Static friction and centripetal force

AI Thread Summary
The discussion focuses on determining the maximum angular velocity of a block on a turntable without slipping, given the coefficient of static friction (\mu_s) and the radius (r). The relevant equations include the static friction force and the centripetal force, leading to the relationship that the static friction force equals the centripetal force for equilibrium. The solution derived is \omega = \sqrt{{\mu_s g}/{r}}, which seems to be confirmed as correct by participants. The problem is being used as a reference to clarify misunderstandings related to another question. The final expression for angular velocity appears satisfactory to the contributors.
Xamien
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Homework Statement


"If the coefficient of static friction between the block of mass m and the turntable is \mu_s, determine the maximum constant angular velocity of the platform without causing the block to slip." I'm actually using this problem to understand why I'm getting the wrong answer to another question, unfortunately the book doesn't actually say if I have the right answer to THIS one. r is the radius, of course.


Homework Equations


f = \mu_s F_{n}\<br /> <br /> F = m r \omega^2 \<br /> <br /> \sum F = F_1 + F_2<br /> <br /> F_{n} = m g


The Attempt at a Solution


Using summation, I set the static friction force equal to the centripetal force for equilibrium and solved for \omega.
\omega = \sqrt{{\mu_s g}/{r}}
 
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Xamien said:

Homework Statement


"If the coefficient of static friction between the block of mass m and the turntable is \mu_s, determine the maximum constant angular velocity of the platform without causing the block to slip." I'm actually using this problem to understand why I'm getting the wrong answer to another question, unfortunately the book doesn't actually say if I have the right answer to THIS one. r is the radius, of course.


Homework Equations


f = \mu_s F_{n}\<br /> <br /> F = m r \omega^2 \<br /> <br /> \sum F = F_1 + F_2<br /> <br /> F_{n} = m g


The Attempt at a Solution


Using summation, I set the static friction force equal to the centripetal force for equilibrium and solved for \omega.
\omega = \sqrt{{\mu_s g}/{r}}

That final expression looks OK.
 
Thanks!
 
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