# Angular momentum conservation when a particle collides with a rod

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.

• The rod's rotational inertia depends on whether it is pivoted at the end or in the middle. – Richard H Downey Jun 20 '17 at 17:26
• @RichardHDowney Yes, but that is irrelevant here as the rod is pivoted nowhere. It is free. – Dvij Mankad Jun 20 '17 at 20:16
• Here’s a MathJax tutorial – garyp Jun 20 '17 at 20:16
• The comment by @RichardHDowney is relevant. – garyp Jun 20 '17 at 20:17
• @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. – Dvij Mankad Jun 20 '17 at 23:21