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</o:shapelayout></xml><![endif]--></head><body lang=EN-US link=blue vlink=purple><div class=WordSection1><p class=MsoNormal><span style='font-size:11.0pt;font-family:"Calibri","sans-serif";color:#1F497D'><o:p> </o:p></span></p><p class=MsoNormal><span style='font-size:11.0pt;font-family:"Calibri","sans-serif";color:#1F497D'><o:p> </o:p></span></p><p class=MsoNormal><b><span style='font-size:10.0pt;font-family:"Tahoma","sans-serif"'>From:</span></b><span style='font-size:10.0pt;font-family:"Tahoma","sans-serif"'> extropy-chat-bounces@lists.extropy.org [mailto:extropy-chat-bounces@lists.extropy.org] <b>On Behalf Of </b>Dave Sill<br><b>Sent:</b> Saturday, March 30, 2013 10:53 AM<br><b>To:</b> ExI chat list<br><b>Subject:</b> Re: [ExI] underwater sprinkler, was: RE: Musical instruments in space<o:p></o:p></span></p><p class=MsoNormal><o:p> </o:p></p><div><div><p class=MsoNormal>On Fri, Mar 8, 2013 at 9:26 AM, spike <<a href="mailto:spike@rainier66.com" target="_blank">spike@rainier66.com</a>> wrote:<o:p></o:p></p><div><blockquote style='border:none;border-left:solid #CCCCCC 1.0pt;padding:0in 0in 0in 6.0pt;margin-left:4.8pt;margin-right:0in'><div id=":2m3"><p class=MsoNormal>An example of the kinds of stuff we used to do is from a Feynman question. You have seen those S-shaped lawn sprinklers, turn on the water, they spin. What would happen if you took one of those, submerged it in water, connected the hose to a pump and pulled water thru it backwards? Would it turn the opposite direction, since the tips of the arms would act as nozzles in reverse? Or would it turn the same direction as before, since the water is creating centrifugal force as it goes around the bend? Or would the forces exactly cancel and not turn? Or a fourth possibility that blows your mind?<br><br>I know the answer from experiment. Can you figure it out from math and thought experiment?<o:p></o:p></p></div></blockquote></div><p class=MsoNormal><br>So what's the answer?<o:p></o:p></p></div><div><p class=MsoNormal><o:p> </o:p></p></div><div><p class=MsoNormal>-Dave<o:p></o:p></p><p class=MsoNormal><span style='font-size:11.0pt;font-family:"Calibri","sans-serif";color:#1F497D'><o:p> </o:p></span></p><p class=MsoNormal><span style='font-size:11.0pt;font-family:"Calibri","sans-serif";color:#1F497D'><o:p> </o:p></span></p><p class=MsoNormal><span style='font-size:11.0pt;font-family:"Calibri","sans-serif";color:#1F497D'>Dave, this one still blows my mind. As I think John Clark pointed out, conservation of momentum requires that there is no net rotation. Otherwise you would be creating angular momentum out of nothing. However, the concept is often misunderstood, and John missed something important.<o:p></o:p></span></p><p class=MsoNormal><span style='font-size:11.0pt;font-family:"Calibri","sans-serif";color:#1F497D'><o:p> </o:p></span></p><p class=MsoNormal><span style='font-size:11.0pt;font-family:"Calibri","sans-serif";color:#1F497D'>You have perhaps heard that if you hold a cat by her feet and release, the cat will land on her feet. This seems to violate the concept of conservation of angular momentum, but it doesn’t. Cats are clever in that way, but not magic. They borrow some angular momentum, get feet downward, then pay back. You can google on how cats do this, or check out the video and move to about 1:15 in this excellent presentation:<o:p></o:p></span></p><p class=MsoNormal><span style='font-size:11.0pt;font-family:"Calibri","sans-serif";color:#1F497D'><o:p> </o:p></span></p><p class=MsoNormal><span style='font-size:11.0pt;font-family:"Calibri","sans-serif";color:#1F497D'><a href="http://www.huffingtonpost.com/2012/08/24/why-cats-always-land-on-their-feet-_n_1828748.html">http://www.huffingtonpost.com/2012/08/24/why-cats-always-land-on-their-feet-_n_1828748.html</a><o:p></o:p></span></p><p class=MsoNormal><span style='font-size:11.0pt;font-family:"Calibri","sans-serif";color:#1F497D'><o:p> </o:p></span></p><p class=MsoNormal><span style='font-size:11.0pt;font-family:"Calibri","sans-serif";color:#1F497D'>Regarding the underwater sprinkler, during the time in which the flow rate is increasing, there is a counter-rotation. During the time in which the flow rate is decreasing, the borrowed angular momentum must be paid back, exactly with no interest. So during that phase, there is a rotation in the same direction as the above-water sprinkler.<o:p></o:p></span></p><p class=MsoNormal><span style='font-size:11.0pt;font-family:"Calibri","sans-serif";color:#1F497D'><o:p> </o:p></span></p><p class=MsoNormal><span style='font-size:11.0pt;font-family:"Calibri","sans-serif";color:#1F497D'>Some of you mathematical hotties, do explain that observation please, using differential equations, or whatever is your favorite mathematical technology, including even a digital model or a Matlab sim. If you manage it, the grand prize will be yours: my sincere everlasting admiration.<o:p></o:p></span></p><p class=MsoNormal><span style='font-size:11.0pt;font-family:"Calibri","sans-serif";color:#1F497D'><o:p> </o:p></span></p><p class=MsoNormal><span style='font-size:11.0pt;font-family:"Calibri","sans-serif";color:#1F497D'>spike<o:p></o:p></span></p></div></div></div></body></html>