaces, but the necessary conditions for the block to leave (or not) the surface deserve a detailed treatment. In this article, using basic differential geometry, we generalize this problem to an arbitrary surface, including the effects of friction, providing a general expression to determine under which conditions the particle leaves the surface. An explicit integral form for the velocity is given, which is analytically integrable for some cases, and we find general criteria to determine the critical velocity at which the particle immediately leaves the surface. Five curves, a circle, ellipse, parabola, catenary and cycloid, are analyzed in detail. Several interesting features appear, for instance, in the absense of friction, a particle moving on a catenary leaves the surface just before touching the floor, and in the case of the parabola, the particle never leaves the surface, regardless of the friction. A phase diagram that separates the conditions that lead to a particle stopping in the surface to those that lead to a particle departuring from the surface is also discussed.

I. INTRODUCTION A natural question that arises from studying this situation is how to incorporate friction on the circular surface. In this case, the energy is no longer conserved and it has A well studied problem in elementary mechanics, in 3–5 one of its many formulations, is the following1 (Chap. 8): been already solved in previous works . A more general A boy/girl of mass m is seated on a hemispherical mound version of the problem is to study the point of departure of ice. If he/she starts sliding from rest (assume the ice to in an arbitrary, frictionless surface, described by a func- be frictionless), where is the point P at which the boy/girl tion y = f(x). Here, the conservation of energy does leaves the mound? In terms of classical mechanics, it apply, and allows to write an equation in terms of f and its derivatives, whose solution gives the point of depar- corresponds to the motion of a point particle slipping 6 from the top of a semicircle under the action of gravity, ture . Although the treatment is general, it assumes that as shown Fig.1. the y coordinate can be expressed as an explicit function of x, which is not always possible (like a cycloid or a y general conic section), and is also limited to the cases g where the mechanical energy is conserved (no friction). The case of rough surfaces (with friction) is only brieﬂy discussed for the cases in which f(x) = −αxk, without R giving an equation to determine the point of departure, ϕ but providing some criteria to determine if the particle

x leaves the surface or not. Our goal is to solve the following general problem: to FIG. 1. Particle sliding from the top of a semicircle under the action of gravity. find the point of departure of a particle sliding in an arbitrary surface with friction, providing an expression for the velocity of the particle, determining under which con- It is difficult to know what is the original version of ditions the particle stops on surface, and when it always this problem, but we can suppose that it comes after the remains on the surface. We also determine the maximum 2 arXiv:1512.00515v1 [cond-mat.soft] 7 Aug 2015 influential books of Francis Weston Sears . What is in- initial velocity, above which the particle leaves the sur- teresting about the problem is that it combines many face immediately, and identify for which curves this max- topics. The usual solution combines conservation of me- imum velocity is zero, i.e., the particle leaves the surface chanical energy with elements from circular motion (ra- immediately, regardless of the initial velocity given. dial and tangential acceleration), together with the key condition to the problem: the release point takes place We pose the problem in terms of basic differential ge- when the normal force is zero. The solution is indepen- ometry7,8, which gives us a systematic way to deal with dent of both the particle mass m and the gravitational curves and allows us to obtain the velocity of the par- acceleration g, resulting in the particle leaving the cir- ticle as a general explicit expression. We consider some cular surface when cos ϕ = 2/3. This release point does particular famous curves, like parabola and ellipse, and not depend on the initial height either, and its value is later we provide a classification of the curves that show ϕ = arccos(2/3) ≈ 48.19◦. the same behaviour. 2

II. THEORETICAL FRAMEWORK y

In this section, we review some basic concepts of dif- b ferential geometry and present the equations of motion of a particle sliding over an arbitrary surface written in t n the Frenet–Serret frame, which are used to derive an ex- n pression for the velocity of the particle at any point in the surface. t y r b A. Kinematics z x x Consider a particle moving on the XY plane, describ- FIG. 2. Tangent ˆt, normal ˆn, and binormal bˆ vectors of the ing an arbitrary trajectory. Its position at any instant Frenet–Serret frame, viewed from the Cartesian system. of time t is given by r(t) = (x(t), y(t)), the velocity by v(t) = dr/dt, and its acceleration by a = dv/dt. The trayectory is the collection of all points r(t), which means If we introduce the notation v(s) ≡ v(t(s)) and a(s) = that for a time interval [t ,T ], the function r:[t ,T ] → 0 0 a(t(s)), these equations take the form R2 is a parametrized curve whose image corresponds to the trajectory of the particle. Conversely, if the particle v(s) = v(s)ˆt(s) (5) is sliding on a surface, a parametrization of the surface dv(s) describes its trayectory along the surface. a(s) = v(s) ˆt(s) + v2(s)κ(s)ˆn(s). (6) A curve r can always be reparametrized by its ar- ds clength Z t dr(τ) B. Dynamics in the Frenet–Serret Frame s(t) = dτ, where t ∈ [t0,T ], (1) t0 dτ as long as its derivative dr(t)/dt 6= 0 in the interval, since The forces acting over a particle of mass m sliding on a ds(t)/dt = kr0(t)k = v(t) > 0 for all t implies that s is surface are its weight W = mg = m (g · t)ˆt + (g · n)ˆn , an increasing diferentiable function and, therefore, has the normal force N and the friction fr = −µNˆt, where a differentiable inverse function t = t(s). The arclength µ is the friction coefficient and N = kNk. The normal also allows us to define the so called Frenet–Serret frame, force is given by N = (bˆ · zˆ)N ˆn, resulting in N being which is a coordinate system (orthonormal basis) formed parallel or antiparallel to ˆn. by three orthonormal vectors associated to any regular Using the Newton’s laws, F = ma = W + f + N leads ˆ r curve: the tangent ˆt, the normal ˆn and binormal b. They to are defined by dv(s) dr mv(s) = mg · ˆt(s) − µN(s) (7) ˆt ≡ (2a) ds ds 2 ˆ 1 dˆt mκ(s)v (s) = mg · ˆn(s) + (b · zˆ)N(s). (8) ˆn ≡ (2b) κ(s) ds Eliminating N(s) from the equations, we obtain a differ- bˆ ≡ ˆt × ˆn, (2c) ential equation for v2(s): where κ(s) ≡ dˆt/ds is the curvature of the curve at 1 d 2 µκ(s) 2 g · ˆn(s) s. Notice that the normal vector ˆn is not defined when v (s) + v (s) = g · ˆt(s) + µ . (9) 2 ds ˆ ˆ κ(s) = 0, which is the case of flat surfaces and inflection b · zˆ b · zˆ ˆ points of functions. The vector b is out of the plane XY, Multiplying by the integrating factor and is parallel to the unit vector zˆ of the Cartesian co- s exp 2µ R κ(s0)ds0 , the above equation can be ordinates. Thus, bˆ · zˆ = ±1, depending on the concavity bˆ·zˆ 0 of the curve (see Fig.2). rewritten as The arclength s corresponds to the distance that the d h 2 2µ Φ(s)i 2µ Φ(s) g · ˆn(s) particle has travelled along the surface, and can be used v (s) e bˆ·zˆ = 2 e bˆ·zˆ g · ˆt(s) + µ , to describe the velocity and acceleration of the particle: ds bˆ · zˆ (10) dr ds v(t(s)) = = v(t(s))ˆt(s) (3) where we have defined ds dt Z s dv(t(s)) ds dˆt(s) ds 0 0 a(t(s)) = ˆt(s) + v(t(s)) . (4) Φ(s) ≡ κ(s )ds . (11) ds dt ds dt 0 3

2 Integrating from 0 to s at both sides of Eq. (10), we v (s)/gR > H(s). Therefore the initial velocity must 2 obtain satisfy v0/gR > H(0). Since g · ˆn(0) < 0 in this case, we 2 2µ 2µ have H(0) 0, therefore v /gR H(0) for all concave- 2 2 − ˆ Φ(s) − ˆ Φ(s) 6 0 > v (s) = v0 e b·zˆ + 2 e b·zˆ upwards surfaces. This means that the initial velocity has Z s 0 0 g · ˆn(s ) 2µ Φ(s0) 0 no restrictions (the particle always stays in the surface if × g · ˆt(s ) + µ e bˆ·zˆ ds . (12) 0 bˆ · zˆ the surface is concave upwards). where we have used that Φ(0) = 0 and v(0) ≡ v0. This is the most general equation for the velocity, since s is always defined, increasing, and unique. However, an anal- C. Tangential Angle lytical expression for the arclength s is impossible to obtain in most of the particular cases, since the integral in We are interested in “hill-down” curves (concave down- Eq. (1) cannot be solved. We will overcome this problem wards), which means that the vertical component of the using the tangential angle, described in the next section, position decreases monotonically. For these curves, ˆt is a acknowledging its restrictions, but being sufficient for our vector with a positive x component and negative y com- purposes. ponent. Then, a useful way of decribing the tangent vec- The magnitude of the normal force can be obtained tor is through the Whewell angle7–9 from Eq. (8): ˆt = cos ϕxˆ − sin ϕyˆ, (18) κ(s) g · ˆn(s) N(s) = mv2(s) − m , (13) ˆ ˆ b · zˆ b · zˆ where ϕ is the angle that ˆt forms with the horizontal (the where v(s) is given by Eq. (12). The release condition is same that ˆn forms with the vertical, see Fig.3). With N(s) = 0, which leads to this definition, the normal becomes

v2(s) 1 g · ˆn(s) (dϕ/ds) = (14) ˆn = − [sin ϕxˆ + cos ϕyˆ]. (19) gR gR κ(s) |dϕ/ds| for some s, where R is the height from which the particle The curvature is deﬁned by κ(s) = kdˆt/dsk, there- starts moving. Both sides of the above equation are di- fore κ(s) = |dϕ/ds|. If ϕ is monotonically increasing mensionless, and we will call the right-hand side of this (the curve is concave downwards), dϕ/ds > 0, therefore equation the horizon of the curve: κ(s) = dϕ/ds. This means that R κ(s)ds = ϕ(s), and 1 g · ˆn(s) therefore Φ in Eq. (11) becomes Φ(s) = ϕ(s) − ϕ0. We H(s) ≡ . (15) ˆ ˆ gR κ(s) also have b = −zˆ, then b · zˆ = −1. This allows us to write Eq. (12) as The particle remains in the surface as long as the normal 2 2 2µ(ϕ−ϕ0) 2µϕ is greater than zero. For surfaces that are concave down- v (ϕ) = v0 e + 2g e wards (such as the left portion of the curve in Fig.2), 0 Z ϕ e−2µϕ bˆ · zˆ = −1, and the condition N(s) 0 becomes, using × (sin ϕ0 − µ cos ϕ0) dϕ0, (20) > κ(ϕ0) Eq. (13), ϕ0

v2(s) where we have used g = −gyˆ. The horizon in Eq. (15) H(s). (16) gR 6 can also be written in terms of ϕ: The horizon, then, represents the frontier that the ve- 1 g · ˆn cos ϕ H(ϕ) = = . (21) locity has to reach in order to release the particle from gR κ(ϕ) Rκ(ϕ) 2 the surface. In particular, for s = 0 we have v0/gR 6 H(0) = g·ˆn/(gRκ(0)). If both sides are equal (v0 touches When ϕ is a decreasing function (the curve is concave the horizon), the particle leaves the surface immediately, upwards), κ = −dϕ/ds and (bˆ · zˆ)= 1, then we have ﬂying in a parabolic trajectory above the surface. There- fore, the maximum initial velocity that can be given to v2(ϕ) = v2 e2µ(ϕ−ϕ0) − 2g e2µϕ the particle is given by 0 Z ϕ −2µϕ0 0 0 e 0 (max) × (sin ϕ − µ cos ϕ ) dϕ . (22) v2 g · ˆn(0) κ(ϕ0) 0 ≡ H(0) = . (17) ϕ0 gR gRκ(0) In the following section, we will only analyze curves which When the surface is concave upwards (like the right por- are concave downwards, since we are interested in the tion of the curve in Fig.2), then bˆ · zˆ = 1, and the point of departure of a particle slipping on its surface, remain-on-the-surface condition, N(s) > 0, implies that which does not appear in concave upwards curves. 4

III. RESULTS that can be given to the particle is

v2 (max) We will analyze some particular curves in the frame- 0 = 1. (26) work presented, characterizing the movement of a par- gR ticle falling on their surface. In all cases, the particle For the case µ = 0, we recover the well-known expression starts its motion from the top of a curve, at a position found in text books: r(0) = Ryˆ. 2 2 v (ϕ) = v0 + 2gR(1 − cos ϕ), A. Departure from the Circle which can be directly obtained from conservation of energy. In this case, the particle leaves the surface (or We can parametrize the circle using the tangential an- touches the horizon) when gle ϕ by r(ϕ) = (R sin ϕ, R cos ϕ), where ϕ ∈ [0, π/2] also 2 v2 corresponds to the angle between r(ϕ) and the vertical cos ϕ = + 0 , axis, as shown in Fig.3. Note that the circle is the only 3 3gR recovering the already known departure angle for the fric- o tionless circle with v0 = 0, ϕ = arccos(2/3) ≈ 48.19 . In Fig.4 we show the evolution of v2(ϕ)/gR as it ap- y proaches the horizon H(ϕ) = cos ϕ. The thin, black line

R 1.0 µ =0 (v0 =0) ϕ µ =0.0 t 0.8 µ =1.0 y r µ =1.16 ϕ µ =1.4 x 0.6 Horizon /gR

x R ) ϕ (

2 0.4 v

0.2

FIG. 3. A circle parametrized by r(ϕ) = (R sin ϕ, R cos ϕ). 0.0 0 15 30 45 60 75 90 ϕ curve in which the tangential angle ϕ coincides with the angle that the vector r forms with the vertical, which is FIG. 4. Squared velocity as a function of ϕ and the horizon 2 not the case of other curves, as we will see in the follow- of the circle, H(ϕ) = cos ϕ for v0 /gR = 0.6. As pointed out by Mungan3, the horizon is equal to v2(ϕ)/gR for the case ing sections. For the circle, we have κ(ϕ) = 1/R, and 2 since ϕ is increasing, Eq. (20) leads to v0 /gR → 1 and µ → ∞.

2 2 Z ϕ v (ϕ) v0 2µϕ 2µϕ 0 0 −2µϕ0 0 corresponds to the velocity of a particle slipping in a = e +2 e (sin ϕ −µ cos ϕ ) e dϕ . frictionless, circular surface. We observe that the inter- gR gR 0 (23) section with the horizon (thick, black line) takes place o This integral can be easily solved by replacing sin ϕ = at the already predicted angle ϕ ≈ 48.19 , which is the ( eiϕ − e−iϕ)/2i and cos ϕ = ( eiϕ + e−iϕ)/2, leading to point at which the particle leaves the surface. The evo- an explicit form for the velocity: lution of the velocity is also shown for different values of 2 µ for an initial velocity of v0/gR = 0.6: with no friction v2(ϕ) v2 (2 − 4µ2) e2µϕ − cos ϕ − 6µ sin ϕ (blue line), µ = 1.0 (green) and µ = 1.16 (cyan). In = 0 e2µϕ + . gR gR 1 + 4µ2 all these cases, we observe that the friction decreases the (24) velocity up to certain point, in which the gravity over- This equation is identical to the one found by Mungan3. comes friction and the velocity starts increasing again, For the circle, the horizon in Eq. (21) becomes H(ϕ) = until it reaches the horizon (the point at which the parti- cos ϕ, which means that, before leaving the surface, the cle leaves the surface). Notice that the more we increase velocity satisfies the friction coefficient for a given initial velocity, the far- ther the particle travels on the surface, departuring at v2(ϕ) even greater angles. cos ϕ, (25) gR 6 However, if the friction coefficient is too large, it leads to solutions in which the particle comes to rest in the sur- and, according to Eq. (17), the maximum initial velocity face (red line Fig.4). This phenomenon suggests that, 5 for a given curve, there must be a critical friction coef- and its value in ϕ = 0 gives us the maximum initial ficient for every initial velocity (or a critical velocity for velocity: every friction coefficient) such that the particle will get (max) stuck on the surface for any value of µ greater than this v2 0 = γ2. (31) critical coefficient. Conversely, for a given friction coeffi- gR cient, there must be a critical initial velocity such that, for initial velocities lower than the critical one, the parti- For µ = 0, the velocity takes the form cle will always get stuck on the surface. We will analyze 2 2 v (ϕ) v0 2 cos ϕ this phenomenon in detail in sectionIIIF. = + 2 − p , (32) gR gR γ2 sin2 ϕ + cos2 ϕ which, in terms of α, becomes B. Departure from an Ellipse v2(α) v2 = 0 + 2(1 − cos α). (33) We can parametrize the ellipse by gR gR r(α) = (Rγ sin α, R cos α), with α ∈ [0, π/2] and γ > 0, which represents the curve shown in Fig.5. The When this velocity reaches the horizon, which in terms of α becomes H(α) = cos α(γ2 cos2 α+sin2 α), we obtain y v2 R 0 + 2 = 3 cos α + (γ2 − 1) cos3 α, (34) gR y ϕ r which give us the departure angle α. We can then obtain θ x it in terms of ϕ using Eq. (27). This equation always has x Rγ 2 2 a solution, as long as the condition v0/gR ≤ γ (v0 ≤ (max) v0 ). The solution also reobtains the departure angle for the circle setting γ = 1, where α = ϕ. 2 FIG. 5. An ellipse represented by r(α) = (Rγ sin α, R cos α). In Fig.6 we show the evolution of v (ϕ)/gR as it 2 2 2 Notice that ϕ is not equal to θ, the angle that r forms with approaches the horizon H(ϕ) = γ cos ϕ/(γ sin ϕ + 2 3/2 the vertical, and both of them are different from α. They are cos ϕ) . related by tan θ = γ tan α = γ2 tan ϕ.

1.0 tangent vector in this case is µ =0 (v0 =0) µ =0.0 0.8 µ =1.0 ˆ (Rγ cos αxˆ − R sin αyˆ) t = p , µ =1.7 2 2 2 2 2 µ =1.8 R γ cos α + R sin α 0.6 Horizon /gR )

ˆ ϕ and since t = cos ϕxˆ − sin ϕyˆ, we obtain the relationship (

2 0.4 v 1 tan ϕ = tan α. (27) γ 0.2

This allows us to calculate the curvature in terms of ϕ: 0.0 0 15 30 45 60 75 90 ϕ 2 2 2 3 (γ sin ϕ + cos ϕ) 2 κ(ϕ) = . (28) γ2R FIG. 6. Squared velocity as a function of ϕ and the horizon of the ellipse, H(ϕ) = γ2 cos ϕ/(γ2 sin2 ϕ+cos2 ϕ)3/2 for γ = 0.6, 2 Then, according to Eq. (20), the velocity becomes v0 /gR = 0.3. v2(ϕ) v2 = 0 e2µϕ + 2 e2µϕ gR gR ϕ 2 0 0 Z γ [sin ϕ − µ cos ϕ ] 0 C. Departure from a Parabola × e−2µϕ dϕ0, (29) 2 2 2 3 0 (γ sin ϕ + cos ϕ) 2 Using the parametrization r(α) = (Rγα, R(1 − α2)) which is not integrable analitically. However, we can still with α ∈ [0, ∞) and γ > 0, we can describe the curve determine the maximum initial velocity like we did in the shown in Fig.7. For this parametrization, we obtain a last section: the horizon of the ellipse is tangent vector

2 γ cos ϕ ˆ (γxˆ − 2αyˆ) H(ϕ) = , (30) t = p , (γ2 sin2 ϕ + cos2 ϕ)3/2 γ2 + 4α2 6

y Then, the particular value for which the velocity no longer depends on the friction is actually the maximum R initial velocity. If the particle is given an initial velocity ϕ (max) r v0 ≥ v0 , it will ﬂy on a trayectory given by t y 1 x(t) = v t, y(t) = R − gt2. 0 2 Rγ x x Replacing t = x/v0 for y(t), we obtain

g x2 x2 y = R − 2 = R 1 − 2 2 , (41) 2 v0 2R (v0/gR) 2 FIG. 7. A parabola represented by r(α) = (Rγα, R(1 − α )). In our parametrization, x = Rγα and y = R(1 − α2), then which is equal to ˆt = cos ϕxˆ − sin ϕyˆ, whereof the rela- x2 y = R(1 − α2) = R 1 − . (42) tionship 2R2(γ2/2)

2 tan ϕ = 2α/γ (35) Then, we can clearly see that as the velocity v0/gR, given to the particle to leave the surface, approaches to γ2/2, is obtained. From this, the curvature in tems of ϕ is the maximum velocity allowed in the surface, the two 2 parabolic trajectories become equal. This means that 3 the particle ﬂies over the surface at all times, which is κ(ϕ) = 2 cos ϕ. (36) γ R the reason why this velocity eliminates the dependence Then, according to Eq. (20), the velocity becomes on µ in Eq. (37). The second interesting fact in this case is that when 2 2 Z ϕ −2µϕ0 the second term of Eq. (37) is positive; that is, in the v (ϕ) v0 2µϕ 2 2µϕ 0 0 e 0 2 2 = e +γ e [sin ϕ −µ cos ϕ ] 3 0 dϕ . case that v0/gR < γ /2, we have gR gR 0 cos ϕ v2(ϕ) γ2 An interesting fact is that < sec2 ϕ = H(ϕ), (43) gR 2 −2µϕ d −2µϕ 2 2 e [ e sec ϕ] = 3 [sin ϕ − µ cos ϕ], for any value of µ and v0. We have concluded, then, that dϕ cos ϕ the velocity never reaches the horizon, which means that which makes the integral fully solvable, thus obtaining the particle never leaves the parabolic surface, i.e., there an explicit solution for the velocity: is no departure angle. For µ = 0, the velocity becomes v2(ϕ) γ2 γ2 v2 = sec2 ϕ − − 0 e2µϕ. (37) v2(ϕ) v2 γ2 gR 2 2 gR = 0 + tan2 ϕ (44) gR gR 2 We observe that the particular velocity v2/gR = γ2/2 0 (max) p 2 anihilates all dependence on the friction, leading to and if v0 = v0 = γ gR/2, we recover v (ϕ)/gR = 1 2 2 2 γ sec ϕ = H(ϕ) obtained in Eq. (38). 2 2 2 v (ϕ) γ 2 In Fig.8 we show the evolution of v (ϕ)/gR as it ap- = sec ϕ (38) 1 2 2 gR 2 proaches the horizon H(ϕ) = 2 γ sec ϕ, never reaching it. for any value of µ. This is surprising, since no other curve shows this phenomenon. The explanation is related to the fact that the trajectory of the particle in free space D. Departure from a Catenary under the action of gravity, is precisely a parabola. The horizon of the parabola is The parametrization r(α) = (Rα, R(2 − cosh α)) represents the catenary shown in Fig.9. Under this γ2 H(ϕ) = sec2 ϕ (39) parametrization, 2 ˆt = sech(α)xˆ − tanh αyˆ, and the limit for the initial velocity in this case is which leads to v2 (max) γ2 0 = . (40) gR 2 tan ϕ = sinh α. (45) 7

The horizon is 2.0

µ =0 (v0 =0) µ =0.0 H(ϕ) = sec ϕ (48) µ =0.5 1.5 µ =1.0 which bounds the initial velocity to µ =1.8 Horizon 2 (max) /gR v0 ) 1.0 = H(0) = 1, (49) ϕ ( gR 2 v

0.5 the same as the circle, since κ(0) = 1/R in both cases. For µ = 0, the velocity is

2 2 0.0 v (ϕ) v0 0 15 30 45 60 75 90 = + 2(sec ϕ − 1), (50) ϕ gR gR which touches the horizon when FIG. 8. Squared velocity as a function of ϕ and the horizon 1 2 2 2 2 of the parabola. H(ϕ) = γ sec ϕ for γ = 0.8, v0 /gR = v 2 0 + 2(sec ϕ − 1) = sec ϕ, 0.2. For some conditions, the particle sticks to the surface, gR but when it does not stop, it never touches the horizon (no departure from the surface). or v2 y sec ϕ = 2 − 0 . (51) gR R ϕ If the particle starts with zero velocity, it leaves the sur- r t y face at a departure angle ϕ0 that satisfies sec ϕ0 = 2, o that is, when ϕ0 = 60 . Therefore, the particle leaves x the frictionless catenary with a steeper slope than in the x Rα0 frictionless circle, when the particle starts with zero velocity. In terms of the parameter α, this means that the particle leaves the catenary for a value α0 that satisfies cos α0 = 2 (since sec ϕ = cosh α). Therefore the particle leaves the surface when r(α ) = Rα xˆ, this is, when it FIG. 9. A catenary represented by r(α) = (Rα, R(2−cosh α)). 0 0 −1 reaches the floor (y = 0). Here, α0 = cosh (2) ≈ 1.317. The long-dashed line corresponds to the catenary. The dotted line is a parabola When an initial velocity is added to the particle, it (shown for reference) that crosses through the points (0,R) slides on the frictionless catenary until, according to 2 and (Rα0, 0). Eq. (51), we have cosh α = 2 − v0/gR. We conclude that the point√ of departure, in terms of α, is given by 2 2 α = ln[ε + ε − 1], where ε = 2 − v0/gR. From this From this relationship, we can obtain the curvature in expression, we√ recover the result for zero initial velocity, −1 o term of ϕ: α0 = ln(2 + 3) = cos (2) (or ϕ = 60 ). In Fig. 10 we show the evolution of v2(ϕ)/gR as it 1 approaches the horizon of the catenary, H(ϕ) = sec ϕ. κ(ϕ) = cos2 ϕ. (46) R

Notice that, although the parabola that intersects the E. Departure from a Cycloid vertical and horizontal axes in the same points as the catenary (see dotted line in Fig.9), is very similar to the The cycloid of height R can be parametrized by r(α) = catenary, their curvatures are not. From Eq. (36), we R (α + sin α, 1 + cos α), which produces the curve shown obtain that the curvature of this particular parabola is 2 2 3 3 in Fig. 11. With this parametrization, the tangent vector κ(ϕ) = (2/α0R) cos ϕ ≈ 1.153 cos ϕ, while the catenary is has a curvature κ(ϕ) = cos2 ϕ/R. Interestingly, both 2 curves have the same curvature when cos ϕ = α0/2 (ϕ ≈ ˆt = cos(α/2)xˆ − sin(α/2)yˆ 29.86o). ˆ If we replace the curvature of the catenary in Eq. (20), which is equal to t = cos ϕxˆ − sin ϕyˆ. Therefore we obtain the velocity of the particle slipping on this 1 ϕ = α. (52) surface, 2

0 v2(ϕ) v2 Z ϕ e−2µϕ From here, = 0 e2µϕ+2 e2µϕ [sin ϕ0−µ cos ϕ0] dϕ0. gR gR cos2 ϕ0 1 0 κ(ϕ) = sec ϕ, (53) (47) 2R 8

twice the velocity that can be given in a circle (since the 4.0 curvature at ϕ = 0 is half that of the circle). For µ = 0, µ =0 (v0 =0) 3.5 µ =0.0 the velocity is µ =0.8 3.0 µ =0.86 2 2 v (ϕ) v0 2 2.5 µ =1.0 = + 2 sin ϕ. (58) Horizon gR gR /gR

) 2.0 ϕ (

2 1.5 In this case, the point of departure is satisﬁed when v 1.0 v2 0 + 2 sin2 ϕ = 2 cos2 ϕ, 0.5 gR

0.0 0 15 30 45 60 75 90 therefore ϕ v2 cos(2ϕ) = 0 . FIG. 10. Squared velocity as a function of ϕ and the horizon of 2gR 2 the catenary. H(ϕ) = sec ϕ. Initial velocity v0 /gR = 0.5. For some conditions, the particle sticks to the surface, but when Notice that this means that v0 = 0 implies ϕ0 = π/4 it does not stop, it never touches the horizon (no departure (α0 = π/2). Since a cycloid is drawn from a point in the from the surface). surface of a circle, which in this case has a radius R/2, this situation concides with the moment in which this y circle has moved 1/4 of the whole turn: R ϕ y r t R 1 R R r(α0) = + π xˆ + yˆ. x 2 2 2 2 x Rπ 2 The slope in this point (ϕ = 45o) is slightly smaller than in the case of the frictionless circle, where ϕ ≈ 48o. FIG. 11. A cycloid represented by r(α) = R (α + sin α, 1 + 2 In Fig. 12 we show the evolution of v2(ϕ)/gR as it cos α). approaches the horizon of the cycloid, H(ϕ) = 2 cos2 ϕ. which gives us the expression for the velocity: 2.0 µ =0 (v =0) 2 2 Z ϕ −2µϕ0 0 v (ϕ) v0 2µϕ 2µϕ 0 0 e 0 µ =0.0 = e +4 e [sin ϕ −µ cos ϕ ] dϕ . µ =0.5 gR gR sec ϕ0 1.5 0 µ =1.0 (54) µ =1.8 As the case of the circle and the parabola, this integral Horizon /gR

) 1.0 can be solved analitically, since ϕ ( 2 −2µϕ v d −2µϕ 2 e [ e (sin ϕ − µ cos ϕ) ] = (sin ϕ − µ cos ϕ) 0.5 dϕ sec ϕ × 2 1 + µ2 0.0 0 15 30 45 60 75 90 from where we obtain ϕ 2 2 2 2 2µϕ v (ϕ) v0 2µϕ 2(sin ϕ − µ cos ϕ) − 2µ e = e + 2 . FIG. 12. Squared velocity as a function of ϕ and the horizon gR gR 1 + µ 2 2 of the cycloid. H(ϕ) = 2 cos ϕ. Initial velocity v0 /gR = 1. (55) We observe the same behaviour as in the other cases: it is 2 Notice that when µ = 1 and v0/gR = 1, the particle stops always possible to stop the particle in the surface for a big exactly at ϕ = π/4 (see Fig. 12). In the circle, as shown enough µ. 3 by Mungan , the particle reaches ϕ = π/√4 with zero 2 2 −π/2 velocity when µ = 1 and v0/gR = 5 1 + 2e ≈ 0.52.

The horizon of the catenary, according to Eq. (21), is F. Phase diagram µ–v0 H(ϕ) = 2 cos2 ϕ, (56) As we increase µ for a particular value of v0 > 0, which sets a maximum initial velocity of there is a moment in which the velocity splits into two branches: the first one physical (red line in all previous (max) v2 figures), where the particle stops in the surface for some 0 = H(0) = 2, (57) gR critical angle ϕc, and the second non-physical (not drawn 9 in the figures), which appears as a mathematical contin- 2 4.0 uation of the function v (ϕ) after it becomes negative. Circle 5 The critical angle ϕc is obtained, as de Lange et. al did 3.5 Cycloid Parabola (γ =1.2) for the circle, when the velocity satisfies the following two 3.0 Catenary conditions: Particle 2.5 Ellipse (γ =1.5) leaves

d R 2 2 g v (ϕc) = 0, v (ϕ) = 0. (59) / 2.0 2 dϕ 0 ϕ v c 1.5

By applying these conditions to the velocity equation (9), 1.0 we see that the left-hand side of the equation vanishes, 0.5 Particle and a general result is obtained: sticks 0.0 0 1 2 3 4 5 g · ˆt(sc) µ = −(bˆ · zˆ) = tan ϕc (60) µ g · ˆn(sc) ˆ FIG. 13. Phase diagram separating the conditions µ–v0 that regardless of the sign of (b · zˆ) = ±1. Here, sc = s(ϕc). lead to a particle stuck in the surface, to those conditions This result is particulary meaningful, since when a par- that allow it to move without stop. The curves are plots of ticle moves in an inclined rough plane (where κ(ϕ) = 0) equations (62) (circle), (63) (parabola), (64) (cycloid), (65) with a friction coeﬃcient equal to the slope of the plane (catenary) and (66) (ellipse). As µ increases, the velocity (µ = tan ϕ), the particle travels at constant velocity. If converges to 1 in the case of the circle and catenary, to 2 for −1 we substitute this result (ϕc = tan µ) in Eq. (20), us- the cycloid. For the chosen values of γ, the velocity converges 2 2 ing the conditions in Eq. (59), we obtain v0 as a function to γ /2 = 0.72 in the case of the parabola, and to γ = 2.25 of µ: for the ellipse.

2 Z tan−1 µ −2µϕ0 v0(µ) 0 0 e 0 = 2 (µ cos ϕ − sin ϕ ) 0 dϕ . (61) G. Curves with no point of departure gR 0 Rκ(ϕ ) This curve defines a phase diagram for the initial con- A general behavior observed in the analysis performed ditions µ, v0, that separates those that lead to a particle is that, if the friction coefficient does not stop the parti- that sticks in the surface from those that allow it to move cle, the velocity always starts increasing in some point. It on and reach the horizon. For the circle, parabola and is clear that a flat or decreasing horizon (constant or in- the cycloid this relationship is: creasing curvature, such as the circle, the ellipse and the −1 v2(µ) 2(2µ2 − 1 + e−2µ tan µp1 + µ2) cycloid) will always be intersected by the velocity, that is, 0 = , (62) gR 1 + 4µ2 the particle will always leave the surfaces characterized 2 2 by a decreasing horizon. But if the horizon increases, it v (µ) γ −1 0 = 1 − (1 + µ2) e−2µ tan µ , (63) is not clear whether the velocity will reach the horizon gR 2 or not. Some horizons do not increase fast enough, and v2(µ) 2µ2 the velocity intersects them (such as the catenary); but 0 = , (64) gR 1 + µ2 some horizons increase faster than the velocity and the intersection with it never takes place, as we saw in the respectively. As µ increases, these velocities tend to 1, case of the parabola. γ2/2 and 2, respectively. Since the integral term of the Since the behavior of the horizon depends on the cur- velocity cannot be calculated explicitly neither for the vature, it is necessary to study different curvatures. For catenary, nor the ellipse, the substitution ϕ = tan µ also c a given curve of known curvature, the relationship leads to non integrable expressions:

−1 0 dr dr ds ˆt ˆt(ϕ) v2(µ) Z tan µ e−2µϕ = = = , (68) 0 0 0 0 dϕ ds dϕ dϕ/ds κ(ϕ) = 2 [µ cos ϕ − sin ϕ ] 2 0 dϕ (65) gR 0 cos ϕ ˆ for the catenary, and where t = cos ϕxˆ−sin ϕyˆ, allows to build a parametrization r(ϕ) = (x(ϕ), y(ϕ)) for the curve, since it implies 2 tan−1 µ 2 0 0 v (µ) Z γ [µ cos ϕ − sin ϕ ] 0 that 0 = 2 e−2µϕ dϕ0 2 3 gR (γ2 sin ϕ + cos2 ϕ) 2 cos ϕ sin ϕ 0 (x0(ϕ), y0(ϕ)) = , − , (66) κ(ϕ) κ(ϕ) for the ellipse, which tends to 1 and γ2, respectively. As we see, these limits correspond to the maximum initial from where Z ϕ 0 velocity in Eq. (17): cos ϕ 0 x(ϕ) = x0 + 0 dϕ (69a) (crit) max 0 κ(ϕ ) lim v0 (µ) = v0 . (67) Z ϕ 0 µ→∞ sin ϕ 0 y(ϕ) = y0 − 0 dϕ , (69b) All these curves are shown in Fig. 13. 0 κ(ϕ ) 10 where x0 = 0 and y0 = R represent the departure maximum initial velocity. The cases n = 1 and n = 2 point we have used in all cases. Notice that all the are the inclined plane and the parabola, in which we al- curvatures of the curves we have presented can be ex- ready know that the particle will remain always in the pressed by κ(ϕ) = cosn ϕ/R. For example, the case of surface. When 0 < n < 1, the concavity changes and the parabola corresponds to n = 3 (see Eq. (36)), and bˆ · zˆ = 1, which means that the particle always stays on from the above equations we obtain x(ϕ) = R tan ϕ, the surface. y(ϕ) = R − R(1 − sec2 ϕ/2), which indeed represents 1 2 the parabola y(x) = 2 (R − x /R). In similar fash- ion, it represents a catenary if n = 2, the function IV. SUGGESTED PROJECTS y(x) = R + R ln cos(x/R) if n = 1, a circle if n = 0, and a cycloid if n = −1. The velocity intersects the hori- We have developed a general treatment to find the ve- zon H(ϕ) = cos ϕ/Rκ(ϕ) in these four cases, and also for locity of a particle sliding in an arbitrary surface with every n < 3, since the horizon will be decreasing. friction, providing a framework to find the point of de- Any curve such that H(π/2) = const. 6= 0 will also parture. For each surface, a maximum initial velocity is be intersected by the increasing velocity in some point, found, above which the particle leaves the surface imme- meaning that the particle will departure from the sur- 0 diately. We introduced the concept of the horizon of a face. This happens when κ(π/2) = 0 and κ (π/2) 6= curve, which has proven to be fundamental to find the 0, for example κ(ϕ) = (1 − 2ϕ/π)/R, in which case 0 π π point of departure, given by the intersection between the limϕ→ π H(ϕ) = −1/κ ( ) = exists, and the horizon 2 2 2 velocity and the horizon. A summary of all the curves is a slowly increasing function. But not all curvatures analyzed is shown in TableI. that are zero at ϕ = π produce a constant value of H 2 Here we propose a set of possible projects with different at this angle. This is the case of the catenary and the π levels of complexity. parabola, whose curvature is zero at ϕ = 2 and the horizon diverges at this point. However, a divergent horizon (which is obtained when κ( π ) = κ0( π ) = 0) is still not 2 2 A. Power Laws sufficient requirement to avoid the velocity to reach it, as the case of the catenary show us. Therefore, we observe that a sufficient condition to avoid an intersection Surfaces described by a decreasing function f : R → R, between the velocity and the horizon (to keep the parti- have a natural parametrization r(x) = (x, f(x)), such k cle always in the curve) is to require κ to be a decreasing as f(x) = −αx . Find their curvature and demonstrate function (κ0(ϕ) < 0) such that that for k > 2 it is not possible to throw the particle from the origin with any bounded initial velocity such that it π π π κ = κ0 = κ00 = 0. (70) immediately leaves the surface. Demonstrate that any 2 2 2 curve with an inflection point at the origin also shows this behavior. We believe that it is necessary to require the monotonically decreasing condition, since curvatures such as κ(ϕ) = ϕ(1−2ϕ/π)3, which increases and then decreases, satisfy Eq. (70) and still allow an intersection between B. Nephroid the horizon and a the velocity. In this way, the fam- n ily of curves κ(ϕ) = cos ϕ/R, with n > 3 have non- Another interesting, famous curve is the nephroid. A intersectable horizons, with the parabola being only a possible parametrization for this curve is particular case. Other examples are easy to find, such as n R R the family of functions κ(ϕ) = (1 − 2ϕ/π) /R for n > 3, r(α) = (3 sin(α) + sin(3α)), (3 cos(α) + cos(3α)) , in which the particle also stays always in the surface. 4 4 We would also like to remark that the family of func- (71) tions f(x) = R(1 − αxn) have a curvature κ(ϕ) = which we show in Fig. 14. Obtain the velocity of the 1 n−2 particle in terms of the tangential angle ϕ, the horizon (αn) n−1 |n − 1| cos3(ϕ) tan n−1 (ϕ), which is zero at ϕ = 0 π of the curve and the maximum initial velocity. Plot the and ϕ = 2 for n > 2. This means that the horizon diverges in these two points, and despite of the fact that velocity as a function of ϕ for different initial velocities this curvature satisfies the conditions in Eq. (70), an in- and friction coefficients. What can be said about the tersection between the horizon and the velocity does ex- departure angle? ist. When 1 < n < 2, the curvature diverges at ϕ = 0 and limϕ→0 H(ϕ) = 0. This means that, according to Eq. (17), the maximum velocity that can be given to the C. Other forces particle is zero, which in turns means that at the slightest 2 impulse provided to the particle, it will leave the surface. A dragging force FD = −βv (s)ˆt, where β is a constant This result has been previously explained6, but here we and v the velocity of the particle, can be incorporated to have presented it from the point of view of a quantifiable the model. This force competes with the friction. Show 11

Circle Ellipse Parabola Catenary Cycloid 2 R r(α) (R sin α, R cos α) (Rγ sin α, R cos α) (Rγα, R(1 − α )) (Rα, R(2 − cosh α)) 2 (α+sin α, 1+cos α) 1 ϕ(α) ϕ = α tan ϕ = (1/γ) tan α tan ϕ = 2α/γ tan ϕ = sinh α ϕ = 2 α 3 (γ2 sin2 ϕ+cos2 ϕ) 2 2 3 1 2 1 κ(ϕ) 1/R 2 cos ϕ cos ϕ sec ϕ γ2R γ R R 2R 2 2 2 2 v0 2µϕ v 2µϕ v 2µϕ v0 2µϕ e + 0 e + 2 0 e + e + 2 gR gR γ sec2 ϕ − gR gR v (ϕ) (2−4µ2) e2µϕ−cos ϕ 2 2µϕ R ϕ 0 2 2µϕ R ϕ 0 2(sin ϕ−µ cos ϕ)2 ( ) 2 R e 0 [sin ϕ − 2 v 2 e [sin ϕ − 2 − 1+4µ2 γ 0 2µϕ 0 1+µ gR −2µϕ0 − e −2µϕ0 6µ sin ϕ µ cos ϕ0] e dϕ0 2 gR µ cos ϕ0] e dϕ0 2µ2 e2µϕ − 1+4µ2 κ(ϕ0) cos2 ϕ0 1+µ2 2 2 v (ϕ) v0 for v2 gR + 2 − v2 2 v2 v2 0 + 2(1 − cos ϕ) 0 + γ tan2 ϕ 0 + 2(sec ϕ − 1) 0 + 2 sin2 ϕ gR gR √ 2 cos ϕ gR 2 gR gR µ = 0 γ2 sin2 ϕ+cos2 ϕ v2 (max) 0 1 γ2 1 γ2 1 2 gR 2 2 Horizon cos ϕ γ cos ϕ γ2 2 sec ϕ 2 cos2 ϕ (γ2 sin2 ϕ+cos2 ϕ)3/2 2 sec ϕ

TABLE I. Summary of the curves.

y V. CONCLUSIONS

R We have found analytical expressions for the velocity of the particle when it slides on a circle, a parabola and on a cycloid, including the effects of friction, explaining in detail why a particle in a parabola does never leave that surface. Explicit departure angles have been found for yˆ frictionless surfaces when the particle starts from rest: xˆ R x o o 2 ϕ = arccos(2/3) ≈ 48 for the circle, ϕ = 45 for the cycloid, and ϕ = 60o for the catenary, an angle that coincides with the moment the particle reaches the floor. In the roughless ellipse, we do not have a determined departure angle, since it depends on the parameter γ. The departure angle decreases as γ increases, and it corresponds to the circle when γ = 1. For concave upwards surfaces, there is no departure angle. FIG. 14. A nephroid. Two important, general results obtained are the relationship between the critical angle ϕc and the friction coefficient µ, given by Eq. (60), and, as a consequence, the Eq. (61). This equation relates the friction coefficient that the equations of motion in this case are with the initial velocity and allows us to build the phase diagram in Fig. 13, which separates the initial conditions dv(s) µ–v that lead to a particle stuck on the surface to those mv(s) = mg · ˆt(s) − µN(s) − mβv2(s) (72) 0 ds conditions that allow it to move until it takes off (if the mκ(s)v2(s) = mg · ˆn(s) + (bˆ · zˆ)N(s). (73) intersection with the horizon exists). Regarding the maximum velocity that can be given initially to the particle, we can say that if the particle is Eliminate the normal force from the equations, and ob- pushed from the top of a curve with initially horizontal tain tangent (ˆn(0) = −yˆ), according to Eq. (17) we have

(max) d v2 g · ˆn(0) 1 v2(s) + 2 (β ± µκ(s)) v2(s) = 2(g · ˆt(s) ± µg · ˆn(s)). 0 = = , ds g gκ(0) κ(0) (74) Note that this equation already appears in the classi- which means that any curve that is locally ﬂat at the top cal book of Appel10, Chap. XII, § 251. Show that (κ(0) = 0) has inﬁnite maximum initial velocity. In other the the integrating factor that solves this equation is words, if the particle is kicked from a point of zero cur- exp [2βs(ϕ) ± 2µ(ϕ − ϕ0)]. Following the treatment pre- vature, there is no initial velocity for which the particle sented in this paper, analyze the case of a particle sliding leaves the surface immediately. on a circle with friction and a dragging force and compare Finally, although a formal demonstration is still lack- to the case presented inIIIA. ing, we have found a general criteria (κ0(ϕ) < 0 plus 12

Eq. (70)) to discriminate in which curves the particle cle is initially kicked from a point of zero curvature (like will always stay in the surface, and in which curves the an inflection point), there is no initial velocity for which particle takes off in some point. We have shown fam- the particle leaves the surface immediately. ilies of functions in which the particle never takes off (κ(ϕ) = (1 − 2ϕ/π)n/R, κ(ϕ) = cosn ϕ/R), and we have given the tools to analyze the particular case of power ACKNOWLEDGMENTS laws, in which the particle exhibits all the behaviors, depending on the exponent: it leaves the surface immedi- F.G.-C. acknowledges CONICYT PhD fellowship No. ately, in some point on the surface, or never leaves the 201090712 short-term fellowship from Universidad de surface. In addition, we have concluded that if the parti- Chile. G.G. thanks project Anillo ACT-1115.

∗ [email protected] [Am. J. Phys. 75 (5), 423426 (2007)]. American Journal of † gonzalo@fisica.ciencias.uchile.cl Physics 76, 92 (2008). ‡ [email protected] 6 Aghamohammadi, A. The point of departure of a particle 1 Alonso, M. & Finn, E. J. Fundamental University Physics: sliding on a curved surface. European Journal of Physics Mechanics, vol. 1 (Addison-Wesley, 1967). 33, 1111–1117 (2012). 2 Holbrow, C. H. Archaeology of a Bookstack: Some Major 7 do Carmo, M. P. Differential Geometry of Curves and Introductory Physics Texts of the Last 150 Years. Physics Surfaces (Pearson, 1976). Today 52, 50 (1999). 8 Shifrin, T. Differential Geometry: A First Course in 3 Mungan, C. E. Sliding on the Surface of a Rough Sphere. Curves and Surfaces (2015). The Physics Teacher 41, 326 (2003). 9 Whewell, W. Of the Intrinsic Equation of a Curve, and its 4 Prior, T. & Mele, E. J. A block slipping on a sphere with Application, vol. VIII (Cambridge Philosophical Transac- friction: Exact and perturbative solutions. American Jour- tions, 1849). nal of Physics 75, 423 (2007). 10 Appell, P. E. Traitéde mécanique rationnelle: Statique. 5 de Lange, O. L., Pierrus, J., Prior, T. & Mele, E. J. Com- Dynamique du point, vol. 1 (Gauthier-Villars et cie, Paris, ment on A block slipping on a sphere with friction: Exact 1893). and perturbative solutions, by Tom Prior and E. J. Mele