# Background: Equation of Motion

Okay. First I want to see if my "Newtonian Mechanics" lens of the problem is correct.

Let the particle's path be given by $$\vec{r}(t) = (x(t), y(t))$$ and just as in figure 1, WLOG, assume that the path lies in the first quadrant; according to figure 1, we identify the following forces on the point mass:

$$\vec{w} = (0, mg)$$

$$\vec{N} = (N_x, -N_y)$$

$$\text{where m, g, N_x, N_y \in \mathbb{R^{+}} := [0, \infty)}$$

Thus, in Cartesian coordinates $$\vec{F} = m\vec{a} = \vec{w} + \vec{N} \implies (\ddot{x}, \ddot{y}) = (N_x, mg - N_y)$$

Let $$v^2 := \dot{x}^2 + \dot{y}^2$$; then, utilizing the fact that the tangential $$\mathbf{\hat{T}}$$ and normal $$\mathbf{\hat{N}}$$ unit vectors to the curve are

$$\mathbf{\hat{T}} = \frac{1}{v} (\dot{x}, \dot{y})$$

$$\mathbf{\hat{N}} = \frac{1}{v} (-\dot{y}, \dot{x})$$

Let $$N^2 := {N_x}^2 + {N_y}^2$$; we project the force vectors onto the unit tangent and unit normal via

$$\vec{w} = (\vec{w}\cdot\mathbf{\hat{T}})\mathbf{\hat{T}} + (\vec{w}\cdot\mathbf{\hat{N}})\mathbf{\hat{N}}$$

$$\vec{N} = 0 \mathbf{\hat{T}} - N\mathbf{\hat{N}}$$

since the particle is constrained to move along the curve, we apply the normal condition

$$-N\mathbf{\hat{N}} = (\vec{w}\cdot\mathbf{\hat{N}})\mathbf{\hat{N}}$$

it follows that

$$\vec{w} = \frac{mg}{v} (\dot{y} \mathbf{\hat{T}} + \dot{x} \mathbf{\hat{N}})\tag{1}$$

$$\vec{N} = 0 \mathbf{\hat{T}} - \frac{mg}{v}\dot{x}\mathbf{\hat{N}}\tag{2}$$

Using the fact that the acceleration of the particle is given by

$$\ddot{\vec{r}} = \dot{v}\mathbf{\hat{T}} + \frac{\ddot{x}\ddot{y} - \dot{y}\ddot{x}}{v} \mathbf{\hat{N}}$$

we find that by the normal condition, the normal component is zero so we are left with only tangential acceleration and hence:

$$m\dot{v} = \frac{mg\dot{y}}{v}\tag{3}$$

where by the ODE given by equation 3, using separation of variables, we get:

$$v = \sqrt{2gy}$$

Let $$s := ||{\vec{r}}||$$, and suppose $$y$$ is paremetrised by $$x$$, so that now $$\vec{r} = (x, y(x))$$. Then, taking into account that $$v = ds/dt$$ where we denote $$T$$ as the time it takes for the particle to reach point $$x(T) = p$$ from the origin:

$$T = \int_{0}^{T} dt = \int_{0}^{p} \sqrt{\frac{1 + y'(x)^2}{2gy(x)}} dx\tag{4}$$

# Calculus of Variations Stuff

This process is well documented so I will be omitting a lot of steps here. Using the integrand from equation 4, define:

$$f(y, y', x) = \sqrt{\frac{1 + y'(x)^2}{2gy(x)}}\tag{5}$$

and by the Euler-Lagrange equation, we end up with the following differential equation:

$$y'(x) = \sqrt{\frac{c - y(x)}{y(x)}}$$

Here, $$c$$ is a constant where, according to how we aligned our coordinate system and how we assumed the particle's path (i.e. $$x > 0$$ and $$y > 0$$), it necessarily follows that $$c > 0$$.

Using the paremetrization $$y = c \sin^2\big(\tilde{t}\big)$$ where it necessarily follows that $$\tilde{t} \in [0, \pi / 2)$$, we obtain the following solutions:

$$x(\tilde{t}) = c(t - \frac{1}{2} \sin(2\tilde{t}))$$

$$y(\tilde{t}) = c\bigg(\frac{1}{2} - \frac{1}{2} \cos(2\tilde{t})\bigg)$$

Now, in order to check the solution, let's express the forces given by eq 1 and eq 2 in cartesian coordinates and write out the corresponding equation of motion; working out all the details yields:

$$\vec{F} = \vec{N} + \vec{w}$$

$$\Rightarrow m(\ddot{x}, \ddot{y}) = \frac{mg\dot{y}}{v^2} \big(\dot{x} , \dot{y})$$

Perhaps this was where I made a mistake: I assumed that we can treat $$\tilde{t}$$ as $$t$$, so we can check if the solution given by (5) satisfies eq (6). However, if we assume that, then the RHS of (6) ends up as:

$$mg\bigg(\frac{1}{2} \sin(2t), \frac{1}{2} + \frac{1}{2}\cos(2t)\bigg)$$

whereas the LHS of (6) yields:

$$m(2c \sin(2t), 2c \cos(2t))$$

Looking at only the first entries of (7) and (8), it follows that

$$c = \frac{g}{4}$$

HOWEVER, if we assume this value of $$c$$, then, using the second entries of (7) and (8), it necessarily follows that:

$$\boxed{\frac{g}{2}\cos(2t) = \frac{g}{2} + \frac{g}{2} \cos(2t)}$$

Now, if you ask me, this is a contradiction. However, if we are generous with our estimates and consider the y coordinate acceleration yielding a discrepancy of $$\frac{g}{2} \approx 4.9 \ \frac{m}{s^2}$$ negligible, then I guess we don't have a contradiction..?