# Circular motion of a small block

#### Priscilla

1. The problem statement, all variables and given/known data
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).

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

3. 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.

#### Attachments

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#### ehild

Homework Helper
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?

Homework Helper
Yes!

ehild

#### Priscilla

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

Homework Helper
Well done!

ehild

Thanks alot!

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