GR0177 #2



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Comments 
buaasyh 20141016 20:14:04  ,
take the conventional approximation , actually it's pretty fast to get , which is a good approximation.   walczyk 20121006 15:52:47  the fastest way i found was this: mrw^2 = mg*mu, approx 2pi/60 as 1/10, so w=3.33 rad/s. then assume w^2 is 10. so r=g*mu/w^2 = mu which is 0.3.   wittensdog 20091008 16:05:46  Here's a pretty huge coincidence that I came across when doing out this problem. Notice that, roughly,
g = 9.8067
Pi*Pi = 9.8696
Not deadon, but when you have a problem where the answer choices are separated by factors of two, it's pretty reasonable to take:
g = Pi*Pi = 10
walczyk 20121029 17:34:08 
definitely the best way to speed this up!

  mirabzad 20090803 20:53:13  want the exact answer instead of an approximation?
The revolutions per minute is the frequency f=1/T and the period T=2*pi/w therefore w=2*pi*frequency. Setting the centripetal force equal to the frictional force gives you (mv^2)/r = umg . from v=rw => mrw^2 = umg. solving for r leads to r=ug/[(f*2*pi)^2]. using u=0.30 g=9.8 f=33.3/60 makes r=.2417 or .242 (D) which is the answer. sorry, i don't know how to use latex.   cherianjudy 20061103 02:42:25  Isn\\\'t it frequency(f) that is given, rather than w. Using frequency and then changing it to angular frequency will give give you a very approximate answer
a19grey2 20081105 20:10:09 
If you look at the units, I believe yosun does convert from frequency to angular frequency with the 2*pi factor.

  cherianjudy 20061103 02:37:53   

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