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The problem

A ball with radius $R$ slips on a surface with speed $v_{cm}=v_0$, so that the velocity of the center of mass, decreases and the angular velocity $\omega$ increases. The ball starts a pure rolling motion when $v_{cm}=\omega R$.

The question

My question regards the conservation of angular momentum, and the choice of the point to compute it. If I choose the contact point (or any point on the surface), to compute the angular momentum, then it is conserved.

Instead, if I choose the center of mass of the ball, it turns out that I can't use conservation of angular momentum, since there's an external torque due to friction.

Can someone clarify me this point? Isn't the conservation of angular momentum independent of the point chosen to compute it?

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    $\begingroup$ You're right about conservation being independent of the point but you're forgetting the fact that conserving angular momentum is possible only when there is no torque acting about that point. When you conserve at the point of contact, the frictional force vector acts on the point, hence making the frictional torque zero. But this would not be the case for any random point. $\endgroup$
    – Atgytg 123
    Jan 25 at 12:08
  • $\begingroup$ Am I correct if I say the following ? In this system the angular momentum is not conserved but it turns out that if I smartly choose the axis I can take advantage of the fact that friction has zero momentum and then angular momentum stays constant. Then $\endgroup$
    – semola
    Jan 25 at 14:02
  • $\begingroup$ Almost, you can say that you could take the advantage that friction has zero moment of force not momentum. Due to this, it helps in calculations if we smartly choose that particular axis. $\endgroup$
    – Atgytg 123
    Jan 25 at 18:29
  • $\begingroup$ Sliding friction causes a torque and hence angular momentum is not conserved. $\endgroup$ Jan 25 at 21:26
  • $\begingroup$ @Atgytg123 sorry, I should have written "friction has zero torque". Thanks. $\endgroup$
    – semola
    Jan 28 at 15:53

1 Answer 1

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System: ball, mass $m$ and moment of inertia about centre of mass $I_{\rm C}$.

The horizontal force diagram looks like this.

enter image description here

$F$ is the kinetic frictional force acting on the ball due to the ground which does two things:

It decreases the linear velocity of the centre of mass of the ball, $C$, to the right. $(-F = m\frac {dv}{dt})$
It provide a clockwise torque $FR$ about the centre of mass which increases the clockwise angular velocity $\omega$ of the ball. $(FR = I_{\rm C} \frac {d\omega }{dt})$.

The kinetic frictional force is trying to reduce the relative motion at the point of contact between the ball and the ground to reach the no slipping condition $v=R\omega$ and then there will be no frictional force acting on the ball.

What might seem strange is that by choosing position $A$, the point of contact between the ball and the ground, the net torque on the ball is zero and so the angular momentum of the ball, $I_\rm C\omega + mvR$ clockwise, stays constant with $v$ decreasing which correspondingly $\omega$ increasing until the condition $v=R\omega$ is reached.

Isn't the conservation of angular momentum independent of the point chosen to compute it?
I think that this example shows that the answer is, No.

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  • $\begingroup$ I feel like you have missed something while stating this. It is actually independent only when an external torque is not acting about the point you're conserving momentum. It is pretty similar to how linear momentum can be conserved only when there is no external force involved. $\endgroup$
    – Atgytg 123
    Jan 25 at 18:32
  • $\begingroup$ @Atgytg In this example there is an external torque acting except about the point of contact. $\endgroup$
    – Farcher
    Jan 25 at 22:36
  • $\begingroup$ Yeah exactly, when it acts you can't calculate it that way. Hence, you need a different approach if you're going about a different point. $\endgroup$
    – Atgytg 123
    Jan 26 at 6:54

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