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If a rod of length l is placed on a frictionless horizontal surface and an impulse P is applied along the plane of the horizontal surface on the rod at a point O which is at a distance d (d≠0) from the center of mass of the rod then is the angular momentum of the rod conserved about the point of application of impulse?

Initially the rod is at rest so the angular momentum about O is zero but when the impulse is applied there must be some angular acceleration and subsequently angular velocity , so rod will have some angular momentum. But about point O the torque on the rod is zero since the line of action of impulse is passing through O, therefore angular momentum should be conserved but following the arguments in the beginning of the paragraph we can see that it is not conserved.

I just can't figure out the flaw in the reasoning. I would be grateful if you could figure out where I am going wrong and provide the correct explanation on whether we can apply conservation of angular momentum in this case or not.

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  • $\begingroup$ Initially the rod is at rest so the angular momentum about O is zero but when the impulse is applied there must be some angular acceleration and subsequently angular velocity , so rod will have some angular momentum. Not about point $O$ though. $\endgroup$ – BioPhysicist Oct 19 '19 at 4:28
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    $\begingroup$ It's torque about the center of mass that matters. $\endgroup$ – John Alexiou Oct 19 '19 at 17:36
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The magnitude of the angular momentum of the rod about its center of mass after the impulse will be $Pd$. The linear momentum will be $P$, so the magnitude of the contribution of the center-of-mass motion to the angular momentum about point $O$ will also be $Pd$, but in the opposite direction. Thus the total angular momentum about $O$ is zero both before and after the impulse.

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