GREPhysics.NET
GR | # Login | Register
   
  GR0177 #2
Problem
GREPhysics.NET Official Solution    Alternate Solutions
This problem is still being typed.
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).

See below for user comments and alternate solutions! See below for user comments and alternate solutions!
Alternate Solutions
There are no Alternate Solutions for this problem. Be the first to post one!
Comments
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 trimmingNEC
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.
NEC
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. NEC
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
walczyk
2012-10-29 17:34:08
definitely the best way to speed this up!
NEC
mirabzad
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.
NEC
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
a19grey2
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.
NEC
cherianjudy
2006-11-03 02:37:53
NEC

Post A Comment!
You are replying to:
This seems so obvious after seeing the answer. It\'s hard to get out of the crunch mode. tree trimming

Username:
Password:
Click here to register.
This comment is best classified as a: (mouseover)
 
Mouseover the respective type above for an explanation of each type.

Bare Basic LaTeX Rosetta Stone

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