Point mass sliding down quarter circle Consider the problem of determining the equations of motion in 2D for a point mass sliding down the quarter unit circle lying in the 3rd quadrant.
That is, at $t=0$, it is at position $(-1,0)$, and we wish to determine its position $x(t)$ for $t>0$, as it slides to $(0,-1)$.
Let $\alpha(t)$ denote the angle between the tangent line to the quarter circle at position $x(t)$, and the x-axis, so that essentially $\alpha(0) = \frac{\pi}{2}$.
Then $x(t) = -(\sin \alpha(t), \cos \alpha(t))$.
Moreover, the instantaneous force acting on the particle is
$F = m \ddot{x} = mg \sin \alpha (\cos \alpha, -\sin \alpha)$.
Edit: The calculations from here onwards are not correct, as pointed out in the answers, although coincidentally the end result and the plot are. See my answer below.
Differentiating $x$ twice and equating it to $\frac{1}{m} F$ gives
$\begin{pmatrix}\sin \alpha & -\cos \alpha \\ \cos \alpha & \sin \alpha\end{pmatrix}
%
\begin{pmatrix} \dot{\alpha} \\ \ddot{\alpha} \end{pmatrix}
=
g \begin{pmatrix} \sin \alpha \cos \alpha \\ -\sin^2 \alpha \end{pmatrix}
$
so that by assuming $\alpha$ is never exactly $\frac{\pi}{2}$, hence being able to invert the matrix, after some cancellations we get
$\begin{pmatrix} \dot{\alpha} \\ \ddot{\alpha} \end{pmatrix}
%
= \begin{pmatrix} 0 \\ -g \sin \alpha \end{pmatrix}$ .
Now solving $\ddot{\alpha} = -g \sin \alpha$ alone using Euler integration with initial conditions $\alpha(0) = \frac{\pi}{2}$ and $\dot{\alpha}(0) = 0$, and then plotting $x(t)$, produces something that looks reasonable: the spacings increase as the particle slides down and gains speed.

However, $\dot{\alpha} = 0$ forces $\alpha$ to be constant, and $\ddot{\alpha}$ to be zero, constraining away the expected trajectories.
Why does this problem occur with this solution?
 A: 
$$F=m\ddot{x}=g\sin \alpha (\cos \alpha, -\sin \alpha)$$

You are not simulating a particle sliding down a circle with this equation.
The net force on the particle will be the sum of gravitational and Normal force applied by the surface.
$$\vec{F}=\vec{F_g}+\vec{N}$$
$$\begin{equation}
\vec{F}=mg(0,-1)+N(\sin \alpha, \cos \alpha)=m\ddot{x}
\end{equation}$$
Morever, since $x = (-\sin \alpha(t), \cos \alpha(t))$,
$$\dot{x} = -\dot{\alpha}(t) (\cos \alpha, \sin \alpha)$$
and
$$\ddot{x}=-\ddot{\alpha}(\cos \alpha, \sin \alpha)-\dot{\alpha}^2(-\sin \alpha, \cos \alpha)$$
The next step should be to eliminate $N$ from the equation by subsitituting this $\ddot{x}$ in the net force equation. The final result will be some nasty looking pair of non linear ODE's in $\alpha(t)$ which you will have to solve.
A: Alright, here's how to do this properly:
One thing to notice is that this system is completely equivalent to the pendulum, with the normal force replacing the role of tension.
We can model the problem by considering the point mass to be at a fixed position $(0,-r)$ with respect to moving coordinate axes rotating around the origin of the standard fixed axes.
Following Classical Dynamics of Particles and Systems by Thornton and Marion, Netwon's equation for a non-inertial reference frame becomes:
$\boldsymbol{F} = m \boldsymbol{\ddot{R}_f} + m \boldsymbol{a_r} + m \boldsymbol{\dot{\omega} \times r} + m \boldsymbol{\omega \times (\omega \times r)} + 2 m \boldsymbol{\omega \times v_r}$
where $\boldsymbol{R_f}$ is the position of the moving origin,
$\boldsymbol{r}$ is the position with respect to the moving axes,
$\boldsymbol{a_r}$ and $\boldsymbol{v_r}$ are the acceleration and velocity relative to the moving axes,
and $\boldsymbol{\omega}$ is the angular velocity of the rotating axes.
For this setup, $m \boldsymbol{\ddot{R}_f}$, $m \boldsymbol{a_r}$, and the Coriolis term $2 m \boldsymbol{\omega \times v_r}$ cancel out, as the origin of the rotating axis is fixed, and the particle is also fixed with respect to the rotating axes.
The total external force acting on the particle in the fixed frame is $\boldsymbol{F} = F_g + N$, the sum of the gravitational force and the normal force pointing to the center of rotation, so
$F_g + N = m \boldsymbol{\dot{\omega} \times r} + m \boldsymbol{\omega \times (\omega \times r)}$.
As the particle maintains a circular trajectory, the radial components must cancel out, so that we are left with
$mg \sin \theta = m \dot{\omega} r \ $, that is,  $\ \ddot{\theta} = \frac{g}{r} \sin \theta$.
