A uniform rod of length $L$ lies on a smooth horizontal table. A particle moving on the table strikes the rod perpendicularly at an end and stops. Suppose I have to find final angular velocity of rod.

If I apply angular conservation at 'com' $$mv\frac{l}{2} = \omega \frac{M l^2}{12}$$ where $\omega$ = angular velocity , $m$ = mass of particle ,$M$ = mass of rod.

But when I apply angular conservation at one end $$mvl=\omega \frac{ml^2}{3}.$$ In both cases $\omega$ is different. Why? What am I missing? Because $\omega$ is different in both cases, the time taken by rod for a particular angular displacement is also different.

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    $\begingroup$ The rod's rotational inertia depends on whether it is pivoted at the end or in the middle. $\endgroup$ – Richard H Downey Jun 20 '17 at 17:26
  • $\begingroup$ @RichardHDowney Yes, but that is irrelevant here as the rod is pivoted nowhere. It is free. $\endgroup$ – Dvij Mankad Jun 20 '17 at 20:16
  • $\begingroup$ Here’s a MathJax tutorial $\endgroup$ – garyp Jun 20 '17 at 20:16
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    $\begingroup$ The comment by @RichardHDowney is relevant. $\endgroup$ – garyp Jun 20 '17 at 20:17
  • $\begingroup$ @garyp Oh. Sorry I didn't realize. I agree with the fact that it will depend on where it is hinged but isn't it free in this set up? So it will depend on what axis we choose not on where it is hinged. $\endgroup$ – Dvij Mankad Jun 20 '17 at 23:21

I'm a bit confused by the typography, but I'll just assume the particle stops on the table, implying a small amount of elasticity. The momentum carried by the particle is mv; the moment of inertia of the rod about its center (where its center of mass is, around which it will rotate) is Ml^2/12. The linear impulse on the end of the rod is mv, so the angular impulse is 2mv/l. Therefore the angular velocity is angular momentum/moment of inertia, so 24mv/Ml^3.

  • $\begingroup$ You meant to say the angular impulse is mvl/2. The dimensionality of the angular velocity should be inverse time. $\endgroup$ – Richard H Downey Jun 23 '17 at 12:58

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