How to predict the motion of a cylinder on top of inclined plane in scenario with different coefficients of friction? Consider an inclined plane with angle $x$. Now put a cylinder on top of it and consider three different cases.


*

*A the limiting frictional force $f$ is greater than $mg\sin(x)$ where $m$ is the mass of cylinder.

*B the limiting friction $f$ is exactly equal to $mg\sin(x)$

*C it is less than $mg\sin(x)$.


What will happen in each case?
In A and B I suppose $f$ will provide torque and then friction will change direction to oppose the angular velocity maybe.
In C it will start moving and then provide torque but the friction should vanish although it stays there and apparently it may or may not cause slipping. What is meaning of slipping?
I am not sure about these cases. Can someone explain. I just need physical intuitive explanation.
 A: The no slipping condition is that the angular acceleration of the cylinder $\alpha$ and the translational acceleration of the centre of mass of the cylinder $a_{\rm c}$ are related as follows $a_{\rm c} = r \alpha$ where $r$ is the radius of the cylinder.  
You have to decide whether or not this condition can be satisfied for differ values of the limiting frictional force.
I assume that by limiting frictional force this means the maximum static frictional force and that once slipping occurs the kinetic frictional force will be less than the limiting frictional force.
The free body diagram with the weight of the cylinder $mg$ resolved into two components looks something like this.

The forces $mg \sin x$ and $f$ are going to determine the translational acceleration of the centre of mass of the cylinder, $(F=ma_{\rm c})$, and the frictional force $f$ is going to determine the angular acceleration of the cylinder if the torque is about the centre of mass of the cylinder, $(\tau_{\rm c}= I_{\rm c}\alpha)$.  
A: Consider a ramp that goes downhill to the left and uphill to the right. A coordinate system is placed on the cylinder center, pointing uphill and away from the ramp surface.
A free-body diagram and force consideration along the ramp and perpendicular to the ramp produces the following three equations of motion:
$$ \begin{aligned}
  T - m g \sin x & = m \dot{v} \\
 N - m g \cos x & = 0 \\
  T r & = I \dot{\omega}
\end{aligned} $$
Where $T$ is the friction force, $N$ the normal force, $r$ is the cylinder radius, $v$ the velocity uphill, $\omega$ the counter-clockwise rotation of the cylinder. Also $m$ is the mass and $I = \tfrac{m}{2} r^2$ the mass moment of inertia of the cylinder.
First, consider the no-slip conditions, and find the friction force $T$ and compare it to the normal force. IF $|T| \leq \mu N$ then this condition can continue (here $\mu$ is the coefficient of friction).

*

*Rolling with ${v} + {\omega}\, r = 0$ and $\dot{v} + \dot{\omega}\, r = 0$.
$$ \begin{aligned} N & = m g \cos x \\ T & = \tfrac{m g}{3} \sin x \\ \dot{v} & = -\tfrac{2 g}{3} \sin x \\ \dot{\omega} &= \tfrac{2 g}{3 r} \sin x \end{aligned}$$ which is maintained as long as $|T| \leq \mu N$. Using the values above this condition becomes $$ \mu \ge \tfrac{1}{3} \tan x $$


*Slipping uphill with $T = \mu N$
$$ \begin{aligned} N & = m g \cos x \\ T & = \mu m g \cos x \\ \dot{v} &= -g \sin x + \mu g \cos x \\ \dot{\omega} &= \mu \tfrac{2 g}{r} \cos x\end{aligned}$$ which is maintained as long as $v \ge -\omega\,r$.


*Slipping downhill with $T = -\mu N$
$$ \begin{aligned} N & = m g \cos x \\ T & = -\mu m g \cos x \\ \dot{v} &= -g \sin x - \mu g \cos x \\ \dot{\omega} &= -\mu \tfrac{2 g}{r} \cos x\end{aligned}$$ which is maintained as long as $v \le -\omega\,r$.
