Love this curtain-ring series -- especially the last one with the pen tip! My gut tells me that the spin time isn't going to end up proportional to ring density, but I'll think about it.

My gut also told me that it should be something more complicated than linear, but that's not what I got.

If we assume the air drag is proportional to the spin w then

we previously got an exponential decay of the spin:

w = w

_{0} exp(-k/I t)

where I is the moment of inertia, w

_{0} the initial spin and the drag torque = k w.

If we set w to the critical spin at which the top falls, w

_{c}, and we solve for the time t

_{c} to get there:

t

_{c}= I 1/k ln(w

_{0}/w

_{c})

So the time it takes to fall is indeed proportional to the moment of inertia and therefore to the density, if you keep the dimensions constant and you also manage to keep the initial spin constant.

Edit: This seems to be a general result, not just for an exponential decay. We found before: I dw/dt = - T, where T is the torque produced by air drag and tip friction and may be an arbitrary function of the spin, T(w), but does not depend on I. It can be re-written as - 1/T(w) dw = 1/I dt. Integrating both sides: F(w) = 1/I t, where F(w) can be a complicated function of w, but does not depend on I or t. Then t

_{c} = I F(w

_{c}).