Why is angular momentum conserved here?

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Angular momentum is conserved in the scenario of a spinning cylinder transitioning to rolling without slipping because the angular momentum is calculated about a point on the surface where the friction force acts. Although friction exerts a torque, the angular momentum about that specific point remains constant. If calculated about the center of the cylinder, the final angular momentum would differ due to the external torque. The discussion highlights that conservation of angular momentum applies when considering the appropriate reference point. This understanding clarifies the behavior of the cylinder during the transition from skidding to rolling.
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A cylinder of radius R spins with angular velocity w_0 . When the cylinder is gently laid on a plane, it skids for a short time and eventually rolls without slipping. What is the final angular velocity, w_f?

The solution follows from angular momentum conservation. $$L_i = I \omega_0 = L_f = mv_fr + I\omega_f$$
My question is, why is angular momentum conserved? Friction exerts a torque on the ball, so shouldn't its angular momentum be changing with time?
 
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cpgp said:
A cylinder of radius R spins with angular velocity w_0 . When the cylinder is gently laid on a plane, it skids for a short time and eventually rolls without slipping. What is the final angular velocity, w_f?

The solution follows from angular momentum conservation. $$L_i = I \omega_0 = L_f = mv_fr + I\omega_f$$
My question is, why is angular momentum conserved? Friction exerts a torque on the ball, so shouldn't its angular momentum be changing with time?
AM is always calculated about a point. In this case, the AM in that equation is taken about a point on the surface. As the force acts in the direction of the surface, the AM about that point is conserved.

You can also solve this problem using AM about the initial position of the centre of the cylinder. In that case, the final AM is different from the initial AM because of the external torque about that point.

In general, if you have an external force, then the AM about a point on the line of that force is conserved.
 
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Thanks, I understand now.
 
cpgp said:
Thanks, I understand now.
When I first did this problem I used the second method (AM about the centre of the cylinder). Then I saw the solution using conservation of AM about the surface and I thought it was very clever!
 
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I remember a nice fact in this regard. There is a horizontal rough enough table. There is a paper sheet on the table.
Then you launch a homogeneous ball to roll on the table. The ball rolls without slipping, its angular velocity ##\boldsymbol \omega## and velocity of its center of mass ##\boldsymbol v## are the constant vectors.

When the ball rolls on the paper sheet you begin to jerk the paper as you wish in any horizontal direction. The ball may slide. When the ball leaves the paper it once stops to slide and rolls without slipping again.

It hard to believe but its angular velocity and the velocity of its center of mass restore to the initial vectors ##\boldsymbol \omega,\quad## ##\boldsymbol v## respectively.
 
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