# 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.

If there is slipping and friction you now have to deal with an accelerating (non-inertial) frame of reference and axis of rotation.
To use Newton's laws of motion in the non-inertial frame of reference a pseudo-force has to be introduced which acts at the centre of mass, has a magnitude equal to the frictional force and acts in the opposite direction to the frictional force.
That pseudo-force produces a torque about the point of contact with the ground which changes the angular momentum of the disc about that point.

• ohh that makes sense. thanks a lot :) Mar 12, 2019 at 10:17
• Where this pseudo-force comes from?
– Eli
Mar 12, 2019 at 13:51
• @Eli It is introduced so that Newton’s laws of motion can be used in a non-inertial frame of reference. It has no origin other than to make Newton’s laws work. Mar 12, 2019 at 13:57
• @LeloucheLamperouge I answer your question 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. You subsequently added an edit which has been answered by BillWatts. Mar 21, 2019 at 6:59

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.

• How does this analysis include that there is slipping? I would've used the same analysis for a pure rolling case too. Mar 20, 2019 at 8:11
• So the answer by @Farcher is wrong? Mar 20, 2019 at 14:49
• @BillWatts What is your “whole system”? Mar 20, 2019 at 17:20
• @Lelouche Lamperouge I will stand by my analysis, however, because I have worked numerous problems with forward and reverse english wrt billiard balls by this same analysis and they are all correct. Mar 21, 2019 at 7:22
• It absolutly does. If you hit a cue ball horizontally dead center, it will initially have a linear velocity but its angular velocity will be zero. Friction will slow the linear velocity and increase the angular velocity until $v = R\ \omega$. The height of the bank on a pool table is designed such that when a ball hits it, it will impart neither over or under spin so that the angle of incidence equals the angle of reflection. Surely you have seen overspin on a pool table and watched the ball accelerate. Mar 21, 2019 at 8:29