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Say we've got a rod floating around in space, with two masses of $m_0$, one attached at each end. Let's say the rod has a length of $l$.

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There's another mass, $m_1$, moving at some velocity $v$ towards one of the masses.

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$m_1$ collides and sticks instantaneously with $m_2$. In the picture below I drew the collided masses as one big blob.

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After the collision, the rod will have some angular velocity $w$, and some linear velocity $v_f$.

My question is this...will $v_f$ be constant?

For the linear velocity, I want to say:

"Well, linear momentum is conserved, so..."

$m_1v_0=(2m_0 + m_1)v_f$

$v_f=\frac{m_1v_0}{(2m_0 + m_1)}$

However, now I'm doubting how the linear velocity of the entire rod $v_f$ can be constant at all, and thinking that the situation is a lot more complicated.

This is because if it is constant, it seems to me that linear momentum isn't being conserved as the rod spins!

Consider the case when the rod is vertical, the heavier side is moving left, and the lighter side is moving right, versus the case when the rod is vertical, the heavier side is moving right, and the lighter side is moving left.

If the velocity of the center of mass of the rod is constant, then there's more net momentum when the rod is vertical and the heavy side is moving right than when the rod is vertical and the heavier side is moving left...!!!

Which would...disagree with the conservation of linear momentum?

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  • $\begingroup$ Is the bar massless? Also, is the collision perfectly inelastic? $\endgroup$ – nicoguaro Jun 21 '19 at 22:35
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Due to symmetry we can treat the problem as 2d (which is handy, we can draw everything on paper). You correctly computed the final velocity of the center of mass. Linear momentum is conserved here, too.

As the bar is floating freely in space, the translational and rotational degrees of freedom are completely independent after the impact. You can just imagine the constant rotation around the center of mass being superimposed with the constant translational movement of the center of mass.

The problem with your counter-argument is just that you haven't considered that the center of mass of the structure is not in the middle any more after the impact, but more towards the heavier mass blob. Thus, while the angular velocity of the rod is constant, the velocity of the light side is higher w.r.t. the center of mass than the velocity of the heavier side ($v = \omega r$ with r the distance from the rotational centre). Thus in the center of mass reference frame, the momenta of the two blobs always cancel.

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The linear momentum is determined from the velocity of the center of mass, and uses the entire mass of the rod. The mass of the rod in total will not change, and thus we need the velocity to be constant as well. The center of mass in this case will lie somewhere closer to the heavier side of the rod, but when you are discussing the linear momentum it is easiest to just think of the rod as a point of mass located entirely at the center of mass.

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The shape of the object does not matter, whether it is a rod with different masses at each end, a round ball, or any other object, once set in motion, the rotation will be a constant RPM around the center of mass, and the linear motion of the center of mass will remain constant, unless acted upon by another force.

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