Conservation of Angular Momentum and linear velocity

I have a problem where a cylinder is moving on a horizontal surface, starting with velocity $v_0$. It is given that its radius is $10\text{cm}$, its mass is $200\text g$ and the coefficient of friction with the surface is $\mu = 0.25$. I have to find the final velocity.

First I tried to determine torque about the cylinder's contact point with the surface. The normal reaction force and the weight have the same magnitude and arm, but opposite direction, so their torque equals zero. The friction force has no arm, so total torque is $0$. This tells me that there is conservation of angular momentum, $L_0 = L_f$. Since $L = rmv$, and $r$ and $m$ are constants, $v$ has to be conserved so that $L$ can be conserved as well.

This is the conclusion I reached, however, I think I must be missing something, because if there is a friction force, the velocity can't be conserved(?).

I would appreciate any input.

• Comments to the question (v3): What part of the cylinder is in contact with ground? How is the cylinder oriented relative to the initial velocity vector $v_0$? Is it sliding or only rolling? A picture might come handy. Dec 29, 2013 at 10:05

Here there is an error in your angular momentum calculation.$L=mvr$ is valid only if that is a point size particle.Here it is a cylinder,so angular momentum $L$ about your point of contact should be $$L=L_{cm} +I_{cm}\omega$$ where
$L_{cm}=$angular momentum of CM about your orgin $$and$$ $I_{cm}=$moment of inertia about CM
It looks like the friction force will decelerate the cylinder until (at some time t) there is no sliding, just pure rolling on the surface. Until then the friction force will equal $\mu mg$ (the sign depends on the direction of the axis). Therefore, $\mu m g t=m(v_0-v_f)$, where $v_f$ is the final velocity of the center of mass. On the other hand, we can write the equation for the change of angular momentum with respect to the center of mass: $\mu m g t R=J\frac{v_f}{R}$, where $J$ is the moment of inertia with respect to the center of mass (as far as I remember, it equals $\frac{1}{2}m R^2$ for a homogeneous cylinder).