# Position of a particle sliding down an arbitrary curve as a function of time

Given a curve in a frictionless environment with parameterization $$\displaystyle \mathbf{r}(\theta)=x(\theta)\hat{\mathbf{i}}+y(\theta)\hat{\mathbf{j}}$$ for $$\theta\in[0,\theta_f]$$, how can I find the position of a particle, which starts at $$\mathbf{r}(0)$$ and which slides down $$\mathbf{r}$$ under only the force of gravity, as a function of time? Furthermore, what if the particle has an initial velocity $$v_i$$ in the direction of travel?

I attempted the first part, but as I am not well-versed in physics I was unsure how to do the second, and I am not even sure if my work for the first part is right.

I did some hand-waving and said $$\displaystyle v=\sqrt{2gy(\theta)}$$ from the conversion of PE to KE, and from the curve parameterization we have $$\displaystyle v=\sqrt{{[x'(\theta)]}^2+{[y'(\theta)]}^2}\,\frac{d\theta}{dt}$$. So simply solve $$\displaystyle \frac{d\theta}{dt}=\frac{\sqrt{2gy(\theta)}}{\sqrt{{[x'(\theta)]}^2+{[y'(\theta)]}^2}}$$ for $$\theta$$ in terms of $$t$$ and substitute this back into the parameterization of $$\mathbf{r}$$.

Is there any better way of doing this? For one, this method rarely results in closed-form solutions (edit: which is not a requirement, but would be nice if other methods did have closed-form solutions), for another, I don't even know if it's right. I was then unsure how to do the second part because it would change the KE-PE equation and as I was already hand-waving I wasn't sure if I would need to use $$\displaystyle \Delta v$$ and $$\Delta y$$ or what.

• Why do you expect elementary, closed-form solutions for arbitrary curves? The simple pendulum is equivalent to motion along a circular curve (it doesn't get much simpler), and the general solution is not very pleasant. – J. Murray Jan 6 at 21:35
• I don't expect closed-form solutions for arbitrary curves, but for at least for some simple curves like a circular curve, for which my method gave me a nasty elliptic integral. For context I am doing this for an animation, unfortunately Geogebra does not do physics simulations so I have to explicitly input formulas for the position (not necessarily closed-form) – legendariers Jan 6 at 21:42
• As I said, the solution to that problem is very nasty. Even though the curve is simple, the resulting force is an extremely non-linear function of $\theta$. – J. Murray Jan 6 at 21:49
• I may have had a lapse in thought here, I don't suppose that a closed-form solution exists for one equation but does not for another which represents the same problem. Oops. I still am unsure how to proceed with the second part, as I said I arrived at my first equation through some hand-waving. – legendariers Jan 6 at 21:50
• Note that a circular curve is the same problem as a large amplitude pendulum, which has no closed form solution. – Bill Watts Jan 16 at 23:07

Since both Cartesian coordinates are parameterized by $$\theta$$, one can write the Lagrangian of the system, with a single generalized coordinate $$\theta$$.
$$L=\frac m2(x'^2+y'^2)\dot\theta^2 -mgy$$
Use the Euler-Lagrange equations to find a solution. Be careful about the derivatives, as everything depends on $$\theta$$!.