How Far Can a Coin Be Placed on a Rotating Record Without Slipping?

AI Thread Summary
The discussion focuses on determining how far a coin can be placed on a rotating record without slipping, given a rotation speed of 33.3 rpm and a static friction coefficient of 0.1. Initial calculations incorrectly suggested a distance of 0.795 meters, but further analysis revealed an error in the angular velocity conversion, which should be 3.487 rad/sec. Correcting this led to the accurate distance of 0.08 meters. The relationship between frictional force and centripetal force is central to solving the problem. Ultimately, the correct distance for the coin placement is approximately 0.08 meters.
pucr
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Homework Statement


A small coin is placed on a record that is rotating at 33.3 rpm. If the coefficient of static friction between the coin and the record is 0.1, how far from the center of the record can the coin be placed without having it slipped off?


Homework Equations


w = 33.3 rpm = 1.11 rad/sec
Force due to friction = Centripetal force


The Attempt at a Solution


I tried this...
Force due to friction = .1mg = .98m
Centripetal force = ma = m((r*w^2)/r) = mrw^2 = 1.2321mr
soo...
.98m = 1.2321mr
=> .98 = 1.2321r
=> r = .795 m

Unfortunately, the answer should be .08 meters.
I thought at first that I messed up a decimal place somewhere but I can't see where. :-\
I may be going insane...
 
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Welcome to PF.

What acceleration will the coin tolerate without slipping?

If the coefficient of friction is u and the mass is m then

u*m*g = m*v2/r

v = w*r =2*π*f*r

u*g/w2 = r = u*g/(2*π*f)2
 
pucr said:

Homework Equations


w = 33.3 rpm = 1.11 rad/sec
Fix that conversion.
 
ooh thanks.
It should be 3.487 rad/sec making the answer .0806 meters
 

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