# Yo-yo performance in zero g **URL:** \ **Category:** Cecil's Columns/Staff Reports\ **Created:** [July 26, 2001, 4:29pm UTC](https://boards.straightdope.com/t/yo-yo-performance-in-zero-g/73866 "2001-07-26T16:29:21Z")\ **Posts on this page:** 4\ **Page:** 1
**Author:** ![minty\_green](https://avatars.discourse-cdn.com/v4/letter/m/f6c823/32.png) [@minty\_green](https://boards.straightdope.com/u/minty_green)\ **Post date:** [July 26, 2001, 4:29pm UTC](https://boards.straightdope.com/t/yo-yo-performance-in-zero-g/73866/1 "2001-07-26T16:29:21Z")
Loved today’s Staff Report on [yo-yo’s](http://www.straightdope.com/mailbag/myo-yo.html). My curiosity was piqued, however, by the mention of NASA’s yo-yo experiments in space. In zero g (yeah, yeah, I really mean “free fall”), the yo-yo’s apparently do not “sleep” at the end of their strings. Since they do so quite readily here on earth, it seems that the gravitational pull of earth is what causes them to spin on their access at the end of the string rather than zip back up to the hand. Basically, my hypothesis is that the pull of gravity is stronger than the “rebound” force necessary to get it moving back up the string. In orbit, it’s all rebound and no gravity, so no sleep. If my physics are crap, I apologize for being an English major. So on to my question. Since gravity and acceleration are pretty much indistinguishable, shouldn’t the astronuats have been able to get the yo-yo to sleep by doing the old around-the-world trick? That’s where you fling the yo-yo out so that it flies around in a circle at the end of its string. Add the angular momentum (yeah, yeah, I _don’t_ mean “centrifugal force”) back into the equation, and you should be able to overcome the “rebound” that otherwise dominates yo-yo physics in the space shuttle. Because if you can’t play with yo-yo’s properly in space, I ain’t goin’. 🙂 ---
**Author:** ![Chronos](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/chronos/32/134_2.png) [@Chronos](https://boards.straightdope.com/u/Chronos)\ **Post date:** [July 26, 2001, 5:30pm UTC](https://boards.straightdope.com/t/yo-yo-performance-in-zero-g/73866/2 "2001-07-26T17:30:55Z")
> [@](#): > > (yeah, yeah, I _don’t_ mean “centrifugal force”) Actually, you probably do, or at least, you should. Centrifugal force is the force which would be pulling the yo-yo out on its string. The “orbital” angular momentum from the toy whirling about your head wouldn’t really be significant here (although the spin angular momentum is). Yes, I’m quite aware that the centrifugal force is a fictitious force. Nonetheless, it remains a very convenient concept for solving some problems, including the free-fall yo-yo. To answer your original question: Yes, that should work, but it might take more coordination than a similar trick on Earth’s surface. ---
**Author:** ![Podkayne](https://avatars.discourse-cdn.com/v4/letter/p/edb3f5/32.png) [@Podkayne](https://boards.straightdope.com/u/Podkayne)\ **Post date:** [July 26, 2001, 7:27pm UTC](https://boards.straightdope.com/t/yo-yo-performance-in-zero-g/73866/3 "2001-07-26T19:27:52Z")
> [@](#): > > \*Originally posted by Chronos \* > To answer your original question: Yes, that should work, but it might take more coordination . . . And space–how much room did they have to swing that thing around? Wouldn’t want to clock a fellow astronaut on the noggin or break the glass on the ant farm. . . ---
**Author:** ![minty\_green](https://avatars.discourse-cdn.com/v4/letter/m/f6c823/32.png) [@minty\_green](https://boards.straightdope.com/u/minty_green)\ **Post date:** [July 27, 2001, 3:20am UTC](https://boards.straightdope.com/t/yo-yo-performance-in-zero-g/73866/4 "2001-07-27T03:20:06Z")
All right, **Chronos** , maybe I _did_ mean “zero g.” 😉 Thanks for the confirmation of my little theory, though.