How Far Can a Bug Crawl on a Spinning CD Before Slipping?

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SUMMARY

The discussion focuses on calculating how far a bug can crawl on a CD spinning at 200 revolutions per minute (RPM) before slipping due to static friction. The static friction coefficient between the bug and the CD is 1.2. The key equations used include the centripetal acceleration formula, a = ((2*pi)/T)^2*r, and the static friction force equation, f <= "mu"*N. By equating the centrifugal force experienced by the bug with the frictional force, participants concluded that solving for the distance r will yield the maximum distance the bug can crawl before slipping.

PREREQUISITES
  • Understanding of centripetal acceleration and its formula.
  • Knowledge of static friction and its coefficient.
  • Basic algebra for solving equations.
  • Familiarity with rotational motion concepts.
NEXT STEPS
  • Calculate the maximum distance r using the equations provided.
  • Explore the effects of varying the static friction coefficient on the bug's crawling distance.
  • Investigate the relationship between RPM and the maximum distance before slipping.
  • Learn about real-world applications of centripetal force in engineering and physics.
USEFUL FOR

Students studying physics, particularly those focusing on rotational motion and friction, as well as educators looking for practical examples of these concepts in action.

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Homework Statement


A bug crawls outward from the center of a CD spinning at 200 revolutions per min. The static friction coefficient is 1.2 between the bug and cd. How far does the bug get from the center before slipping? https://www.physicsforums.com/Nexus/editor/menupop.gif
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Homework Equations


a = ((2*pi)/T)^2*r //where T is number of seconds it takes to complete a revolution.
f <= "mu"*N // static equation.


The Attempt at a Solution


I started out by trying to find f by using "mu" * g and then trying to solve for r when i have f. So that f = "mu"*a , where a = ((2*pi)/T)^2*r. I don't know if I am even on the right track.
 
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Centrifugal force experienced by the bug when it is at a distance r fron the centre =
m(2*pi)/T)^2*r ( m = mass of the bug).[This force increases as r increases]

Friction force acting on the bug = mu*mg [This force remains constant]

Equate the above two (m gets canceled out) and solve for r. You are on the right track.
 

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