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A disk of mass M and radius R is located on a frictionless table and pivoted at its center, and initially at rest. A point mass of m with an initial speed v0 hits and scatters from the disk. Since there is no external torque, angular momentum about B is conserved. But why is the angular momentum about A not conserved? I don't see any external torque about there as well. Although I can't prove it with the rules of angular momentum conservation, I know the angular momentum about A is not conserved since $L_1=0$ and $L_2=I\omega$ thus making $L_1$ different from $L_2$. Why does this happen?

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  • $\begingroup$ I'm not sure on this but I think the constraining force that holds the disk at $B$ has a non-zero torque about $A$ $\endgroup$
    – AfterShave
    Commented Feb 20, 2022 at 10:55
  • $\begingroup$ What makes you think there post-collision rotation. You don't mention friction at all. $\endgroup$
    – Gert
    Commented Feb 20, 2022 at 11:11
  • $\begingroup$ It is said that the table is frictionless and there is angular velocity present for the table post-collision, as shown with ω in the image. $\endgroup$
    – KayB
    Commented Feb 20, 2022 at 11:18
  • $\begingroup$ Sorry but your problem is pretty badly posed. Good luck anyway. $\endgroup$
    – Gert
    Commented Feb 20, 2022 at 11:24

1 Answer 1

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Form the change in the direction of the momentum of the point mass you can infer the direction of the impulsive force on the point mass due to the disc (blue).

enter image description here

With the application of Newton's third law that in turn enables you to find the direction of the impulsive force on the disc due to the point mass (red).

And then the direction of the impulsive force on the disc due to the pivot as the centre of mass of the disc does not move (green).

Since there is no external torque, angular momentum about B is conserved.??
From the diagram you can see that there are impulsive external torques about both $A$ and $B$ and so angular momentum is not conserved about $A$ or $B$.

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