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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. $\endgroup$
    – Rich006
    Commented Jun 20, 2017 at 17:26
  • $\begingroup$ @RichardHDowney Yes, but that is irrelevant here as the rod is pivoted nowhere. It is free. $\endgroup$
    – user87745
    Commented Jun 20, 2017 at 20:16
  • $\begingroup$ Here’s a MathJax tutorial $\endgroup$
    – garyp
    Commented Jun 20, 2017 at 20:16
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    $\begingroup$ The comment by @RichardHDowney is relevant. $\endgroup$
    – garyp
    Commented Jun 20, 2017 at 20:17
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    $\begingroup$ You can describe the rod's motion as a combination of rotation about a point plus translation of the same point. This works for any point: the center of mass, the end where the particle hits, the far end, etc. The point you choose determines the rotational inertia coefficient (e.g. 1/3 or 1/12). Any choice should lead to the correct answer, but some choices may involve more steps than others. $\endgroup$
    – Rich006
    Commented Jun 21, 2017 at 19:26

5 Answers 5

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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. $\endgroup$
    – Rich006
    Commented Jun 23, 2017 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
    Commented Aug 29, 2020 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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    $\begingroup$ I'm not sure about your first paragraph. One can choose the point about which to calculate angular momentum, including a point along the path of the incoming particle. The reality is that after the collision the object is translating and rotating. The net angular momentum of these motions about the chosen point should be zero. $\endgroup$
    – Not_Einstein
    Commented Feb 2, 2022 at 18:22
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    $\begingroup$ Angular momentum about the end of the rod will give you misleading results. I do not think this is true as I have shown in my answer with $x = - \ell/2$ $\endgroup$
    – Farcher
    Commented Dec 2, 2023 at 14:31
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This in old post which I found in my travels and being so old I do not think it should be viewed as a current homework question but is still worthy of answering.

enter image description here

$C$ is the centre of mass of the rod and the moment of inertia of the rod about $C$ is $\dfrac{M\ell^2}{12}$.
$P$ is any position with a displacement $x$ as shown in the diagram.

Angular momentum can be though of as having two components, orbital, and spin about the centre of mass.

As there are no external force for the rod and particle system conservation of linear momentum means that after collision the centre of mass of the rod has a velocity $\frac {mv}{M}$.

As there are no external torques angular momentum is conserved about point $P$.

$mv(\dfrac \ell 2 +x) = \underbrace{M\,\dfrac {mv}{M} x}_{\rm orbital} + \underbrace{\dfrac{M\ell^2}{12}\omega}_{\rm spin} \Rightarrow \omega = \dfrac {6mv}{M\ell}$

Thus, as expected, $\omega$ is independent of position $P$.

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  • $\begingroup$ See my recent answer to a related question on how to handle 2D collisions, that preserve momentum by defining the exchange of momentum between objects via impulses. $\endgroup$
    – John Alexiou
    Commented Dec 2, 2023 at 23:34
  • $\begingroup$ @Alexiou Your recent answer is an excellent answer but I did find your first paragraph in this answer might have caused confusion. Your impulse at the end of the rod can be shown to be equivalent t5o an impulsive force acting at the centre of mass and an impulsive couple. This sometimes makes the analysis of a particular situation easier? $\endgroup$
    – Farcher
    Commented Dec 3, 2023 at 9:00
  • $\begingroup$ I see your point. Thank you. $\endgroup$
    – John Alexiou
    Commented Dec 3, 2023 at 17:22

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