Accelerated pure rolling and friction I know that if a body is pure rolling with no acceleration then friction acting on it is zero.
But what about if the CM has an acceleration? Then can we say that friction is present? 
Suppose that there is another force acting on the rolling body. Can there be a case such that there is no external force, but on a rough surface a body is pure rolling. Is it possible for the body to have A(cm) without another force? And can friction be non-zero in such a case? (I think that both these cases if they occur at all then occur together, but are they possible?)
 A: Consider the diagram below of a ball on a horizontal surface:

Newton's Laws tell us that if no net force acts on the ball it will remain in a constant state of motion ($v=0$ or $v=\text{constant}$). Consequently, if no net torque acts on the ball its state of rotation will also remain constant ($\omega=0$ or $\omega=\text{constant}$).
Where friction does play a part, it is usually somewhat simplistically modelled as $F_f=\mu F_n$ with $\mu$ some friction coefficient and $F_n$ the normal force (here simply $mg$). But in the case of a ball rolling with constant $v$ and constant $\omega$ and no external force (say $F$) acting on it, this would lead to deceleration according to :
$$ma=-F_f$$
But then $F_f$ would also provide torque leading to angular acceleration according to:
$$I\dot{\omega}=F_f R,$$
with $I$ the inertial moment, $R$ the radius and $\dot{\omega}=\frac{d\omega}{dt}$ the angular acceleration.
This would be the case where you launch a ball with initial speed $v$ but no angular momentum ($\omega=0$) onto a surface that provides much friction: the ball would start spinning ($\dot{\omega} > 0$ but also start decelerating ($a < 0$) and translational kinetic energy would be converted to rotational kinetic energy. If the surface can provide enough friction that process would continue until $v=\omega R$: rolling without slipping.
To keep the ball moving without any deceleration we would have to supply an external force, so that:
$$F_f=F,$$
and $a=0$.
Critical coefficient of friction $\mu_c$:
Assume a ball with $v=0, \omega=0$ at $t=0$. We now apply a horizontal force $F$, so that $a>0$.
$$F_f=\mu F_n=\mu mg$$
$$I\dot{\omega}=F_fR=\mu mgR$$
After integration we get:
$$\omega=\mu\frac{mgR}{I}t$$
In that same amount of time the ball has also acquired translational speed:
$$F-F_f=ma$$
After integration we get:
$$v=\frac{F-\mu mg}{m}t$$
Without slipping we have:
$$v=\omega R$$
$$\frac{F-\mu mg}{m}t=\mu\frac{mgR^2}{I}t$$
Reworking and isolating $\mu$ we get:
$${\mu_c=\frac{FI}{mg(I+mR^2)}}$$
This is the minimum value for the coefficient of friction in order to achieve rolling without slipping when a horizontal force $F$ is applied.
