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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 $\omega$ obtained 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 D.C. 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 D.C. Jun 20 '17 at 23:21
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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.

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  • $\begingroup$ 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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The $\omega $ obtained by Conservation of angular momentum (COAM) about an axis is the $\omega$ about that respective axis. The $\omega $ obtained by COAM about COM is $\omega $ about COM and the other $\omega $ is about the end.

When the ball hits the rod the both linear and angular momentum are conserved. This implies that COM of rod is not at rest but moves forward with a constant velocity and rod rotates about COM with constant a $\omega $.

verification

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  • $\begingroup$ The given problem does not include a hing. $\endgroup$ – R.W. Bird Aug 29 at 15:58
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The main point to learn here is to be clear about when a physical property is a property of one thing, and when it is a property of a pair of things.

For example, the mass of a ball is a property of that ball. It does not depend on anything else. The momentum of a ball, on the other hand, is a statement about how the ball is moving relative to some agreed reference frame. It is not quite right to say it is a property of the ball alone; rather, it is a property of the ball when regarded from the point of view of an agreed inertial frame of reference. (When one learns more advanced methods one can make a further statement, and then the 4-momentum can be regarded as a more absolute kind of thing, but I don't want to get into that here).

Rotational properties require a further piece of information. They require also that an axis be specified. The following properties:

  1. angular momentum
  2. torque (also called moment)
  3. moment of inertia

are not properties of a body alone; they are properties of a body and an axis of rotation (all within some agreed inertial frame of reference)

Thus one should not say "the body has angular momentum $J$", one should say "the body has angular momentum $J$ about axis $A$" or something like that. The same goes for torque and moment of inertia. Once you have got used to this then questions such as the one asked here answer themselves (see other answers for more details in this specific case).

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Angular momentum about the end of the rod will give you misleading results. For one the angular momentum of the incoming particle is zero about any point on its path. So after the impact, if you use this value, it means the resulting angular momentum is also going to be zero, which is inconsistent with reality.

What happens is that a certain amount of linear momentum is exchanged at the point of contact, and along the contact normal direction. This amount $J$ is called an impulse, and it is in units of $\text{[Newton second]}$. This exchange is just what is needed to translate and rotated the contacting bodies such that the contact conditions is met. In this case the particle has zero velocity after the impact.

So if the particle has initial velocity $v$, it carries momentum $p=m v$ which will be wholly transferred to the rod.

Now consider the rod, which as an impuse $J=m v$ applied to it on one end. This changes the momentum of the rod by $J$ and the angular momentum of the rod about the center of mass by $(\ell/2) J$. So the change in linear a rotational speed of the rod due to the impact is

$$ \begin{aligned} \Delta v_{\rm rod} & = \tfrac{J}{M} = \tfrac{m}{M} v \\ \Delta \omega_{\rm rod} & = \frac{(\ell/2) J}{I} = \tfrac{\tfrac{\ell}{2} m v}{\tfrac{1}{12} M \ell^2} = \frac{6 m v}{M \ell} \end{aligned} $$

To make a long story short, you need to state the equations of motion (result of impulse) at the center of mass, so when you are conserving angular momentum, it needs to be also specified at the center of mass.

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