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

[pooface]

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?

 

[hotcommodity]

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.

 

[ogb p]

Your approach to solving this problem is correct. However, there are a few things to consider. First, the coefficient of static friction is typically represented by the symbol "μ," not "mu." Second, you should use the radius of the circular platform (2.0m) in your calculations, not the distance from the axis to the box. Finally, the frequency should be expressed in units of Hz, not rev/s. So the correct solution would be:

μ = 0.25
a = μg = 0.25(9.8m/s^2) = 2.45m/s^2
a = v^2/r
Frequency = √(a*r)/2π*r = √(2.45m/s^2 * 2.0m)/2π * 2.0m = 0.176 Hz

Overall, your approach is correct, but make sure to use the correct symbols and units in your calculations.