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.

1. A the limiting frictional force $f$ is greater than $mg\sin(x)$ where $m$ is the mass of cylinder.
2. B the limiting friction $f$ is exactly equal to $mg\sin(x)$
3. 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.

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, $(\tau_{\rm c}= I_{\rm c}\alpha)$.

• What about motion in case 1 ? – Matt May 6 '17 at 11:25
• It says limiting friction is greater than $mg\sin x$. It does not say limiting friction is equal to $mg\sin x$. Read the second sentence of my second paragraph which implies that the static frictional force can be any value from zero up the the maximum (limiting) value. – Farcher May 6 '17 at 11:28
• Yes but in this case what will happen to the cylinder will it stay at rest or just rotate at one place without any translatios as there is torque due to friction – Matt May 6 '17 at 11:30
• Just look at the diagram. Does it look like an equilibrium situation? It might be easier to see if you replace the two components of weight with just $mg$. – Farcher May 6 '17 at 11:50
• There must be a translational equilibrium because a = 0 – Matt May 6 '17 at 11:52