Consider a uniform rod which is spinning about an axis that goes through its centre, perpendicular to the rod itself. Two small rings are attached on the rod at equal distances from the centre. As the rod is spinning at its initial angular velocity, the rings get released and proceed towards their respective ends of the rod (and then eventually fly off).

The problem I'm doing requires me to calculate the angular velocity of the system at the instant the rings reach the end of the rod.

After some research, I see that we have to use conservation of angular momentum, i.e. $$ \omega_f=\frac{I_i\omega_i}{I_f} $$

but I originally though that kinetic energy was conserved. So I though that since $K_i=K_f$, I could say that $$ \omega_f=\sqrt\frac{I_i\omega_i^2}{I_f} $$

Why is this wrong? After some reflection I thought that perhaps potential energy is stored due to the rings being attached to the rod, somehow. But then surely this would be converted, and $K_f$ would end up being greater than $K_i$.

Edit: I've now explicitly shown myself using a calculator that the final kinetic energy of the system is much lower than the initial kinetic energy - so a side question would be: where has this energy gone?

Edit #2: From another question about ballerinas:

Going by either the bead or spring model, the rotational energy gets converted into kinetic energy of the arms, accelerated by the centrifugal force in direction of the radial work variable and ultimately dissipating via vibrations when the arms abruptly reach maximal extension.

Quoting: Christoph

So I guess this is all to do with rotational energy being converted to translational kinetic energy of the rings, as mentioned in an answer below.

Any help is appreciated. Thank you.

  • $\begingroup$ Did you take into account the fact that the moment of inertia changes as the rings slide outward? $\endgroup$
    – march
    Dec 23, 2015 at 22:30
  • $\begingroup$ Yes, but I think the reason that final rotational kinetic energy < initial rotational kinetic energy is because some rotational kinetic energy is transferred to translational kinetic energy of the ring. $\endgroup$
    – mathphys
    Dec 23, 2015 at 22:49

1 Answer 1


In your energy conservation equation, your are assuming that both the initial system and the final system have kinetic energy due just to the rotation around the axis. However, there is also some kinetic energy due to the rings translating away from the axis.

In other words, the velocity vector of a ring is not parallel to the velocity vector of the point of the rod it touches.


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