Falling yoyo experiment I was playing with a yoyo and a spring dynamometer and realized the tension during the fall is not constant but clearly has a small oscillatory component. This experimental result contradicts the classical  textbook formula
$a=\frac{g}{\left(1+\frac{I}{mR^2}\right)}$. Has anyone any idea how to explain it? Tomorrow I am going to change the string and use a nylon one instead of the current two threads one and see if I can reproduce the result.
 A: This is absolutely an experiment you should do, and report your results back here as an answer. But it's worth taking some time on the front end to make some hypotheses.  Be strategic about it, and make a list.  What are all of the components of a falling, unrolling yo-yo?

*

*a support at the top of the string (usually a finger, in your case some force meter)

*a knotted loop in the string attached to the support

*a mostly-straight section of string

*a section of string wound around the axle of the yo-yo

*the axle of the yo-yo

*the left half of the yo-yo

*the right half of the yo-yo

You have a comment suggesting the string might be stretchy, or that its mass might be a non-negligible factor — so measure those things.  Both are easier to measure with a long string. I've bought replacement yo-yo strings in packages that weigh several of grams, enough for my kitchen scale.  A place were I taught used to do an elasticity measurement lab where we'd close off a stairwell and hang weights from a copper wire that was two or three stories long; it would stretch elastically for several centimeters before it snapped.
My suspicion is that the yo-yo is not quite radially symmetric about its axle, in which case the frequency of the oscillations would match the frequency of the yo-yo's rotation.  Think of things you can measure.  How many times does the yo-yo turn on its way down?  How fast is it going at each point on its fall?  That gives you a rotational frequency.  Compare to when you're having a problem with a car's tire, and you get some wub-wub-wub noise that speeds up when you're driving fast but slows down when you're driving slowly.
Even if the yo-yo itself has good cylindrical symmetry, the loop of string about the axle doesn't quite: there is always a little bunchy place somewhere, which is what you create when you "wake" a yo-yo that is "sleeping."  You might estimate the size of this irregularity compared the the bare axle and the completely-wound loop, and make a prediction about which part of the fall might have a bigger or smaller effect.
There are also lots of secondary wobbles in rotating systems that might have different frequencies.  The tennis racket effect should not apply to a yo-yo, but maybe there are some other things.
A: I have been thinking more about this problem and came to the following conclusion. I am using a spring dynamometer. Before release, the force pulling down the spring is T=mg and after release T= 1/3.m.g. Thus, the spring in the dynamometer receives a kick at t=o.
If we write the equation of motion for the spring, one must introduce a step function to model T(t). When the differential equation is properly solved, this step function generates an extra cos(w.t) into the solution.
$$T(t)-k.y(t)=m \frac{d^{2}y(t) }{d t^{2} } $$
Before releasing the yoyo, let's assume that, because of the weight of the yoyo, the spring is extended. The downward force on the spring is just mg.
When the yoyo falls, we have, as explained in every textbook:
$$ T= \frac{1}{2} m.a$$  and $$ a= \frac{2}{3} g $$
So during the fall, the downward force on the tip of the spring is:
$$ T= \frac{m.g}{3} $$
The downward force is thus modeled by:
$$ T(t)=\begin{cases}mg & t<0\\ \frac{mg}{3} & t \geq  0\end{cases} $$
Physically, the brusque variation of the force acting on the spring triggers the oscillation. Mathematically, the force acting on the spring is discontinuous at t=0 and one must solve the pendulum equation with the Laplace transform method. All things considered, solving the above differential equation with the appropriate piecewise function T(t) gives:
$$y(t) =\begin{cases} \frac{mg}{k}  & t < 0\\ \frac{mg}{3k} \big(1+2.cos( \omega t))\big)   & t  \geq  0\end{cases}$$
With as usual:  $$ \omega = \sqrt{ \frac{k}{m} }  $$
