Rolling down a slope I am having some trouble understanding rotational dynamics. If we have a cylinder that we give an initial velocity and rotation such that it satisfies the non slip condition ($ v_{cm} = \omega R$) and let it roll down a slope where its weight component down the slope equals the static friction up the slope then acceleration is equal to 0 and it will have a constant velocity of vcm. But because the friction provides a torque there is an angular acceleration so surely the cylinder is actually accelerating since $a = \alpha r$ even though the forces of friction and weight component down the slope are equal?
I know that when there is a constant velocity there is no friction but considering $m g \sin \theta - F_{static} = m a$ wouldn't that make $m g \sin \theta = 0$?
 A: The issue is that for a no slip condition, the static friction cannot be equal to the component of the weight parallel to the incline.
Setting up Newton's second law (for linear and rotational motion), we have$^*$
$$mg\sin\theta-f=ma$$
$$fR=I\alpha$$
Imposing the no slip condition $a=R\alpha$, we can determine that
$$f=\frac{mg\sin\theta}{1+mR^2/I}$$
So, the only way the magnitude of the static friction force $f$ can be equal to the component of weight down the incline $mg\sin\theta$, it must be that $mR^2/I=0$. This could be obtained when $I\to\infty$ so that we have an object that essentially cannot be rotated, and then in this case the object would in fact remain at rest, since we are imposing a no slip condition on an object that cannot rotate, and hence cannot translate either.

$^*$ Sign conventions have been chosen so that the linear acceleration and angular acceleration always have the same sign.
A: There is a condition for object to be rolling without slipping. As show above by BioPhysicst, the static firctional force $f$ equals to (adopts the inertial moment for a cylinder of mass $M$ and radius $R$ : $ I = \frac{1}{2} M R^2$,
$$
   f = \frac{1}{3} M g \sin \theta \tag{1}
$$
and this frictional force is arised from the normal froce of $M$ and the static frictional coefficient $\mu$.
$$
   f \leq \mu M g \cos \theta \tag{2}
$$
These two equations together give a conrtrain for the rolling without slipping. Under this contrain, the firtional force is at greatest equal to $1/3$ of the incline downward gravitation components.
 For the case that you assume $ f = M g \sin \theta$, It is not posiible to be a rolling motion.
