Bug on spinning disk -- What distance does the bug fly off?

AI Thread Summary
A bug crawls outward on a spinning disc at 210 RPM, with a static friction coefficient of 1.1. The goal is to determine how far the bug can move from the center before slipping occurs. The correct formula for centripetal force is Fc = m v²/r, not the incorrect expressions previously used. After correcting the calculations, a distance of approximately 0.022 meters is determined as the point where the bug slips. Accurate application of the equations is crucial for solving this problem.
reanamate
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Homework Statement


A bug crawls outward from the center of a compact disc spinning at 210 revolutions per minute. The coefficient of static friction between the bug's sticky feet and the disc surface is 1.1.

Homework Equations


(static friction)(Gravity) = (centripetal force)
This may be wrong but I tried setting (Centripetal Force)= (Radius)(2(pi)(rps))^2

The Attempt at a Solution


I have been getting .02 m but the website I'm using is saying that it's wrong.
 
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I forgot to add that I am trying to find How far the bug gets from the center before slipping?
 
You have the wrong expression for the centripetal force. The correct expression is Fc = m v2/r. What you have is Fc = v2.
 
kuruman said:
What you have is Fc = v2.
No, @reanamate has rω2. Changing "centripetal force" to "centripetal acceleration" everywhere fixes all the issues.
I get the same answer - well, 0.022m to be a little more precise. Maybe that's the problem.
 
That worked I just need .022 thanks for the help.
 
haruspex said:
No, @reanamate has rω2.
Having misread the parentheses, I stand corrected.
 
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