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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.

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    $\begingroup$ 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. $\endgroup$ – J. Murray Jan 6 at 21:35
  • $\begingroup$ 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) $\endgroup$ – legendariers Jan 6 at 21:42
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    $\begingroup$ 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 $\theta$. $\endgroup$ – J. Murray Jan 6 at 21:49
  • $\begingroup$ 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. $\endgroup$ – legendariers Jan 6 at 21:50
  • $\begingroup$ Note that a circular curve is the same problem as a large amplitude pendulum, which has no closed form solution. $\endgroup$ – Bill Watts Jan 16 at 23:07
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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 $\theta$!.

Note that the Euler-Lagrange equation is a second order differential equation, so solving it results in 2 arbitrary constants, which can be associated with the initial velocity and position of the particle. Your second question can be solved by correctly identifying the initial velocity after solving the differential equation.

I did not solve it by hand myself, since it's a bit messy. Even so, if a closed form solution does exist, this is certainly a possible way of obtaining it.

The equation that you used for the velocity is that of a free falling object. Certainly, in your problem, the object is not free falling, otherwise it wouldn't have a parameterized x-coordinate. Basically, you're constraining the object to move on a particular parameterized trajectory, and this constraint has to come from somewhere (a force other than gravity).

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