# Measuring the mass using angular velocities

I have 2 objects which are intially connected together, $O_1$ and $O_2$. When they are connected together, they have a rotation rate about their center of mass of $w_1$. $O_2$ is cleanly released from the connected system, and $O_1$ is now rotating at a rate of $w_2$.

Given that any mass properties of the system can be measured, and that the mass of $O_2$ is known, how can one find the mass of $O_1$, given that this is the one property of the system which cannot be directly measured? Also, what will happen to $O_2$ after release?

One can assume that there is absolutely no friction/resistance.

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## 1 Answer

If there is no friction or resistance, then when you cut the masses free they will both simply move in straight lines maintaining the velocity they had at the moment of the cut. Neither one will rotate around a center any more, except that they will both rotate around their own centers at angular velocity $\omega_1$, just as they were doing when they were cut.

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They still will rotate about their own centers, especially if the masses are not equal, right? –  PearsonArtPhoto Sep 1 '11 at 23:33
Yes, but that rotation rate also will not change. It is just $\omega_1$, assuming I understand the situation correctly. I'm thinking of hockey pucks connected by a string. –  Mark Eichenlaub Sep 1 '11 at 23:36
Think more of a skater holding a weight, which is released. –  PearsonArtPhoto Sep 2 '11 at 0:08
@Pearson Same thing. Skater and weight both translate across the ice at their velocities at the moment of release and both continue rotating about their own centers at the angular velocity they shared at the moment of release. This is just by conservation of angular momentum. –  Mark Eichenlaub Sep 2 '11 at 0:12
If they are both rotating around a common center of mass, then separate, then the angular momentum of the combined system must be parsed out into each of the sub systems. I'm just struggling to come up with a method of analysis... –  PearsonArtPhoto Sep 2 '11 at 0:16
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