# 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 $$m g \sin \theta = 0$$?

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.

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.