Question about frequency?

Question

A coin of mass m rests on a turn table a distance r from the axis rotation. The turntable rotates with a frequency of f. What is the minimum coefficient of static friction between the turntable and the coin if the coin is not to slip?
A) (4(pi^2)f^2r)/g
B) (4(pi^2)fr^2)/g
C) (4(pi)f^2r)/g
D) (4(pi)fr^2)/g

Answer 1

The force that will act on the coin to make it slip is
m*w^2*R

the resistant force is friction, when max is
m*g*u

so
w^2*R=g*u
u=w^2*R/g

since f=w/(2*pi)
w=f*2*pi
plugging in
u=4*pi^2*f^2*R/g

A is the correct answer

j

Answer 2

A.

mw²r=mgu
u = w²r/g = 4pi²f²r/g