How Fast Must a Turntable Spin to Slide a Bug Off?

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SUMMARY

The discussion centers on calculating the required RPM of a turntable to slide a bug off, given a coefficient of friction of 0.55 and a distance of 25 cm from the center. The user initially struggled with the relationship between centripetal force and frictional force but later resolved the issue. The formula for centripetal force, Fc = m(4π²r)/T², is critical in determining the necessary speed to overcome the frictional force. The user successfully derived the solution after clarifying the role of friction in the equation.

PREREQUISITES
  • Understanding of centripetal force and its formula.
  • Knowledge of friction coefficients and their implications.
  • Basic algebra for manipulating equations.
  • Familiarity with rotational motion concepts.
NEXT STEPS
  • Research the derivation of centripetal force equations.
  • Learn about the effects of friction on rotational motion.
  • Explore practical applications of friction coefficients in physics.
  • Study the relationship between RPM and linear velocity in circular motion.
USEFUL FOR

Physics students, mechanical engineers, and anyone interested in the dynamics of rotational motion and frictional forces.

neoking77
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If the coefficient of friction between a bug and the turntable is 0.55 and the bug is 25 cm from the centre, how fast (in RPM) does the turntable have to spin to cause the bug to slide off?

All I know so far is that the frictional force must be 0 (?)
Fc = m4pi^2r/T^2
but this obviously can't work if Ff is 0...so what am I doing wrong?
thanks in advance
 
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hey nvm i got it :)
 

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