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GR0177 #2
Problem
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Mechanics$\Rightarrow$}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).

Alternate Solutions
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fredluis
2019-08-08 06:51:12
This seems so obvious after seeing the answer. It\'s hard to get out of the crunch mode. tree trimming
buaasyh
2014-10-16 20:14:04
$m\omega^2\,R=\mu\,mg$,
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.
walczyk
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.
wittensdog
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
 walczyk2012-10-29 17:34:08 definitely the best way to speed this up!
2009-08-03 20:53:13
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
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
 a19grey22008-11-05 20:10:09 If you look at the units, I believe yosun does convert from frequency to angular frequency with the 2*pi factor.
cherianjudy
2006-11-03 02:37:53

LaTeX syntax supported through dollar sign wrappers $, ex.,$\alpha^2_0$produces $\alpha^2_0$. type this... to get...$\int_0^\infty$$\int_0^\infty$$\partial$$\partial$$\Rightarrow$$\Rightarrow$$\ddot{x},\dot{x}$$\ddot{x},\dot{x}$$\sqrt{z}$$\sqrt{z}$$\langle my \rangle$$\langle my \rangle$$\left( abacadabra \right)_{me}$$\left( abacadabra \right)_{me}$$\vec{E}$$\vec{E}$$\frac{a}{b}\$ $\frac{a}{b}$