Circular motion of a small block

[Priscilla]

Homework Statement

A small block sits 0.15 m from the center of a horizontal turntable whose frequency of rotation can be smoothly increased. If the coefficient of static friction between the block and the turntable is 0.55, at what frequency will the block begin to slide off? (Draw a force diagram).

Homework Equations

f_s = U_s F_n
F = ma
a = v^2/R

The Attempt at a Solution

f_s = U_s F_n
-f_s = ma
a = -f_s/m = -(U_s mg)/m = -U_s g
a = v^2/R
-U_s g = v^2/R
v = Sq rt(-R u_s g)

I am not sure what the question is really asking. What's frequency? Is it velocity?
I attached a force diagram.

 

[ehild]

The rotation of the turntable is a periodic motion. It makes a turn in T time, T is the time period. Frequency is the reciprocal of the time period f=1/T: here it is the same as revolutions per second.

First you need to find out the relation between the speed of the block and the frequency of the turntable.

ehild

 

[Priscilla]

o~
So,
v = rw = r(2[pi]/T) f=1/T
v=2[pi]rf

Right?

 

[ehild]

Yes!

ehild

 

[Priscilla]

Then f_s = F_c
And then I can solve for f!

 

[ehild]

Well done!

ehild

 

[Priscilla]

Thanks alot!