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I am working on an old AP physics problem before showing it to my class. The problem involves a hinged rod of length $D$, mass $M_1$. It has angular velocity $ω$ right before hitting a ball (mass $M_2$) at its lowest position. As in this picture:

enter image description here

The rod comes to full rest after the collision.

To solve for the speed of the ball, you can use angular momentum conservation: $$I\omega = mvD$$ However, by dividing through by $D$, you have units of linear momentum. Further expansion gives $$\begin{align} \frac{I\omega}{D} &= \frac{\bigl[\int r^2\ \mathrm{d}m\bigr]\omega}{D} \\ &= \frac{\bigl[\int_0^D r^2\frac{M_1}{D}\mathrm{d}r\bigr]\omega}{D} \\ &= \frac{\omega M_1}{D^2}\int_0^D r^2\ \mathrm{d}r \\ &= \frac{\omega M_1}{D^2}\times\frac{1}{3}D^3 \\ &= \frac{1}{3}M_1\omega D \\ &= \frac{1}{3}M_1 v_{\text{tan}} \\ &= \frac{2}{3}M_1 v_{\text{cm}} \end{align}$$ And finally: $$\frac{2}{3}M_1 v_{\text{cm}} = M_2 v_{f}$$

So this is saying 2/3 of the rod's linear momentum goes into the linear momentum of the ball? I understand angular momentum is conserved just fine, but I would really like some insight on what the last equation might mean physically. Many of my students wish to use linear momentum conservation over angular, so I want to explain to them exactly why they can't (or can?).

Thanks!

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  • $\begingroup$ Hi Christopher, we!come to Physics.SE. Many users don't like going off site to view images, I would recommend you include them in your post. Also, here is a straightforward guide to mathjax, for your last image. math.meta.stackexchange.com/questions/5020/…. I would also include the homework tag, as this covers homework like questions, which is what this question will be treated as if you were actually a student yourself, but aren't we all, really :) $\endgroup$ – user184116 Feb 10 '18 at 23:19
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    $\begingroup$ Thanks @Countto10, good advice; I handled that in my edit since I figured this post could use some fixing up. $\endgroup$ – David Z Feb 10 '18 at 23:32
  • $\begingroup$ I really appreciate the positive feedback and the tidying up of my question. Looks great. Thanks @DavidZ $\endgroup$ – Christopher Allen Feb 10 '18 at 23:37
  • $\begingroup$ Supposing you did it right, try to rearrange it so that $\frac{2}{3} \frac{M_1}{M_2}=\frac{v_f}{v_{cm}}$. It's giving you the ratio between $v_{cm} and $v_f$, and the factor ${2}{3}$ appears due the moment of "inertia". It works like the "effective mass" of the rod is 3/2 of a point mass that was to impact the same way. $\endgroup$ – FGSUZ Feb 10 '18 at 23:51
  • $\begingroup$ Thanks @FGSUZ I have ran through it a few times. What do you mean because of the moment of inertia? Physically what does this mean in regards to linear momentum? Is it fair to say that linear momentum is NOT conserved? $\endgroup$ – Christopher Allen Feb 11 '18 at 0:00
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You can not use linear momentum conservation as long as the rod is fixed. This produces an impulse transfer from the rod to the point of fixation, increasing the normal force exerted over the rod. As long as $\sum \vec{F} \neq 0$, there is no linear momentum conservation for the system rod + ball, only angular momentum conservation in relation to the axis that passes through the fixed point of the rod. This is because the torque generated in the rod for the increasing normal force is zero (as long as the distance to the considered axis is zero), so the torque exchange (rod-ball) is a kind of quid-pro-quo, which is demonstrated in your equations.

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  • $\begingroup$ Thanks for your comment. While waiting for some replies, I found some info on the idea of center of percussion, which is related to this. I guess the right answer is, like you said, there is a net force on the hinge that causes linear momentum to not preserve. However, if one hits the ball at 2/3D, apparently it does. Thanks again. $\endgroup$ – Christopher Allen Feb 11 '18 at 0:45

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