Simple pendulum using plane polar coordinates: where did I miss a minus sign? Suppose we want to write the equation of a simple pendulum using plane polar $(r,\theta)$ coordinates with the point of suspension as the origin and with $\theta$ increasing anticlockwise from the equilibrium position. As usual $\hat{r}$ is taken to be radially outward from the point of suspension and $\hat{\theta}$ along the direction of increasing $\theta$.
Let us use the formula $$\vec{\tau}={\vec r}\times{\vec F}=I{\vec \alpha}$$ where ${\vec F}=-mg\sin\theta\hat{\theta}$ and ${\vec r}=\ell\hat{r}$, $\ell$ being the length of the pendulum. The negative sign in the expression for force tells that the force is directed opposite to the direction of $\hat{\theta}$ (i.e. the direction of increasing $\theta$).
Therefore, torque expression simplifies to $$\vec\tau={\vec r }\times{\vec F}=\ell\hat{r}\times{-mg\sin\theta}\hat{\theta}=-mg\ell\sin\theta{\hat z}$$ where ${\hat z}$ is a unit vector point out of the plane of motion and perpendicular to it.
Now, since the linear accleration $\vec a$ and the angular acceleration $\vec \alpha$ are related as ${\vec a}=\vec{\alpha}\times{\vec r}$, and $\vec a=-a\hat{\theta}$, we have, $$-a\hat{\theta}=\ell\vec{\alpha}\times\hat{r}$$ which implies that $\vec\alpha=-\alpha\hat{z}=-\frac{d^2\theta}{dt^2}\hat{z}$. Therefore, we get, $$\vec\tau=I\vec\alpha\Rightarrow I\frac{d^2\theta}{dt^2}=mg\ell\sin\theta.$$
Where did I miss the minus sign? I think that the minus sign for the restoring force should have arose automatically.
 A: 
Where did I miss the minus sign? I think that the minus sign for the restoring force should have arose automatically.

You defined relationship between linear (tangential) and rotational acceleration as
$$-a \hat \theta = \ell \vec \alpha \times \hat r$$
I am not sure where you got the minus sign from. If you lose that minus sign, you will get the expected result, i.e. minus sign with the restoring force. I show below why there should be positive sign next to the (tangential) acceleration.

Note that for circular motion with constant radius $R$, linear velocity $\vec v$ and distance $\vec r$ vectors are always perpendicular
$$x^2 + y^2 = R^2, \qquad x v_x + y v_y = 0, \qquad \boxed{\vec v \cdot \vec r = 0}$$
Here is the definition of the angular velocity vector from the "Angular velocity" Wiki article

In three-dimensional space, we again have the position vector $\textbf{r}$ of a moving particle. Here, orbital angular velocity is a pseudovector whose magnitude is the rate at which $\textbf{r}$ sweeps out angle, and whose direction is perpendicular to the instantaneous plane in which $\textbf{r}$ sweeps out angle (i.e. the plane spanned by $\textbf{r}$ and $\textbf{v}$). However, as there are two directions perpendicular to any plane, an additional condition is necessary to uniquely specify the direction of the angular velocity; conventionally, the right-hand rule is used.

(Italics mine)
The angular velocity vector is then
$$\vec \omega = \frac{d\phi}{dt} \hat\omega = \frac{v \sin\theta}{r} \hat\omega = \frac{v}{r} \hat\omega$$
where $\theta$ is the angle between $\vec r$ and $\vec v$ which is $90^\circ$ in your example as discussed above, $\hat\omega$ is the unit vector perpendicular to the plane spanned by $\vec r$ and $\vec v$ and with the right-hand rule satisfied
$$\vec \omega = \frac{\vec r \times \vec v}{|\vec r|^2}$$
For positive rotation (counter-clockwise), if position is $\vec r = |\vec r| \angle\phi$ then linear velocity is $\vec v = |\vec v| \angle\phi+\frac{\pi}{2}$ and the above expression becomes
$$\vec \omega = 0 \hat\imath + 0 \hat\jmath + \frac{|\vec v|}{|\vec r|} \hat k$$
From here it is obvious that for positive rotation (counter-clockwise), the angular velocity vector points in the direction of positive $\hat k$ axis. The angular acceleration in that case is
$$\vec \alpha = \frac{d \vec \omega}{dt} = \frac{1}{|\vec r|} \frac{d |\vec v|}{dt} \hat k = \frac{a_\parallel}{|\vec r|} \hat k$$
where $a_\parallel$ is tangential acceleration. Rotational acceleration is positive for positive linear (tangential) acceleration, hence
$$\boxed{+a \hat \theta = \ell \vec \alpha \times \hat r}$$
The above cross product can be expanded as
$$
\ell \vec\alpha \times \hat r =
\left|
\begin{array}{ccc}
\hat\imath & \hat\jmath & \hat k \\
0 & 0 & \alpha_z \\
\cos\theta & \sin\theta & 0
\end{array}
\right|
=
\ell (-\alpha_z \sin\theta \hat\imath + \alpha_z \cos\theta \hat\jmath + 0 \hat k) = +\ell \alpha_z \hat\theta = +a_\parallel \hat\theta
$$
where $\hat\theta = -\sin\theta \hat\imath + \cos\theta \hat\jmath$ for positive rotation (counter-clockwise) direction.
A: 
with
$$\vec r=
  l\,\begin{bmatrix}
   \sin(\theta)\\
   -\cos(\theta) \\
   0 \\
 \end{bmatrix}\\
\vec F=
  \begin{bmatrix}
   0\\
   -m\,g \\
   0 \\
 \end{bmatrix}\\
\vec \tau_A=\vec r\times\vec F=m\,g\,l\sin(\theta)\,(-\vec e_z)$$
the equation of motion is now
$$m\,l^2\ddot\theta+m\,g\,l\sin(\theta)=0$$


from the figure is the torque
$$\vec \tau=F\,a\,\vec e_z$$
with cross product
$$\vec\tau=\vec r\times\vec F=\begin{bmatrix}
   a\\
   0 \\
   0 \\
 \end{bmatrix}\times \begin{bmatrix}
   0\\
   F \\
   0 \\
 \end{bmatrix}=\begin{bmatrix}
   0\\
   0 \\
   F\,a \\
 \end{bmatrix}$$
