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Suppose a cylinder is slipping rigidly on a frictionless horizontal surface. Then, at $t=0$ it reaches a different ground. The coefficient of friction between the cylinder and this new type of ground is $\mu$. What happens? - assuming usual high-school friction to be the only horizontal force and a homogeneous cylinder with moment of inertia $I=MR^2/2$.

I have come up with this question myself, it is not homework.

Once the cylinder starts slipping on the new ground, kinetic friction will dissipate energy, slowing it down. It will also produce torque and make it roll.

Eventually, the velocity of the contact point will vanish and the cylinder will start rolling without slipping (RWS). This happens when $v=\omega R$ where $v$ is the velocity of the center of mass and $\omega$ is the angular velocity with respect to the center of mass.

Equations of motion are $FR=I\dot{\omega}$ and $F=M\dot{v}$. With $v(0)=v_0$ and $\omega(0)=0$ this leads to the conclusion - using that friction is $F=\mu mg$ and if I made no mistake - that RWS will start at $t=v_0/(3\mu g)$.

Now here comes the actual question. Once the cylinder is RWS, the friction force must actually be zero (otherwise, it would slow down the cylinder without performing work, which is impossible).

However, the friction force is not zero immediately before RWS (it is $\mu mg$). Hence the question: is the friction force dicontinuous? Or have I made some mistake?

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  • $\begingroup$ How does the cylinder reach other ground? Does it have some horizontal velocity to get it there? $\endgroup$ – Mephistopheles Aug 9 '18 at 14:54
  • $\begingroup$ "Suppose a cylinder is slipping rigidly on a frictionless horizontal surface" - meaning all its points have the same velocity, which is horizontal $\endgroup$ – thedude Aug 9 '18 at 14:56
  • $\begingroup$ Ok. So initially it is not rotating about its central axis? $\endgroup$ – Mephistopheles Aug 9 '18 at 15:06
  • $\begingroup$ @Mephistopheles No $\endgroup$ – thedude Aug 9 '18 at 15:26
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is the friction force discontinuous?

Your analysis is correct, but it includes the assumption that the cylinder is rigid. As a result, the friction force in theory drops from a finite value to zero as soon as pure rolling starts. You can actually see this happen in bowling, or billiards where initially the slipping ball is slowing down until it starts rolling. Then it moves in more or less constant velocity.

In real life through the elasticity of the objects, and the elasticity of the asperities on the contacting surfaces causes a transient region near pure slipping and also a non-zero rolling friction. An extreme example is a tire whose rolling friction depends on the amount of deformation (due to load or air pressure).

The specifics of the transient region, are just that, specifics. Meaning they change with each situation and rather hard to model analytically.

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If you analyze a (non-rotating) block in the same situation, you will see similar behavior.

As the block reaches the surface with friction, the frictional force appears and the block decelerates. When the block's speed becomes zero, the modeled frictional force becomes zero.

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You can simplify the problem by looking at a box sliding on rough table, slowing to full stop. When sliding, friction is constant $\mu mg$. Once stopping, friction is zero. However, the transition is actually from constant kinetic friction force to gradually reducing static friction force. A rolling / sliding cylinder goes through the same phases.

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