# Weird rotational movement of a cylinder

I have had this question for a while. Let's suppose we have a disk/cylinder with a radius $$R$$ and mass $$M$$ placed along its length on a flat surface. We press one of its sides with enough force, and the cylinder will move forward (in $$x>0$$) but rotate backward. Under certain conditions, the object will change its translational direction after covering a short distance and return while rotating.

What I believe happens is as follows: The force applied to the cylinder has two components: one pointing forward and another one downward, with the latter exerting torque and causing the cylinder to rotate. The force pushing the center of mass displaces the cylinder from equilibrium, making it move forward while simultaneously rotating and sliding. As the cylinder rotates, it experiences kinetic friction with the surface. After traveling a certain distance, the kinetic friction decelerates the center of mass and eventually stops it. However, since it still possesses rotational kinetic energy, the disk returns by rolling.

I havenig some problems to write the equations involved. I know we must set, for example, the external force $$F$$ (finger) and where it is applied. Suppose it forms some angle $$\theta$$, with, say, $$x>0$$. Also, we need some friction coefficients, say $$\mu_s$$ and $$\mu_k$$.

What should be the system I need to solve in order to obtain, for example, the position of the CM?

Forget about the force exerted by the finger and concentrate of the condition of the cylinder the instant the finger is removed at time $$t=0$$ with the translational velocity to the left being $$v_0$$ and the rotational velocity counterclockwise of $$\omega_0$$.

Because there are no eternal torques acting on the cylinder, angular momentum is conserved.
Consider the angular momentum about $$P$$.
The initial angular momentum is $$I_{\rm cm} \omega_0\hat z -mv_0r \hat z = \frac 12mr^2\omega_0\hat z -mv_0r \hat z$$.

To find the final state of the system one only needs to compare the two terms.

If $$\frac 12mr^2\omega_0>mv_0r$$ then the final angular momentum will be in the $$+\hat z$$ direction and finally the cylinder will be moving to the left, $$-\hat x$$ direction whereas if $$\frac 12mr^2\omega_0 then the final angular momentum will be in the $$-\hat z$$ direction and finally the cylinder will be moving to the right, $$+\hat x$$ direction,

So it a matter of whether or not the initial rotational motion "defeats" the initial translational motion.

For the final state the angular momentum is $$\frac 12mr^2\omega_{\rm f}\hat z +mv_{\rm f}r \hat z$$ and $$v_{\rm f} = r\omega _{\rm f}$$, the no slipping condition.

Equating the initial and final angular momenta will enable you to find $$v_{\rm f}$$ and $$\omega _{\rm f}$$.

If you want to investigate the motion between the initial and final states the you need to set up two equations.

The "aim" of the frictional force between cylinder and surface surface, $$f_{\rm k}$$ in your diagram, is to reduce to zero the relative motion between the cylinder and surface so that the no slip, $$v=r\omega$$, condition is reached.

As shown in your second diagram the frictional force $$f_{\rm k}$$ will act towards the left.
That frictional force will try and change the translational velocity whilst at the same time also change the rotational velocity.
The two equations of motion are $$m\dot v = -f_{\rm k} = \mu m g\Rightarrow \dot v = -\mu g$$ for translation and $$I\dot \omega = -f{\rm k}\,r=- \mu mg\,r \Rightarrow r \dot \omega = -2\mu g$$ for rotational motion where $$I=mr^2/2$$ and $$r$$ is the radius of the cylinder.

Using the initial conditions you can now solve for $$v(t)$$ and $$r\omega(t)$$ and when these are equal you have the final state of rolling without slipping when the two are equal.