Imagine a homogenous rod with total mass $M$ and length $l$ floating in free space without any force or constraint acting on it. Then, think about two possible scenarios. In the first, a particle with lineal momentum $mv$ impacts right at the center of the rod. In the second, the same particle with same lineal momentum impacts right at one of the ends of the rod.

**First case**

In the first case there will be no rotation, only translation. We apply conservation of momentum (where I think there are two cases, one with the particle having final lineal momentum $v_f$ and the other with $-v_f$, but i will omit the first one).

$$mv = MV – mv_f quad quad V= frac{m}{M}(v+v_f)$$

Then the final energy is:

$$E_f = frac{1}{2} MV^2 + frac{1}{2} mv_f^2 = frac{m^2}{2M} (v+v_f)^2 + frac{1}{2} mv_f^2$$

**Second case**

In the second case there will only be rotation (or so I think, but I’m getting really confused so that’s why I’ve come up with this scenarios to try to prove it) without translation. Again we apply conservation of momentum:

$$mv = MV – mv_f quad quad V= frac{m}{M}(v+v_f)$$

Let me explain what $V$ represents in this case. Since the rod rotates all the points in it don’t have the same velocity, but, since the velocity increases linearly with the radius and the rod is homogenous, if the velocity of the ends is $2V$ (which I compell it to be this way) then the average speed of thte whole rod is $V$. Then we proceed (the cross product of $L$ doesn’t matter):

$$I omega = L = pr = MV l/4 = frac{ml(v+v_f)}{4}$$

So the final energy of the whole system is (having $I=frac{1}{12}Ml^2$):

$$E_f= frac{1}{2} frac{(Iomega)^2}{I} + frac{1}{2} mv_f^2 = frac{3m^2}{8M} (v+v_f)^2 + frac{1}{2} mv_f^2$$

So comparing the two cases, there’s a difference in the final energy, specifically a difference of $frac{m^2}{8M} (v+v_f)^2$, which doesn’t make sense.

**Conclusions**

This difference can mean various things, and I don’t know which of this ones is, or even if it implies something else I haven’t thought about. But before, there’s something I don’t understand and then there would be no need of other explanations.

As I’ve said the average velocity of the rod is $V$ so why should it have different speed when rotating than if the whole rod is advancing? In the end, the product mass times velocity is the same. So that’s one thing I don’t understand.

If that’s not the case and the kinetic energy is different for some reason here’s the explanations I’ve thought about for this difference in the energies:

- In the second case there’s also kinetic energy in the form of translation of the rod, and not only rotation, specifically with a translational kinetic energy of $frac{m^2}{8M} (v+v_f)^2$, but why this specific quantity?
- Actually, apart from the rotation, the rod has the same translational kinetic energy as in the first case, and $v_f$ in the second case is what causes the difference, being much smaller, so in this case the rod would have much more kinetic energy in the end when having the particle impact at one of the ends than having it impact at the very center.
- There’s no translation, which would agree with my hypothesis, and $v_f$ would be bigger in the second case than in the first.