Question

A penny is placed at the outer edge of a disk (radius = 0.159 m) that rotates about an axis perpendicular to the plane of the disk at its center. The period of the rotation is 1.50 s. Find the minimum coefficient of friction necessary to allow the penny to rotate along with the disk

Answer 1

Hey, Do your homework. Look at your notes and read the book. We are watching you.
Dr. C will not be happy.

Answer 2

Static Friction Coefficient = Cfs
For of Static Friction = Fs
Normal Force = Fn
Centripetal Force = Fc
mass = m
radius = r
speed = v
period of time = T

Fs = Cfs * Fn so Cfs = Fs/Fn

Fc = Fs and since Fs = mv^2/r

Cfs = Fs/Fn = mv^2/r*Fn

v = 2πr/T and Fn = mg



so....

mv^2 (2πr/T)^2 4(π^2)r
Cfs = ------ = ----------- = --------
rFn r(g) g(T^2)

4(π^2)(0.159 m)
Cfs = ------------------- = 0.28467
(9.8 m/s^2)(1.50 s)^2

*** Remember there are no units on coefficients***

Answer 3

well that doesn't look readable. the answer should still be right but I'll rewrite how I did the math

Cfs = mv^2/rFn = ((2πr/T)^2)/r(g) = (4(π^2)r)/g(T^2)

Cfs = 4(π^2)(0.159 m) / (9.8 m/s^2)(1.50 s)^2 =
0.28467