Well, this is possibly a dumb thought experiment that can't be realized in practice, but anyways:
Attach to a vertical rod L a spring K standing normal to L.
K is sufficiently stiff so that it can't bend downwards much.
Attach to K at the end an object O of mass m.
Apply an external torque T to L, such that the system rotates around the axis L with angular velocity w0.
Due to the centrifugal force on O, K will stretch a bit in its length direction.
Now remove T, and assume that L is free to rotate about the vertical but that it's centerline doesn't move (for example, placing the lower part of L in a closely fitting container.
We assume that the friction between the container wall and L is negligible, at least over the duration period of the experiment.
MECHANISM OF CONSERVATION OF ANG. MOM::
Since we neglect the effect of friction on L from the container, changes in angular velocity w(t), should come from the dynamics associated with O.
The only forces acting on O is the centrifugal force, the spring force, and gravity.
I will assume that the motion of O remains 2-dimensionsial, in the plane normal to L.
(This is probably incorrect; I suspect rod L would jump up and down somewhat, being correlated with the interplay of gravity and the spring force's component normal to K's length direction, making O oscillate somewhat in the vertical.)
We look at the component of angular momentum along L.
In the 2-D approximation, we see:
a) The forces on O (centrifugal and spring force) are radial (along K), and, so the rate of change of angular momentum along L should be zero.
b) The 2-D approximation implies therefore that m*r(t)^(2)w'(t)=constant.
(r(t) is K's length at t)
The prediction should therefore be:
When K contracts, w(t) should increase, when K lengthens, w(t) should decrease..
