Optimizing Grip on a Rotating Vinyl Record: Bug's Position and Weight

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SUMMARY

The discussion centers on the physics of a bug sitting on a rotating vinyl record, specifically regarding its grip. The correct answer to the problem is option d, which states that the bug should sit closer to the center and lose weight to maintain its grip. The centripetal force acting on the bug decreases as it moves closer to the center, while the friction force, which is the centripetal force, also decreases but remains effective as long as it is less than or equal to the normal force multiplied by the coefficient of friction (uN). The mass of the bug does not influence the solution, as the relationship between centripetal force and friction remains constant.

PREREQUISITES
  • Understanding of centripetal force (F=mω²R)
  • Knowledge of friction and its formula (F=μmg)
  • Basic principles of rotational motion
  • Concept of static friction and its role in maintaining grip
NEXT STEPS
  • Study the relationship between centripetal force and friction in rotating systems
  • Explore the effects of mass on friction and centripetal force
  • Learn about the coefficient of friction and its impact on grip in various scenarios
  • Investigate real-world applications of rotational dynamics in engineering
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Physics students, educators, and anyone interested in understanding the dynamics of rotational motion and friction in practical scenarios.

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Homework Statement


A bug is sitting on a rotating vinyl record. Between the bug to the record acts static force. for the bug not to lose it's grip and be thrown outside of the record, he should:
a. sit closer to the center.
b. sit further from the center.
c. lose weight.
d. both a and c are correct.


Homework Equations


Centripetal force: F=mw2R
Friction: F=μmg


The Attempt at a Solution


I think the changing the radius will reduce the centripetal force and will not affect the friction. this it will make the bug's grip more firm and will prevent it from losing grip.
although, losing weight will decrease both the centripetal force and the friction at the same ratio. the ratio between them will stay the same. my teacher said the answer is d.
Who is correct??
Thanks for the help!
 
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It looks like you are, DJ (assuming you chose 'a' and not 'b'!), but your reasoning is not quite correct. When the bug moves closer to the center, the centripetal force decreases, and the friction force also decreases to a value less than uN. That's because the friction force IS the centripetal force. And as long as it is less than or equal to uN, the bug won't slip. In any case, decreasing the mass doesn't matter, because the solution is independent of it.

Welcome to PF!
 

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