# 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 real string is not actually inextensible, nor massless, so if you are careful enough, you will measure deviations from the simplified models typically presented in textbooks. Those oscillations could be due to the stretchiness of the string. It might also be due to the wobbliness of whatever the string is attached to. Commented Aug 5, 2022 at 0:38
• It would help if you published a graph of your data. Commented Aug 5, 2022 at 1:30
• I was thinking the same thing. That is why I am going to change the string. Commented Aug 5, 2022 at 8:08
• What's the period of this oscillation? Is it (roughly) proportional to the current string length? Strings have a few vibrational modes (consider a guitar or violin string), but at yo-yo tension I expect the string frequencies to be sub-audible. Commented Aug 5, 2022 at 9:41
• I have calculated it to be (pulsation) w=sqrt(k/m) where k is the spring constant and m the mass of the yoyo. (see below my follow up self answer). I am going to try to set up an experiment to measure it. Though, my experimental apparatus are quite limited and might not be adequate. Commented Aug 5, 2022 at 11:45

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.

• "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." +1, This is my suspicion also. Commented Aug 5, 2022 at 2:33

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: $$\omega = \sqrt{ \frac{k}{m} }$$

• Physics comment: to vary the internal oscillation frequency of your dynamometer, attach extra mass to the part of it that moves. (Or remove mass, if there’s a such way to modify the device.)
– rob
Commented Aug 5, 2022 at 13:55