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Imagine you had a long heavy rod in space with no significant gravity acting upon it. And a projectile is flying towards it, perpendicular to the orientation of the rod, with the impact some between the centre of mass of the rod and an end of the rod. Upon impact there is no deflection (for whatever reason the bodies fuse together).

Basic preservation of momentum says the centre of mass this fused body will continue to travel in the same direction of the original projectile. But will it also be rotating? And at what rate? If there were a fixed point of rotation I would know how to calculate the rate of rotation. But since it is moving I am confused.

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Consider the conservation of angular momentum. – leongz Jun 23 '13 at 23:47
you should learn about conservation of angular momentum – raindrop Jun 24 '13 at 2:07
these comments suggest there would be no rotation because there was no angular momentum prior to the collision. Is this your intention? I can accept this but it seems like quite a claim. – Steven Noble Jun 24 '13 at 4:32
@StevenNoble The angular momentum prior to the collision is nonzero. – leongz Jun 24 '13 at 4:48
up vote 1 down vote accepted

Just as you said, the problem is doable if only you knew the point around which the system would rotate. Well, there is only one point about which a rigid system can freely rotate: the center of mass - rotation about any other point would accelerate the center of mass, and so require an external force on the system. So the first step is to transform into a center-of-momentum frame. (Some people will say the center-of-mass frame, others will say the latter is a special case of the former where the center of mass coincides with the origin; it doesn't matter.)

Step two is to calculate the angular momentum $\vec{L}$ in this frame with respect to the center of mass. It will be conserved, regardless of how inelastic the collision is.

Step three is to calculate the moment of inertia $I$ of the fused body about the same point.

Finally, the angular velocity about that point will be $\vec{\omega} = \vec{L}/I$.

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