Cylinder on an accelerated rough plane 
The following figure shows a cylinder resting on a rough horizontal
  rug that is pulled out from under it with acceleration $a$
  perpendicular to the axis of the cylinder. What is the motion of the
  cylinder, assuming it does not slip?


The only horizontal force on the cylinder is that of friction at $P$. Therefore let us take the moments about $P$. The forces of gravity and reaction at the surface pass through the point $P$, as does also the friction force. Also, we know* that:
$$\dot{\mathbf{L}}=\mathbf{M}-m\mathbf{v}_P \times \mathbf{v}_O$$
Where $\mathbf{L}$ - the angular momentum of the system, $\mathbf{M}$ - the net external torque on the system, $m$ - the mass of the system, $\mathbf{v}_P$ - the velocity of point $P$ (as viewed from an inertial system - say, ground) and $\mathbf{v}_O$ - the velocity of the center of mass. 
Because $\mathbf{v}_O\parallel \mathbf{v}_P$ we have $\dot{\mathbf{L}}=\mathbf{M}$. So the net torque about $P$ is $0$. Therefore:
$$\dot{\mathbf{L}}=0 \Rightarrow I_P \omega = \text{const} \Rightarrow \omega = \text{const}$$
which is wrong. Only by taking the moments about $O$ we obtain $\omega \propto a$.
Why my approach was wrong in the first place?

*Let $\mathbf{v}$ and $\mathbf{p}=m \mathbb{v}$ be the velocity and the momentum of a particle with respect to a stationary inertial system $S$, and $\mathbf{r}$ - the radius-vector of the particle with respect to the moving point $P$. The motion of $P$ can be either uniform or nonuniform (accelerating). Suppose $\mathbf{v}_P$ is the velocity of $P$. The angular momentum is $\mathbf{L}=\mathbf{r} \times \mathbf{p}$. Taking the derivative we get $\dot{\mathbf{L}}=\dot{\mathbf{r}} \times \mathbf{p} + \mathbf{r} \times \dot{\mathbf{p}}$. Because $\dot{\mathbf{r}}=\mathbf{v}-\mathbf{v}_P$ we get $\dot{\mathbf{L}}=(\mathbf{v}-\mathbf{v}_P) \times \mathbf{p} + \mathbf{r} \times \dot{\mathbf{p}}$. However $\dot{\mathbf{p}}=\mathbf{F}$ and thus we obtain:
$$\dot{\mathbf{L}}=\mathbf{M}-\mathbf{v}_P \times \mathbf{p}$$
However we can express the momentum $\mathbf{p}$ of the system in terms of $\mathbf{v}_O$ - the velocity of the center of mass: $\mathbf{p}=m\mathbf{v}_O$. Thus:
$$\dot{\mathbf{L}}=\mathbf{M}-m\mathbf{v}_P \times \mathbf{v}_O$$
 A: You may have over-complicated the problem. Take the momentum definition at the center of mass, differentiate it and equate it with the applied loading to find the equations of motion.
$$\require{cancel} \begin{align} 
  \boldsymbol{p} & = m \boldsymbol{v}_O & \boldsymbol{L}_O & = \mathcal{I}_O \boldsymbol{\omega}  \\
\boldsymbol{F}_{fric} & =\dot{\boldsymbol{p}} = m \dot{\boldsymbol{v}}_O & (\boldsymbol{r}_P-\boldsymbol{r}_O) \times \boldsymbol{F}_{fric}& = \dot{\boldsymbol{L}}_O = \mathcal{I}_O \dot{\boldsymbol{\omega}}+\cancel{\boldsymbol{\omega} \times \mathcal{I}_O \boldsymbol{\omega}} \\
F_{fric} & = m \dot{v} & R\; F_{fric} & = \mathtt{I}_O \dot{\omega}
\end{align} $$
Now for the no-slip condition
$$ \hat{\boldsymbol{i}} \cdot (\boldsymbol{v}_{rug} - \boldsymbol{v}_P) =  0$$
$$ \hat{\boldsymbol{i}} \cdot (\boldsymbol{v}_{rug} - \boldsymbol{v}_O+\boldsymbol{\omega}\times(\boldsymbol{r}_O-\boldsymbol{r}_P)) =  0$$
$$\begin{align} \omega &= \frac{v_{rug}-v}{R} & \dot{\omega} & = \frac{a-\dot{v}}{R}  \end{align}$$
Now you have three equations with three unknowns $F_{fric}$, $\dot{v}$ and $\dot{\omega}$:
$$\begin{align} F_{fric} & = m \dot{v} & R\;F_{fric} &= \mathtt{I}_O \dot{\omega} & \dot{\omega} & = \frac{a-\dot{v}}{R} \end{align} $$
