What Frequency Causes the Box to Slide on a Rotating Platform?

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SUMMARY

The discussion centers on calculating the frequency at which a box begins to slide on a rotating platform. The box is positioned 2.0 meters from the axis, with a static friction coefficient of 0.25. The centripetal acceleration is determined to be 2.45 m/s², leading to a calculated frequency of 0.176 revolutions per second. The solution emphasizes the importance of significant figures, as both the coefficient of static friction and the radius are given with two significant figures.

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  • Understanding of centripetal acceleration
  • Knowledge of static friction coefficients
  • Familiarity with rotational motion equations
  • Ability to perform calculations with significant figures
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Homework Statement


A box rests at a point 2.0m from the axis of a horizontal circular platform. The coefficient of static friction between box and platform is 0.25. As the rate of rotation of the platform is slowly increased from zero, at what frequency will the box first slide?



Homework Equations





The Attempt at a Solution



a.centripetal = mu(g)

=0.25(g) = 2.45m/s^2

a.centripetal=v^2/r

frequency=sqrt(a.centripetal*r)/2*pi*r = 0.176 rev/s

Have i done this correctly?
 
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I arrived at the same answer, but you'll probably want to use two sig figs because the coefficient of static friction and radius have two sig figs.
 

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