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.
