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Mechanics}Centripetal Force

Frictional force is given by F_f = \mu N, where N is the normal force.

Centripetal force gives the net force to be, \sum F = mv^2/r = \mu mg. The m's cancel out, and one gets r=v^2/(\mu g).

The revolutions per minute \omega given in the problem can be converted to velocity by v=r\omega=r(33.3 rev/min)(2Pi rad/rev)(min/60s)\approx \pi r m/s. Plugging this into the equation for r above, one has r= \pi^2r^2/(\mu g)\Rightarrow r=\mu g/pi^2 \approx 3/\pi^2 <\approx 1/3, which would be choice (D).

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2014-10-16 20:14:04
take the conventional approximation \omega=\frac{100}{3},\,\pi=3,\,g=10, actually it's pretty fast to get R=\frac{3*3*3}{100}=0.27, which is a good approximation.
2012-10-06 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. NEC
2009-10-08 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 dead-on, 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
2012-10-29 17:34:08
definitely the best way to speed this up!
2009-08-03 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.
2006-11-03 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
2008-11-05 20:10:09
If you look at the units, I believe yosun does convert from frequency to angular frequency with the 2*pi factor.
2006-11-03 02:37:53

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