Given a continuous, differentiable convex curve such as a parabola, hyperbola, an ellipse, cycloid, catenary, semicircle, etc (with the domains accordingly restricted),

What is the time taken for a point particle to descend down the curve?

Note: this is a problem that I came up with myself and am wondering about, so it is possible that it is not solvable with the given information. If that is the case, please comment what other details I need to specify to make this solvable.


  • There is static friction
  • It is a particle, not a ball, so assume no rolling and hence no rolling resistance/slipping
  • Acceleration due to gravity is simply $-9.8ms^{-2}$
  • Air resistance is negligible
  • Please use any value for the coefficient of friction, or simply leave it as $\mu$, as well as any other quantities that I did not specify here.


If we place a blue point particle on the curves below (parabola and ellipse),

enter image description here enter image description here

It may look something like this: enter image description here

The ball will descend the ramp, then climb to the right side of the curve, then go down, climb the left side... etc until it stops at the minimum point.

Things I tried

I tried using the concept of energy (gravitational potential/kinetic) and work done by friction, but to no avail.

For the case of the parabola, I parametrised it as $x=t, y=t^2$ and then parametrised it with respect to arc length by taking $r' = i + 2tj$ so $|r'|=\sqrt{1+4t^2}$ and $s(t)$ is just given by the integral of this, which Wolfram gives as $$1/4\Big(2x\sqrt{4t^2+1}+sinh^{-1}{(2x)}\Big)$$

However, $x$ here is a function of time i.e. $x(t)$. I wasn't sure how to find this, and even if I did, I wasn't sure how to proceed.

  • $\begingroup$ The motion will never stop, the speed might go down to a Planck length per duration of the universe, but it will never got to zero. This is just a consequence of the uniqueness property of the Picard-Lindelöf theorem. The only solution that has zero velocity at the minimum point is the stationary one, forward and backward in time. $\endgroup$ Aug 1, 2021 at 9:12
  • 1
    $\begingroup$ That theorem is fascinating, thank you for giving me that insight. How can I show that the Picard-Lindelöf theorem applies in this context, and the limiting value of the speed (possibly the Plank length)? And is there a condition I can impose to the problem so that the particle will indeed stop? $\endgroup$
    – SoySoy4444
    Aug 1, 2021 at 11:38
  • $\begingroup$ No speed limit, that was just an absurd example of how slow the exact solution will become. You only need that the rhs of the ODE is continuously differentiable to get the uniqueness property, physical models in general satisfy this condition, one would have to work rather hard to construct a counter-example. $\endgroup$ Aug 1, 2021 at 12:17
  • 1
    $\begingroup$ What is the ODE? I haven't managed to come up with an ODE yet. And why is the RHS of that ODE not continuous? If I had some other curve, would it be continuous, or would it always be not? Can you think off the top of your head a modification to this question that would have a solution? (perhaps I should make it a rolling ball instead of a particle? or something else?) $\endgroup$
    – SoySoy4444
    Aug 1, 2021 at 12:34
  • 1
    $\begingroup$ @LutzLehmann: If we use the standard assumptions about kinetic friction from introductory mechanics classes, the frictional force is not Lipschitz continuous when $v = 0$. In fact, it's discontinuous, with a positive value as $v \to 0_-$ and the opposite negative value as $v \to 0_+$. So the assumptions of Picard-Lindelöf do not hold if the block ever stops. (And I didn't have to work that hard to find this counterexample.) $\endgroup$ Aug 1, 2021 at 14:07

3 Answers 3


For a material point $p=(x(t),y(t))$ with mass $m$ under gravity, and constrained to a curve $f(x,y)=0$ in presence of a viscous dissipation force $\mu \dot p$ the mechanical energy evolution can be described by

$$ \frac{d}{dt}\left(\frac{\partial L}{\partial\dot p}\right)-\frac{\partial L}{\partial p} = -\mu \dot p $$


$$ L = \frac 12 m\|\dot p\|^2-m g p\cdot e_y+\lambda f(p),\ \ e_y = (0,1) $$

Ex. For $f(x,y) = y-ax^2$ we have the movement equations

$$ \cases{ m\ddot x +\mu\dot x+2\lambda a x = 0\\ m\ddot y +\mu\dot y-\lambda + m g =0\\ y-a x^2 = 0 } $$

and after deriving twice the last equation we have

$$ \cases{ m\ddot x +\mu\dot x+2\lambda a x = 0\\ m\ddot y +\mu\dot y-\lambda + m g =0\\ \ddot y-2a \dot x^2-2a x\ddot x = 0 } $$

solving for $\ddot x,\ddot y,\lambda$ we obtain

$$ \left\{ \begin{array}{rcl} m\ddot x & = & -\frac{2 a x \left(2 a m \dot x^2+g m+\mu \dot y\right)+\mu \dot x}{4 a^2 x^2+1} \\ m\ddot y & = & -\frac{2 a \left(2 a x^2 \left(g m+\mu \dot y\right)-m \dot x^2+\mu x \dot x\right)}{4 a^2 x^2+1} \\ \lambda & = & \frac{2 a m \dot x^2-2 a \mu x \dot x+g m+\mu \dot y}{4 a^2 x^2+1} \\ \end{array} \right. $$


$\lambda$ is a lagrange multiplier and after solving it gives the normal reaction component on the curve. Solving for $a=1,g=10,m=1,\mu=1,x(0)=-1,\dot x(0) = 0$ we obtain for $x$ in light blue and $y$ light orange

enter image description here

The plot was generated with the help of the MATHEMATICA script

tmax = 10;
p = {x[t], y[t]};
solp = NDSolve[{D[x[t], t, t]== -((Derivative[1][x][t] + 2 x[t] (10 + 2 Derivative[1][x][t]^2 + Derivative[1][y][t]))/(1 + 4 x[t]^2)), 
                D[y[t], t, t] == -((2 (x[t] Derivative[1][x][t] - Derivative[1][x][t]^2 + 2 x[t]^2 (10 + Derivative[1][y][t])))/(1 + 4 x[t]^2)), 
                x[0] == -1, y[0] == 1, x'[0] == y'[0] == 0}, {x, y}, {t, 0, tmax}]
Plot[Evaluate[p /. solp], {t, 0, tmax}, PlotRange -> All]
  • $\begingroup$ Thank you for the answer. So, the ball will never stop? Could you possibly think of a modification to the question that would make the ball stop? And in that case, what would be the expression for the time of the ball's movement? Also, how would I work with a parametric curve? $\endgroup$
    – SoySoy4444
    Aug 2, 2021 at 3:13
  • $\begingroup$ To say that the ball will never stop is a questionable affirmation. Kind of Zeno's paradox. We are handling physical models.... $\endgroup$
    – Cesareo
    Aug 2, 2021 at 8:46
  • $\begingroup$ Ok, so suppose I adopt the theoretical model that the ball will never stop. I will define the ball "stopping" as some small negligible value (e.g. on the order of 10^-3). How would I find the amount of time it will take for both the ball's x and y component to oscillate less than 10^-3 units? Also, how can I calculate this value for a parametrically defined function? $\endgroup$
    – SoySoy4444
    Aug 2, 2021 at 9:38
  • $\begingroup$ When starting from rest, it may be a simpler case. $\endgroup$
    – Narasimham
    Aug 2, 2021 at 16:01
  • 1
    $\begingroup$ @SoySoy4444 I attached to the NOTE, the MATHEMATICA script used do produce the graphics. $\endgroup$
    – Cesareo
    Aug 22, 2021 at 13:37

Here's what I was able to come up with using energy methods. Ultimately it involves solving a system of coupled non-linear ODEs several times over, which almost certainly have to be handled using numerical techniques. For brevity I will just give a summary, but if there are points you're unclear on and can't figure out my logic, feel free to ask for clarification in the comments.

We assume that the curve is known in terms of some parametrization $x(\lambda), y(\lambda)$. Let $s$ be a measure of the arc length at any point along the curve, measured relative to some fixed starting location, and such that $ds/d\lambda > 0$. Let $r$ denote the radius of curvature of the curve at any point, and $\theta$ denote the angle it makes with the horizontal at any point. Let $N$ be the normal component of the force between the particle and the curve, and let $\mu$ be the coefficient of kinetic friction, such that the frictional force on this particle is $\mu N$. Denote the kinetic energy of the particle as $K$.

Assume initially that the particle is moving in the direction of increasing $s$. Along any infinitesimal segment of the curve $ds$, the change $dK$ in the particle's kinetic energy is $dK = - (\mu N + m g \sin \theta) ds$ according to the work-energy theorem. Now, the normal force exerted on the particle will be given by $$ N - m g \cos \theta = \frac{mv^2}{r} = \frac{2 K}{r} $$ and so $$ N = 2 K \frac{ x'y''- y'x''}{({x'}^2 + {y'}^2)^{3/2}} + m g \frac{dx}{ds}, $$ where we have used the standard equation for the curvature of a plane curve and the fact that $\cos \theta = dx/ds$. (Primes denote derivatives with respect to $\lambda$; all other derivatives will be written out explicitly.) Noting that $$ \frac{ds}{d\lambda} = \sqrt{{x'}^2 + {y'}^2} $$ we can therefore conclude that \begin{align*} dK &= - \left[ \mu \left( 2 K \frac{ x'y''- y'x''}{{x'}^2 + {y'}^2} \frac{d\lambda}{ds} + m g \frac{dx}{d\lambda} \frac{d\lambda}{ds} \right) + m g \frac{dy}{d\lambda} \frac{d\lambda}{ds} \right] ds \\ &= - \left[ 2 \mu K \frac{ x'y''- y'x''}{{x'}^2 + {y'}^2} + m g \left( \mu x' + y'\right) \right] d\lambda \end{align*} and so $$ \frac{dK}{d\lambda} = - \left[ 2 \mu K \frac{ x'y''- y'x''}{{x'}^2 + {y'}^2} + m g \left( \mu x' + y'\right) \right]. \tag{1} $$ The solution to this differential equation, subject to the initial condition $K(\lambda_0) = 0$ (assuming the particle is released from rest at parameter value $\lambda = \lambda_0$) will give the kinetic energy as a function of $\lambda$ along the curve.

Once we have this, we can infer the amount of time taken to move along the curve via the fact that $$ K = \frac{1}{2} m\left( \frac{ds}{dt} \right)^2 = \frac{1}{2} m \left( \frac{ds}{d\lambda} \right)^2 \left( \frac{d\lambda}{dt} \right)^2 = \frac{1}{2} m ({x'}^2 + {y'}^2) \left( \frac{d\lambda}{dt} \right)^2 $$ and so $$ \frac{dt}{d\lambda} = \sqrt{ \frac{m ({x'}^2 + {y'}^2)}{2 K}}. \tag{2} $$ We can also define $t$ such that $t(\lambda_0) = 0$.

Equations (1) and (2) form a set of coupled first-order ODEs in terms of the unknown functions $K(\lambda)$ and $t(\lambda)$, subject to the conditions that $K(\lambda_0) = t(\lambda_0) = 0$. By conservation of energy, there will be some other parameter value $\lambda_1 > \lambda_0$ at which $K(\lambda_1) = 0$ as well. This will correspond to the point at which the particle first comes instantaneously to rest (i.e., the motion from $\lambda_0$ to $\lambda_1$ corresponds to the purple arc labelled (1) in your sketch); and $t_1 = t(\lambda_1)$ gives the amount of time this takes. We can then "reset" the problem to find the amount of time taken for arc (2) in your diagram by solving the same set of ODEs with $K(\lambda_1) = 0$, $t(\lambda_1) = t_1$, and (important!) flipping the sign of the friction terms in our derivation, which effectively amounts to switching the sign of $\mu$ in (1).

This will yield a $\lambda_2$ and a $t_2$, and we can repeat this process ad nauseam. However, if we use our standard "introductory mechanics" assumptions about friction, we know that static friction can be greater than kinetic friction, and so eventually the particle will slide up to some point on the curve and static friction will hold it in place there. This will happen when $\mu_s > \tan \theta = (y'/x')$ at the point where the particle comes to rest, and this condition could (in principle) be checked at each of the turning points.

This whole process would be somewhere between horrendously awful and straight-up impossible to solve analytically. It could, however, be programmed into a computer to yield a numerical solution. But it would be somewhat complicated, and if you're not familiar with a computer language and/or libraries that can solve ODEs numerically, this is not the problem I would start with.

A few last notes:

  • Note that if we have a solution for $t(\lambda)$ and $K(\lambda)$ for a given value of $m$, we can obtain the solution for a different value of $m \to \alpha m$ by setting $K \to \alpha K$ and leaving $t$ untouched. This makes sense physically, and allows us to basically set $m$ equal to 1 with the understanding that we're "really" solving for $K/m$.
  • This process works for convex curves as well so long as we always have $N > 0$. Alternately, we could use this same process for a bead on a wire, for which a normal force could be exerted in either direction.
  • $\begingroup$ Can this be considered as a Newtonian solution whereas the other answer can be considered as a Lagrangian solution? $\endgroup$
    – SoySoy4444
    Aug 4, 2021 at 8:45
  • $\begingroup$ @SoySoy4444: Sort of? The modified Lagrange equations (including the dissipative terms which have to be put in by hand) are equivalent to Newton's Second Law, after all, so there's not a huge distinction. The main difference is really that this answer uses the energy as a configuration variable while the other one uses the coordinates. $\endgroup$ Aug 4, 2021 at 12:15
  • $\begingroup$ Using Python, I seem to have gotten a graph for kinetic energy (from the first differential equation) which starts at 0 when $\lambda = -1$ (I tried the parabola above), increases until around $\lambda = -0.3$, then becomes 0 at $\lambda = 0.417$, then becomes negative after that. Does it make sense that the kinetic energy becomes negative? imgur.com/I2b58rt $\endgroup$
    – SoySoy4444
    Aug 6, 2021 at 5:30
  • $\begingroup$ Should I just disregard the negative parts since it is part of arc 2 so I would need to flip the coefficient of friction? $\endgroup$
    – SoySoy4444
    Aug 6, 2021 at 6:15
  • $\begingroup$ @SoySoy4444: Yes, that's correct. Note that the equation for $dt/d\lambda$ requires that $K > 0$. $\endgroup$ Aug 6, 2021 at 15:36

The Equations of motion

Euler Lagrange \begin{align*} &\mathcal{L} =T-U\\ &\frac{d}{dt}\left(\frac{\partial \mathcal{L}}{\partial \dot{\mathbf{q}}}\right)^T- \left(\frac{\partial \mathcal{L}}{\partial \mathbf{q}}\right)^T =\left[\frac{\partial \mathbf{R}}{\partial \mathbf{q}}\right]^T\mathbf{f}_a\tag 1 \end{align*}


  • $T$ kinetic energy
  • $U$ potential energy - $\mathbf{q}$ generalized coordinates
  • $\mathbf{R}$ Position vector
  • $\mathbf{f}_a$ extern forces

The position vector to the mass point is: \begin{align*} &\mathbf R= \begin{bmatrix} x(s) \\ y(s) \\ \end{bmatrix} \end{align*} where s is the curve parameter

from here you obtain the kinetic and potential energy \begin{align*} &T=\frac{m}{2}\mathbf v\cdot \mathbf{v}\\ &\text{where}\\&\mathbf v=\dot{\mathbf R}=\frac{\partial\mathbf{R}}{\partial s}\,\dot{s}\\ &U=m\,g\,\mathbf R_y \end{align*}

External forces:

the friction force $F_\mu~$ is take in account as external force components $~\mathbf{f}_a$. \begin{align*} &\mathbf{f}_a=F_\mu\,\mathbf{t}\\ &\text{where $~\mathbf{t}~$ is the tangential vector} \\ &\mathbf{t}=\frac{1}{\sqrt{x^2(s)+y^2(s)}}\,\frac{\partial\mathbf{R}}{\partial s}\\ &\text{and}\\ &F_\mu=\text{signum}(-v_t)\,\mu\,|N| \end{align*}

  • $\mu~$ friction coefficient
  • $N~$ normal force
  • $v_t~$ tangential velocity $v_t=\mathbf v\cdot \mathbf t$

with equation (1) and $\mathbf{q}=s$ the equation of motion \begin{align*} &\ddot{s}+{\frac { \left( \left( {\frac {d}{ds}}x \left( s \right) \right) { \frac {d^{2}}{d{s}^{2}}}x \left( s \right) + \left( {\frac {d}{ds}}y \left( s \right) \right) {\frac {d^{2}}{d{s}^{2}}}y \left( s \right) \right) {{\dot{s}}}^{2}}{ \left( {\frac {d}{ds}}x \left( s \right) \right) ^{2}+ \left( {\frac {d}{ds}}y \left( s \right) \right) ^{2}}}\\&-{\frac {F\mu }{\sqrt { \left( {\frac {d}{ds}}x \left( s \right) \right) ^{2}+ \left( {\frac {d}{ds}}y \left( s \right) \right) ^{2}}m}}+{\frac { \left( {\frac {d}{ds}}y \left( s \right) \right) g}{ \left( {\frac {d}{ds}}x \left( s \right) \right) ^{2}+ \left( {\frac {d}{ds}}y \left( s \right) \right) ^{2}}} =0\tag 2 \end{align*}

The normal force $~N$

To obtain the normal force we open a gap $~\varkappa~$ towards the normal vector $\mathbf n$, thus the position vector is now \begin{align*} &\mathbf{R}\mapsto\mathbf{R}+\varkappa\,\mathbf{n}\\ &\text{where}\\ &\mathbf{n}=\begin{bmatrix} -\mathbf t_y \\ \mathbf t_x \\ \end{bmatrix}\\ &\text{the kinetic energy}\\ &T=\frac{m}{2}\mathbf{v}\cdot\mathbf{v}+N\,\varkappa\\ &\mathbf v=\frac{\partial \mathbf R}{\partial\mathbf q}\,\mathbf{\dot{q}}\qquad \text{with}~ \mathbf{q}=\begin{bmatrix} s \\ \varkappa \\ \end{bmatrix}\\ &\text{the potential energy}\\ &U=m\,g\,\mathbf{R}_y \end{align*}

with EL equation (1) and the holonomic constraint $~\varkappa=0~\Rightarrow~\dot{\varkappa}=0~,\ddot{\varkappa}=0 ~$ you obtain equation of motion (equation (2) and \begin{align*} &\ddot{\varkappa}=0=-{\frac { \left( - \left( {\frac {d}{ds}}x \left( s \right) \right) { \frac {d^{2}}{d{s}^{2}}}y \left( s \right) + \left( {\frac {d^{2}}{d{s }^{2}}}x \left( s \right) \right) {\frac {d}{ds}}y \left( s \right) \right) {{\dot{s}}}^{2}}{\sqrt { \left( {\frac {d}{ds}}x \left( s \right) \right) ^{2}+ \left( {\frac {d}{ds}}y \left( s \right) \right) ^{2}}}}+{\frac { \left( {\frac {d}{ds}}x \left( s \right) \right) g}{\sqrt { \left( {\frac {d}{ds}}x \left( s \right) \right) ^{2}+ \left( { \frac {d}{ds}}y \left( s \right) \right) ^{2}}}} -{\frac {N}{m}}\\ &N=-{\frac {m \left( - \left( {\frac {d}{ds}}x \left( s \right) \right) {\frac {d^{2}}{d{s}^{2}}}y \left( s \right) + \left( {\frac {d^{2}}{d{ s}^{2}}}x \left( s \right) \right) {\frac {d}{ds}}y \left( s \right) \right) {{\dot{s}}}^{2}}{\sqrt { \left( {\frac {d}{ds}}x \left( s \right) \right) ^{2}+ \left( {\frac {d}{ds}}y \left( s \right) \right) ^{2}}}}+ {\frac {mg{\frac {d}{ds}}x \left( s \right) }{\sqrt { \left( {\frac {d}{ds}}x \left( s \right) \right) ^{2}+ \left( {\frac {d}{ds}}y \left( s \right) \right) ^{2}}}} \end{align*}


\begin{align*} &\textbf{Parabola}\\ &x(s)=s\\ &y(s)=a\,s^2\\\\ &\ddot s+4\,{\frac {{a}^{2}s{{\dot s}}^{2}}{1+4\,{a}^{2}{s}^{2}}}-{\frac {F\mu }{\sqrt {1+4\,{a}^{2}{s}^{2}}m}}+2\,{\frac {a\,s\,g}{1+4\,{a}^{2}{s}^{2}}} =0\\ &v_t=\sqrt {1+4\,{a}^{2}{s}^{2}}{\dot{s}}\\ &N=2\,{\frac {a{{\dot{s}}}^{2}m}{\sqrt {1+4\,{a}^{2}{s}^{2}}}}+{\frac {m\,g} {\sqrt {1+4\,{a}^{2}{s}^{2}}}}\\\\ &\textbf{Circle}\\ &x(s)=r\,\cos(s/r)\\ &y(s)=r\,\sin(s/r)\\\\ &\ddot{s}+ \left( -F\mu +\cos \left( {\frac {s}{r}} \right) gm \right) {m}^{-1}\\ &v_t=\dot{s}\\ &N={\frac {{{\dot s}}^{2}m}{r}}-m\sin \left( {\frac {s}{r}} \right) g \end{align*}

Simulation: Parabola

$x(0)=1~,\dot x(0)=1$


enter image description here

enter image description here

enter image description here

  • $\begingroup$ Thank you for the answer. Could you please explain more on how you arrived at the expressions for the equation of motion, the normal force $N$ and the tangential velocity $v_t$? And in your graph, what does $\tau$ represent? $\endgroup$
    – SoySoy4444
    Aug 6, 2021 at 23:04
  • $\begingroup$ i will put more information, $\tau$ is the time t $\endgroup$
    – Eli
    Aug 7, 2021 at 6:36
  • $\begingroup$ Thank you very much! If you don't mind, could you please post the code that generated the gif of the ball moving? (I'm assuming you programmed it..?) $\endgroup$
    – SoySoy4444
    Aug 7, 2021 at 23:54
  • 1
    $\begingroup$ I use Maple , I can post you Maple code $\endgroup$
    – Eli
    Aug 8, 2021 at 6:29
  • $\begingroup$ Yes please, I'd appreciate the Maple code, thank you $\endgroup$
    – SoySoy4444
    Aug 8, 2021 at 6:34

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