# Point particle sliding down an arbitrary curve (no departure)

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.

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

$$\cos(\theta)=\cos\Big(\tan^{-1}\Big(\frac{dy}{dx}\Big)\Big)=\frac{\frac{dy}{dx}}{\sqrt{\Big(\frac{dy}{dx}\Big)^2+1}}$$

and differentiation rules such as

$$\frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}$$

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?

• 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. Feb 25, 2021 at 17:11

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

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

$$U=-m\,g\,f(x)$$

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 $$~\mathbf F_e$$

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

$$\Rightarrow$$

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.

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

Example:

$$f(x)=a\,x^2$$

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

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

Simulation with friction

The Euler Langrage equation:

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