Consider a point-particle $p$ on an arbitrary curve defined by the function $f(x)$. It's constrained to always live within the curve. I.e; it always has coordinates $(x, f(x))$ and hence cannot "fly off". It experiences gravity and friction.

point particle sliding down a curve

I'd be surprised if this hasn't been asked yet, but I'm baffled that not only can I not seem to derive a differential equation that describes the system, but I cannot find an answer that includes both friction and the constraint that it's bounded to the curve.

This question describes a similar scenario, however that system is frictionless and (as far as I can tell) doesn't constrain the particle to the curve.

My attempt

First things first, the kinematics is highly dependent on the instantaneous angle of the curve $\theta$, which can be defined through the identity $\tan\theta=\frac{dy}{dx}$, relating it to the slope. Before that's used, it's relatively trivial to find the net acceleration in each the $x$ and $y$ directions.

$$\Sigma_xF=|F_g|\sin\theta-|F_f|\cos\theta$$ $$\Sigma_xF=mg\sin\theta-\mu mg\cos\theta$$ $$\implies a_x=g(\sin\theta-\mu \cos\theta)$$

$$\Sigma_yF=|F_n|\cos\theta-|F_g|$$ $$\Sigma_yF=mg\cos^2\theta-mg$$ $$\implies a_y=-mg\sin^2\theta$$

(Where $F_g$ is the force of gravity, $g$ is the acceleration of gravity, $F_f$ is the force of friction, $\mu$ is the coefficient of friction, $m$ is the mass, and $F_n$ is the normal force.)

I assume the next steps are to substitute trigonometric identities such as


and differentiation rules such as


to finally find a time-dependent equation for the motion of $p$ along the $x$-axis (because it's vertical coordinate is dependent on it's $x$-coordinate, by the curve-bounding constraint). However it's at this point my brain shuts off and I can't remember basic calculus and kinematics; especially with the added difficulty of not allowing the particle to fly off the curve. Beyond the scope of my question, is this still solvable if $f$ is also dependent on time?

  • 1
    $\begingroup$ Two comments: (1) This would be more straightforward if the curve was given parametrically: $(x(s), y(s))$, where $s$ is the arc length from some point. (2) Even then, the resulting ODEs will almost certainly not be solvable, since the frictional force will not be an analytic function of $s, \dot{s}$, etc.. The normal force can point in or out, and the friction opposes of $\dot{s}$, so the frictional force along the arc will be something like $- \mu \text{sgn}(\dot{s}) |N|$. Having signum & absolute value functions in your ODEs is not often amenable to closed-form solutions. $\endgroup$ Commented Feb 25, 2021 at 17:11

1 Answer 1


Start with the position vector to the particle

$$\mathbf R=\left[ \begin {array}{c} x\\ f \left( x \right) \end {array} \right] $$

from here

the velocity

$$\mathbf v=\left[ \begin {array}{c} {\dot x}\\ \left( {\frac {d}{dx}}f \left( x \right) \right) {\dot x}\end {array} \right] $$

the kinetic energy

$$T=\frac m2 (v_x^2+v_y^2)$$

the potential energy


and with Euler Langrage the equation of motion

$$\ddot x=-{\frac { \left( {\frac {d}{dx}}f \left( x \right) \right) \left( {{ \dot x}}^{2}{\frac {d^{2}}{d{x}^{2}}}f \left( x \right) +g \right) }{1 + \left( {\frac {d}{dx}}f \left( x \right) \right) ^{2}}} $$

with friction:

the friction force $~F_\mu$ interact toward the tangential vector which is

$${\mathbf{t}}_g=\frac{\partial \mathbf R}{\partial x}=\left[ \begin {array}{c} 1\\{\frac {d}{dx}}f \left( x \right) \end {array} \right] $$

you can obtain the equation of motion with EL and external force $~\mathbf F_e$

$$ \mathbf F_e= F_\mu\,\frac{\mathbf t_g}{|\mathbf t_g|}$$


the EOM

$$\ddot x={\frac {F_{{\mu}}}{\sqrt {1+ \left( {\frac {d}{dx}}f \left( x \right) \right) ^{2}}m}}-{\frac { \left( {\frac {d}{dx}}f \left( x \right) \right) \left( {{\dot x}}^{2}{\frac {d^{2}}{d{x}^{2}}}f \left( x \right) +g \right) }{1+ \left( {\frac {d}{dx}}f \left( x \right) \right) ^{2}}} $$

with $~F_\mu=-\mu\,|N|\,\text{sgn}(\mathbf v\,\cdot \mathbf t_g)~$ and $~N~$ the normal force. the "sgn" function causes that the friction force act always to the opposite direction of the tangential velocity.

$$N={\frac { \left( {{\dot x}}^{2}{\frac {d^{2}}{d{x}^{2}}}f \left( x \right) +g \right) m}{\sqrt {1+ \left( {\frac {d}{dx}}f \left( x \right) \right) ^{2}}}} $$



$$\ddot x={\frac {F_{{\mu}}}{\sqrt {1+4\,{a}^{2}{x}^{2}}m}}-2\,{\frac {a\,x \left( 2\,{{\dot x}}^{2}a+g \right) }{1+4\,{a}^{2}{x}^{2}}} $$

$$N={\frac { \left( 2\,{{\dot x}}^{2}a+g \right) m}{\sqrt {1+4\,{a}^{2}{x} ^{2}}}} $$

Simulation with friction enter image description here

The Euler Langrage equation:

$$\frac{d}{dt}\left(\frac{\partial \mathcal L}{\partial \dot x}\right)- \left(\frac{\partial \mathcal L}{\partial x}\right)=\left(\frac{\partial \mathbf R}{\partial x}\right)^T\,\mathbf F_e$$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.