newtonian mechanics - Is Angular momentum conserved in Impure rolling

In a situation where a disk is rolling WITH slipping on the ground i.e velocity of centre of mass is greater than $r\omega$, is angular momentum conserved about a point on the ground.

What confuses me is that friction decelerates the disk to make $v_{com} = r\omega$ and in this process some velocity is lost so according to formula of angular momentum $L=mvr$,about a point on the ground $L$ decreases as $v$ decreases.

But since friction is also acting about a point on the ground, torque about a point on the ground is zero, so then how would angular momentum change.

edit:this question is about applying conservation of angular momentum in rolling with slipping.

Answer

It is easy to show that total angular momentum is conserved in this case. The case is for slipping where the friction force is retarding the translational velocity of the disk while increasing the rotation of the disk.

For disk of radius $R$, mass $m$, moment of inertia $I$ and velocity $v$, friction force $f$.

The angular momentum about a point on the ground is:

$L1=m\ v\ R$

The rate of change due to the friction force:

$\frac{dL1}{dt}=-m\frac{dv}{dt}R=-f\ R$

since the friction force $f$ is negative to the velocity.

Additionally the disk's spin angular momentum is increasing due to the torque created by $f$. The spin angular momentum:

$L2=I\ \omega$

and the rate of change:

$\frac{dL2}{dt}=I\frac{ d \omega }{dt}=\tau =f\ R$

where $\tau$ is the torque due to the friction force.

So we end up with

$\frac{dL1}{dt}+\frac{dL2}{dt}=0$

and the total angular momentum is constant.

Treating the whole system this way makes the friction force an internal force rather than an external one.